04 Solucionario Inteligencia Artificial 2da Edicion Stuart Russell, Peter Norvig - Redação (2024)

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04 Solucionario Inteligencia Artificial 2da Edicion Stuart Russell, Peter Norvig - Redação (2)

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04 Solucionario Inteligencia Artificial 2da Edicion Stuart Russell, Peter Norvig - Redação (5)

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04 Solucionario Inteligencia Artificial 2da Edicion Stuart Russell, Peter Norvig - Redação (8)

04 Solucionario Inteligencia Artificial 2da Edicion Stuart Russell, Peter Norvig - Redação (9)

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LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS GRATIS EN DESCARGA DIRECTASIGUENOS EN:VISITANOS PARA DESARGALOS GRATIS.http://librosysolucionarios.nethttps://twitter.com/Libros_y_Soluhttps://www.facebook.com/pages/Solucionarios-de-Libros/345772498866324https://plus.google.com/b/113394888343830071226/113394888343830071226Instructor’sManual:ExerciseSolutionsforArtificial IntelligenceA ModernApproachSecondEditionStuartJ.RussellandPeterNorvigUpperSaddleRiver, New Jersey 07458http://librosysolucionarios.netLibrary of CongressCataloging-in-PublicationDataRussell,StuartJ. (StuartJonathan)Instructor’s solutionmanualfor artificial intelligence: a modernapproach(secondedition)/ StuartRussell,PeterNorvig.Includesbibliographicalreferencesandindex.1. Artificial intelligence I. Norvig, Peter. II. Title.Vice PresidentandEditorialDirector, ECS:Marcia J. HortonPublisher:Alan R.AptAssociateEditor: Toni DianneHolmEditorialAssistant:Patrick LindnerVice PresidentandDirectorof ProductionandManufacturing,ESM: DavidW. RiccardiExecutiveManagingEditor: VinceO’BrienManagingEditor: CamilleTrentacosteProductionEditor: Mary MasseyManufacturingManager:Trudy PisciottiManufacturingBuyer: Lisa McDowellMarketingManager:PamelaShafferc�2003PearsonEducation,Inc.PearsonPrenticeHallPearsonEducation,Inc.UpperSaddleRiver, NJ07458All rightsreserved.No partof this manualmaybereproducedin any form or by any means,withoutpermissionin writing from thepublisher.PearsonPrenticeHall R�is a trademarkof PearsonEducation,Inc.Printedin theUnitedStatesof America10 9 8 7 6 5 4 3 2 1ISBN: 0-13-090376-0PearsonEducationLtd., LondonPearsonEducationAustraliaPty. Ltd., SydneyPearsonEducationSingapore,Pte.Ltd.PearsonEducationNorthAsiaLtd., HongKongPearsonEducationCanada,Inc.,TorontoPearsonEducacíondeMexico, S.A.deC.V.PearsonEducation—Japan,TokyoPearsonEducationMalaysia,Pte.Ltd.PearsonEducation,Inc.,UpperSaddleRiver, New Jerseyhttp://librosysolucionarios.netPrefaceThis Instructor’s SolutionManualprovidessolutions(or at leastsolutionsketches)foralmostall of the 400 exercisesin Artificial Intelligence: A ModernApproach (SecondEdi-tion). Weonly giveactualcodefor a few of theprogrammingexercises;writing a lot of codewouldnotbethathelpful, if only becausewedon’t know whatlanguageyouprefer.In many cases,we give ideasfor discussionand follow-up questions,and we try toexplainwhywe designedeachexercise.Thereis moresupplementarymaterialthat we want to offer to the instructor, but wehave decidedto do it throughthemediumof theWorld Wide Webratherthanthrougha CDor printedInstructor’s Manual.Theideais thatthissolutionmanualcontainsthematerialthatmustbekeptsecretfrom students,but theWebsitecontainsmaterialthatcanbeupdatedandaddedto in amoretimely fashion.Theaddressfor thewebsiteis:http://aima.cs.b er ke le y. eduandtheaddressfor theonlineInstructor’s Guideis:http://aima.cs.b er ke le y. edu /i ns tr uc tor s. ht mlThereyouwill find:� Instructionson how to join the aima-instructors discussionlist. We stronglyrecom-mendthat you join so that you can receive updates,corrections,notification of newversionsof this SolutionsManual,additionalexercisesandexam questions,etc., in atimely manner.� Sourcecodefor programsfrom thetext. We offer codein Lisp, Python,andJava, andpoint to codedevelopedby othersin C++ andProlog.� Programmingresourcesandsupplementaltexts.� Figuresfrom thetext; for overheadtransparencies.� Terminologyfrom theindex of thebook.� Othercoursesusingthebookthathave homepageson theWeb. You canseeexamplesyllabi andassignmentshere. Pleasedo not put solutionsetsfor AIMA exercisesonpublicwebpages!� AI Educationinformationon teachingintroductoryAI courses.� OthersitesontheWebwith informationonAI. Organizedby chapterin thebook;checkthis for supplementalmaterial.We welcomesuggestionsfor new exercises,new environmentsandagents,etc. Thebookbelongsto you, theinstructor, asmuchasus. We hopethatyou enjoy teachingfrom it,thatthesesupplementalmaterialshelp,andthatyouwill shareyour supplementsandexperi-enceswith otherinstructors.iiihttp://librosysolucionarios.nethttp://librosysolucionarios.netSolutionsfor Chapter1Introduction1.1a. Dictionary definitionsof intelligence talk about “the capacityto acquireand applyknowledge” or “the faculty of thoughtandreason”or “the ability to comprehendandprofit from experience.” Theseareall reasonableanswers,but if we want somethingquantifiablewe would usesomethinglike “the ability to apply knowledgein ordertoperformbetterin anenvironment.”b. We defineartificial intelligenceasthestudyandconstructionof agentprogramsthatperformwell in agivenenvironment,for agivenagentarchitecture.c. We defineanagentasanentity that takesactionin responseto perceptsfrom anenvi-ronment.1.2 Seethesolutionfor exercise26.1for somediscussionof potentialobjections.Theprobabilityof fooling aninterrogatordependson just how unskilledtheinterroga-tor is. Oneentrantin the2002Loebnerprizecompetition(which is not quite a real TuringTest)did fool one judge,althoughif you look at the transcript,it is hard to imaginewhatthat judgewasthinking. Therecertainlyhave beenexamplesof a chatbotor otheronlineagentfooling humans. For example,seeSeeLenny Foner’s accountof the Julia chatbotat foner.www.media.mit.edu/people/foner/Julia/. We’d say the chancetoday is somethinglike 10%,with the variationdependingmoreon theskill of the interrogatorratherthantheprogram.In 50 years,we expectthattheentertainmentindustry(movies,videogames,com-mercials)will have madesufficient investmentsin artificial actorsto createvery credibleimpersonators.1.3 The2002Loebnerprize(www.loebner.net)went to Kevin Copple’s programELLA . Itconsistsof a prioritizedsetof pattern/actionrules: if it seesa text stringmatchinga certainpattern,it outputsthe correspondingresponse,which may includepiecesof the currentorpastinput. It alsohasa large databaseof text andhasthe Wordnetonline dictionary. It isthereforeusingratherrudimentarytools,andis not advancingthetheoryof AI. It is provid-ing evidenceon the numberandtype of rulesthat aresufficient for producingonetype ofconversation.1.4 No. It meansthatAI systemsshouldavoid trying to solveintractableproblems.Usually,thismeansthey canonly approximateoptimalbehavior. Noticethathumansdon’t solve NP-completeproblemseither. Sometimesthey aregoodatsolvingspecificinstanceswith a lot of1http://librosysolucionarios.net2 Chapter 1. Introductionstructure,perhapswith theaid of backgroundknowledge. AI systemsshouldattemptto dothesame.1.5 No. IQ testscorescorrelatewell with certainothermeasures,suchassuccessin college,but only if they’re measuringfairly normalhumans.TheIQ testdoesn’t measureeverything.A programthat is specializedonly for IQ tests(andspecializedfurtheronly for theanalogypart)would very likely performpoorly on othermeasuresof intelligence.SeeTheMismea-sure of Man by StephenJayGould, Norton, 1981or Multiple intelligences: the theory inpracticeby HowardGardner, BasicBooks,1993for moreon IQ tests,what they measure,andwhatotheraspectsthereareto “intelligence.”1.6 Justasyou areunawareof all the stepsthat go into makingyour heartbeat,you arealsounawareof mostof whathappensin your thoughts.You do have aconsciousawarenessof someof your thoughtprocesses,but themajority remainsopaqueto your consciousness.Thefield of psychoanalysisis basedon the ideathat oneneedstrainedprofessionalhelp toanalyzeone’s own thoughts.1.7a. (ping-pong)A reasonablelevel of proficiency wasachievedby Andersson’s robot(An-dersson,1988).b. (driving in Cairo)No. Althoughtherehasbeena lot of progressin automateddriving,all suchsystemscurrently rely on certainrelatively constantclues: that the roadhasshouldersanda centerline, thatthecaraheadwill travel a predictablecourse,thatcarswill keepto their sideof theroad,andsoon. To ourknowledge,noneareableto avoidobstaclesor othercarsor to changelanesasappropriate;theirskillsaremostlyconfinedto stayingin onelaneatconstantspeed.Driving in downtown Cairois toounpredictablefor any of theseto work.c. (shoppingat themarket)No. No robotcancurrentlyput togetherthetasksof moving inacrowdedenvironment,usingvision to identify awidevarietyof objects,andgraspingtheobjects(includingsquishablevegetables)without damagingthem. Thecomponentpiecesarenearlyableto handletheindividual tasks,but it would take a major integra-tion effort to put it all together.d. (shoppingon the web) Yes. Softwarerobotsarecapableof handlingsuchtasks,par-ticularly if thedesignof thewebgroceryshoppingsitedoesnot changeradicallyovertime.e. (bridge)Yes.ProgramssuchasGIB now playatasolid level.f. (theoremproving) Yes. For example,theproof of Robbinsalgebradescribedon page309.g. (funny story) No. While somecomputer-generatedproseand poetry is hystericallyfunny, this is invariably unintentional,except in the caseof programsthat echobackprosethatthey have memorized.h. (legal advice)Yes,in somecases.AI hasa long history of researchinto applicationsof automatedlegal reasoning.Two outstandingexamplesaretheProlog-basedexperthttp://librosysolucionarios.net3systemsusedin theUK to guidemembersof thepublic in dealingwith theintricaciesofthesocialsecurityandnationalitylaws. Thesocialsecuritysystemis saidto havesavedtheUK governmentapproximately$150million in its first yearof operation.However,extensioninto morecomplex areassuchascontractlaw awaitsa satisfactoryencodingof thevastwebof common-senseknowledgepertainingto commercialtransactionsandagreementandbusinesspractices.i. (translation)Yes. In a limited way, this is alreadybeingdone. SeeKay, Gawron andNorvig (1994)andWahlster(2000)for an overview of thefield of speechtranslation,andsomelimitationson thecurrentstateof theart.j . (surgery) Yes. Robotsareincreasinglybeingusedfor surgery, althoughalwaysunderthecommandof adoctor.1.8 Certainlyperceptionandmotorskills areimportant,andit is agoodthingthatthefieldsof vision androboticsexist (whetheror not you want to considerthempart of “core” AI).But given a percept,an agentstill hasthe taskof “deciding” (eitherby deliberationor byreaction)which action to take. This is just as true in the real world as in artificial micro-worldssuchaschess-playing.Socomputingtheappropriateactionwill remainacrucialpartof AI, regardlessof theperceptualandmotorsystemtowhichtheagentprogramis “attached.”On theotherhand,it is true thata concentrationon micro-worldshasled AI away from thereally interestingenvironments(seepage46).1.9 Evolution tendsto perpetuateorganisms(and combinationsand mutationsof organ-isms) that aresuccesfulenoughto reproduce.That is, evolution favors organismsthat canoptimizetheirperformancemeasureto at leastsurvive to theageof sexualmaturity, andthenbeableto win a mate.Rationalityjust meansoptimizingperformancemeasure,so this is inline with evolution.1.10 Yes, they are rational, becauseslower, deliberative actionswould tend to result inmoredamageto the hand. If “intelligent” means“applying knowledge” or “using thoughtandreasoning”thenit doesnot requireintelligenceto make a reflex action.1.11 Thisdependsonyourdefinitionof “intelligent” and“tell.” In onesensecomputersonlydo what theprogrammerscommandthemto do,but in anothersensewhat theprogrammersconsciouslytellsthecomputerto dooftenhasverylittle to dowith whatthecomputeractuallydoes. Anyonewho haswritten a programwith an ornerybug knows this, asdoesanyonewho haswritten a successfulmachinelearningprogram.So in onesenseSamuel“told” thecomputer“learn to play checkers betterthan I do, and thenplay that way,” but in anothersensehetold thecomputer“follow this learningalgorithm” andit learnedto play. Sowe’releft in thesituationwhereyoumayor maynotconsiderlearningto playcheckersto bessignof intelligence(or you maythink that learningto play in theright way requiresintelligence,but not in this way), andyou maythink the intelligenceresidesin theprogrammeror in thecomputer.1.12 Thepointof thisexerciseis to noticetheparallelwith thepreviousone.Whateveryoudecidedaboutwhethercomputerscouldbeintelligentin 1.9,youarecommittedto makingthehttp://librosysolucionarios.net4 Chapter 1. Introductionsameconclusionaboutanimals(includinghumans),unlessyourreasonsfor decidingwhethersomethingis intelligent take into accountthe mechanism(programmingvia genesversusprogrammingvia a humanprogrammer).Note thatSearlemakesthis appealto mechanismin his ChineseRoomargument(seeChapter26).1.13 Again, thechoiceyoumake in 1.11drivesyour answerto thisquestion.http://librosysolucionarios.netSolutionsfor Chapter2IntelligentAgents2.1 Thefollowing arejust someof themany possibledefinitionsthatcanbewritten:� Agent: an entity thatperceivesandacts;or, onethat canbe viewedasperceiving andacting.Essentiallyany objectqualifies;thekey point is theway theobjectimplementsan agentfunction. (Note: someauthorsrestrict the term to programsthat operateonbehalfof a human,or to programsthat cancausesomeor all of their codeto run onothermachineson anetwork, asin mobile agents.)������� ���������� Agentfunction: a functionthatspecifiestheagent’s actionin responseto everypossibleperceptsequence.� Agent program: that programwhich, combinedwith a machinearchitecture,imple-mentsan agentfunction. In our simpledesigns,the programtakesa new perceptoneachinvocationandreturnsanaction.� Rationality: a propertyof agentsthatchooseactionsthatmaximizetheir expectedutil-ity, giventheperceptsto date.� Autonomy: a propertyof agentswhosebehavior is determinedby their own experienceratherthansolelyby their initial programming.� Reflex agent: anagentwhoseactiondependsonly on thecurrentpercept.� Model-basedagent: an agentwhoseactionis derived directly from an internalmodelof thecurrentworld statethatis updatedover time.� Goal-basedagent: an agentthat selectsactionsthat it believeswill achieve explicitlyrepresentedgoals.� Utility-basedagent: an agentthat selectsactionsthat it believes will maximizetheexpectedutility of theoutcopmestate.� Learningagent: anagentwhosebehavior improvesover time basedon its experience.2.2 A performancemeasureis usedby an outsideobserver to evaluatehow successfulanagentis. It is a functionfrom historiesto arealnumber. A utility functionis usedby anagentit*elf to evaluatehow desirablestatesor historiesare. In our framework, theutility functionmaynotbethesameastheperformancemeasure;furthermore,anagentmayhavenoexplicitutility functionatall, whereasthereis alwaysaperformancemeasure.2.3 Although thesequestionsarevery simple,they hint at somevery fundamentalissues.Ouranswersarefor thesimpleagentdesignsfor staticenvironmentswherenothinghappens5http://librosysolucionarios.net6 Chapter 2. IntelligentAgentswhile the agentis deliberating;the issuesget even more interestingfor dynamicenviron-ments.a. Yes;take any agentprogramandinsertnull statementsthatdonotaffect theoutput.b. Yes; the agentfunction might specify that the agentprint ������� whenthe perceptis aTuring machineprogramthat halts,and��������� otherwise. (Note: in dynamicenviron-ments,for machinesof lessthaninfinite speed,therationalagentfunctionmaynot beimplementable;e.g.,theagentfunction thatalwaysplaysa winning move, if any, in agameof chess.)c. Yes;theagent’s behavior is fixedby thearchitectureandprogram.d. Thereare "! agentprograms,althoughmany of thesewill not run at all. (Note: Anygivenprogramcandevoteat most # bits to storage,so its internalstatecandistinguishamongonly "! pasthistories.Becausetheagentfunctionspecifiesactionsbasedonper-cepthistories,therewill bemany agentfunctionsthatcannotbeimplementedbecauseof lackof memoryin themachine.)2.4 Noticethatfor our simpleenvironmentalassumptionswe neednotworry aboutquanti-tative uncertainty.a. It sufficesto show thatfor all possibleactualenvironments(i.e.,all dirt distributionsandinitial locations),thisagentcleansthesquaresat leastasfastasany otheragent.This istriviallytruewhenthereis no dirt. Whenthereis dirt in theinitial locationandnoneintheotherlocation,theworld is cleanafteronestep;no agentcando better. Whenthereis nodirt in theinitial locationbut dirt in theother, theworld is cleanaftertwo steps;noagentcando better. Whenthereis dirt in both locations,theworld is cleanafter threesteps;noagentcandobetter. (Note: in general,theconditionstatedin thefirst sentenceof thisansweris muchstricterthannecessaryfor anagentto berational.)b. Theagentin (a) keepsmoving backwardsandforwardseven after theworld is clean.It is betterto do $&%('*) oncethe world is clean(the chaptersaysthis). Now, sincethe agent’s perceptdoesn’t saywhetherthe othersquareis clean,it would seemthatthe agentmusthave somememoryto saywhetherthe othersquarehasalreadybeencleaned. To make this argumentrigorousis more difficult—for example,could theagentarrangethingssothatit wouldonly bein acleanleft squarewhentheright squarewasalreadyclean? As a generalstrategy, an agentcan usethe environmentit*elf asa form of external memory—a commontechniquefor humanswho usethings like,+-�.�/0���01������2/-3appointmentcalendarsandknotsin handkerchiefs.In thisparticularcase,however, thatis not possible.Considerthereflex actionsfor 4 5&6879�����;:=< and 4 >?6879�����;:@< . If eitheroftheseis $&%('*) , thentheagentwill fail in thecasewherethat is the initial perceptbutthe othersquareis dirty; hence,neithercanbe $A%B'*) andthereforethe simplereflexagentis doomedto keepmoving. In general,theproblemwith reflex agentsis thattheyhave to do the samething in situationsthat look the same,even whenthe situationsareactuallyquite different. In the vacuumworld this is a big liability, becauseeveryinterior square(excepthome)looks either like a squarewith dirt or a squarewithoutdirt.http://librosysolucionarios.net7AgentType PerformanceMeasureEnvironment Actuators SensorsRobotsoccerplayerWinning game,goalsfor/againstField,ball, ownteam,otherteam,own bodyDevices(e.g.,legs)forlocomotionandkickingCamera,touchsensors,accelerometers,orientationsensors,wheel/jointencodersInternetbook-shoppingagentObtainre-quested/interestingbooks,minimizeexpenditureInternet Follow link,enter/submitdatain fields,displayto userWebpages,userrequestsAutonomousMarsroverTerrainexploredandreported,samplesgatheredandanalyzedLaunchvehicle,lander, MarsWheels/legs,samplecollectiondevice,analysisdevices,radiotransmitterCamera,touchsensors,accelerometers,orientationsensors,,wheel/jointencoders,radioreceiverMathematician’stheorem-provingassistantFigureS2.1 AgenttypesandtheirPEASdescriptions,for Ex. 2.5.c. If we considerasymptoticallylong lifetimes, then it is clear that learninga map (insomeform) confersan advantagebecauseit meansthat the agentcanavoid bumpinginto walls. It canalso learn wheredirt is most likely to accumulateandcan devisean optimal inspectionstrategy. The precisedetailsof the explorationmethodneededto constructa completemap appearin Chapter4; methodsfor deriving an optimalinspection/cleanupstrategy arein Chapter21.2.5 Somerepresentative, but notexhaustive,answersaregivenin FigureS2.1.2.6 Environmentpropertiesaregivenin FigureS2.2.Suitableagenttypes:a. A model-basedreflex agentwould suffice for mostaspects;for tacticalplay, a utility-basedagentwith lookaheadwouldbeuseful.b. A goal-basedagentwould be appropriatefor specificbook requests.For moreopen-endedtasks—e.g.,“Find mesomethinginterestingto read”—tradeoffs areinvolvedandtheagentmustcompareutilities for variouspossiblepurchases.http://librosysolucionarios.net8 Chapter 2. IntelligentAgentsTaskEnvironment Observable Deterministic Episodic Static Discrete AgentsRobotsoccer Partially Stochastic SequentialDynamic Continuous MultiInternetbook-shopping Partially DeterministicC Sequential StaticC Discrete SingleAutonomousMarsrover Partially Stochastic SequentialDynamic Continuous SingleMathematician’sassistant Fully Deterministic Sequential Semi Discrete MultiFigureS2.2 Environmentpropertiesfor Ex. 2.6.c. A model-basedreflex agentwould suffice for low-level navigationandobstacleavoid-ance;for routeplanning,explorationplanning,experimentation,etc.,somecombinationof goal-basedandutility-basedagentswouldbeneeded.d. For specificproof tasks,a goal-basedagentis needed.For “exploratory” tasks—e.g.,“Prove someusefullemmataconcerningoperationson strings”—autility-basedarchi-tecturemight beneeded.2.7 Thefile "agents/environm ents /v ac uum.l is p" in thecoderepositoryimple-mentsthe vacuum-cleanerenvironment. Studentscaneasilyextendit to generatedifferentshapedrooms,obstacles,andsoon.2.8 A reflex agentprogramimplementingtherationalagentfunctiondescribedin thechap-ter is asfollows:(defun reflex-rational -v ac uum-agent (percept)(destructuring- bi nd (location status) percept(cond ((eq status ’Dirty) ’Suck)((eq location ’A) ’Right)(t ’Left))))For states1, 3, 5, 7 in Figure3.20, the performancemeasuresare1996,1999,1998,2000respectively.2.9 Exercises2.4,2.9,and2.10maybemergedin futureprintings.a. No; seeanswerto 2.4(b).b. Seeanswerto 2.4(b).c. In this case,a simplereflex agentcanbe perfectlyrational. The agentcanconsistofa tablewith eightentries,indexed by percept,that specifiesanactionto take for eachpossiblestate.After theagentacts,theworld is updatedandthenext perceptwill telltheagentwhat to do next. For larger environments,constructinga tableis infeasible.Instead,theagentcould run oneof theoptimalsearchalgorithmsin Chapters3 and4andexecutethefirst stepof thesolutionsequence.Again,no internalstateis required,but it wouldhelpto beableto storethesolutionsequenceinsteadof recomputingit foreachnew percept.2.10http://librosysolucionarios.net9Figure S2.3 An environmentin which randommotion will take a long time to cover allthesquares.a. Becausetheagentdoesnot know thegeographyandperceivesonly locationandlocaldirt, andcanotrememberwhat just happened,it will get stuckforever againsta wallwhenit triesto move in adirectionthatis blocked—thatis, unlessit randomizes.b. Onepossibledesigncleansup dirt andotherwisemovesrandomly:(defun randomized-refle x- va cu um-agent (percept)(destructuring -b ind (location status) percept(cond ((eq status ’Dirty) ’Suck)(t (random-elemen t ’(Left Right Up Down))))))This is fairly closeto whattheRoombaDFE vacuumcleanerdoes(althoughtheRoombahasa bumpsensorandrandomizesonly whenit hits anobstacle).It worksreasonablywell in nice,compactenvironments.In maze-like environmentsor environmentswithsmallconnectingpassages,it cantake avery long time to cover all thesquares.c. An exampleis shown in FigureS2.3.Studentsmayalsowish to measureclean-uptimefor linearor squareenvironmentsof differentsizes,andcomparethoseto theefficientonlinesearchalgorithmsdescribedin Chapter4.d. A reflex agentwith statecanbuild a map(seeChapter4 for details).An onlinedepth-first exploration will reachevery statein time linear in the size of the environment;therefore,theagentcando muchbetterthanthesimplereflex agent.Thequestionof rationalbehavior in unknown environmentsisacomplex onebut it isworthencouragingstudentsto think aboutit. Weneedto have somenotionof thepriorhttp://librosysolucionarios.net10 Chapter 2. IntelligentAgentsprobaility distribution over the classof environments;call this the initial belief state.Any actionyields a new perceptthat canbe usedto updatethis distribution, movingtheagentto anew belief state.Oncetheenvironmentis completelyexplored,thebeliefstatecollapsesto a single possibleenvironment. Therefore,the problemof optimalexplorationcanbe viewedasa searchfor an optimalstrategy in thespaceof possiblebelief states.This is a well-defined,if horrendouslyintractable,problem. Chapter21discussessomecaseswhereoptimalexplorationis possible.Anotherconcreteexampleof explorationis theMinesweepercomputergame(seeExercise7.11). For very smallMinesweeperenvironments,optimal exploration is feasiblealthoughthebelief stateupdateis nontrivial to explain.2.11 Theproblemappearsat first to bevery similar; themaindifferenceis that insteadofusingthelocationperceptto build themap,theagenthasto “invent” its own locations(which,afterall, arejust nodesin a datastructurerepresentingthestatespacegraph).Whena bumpis detected,theagentassumesit remainsin thesamelocationandcanadda wall to its map.For grid environments,theagentcankeeptrackof its G�H96�IKJ locationandsocantell whenithasreturnedto anold state.In thegeneralcase,however, thereis no simpleway to tell if astateis new or old.2.12a. For areflex agent,thispresentsnoadditionalchallenge,becausetheagentwill continueto LF��M�N as long as the currentlocationremainsdirty. For an agentthat constructsasequentialplan, every LF��MON actionwould needto be replacedby “ L=�8MON until clean.”If the dirt sensorcanbe wrong on eachstep,thenthe agentmight want to wait for afew stepsto getamorereliablemeasurementbeforedecidingwhetherto LF��M�N or moveon to a new square. Obviously, thereis a trade-off becausewaiting too long meansthatdirt remainson thefloor (incurringa penalty),but actingimmediatelyriskseitherdirtying a cleansquareor ignoringa dirty square(if the sensoris wrong). A rationalagentmustalsocontinuetouring andcheckingthesquaresin caseit missedoneon aprevious tour (becauseof badsensorreadings).it is not immediatelyobvioushow thewaiting time at eachsquareshouldchangewith eachnew tour. Theseissuescanbeclarified by experimentation,which may suggesta generaltrend that canbe verifiedmathematically. This problemis a partially observableMarkov decisionprocess—seeChapter17. Suchproblemsarehard in general,but somespecialcasesmay yield tocarefulanalysis.b. In this case,theagentmustkeeptouring thesquaresindefinitely. Theprobability thata squareis dirty increasesmonotonicallywith thetime sinceit waslastcleaned,sotherationalstrategy is, roughlyspeaking,to repeatedlyexecutetheshortestpossibletourofall squares.(Wesay“roughly speaking”becausetherearecomplicationscausedby thefact that theshortesttour mayvisit somesquarestwice, dependingon thegeography.)Thisproblemis alsoapartially observableMarkov decisionprocess.http://librosysolucionarios.netSolutionsfor Chapter3SolvingProblemsby Searching3.1 A state is asituationthatanagentcanfind itself in. Wedistinguishtwo typesof states:world states(theactualconcretesituationsin therealworld) andrepresentationalstates(theabstractdescriptionsof therealworld thatareusedby theagentin deliberatingaboutwhattodo).A state spaceis a graphwhosenodesare the set of all states,and whoselinks areactionsthattransformonestateinto another.A search treeis a tree(a graphwith no undirectedloops)in which therootnodeis thestartstateandthesetof childrenfor eachnodeconsistsof thestatesreachableby takinganyaction.A search node is a nodein thesearchtree.A goal is astatethattheagentis trying to reach.An action is somethingthattheagentcanchooseto do.A successorfunction describedthe agent’s options: given a state,it returnsa setof(action,state)pairs,whereeachstateis thestatereachableby takingtheaction.Thebranching factor in a searchtreeis thenumberof actionsavailableto theagent.3.2 In goal formulation,we decidewhich aspectsof the world we are interestedin, andwhich canbe ignoredor abstractedaway. Thenin problemformulationwe decidehow tomanipulatetheimportantaspects(andignoretheothers).If wedid problemformulationfirstwewouldnot know whatto includeandwhatto leaveout. Thatsaid,it canhappenthatthereis a cycle of iterationsbetweengoal formulation,problemformulation,andproblemsolvinguntil onearrivesatasufficiently usefulandefficient solution.3.3 In Pythonwe have:#### successor_fn defined in terms of result and legal_actionsdef successor_fn(s) :return [(a, result(a, s)) for a in legal_actions(s )]#### legal_actions and result defined in terms of successor_fndef legal_actions(s ):return [a for (a, s) in successor_fn(s) ]def result(a, s):11http://librosysolucionarios.net12 Chapter 3. SolvingProblemsby Searchingfor (a1, s1) in successor_fn(s) :if a == a1:return s13.4 From http://www.cut-the-knot.com/pythagoras/fifteen.shtml, this proof appliesto thefifteenpuzzle,but thesameargumentworksfor theeightpuzzle:Definition: Thegoalstatehasthenumbersin acertainorder, whichwewill measureasstartingat theupperleft corner, thenproceedingleft to right, andwhenwe reachtheendof arow, goingdown to theleftmostsquarein therow below. For any otherconfigurationbesidesthegoal,whenever a tile with a greaternumberon it precedesa tile with a smallernumber,thetwo tilesaresaidto be inverted.Proposition: For agivenpuzzleconfiguration,let P denotethesumof thetotalnumberof inversionsandtherow numberof theemptysquare.Then G�PRQTSVU; WJ is invariantunderanylegal move. In otherwords,after a legal move an odd P remainsodd whereasan even Premainseven. Thereforethegoalstatein Figure3.4,with no inversionsandemptysquareinthefirst row, has PYX[Z , andcanonly bereachedfrom startingstateswith odd P , not fromstartingstateswith even P .Proof: First of all, sliding a tile horizontallychangesneitherthe total numberof in-versionsnor the row numberof the emptysquare. Thereforelet us considersliding a tilevertically.Let’s assume,for example,that the tile 5 is locateddirectly over the emptysquare.Sliding it down changestheparity of therow numberof theemptysquare.Now considerthetotal numberof inversions.Themoveonly affectsrelative positionsof tiles 5 , > , \ , and ] .If noneof the > , \ , ] causedaninversionrelative to 5 (i.e.,all threearelargerthan 5 ) thenafter sliding onegetsthree(an odd number)of additionalinversions. If oneof the threeissmallerthan 5 , thenbeforethemove > , \ , and ] contributeda singleinversion(relative to5 ) whereasafterthemove they’ ll becontributing two inversions- achangeof 1, alsoanoddnumber. Two additionalcasesobviously leadto thesameresult.Thusthechangein thesumP is alwayseven.This is preciselywhatwe have setout to show.Sobeforewe solve apuzzle,weshouldcomputethe P valueof thestartandgoalstateandmake surethey have thesameparity, otherwiseno solutionis possible.3.5 Theformulationputsonequeenpercolumn,with a new queenplacedonly in a squarethatis not attackedby any otherqueen.To simplify matters,we’ll first considerthe # –rooksproblem.Thefirst rook canbeplacedin any squarein column1, thesecondin any squareincolumn2 exceptthesamerow thatasthe rook in column1, andin generaltherewill be #_^elementsof thesearchspace.3.6 No, a finite statespacedoesnot always leadto a finite searchtree. Considera statespacewith two states,bothof whichhaveactionsthatleadto theother. Thisyieldsaninfinitesearchtree,becausewe cango backandforth any numberof times. However, if the statespaceis afinite tree,or in general,afinite DAG (directedacyclic graph),thentherecanbenoloops,andthesearchtreeis finite.3.7http://librosysolucionarios.net13a. Initial state:No regionscolored.Goaltest:All regionscolored,andno two adjacentregionshave thesamecolor.Successorfunction:Assignacolor to a region.Costfunction:Numberof assignments.b. Initial state:As describedin thetext.Goaltest:Monkey hasbananas.Successorfunction: Hop on crate;Hop off crate;Pushcratefrom onespotto another;Walk from onespotto another;grabbananas(if standingon crate).Costfunction:Numberof actions.c. Initial state:consideringall input records.Goaltest:consideringasinglerecord,andit gives“illegal input” message.Successorfunction: run againon thefirst half of therecords;run againon thesecondhalf of therecords.Costfunction:Numberof runs.Note: This is a contingencyproblem; you needto seewhethera run givesan errormessageor not to decidewhatto do next.d. Initial state:jugshave values 4 `W6"`W6"`�< .Successorfunction: givenvalues 4 H96�IK6"a< , generate4 ZW W6�IK6"a< , 4 Hb6"cW6"a< , 4 Hb6�Id6"e�< (by fill-ing); 4 `W6�Id6"a< , 4 H96"`W6"a< , 4 H96�IK6"`�< (by emptying);or for any two jugswith currentvaluesH and I, pour I into H ; this changesthe jug with H to theminimumof HgfhI andthecapacityof thejug, anddecrementsthejug with I by by theamountgainedby thefirstjug.Costfunction:Numberof actions.12 34 5 6 78 9 10 1211 13 14 15FigureS3.1 Thestatespacefor theproblemdefinedin Ex. 3.8.3.8a. SeeFigureS3.1.b. Breadth-first:1 2 3 4 5 6 7 8 9 10 11Depth-limited:1 2 4 8 9 5 10 11Iterative deepening:1; 1 2 3; 1 2 4 5 3 6 7; 1 2 4 8 9 5 1011http://librosysolucionarios.net14 Chapter 3. SolvingProblemsby Searchingc. Bidirectionalsearchis veryuseful,becausetheonly successorof # in thereversedirec-tion is i�G�#_jW WJ�k . Thishelpsfocusthesearch.d. 2 in theforwarddirection;1 in thereversedirection.e. Yes;startat thegoal,andapplythesinglereversesuccessoractionuntil you reach1.3.9a. Hereis onepossiblerepresentation:A stateis asix-tupleof integerslisting thenumberof missionaries,cannibals,andboatson the first side,andthenthe seondsideof theriver. Thegoal is a statewith 3 missionariesand3 cannibalson thesecondside. Thecostfunctionis oneperaction,andthesuccessorsof a stateareall thestatesthatmove1 or 2 peopleand1 boatfrom onesideto another.b. The searchspaceis small, so any optimal algorithmworks. For an example,seethefile "search/domains/ ca nnib al s.l is p" . It sufficesto eliminatemovesthatcircle backto thestatejust visited. From all but thefirst andlast states,thereis onlyoneotherchoice.c. It is not obvious thatalmostall movesareeitherillegal or revert to theprevious state.Thereis a feelingof a largebranchingfactor, andno clearway to proceed.3.10 For the8 puzzle,thereshouldn’t bemuchdifferencein performance.Indeed,thefile"search/domains /p uz zl e8. li sp " shows thatyoucanrepresentan8 puzzlestateasa single32-bit integer, so the questionof modifying or copying datais moot. But for the#mlg# puzzle,as # increases,the advantageof modifying ratherthancopying grows. Thedisadvantageof a modifying successorfunction is that it only workswith depth-firstsearch(or with avariantsuchasiterative deepening).3.11 a. The algorithmexpandsnodesin orderof increasingpathcost; thereforethe firstgoalit encounterswill bethegoalwith thecheapestcost.b. It will be the sameas iterative deepening,U iterations,in which noG�p2q�J nodesaregenerated.c. U;jWrd. Implementationnot shown.3.12 If thereare two pathsfrom the startnodeto a given node,discardingthe moreex-pensive one cannoteliminateany optimal solution. Uniform-costsearchand breadth-firstsearchwith constantstepcostsbothexpandpathsin orderof s -cost.Therefore,if thecurrentnodehasbeenexpandedpreviously, thecurrentpathto it mustbe moreexpensive thanthepreviously foundpathandit is correctto discardit.For IDS, it iseasytofindanexamplewith varyingstepcostswherethealgorithmreturnsasuboptimalsolution:simplyhave two pathsto thegoal,onewith onestepcosting3 andtheotherwith two stepscosting1 each.3.13 Consideradomainin whicheverystatehasasinglesuccessor, andthereis asinglegoalat depth# . Thendepth-firstsearchwill find thegoal in # steps,whereasiterative deepeningsearchwill take Ztfu vfwevf*ck"x"x;fy#TXznoG�#|{WJ steps.http://librosysolucionarios.net153.14 As an ordinaryperson(or agent)browsing the web, we canonly generartethe suc-cessorsof a pageby visiting it. We canthendo breadth-firstsearch,or perhapsbest-searchsearchwheretheheuristicis somefunctionof thenumberof wordsin commonbetweenthestartandgoalpages;thismayhelpkeepthelinks ontarget.Searchengineskeepthecompletegraphof theweb,andmayprovide theuseraccessto all (or at leastsome)of thepagesthatlink to apage;thiswould allow usto do bidirectionalsearch.3.15a. If weconsiderall G�H96�IKJ points,thenthereareaninfinite numberof states,andof paths.b. (For this problem,we considerthe startandgoal pointsto be vertices.) The shortestdistancebetweentwo points is a straightline, and if it is not possibleto travel in astraightline becausesomeobstacleis in the way, thenthe next shortestdistanceis asequenceof line segments,end-to-end,that deviate from the straightline by as littleas possible. So the first segmentof this sequencemust go from the start point to atangentpoint on an obstacle– any paththat gave theobstaclea wider girth would belonger. Becausetheobstaclesarepolygonal,the tangentpointsmustbe at verticesoftheobstacles,andhencetheentirepathmustgo from vertex to vertex. Sonow thestatespaceis thesetof vertices,of which thereare35 in Figure3.22.c. Codenot shown.d. Implementationsandanalysisnot shown.3.16 Codenot shown.3.17a. Any path,no matterhow badit appears,might leadto anarbitraily largereward(nega-tivecost).Therefore,onewouldneedto exhaustall possiblepathsto besureof findingthebestone.b. Supposethegreatestpossiblerewardis } . Thenif wealsoknow themaximumdepthofthestatespace(e.g.whenthestatespaceisatree),thenany pathwith U levelsremainingcanbeimprovedby atmost }U , soany pathsworsethan }U lessthanthebestpathcanbepruned.For statespaceswith loops,this guaranteedoesn’t help,becauseit is possibleto go arounda loopany numberof times,picking up } rewwardeachtime.c. The agentshouldplan to go aroundthis loop forever (unlessit canfind anotherloopwith evenbetterreward).d. The valueof a scenicloop is lessenedeachtime onerevisits it; a novel scenicsightis a greatreward,but seeingthesameonefor thetenthtime in anhour is tedious,notrewarding. To accomodatethis, we would have to expandthestatespaceto includeamemory—astateis now representednot just by the currentlocation,but by a currentlocationanda bagof already-visitedlocations.Therewardfor visiting a new locationis now a(diminishing)functionof thenumberof timesit hasbeenseenbefore.e. Realdomainswith loopingbehavior includeeatingjunk food andgoingto class.3.18 Thebeliefstatespaceis shown in FigureS3.2.No solutionis possiblebecausenopathleadsto abeliefstateall of whoseelementssatisfythegoal. If theproblemis fully observable,http://librosysolucionarios.net16 Chapter 3. SolvingProblemsby SearchingLRL RS SSFigureS3.2 Thebeliefstatespacefor thesensorlessvacuumworld underMurphy’s law.theagentreachesagoalstateby executingasequencesuchthat ~K�9}� is performedonly in adirty square.Thisensuresdeterministicbehavior andeverystateis obviouslysolvable.3.19 Codenot shown, but a good start is in the coderepository. Clearly, graphsearchmustbeused—thisis a classicgrid world with many alternatepathsto eachstate.Studentswill quickly find thatcomputingtheoptimalsolutionsequenceis prohibitively expensive formoderatelylarge worlds,becausethestatespacefor an #�lo# world has#|{Ax� "! states.Thecompletiontimeof therandomagentgrowslessthanexponentiallyin # , sofor any reasonableexchangeratebetweensearchcostadpathcosttherandomagentwill eventuallywin.http://librosysolucionarios.netSolutionsfor Chapter4InformedSearchandExploration4.1 Thesequenceof queuesis asfollows:L[0+244=244]M[70+241=311],T[111+329=440]L[140+244=384],D[145+242=387],T[111+329=440]D[145+242=387],T[111+329=440],M[210+241=451],T[251+329=580]C[265+160=425],T[111+329=440],M[210+241=451],M[220+241=461],T[251+329=580]T[111+329=440],M[210+241=451],M[220+241=461],P[403+100=503],T[251+329=580],R[411+193=604],D[385+242=627]M[210+241=451],M[220+241=461],L[222+244=466],P[403+100=503],T[251+329=580],A[229+366=595],R[411+193=604],D[385+242=627]M[220+241=461],L[222+244=466],P[403+100=503],L[280+244=524],D[285+242=527],T[251+329=580],A[229+366=595],R[411+193=604],D[385+242=627]L[222+244=466],P[403+100=503],L[280+244=524],D[285+242=527],L[290+244=534],D[295+242=537],T[251+329=580],A[229+366=595],R[411+193=604],D[385+242=627]P[403+100=503],L[280+244=524],D[285+242=527],M[292+241=533],L[290+244=534],D[295+242=537],T[251+329=580],A[229+366=595],R[411+193=604],D[385+242=627],T[333+329=662]B[504+0=504],L[280+244=524],D[285+242=527],M[292+241=533],L[290+244=534],D[295+242=537],T[251+329=580],A[229+366=595],R[411+193=604],D[385+242=627],T[333+329=662],R[500+193=693],C[541+160=701]4.2 ��X�` gives �KG�#_J�X��sBG�#_J . This behavesexactly like uniform-costsearch—thefactorof two makesnodifferencein theorderingof thenodes.��XzZ givesA� search.��Xz gives�KG�#_J�Xz W�BG�#_J , i.e.,greedybest-firstsearch.Wealsohave�KG�#_J�X�G� v�y�?J�4 s(G�#_J*f � v��� �=G�#_J�<which behavesexactly like A� searchwith a heuristic �{�� � �BG�#_J . For ����Z , this is alwayslessthan �=G�#_J andhenceadmissible,provided �BG�#_J is itself admissible.4.3a. Whenall stepcostsareequal, sBG�#_J������1)B���(G�#_J , so uniform-costsearchreproducesbreadth-firstsearch.b. Breadth-firstsearchis best-firstsearchwith �KG�#_J�X��;��)B���BG�#_J ; depth-firstsearchisbest-firstsearchwith �KG�#_J�X��?����)B���BG�#_J ; uniform-costsearchis best-firstsearchwith17http://librosysolucionarios.net18 Chapter 4. InformedSearchandExploration�KG�#_J9XwsBG�#_J .c. Uniform-costsearchis A� searchwith �=G�#_J_Xz` .S�A� GBh=7�h=5�h=1 h=0�2 14¡4¡Figure S4.1 A graphwith an inconsistentheuristicon which GRAPH-SEARCH fails toreturnthe optimal solution. The successorsof ¢ are £ with ¤¦¥¨§ and © with ¤v¥�ª . £ isexpandedfirst, so the pathvia © will be discardedbecause£ will alreadybe in the closedlist.4.4 SeeFigureS4.1.4.5 Going betweenRimnicu Vilcea and Lugoj is one example. The shortestpath is thesouthernone,throughMehadia,DobretaandCraiova. But agreedysearchusingthestraight-line heuristicstartingin RimnicuVilceawill startthewrongway, headingto Sibiu. Startingat Lugoj, theheuristicwill correctlyleadusto Mehadia,but thena greedysearchwill returnto Lugoj, andoscillateforever betweenthesetwo cities.4.6 Theheuristic��X��B«bfw� { (addingmisplacedtilesandManhattandistance)sometimesoverestimates.Now, suppose�BG�#_Jg���B�G�#_J¦f�} (as given) and let ¬ { be a goal that issuboptimalby morethan } , i.e., s(G�¬ { J¦­[\ � f®} . Now considerany node# on a pathto anoptimalgoal.Wehave�KG�#_J�X�s(G�#_J|fu�=G�#_J��s(G�#_J|fu� � G�#_JKf®}�¯\ � fu}��s(G�¬ { Jso ¬ { will never beexpandedbeforeanoptimalgoalis expanded.4.7 A heuristicis consistentiff, for every node# andevery successor#|° of # generatedbyany action ± ,�BG�#_J²�z}G�#_6"±86�# ° J*fu�=G�# ° JOnesimpleproof is by inductionon thenumber� of nodeson theshortestpathto any goalfrom # . For �³X´Z , let # ° be the goal node; then �BG�#_J��´}G�#_6"±�6�# ° J . For the inductivecase,assume# ° is on theshortestpath � stepsfrom thegoalandthat �=G�# ° J is admissiblebyhypothesis;then�BG�#_J²�z}G�#_6"±86�# ° J*fu�=G�# ° J²�z}G�#_6"±86�# ° JKf®� � G�# ° J²Xz� � G�#_Jhttp://librosysolucionarios.net19so �=G�#_J at �ofuZ stepsfrom thegoalis alsoadmissible.4.8 Thisexercisereiteratesasmallportionof theclassicwork of Held andKarp (1970).a. TheTSPproblemis to find a minimal (total length)paththroughthecities that formsa closedloop. MST is a relaxed versionof that becauseit asksfor a minimal (totallength)graphthatneednot bea closedloop—it canbeany fully-connectedgraph.Asaheuristic,MST is admissible—itis alwaysshorterthanor equalto aclosedloop.b. The straight-linedistanceback to the start city is a ratherweak heuristic—it vastlyunderestimateswhentherearemany cities. In thelaterstageof asearchwhenthereareonly a few citiesleft it is not sobad.To saythatMST dominatesstraight-linedistanceis to saythatMST alwaysgivesa highervalue. This is obviously truebecausea MSTthatincludesthegoalnodeandthecurrentnodemusteitherbethestraightline betweenthem,or it mustincludetwo or morelines thataddup to more. (This all assumesthetriangleinequality.)c. See"search/domain s/t sp .l is p" for astartat this. Thefile includesaheuristicbasedon connectingeachunvisited city to its nearestneighbor, a closerelative to theMST approach.d. See(Cormenet al., 1990,p.505)for analgorithmthatrunsin n¨G�µ·¶ ¸W¹ºµoJ time,whereµ is the numberof edges. The coderepositorycurrently containsa somewhat lessefficientalgorithm.4.9 Themisplaced-tilesheuristicis exactfor theproblemwhereatile canmovefrom squareA to squareB. As this is a relaxationof theconditionthata tile canmove from squareA tosquareB if B is blank,Gaschnig’s heuristiccanotbe lessthanthemisplaced-tilesheuristic.As it is alsoadmissible(beingexact for a relaxationof the original problem),Gaschnig’sheuristicis thereforemoreaccurate.If wepermutetwo adjacenttiles in thegoalstate,wehaveastatewheremisplaced-tilesandManhattanbothreturn2, but Gaschnig’s heuristicreturns3.To computeGaschnig’s heuristic,repeatthe following until the goal stateis reached:let B be the currentlocationof the blank; if B is occupiedby tile X (not the blank) in thegoalstate,moveX to B; otherwise,moveany misplacedtile to B. Studentscouldbeaskedtoprove thatthis is theoptimalsolutionto therelaxedproblem.4.11a. Localbeamsearchwith �RX�Z is hill-climbing search.b. Local beamsearchwith �»X½¼ : strictly speaking,this doesn’t make sense.(Exercisemay be modified in future printings.) The idea is that if every successoris retained(because� is unbounded),thenthesearchresemblesbreadth-firstsearchin thatit addsonecompletelayerof nodesbeforeaddingthenext layer. Startingfrom onestate,thealgorithmwouldbeessentiallyidenticalto breadth-firstsearchexceptthateachlayerisgeneratedall at once.c. Simulatedannealingwith ¾�Xz` atall times:ignoringthefactthattheterminationstepwouldbetriggeredimmediately, thesearchwouldbeidenticalto first-choicehill climb-http://librosysolucionarios.net20 Chapter 4. InformedSearchandExplorationing becauseevery downwardsuccessorwouldberejectedwith probability1. (Exercisemaybemodifiedin futureprintings.)d. Geneticalgorithmwith populationsize P¿XÀZ : if the populationsize is 1, then thetwo selectedparentswill bethesameindividual; crossover yieldsanexactcopy of theindividual; then thereis a small chanceof mutation. Thus, the algorithmexecutesarandomwalk in thespaceof individuals.4.12 If we assumethecomparisonfunctionis transitive, thenwe canalwayssortthenodesusing it, and choosethe nodethat is at the top of the sort. Efficient priority queuedatastructuresrely only on comparisonoperations,sowe losenothingin efficiency—exceptforthefactthatthecomparisonoperationonstatesmaybemuchmoreexpensive thancomparingtwo numbers,eachof whichcanbecomputedjustonce.A� reliesonthedivisionof thetotalcostestimate�KG�#_J into thecost-so-far andthecost-to-go. If we have comparisonoperatorsfor eachof these,thenwe canprefer to expandanodethat is betterthanothernodeson bothcomparisons.Unfortunately, therewill usuallybeno suchnode.Thetradeoff betweensBG�#_J and �BG�#_J cannotberealizedwithout numericalvalues.4.13 The spacecomplexity of LRTA� is dominatedby the spacerequiredfor �Á�O�����Â�W4 ±86"Ã�< ,i.e., theproductof thenumberof statesvisited(# ) andthenumberof actionstried perstate(Q ). The time complexity is at least n¨G�#KQ�{VJ for a naive implementationbecausefor eachaction taken we computean Ä value, which requiresminimizing over actions. A simpleoptimizationcanreducethis to noG�#KQTJ . This expressionassumesthateachstate–actionpairis tried at mostonce,whereasin factsuchpairsmaybetried many times,astheexampleinFigure4.22shows.4.14 Thisquestionis slightly ambiguousasto whattheperceptis—eithertheperceptis justthe location,or it givesexactly the setof unblocked directions(i.e., blocked directionsareillegal actions). We will assumethe latter. (Exercisemay be modifiedin future printings.)Thereare12 possiblelocationsfor internalwalls, so thereare « {²XÆÅ=`WÇWÈ possibleenviron-mentconfigurations.A belief statedesignatesa subsetof theseaspossibleconfigurations;for example,beforeseeingany perceptsall 4096configurationsarepossible—thisis asinglebelief state.a. Wecanview thisasacontingency problemin belief statespace.Theinitial belief stateis thesetof all 4096configurations.The total belief statespacecontains É�Ê�Ë�Ì beliefstates(onefor eachpossiblesubsetof configurations,but mostof thesearenot reach-able. After eachactionandpercept,the agentlearnswhetheror not an internalwallexistsbetweenthecurrentsquareandeachneighboringsquare.Hence,eachreachablebelief statecan be represntedexactly by a list of statusvalues(present,absent,un-known) for eachwall separately. That is, thebelief stateis completelydecomposableand thereareexactly e « { reachablebelief states. The maximumnumberof possiblewall-perceptsin eachstateis 16 ( É ), soeachbelief statehasfour actions,eachwith upto 16 nondeterministicsuccessors.http://librosysolucionarios.net21b. Assumingtheexternalwalls areknown, therearetwo internalwalls andhence { XÍÅpossiblepercepts.c. Theinitial null actionleadsto four possiblebeliefstates,asshown in FigureS4.2.Fromeachbeliefstate,theagentchoosesasingleactionwhichcanleadto upto 8 beliefstates(on enteringthemiddlesquare).Given thepossibility of having to retraceits stepsata deadend, the agentcan explore the entire mazein no more than 18 steps,so thecompleteplan (expressedasa tree)hasno morethan c «�Î nodes. On the otherhand,therearejust e « { , so theplancouldbe expressedmoreconciselyasa tableof actionsindexedby belief state(apolicy in theterminologyof Chapter17).?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï ?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï ?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?Ï?ÏNoOpRightFigure S4.2 The ÐAÑÒÐ mazeexplorationproblem:the initial state,first percept,andoneselectedactionwith its perceptualoutcomes.4.15 Hereis onesimplehill-climbing algorithm:� Connectall thecitiesinto anarbitrarypath.� Pick two pointsalongthepathat random.� Split thepathat thosepoints,producingthreepieces.� Try all six possiblewaysto connectthethreepieces.� Keepthebestone,andreconnectthepathaccordingly.� Iteratethestepsabove until no improvementis observedfor a while.4.1 4.16Codenot shown.http://librosysolucionarios.net22 Chapter 4. InformedSearchandExploration4.17 Hillclimbing is surprisinglyeffectiveatfindingreasonableif notoptimalpathsfor verylittle computationalcost,andseldomfails in two dimensions.a. It is possible(seeFigureS4.3(a))but veryunlikely—theobstaclehasto haveanunusualshapeandbepositionedcorrectlywith respectto thegoal.b. With convex obstacles,getting stuck is much more likely to be a problem(seeFig-ureS4.3(b)).c. Noticethatthis is justdepth-limitedsearch,whereyouchooseastepalongthebestpathevenif it is notasolution.d. Set � to themaximumnumberof sidesof any polygonandyoucanalwaysescape.CurrentÓpositionGoalÔ(a) (b)CurrentÓpositionGoalÔFigureS4.3 (a)Gettingstuckwith aconvex obstacle.(b) Gettingstuckwith anonconvexobstacle.4.18 The studentshouldfind that on the 8-puzzle,RBFS expandsmore nodes(becauseit doesnot detectrepeatedstates)but haslower costper nodebecauseit doesnot needtomaintaina queue. The numberof RBFS nodere-expansionsis not too high becausethepresenceof many tiedvaluesmeansthatthebestpathchangesseldom.Whentheheuristicisslightly perturbed,thisadvantagedisappearsandRBFS’sperformanceis muchworse.For TSP, thestatespaceis a tree,sorepeatedstatesarenotanissue.On theotherhand,theheuristicis real-valuedandthereareessentiallyno tied values,soRBFSincursa heavypenaltyfor frequentre-expansions.http://librosysolucionarios.netSolutionsfor Chapter5ConstraintSatisfactionProblems5.1 A constraint satisfactionproblem is a problemin which thegoalis to choosea valuefor eachof a setof variables,in suchaway thatthevaluesall obey asetof constraints.A constraint is arestrictionon thepossiblevaluesof two or morevariables.For exam-ple,aconstraintmight saythat 5ÕX�± is notallowedin conjunctionwith >[Xzp .Backtracking search is a form depth-firstsearchin which thereis a singlerepresenta-tion of thestatethatgetsupdatedfor eachsuccessor, andthenmustberestoredwhena deadendis reached.A directedarc from variable 5 to variable > in a CSPis arc consistentif, for everyvaluein thecurrentdomainof 5 , thereis someconsistentvalueof > .Backjumping is awayof makingbacktrackingsearchmoreefficient,by jumpingbackmorethanonelevel whenadeadendissreached.Min-conflicts is a heuristicfor usewith local searchon CSPproblems.Theheuristicsaysthat, whengiven a variableto modify, choosethe valuethat conflictswith the fewestnumberof othervariables.5.2 Thereare18 solutionsfor coloring Australiawith threecolors. Startwith SA, whichcanhave any of threecolors. Thenmoving clockwise,WA canhave eitherof theothertwocolors,andeverythingelseis strictly determined;thatmakes6 possibilitiesfor themainland,times3 for Tasmaniayields18.5.3 The most constrainedvariablemakes sensebecauseit choosesa variablethat is (allother things beingequal) likely to causea failure, and it is more efficient to fail as earlyaspossible(therebypruning large partsof the searchspace).The leastconstrainingvalueheuristicmakes sensebecauseit allows the most chancesfor future assignmentsto avoidconflict.5.4 a. Crossword puzzleconstructioncan be solved many ways. One simple choiceisdepth-firstsearch.Eachsuccessorfills in a word in thepuzzlewith oneof thewordsin thedictionary. It is betterto go onewordata time, to minimizethenumberof steps.b. As a CSP, thereareevenmorechoices.You couldhave a variablefor eachbox inthecrosswordpuzzle;in thiscasethevalueof eachvariableis a letter, andtheconstraintsarethat the lettersmustmake words. This approachis feasiblewith a most-constrainingvalueheuristic.Alternately, we couldhave eachstringof consecutive horizontalor verticalboxesbea singlevariable,andthedomainof thevariablesbewordsin thedictionaryof the right23http://librosysolucionarios.net24 Chapter 5. ConstraintSatisfactionProblemslength.Theconstraintswouldsaythattwo intersectingwordsmusthavethesameletterin theintersectingbox. Solvingaproblemin this formulationrequiresfewersteps,but thedomainsarelarger(assuminga big dictionary)andtherearefewer constraints.Both formulationsarefeasible.5.5 a. For rectilinearfloor-planning,onepossibility is to have a variablefor eachof thesmall rectangles,with the valueof eachvariablebeinga 4-tupleconsistingof the H and Icoordinatesof the upperleft and lower right cornersof the placewherethe rectanglewillbe located.Thedomainof eachvariableis thesetof 4-tuplesthatarethe right sizefor thecorrespondingsmallrectangleandthatfit within thelargerectangle.Constraintssaythatnotwo rectanglescanoverlap;for exampleif thevalueof variableÖ׫ is 4 `W6"`W6"ØW6"c�< , thennoothervariablecantake on avaluethatoverlapswith the `W6"` to ØW6"c rectangle.b. Forclassscheduling,onepossibilityis to havethreevariablesfor eachclass,onewithtimesfor values(e.g. MWF8:00,TuTh8:00,MWF9:00,...), onewith classroomsfor values(e.g. Wheeler110,Evans330,...) andonewith instructorsfor values(e.g. Abelson,Bibel,Canny, ...). Constraintssaythatonly oneclasscanbein thesameclassroomat thesametime,andan instructorcanonly teachoneclassat a time. Theremaybeotherconstraintsaswell(e.g.aninstructorshouldnothave two consecutive classes).5.6 The exact stepsdependon certainchoicesyou are free to make; hereare the onesImade:a. Choosethe ÙRÚ variable.Its domainis ÛW`W6"ZWÜ .b. Choosethevalue1 for Ù Ú . (We can’t choose0; it wouldn’t survive forwardchecking,becauseit would force Ý to be0, andtheleadingdigit of thesummustbenon-zero.)c. ChooseÝ , becauseit hasonly oneremainingvalue.d. Choosethevalue1 for Ý .e. Now Ù { andX « aretied for minimumremainingvaluesat2; let’s chooseÙ { .f. Eithervaluesurvivesforwardchecking,let’s choose0 for Ù { .g. Now Ùg« hastheminimumremainingvalues.h. Again,arbitrarily choose0 for thevalueof Ùg« .i. The variable n mustbe an even number(becauseit is the sumof ¾�fw¾ lessthan5(becausen®fun�XzÖwfuZW`¨ly` ). Thatmakesit mostconstrained.j . Arbitrarily choose4 asthevalueof n .k. Ö now hasonly 1 remainingvalue.l. Choosethevalue8 for Ö .m. ¾ now hasonly 1 remianingvalue.n. Choosethevalue7 for ¾ .o.Þ mustbeanevennumberlessthan9; chooseÞ .p. Theonly valuefor Þ thatsurvivesforwardcheckingis 6.q. Theonly variableleft is ß .r . Theonly valueleft for ß is 3.http://librosysolucionarios.net25s. This is asolution.This is a rathereasy(under-constrained) puzzle,so it is not surprisingthat we arrive at asolutionwith no backtracking(giventhatwe areallowedto useforwardchecking).5.7 Thereare implementationsof CSPalgorithmsin the Java, Lisp, andPythonsectionsof theonlinecoderepository;theseshouldhelpstudentsgetstarted.However, studentswillhave to addcodeto keepstatisticson theexperiments,andperhapswill want to have somemechanismfor making an experimentreturn failure if it exceedsa certain time limit (ornumber-of-stepslimit). Theamountof codethatneedsto bewritten is small; theexerciseismoreaboutrunningandanalyzinganexperiment.5.8 We’ll tracethrougheachiterationof thewhile loop in AC-3 (for onepossibleorderingof thearcs):a. Remove ~²5h�uß[5 , deleteÖ from ~²5 .b. Remove ~²5h�uà , delete> from ~²5 , leaving only ¬ .c. Remove Pá¾®�uß[5 , deleteÖ from PR¾ .d. Remove Pá¾®�u~²5 , delete¬ from PR¾ , leaving only > .e. Remove P�~²ßâ�u~²5 , delete¬ from P�~²ß .f. Remove P�~²ßâ�uà , delete> from Pg~²ß , leaving only Ö .g. Remove ãÕ�uPR¾ , delete> from ã .h. Remove ãÕ�u~²5 , delete¬ from ã .i. remove ã®�®Pg~²ß , deleteÖ from ã , leaving nodomainfor ã .5.9 On a tree-structuredgraph,no arc will be consideredmore than once,so the AC-3algorithmis noG�µ¨]�J , where µ is the numberof edgesand ] is the sizeof the largestdo-main.5.10 Thebasicideais to preprocesstheconstraintsso that, for eachvalueof Ùáä , we keeptrackof thosevariablesÙáå for whichanarcfrom Ùáå to Ù ä is satisfiedby thatparticularvalueof Ù ä . This datastructurecanbe computedin time proportionalto thesizeof theproblemrepresentation.Then,whena valueof Ù ä is deleted,we reduceby 1 thecountof allowablevaluesfor each G�Ùáå86�Ù ä J arc recordedunderthat value. This is very similar to the forwardchainingalgorithmin Chapter7. See? (?) for detailedproofs.5.11 Theproblemstatementsetsout thesolutionfairly completely. To expresstheternaryconstrainton 5 , > and \ that 5mfz>�Xæ\ , we first introducea new variable, 5A> . If thedomainof 5 and > is the setof numbersP , thenthe domainof 5A> is the setof pairsofnumbersfrom P , i.e. P�l·P . Now therearethreebinaryconstraints,onebetween5 and5A> sayingthatthevalueof 5 mustbeequalto thefirst elementof thepair-valueof 5A> ; onebetween> and 5&> sayingthat thevalueof > mustequalthesecondelementof thevalueof 5A> ; andfinally onethatsaysthat thesumof thepair of numbersthat is thevalueof 5&>mustequalthevalueof \ . All otherternaryconstraintscanbehandledsimilarly.Now that we canreducea ternaryconstraintinto binary constraints,we canreducea4-ary constrainton variables5A6">ç6"\v6"] by first reducing 5&6">ç6"\ to binary constraintsashttp://librosysolucionarios.net26 Chapter 5. ConstraintSatisfactionProblemsshown above, thenaddingback ] in a ternaryconstraintwith 5&> and \ , andthenreducingthis ternaryconstraintto binaryby introducing \o] .By induction,we canreduceany # -ary constraintto an G�#»�®ZWJ -ary constraint.We canstopat binary, becauseany unaryconstraintcanbedropped,simply by moving theeffectsoftheconstraintinto thedomainof thevariable.5.12 A simplealgorithmfor finding a cutsetof no morethan � nodesis to enumerateallsubsetsof nodesof size ZW6" W6"è"è"èé6"� , andfor eachsubsetcheckwhethertheremainingnodesform a tree.Thisalgorithmtakestimeêë�ì�í !! å , which is noG�# å J .Becker andGeiger(1994;http://citeseer.nj.nec.com/becker94approximation.html) giveanalgorithmcalledMGA (modifiedgreedyalgorithm)thatfindsacutsetthatis nomorethantwice thesizeof theminimalcutset,usingtime n¨G�µ·f·à·¶ ¸W¹KG�à?J�J , whereµ is thenumberofedgesand à is thenumberof variables.Whetherthis makes the cycle cutsetapproachpracticaldependsmore on the graphinvolved thanon theagorithmfor finding a cutset.That is because,for a cutsetof size } , westill have anexponentialG�U�î�J factorbeforewe cansolve theCSP. Soany graphwith a largecutsetwill beintractibleto solve,evenif we couldfind thecutsetwith no effort atall.5.13 The “Zebra Puzzle”canbe representedasa CSPby introducinga variablefor eachcolor, pet,drink,countryandcigaretbrand(atotalof 25variables).Thevalueof eachvariableis anumberfrom 1 to 5 indicatingthehousenumber. This is agoodrepresentationbecauseiteasyto representall theconstraintsgivenin theproblemdefinitionthis way. (We have doneso in the Pythonimplementationof the code,andat somepoint we may reimplementthisin theotherlanguages.)Besideseaseof expressinga problem,theotherreasonto choosearepresentationis theefficiency of finding a solution. herewe have mixed results—onsomeruns,min-conflictslocal searchfindsa solutionfor this problemin seconds,while on otherrunsit fails to find asolutionafterminutes.Anotherrepresentationis to have fivevariablesfor eachhouse,onewith thedomainofcolrs,onewith pets,andsoon.http://librosysolucionarios.netSolutionsfor Chapter6AdversarialSearch6.1 FigureS6.1showsthegametree,with theevaluationfunctionvaluesbelow theterminalnodesandthebacked-upvaluesto theright of thenon-terminalnodes.Thevaluesimply thatthebeststartingmove for X is to take thecenter. Theterminalnodeswith a bold outlinearetheonesthatdo notneedto beevaluated,assumingtheoptimalordering.xxxx o xox o xoxoxoxox o xoxoxoxo1−1 1 −21 −1 0 0 1 −1 −2 0 −1 01 2FigureS6.1 Partof thegametreefor tic-tac-toe,for Exercise6.1.6.2 Considera MIN nodewhosechildrenare terminalnodes. If MIN playssuboptimally,thenthevalueof thenodeis greaterthanor equalto thevalueit would have if MIN playedoptimally. Hence,the value of the MAX nodethat is the MIN node’s parentcan only beincreased.This argumentcanbe extendedby a simpleinductionall theway to the root. Ifthesuboptimalplay by MIN is predictable, thenonecando betterthana minimax strategy.For example,if MIN always falls for a certainkind of trap and loses,thensettingthe trapguaranteesa win even if thereis actuallya devastatingresponsefor MIN. This is shown inFigureS6.2.6.3a. (5) Thegametree,completewith annotationsof all minimaxvalues,is shown in Fig-ureS6.3.b. (5) The“?” valuesarehandledby assumingthatanagentwith a choicebetweenwin-ning thegameandenteringa “?” statewill alwayschoosethewin. That is, min(–1,?)is –1 andmax(+1,?)is +1. If all successorsare“?”, thebacked-upvalueis “?”.27http://librosysolucionarios.net28 Chapter 6. AdversarialSearchMAXMINa1AïB D−101000 10002ðbñ 3òbñ1bñ2ðdó1dó3òdó−5−5−5a2ôFigureS6.2 A simplegametreeshowing thatsettingatrapfor MIN by playing õ"ö is awinif MIN falls for it, but mayalsobedisastrous.Theminimaxmove is of courseõ÷ , with valueø § .(1,4)ù(2,4)ù(2,3)ù(1,3)ù(1,2)ù(3,2)ù(3,4)ù(4,3)ù(3,1)ù(2,4)ù(1,4)ù+1−1???−1−1−1+1+1+1Figure S6.3 Thegametreefor thefour-squaregamein Exercise6.3. Terminalstatesarein singleboxes,loop statesin doubleboxes.Eachstateis annotatedwith its minimaxvaluein a circle.c. (5) Standardminimax is depth-firstandwould go into an infinite loop. It canbefixedby comparingthecurrentstateagainstthestack;andif thestateis repeated,thenreturna “?” value. Propagationof “?” valuesis handledasabove. Althoughit works in thiscase,it doesnot alwayswork becauseit is not clearhow to compare“?” with a drawnposition;nor is it clearhow to handlethecomparisonwhentherearewins of differentdegrees(asin backgammon).Finally, in gameswith chancenodes,it is unclearhow tohttp://librosysolucionarios.net29computetheaverageof a numberanda “?”. Notethatit is not correctto treatrepeatedstatesautomaticallyasdrawn positions;in this example,both(1,4) and(2,4) repeatinthetreebut they arewonpositions.Whatis really happeningis thateachstatehasa well-definedbut initially unknownvalue.Theseunknown valuesarerelatedby theminimaxequationatthebottomofpage163. If thegametreeis acyclic, thentheminimaxalgorithmsolvestheseequationsbypropagatingfrom theleaves.If thegametreehascycles,thena dynamicprogrammingmethodmustbeused,asexplainedin Chapter17. (Exercise17.8studiesthisprobleminparticular.) Thesealgorithmscandeterminewhethereachnodehasa well-determinedvalue(asin this example)or is really aninfinite loop in thatbothplayerspreferto stayin theloop(or havenochoice).In suchacase,therulesof thegamewill needto definethevalue(otherwisethegamewill neverend).In chess,for eaxmple,astatethatoccurs3 times(andhenceis assumedto bedesirablefor bothplayers)is adraw.d. This questionis a little tricky. Oneapproachis a proof by inductionon thesizeof thegame.Clearly, thebasecase#áX�e is a lossfor A andthebasecase#áXÍÅ is a win forA. For any #T­wÅ , theinitial movesarethesame:A andB bothmove onesteptowardseachother. Now, we canseethat they areengagedin a subgameof size #y�ú on thesquares4 W6"è"è"è�6�#��³Z�< , exceptthat thereis anextra choiceof moveson squares and#��úZ . Ignoringthis for a moment,it is clearthat if the“ #y�m ” is won for A, thenAgetsto the square#û�zZ beforeB getsto square (by the definition of winning) andthereforegetsto # beforeB getsto Z , hencethe “ # ” gameis won for A. By thesameline of reasoning,if “ #��Õ ” is won for B then“ # ” is won for B. Now, thepresenceoftheextramovescomplicatestheissue,but not too much.First, theplayerwho is slatedto win thesubgame4 W6"è"è"è�6�#y�ÕZ�< never movesbackto his homesquare.If theplayerslatedto losethesubgamedoesso,thenit is easyto show thathe is boundto losethegameitself—theotherplayersimply movesforwardanda subgameof size #��m W� isplayedonestepcloserto theloser’s homesquare.6.4 See"search/algorit hms/ games .l is p" for definitionsof games,game-playingagents,andgame-playingenvironments."search/algorit hms/ mi nim ax .l is p" con-tainsthe minimax andalpha-betaalgorithms. Notice that the game-playingenvironmentisessentiallya genericenvironmentwith theupdatefunctiondefinedby therulesof thegame.Turn-takingis achievedby having agentsdo nothinguntil it is their turn to move.See"search/domains /co gnac .l is p" for thebasicdefinitionsof asimplegame(slightly morechallengingthanTic-Tac-Toe). Thecodefor this containsonly a trivial eval-uation function. Studentscan useminimax and alpha-betato solve small versionsof thegameto termination(probablyup to Å�lwe ); they shouldnoticethat alpha-betais far fasterthanminimax,but still cannotscaleupwithoutanevaluationfunctionandtruncatedhorizon.Providing an evaluationfunction is an interestingexercise. From the point of view of datastructuredesign,it is alsointerestingto look at how to speedup thelegal move generatorbyprecomputingthedescriptionsof rows,columns,anddiagonals.Very few studentswill have heardof kalah, so it is a fair assignment,but the gameis boring—depth6 lookaheadanda purely material-basedevaluationfunction areenoughhttp://librosysolucionarios.net30 Chapter 6. AdversarialSearchto beatmosthumans.Othello is interestingandaboutthe right level of difficulty for moststudents.Chessandcheckers aresometimesunfair becauseusually a small subsetof theclasswill beexpertswhile therestarebeginners.6.5 This questionis not ashardasit looks. Thederivationbelow leadsdirectly to a defini-tion of ü andý values.Thenotation# ä refersto (thevalueof) thenodeatdepthþ on thepathfrom theroot to theleafnode#*ÿ . Nodes# ä «(è"è"è�# ä�� ë arethesiblingsof nodeþ .a. Wecanwrite # { X������dG�# Ú 6�# Ú�« 6"è"è"èO6�# Ú�� J , giving#�«�X�� � G������dG�#|ÚW6�#|Ú�«6"è"è"è�6�# Ú� J�6�# { «6"è"è"èO6�# { �� JThen#|Ú canbesimilarly replaced,until we have anexpressioncontaining#*ÿ itself.b. In termsof the � and� values,we have# « X�� � G�� { 6������|G�� Ú 6�# Ú 6�� Ú J�6�� { JAgain, #|Ú can be expandedout down to #*ÿ . The most deeplynestedterm will be� � G�� ÿ 6�# ÿ 6�� ÿ J .c. If # ÿ is a maxnode,thenthe lower boundon its valueonly increasesasits successorsareevaluated.Clearly, if it exceeds�Âÿ it will haveno furthereffecton #�« . By extension,if it exceeds� � G�� { 6�� É 6"è"è"è 6�� ÿOJ it will have no effect. Thus,by keepingtrack of thisvaluewe candecidewhento prune#*ÿ . This is exactlywhat ü -ý does.d. Thecorrespondingboundfor min nodes#*å is �����KG�� Ú�6�����6"è"è"è 6�� å8J .6.7 Thegeneralstrategy is to reducea generalgametreeto a one-plytreeby inductiononthedepthof thetree. Theinductive stepmustbedonefor min, max,andchancenodes,andsimply involvesshowing thatthetransformationis carriedthoughthenode.Supposethatthevaluesof thedescendantsof anodeareH « è"è"è2H ! , andthatthetransformationis ± H×f»p , where± is positive. Wehave� � G�± H « fup6"± H { fup6"è"è"è 6"±�H ! fupJ�X¯±�� � G�H « 6�H { 6"è"è"è�6�H ! JKfup�����KG�± H�«bfup6"± H { fup6"è"è"è 6"±�H ! fupJ�X¯±�� � G�Hb«6�H { 6"è"è"è�6�H ! JKfup� « G�±�H « f®p�J*f � { G�± H { fupJKf*ck"x"x;f � ! G�±�H ! fupJ�X¯±�G � « H « f � { H { f*ck"x"x � ! H ! J|fupHencetheproblemreducestoaone-plytreewheretheleaveshavethevaluesfrom theoriginaltreemultiplied by thelinear transformation.SinceH·­úI��¯± H�fmpº­�± IáfÕp if ±�­�` , thebestchoiceat theroot will bethesameasthebestchoicein theoriginal tree.6.8 This procedurewill give incorrectresults. Mathematically, the procedureamountstoassumingthat averagingcommuteswith min andmax, which it doesnot. Intuitively, thechoicesmadeby eachplayerin thedeterministictreesarebasedon full knowledgeof futuredice rolls, andbearno necessaryrelationshipto the movesmadewithout suchknowledge.(Noticetheconnectionto thediscussionof cardgameson page179andto thegeneralprob-lem of fully andpartially observableMarkov decisionproblemsin Chapter17.) In practice,themethodworksreasonablywell, andit might beagoodexerciseto have studentscompareit to thealternative of usingexpectiminimaxwith sampling(ratherthansummingover) dicerolls.http://librosysolucionarios.net316.9 Codenot shown.6.10 Thebasicphysicalstateof thesegamesis fairly easyto describe.Oneimportantthingto rememberfor Scrabbleandbridgeis that thephysicalstateis not accessibleto all playersandsocannotbeprovideddirectly to eachplayerby theenvironmentsimulator. Particularlyin bridge,eachplayerneedsto maintainsomebestguess(or multiple hypotheses)asto theactualstateof theworld. We expectto beputtingsomeof thegameimplementationsonlineasthey becomeavailable.6.11 Onecan think of chanceeventsduring a game,suchasdice rolls, in the samewayashiddenbut preordainedinformation(suchasthe orderof the cardsin a deck). The keydistinctionsarewhethertheplayerscaninfluencewhat informationis revealedandwhetherthereis any asymmetryin theinformationavailableto eachplayer.a. Expectiminimaxis appropriateonly for backgammonand Monopoly. In bridge andScrabble,eachplayerknows thecards/tilesheor shepossessesbut not theopponents’.In Scrabble,thebenefitsof a fully rational,randomizedstrategy thatincludesreasoningabouttheopponents’stateof knowledgeareprobablysmall,but in bridgethequestionsof knowledgeandinformationdisclosurearecentralto goodplay.b. None,for thereasonsdescribedearlier.c. Key issuesincludereasoningabouttheopponent’s beliefs,theeffect of variousactionson thosebeliefs,andmethodsfor representingthem. Sincebelief statesfor rationalagentsareprobabilitydistributionsoverall possiblestates(includingthebeliefstatesofothers),this is nontrivial.function MAX-VALUE( �������� ) returns �"!#�%$'&($'��)+*,�-&(!# if TERMINAL-TEST( �.������ ) then return UTIL ITY( �.������ )*0/ ø21for �435� in SUCCESSORS( �.�%�-�% ) doif WINNER( � ) = MAXthen *0/ MAX(v, MAX-VALUE( � ))else *0/ MAX(v, M IN-VALUE( � ))return *FigureS6.4 Partof themodifiedminimaxalgorithmfor gamesin which thewinnerof theprevioustrick playsfirst on thenext trick.6.12 (In thefirst printing,thisexericserefersto WINNER( ���,6�M�N ); subsequentprintingsreferto WINNER( � ), denotingthewinnerof thetrick justcompleted(if any), ornull.) Thisquestionis interpretedasapplyingonly to theobservablecase.a. Themodificationto MAX-VALUE is shown in FigureS6.4.If MAX hasjustwonatrick,MAX getsto playagain,otherwiseplayalternates.Thus,thesuccessorsof aMAX nodehttp://librosysolucionarios.net32 Chapter 6. AdversarialSearch MAX MINS 2H 6 4D 6 C 9,8 10,5 MAX MINS 2H 4D 6 C 9,8 10,5 MAX MINS 2H 6 4D C 9,8 10,5 MAX MINS 2H 6 4D 6 C 9 10,5 MAX MINS 2H D 6 C 9,8 10,5 MAX MINS 2H D 6 C 9 10,5 MAX MINS 2H D C 9,8 10,5 MAX MINS H D C 9,8 10,5 MAX MINS 2H D C 9,8 10 MAX MINS 2H D C 9,8 5 MAX MINS H D C 9 10,5 MAX MINS 2H D C 9 10 MAX MINS 2H D C 9 5 MAX MINS H D C 9 5 MAX MINS H D C 9 10 MAX MINS H D C 9 MAX MINS H D C 10 MAX MINS 2H D C 9 MAX MINS 2H D C 9 MAX MINS H D C 9 MAX MINS 2H D C MAX MINS 2H D 6 C 9 10 MAX MINS 2H D 6 C 9 5 MAX MINS 2H D C 9 10 MAX MINS 2H D 6 C 10 MAX MINS H D C 9 10 MAX MINS 2H D C 9 MAX MINS H D C 10 MAX MINS 2H D C MAX MINS H D 6 C MAX MINS 2 H D 6 C MAX MINS H D 6 C 9 5 MAX MINS 2H D 6 C 9 MAX MINS H D C 9 5 MAX MINS H D 6 C 5 MAX MINS H D C 9 MAX MINS H D 6 C MAX MINS 2H D 6 C MAX MINS 2H D C +2 +2 07+4 +2 +4 0707−2 +2+2 +2 07+4 +2 +4 0707−2 +2+2 +2 07+4 +2 +4 0707−2 +2+2+207+407+407+2 0707+207+2070707FigureS6.5 Ex. 6.12:Part of thegametreefor thecardgameshown on p.179.canbe a mixture of MAX andM IN nodes,dependingon the variouscardsMAX canplay. A similar modificationis neededfor M IN-VALUE.b. Thegametreeis shown in FigureS6.5.6.13 The naive approachwould be to generateeachsuchposition,solve it, andstoretheoutcome.Thiswouldbeenormouslyexpensive—roughlyontheorderof 444billion seconds,or 10,000years,assumingit takes a secondon averageto solve eachposition (which isprobablyvery optimistic). Of course,we can take advantageof already-solved positionswhen solving new positions,provided thosesolved positionsare descendantsof the newpositions.To ensurethatthis alwayshappens,we generatethefinal positionsfirst, thentheirpredecessors, andsoon. In this way, theexactvaluesof all successorsareknown wheneachstateis generated.Thismethodis calledretrogradeanalysis./0 �./����2/��980�0���� 3;:�� :http://librosysolucionarios.net336.14 Themostobviouschangeis thatthespaceof actionsis now continuous.For example,in pool, thecueingdirection,angleof elevation,speed,andpoint of contactwith thecueballareall continuousquantities.Thesimplestsolutionis just to discretizetheactionspaceandthenapplystandardmeth-ods.This might work for tennis(modelledcrudelyasalternatingshotswith speedanddirec-tion), but for gamessuchas pool and croquetit is likely to fail miserablybecausesmallchangesin directionhave large effects on actionoutcome. Instead,onemust analyzethegameto identify a discretesetof meaningfullocal goals,suchas“potting the4-ball” in poolor “laying up for thenext hoop” in croquet.Then,in thecurrentcontext, a localoptimizationroutinecanwork out thebestwayto achieveeachlocalgoal,resultingin adiscretesetof pos-siblechoices.Typically, thesegamesarestochastic,sothebackgammonmodelis appropriateprovidedthatweusesampledoutcomesinsteadof summingover all outcomes.Whereaspool andcroquetaremodelledcorrectlyasturn-takinggames,tennisis not.While oneplayeris moving to theball, theotherplayeris moving to anticipatetheopponent’sreturn.This makestennismorelike thesimultaneous-actiongamesstudiedin Chapter17. Inparticular, it maybereasonableto derive randomizedstrategiesso that theopponentcannotanticipatewheretheball will go.6.15 The minimax algorithm for non-zero-sumgamesworks exactly as for multiplayergames,describedon p.165–6;that is, the evaluationfunction is a vectorof values,oneforeachplayer, andthebackupstepselectswhichever vectorhasthehighestvaluefor theplayerwhoseturn it is to move. Theexampleat theendof Section6.2(p.167)shows thatalpha-betapruning is not possiblein generalnon-zero-sumgames,becausean unexaminedleaf nodemight beoptimalfor bothplayers.6.16 With 32 pieces,eachneeding6 bits to specifyits positionon oneof 64 squares,weneed24bytes(6 32-bitwords)tostoreaposition,sowecanstoreroughly20million positionsin thetable(ignoringpointersfor hashtablebucket lists). This is aboutone-ninthof the180million positionsgeneratedduringa three-minutesearch.Generatingthe hashkey directly from an array-basedrepresentationof the positionmight be quite expensive. Modernprograms(see,e.g.,Heinz, 2000)carry alongthe hashkey andmodify it aseachnew positionis generated.Supposethis takeson theorderof 20operations;thenon a 2GHz machinewherean evaluationtakes2000operationswe candoroughly100lookupsperevaluation.Usinga roughfigureof onemillisecondfor adisk seek,we coulddo 1000evaluationsper lookup. Clearly, usinga disk-residenttableis of dubiousvalue,evenif wecangetsomelocality of referenceto reducethenumberof disk reads.http://librosysolucionarios.netSolutionsfor Chapter7AgentsthatReasonLogically7.1 The wumpusworld is partially observable,deterministic,sequential(you needto re-memberthestateof onelocationwhenyou returnto it on a later turn), static,discrete,andsingleagent(thewumpus’s soletrick—devouring anerrantexplorer—is not enoughto treatit asanagent).Thus,it is a fairly simpleenvironment.Themaincomplicationis thepartialobservability.7.2 To save space,we’ll show the list of modelsasa tableratherthana collectionof dia-grams.Thereareeightpossiblecombinationsof pits in thethreesquares,andfour possibili-tiesfor thewumpuslocation(includingnowhere).We can seethat <y>>= X ü { becauseevery line where <y> is true also has ü { true.Similarly for üBÚ .7.3a. Thereis a pl true in the Pythoncode,anda versionof ask in the Lisp codethatservesthesamepurpose.TheJava codedid nothave this functionasof May 2003,butit shouldbeaddedsoon.)b. Thesentences¾?�V�A@ , BDCFEGB , and BDHIEJB canall bedeterminedto betrueor falseinapartialmodelthatdoesnot specifythetruth valuefor B .c. It is possibleto createtwo sentences,eachwith � variablesthatarenot instantiatedinthepartialmodel,suchthatoneof themis truefor all å possiblevaluesof thevariables,while theothersentenceis falsefor oneof the å values.Thisshowsthatin generalonemustconsiderall å possibilities.Enumeratingthemtakesexponentialtime.d. The python implementationof pl true returnstrue if any disjunctof a disjunctionis true,andfalseif any conjunctof a conjunctionis false. It will do this even if otherdisjuncts/conjunctscontainsuninstantiatedvariables.Thus,in thepartialmodelwhereB is true, BKCçã returnstrue,and EJB�Hçã returnsfalse.But thetruthvaluesof ãLCME_ã ,ãDC?¾?�V�A@ , and ãDHNE²ã arenotdetected.e. Ourversionof tt entails alreadyusesthismodifiedpl true . It wouldbeslowerif it did not.7.4 Remember, üO= X ý if f in very modelin which ü is true, ý is alsotrue.Therefore,a. A valid sentenceis onethatis truein all models.Thesentence¾?�V�A@ is alsovalid in allmodels.Soif±P� � �=± is valid thentheentailmentholds(becauseboth ¾?�V�A@ and ü hold34http://librosysolucionarios.net35Model Q�© R8÷ RPST�UWVYXZY[.\ S T�UWVYXZ ÷ \ ÷Z S \ [ T�UWVYXZY[.\ S , Z ÷ \ ÷Z ÷ \ ÷ , Z S \ [Z S \ [ , ZY[.\ S T�UWVYXZY[.\ S , Z S \ [ , Z ÷ \ ÷]N[.\ S T�UWVYX T�UWV^X]N[.\ S , ZY[.\ S T�UWVYX T�UWV^X]N[.\ S , Z ÷ \ ÷ T�UWV^X]N[.\ S , Z S \ [ T�UWVYX T�UWVYX T�UWV^X] [.\ S , Z [.\ S , Z ÷ \ ÷ T�UWV^X] [.\ S , Z ÷ \ ÷ , Z S \ [ T�UWV^X] [.\ S , Z S \ [ , Z [.\ S T�UWVYX T�UWV^X] [.\ S , Z [.\ S , Z S \ [ , Z ÷ \ ÷ T�UWV^X] S \ [ , T�UWVYX] S \ [ , Z [.\ S T�UWVYX] S \ [ , Z ÷ \ ÷] S \ [ , Z S \ [ T�UWVYX] S \ [ , Z [.\ S , Z ÷ \ ÷] S \ [ , Z ÷ \ ÷ , Z S \ [] S \ [ , Z S \ [ , ZY[.\ S T�UWVYX] S \ [ , ZY[.\ S , Z S \ [ , Z ÷ \ ÷] ÷ \ ÷ T�UWVYX] ÷ \ ÷ , ZY[.\ S T�UWVYX] ÷ \ ÷ , Z ÷ \ ÷] ÷ \ ÷ , Z S \ [ T�UWVYX] ÷ \ ÷ , ZY[.\ S , Z ÷ \ ÷] ÷ \ ÷ , Z ÷ \ ÷ , Z S \ [] ÷ \ ÷ , Z S \ [ , ZY[.\ S T�UWVYX] ÷ \ ÷ , ZY[.\ S , Z S \ [ , Z ÷ \ ÷Figure7.1 A truth tableconstructedfor Ex. 7.2.Propositionsnot listedastrueonagivenline areassumedfalse,andonlyT�UWVYXentriesareshown in thetable.in every model),andif theentailmentholdsthen ü mustbe valid, becauseit mustbetruein all models,becauseit mustbetruein all modelsin which ¾?�V�A@ holds.b. Ý ±_�.Ã`@ doesn’t hold in any model,so ü trivially holdsin every modelthat Ýo±_�.Ã`@ holdsin.c. üa� ý holdsin thosemodelswhereý holdsor where E_ü holds.Thatis preciselythecaseif üb� ý is valid.d. This follows from applyingc in bothdirections.http://librosysolucionarios.net36 Chapter 7. AgentsthatReasonLogicallye. This reducesto c, becauseücHFEKý is unsatisfiablejustwhen üb� ý is valid.7.5 Thesecanbe computedby countingthe rows in a truth tablethat comeout true. Re-memberto countthepropositionsthatarenot mentioned;if a sentencementionsonly 5 and> , thenwe multiply thenumberof modelsfor ÛW5A6">?Ü by { to accountfor \ and ] .a. 6b. 12c. 47.6 A binary logical connective is definedby a truth tablewith 4 rows. Eachof the fourrows may be true or false,so thereare ÉgX ZWÈ possibletruth tables,andthus16 possibleconnectives. Six of thesearetrivial onesthat ignoreoneor both inputs; they correspondto¾?�V�A@ , Ý ±P�.Ã-@ , B , ã , EJB and E²ã . Four of themwe have alreadystudied: H²6�C_6d� 6fe .The remainingsix arepotentiallyuseful. Oneof themis reverseimplication ( g insteadof� ), andtheotherfive arethenegationsof H_6�C²6D� 6he and g . (Thefirst two of thesearesometimescallednandandnor.)7.7 Weusethetruth tablecodein Lisp in thedirectorylogic/prop.lisp to show eachsentenceis valid. We substituteP, Q, R for ü 6�ý�6�i becauseof thelack of GreeklettersinASCII. To save spacein this manual,weonly show thefirst four truth tables:> (truth-table "P ˆ Q <=> Q ˆ P")--------------------------------------- --P Q P ˆ Q Q ˆ P (P ˆ Q) <=> (Q ˆ P)--------------------------------------- --F F F F \(true\)T F F F TF T F F TT T T T T--------------------------------------- --NIL> (truth-table "P | Q <=> Q | P")--------------------------------------- --P Q P | Q Q | P (P | Q) <=> (Q | P)--------------------------------------- --F F F F TT F T T TF T T T TT T T T T--------------------------------------- --NIL> (truth-table "P ˆ (Q ˆ R) <=> (P ˆ Q) ˆ R")--------------------------------------- ------ ------ ------ ------ ------ --P Q R Q ˆ R P ˆ (Q ˆ R) P ˆ Q ˆ R (P ˆ (Q ˆ R)) <=> (P ˆ Q ˆ R)--------------------------------------- ------ ------ ------ ------ ------ --F F F F F F TT F F F F F TF T F F F F Thttp://librosysolucionarios.net37T T F F F F TF F T F F F TT F T F F F TF T T T F F TT T T T T T T--------------------------------------- ------ ------ ------ ------ ------ --NIL> (truth-table "P | (Q | R) <=> (P | Q) | R")--------------------------------------- ------ ------ ------ ------ ------ --P Q R Q | R P | (Q | R) P | Q | R (P | (Q | R)) <=> (P | Q | R)--------------------------------------- ------ ------ ------ ------ ------ --F F F F F F TT F F F T T TF T F T T T TT T F T T T TF F T T T T TT F T T T T TF T T T T T TT T T T T T T--------------------------------------- ------ ------ ------ ------ ------ --NILFor the remainingsentences,we just show that they arevalid accordingto the validityfunction:> (validity "˜˜P <=> P")VALID> (validity "P => Q <=> ˜Q => ˜P")VALID> (validity "P => Q <=> ˜P | Q")VALID> (validity "(P <=> Q) <=> (P => Q) ˆ (Q => P)")VALID> (validity "˜(P ˆ Q) <=> ˜P | ˜Q")VALID> (validity "˜(P | Q) <=> ˜P ˆ ˜Q")VALID> (validity "P ˆ (Q | R) <=> (P ˆ Q) | (P ˆ R)")VALID> (validity "P | (Q ˆ R) <=> (P | Q) ˆ (P | R)")VALID7.8 We usethevalidity function from logic/prop.lis p to determinethevalidityof eachsentence:> (validity "Smoke => Smoke")VALID> (validity "Smoke => Fire")SATISFIABLE> (validity "(Smoke => Fire) => (˜Smoke => ˜Fire)")SATISFIABLE> (validity "Smoke | Fire | ˜Fire")VALIDhttp://librosysolucionarios.net38 Chapter 7. AgentsthatReasonLogically> (validity "((Smoke ˆ Heat) => Fire) <=> ((Smoke => Fire) | (Heat => Fire))")VALID> (validity "(Smoke => Fire) => ((Smoke ˆ Heat) => Fire)")VALID> (validity "Big | Dumb | (Big => Dumb)")VALID> (validity "(Big ˆ Dumb) | ˜Dumb")SATISFIABLEMany peoplearefooledby (e) and(g) becausethey think of implicationasbeingcau-sation,or somethingcloseto it. Thus,in (e), they feel that it is the combinationof SmokeandHeatthat leadsto Fire, andthusthereis no reasonwhy oneor the otheraloneshouldleadto fire. Similarly, in (g), they feel that thereis no necessarycausalrelationbetweenBigandDumb, so the sentenceshouldbe satisfiable,but not valid. However, this reasoningisincorrect,becauseimplication is not causation—implicationis just a kind of disjunction(inthesensethat B � ã is thesameas EJBjC�ã ). So >&þ0skC�]o�=QTplCyG�>&þ0sm� ] �BQTpJ isequivalentto >&þ0snC�]o�=QTp5CFE²>&þ0snC�]o�=QTp , which is equivalentto >&þ0snCFE²>&þ0snC»]o�=QTp ,which is truewhether>tþ s is trueor false,andis thereforevalid.7.9 Fromthefirst two statements,weseethatif it is mythical,thenit is immortal;otherwiseit is a mammal.Soit mustbeeitherimmortalor a mammal,andthushorned.Thatmeansitis alsomagical.However, we can’t deduceanything aboutwhetherit is mythical. Usingthepropositionalreasoningcode:> (setf kb (make-prop-kb))#S(PROP-KB SENTENCE(AND))> (tell kb "Mythical => Immortal")T> (tell kb "˜Mythical => ˜Immortal ˆ Mammal")T> (tell kb "Immortal | Mammal => Horned")T> (tell kb "Horned => Magical")T> (ask kb "Mythical")NIL> (ask kb "˜Mythical")NIL> (ask kb "Magical")T> (ask kb "Horned")T7.10 Eachpossibleworld canbe written asa conjunctionof symbols,e.g. G�5DH�\�HyµoJ .Assertingthatapossibleworld is not thecasecanbewrittenby negatingthat,e.g. E²G�5NH?\IHµoJ , which canberewritten as G�E_5oCpE²\OCpE²µoJ . This is theform of aclause;a conjunctionof theseclausesis aCNFsentence,andcanlist all thepossibleworldsfor thesentence.7.11http://librosysolucionarios.net39a. This is a disjunctionwith 28 disjuncts,eachonesayingthat two of the neighborsaretrueandtheothersarefalse.Thefirst disjunctisÙ {-q { H?Ùg« q { HFEdÙ Ê-q { HNEKÙ Ê-q «0HFEdÙ {-q «AHFEdÙ Ê-q Ê HFEdÙg« q Ê HNEKÙ {-q ÊTheother27 disjunctseachselecttwo differentÙ ä q ÿ to betrue.b. Therewill be ! å disjuncts,eachsayingthat � of the # symbolsaretrueandtheothersfalse.c. For eachof thecellsthathavebeenprobed,take theresultingnumber# revealedby thegameandconstructa sentencewith ! Î disjuncts.Conjoin all thesentencestogether.ThenuseDPLL to answerthe questionof whetherthis sentenceentails Ù ä q ÿ for theparticularþ 6�r pair youareinterestedin.d. To encodethe global constraintthat thereare s minesaltogether, we canconstructa disjunctwith t u disjuncts,eachof size P . Remember, tuwv tFx y{z(t � uw| x . So fora Minesweepergamewith 100 cells and20 mines,this will be morrethan ZW` Ú Ë , andthus cannotbe representedin any computer. However, we can representthe globalconstraintwithin the DPLL algorithmitself. We add the parametermin and max totheDPLL function; theseindicatetheminimumandmaximumnumberof unassignedsymbolsthatmustbetruein themodel.For anunconstrainedproblemthevalues0 andP will be usedfor theseparameters.For a mineseeperproblemthe value s will beusedfor bothmin andmax. Within DPLL, we fail (returnfalse)immediatelyif min islessthanthenumberof remainingsymbols,or if maxis lessthan0. For eachrecursivecall to DPLL, we updatemin andmaxby subtractingonewhenwe assigna truevalueto asymbol.e. No conclusionsare invalidatedby addingthis capability to DPLL and encodingtheglobalconstraintusingit.f. Considerthis stringof alternating1’s andunprobedcells(indicatedby adash):|-|1|-|1|-|1|- |1 |- |1 |-| 1| -| 1| -|Thereare two possiblemodels: either thereare minesunderevery even-numbereddash,or underevery odd-numbereddash.Making a probeat eitherendwill determinewhethercellsat thefarendareemptyor containmines.7.12a. Bm� ã is equivalentto EGB}Cáã by implicationelimination(Figure7.11),and E²G�Bb«YHx"x"x~HKB��oJ is equivalentto G�EJBb«�Cyx"x"x~CLEJB��oJ by de Morgan’s rule, so G�EJB9«lCyx"x"x~CEJB��DC�ãoJ is equivalentto G�B9«0H�x"x"x�HFB��¨JF� ã .b. A clausecanhave positive andnegative literals;arrangethemin theform G�EJB « CÆx"x"x�CB���CÆ㨫0C�x"x"x4C»ã ! J . Then,setting ã�Xz㨫AC�x"x"x4C»ã ! , we haveG�EJBb«0C�x"x"x4CFB��DC»ã¨«AC�x"x"x4C�ã ! Jis equivalenttoG�Bb«0H�x"x"x4HFB��oJF� 㨫0CÆx"x"x�C»ã !http://librosysolucionarios.net40 Chapter 7. AgentsthatReasonLogicallyc. For atoms� ä 6�� ä 6�� ä 6"à ä whereUNIFY G � ÿ�6���å�J�X�� :� « H»è"è"è � ÿ è"è"è�H � ! í � � « C�è"è"è.� ! �ÃV«0HÆè"è"è�HÆà ! � �O«0CÆè"è"èl��åûè"è"è4CF� !��SUBST z(� q z�� í��_����� ����� í � �W��� í � ��� í ���"í��P����� � � n��� í��_����� � � � �{�"í��_�����.� ê � í �{� ê � í �_����� �{� � � |�|7.13a. 5J���FS �+�Le 5G���@S �w� � « HFE²~²�BSéS,���b. Ý ±�}Áþ0#Ks(Ö�þ0sB�#� � e G�Ý ±�}Áþ0#Ks(Ö�þ0sB�#� � � « HIEd¾v���V#_Ö�þ0s(�#� � HIEd¾v���V#G��@��P� � JC²G�Ý ±8} þ #KsBÞ � � � « H ¾v���V#_Ö�þ0sB�4�.�C²G�Ý ±8} þ #KsB]�S �&# � � « Hç¾v���V#G��@��P� �c. Theseformulaearethesameas(7.7)and(7.8),exceptthat the B for pit is replacedbyß for wumpus,and > for breezyis replacedby ~ for smelly.<yG�E²ß É-q É J � e <yG�E²~FÚ q É J � CN<yG�E²~KÅ=6�ÅBJ �<yG�ß É-q É J��}e <yG�~@Ú q É J��YHF<yG�E²ß {-q É J��YHF<yG�E²ß�Ú q ÚVJ��C_G�<�G�~ É-q ÚVJ � HI<�G�E_ß É-q { J � HF<yG�E²ßÆÚ q ÚVJ �7.14 Optimal behavior meansachieving an expectedutility that is as good as any otheragentprogram.ThePL-WUMPUS-AGENT is clearlynon-optimalwhenit choosesa randommove (andmaybenon-optimalin otherbraqnchesof its logic). Oneexample:in somecaseswhentherearemany dangers(breezesandsmells)but no safemove, the agentchoosesatrandom.A morethoroughanalysisshouldshow whenit is betterto do that,andwhenit isbetterto go homeandexit thewumpusworld, giving up on any chanceof finding thegold.Evenwhenit is bestto gambleon anunsafelocation,our agentdoesnot distinguishdegreesof safety– it shouldchoosetheunsafesquarewhich containsa dangerin thefewestnumberof possiblemodels.Theserefinementsarehardto stateusinga logicalagent,but we will seein subsequentchaptersthataprobabilisticagentcanhandlethem.does7.15 PL-WUMPUS-AGENT keepstrackof 6 staticstatevariablesbesidesKB. Thedifficultyis thatthesevariableschange—wedon’t justaddnew informationaboutthem(aswedowithpitsandbreezylocations),wemodify exisitng information.Thisdoesnot sit well with logic,which is designedfor eternaltruths. Sotherearetwo alternatives. Thefirst is to superscripteachpropositionwith the time (aswe did with the circuit agents),and thenwe could, forexample,doTELL G�<y>ç6"5 Ú « q « J to saythattheagentis atlocation ZW6"Z attime3. Thenattime4,wewouldhaveto copy overmany of theexistingpropositions,andaddnew ones.Thesecondpossibilityis to treateverypropositionasatimelessone,but to removeoutdatedpropositionsfrom the KB. That is, we could do RETRACT G�<y>?6"5º« q «J andthenprogTell G�<y>ç6"5A« q { J toindicatethat theagenthasmovedfrom ZW6"Z to ZW6" . Chapter10 describesthesemanticsandimplementationof RETRACT.NOTE: Avoid assigningthis problemif you don’t feel comfortablerequiringstudentsto think aheadaboutthepossibilityof retraction.7.16 It will take time proportionalto thenumberof puresymbolsplus thenumberof unitclauses.We assumethat <y> � ü is false,andprove a contradiction. E²G�<y> � ü J isequivalent to <y>�H�E²ü . From this sentencethe algorithmwill first eliminateall the purehttp://librosysolucionarios.net41symbols,thenit will work on unit clausesuntil it choosesü (which is a unit clause);at thatpoint it will immediatelyrecognizethat eitherchoice(true or false)for ü leadsto failure,whichmeansthattheoriginalnon-negatedassertionis true.7.17 Codenot shown.http://librosysolucionarios.netSolutionsfor Chapter8First-OrderLogic8.1 This questionwill generatea wide variety of possiblesolutions. The key distinctionbetweenanalogicaland sententialrepresentationsis that the analogicalrepresentationau-tomaticallygeneratesconsequencesthat canbe “read off ” whenever suitablepremisesareencoded.Whenyou get into the details,this distinction turns out to be quite hard to pindown—for example,what does“read off ” mean?—but it canbe justified by examiningthetime complexity of variousinferenceson the “virtual inferencemachine”provided by therepresentationsystem.a. Dependingon the scaleandtype of the map,symbolsin the map languagetypicallyincludecity andtown markers,roadsymbols(varioustypes),lighthouses,historicmon-uments,rivercourses,freeway intersections,etc.b. Explicit andimplicit sentences:thisdistinctionis alittle tricky, but thebasicideais thatwhenthemap-drawer plunksa symboldown in a particularplace,hesaysoneexplicitthing (e.g. thatCoit Tower is here),but theanalogicalstructureof themaprepresenta-tion meansthatmany implicit sentencescannow bederived. Explicit sentences:thereis amonumentcalledCoit Towerat this location;LombardStreetruns(approximately)east-west;SanFranciscoBay exists andhasthis shape.Implicit sentences:VanNessis longerthanNorth Willard; Fisherman’s Wharf is north of theMission District; theshortestdrivableroutefrom Coit Tower to Twin Peaksis thefollowing è"è"è .c. Sentencesunrepresentablein themaplanguage:TelegraphHill is approximatelyconi-cal andabout430feethigh (assumingthemaphasno topographicalnotation);in 1890therewasno bridgeconnectingSanFranciscoto Marin County(mapdoesnot repre-sentchanginginformation);Interstate680runseithereastor westof WalnutCreek(nodisjunctive information).d. Sentencesthat are easierto expressin the map language:any sentencethat can bewritten easily in English is not going to be a goodcandidatefor this question. Anylinguistic abstractionfrom thephysicalstructureof SanFrancisco(e.g. SanFranciscois on theendof a peninsulaat themouthof a bay)canprobablybeexpressedequallyeasily in the predicatecalculus,sincethat’s what it wasdesignedfor. Factssuchasthe shapeof the coastline,or the pathtaken by a road,arebestexpressedin the maplanguage.Eventhen,onecanarguethatthecoastlinedrawn onthemapactuallyconsistsof lots of individual sentences,onefor eachdot of ink, especiallyif themapis drawn42http://librosysolucionarios.net43usinga digital plotter. In this case,the advantageof the mapis really in the easeofinferencecombinedwith suitability for human“visual computing”apparatus.e. Examplesof otheranalogicalrepresentations:� Analog audiotaperecording.Advantages:simplecircuits canrecordandrepro-ducesounds.Disadvantages:subjectto errors,noise;hardto processin ordertoseparatesoundsor remove noiseetc.� Traditionalclock face.Advantages:easierto readquickly, determinationof howmuchtimeis availablerequiresnoadditionalcomputation.Disadvantages:hardtoreadprecisely, cannotrepresentsmallunitsof time (ms)easily.� All kindsof graphs,barcharts,pie charts.Advantages:enormousdatacompres-sion, easytrendanalysis,communicateinformation in a way which we can in-terpreteasily. Disadvantages:imprecise,cannotrepresentdisjunctiveor negatedinformation.8.2 The knowledgebasedoesnot entail �vHOB G�H9J . To show this, we mustgive a modelwhereB G�±�J and B G�pJ but �vHcB G�H9J is false.Considerany modelwith threedomainelements,where ± and p referto thefirst two elementsandtherelationreferredto by B holdsonly forthosetwo elements.8.3 ThesentencebH96�I H?X I is valid. A sentenceis valid if it is true in every model.Anexistentiallyquantifiedsentenceis truein amodelif it holdsunderany extendedinterpretationin which its variablesareassignedto domainelements.Accordingto thestandardsemanticsof FOL asgiven in the chapter, every modelcontainsat leastonedomainelement,hence,for any model,thereis an extendedinterpretationin which H and I areassignedto thefirstdomainelement.In suchaninterpretation,HçX I is true.8.4 �vHb6�I H?X I stipulatesthat thereis exactly oneobject. If therearetwo objects,thenthereis anextendedinterpretationin which H and I areassignedto differentobjects,so thesentencewould befalse.Somestudentsmayalsonoticethatany unsatisfiablesentencealsomeetsthecriterion,sincetherearenoworldsin which thesentenceis true.8.5 We will usethesimplestcountingmethod,ignoring redundantcombinations.For theconstantsymbols,thereare ] î assignments.Eachpredicateof arity � is mappedontoa � -aryrelation,i.e.,asubsetof the ] å possible� -elementtuples;thereare �¡ ê suchmappings.Eachfunctionsymbolof arity � is mappedontoa � -ary function,which specifiesa valuefor eachof the ] å possible� -elementtuples.Includingtheinvisibleelement,thereare ]ÕfûZ choicesfor eachvalue,sothereare G�]�fhZWJ ¡ ê functions.Thetotalnumberof possiblecombinationsis therefore] î x¢å v « ¡êx¢å v « G�][fuZWJ ¡êèTwo things to note: first, the numberis finite; second,the maximumarity 5 is the mostcrucialcomplexity parameter.8.6 In this exercise,it is bestnot to worry aboutdetailsof tenseandlarger concernswithconsistentontologiesandso on. The main point is to make surestudentsunderstandcon-http://librosysolucionarios.net44 Chapter 8. First-OrderLogicnectivesandquantifiersandtheuseof predicates,functions,constants,andequality. Let thebasicvocabulary beasfollows:¾¨±8�~@�ÃéG�H96"}6"ÃéJ : studentH takescourse} in semesterà ;B ±8ÃOÃ`@�ÃéG�H96"}6"ÃéJ : studentH passescourse} in semesterà ;~²}S,�£@�G�H96"}6"ÃéJ : thescoreobtainedby studentH in course} in semesterà ;H�­wI : H is greaterthanI ;Ý and ¬ : specificFrenchandGreekcourses(onecouldalsointerpretthesesentencesasre-ferring to anysuchcourse,in which caseonecouldusea predicate~d�bp�rP@�}��"G�}6"�KJ meaningthatthesubjectof course} is field � ;>&�=IKÃOG�Hb6�Id6"a;J : H buys I from a (usinga binarypredicatewith unspecifiedselleris OK butlessfelicitous);~J@-���.ÃéG�H96�IK6"a�J : H sellsI to a ;~²�B±{¤^@�ÃéG�H96�IKJ : personH shavespersonI>çS,�é#_G�Hb6"}J : personH is bornin country } ;B ±��£@�#Y�"G�H96�IKJ : H is aparentof I ;\Òþ¥�Áþ a�@�#_G�H96"}6��FJ : H is acitizenof country } for reason� ;Ö"@�Ã�þ U_@�#Y�"G�H96"}J : H is a residentof country } ;Ý SVS��.ÃOG�Hb6�Id6��"J : personH fools personI at time � ;~^���9U_@�#Y�"G�H9J , B¦@��FÃOS #_G�HbJ , s�±�#_G�H9J , >ç±{�FpW@��FG�H9J , µÒH � @�#_Ã�þ9¤�@OG�H9J , 5_s�@1#Y�G�HbJ , § #_Ã1���~@OU;G�HbJ ,~dQT±����G�HbJ , B S�� þ¥�Áþ }Áþ ±�#_G�H9J : predicatessatisfiedby membersof thecorrespondingcategories.a. SomestudentstookFrenchin spring2001.bH ~^���9U_@�#Y�"G�H9JYHç¾?±��~@OÃOG�Hb6"ݲ6"~ � �Vþ0#KsB W`W`WZWJ .b. Everystudentwho takesFrenchpassesit.�vH96"Ãm~^���9U�@1#Y�G�HbJYH羨±8�~@�ÃéG�H96"ݲ6"ÃéJI� B ±�ÃéÃ`@�ÃéG�H96"ݲ6"ÃéJ .c. Only onestudenttook Greekin spring2001.bH ~^���9U_@�#Y�"G�H9J�Hd¾?±��~@OÃOG�Hb6"¬ 6"~ � �Vþ0#KsB W`W`WZWJ�HY�vI I�¨X HK�©EK¾¨±��~@OÃOG�IK6"¬ 6"~ � �éþ #KsB W`W`WZWJ .d. Thebestscorein Greekis alwayshigherthanthebestscorein French.�¨ÃDbHª�vI ~²}S«�£@OG�H96"¬ 6"ÃéJ�­³~_}�S,�£@�G�IK6"ݲ6"ÃéJ .e. Everypersonwhobuysapolicy is smart.�vH¬B­@�FÃOS #_G�H9JYHÆG��Id6"afB S�� þ�} IKG�IKJ®HÆ>&�=IKÃOG�Hb6�Id6"a;J�JI� ~dQT±����G�HbJ .f. No personbuysanexpensive policy.�vH96�IK6"a�B­@�FÃOS #_G�H9JYHNBoS`�Âþ }ÁIKG�IdJ¯HƵÒH � @�#_Ã�þ9¤�@OG�IdJp� E²>&�=IKÃéG�H96�IK6"a�J .g. Thereis anagentwhosellspoliciesonly to peoplewhoarenot insured.bH 5_s�@�#Y�"G�H9JYH°�vIK6"a�BoS`�Âþ }ÁIKG�IdJYH ~J@-�;��ÃéG�H96�IK6"a�JF� G�B¦@��FÃOS #_G�a�J^HNEJ§ #_Ã1���£@�U;G�a;J�J .h. Thereis abarberwhoshavesall menin town who donot shave themselves.bH >ç±��@pW@�@G�HbJYHM�vIas�±�#_G�IdJ®HFE²~²�B±{¤^@�ÃéG�Id6�IKJF� ~²�B±{¤^@�ÃéG�H96�IKJ .i. A personbornin theUK, eachof whoseparentsis a UK citizenor a UK resident,is aUK citizenby birth.�vH¬B­@�FÃOS #_G�H9J�Ht>çS,�é#_G�Hb6"Þ¦<yJ{H&G��vI±B ±��~@1#Y�G�IK6�H9J²� G�G�0��\vþ9�Áþ�a4@�#_G�IK6"Þ¦<Í6��FJ�J{CÖ"@�Ã�þ U�@1#Y�G�IK6"Þ¦<yJ�J�JF� \Òþ9��þ a4@1#_G�H96"Þ¦<Í6">&þ9����BJ .j . A personbornoutsidetheUK, oneof whoseparentsis a UK citizenby birth, is a UKhttp://librosysolucionarios.net45citizenby descent.�vH³B­@��FÃéS�#_G�HbJYHFE²>çS,�V#_G�H96"Þ¦<yJ^H�G�bIaB ±��£@�#Y�"G�Id6�HbJ´HÆ\Òþ9��þ a4@1#_G�Id6"Þ°<�6">&þ9���"�=J�J� \Òþ9��þ a4@1#_G�H96"Þ¦<Í6"]k@�Ãé}�@1#Y�J�èk. Politicianscanfool someof thepeopleall of thetime,andthey canfool all of thepeoplesomeof thetime,but they can’t fool all of thepeopleall of thetime.�vH³B S�� þ¥�Áþ }Áþ ±�#_G�H9Jp�G��If�?�dB­@�@ÃéS #_G�IdJYH»Ý SVS`�.ÃéG�H96�IK6��J�J®HG�5�p�vIaB­@�@ÃéS #_G�IdJN� Ý SVS`�.ÃéG�H96�IK6��J�J¯HE_G��?�p�tIbB­@�FÃOS #_G�IKJN� Ý SVS���ÃéG�H96�IK6��"J�J8.7 Thekey ideais to seethattheword“same”is referringto everypair of Germans.Thereareseverallogically equivalentformsfor this sentence.Thesimplestis theHornclause:�tH96�IK6��Ƭ­@��VQT±�#_G�H9J´HƬ­@�VQT± #_G�IKJ´H�~ � @�±8�@ÃOG�Hb6���JI� ~ � @O±��@ÃéG�Id6��.Jbè8.8 �vHb6�I ~ � S �bÃ-@OG�H96�IKJ�Hos�±P�;@�G�HbJa� Ý­@�QT±P�;@�G�IKJ . This axiom is no longer true incertainstatesandcountries.8.9 This is a very educationalexercisebut also highly nontrivial. Once studentshavelearnedaboutresolution,ask them to do the proof too. In most cases,they will discovermissingaxioms. Our basicpredicatesare ÄK@�±��FU;G�H96�@O6��J (H heardaboutevent @ at time � );no}}Á�����£@�U�G�@�6��"J (event @ occurredat time � ); 5µ�Âþ9¤^@�G�Hb6��J (H is alive at time � ).5� ÄK@�±��FU;G�ß�6"]¶@O±{�"�=n¨�KG�PgJ�6��"J�tH96�@�6��uÄK@�±��FU;G�H96�@O6��JK� 5+�Âþ9¤�@OG�H96��"J�tH96�@�6�� { ÄK@�±��FU;G�H96�@O6�� { JF� 0� «hno}}Á�����£@�U;G�@O6��"« J´HM�"«¸·d� {�²� « n¨}�}Á���`�£@OU;G�]k@�±���Bno�KG�H9J�6�� « JF� �²� { � « ·¹� { � E²5µ�Âþ9¤^@�G�Hb6�� { J�²�"«�6�� { E²G�� { ·d�"«�JF� G�G�� «¸·¹� { JYC�G�� «¦X¹� { J�J�²�"«�6�� { 6���ÚzG�� «¸·¹� { J®HÆG�G�� { ·d��ÚWJYC�G�� { X¹�2ÚVJ�Jp� G�� «¸·¹��ÚWJ�²�"«�6�� { 6���ÚzG�G�� «¸·¹� { J®C�G��"«¦Xd� { J�J´H�G�� { ·o�2Ú�JF� G�� «¸·¹��ÚWJ8.10 It is notentirelyclearwhichsentencesneedto bewritten,but this is oneof them:�?Ãé«h>µ�£@�@�a"IdG�ÃV«J}e và { 5&U9r8±8}W@�#Y�"G�Ãé«"6"à { J´HFB¦þ¥�G�à { J èThat is, a squareis breezyif andonly if thereis a pit in a neighboringsquare.Generallyspeaking,the sizeof the axiom set is independentof the sizeof the wumpusworld beingdescribed.8.11 Make sureyou write definitionswith e . If you use � , you areonly imposingcon-http://librosysolucionarios.net46 Chapter 8. First-OrderLogicstraints,notwriting a realdefinition.¬2�@±�#_U;\¨�Wþ,�.U�G�}�6"±8Joe vp \o��þ,�.U;G�}6"pJ®H�\¨�Wþ,�.U;G�p6"±8J¬2�~@O±{�"¬¸�F±�#_U�B ±��£@�#Y�"G�±�6"U�J�e vp6"}w\¨�Wþ,�.U;G�U�6"}J®H�\¨�Wþ,�.U;G�}6"pJ®H»\o�Wþ���U�G�p6"±�J>µ�FS,���@��FG�H96�IKJ}e s�±P��@OG�H9J¯HÆ~dþ pW�Âþ0#KsBG�H96�IKJ~Kþ Ã��@��FG�Hb6�IdJ}e Ý­@�QT±P�;@�G�HbJ´H�~dþ pW�Âþ0#KsBG�Hb6�IdJ]ͱ��=sB�4�W@�@G�U�6 � Joe Ý­@�QT±P�;@�G�U�J®H�\¨�Wþ,�.U;G�U�6 � J~_S�#_G�Ãé6 � J}e s�±P�;@�G�ÃéJ´H�\o��þ,�.U;G�Ãé6 � J5_�=#Y�"n2�FÞv#_}��;@OG�±�6"}J}e � \o��þ,�.U;G�}6 � J¯HÆ~dþ pW�Âþ0#KsBG�±86 � J5_�=#Y�"G�±86"}�Jºe Ý­@1QT±_�;@�G�±8J®H»5²�B#Y�n?�@Þv#_}W��@OG�±�6"}JÞv#_}W��@OG��96"}J�e s�±P�;@�G��bJ´H�5_�=#Y�"n2�FÞÒ#_}W�;@�G�±86"}J>µ�FS,���@��£§"#G��± �çG�p6�H9J}e bQ ~ � S �9Ã`@�G�Hb6�QTJYH�>µ�FS«�"��@�FG�p�6�QTJ~Kþ Ã��@��£§ #G�_±��?G�Ãé6�H9J}e bQ ~ � S �9Ã`@�G�Hb6�QTJYH�~dþ ë��@��FG�ÃO6�QTJÝÒþ9�FÃ�"\oS �9Ã�þ0#_G�}6"�@J}e tG � J�5_�=#Y�"n2�FÞÒ#_}W�;@�G � 6"}JYHFBo±{�£@�#Y�"G � 6"�@JA secondcousinis a a child of one’s parent’s first cousin,andin generalan # th cousinis definedas:P��"�=\oS �bÃ1þ0#_G�#_6"}6"�@Jºe � 6"�aB ±��£@�#Y�"G � 6"}JYH�P���B\oS �9Ã�þ0#_G�#»�wZW6"�@6 � J´H»\o�Wþ���U�G��@6"�KJThefactsin the family treearesimple: eacharrow representstwo instancesof \o��þ,�.U(e.g., \o�Wþ���U�G�ßÕþ,�;� þ�± QT6"]oþ ±�#_±�J and \o��þ,�.U;G�ßÕþ,�;�Âþ ±�QT6"\o�=±��£�;@�ÃéJ ), eachnamerepresentsasex proposition(e.g., s�±P��@OG�ßÕþ,�;� þ�± QTJ or ݦ@�QT±P��@OG�] þ�± #_±8J ), eachdoubleline indicatesa~ � S �bÃ-@ proposition(e.g. ~ � S �9Ã`@�G�\o�B±{�£�;@�Ãé6"]oþ ± #_±8J ). Making the queriesof the logicalreasoningsystemis justawayof debuggingthedefinitions.8.12 �vHb6�I G�H�fwIKJ?X G�I�f HbJ . This doesfollow from the Peanoaxioms(althoughweshouldwrite the first axiom for + as �vQ P�±{�"PR�BQTG�QTJh� f�G�`W6�QTJRX Q ). Roughlyspeaking,thedefinitionof + saysthat H fyI X�~´»�G�IdJ²X�~´»4¼¾½�G�`WJ , where ~®» is shorthandforthe ~ functionappliedH times. Similarly, I�fhHÕX ~ ½ G�H9J×X ~ ½-¼¾» G�`WJ . Hence,theaxiomsimply that HgfhI and IÍfhH areequalto syntacticallyidenticalexpressions.This argumentcanbeturnedinto a formalproofby induction.8.13 Although theseaxiomsare sufficient to prove set membershipwhen H is in fact amemberof a given set,they have nothingto sayaboutcaseswhereH is not a member. Forexample,it is notpossibleto provethat H is notamemberof theemptyset.Theseaxiomsmaythereforebesuitablefor a logical system,suchasProlog,thatusesnegation-as-failure.8.14 Herewetranslate�|þ ë�W¿ to mean“properlist” in Lisp terminology, i.e.,aconsstructurewith PRþ�� asthe“rightmost” atom.�|þ ë�W¿éG�PRþ���J�tH96��N�Kþ�Ã��¿OG��.JLe �Kþ ë�W¿éG�\oS #_ÃéG�H96��.J�J�tH96�I�ݦþ9�Fë�G�\¨S�#_ÃéG�H96�IKJ�J@XÍH�tH96�I�Ö"@�ë�G�\¨S #_ÃOG�Hb6�IdJ�J@XÍI�tH 5 �4� @�#_U�G�PRþ���6�HbJ@X�H�²¤(6�H96�IK6"aÀ�Kþ ë�W¿éG�H9JI� G�5 ��� @�#_U�G�H96�IKJ@XTa�e 5 �4� @�#_U;G�\¨S #_ÃOG�¤d6�H9J�6�IKJ@XT\oS #_ÃéG�¤(6"a�J�J�tH¬E²Ý¦þ #_U;G�H96"Páþ,�.J�tH¬�Kþ�Ã��¿OG�a;JF� G�ݦþ #_U;G�H96"\¨S #_ÃOG�IK6"a�J�J�e G�H?XÆIÁCÆݦþ0#_U�G�H96"a�J�Jhttp://librosysolucionarios.net478.15 Thereareseveralproblemswith theproposeddefinition. It allows oneto prove, say,5AU�r8±8}W@�#Y�"G�4 ZW6"Z�< 6;4 ZW6" �< J but not 5AU�r�±�}W@�#Y�"G�4 ZW6" �< 6;4 ZW6"Z�< J ; so we needan additionalsymmetryaxiom. It doesnot allow oneto prove that 5AU�r�±�}W@1#Y�G�4 ZW6"Z�< 6;4 ZW6"e�< J is false,so it needsto bewrittenas�?Ãé«�6"à { e è"è"èFinally, it doesnot work asthe boundariesof the world, so someextra conditionsmustbeadded.8.16 Weneedthefollowing sentences:�?Ãé«h~dQ�@-���ÂIKG�Ãé«J�e và { 5AU�r�±�}W@1#Y�G�ÃV«"6"à { J®HF§"#_G�ßÕ�=Q � �9Ãé6"à { JtÃé«o§"#_G�ßÕ�BQ � �9Ãé6"Ãé«�JYHn�?à { G�ÃV«­¨Xzà { JF� EJ§"#_G�ßÕ�=Q � �9Ãé6"à { J è8.17 Therearethreestagesto go through.In thefirst stage,we definetheconceptsof one-bit and # -bit addition. Then,we specifyone-bitand # -bit addercircuits. Finally, we verifythatthe # -bit addercircuit does# -bit addition.� One-bitadditionis easy. Let 5AU;U « bea functionof threeone-bitarguments(the thirdis the carry bit). The resultof the additionis a list of bits representinga 2-bit binarynumber, leastsignificantdigit first:5AU;U�«G�`W6"`W6"`WJ@Xz4 `W6"`�<5AU;U « G�`W6"`W6"ZWJ@Xz4 `W6"Z�<5AU;U�«G�`W6"ZW6"`WJ@Xz4 `W6"Z�<5AU;U « G�`W6"ZW6"ZWJ@Xz4 ZW6"`�<5AU;U�«G�ZW6"`W6"`WJ@Xz4 `W6"Z�<5AU;U�«G�ZW6"`W6"ZWJ@Xz4 ZW6"`�<5AU;U�«G�ZW6"ZW6"`WJ@Xz4 ZW6"`�<5AU;U�«G�ZW6"ZW6"ZWJ@Xz4 ZW6"Z�<� # -bit additionbuilds on one-bitaddition. Let 5&U�U ! G�Hb«�6�H { 6"pJ bea function that takestwo listsof binarydigitsof length# (leastsignificantdigit first) andacarrybit (initially0), andconstructsa list of length #Tf�Z that representstheir sum. (It will alwaysbeexactly #gf®Z bits long,evenwhentheleadingbit is 0—theleadingbit is theoverflowbit.)5AU;U ! G�4 < 6;4 < 6"pJ_X½4 pÁ<5AU;U�«G�p"«6"p { 6"pJ²X½4 p�ÚW6"p É <¶� 5AU;U ! G�4 p"«4= Hb«Á< 6;4 p { = H { < 6"pJ²X 4 pÁÚ�= 5AU;U ! G�H�«6�H { 6"p É J�<� Thenext stepis to definethestructureof aone-bitaddercircuit, asgivenin Section??.Let 5AU;U « \Òþ¥�@}Á�Bþ9�G�}J be true of any circuit that hasthe appropriatecomponentsandhttp://librosysolucionarios.net48 Chapter 8. First-OrderLogicconnections:�¨} 5AU;U�« \Òþ9�F}Á�=þ9�"G�}J}ebH « 6�H { 6"± « 6"± { 6"S « ¾vI � @�G�H « J@X ¾vI � @�G�H { J@X Ùhn¨ÖH?¾tI � @OG�±8«J@XƾvI � @�G�± { J@Xû5&P�]ÀHç¾vI � @�G�SW«J@Xûn¨ÖHÆ\¨S #K#G@�}��W@OU;G�nv���"G�ZW6�Hb«"J�6�§ #_G�ZW6�H { J�J´H»\oS #K#G@�}���@�U;G�§ #_G�ZW6"}J�6�§"#_G�ZW6�Hb«�J�JHÆ\¨S #K#G@�}��W@OU;G�nv���"G�ZW6�Hb«"J�6�§ #_G� W6"± { J�J®H�\oS #K#G@�}���@�U�G�§"#_G�ZW6"}J�6�§"#_G�ZW6"±�« J�JHÆ\¨S #K#G@�}��W@OU;G�nv���"G�ZW6"± { J�6�§ #_G�ZW6"SW«"J�J´H�\¨S�#K#G@O}��W@OU;G�§"#_G� W6"}J�6�§ #_G� W6�Hb« J�JHÆ\¨S #K#G@�}��W@OU;G�nv���"G�ZW6"± « J�6�§ #_G� W6"S « J�J´H�\¨S�#K#G@O}��W@OU;G�§"#_G� W6"}J�6�§ #_G� W6"± « J�JHÆ\¨S #K#G@�}��W@OU;G�nv���"G�ZW6�H { J�6"nv���"G�ZW6"}�J�JAH�\¨S�#K#G@O}��W@OU;G�§"#_G�eW6"}J�6�§ #_G� W6�H { J�JHÆ\¨S #K#G@�}��W@OU;G�nv���"G�ZW6"SV« J�6"nÒ���"G� W6"}J�J¯H»\oS #K#G@�}���@�U;G�§ #_G�eW6"}J�6�§"#_G�ZW6"± { J�JNotice that this allows the circuit to have additionalgatesandconnections,but theywon’t stopit from doingaddition.� Now we definewhatwe meanby an # -bit addercircuit, following thedesignof Figure8.6. We will needto be careful,becausean # -bit adderis not just an #û�³Z -bit adderplusa one-bitadder;we have to connecttheoverflow bit of the #��mZ -bit adderto thecarry-bitinput of theone-bitadder. Webegin with thebasecase,where#áXT` :�¨} 5AU;U ! \Òþ¥�@}Á�Bþ9�G�}6"`WJºe~dþ0s;#_±_�.G�nÒ���G�ZW6"}J�J@X `Now, for the recursive casewe specifythat the first connectthe “overflow” outputofthe #»�uZ -bit circuit asthecarrybit for thelastbit:�¨}6�# #T­z`O� 4 5&U�U ! \Òþ9�F}Á�=þ9�"G�}6�#_J�ev} { 6"U�5&U�U ! \Òþ9�F}Á�=þ9�"G�} { 6�#»�uZWJ´H�5AU;U8«\Òþ9�F}Á�=þ¥�G�U�JHM�tQ G�QÀ­z`WJ´H�G�Qm·� �#»�uZWJF� § #_G�QT6"}J@X�§"#_G�QT6"} { JHM�tQ G�QÀ­z`WJ´H�G�Qm·w#_JN� H�nv���"G�QT6"}�J@X·nÒ���G�QT6"} { JHÆ\¨S #K#G@�}��W@OU;G�nv���"G�#_6"} { J�6�§ #_G�eW6"U;J�JHÆ\¨S #K#G@�}��W@OU;G�§ #_G� �#g�uZW6"}J�6�§"#_G�ZW6"U�J�JAH»\oS #K#G@�}���@�U;G�§ #_G� �#_6"}J�6�§"#_G� W6"U;J�JHÆ\¨S #K#G@�}��W@OU;G�nv���"G�ZW6"U;J�6"nÒ���G�#_6"}J�J0H�\¨S�#K#G@O}��W@OU;G�nÒ���G� W6"U�J�6"nÒ���G�#yfuZW6"}J�J� Now, to verify thataone-bitaddercircuit actuallyaddscorrectly, weaskwhether, givenany settingof theinputs,theoutputsequalthesumof theinputs:�¨} 5AU;U�« \Òþ9�F}Á�=þ9�"G�}JF��vþ« 6�þ { 6�þ Ú�~dþ0s;#_±_�.G�§"#_G�ZW6"}J�J@Xgþ�«0HÆ~dþ s�#_±_�.G�§ #_G� W6"}�J�J@Xgþ { H»~dþ0s�#_±P�.G�§"#_G�eW6"}J�J@Xgþ Ú� 5&U�U « G�þ « 6�þ { 6�þ Ú J@XÕ4 nÒ���G�ZW6"}J�6"nv���"G� W6"}�J�<If this sentenceis entailedby theKB, thenevery circuit with the 5&U�U�«\vþ9�F} �Bþ9� designis in factanadder. Thequeryfor the # -bit canbewrittenas�¨}6�#Y5AU;U ! \Òþ¥�@}Á�Bþ9�G�}6�#_JF��vH « 6�H { 6�I§"#Y��@��£�;@�±�¤�@�U_§"# � ���>&þ9�"ÃéG�H « 6�H { 6"}J´H�nv��� � ���">&þ9�"ÃOG�IK6"}J� 5&U�U ! G�Hb«6�H { 6�IKJwhere §"#Y��@��£�;@�±�¤�@OU�§"# � ���">tþ¥�à and nv��� � ���">&þ9�"à aredefinedappropriatelyto mapbitsequencesto theactualterminalsof thecircuit. [Note: this logical formulationhasnotbeentestedin a theoremprover andwe hesitateto vouchfor its correctness.]http://librosysolucionarios.net498.18 Strictly speaking,the primitive gatesmust be definedusing logical equivalencestoexcludethosecombinationsnot listedascorrect.If weareusingalogic programmingsystem,we cansimply list thecases.For example,Ã+Ä+ÅÁÆ�Ç#È�Ç#È�Ç#É ÃµÄ+ÅÁÆ�Ç#È�Ê#È�Ç#É Ã+ĵÅ�Æ�Ê#È�Ç#È�Ç#É ÃµÄ+ÅÁÆ�Ê#È�Ê#È�Ê#É5ËFor theone-bitadder, we have̲Í�Î È Í¥Ï È Í¥Ð È�Ñ Î È�Ñ Ï Ã�Ò�Ò#ÓAÔAÕ�Ö�×-ØPÕ�Ù,Æ Í�Î È Í¥Ï È Í¥Ð È�Ñ Î È�Ñ Ï É}ÚÛ ÑWÜ Î È�ÑWÝ Î È�Ñ�Ü ÏfÞMßlà Æ Í Î È Í Ï È�ÑWÜ Î É´á Þ°ßlà Æ�ÑWÜ Î È Í Ð È�Ñ Î ÉápÃ+ĵÅnÆ ÍWÎ ÈÍ¥Ï È�Ñ Ý Î ÉYápÃ+Ä+ÅnÆ Í¥Ð È�Ñ Ü Î È�Ñ Ý Ï Éá ßlà Æ�Ñ Ý Ï È�Ñ Ý Î È�Ñ Ï ÉTheverificationqueryis̲Í�Î È Í¥Ï È Í¥Ð È�Ñ Î È�Ñ Ï�âµã_ã_Î Æ ÍWÎ È Í¥Ï È Í¥Ð É~ä�å È�Ñ Î È�Ñ Ï�æ¶ç ÃlÒ4Ò#ÓAÔAÕ�Ö�×�ؾÕ%Ù,Æ Í�Î È Í�Ï È Í¥Ð È�Ñ Î È�Ñ Ï ÉIt is not possibleto askwhetherparticularterminalsareconnectedin a given circuit, sincetheterminalsis not reified(nor is thecircuit itself).8.19 Theanswersherewill vary by country. Thetwo key rulesfor UK passportsaregivenin theanswerto Exercsie8.6.http://librosysolucionarios.netSolutionsfor Chapter9Inferencein First-OrderLogic9.1 Wewantto show thatany sentenceof theformÌ?èbéentailsany universalinstantiationof thesentence.ThesentenceÌ?èbéis trueiféis truein all possibleextendedinterpretations.But replacingèwith any groundterm ê mustcountasoneof the interpretations,so if theoriginal sentenceis true,thentheinstantiatedsentencemustalsobetrue.9.2 For any sentenceécontaininga groundterm ê andfor any variableènot occuringiné, we have éÛ5èSUBSTÎ Æ�ë ê�ì è�í È é ÉwhereSUBSTÎis a functionthatsubstitutesfor asingleoccurrenceof ê withè.9.3 Both b andc arevalid; a is invalid becauseit introducesthe previously-usedsymbolEverest. Note that c doesnot imply that thereare two mountainsashigh as Everest,be-causenowhereis it statedthat BenNevis is different from Kilimanjaro (or Everest, for thatmatter).9.4 This is aneasyexerciseto checkthatthestudentunderstandsunification.a.ë�î ì â È�ï ì#ð È�ñ ì#ð í (or somepermutationof this).b. No unifier (îcannotbind to bothâand ð ).c.ë�ï ì#ò Ñ�ó#ôGÈ�î ì#ò Ñ�ó4ô í .d. No unifier (becausetheoccurs-checkpreventsunificationofïwith õ­ö�÷ ó�ø�ù£Æ�ïYÉ ).9.5 Employs(Mother(John),Father(Richard))This pageisn’t wide enoughto draw thedia-gramasin Figure9.2,sowe will draw it with indentationdenotingchildrennodes:[1] Employs(x, y)[2] Employs(x, Father(z))[3] Employs(x, Father(Richard))[4] Employs(Mother(w ), Father(Richard ))[5] Employs(Mother (J ohn) , Father(Richard ))[6] Employs(Mother(w ), Father(z))[4] ...[7] Employs(Mother(J ohn) , Father(z))[5] ...50http://librosysolucionarios.net51[8] Employs(Mother( w), y)[9] Employs(Mother(J ohn) , y)[10] Employs(Mother( Jo hn), Father(z)[5] ...[6] ...9.6 We will give theaverage-casetime complexity for eachquery/schemecombinationinthefollowing table.(An entryof theform “Ê#ú�ô” meansthatit is û Æ�Ê#É to find thefirst solutionto thequery, but û Æ�ôGÉ to find themall.) Wemakethefollowing assumptions:hashtablesgiveû Æ�Ê#É access;thereareôpeoplein thedatabase;thereare û Æ�ôGÉ peopleof any specifiedage;every personhasonemother;thereare ü peoplein Houstonand ý peoplein Tiny Town; ýis muchlessthanô; in Q4, thesecondconjunctis evaluatedfirst.Q1 Q2 Q3 Q4S1Ê Ê#ú ü Ê#ú®ô ý ú ýS2Ê ôGú´ô Ê#ú®ô ôGú´ôS3ô ôGú´ô Ê#ú®ô ô Ï ú¯ô ÏS4Ê ôGú´ô Ê#ú®ô ôGú´ôS5Ê Ê#ú ü Ê#ú®ô ý ú ýAnything that is û Æ�Ê#É canbe considered“efficient,” asperhapscananything û Æ ý É . NotethatS1andS5dominatetheotherschemesfor this setof queries.Also notethatindexing onpredicatesplaysnorole in this table(exceptin combinationwith anargument),becausethereareonly 3 predicates(which is û Æ�Ê#É ). It would make a differencein termsof the constantfactor.9.7 This would work if therewereno recursive rulesin theknowledgebase.But supposetheknowledgebasecontainsthesentences:þ ø�ÿ��Wøù~Æ�î0È�å î�� ù æ Éþ ø�ÿ��Wøù~Æ�î0È�ù£É ç þ ø�ÿ��Wø�ù£Æ�î0È�å ï�� ù æ ÉNow take the queryþ øÿ��Wø�ù£Æ��#È�å Ê#È��#È�� æ É, with a backward chainingsystem.We unify thequerywith theconsequentof theimplicationto getthesubstitution ä�ë�î ì �#È�ï ì Ê#È�ù ì å �#È�� æ¥í .We then substitutethis in to the left-handside to getþ øÿ���øù£Æ��#È�å �#È�� æ Éand try to backchainon thatwith the substitution . Whenwe thentry to apply the implication again,weget a failure becauseïcannotbe bothÊand�. In otherwords, the failure to standardizeapartcausesfailure in somecaseswhererecursive ruleswould resultin a solutionif we didstandardizeapart.9.8 Considera3-SAT problemof theformÆ�î Î� Î� î Ï� Î�� î Ð� Î É´áFÆ�î Î� Ï�� î Ï� Ï�� î Ð� Ï É � Ë�Ë�ËWewantto rewrite thisasasingledefineclauseof theformâ á ð á���áNË�Ë�Ë ç � Èalongwith a few groundclauses.Wecando thatwith thedefiniteclauseû ôGø û�� Æ�î Î� Î È�î Ï� Î È�î Ð� Î É®á û ôGø û�� Æ�î Î� Ï È�î Ï� Ï È�î Ð� Ï É´áFË�Ë�Ë ç � Ñ�� è ø ãhttp://librosysolucionarios.net52 Chapter 9. Inferencein First-OrderLogicalongwith thegroundclausesû ôGø û�� Æ ý ù��Aø�È�î0È�ï^Éû ôGø û�� Æ�îAÈ ý ù��Aø-È�ï^Éû ôGø û�� Æ�îAÈ�ïYÈ ý ù��Aø�É9.9 Weuseavery simpleontologyto make theexampleseasier:a. ü Ñ,ù��-ø-Æ�îAÉ ç þ ö ÿNÿ ö �;Æ�î0É�¦Ñ��MÆ�î0É ç þ ö ÿNÿ ö �;Æ�îAÉ� Í ê Æ�î0É ç þ ö ÿNÿ ö �;Æ�îAÉb.ß! #"%$ Ö�Õ'&�(PÆ�îAÈ�ïYÉ´á ü Ñ,ù!�`ø�Æ�ïYÉ ç ü Ñ,ù��-ø-Æ�îAÉc. ü Ñ,ù��-ø-Æ ð �)�0ø*�Wø ö ù ã Éd.� ö ù£øô ÷ Æ ð �+�Aø���ø ö ù ã È��¦ó ö ù�� Í ø�Ée.ß! #"%$ Ö�Õ'&�(PÆ�îAÈ�ïYÉ ç � ö ù£ø�ô ÷ Æ�ïYÈ�îAÉ� ö ù£øô ÷ Æ�î0È�ï^É ç ß! #"%$ Ö�Õ'&�(PÆ�ï^È�î0É(Notewecouldn’t doß! #",$ Ö,Õ'&�(PÆ�î0È�ï^É}Ú � ö ù~øô ÷ Æ�ïYÈ�îAÉ becausethatis not in theformexpectedby GeneralizedModusPonens.)f.þ ö ÿNÿ ö �;Æ�î0É ç � ö ù£ø�ô ÷ Æ�-­Æ�îAÉ�È�î0É (here-is aSkolemfunction).9.10 This questionsdealswith thesubjectof loopingin backward-chainingproofs.A loopis boundto occurwhenever asubgoalarisesthatis asubstitutioninstanceof oneof thegoalson thestack.Not all loopscanbecaughtthisway, of course,otherwisewewouldhaveawayto solve thehaltingproblem.a. Theproof treeis shown in FigureS9.1.Thebranchwithß! #",$ Ö�Õ.&�(~Æ ð �+�0ø*�Wø ö ù ã È�ïYÉ and� ö ù£øô ÷ Æ�ïYÈ ð �)�0ø*�Wø ö ù ã É repeatsindefinitely, sotherestof theproof is never reached.b. Wegetaninfinite loopbecauseof ruleb,ß! #"%$ Ö,Õ'&�(PÆ�îAÈ�ïYÉYá ü Ñ,ù!�`ø�Æ�ïYÉ ç ü Ñ«ù��`ø�Æ�î0É .Thespecificloopappearingin thefigurearisesbecauseof theorderingof theclauses—it wouldbebetterto order ü Ñ«ù��`ø�Æ ð �+�Aø���ø ö ù ã É beforetherule from b. However, a loopwill occurno matterwhichway therulesareorderedif thetheorem-prover is askedforall solutions.c. Oneshouldbeableto prove thatbothBluebeardandCharliearehorses.d. Smith et al. (1986) recommendthe following method. Whenever a “looping” goaloccurs(one that is a substitutioninstanceof a supergoal higher up the stack),sus-pendtheattemptto prove that subgoal.Continuewith all otherbranchesof theprooffor the supergoal, gatheringup the solutions. Then usethosesolutions(suitably in-stantiatedif necessary)assolutionsfor thesuspendedsubgoal,continuingthatbranchof the proof to find additionalsolutionsif any. In the proof shown in the figure, theß! #"%$ Ö�Õ'&�(PÆ ð �)�0ø*�Wø ö ù ã È�ïYÉ is a repeatedgoalandwould besuspended.Sinceno otherway to prove it exists, that branchwill terminatewith failure. In this case,Smith’smethodis sufficient to allow thetheorem-prover to find bothsolutions.9.11 Surprisingly, the hardpart to representis “who is that man.” We want to ask“whatrelationshipdoesthat manhave to someknown person,” but if we representrelationswithhttp://librosysolucionarios.net53Parent(y,Bluebeard)Horse(h)Offspring(h,y)Parent(y,h)Horse(Bluebeard)Yes, {y/Bluebeard, h/Charlie}Offspring(Bluebeard,y)Offspring(Bluebeard,y)FigureS9.1 Partial proof treefor findinghorses.predicates(e.g.,� ö ù£ø�ô ÷ Æ�îAÈ�ïYÉ ) thenwe cannotmake the relationshipbe a variablein first-orderlogic. Soinsteadweneedto reify relationships.Wewill use/ ø*�;Æ�ù#È�îAÈ�ïYÉ to saythatthefamily relationshipùholdsbetweenpeopleîandï. Letþ ødenoteme andþ ù�0denote“that man.” We will alsoneedtheSkolemconstantsõ þ for the fatherofþ øand õ 0 forthefatherofþ ù�0. Thefactsof thecase(put into implicative normalform) are:Æ�Ê#É / ø*�;Æ �^Í �1� Í ô ê È þ ø�È�î0É ç õ­ö �2�`øÆ��#É þ ö �;ø�Æ þ ù�0oÉÆ��#É / ø*�;Æ õ­ö�÷ ó�ø�ù#È õ 0KÈ þ ù�0oÉÆ'3�É / ø*�;Æ õ­ö�÷ ó�ø�ù#È õ þ È þ ø-ÉÆ�4#É / ø*�;Æ � Ñ,ôGÈ õ 0�È õ þ ÉWe want to be ableto show thatþ øis theonly sonof my father, andthereforethatþ øisfatherofþ ù�0, whois male,andthereforethat“that man”is my son.Therelevantdefinitionsfrom thefamily domainare:Æ�5#É / ø*�;Æ � ö ù£ø�ô ÷ È�îAÈ�ïYÉYá þ ö �;ø�Æ�î0É�Ú / ø���Æ õ¦ö{÷ ó�øù�È�î0È�ï^ÉÆ�6#É / ø*�;Æ � Ñ,ôGÈ�îAÈ�ïYÉ}Ú / ø��;Æ � ö ù£øô ÷ È�ïYÈ�îAÉ´á þ ö �;ø�Æ�îAÉÆ�7#É / ø*�;Æ �^Í �1� Í ô ê È�î0È�ï^ÉºÚ î98ä¹ïná Û: / ø*�;Æ � ö ù£ø�ô ÷ È : È�îAÉYá / ø*�;Æ � ö ù~øô ÷ È : È�ï^ÉÆ�;#É / ø*�;Æ õ­ö�÷ ó�ø�ù#È�î Î È�ïYÉ´á / ø��;Æ õ­ö{÷ ó�øù�È�î Ï È�ïYÉ ç î Î ädî Ïandthequerywewantis:Æ�<°É / ø*�;Æ�ù#È þ ù�0�È�ï^ÉWe wantto beableto getbacktheanswerë�ù ì � Ñ,ôGÈ�ï ì þ ø í . Translating1-9 and<into INFhttp://librosysolucionarios.net54 Chapter 9. Inferencein First-OrderLogic(andnegating<andincludingthedefinitionof8ä) we get:Æ�5 ö É / ø��;Æ � ö ù£øô ÷ È�î0È�ï^É´á þ ö ��ø-Æ�îAÉ ç / ø*�;Æ õ­ö�÷ ó�ø�ù#È�îAÈ�ïYÉÆ�5=�WÉ / ø*�;Æ õ­ö�÷ ó�ø�ù#È�îAÈ�ïYÉ ç þ ö ��ø-Æ�îAÉÆ�5=>WÉ / ø*�;Æ õ­ö�÷ ó�ø�ù#È�îAÈ�ïYÉ ç / ø��;Æ � ö ù£øô ÷ È�î0È�ï^ÉÆ�6 ö É / ø��;Æ � Ñ,ôGÈ�îAÈ�ïYÉ ç / ø*�;Æ � ö ù£ø�ô ÷ È�ï^È�î0ÉÆ�6=�WÉ / ø*�;Æ � Ñ,ôGÈ�îAÈ�ïYÉ ç þ ö �;ø-Æ�îAÉ�ÉÆ�6=>WÉ / ø*�;Æ � ö ù~øô ÷ È�ïYÈ�îAÉ®á þ ö �;ø�Æ�î0É ç / ø���Æ � Ñ«ôGÈ�î0È�ï^ÉÆ�7 ö É / ø��;Æ �^Í �1� Í ô ê È�îAÈ�ïYÉ ç î?8ä¹ïÆ�7=�WÉ / ø*�;Æ �^Í �1� Í ô ê È�î0È�ï^É ç / ø*�;Æ � ö ù£ø�ô ÷ È � Æ�îAÈ�ïYÉ�È�îAÉÆ�7=>WÉ / ø*�;Æ �^Í �1� Í ô ê È�î0È�ï^É ç / ø*�;Æ � ö ù£ø�ô ÷ È � Æ�îAÈ�ïYÉ�È�ïYÉÆ�7 ã É / ø���Æ � ö ù£ø�ô ÷ È � Æ�î0È�ï^É�È�î0É®á / ø���Æ � ö ù£ø�ô ÷ È � Æ�î0È�ï^É�È�ïYÉ´ánî98ä¹ï ç / ø*�;Æ �^Í �1� Í ô ê È�î0È�ï^ÉÆ�;#É / ø*�;Æ õ­ö�÷ ó�ø�ù#È�î Î È�ïYÉ´á / ø��;Æ õ­ö{÷ ó�øù�È�î Ï È�ïYÉ ç î Î ädî ÏÆ�@KÉ ý ù��0ø ç î�ä¹ï � î?8ä¹ïÆ�@BA�É�î�ädïnáMîC8ä¹ï ç õ­ö �'�`øÆ�< A É / ø*�;Æ�ù�È þ ù�0KÈ�ïYÉ ç õ­ö �2�`øNote that (1) is non-Horn,sowe will needresolutionto bebe sureof gettinga solution. Itturnsoutwealsoneeddemodulation(page284)to dealwith equality. Thefollowing lists thestepsof theproof,with theresolventsof eachstepin parentheses:Æ�Ê#Ç#É / ø*�;Æ � ö ù£ø�ô ÷ È õ þ È þ ø�É Æ'3�È�5=>WÉÆ�Ê#Ê#É / ø*�;Æ � ö ù£ø�ô ÷ È õ þ È õ 0oÉ Æ�4#È�6 ö ÉÆ�Ê=�#É / ø*�;Æ � ö ù£ø�ô ÷ È õ þ È�ïYÉ´á þ øD8ä¹ï ç / ø*�;Æ �^Í �1� Í ô ê È þ ø�È�ïYɺÆ�Ê#Ç#È�7 ã ÉÆ�Ê=�#É / ø*�;Æ � ö ù£ø�ô ÷ È õ þ È�ïYÉ´á þ øD8ä¹ï ç õ­ö �2�`ø Æ�Ê=�#È�Ê#ÉÆ�ÊE3�É þ øD8ä õ 0 ç õ­ö �2�-ø Æ�Ê=�#È�Ê#Ê#ÉÆ�Ê=4#É þ ø+ä õ 0 Æ�ÊE3�È�@KÉÆ�Ê=5#É / ø*�;Æ õ­ö�÷ ó�ø�ù#È þ ø-È þ ù�0¹É Æ�Ê=4#È��#È ã ø�ÿ�Ñ ã �� ö{÷ Í Ñ,ôGÉÆ�Ê=6#É / ø*�;Æ � ö ù£ø�ô ÷ È þ ø-È þ ù�0oÉ Æ�Ê=5#È�5=>WÉÆ�Ê=7#É / ø*�;Æ � Ñ,ôGÈ þ ùF0�È þ ø-É Æ�Ê=6#È��#È�6=>WÉÆ�Ê=;#É õ­ö �'�`ø²ë�ù ì � Ñ,ôGÈ�ï ì þ ø í Æ�Ê=7#È�< A É9.12 Hereis agoaltree:goals = [Criminal(West)]goals = [American(West), Weapon(y), Sells(West, y, z), Hostile(z)]goals = [Weapon(y), Sells(West, y, z), Hostile(z)]goals = [Missle(y), Sells(West, y, z), Hostile(z)]goals = [Sells(West, M1, z), Hostile(z)]goals = [Missle(M1), Owns(Nono, M1), Hostile(Nono)]goals = [Owns(Nono, M1), Hostile(Nono)]goals = [Hostile(Nono) ]goals = []9.13a. In the following, an indentedline is a stepdeeperin theproof tree,while two linesatthesameindentationrepresenttwo alternative waysto prove thegoalthatis unindentedhttp://librosysolucionarios.net55above it. TheP1 andP2 annotationon a line meanthat thefirst or secondclauseof Pwasusedto derive theline.P(A, [1,2,3]) goalP(1, [1|2,3]) P1 => solution, with A = 1P(1, [1|2,3]) P2P(2, [2,3]) P1 => solution, with A = 2P(2, [2,3]) P2P(3, [3]) P1 => solution, with A = 3P(3, [3]) P2P(2, [1, A, 3]) goalP(2, [1|2, 3]) P1P(2, [1|2, 3]) P2P(2, [2|3]) P1 => solution, with A = 2P(2, [2|3]) P2P(2, [3]) P1P(2, [3]) P2b. P couldbetterbecalledMember; it succeedswhenthefirst argumentis anelementofthelist thatis thesecondargument.9.14 Thedifferentversionsof sort illustratethedistinctionbetweenlogicalandprocedu-ral semanticsin Prolog.a. sorted([]).sorted([X]).sorted([X,Y|L]) :- X<Y, sorted([Y|L]).b. perm([],[]).perm([X|L],M) :-delete(X,M,M1),perm(L,M1).delete(X,[X|L],L). %%deleting an X from [X|L] yields Ldelete(X,[Y|L],[Y|M]) :- delete(X,L,M).member(X,[X|L]).member(X,[_|L]) :- member(X,L).c. sort(L,M) :- perm(L,M), sorted(M).This is aboutascloseto an executableformal specificationof sortingasyou canget—itsaystheabsoluteminimumaboutwhatsortmeans:in orderfor Mto beasortofL, it musthave thesameelementsasL, andthey mustbein order.d. Unfortunately, this doesn’t fareaswell asa programas it doesasa specification. Itis a generate-and-testsort: perm generatescandidatepermutationsoneat a time, andsorted teststhem.In theworstcase(whenthereis only onesortedpermutation,andit is thelastonegenerated),thiswill take û Æ�ôHG'É generations.Sinceeachperm is û Æ�ô Ï Éandeachsorted is û Æ�ôGÉ , thewholesort is û Æ�ôHG ô Ï É in theworstcase.e. Here’s asimpleinsertionsort,which is û Æ�ô Ï É :isort([],[]).isort([X|L],M) :- isort(L,M1), insert(X,M1,M).http://librosysolucionarios.net56 Chapter 9. Inferencein First-OrderLogicinsert(X,[],[X]).insert(X,[Y|L],[X,Y|L]) :- X=<Y.insert(X,[Y|L],[Y|M]) :- Y<X, insert(X,L,M).9.15 Thisexerciseillustratesthepower of pattern-matching,which is built into Prolog.a. Thecodefor simplificationlooksstraightforward,but studentsmayhavetroublefindingthemiddlewaybetweenundersimplifyingandloopinginfinitely.simplify(X,X) :- primitive(X).simplify(X,Y) :- evaluable(X), Y is X.simplify(Op(X)) :- simplify(X,X1), simplify_exp(Op(X1)).simplify(Op(X,Y)) :- simplify(X,X1), simplify(Y,Y1), simplify_exp(Op(X1,Y1)).simplify_exp(X,Y) :- rewrite(X,X1), simplify(X1,Y).simplify_exp(X,X).primitive(X) :- atom(X).b. Hereareafew representativerewrite rulesdrawn fromtheextensivelist in Norvig (1992).Rewrite(X+0,X).Rewrite(0+X,X).Rewrite(X+X,2*X).Rewrite(X*X,Xˆ2).Rewrite(Xˆ0,1).Rewrite(0ˆX,0).Rewrite(X*N,N*X) :- number(N).Rewrite(ln(eˆX),X).Rewrite(XˆY*XˆZ,Xˆ(Y+Z)).Rewrite(sin(X)ˆ2+cos(X)ˆ2,1).c. Herearethe rulesfor differentiation,usingd(Y,X) to representthederivative of ex-pressionY with respectto variableX.Rewrite(d(X,X),1).Rewrite(d(U,X),0) :- atom(U), U /= X.Rewrite(d(U+V,X),d(U,X)+d(V,X)).Rewrite(d(U-V,X),d(U,X)-d(V,X)).Rewrite(d(U*V,X),V*d(U,X)+U*d(V,X)).Rewrite(d(U/V,X),(V*d(U,X)-U*d(V,X))/ (Vˆ2) ).Rewrite(d(UˆN,X),N*Uˆ(N-1)*d(U,X)) :- number(N).Rewrite(d(log(U),X),d(U,X)/U).Rewrite(d(sin(U),X),cos(U)*d(U,X)).Rewrite(d(cos(U),X),-sin(U)*d(U,X)).Rewrite(d(eˆU,X),d(U,X)*eˆU).9.16 Onceyouunderstandhow Prologworks,theansweris easy:solve(X,[X]) :- goal(X).solve(X,[X|P]) :- successor(X,Y), solve(Y,P).We could renderthis in English as “Given a start state,if it is a goal state,then the pathconsistingof just thestartstateis a solution. Otherwise,find somesuccessorstatesuchthatthereis apathfrom thesuccessorto thegoal;thenasolutionis thestartstatefollowedby thatpath.”http://librosysolucionarios.net57Noticethatsolve cannotonly beusedto find apathP thatis asolution,it canalsobeusedto verify thatagivenpathis asolution.If you wantto addheuristics(or evenbreadth-firstsearch),youneedanexplicit queue.Thealgorithmsbecomequitesimilar to theversionswritten in Lisp or Pythonor Java or inpseudo-codein thebook.9.17 This questiontestsboth thestudent’s understandingof resolutionandtheir ability tothink atahighlevel aboutrelationsamongsetsof sentences.Recallthatresolutionallowsoneto show that I�ð � ä éby proving that I�ð áKJ é is inconsistent.Supposethatin generaltheresolutionsystemis calledusingASKÆ I�ð È é É . Now we wantto show thatagivensentence,sayL is valid or unsatisfiable.A sentenceL is valid if it canbeshown to betruewithout additionalinformation. Wecheckthisby calling ASKÆ I�ðNM È L É where I�ðNM is theemptyknowledgebase.A sentenceL that is unsatisfiableis inconsistentby itself. So if we theemptyknowl-edgebaseagainandcall ASKÆ I�ðNM È�J L É theresolutionsystemwill attemptto derive a con-tradiction startingfromJ�J L . If it can do so, then it must be thatJHJ L , and henceL , isinconsistent.9.18 This is a form of inferenceusedto show thatAristotle’s syllogismscouldnot captureall soundinferences.a.Ì î üÑ,ù!�`ø�Æ�î0É ç â ô Í ÿ ö ��Æ�î0ÉÌ îAÈ�ó ü Ñ,ù��-ø-Æ�îAÉ´á ü ø ö ã û�� Æ�ó�È�î0É ç Û ï â ô Í ÿ ö ��Æ�ïYÉYá ü ø ö ã û�� Æ�ó�È�ïYÉb.â ËOJ ü Ñ«ù��`ø�Æ�î0É �Nâ ô Í ÿ ö �;Æ�îAÉð Ë ü Ñ,ù��`ø�Æ�-­É�?Ë ü ø ö ã û�� Æ ü È�-¦ÉP ËQJ â ô Í ÿ ö �;Æ�ïYÉ � J ü ø ö ã û�� Æ ü È�ïYÉ(Hereâ. comesfrom thefirst sentencein a. while theotherscomefrom thesecond.üand-areSkolemconstants.)c. ResolvePand�to yieldJ â ô Í ÿ ö �;Æ�-­É . Resolve this withâto giveJ ü Ñ,ù!�`ø�Æ�-¦É .Resolve thiswith ð to obtainacontradiction.9.19 Thisexerciseteststhestudentsunderstandingof modelsandimplication.a. (A) translatesto “For every naturalnumberthereis someothernaturalnumberthat issmallerthanor equalto it.” (B) translatesto “Thereis a particularnaturalnumberthatis smallerthanor equalto any naturalnumber.”b. Yes,(A) is trueunderthis interpretation.You canalwayspick thenumberitself for the“someother”number.c. Yes, (B) is true underthis interpretation. You can pick 0 for the “particular naturalnumber.”d. No, (A) doesnot logically entail(B).e. Yes,(B) logically entails(A).f. Wewantto try to provevia resolutionthat(A) entails(B). To dothis,wesetourknowl-edgebaseto consistof (A) andthenegationof (B), which we will call (-B), andtry tohttp://librosysolucionarios.net58 Chapter 9. Inferencein First-OrderLogicderiveacontradiction.Firstwehaveto convert(A) and(-B) to canonicalform. For (-B),this involvesmoving theJin pastthe two quantifiers.For bothsentences,it involvesintroducinga Skolemfunction:(A)îSR õ Î Æ�îAÉ(-B)J õ Ï Æ�ï^ÉTR¹ïNow wecantry to resolve thesetwo together, but theoccurscheckrulesout theunifica-tion. It lookslike thesubstitutionshouldbeë�î ì#õ Ï Æ�ï^É�È0ï ì#õ Î Æ�îAÉ í , but thatis equivalenttoë�î ì#õ Ï Æ�ï^É�È�ï ì#õ Î Æ õ Ï Æ�ïYÉ�É í , which fails becauseïis boundto anexpressioncontain-ingï. Sotheresolutionfails,therearenootherresolutionstepsto try, andtherefore(B)doesnot follow from (A).g. To prove that (B) entails(A), we startwith a knowledgebasecontaining(B) andthenegationof (A), whichwewill call (-A):(-A)J õ Î R¹ï(B)îSR õ Ï Æ�îAÉThistimetheresolutiongoesthrough,with thesubstitutionë�î ì#õ Î ÈAï ì#õ Ï Æ õ Î É í , therebyyielding õ­ö �2�`ø , andproving that(B) entails(A).9.20 Oneway of seeingthis is thatresolutionallows reasoningby cases,by which we canprove�by proving that eitherâor ð is true, without knowing which one. With definiteclauses,we alwayshave a singlechainof inference,for which we canfollow thechainandinstantiatevariables.9.21 No. Part of the definition of algorithmis that it mustterminate.Sincetherecanbean infinite numberof consequencesof a setof sentences,no algorithmcangeneratethemall. Another way to seethattheansweris no is to rememberthat entailmentfor FOL issemidecidable.If therewereanalgorithmthatgeneratesthesetof consequencesof a setofsentences�, thenwhengiventhetaskof decidingif ð is entailedby�, onecouldjust checkif ð is in thegeneratedset.But weknow thatthis is notpossible,thereforegeneratingthesetof sentencesis impossible.http://librosysolucionarios.netSolutionsfor Chapter10KnowledgeRepresentation10.1 Shootingthe wumpusmakes it dead,but thereareno actionsthat causeit to comealive. Hencethesuccessor-stateaxiomforâ � Í9è øjust hasthesecondclause:Ì ö È�� â � Í¥è ø�Æ�UV��ÿ : ���`È / ø*�W��� ÷ Æ ö È��`É�É Ú å â � Í9è ø-Æ�îAÈ�ïYÈ��-Éá�JGÆ ö ä � ó�Ñ�Ñ ÷ á üKö �-Æ â ê ø�ô ÷ È â ù`ù£Ñ��MÈ��`Éá õ¦ö > Í ô ê Æ â ê øô ÷ È�UV��ÿ : ���`È��-É�É æwhere õ­ö > Í ô ê Æ ö È��WÈ��-É is definedappropriatelyin termsof the locationsof ö and�andtheorientationof ö . Possessionof the arrow is lost by shooting,andagainthereis no way tomake it true:Ì ö È�� üKö �`Æ â ê ø�ô ÷ È â ù�ù£Ñ��MÈ / ø*�W��� ÷ Æ ö È��`É�É�Úå üpö �`Æ â ê øô ÷ È â ù�ù£ÑW�nÈ��-ɯáNÆ ö 8ä � ó�Ñ�Ñ ÷ É æ10.2ý Í ÿ�ø�Æ � M È�Ç#Éý Í ÿ�ø�Æ / ø����� ÷ Æ��`ø*X�È��-É�È ÷ É ç ý Í ÿ�ø�Æ / ø����� ÷ Æ�å ö � �-ø�X æ È��-É�È ÷HY Ê#ÉNoticethattherecursionneedsnobasecasebecausewe alreadyhave theaxiom/ ø����� ÷ Æ�å æ È��`ÉJäZ�?Ë10.3 This questiontakesthestudentthroughthe initial stagesof developinga logical rep-resentationfor actionsthatincorporatesmoreandmorerealism.Implementingthereasoningtasksin a theorem-prover is alsoa goodidea. Althoughtheuseof logical reasoningfor theinitial task—findinga routeon a graph—mayseemlike overkill, thestudentshouldbe im-pressedthatwe cankeepmakingthesituationmorecomplicatedsimply by describingthoseaddedcomplications,with noadditionsto thereasoningsystem.a.â ÷ Æ / Ñ���Ñ ÷ È â ù ö ã È � M É .b.Û � â ÷ Æ / Ñ��WÑ ÷ È ð �>�ó ö ù£ø�� ÷ È��-É .c. Thesuccessor-stateaxiomshouldbemechanicalby now.Ì ö È�îAÈ�ïYÈ�� :â ÷ Æ / Ñ��WÑ ÷ È�ïYÈ / ø*�W��� ÷ Æ ö È��`É�É Ú å Æ ö ä[-­Ñ�Æ�î0È�ï^Éá P Í ù£ø�> ÷�/ Ñ�� ÷ ø�Æ�î0È�ï^É´á â ÷ Æ / ÑF�WÑ ÷ È�îAÈ��`É�É� Æ â ÷ Æ / ÑF�WÑ ÷ È�ï^È��`ÉáHJJÆ Û ñ ö ä[-­Ñ�Æ�ïYÈ�ñ4É®áFñ\8ä¹ï^É�É æ59http://librosysolucionarios.net60 Chapter 10. KnowledgeRepresentationd. To representthe amountof fuel the robot hasin a given situation,usethe functionõ �Aø���Æ / Ñ��WÑ ÷ È��`É . LetP Í � ÷Wö ôH>Wø�Æ�îAÈ�ïYÉ denotethedistancebetweencitiesîandï, mea-suredin unitsof fuel consumption.Let õ ���2� beaconstantdenotingthefuel capacityofthetank.e. The initial situationis describedbyâ ÷ Æ / Ñ��WÑ ÷ È â ù ö ã È � M ɸá õ �Aø���Æ / Ñ��WÑ ÷ È��`É~ä õ ���2� .The above axiom for location is extendedas follows (note that we do not say whathappensif therobotrunsoutof gas).Ì ö È�î0È�ï^È�� :â ÷ Æ / Ñ��WÑ ÷ È�ïYÈ / ø*�W��� ÷ Æ ö È��`É�É Ú å Æ ö ä�-¦Ñ`Æ�î0È�ï^Éá P Í ù£ø�> ÷�/ Ñ�� ÷ ø�Æ�î0È�ï^É´á â ÷ Æ / Ñ���Ñ ÷ È�îAÈ��`Éá P Í � ÷�ö ôH>Wø�Æ�î0È�ï^ÉH] õ �Aø��;Æ / ÑF�WÑ ÷ È��-É�É� Æ â ÷ Æ / ÑF�WÑ ÷ È�ï^È��`ÉáHJJÆ Û ñ ö ä[-­Ñ�Æ�ïYÈ�ñ4É®áFñ\8ä¹ï^É�É æõ �0ø*�;Æ / Ñ���Ñ ÷ È / ø*�W��� ÷ Æ ö È��`É�É~ä � Ú å Æ ö ä�-¦Ñ`Æ�î0È�ï^Éá P Í ù£ø�> ÷�/ Ñ�� ÷ ø�Æ�î0È�ï^É´á â ÷ Æ / Ñ���Ñ ÷ È�îAÈ��`Éá P Í � ÷�ö ôH>Wø�Æ�î0È�ï^ÉH] õ �Aø��;Æ / ÑF�WÑ ÷ È��-Éá � ä õ �Aø���Æ / Ñ��WÑ ÷ È��`É_^ P Í � ÷Wö ôH>Wø�Æ�î0È�ï^É�É� Æ � ä õ �Aø���Æ / Ñ��WÑ ÷ È��`ÉáHJJÆ Û0è È`� ö ä[-­Ñ�Æ è È`�nɯá è 8äa�MÉ�É æf. Thesimplestway to extendtherepresentationis to addthepredicate- ö � � ÷�ö�÷ Í Ñ,ôGÆ�î0É ,which is true of cities with gasstations. The õ Í �2�+� : action is describedby addinganotherclauseto theabove axiomfor õ �0ø*� , sayingthat � ä õ ��'� when ö ä õ Í �2�2b : .10.4 This questionwasinadvertentlyleft in theexercisesafter thecorrespondingmaterialwasexcisedfrom thechapter. Futureprintingsmayomit or replacethisexercise.10.5 Rememberthat we definedsubstancesso thatU ö{÷ ø�ù is a category whoseelementsareall thosethingsof which onemight say“it’ s water.” Onetricky part is that theEnglishlanguageis ambiguous.Onesenseof theword “water” includesice (“that’s frozenwater”),while anothersenseexcludesit: (“that’s not water—it’s ice”). The sentenceshereseemtousethefirst sense,sowe will stick with that. It is thesensethatis roughlysynonymouswithü Ï û .Theothertricky part is thatwe aredealingwith objectsthatchange(freezeandmelt)over time. Thus, it won’t do to say�aced Í X�� Í,ã, because�(a massof water)might be aliquid at one time anda solid at another. For simplicity, we will usea situationcalculusrepresentation,with sentencessuchas ý Æ'�acDd Í X�� Í,ã È��-É . Therearemany possiblecorrectanswersto eachof these. The key thing is to be consistentin the way that information isrepresented.For example,do not used Í XW� Í,ãasa predicateon objectsifU ö�÷ øù is usedasasubstancecategory.a. “Water is a liquid between0 and 100 degrees.” We will translatethis as “For anywaterandany situation,thewateris liquid iff andonly if thewater’s temperaturein thehttp://librosysolucionarios.net61situationis between0 and100centigrade.”Ì �nÈ����aceU ö{÷ ø�ù çÆ��°ø�ô ÷ Í ê ù ö ã ø�Æ�Ç#Éf] ý øÿ : ø�ù ö�÷ ��ù~ø-Æ'�MÈ��`É�]Z�°ø�ô ÷ Í ê ù ö ã ø�Æ�Ê#Ç#Ç#É�ÉoÚý Æ'�acgd Í X�� Í�ã È��`Éb. “Water boils at 100 degrees.” It is a good idea hereto do sometool-building. Onpage243 we usedþ ø*� ÷ Í ôê � Ñ Í ô ÷ asa predicateapplying to individual instancesofa substance.Here,we will define� ð Ñ Í � Í ô ê � Ñ Í ô ÷ to denotethe boiling point of allinstancesof asubstance.Thebasicmeaningof boiling is thatinstancesof thesubstancebecomesgaseousabove theboiling point:� ð Ñ Í � Í ô ê � Ñ Í ô ÷ Æ�>WÈ�� : É�ÚÌ îAÈ���îhce> çÆ Ì ÷�ý Æ ý ø�ÿ : ø�ù ö{÷ ��ù£ø-Æ�îAÈ ÷ É�È��-É´á ÷�i � :fç ý Æ�îhce- ö �`È��-É�ÉThenwe needonly say� ð Ñ Í � Í ô ê � Ñ Í ô ÷ Æ�U ö�÷ ø�ù#È��¦ø�ô ÷ Í ê ù ö ã ø-Æ�Ê#Ç#Ç#É�É .c. “The waterin John’s waterbottleis frozen.”Wewill usetheconstant@KÑW�to representthesituationin whichthissentenceholds.Notethat it is easyto make mistakesin which oneassertsthatonly someof thewaterin thebottleis frozen.Û � Ì �j�kceU ö�÷ øù­áK�lcgU ö�÷ ø�ù ð Ñ ÷�÷ ��ø��Gá üKö �-Æ ò Ñ�ó#ôGÈ��WÈ�@KÑW�nÉá�m{ôH� Í,ã ø�Æ'�nÈ���È�@KÑ��MÉ ç Æ'�?c � Ñ�� Í,ã È�@KÑ��MÉd. “Perrieris akind of water.”� øù`ù Í øùenZU ö{÷ ø�ùe. “JohnhasPerrierin his waterbottle.”Û � Ì �j�kceU ö�÷ øù­áK�lcgU ö�÷ ø�ù ð Ñ ÷�÷ ��ø��Gá üKö �-Æ ò Ñ�ó#ôGÈ��WÈ�@KÑW�nÉá�m�ôH� Í,ã ø-Æ'�MÈ��WÈ�@KÑW�nÉ ç �ac � ø�ù�ù Í ø�ùf. “All liquidshave a freezingpoint.”Presumablywhat this meansis that all substancesthat are liquid at room temper-aturehave a freezingpoint. If we use /lý d Í X�� Í�ão� ��E� ÷�ö ôH>Wø to denotethis classofsubstances,thenwe haveÌ > /lý d Í X�� Í,ãQ� ���1� ÷Wö ôH>Wø�Æ�>WÉ ç Û ÷ � õ ù~ø-ø�ñ Í ô ê � Ñ Í ô ÷ Æ�>�È ÷ Éwhere� õ ù~ø-ø�ñ Í ô ê � Ñ Í ô ÷ is definedsimilarly to� ð Ñ Í � Í ô ê � Ñ Í ô ÷ . Notethat this state-mentis falsein therealworld: wecaninventcategoriessuchas“blue liquid” whichdonot have a uniquefreezingpoint. An interestingexercisewould be to definea “pure”substanceasoneall of whoseinstanceshave thesamechemicalcomposition.g. “A liter of waterweighsmorethana liter of alcohol.”Ì �nÈ ö �?cDU ö{÷ ø�ù­á ö c â �2>�Ñ�ó�ÑF�~áKpnÑ��+��ÿ�ø-Æ'�MÉ~ä�d Í ÷ ø�ù��-Æ�Ê#ÉáHp¦Ñ��+��ÿ�ø�Æ ö É~ä�d Í ÷ ø�ù��-Æ�Ê#É ç þ ö �F�-Æ'�nÉ i þ ö �F�`Æ ö É10.6 This is a fairly straightforward exercisethatcanbedonein directanalogyto thecor-respondingdefinitionsfor sets.http://librosysolucionarios.net62 Chapter 10. KnowledgeRepresentationa. q îAó ö �� ÷ Í9è ø � ö ù ÷ P ø*>WÑ,ÿ : Ñ�� Í ÷ Í Ñ,ô holds betweena set of partsanda whole, sayingthatanything thatis apartof thewholemustbeapartof oneof thesetof parts.Ì �`È`� q îAó ö �� ÷ Í9è ø � ö ù ÷ P ø*>WÑ,ÿ : Ñ�� Í ÷ Í Ñ,ôGÆ��`È`�MÉ�ÚÆ Ì�: � ö ù ÷�û�� Æ : È`�nÉ ç Û: Ï : Ï ce�lá � ö ù ÷Wû�� Æ : È : Ï É�Éb.� ö ù ÷ � ö ù ÷ Í ÷ Í Ñ,ô holdsbetweenasetof partsandawholewhenthesetis disjointandisanexhaustive decomposition.Ì �`È`� � ö ù ÷ � ö ù ÷ Í ÷ Í Ñ,ôGÆ��-È`�nÉLÚ� ö ù ÷ � Í �-ø P Í ��r¾Ñ Í ô ÷ Æ��`É´á q îAó ö ��� ÷ Í9è ø � ö ù ÷ P ø�>WÑ«ÿ : ÑF� Í ÷ Í Ñ,ôGÆ��-È`�nÉc. A setof partsis� ö ù ÷ � Í �`ø P Í �WrPÑ Í ô ÷ if whenyou take any two partsfrom theset,thereis nothingthatis apartof bothparts.Ì � � ö ù ÷ � Í �-ø P Í �WrPÑ Í ô ÷ Æ��`É}ÚÌ�:�Î È :�Ïa:^Î cg��á :_Ï cg��á :�Î 8ä :�Ïdç J Û�:_Ð � ö ù ÷Wû�� Æ :�Ð È :�Î É´á � ö ù ÷Wû�� Æ :�Ð È :_Ï ÉIt is not the casethat sft Ö,Ù sft Ö�Ù�Õ%Ù�Õ'uv&AÆ��`È*wlØx&£×�y ß!z Æ��`É�É for any�. A set�may consistofphysically overlappingobjects,suchas a handand the fingersof the hand. In that case,w¸Øx&£×*y ß!z Æ��-Éis equalto the hand,but�is not a partition of it. We needto ensurethat theelementsof�arepartwisedisjoint:Ì � � ö ù ÷ � Í �`ø P Í ��r¾Ñ Í ô ÷ Æ��`É ç sHt Ö�Ù sHt Ö�Ù�Õ�Ù�Õ.uv&¯Æ��`È*w¸ØO&�×*y ß!z Æ��-É�ÉAË10.7 For an instanceÍof a substance�with price per pound>andweightôpounds,thepriceofÍwill beô|{}>, or in otherwords:Ì²Í È��-È�ôGÈ�> Í cg�lá � ù Í >Wø � øù~Æ��`È � Ñ���ô ã �-Æ�Ê#É�É~äk~#Æ�>WÉ®á�U�ø Í ê ó ÷ Æ Í É~ä � Ñ���ô ã �-Æ�ôGÉç � ù Í >Wø-Æ Í É~ä�~#Æ�ôa{�>�ÉIf�is thesetof tomatoesin a bag,then ð ��ôH>Wó û�� Æ��WÉ is thecompositeobjectconsistingofall thetomatoesin thebag.Thenwe haveÌ²Í È��-È�ôGÈ�>?��n���á � ù Í >Wø � ø�ù£Æ��`È � Ñ���ô ã �`Æ�Ê#É�É~äk~#Æ�>�Éá�U�ø Í ê ó ÷ Æ ð ��ôH>Wó û�� Æ��WÉ�É~ä � Ñ���ô ã �-Æ�ôGÉç � ù Í >Wø-Æ ð ��ôH>Wó û�� Æ��WÉ�É~ä|~#Æ�ôk{}>WÉ10.8 In theschemein thechapter, aconversionaxiomlookslike this:Ì î Ô��*&�Ù�Õ.����Ù%�-Ö " Æ��#Ë�4E3K{¶î0É~ä|��&£×*yo� " Æ�îAÉAË“50 dollars” is just~#Æ�4#Ç#É, thenameof anabstractmonetaryquantity. For any measurefunc-tion suchas~, we canextendtheuseof i asfollows:Ì îAÈ�ï�î i ï ç ~#Æ�î0É i ~#Æ�ïYÉ0ËSincetheconversionaxiomfor dollarsandcentshasÌ î Ô��*&�Ù " Æ�Ê#Ç#Çh{¶î0É~ä[~#Æ�îAÉit follows immediatelythat~#Æ�4#Ç#É i Ô���&£Ù " Æ�4#Ç#É.In thenew scheme,we mustintroduceobjectswhoselengthsareconverted:Ì î Ô��*&�Ù�Õ.����Ù%�-Ö " ÆW���*&�(�Ù%y�Æ�îAÉ�ÉJä��#Ë�4E3K{[��&£×�yv� " ÆW���*&�(-Ù%y�Æ�îAÉ�ÉAËhttp://librosysolucionarios.net63Thereis no obviousway to referdirectly to “50 dollars” or its relationto “50 cents”.Again,we mustintroduceobjectswhosemonetaryvalueis 50 dollarsor 50cents:Ì îAÈ�ï�~#Æ�p ö �+�Aø-Æ�îAÉ�É~ä|4#Ç?á¹Ô��*&�Ù " Æ�p ö �+�Aø�Æ�ïYÉ�É~äk4#Ç ç ~#Æ�p ö �+�0ø�Æ�îAÉ�É i ~#Æ�p ö �+�0ø�Æ�ïYÉ�É10.9 We will definea function q î�>Wó ö ô ê ø /"ö�÷ ø that takesthreearguments:a sourcecur-rency, a time interval, anda target currency. It returnsa numberrepresentingtheexchangerate.For example,q î�>Wó ö ô ê ø /"ö{÷ ø�Æ�b � P Ñ��2� ö ù#È�Ê=6 õ ø���Ê=;=;=4#È P ö ô Í �`ó I ù£Ñ«ôGø-É~ä94#Ë�7=5=6=6meansthat you canget 5.8677Krone for a dollar on February17th. This wasthe FederalReservebank’sSpotexchangerateasof 10:00AM. It is themid-pointbetweenthebuyingandsellingrates.A morecompleteanalysismight want to includebuying andselling rates,andthepossibility for many differentexchanges,aswell asthecommissionsandfeesinvolved.Notealsothedistinctionbetweena currency suchasb � P ÑF�2� ö ù anda unit of measurement,suchasis usedin theexpressionb � P Ñ��'� ö ù��-Æ�Ê#Ë�;=;#É .10.10 Anotherfun problemfor clearthinkers:a. Rememberthat ý Æ�>WÈ Í É meansthatsomeeventof type>occursthroughouttheintervalÍ: Ì >WÈ Í ý Æ�>�È Í É�Ú Æ Û ø ø�cB>0á P ��ù Í ô ê Æ Í È�ø-É�ÉUsing� �� q è øô ÷ asthequestionrequestsis notsoeasy, becausetheinterval subsumesall eventswithin it.b. A ð Ñ ÷ ó�Æ : È�X�É event is onein which both:andXoccurthroughoutthedurationof theevent.Thereis only onewaythiscanhappen:both:andXhave to persistfor thewholeinterval. Anotherway to sayit:Ì?Í È`r P ��ù Í ô ê Æ'r�È Í É ç å ý Æ : È`rPɯá ý Æ�X�È`rPÉ æis logically equivalenttoå Ì?Í È`r P ��ù Í ô ê Æ'r�È Í É ç ý Æ : È`rPÉ æ áºå Ì?Í È`r P ��ù Í ô ê Æ'r�È Í É ç ý Æ�X�È`rPÉ æwhereasthesameequivalencefails to hold for disjunction(seethenext part).c. ý Æ û ôGø û�� Æ : È�X�É�È Í É meansthata:eventoccursthroughoutÍor aXeventdoes:Ì�: È�X�È Í ý Æ û ôGø û�� Æ : È�X�É�È Í É}Úå Æ Ì r P ��ù Í ô ê Æ'r`È Í É ç ý Æ : È`rPÉ�É � Æ Ì r P ��ù Í ô ê Æ'r�È Í É ç ý Æ�X�È`rPÉ�É æ ËOn the otherhand,ý ÆW�GÕ�Ù%yv��Ö�Æ : È�X�É�È Í É holdsif, at every point inÍ, eithera:or aXishappening:Ì�: È�X�È Í ý Æ û ôGø û�� Æ : È�X�É�È Í É}Ú Æ Ì r P ��ù Í ô ê Æ'r`È Í É ç Æ ý Æ : È`r¾É � ý Æ�X`È`r¾É�É�É5Ëd. ý Æ�@pø è øù~Æ : É�È Í É shouldmeanthat thereis never an event of type:going on in anysubinterval ofÍ, while ý Æ�@pÑ ÷ Æ : É�È Í É shouldmeanthatthereis no singleeventof type:thatspansall ofÍ, eventhoughtheremaybeoneor moreeventsof type:for subinter-valsofÍ:Ì�: È Í ý Æ�@pø è øù~Æ : É�È Í ÉºÚ J Û r P ��ù Í ô ê Æ'r`È Í É¯á ý Æ : È`r¾ÉÌ�: È Í ý Æ�@pÑ ÷ Æ : É�È Í ÉºÚ J ý Æ : È Í Éhttp://librosysolucionarios.net64 Chapter 10. KnowledgeRepresentationOnecould alsoaskstudentsto prove two versionsof deMorgan’s laws usingthe twotypesof negation,eachwith its correspondingtypeof disjunction.10.11 Any objectîis anevent,anddGÑ�> ö{÷ Í Ñ,ôGÆ�îAÉ is theevent that for every subinterval oftime, refersto the placewhereîis. For example,dGÑ�> ö{÷ Í Ñ,ôGÆ � ø ÷ ø�ù£É is the complex eventconsistingof hishomefrom midnightto about9:00today, thenvariouspartsof theroad,thenhis office from 10:00to 1:30,andsoon. To saythatanevent is fixedis to saythatany twomomentsof theeventhave thesamespatialextent:Ì ø õ Í îAø ã Æ�ø�ɺÚÆÌ ö È�� ö c þ Ñ,ÿ�øô ÷ ��á���c þ Ñ,ÿ�øô ÷ �lá � ���Wø è ø�ô ÷ Æ ö È�ø�ÉAá � ���Wø è ø�ô ÷ Æ��WÈ�ø�Éç ��: ö�÷ Í ö � q î ÷ øô ÷ Æ ö É~ä ��: ö{÷ Í ö � q î ÷ ø�ô ÷ Æ��WÉ�É10.12 Wewill omit universallyquantifiedvariables:ð ø � Ñ«ù£ø-Æ Í È`rPÉ Ú Û�� þ ø-ø ÷ Æ Í È � É®á þ ø�ø ÷ Æ � È`r¾Éâ �P÷ ø�ù£Æ Í È`rPÉ Ú ð ø � Ñ«ù£ø-Æ'r`È Í ÉP ��ù Í ô ê Æ Í È`rPÉ Ú Û�� È�ÿ þ ø-ø ÷ Æ � ÷Wö ù ÷ Æ'r¾É�È � ɯá þ ø-ø ÷ Æ � È Í Éá þ ø-ø ÷ Æ Í È�ÿ�É®á þ ø-ø ÷ Æ�ÿ�È q ô ã Æ'r¾É�Éû è øù!� ö : Æ Í È`r¾É Ú Û�� P ��ù Í ô ê Æ � È Í É®á P ��ù Í ô ê Æ � È`rPÉ10.13 d Í ô � Æ'��ù�� Î È`��ù�� Ï É}Úm�ô ý°ö�ê Æ “ ö ” È " Ù�Ö_ÈO�N�-Ù sHt (���Æ'� ÷ � Î É�É´á�m{ôGÆ “ ó#ù£ø � ä “” Y ��ù�� Ï Y “” ”È " Ù�Ö_Éd Í ô � ý øî ÷ Æ'��ù�� Î È`��ù!� Ï È ÷ øî ÷ É}Úm�ô ý°ö�ê Æ “ ö ” È " Ù�Ö_ÈO�N�-Ù sHt (���Æ'� ÷ � Î É�É´á�m{ôGÆ “ ó#ù£ø � ä “” Y ��ù�� Ï Y “” ” Y�÷ øî ÷ È " Ù�Ö_É10.14 Hereis aninitial sketchof oneapproach.(Othersarepossible.)A givenobjectto bepurchasedmayrequire someadditionalparts(e.g.,batteries)to befunctional,andtheremayalsobe optionalextras. We canrepresentrequirementsasa relationbetweenan individualobjectandaclassof objects,qualifiedby thenumberof objectsrequired:Ì î±îhcLÔ�u�uv� $ Õ.���v�v�4ÅwÕ'(�Õ�Ù t ��Ô t ����Ö t ç à �W�-ØPÕ�Ö`� " Æ�î0È�õÃ�w t Ù�Ù%��ÖE�£È`3�É0ËWe alsoneedto know thata particularobjectis compatible,i.e., fills a given role appropri-ately. For example,Ì îAÈ�ï�îhcKÔ�u�uo� $ Õ'�*�o�v�4ÅwÕ.(-Õ%Ù t �;Ô t ���-Ö t ánï�c¶ÅwØ¾Ö t ×��*�'� Ã+Ã�w t Ù�Ù%��ÖE�ç �°Ñ«ÿ : ö{÷ Í �1�;ø�Æ�ïYÈ�îAÈ â+â ðnö�÷�÷ ø�ù�ïYÉThenit is relatively easyto testwhetherthe setof orderedobjectscontainscompatiblere-quiredobjectsfor eachobject.10.15 Pluralscanbehandledby a s � Ø¾Ö t � relationbetweenstrings,e.g.,s � Ø¾Ö t ��Æ “ >WÑ«ÿ : � ÷ ø�ù ” È “ >WÑ,ÿ : � ÷ ø�ù�� ” Éplusanassertionthattheplural (or singular)of anameis alsoanamefor thesamecategory:Ì >�È�� Î È�� Ï Ä t ���_Æ�� Î È�>WÉ´áIÆ s � ØPÖ t ��Æ�� Î È�� Ï É � s � Ø¾Ö t ��Æ�� Ï È�� Î É�É ç Ä t ����Æ�� Ï È�>WÉhttp://librosysolucionarios.net65Conjunctionscanbehandledby sayingthatany conjunctionstringis anamefor acategory ifoneof theconjunctsis anamefor thecategory:Ì >�È��`È�� Ï Ô�uv&���ØO&�×�Ù,Æ�� Ï È��`ÉYápÄ t ���_Æ�� Ï È�>WÉ ç Ä t ���_Æ��-È�>WÉwhereÔ�uv&���ØO&�×�Ùis definedappropriatelyin termsof concatenation.Probablyit would bebetterto redefineà �*���*� t &£Ù«Ô t Ù%��(*u4ÖE��Ä t �h� instead.10.16 Chapter22explainshow to uselogic to parsetext stringsandextractsemanticinfor-mation.Theoutcomeof thisprocessis adefinitionof whatobjectsareacceptableto theuserfor a specificshoppingrequest;this allows theagentto go out andfind offersmatchingtheuser’s requirements.Weomit thefull definitionof theagent,althoughaskeletonmayappearon theAIMA projectwebpages.10.17 Hereis asimpleversionof theanswer;it canbeelaboratedadinfinitum. Let thetermð ��ïYÆ��WÈ�îAÈ��`È : É denotetheeventcategoryof buyer�buyingobjectîfrom seller�for price:.Wewantto sayaboutit that�transfersthemoney to�, and�transfersownershipofîto�.ý Æ ð ��ï^Æ��WÈ�î0È��-È : É�È Í ÉºÚý Æ û �µôH�-Æ��`È�îAÉ�È � ÷Wö ù ÷ Æ Í É�É�áÛ ÿ þ Ñ«ôGøï^Æ�ÿ�É®á : ä�p ö �+�Aø-Æ�ÿ�É®á ý Æ û �²ôH�`Æ��WÈ�ÿ�É�È � ÷Wö ù ÷ Æ Í É�É�áý Æ û �µôH�-Æ��WÈ�î0É�È q ô ã Æ Í É�ɯá ý Æ û �²ôH�`Æ��`È�ÿ�É�È q ô ã Æ Í É�É10.18 Let ý ù ö ã ø�Æ��WÈ�î0È ö È�ïYÉ denotethe classof eventswhereperson�tradesobjectïtopersonö for objectî:ý Æ ý ù ö ã ø�Æ��WÈ�îAÈ ö È�ï^É�È Í ÉoÚý Æ û �µôH�-Æ��WÈ�ïYÉ�È � ÷�ö ù ÷ Æ Í É�É®á ý Æ û �²ôH�`Æ ö È�îAÉ�È � ÷�ö ù ÷ Æ Í É�É�áý Æ û �µôH�-Æ��WÈ�î0É�È q ô ã Æ Í É�ɯá ý Æ û �²ôH�`Æ ö È�ï^É�È q ô ã Æ Í É�ÉNow theonly tricky partaboutdefiningbuying in termsof tradingis in distinguishingaprice(ameasurement)from anactualcollectionof money.ý Æ ð ��ï^Æ��WÈ�î0È ö È : É�È Í ÉºÚ Û ÿ þ Ñ,ôGø�ïYÆ�ÿ�É´á ý ù ö ã ø-Æ��WÈ�îAÈ ö È�ÿ�É®á�p ö �+�0ø�Æ�ÿ�É~ä :10.19 Therearemany possibleapproachesto this exercise.The ideais for thestudentstothink aboutdoingknowledgerepresentationfor real;to considerahostof complicationsandfind someway to representthefactsaboutthem.Someof thekey pointsare:� Ownershipoccursover time,soweneedeitherasituation-calculusor interval-calculusapproach.� Therecanbe joint ownershipandcorporateownership. This suggeststhe owner is agroupof somekind, which in thesimplecaseis a groupof oneperson.� Ownershipprovidescertainrights: to use,to resell,to give away, etc. Much of this isoutsidethe definition of ownershipper se, but a goodanswerwould at leastconsiderhow muchof this to represent.� Own canown abstractobligationsaswell asconcreteobjects.This is the ideabehindthe futuresmarket, andalsobehindbanks:whenyou deposita dollar in a bank,youaregiving up ownershipof thatparticulardollar in exchangefor ownershipof therighthttp://librosysolucionarios.net66 Chapter 10. KnowledgeRepresentationto withdraw anotherdollar later. (Or it could coincidentallyturn out to be the exactsamedollar.) Leasesand the like work this way aswell. This is tricky in termsofrepresentation,becauseit meanswe have to reify transactionsof this kind. That is,U Í ÷ ó ã ù ö �nÆ : ø�ù��`Ñ,ôGÈ�ÿ�Ñ«ôGøï^È�� ö ô � È ÷ Í ÿ�ø�É mustbeanobject,nota predicate.10.20 Mostschoolsdistinguishbetweenrequiredcoursesandelectedcourses,andbetweencoursesinsidethedepartmentandoutsidethedepartment.For eachof these,theremaybere-quirementsfor thenumberof courses,thenumberof units(sincedifferentcoursesmaycarrydifferentnumbersof units),andongradepointaverages.Weshow ourchosenvocabulary byexample:� StudentJones’completecourseof studyfor thewholecollegecareerconsistsof Math1,CS1,CS2,CS3,CS21,CS33andCS34,andsomeothercoursesoutsidethemajor.ý°ö � ø�Æ ò Ñ,ôGø*�`Èë þ ö{÷ ó�Ê#È q�q Ê#È ð Í Ñ��E3�È�� � Ê#È�� � �#È�� � �#È�� � �#Ê#È�� � �=�4È�� � �E3�� Ñ ÷ ó�ø�ù�� í É� Jonesmeetstherequirementsfor amajorin ComputerScienceþ ö r¾Ñ«ù£Æ ò Ñ,ôGø*�`È�� � É� CoursesMath1,CS1,CS2,andCS3arerequiredfor aComputerSciencemajor./ ø*X�� Í ù£ø ã Æ�ë þ ö�÷ ó�Ê#È�� � Ê#È�� � �#È�� � � í È�� � ÉÌ �`È ã / ø�XW� Í ù£ø ã Æ��-È ã É�ÚÆ Ì�:ªÛ Ñ ÷ ó�ø�ù�� þ ö rPÑ,ù£Æ : È ã É ç ý°ö � ø�Æ : È�b?ô Í Ñ,ôGÆ��-È�Ñ ÷ ó�øù!�`É�É�É� A studentmusttake at least18unitsin theCSdepartmentto getadegreein CS.P ø : ö ù ÷ ÿ�øô ÷ Æ�� � Ê#É~ä�� � á P ø : ö ù ÷ ÿ�ø�ô ÷ Æ þ ö�÷ ó�Ê#É~ä þ ö{÷ óMáNË�Ë�Ëb2ô Í ÷ �`Æ�� � Ê#É~ä��2á�b2ô Í ÷ �`Æ�� � �#É~ä�3MáFË�Ë�Ë/ ø*X�� Í ù£ø ã b?ô Í ÷ ��m{ôGÆ�Ê=7#È�� � È�� � ÉÌ �0È ã / ø�XW� Í ù£ø ã b2ô Í ÷ ��m{ôGÆ'�AÈ ã É}ÚÆ Ì�:ªÛ �`È�Ñ ÷ ó�øù�� þ ö r¾Ñ,ù~Æ : È ã É ç ý°ö � ø-Æ : È�b2ô Í Ñ«ôGÆ��`È�Ñ ÷ ó�ø�ù��`É�Éá â �2�2m�ô P ø : ö ù ÷ ÿ�øô ÷ Æ��`È ã ÉYá ý Ñ ÷Wö �2b2ô Í ÷ �-Æ��`ÉRa�Ì �`È ã�â �2�'m{ô P ø : ö ù ÷ ÿ�ø�ô ÷ Æ��`È ã ÉLÚ Æ Ì >?>lcg� ç P ø : ö ù ÷ ÿ�ø�ô ÷ Æ�>�É~ä ã ÉÌ > ý Ñ ÷�ö �2b2ô Í ÷ �`Æ�ë í É~ä�ÇÌ >WÈ�� ý Ñ ÷Wö �2b2ô Í ÷ �-Æ�ë=>v� � í É~ä�b?ô Í ÷ �`Æ�>WÉ Y�ý Ñ ÷�ö �2b?ô Í ÷ �`Æ��`ÉOnecaneasilyimagineotherkindsof requirements;thesejust giveyouaflavor.In this solutionwe took “over anextendedperiod” to meanthatwe shouldrecommenda setof coursesto take, without schedulingthemon a semester-by-semesterbasis. If youwantedto do that,you would needadditionalinformationsuchaswhencoursesaretaught,what is a reasonablecourseload in a semester, andwhat coursesareprerequisitesfor whatothers.For example:ýMö � ê ó ÷ Æ�� � Ê#È õ¦ö �2�;É� ù~øù£ø*X�� Í � Í ÷ ø*�`Æ�ë=� � Ê#È�� � � í È�� � �#ÉýMö � ø�m�ô � øÿ�ø*� ÷ øù~Æ ò Ñ,ôGø*�`È õ­ö �2�';=4#È�ë þ ö�÷ ó�Ê#È�� � Ê#È q ô ê � Í �-ó�Ê#È ü Í � ÷ Ñ,ù`ïYÊ í Éþ ö î��¦Ñ���ù!�`ø*� � øù � ø�ÿ�ø*� ÷ ø�ù£Æ�4#Éhttp://librosysolucionarios.net67The problemwith finding the bestprogramof study is in definingwhat bestmeansto thestudent.It is easyenoughto saythatall otherthingsbeingequal,oneprefersa goodteacherto a badone,or an interestingcourseto a boringone. But how do you decidewhich is bestwhenonecoursehasabetterteacherandis expectedto beeasier, while analternative is moreinterestingandprovidesonemorecredit? Chapter16 usesutility theoryto addressthis. Ifyou canprovide a way of weighingtheseelementsagainsteachother, thenyou canchooseabestprogramof study;otherwiseyoucanonlyeliminatesomeprogramsasbeingworsethanothers,but can’t pick anabsolutebestone. Complexity is a furtherproblem:witha general-purposetheorem-prover it’s hardto do muchmorethanenumeratelegal programsandpickthebest.10.21 This exerciseand the following two are rathercomplex, perhapssuitablefor termprojects.At this point, we want to stronglyurge thatyou do assignsomeof theseexercises(or oneslike them) to give your studentsa feeling of what it is really like to do knowl-edgerepresentation.In general,studentsfind classificationhierarchieseasierthanotherrep-resentationtasks. A recenttwist is to compareone’s hierarchywith online onessuchasyahoo.com .10.22 This is themostinvolvedrepresentationproblem.It is suitablefor agroupprojectof2 or 3 studentsover thecourseof at least2 weeks.10.23 Normally onewould assign10.22in oneassignment,andthenwhenit is done,addthis exercise(posibly varying thequestions).That way, thestudentsseewhetherthey havemadesufficientgeneralizationsin their initial answer, andgetexperiencewith debuggingandmodifying aknowledgebase.10.24 In many AI andProlog textbooks,you will find it statedplainly that implicationssuffice for the implementationof inheritance.This is true in thelogical but not thepracticalsense.a. Herearethreerules,written in Prolog.We actuallywould needmany moreclausesontheright handsideto distinguishbetweendifferentmodels,differentoptions,etc.worth(X,575) :- year(X,1973), make(X,dodge), style(X,van).worth(X,27000) :- year(X,1994), make(X,lexus), style(X,sedan).worth(X,5000) :- year(X,1987), make(X,toyota), style(X,sedan).Tofind thevalueof JB,givenadatabasewith year(jb,1973) ,make(jb,dodge)andstyle(jb,van) wewouldcall thebackwardchainerwith thegoalworth(jb,D) ,andreadthevaluefor D.b. Thetime efficiency of this queryis û Æ�ôGÉ , whereôin this caseis the11,000entriesintheBlue Book. A semanticnetwork with inheritancewould allow us to follow a linkfrom JB to 1973-dodge-van , andfrom thereto follow theworth slot to find thedollarvaluein û Æ�Ê#É time.c. With forwardchaining,assoonaswe aretold thethreefactsaboutJB,we addthenewfact worth(jb,575) . Thenwhenwe get the queryworth(jb,D) , it is û Æ�Ê#É tofind the answer, assumingindexing on the predicateandfirst argument. This makeslogical inferenceseemjust like semanticnetworks except for two things: the logicalhttp://librosysolucionarios.net68 Chapter 10. KnowledgeRepresentationinferencedoesa hashtablelookup insteadof pointerfollowing, andlogical inferenceexplicitly storesworth statementsfor eachindividual car, thuswastingspaceif therearea lot of individual cars. (For this kind of application,however, we will probablywantto consideronly afew individualcars,asopposedto the11,000differentmodels.)d. If eachcategoryhasmany properties—forexample,thespecificationsof all thereplace-mentpartsfor thevehicle—thenforward-chainingon the implicationswill alsobeanimpracticalway to figureout thepriceof a vehicle.e. If we have a ruleof thefollowing kind:worth(X,D) :- year-make-style(X,Yr,Mk,St),year-make-style(Y,Yr,Mk,St), worth(Y,D).togetherwith factsin thedatabaseaboutsomeotherspecificvehicleof thesametypeasJB, then the queryworth(jb,D) will be solved in û Æ�Ê#É time with appropriateindexing, regardlessof how many otherfactsareknown aboutthat typeof vehicleandregardlessof thenumberof typesof vehicle.10.25 Whencategoriesarereified, they canhave propertiesasindividual objects(suchas� ö ù ã�Í ô ö � Í ÷ ï and� � : ø�ù��-ø ÷ � ) thatdonotapplyto theirelements.Without thedistinctionbe-tweenboxedandunboxed links, thesentence� ö ù ã{Í ô ö � Í ÷ ïYÆ �^Í ô ê �;ø ÷ Ñ,ô � ø ÷ �`È�Ê#É might meanthatevery singletonsethasoneelement,or thattheresi only onesingletonset.http://librosysolucionarios.netSolutionsfor Chapter11Planning11.1 Bothproblemsolverandplannerareconcernedwith gettingfrom astartstateto agoalusingasetof definedoperationsor actions.But in planningweopenup therepresentationofstates,goals,andplans,whcihcallows for a wider varietyof algorithmsthatdecomposethesearchspace.11.2 This is aneasyexercise,thepoint of which is to understandthat “applicable”meanssatisfyingthe preconditions,and that a concreteaction instanceis one with the variablesreplacedby constants.Theapplicableactionsare:� ���£Æ � Î È* ��¡ È*¢ � ß É� ���£Æ � Î È* ��¡ È* �¡ É� ���£Æ � Ï È*¢ � ß È* �¡ É� ���£Æ � Ï È*¢ � ß È*¢ � ß ÉA minor point of this is that the action of flying nowhere—fromone airport to itself—isallowableby thedefinitionof� �+�, andis applicable(if notuseful).11.3 For theregularschemawehave:õ ��ï � ù£ø*>WÑ,ô ã Æ : È � È ÷ Ñ�È��`É}Úâ ÷ Æ : È � È��`ÉYá � � ö ôGø�Æ : É´á âJÍ ù : Ñ,ù ÷ Æ � É�á âJÍ ù : Ñ,ù ÷ Æ ÷ Ñ�Éâ ÷ Æ : È�îAÈ / ø����� ÷ Æ ö È��-É�ɺÚÆ â ÷ Æ : È�î0È��-É´áIÆ ö 8ä õ � ï^Æ : È � È�î0É � J õ � ï � ù£ø*>WÑ,ô ã Æ : È � È�î0È��-É�É�É�Fâ ÷ Æ : È � È��`ÉYá ö ä õ � ïYÆ : È � È�îAÉ´á õ � ï � ù£ø�>�Ñ,ô ã Æ : È � È�îAÈ��`ÉWhenwe add £Bt Ö $ �«Ò weget:â ÷ Æ : È�îAÈ / ø����� ÷ Æ ö È��-É�ɺÚÆ â ÷ Æ : È�î0È��-É´áIÆ ö 8ä õ � ï^Æ : È � È�î0ÉYá ö 8ä ý ø*�;ø : Ñ,ù ÷ Æ : È � È�î0É�É� ö ä õ � ïYÆ : È � È�îAÉ®áKJ õ � ï � ù~ø�>WÑ«ô ã Æ : È � È�îAÈ��`É� ö ä ý ø��;ø : Ñ,ù ÷ Æ : È � È�îAÉYá�J ý ø*�;ø : Ñ,ù ÷ � ù£ø*>WÑ,ô ã Æ : È � È�î0È��-É�Fâ ÷ Æ : È � È��`ÉYá ö ä õ � ïYÆ : È � È�îAÉ´á õ � ï � ù£ø�>�Ñ,ô ã Æ : È � È�îAÈ��`É�Fâ ÷ Æ : È � È��`ÉYá ö ä ý ø���ø : Ñ,ù ÷ Æ : È � È�î0É´á ý ø���ø : Ñ,ù ÷ � ù£ø�>�Ñ,ô ã Æ : È � È�îAÈ��`ÉIn general,we(1) createda s Ö`�«×�uv&£Ò predicatefor eachaction,andthen,for eachfluentsuchasÃ?Ù, wecreateapredicatethatsaysthefluentkeepsits old valueif eitheranirrelevantactionis taken,or anactionwhosepreconditionis not satisfied,andit takeson a new valueaccordingto theeffectsof a relevantactionif thataction’s preconditionsaresatisfied.69http://librosysolucionarios.net70 Chapter 11. Planning11.4 This exerciseis intendedasa fairly easyexercisein describinga domain.It is similarto theShakey problem(11.13),soyoushouldprobablyassignonly oneof thesetwo.a. Theinitial stateis:â ÷ Æ þ Ñ,ô � ø�ïYÈ â É´á â ÷ Æ ðnö ô ö ô ö �`È ð É^á â ÷ Æ ð Ñ,îAÈ��¦É®áü ø Í ê ó ÷ Æ þ Ñ,ô � ø�ïYÈ�dGÑ��MÉAá ü ø Í ê ó ÷ Æ ð Ñ,î0È�dGÑ��MÉAá ü ø Í ê ó ÷ Æ ðnö ô ö ô ö �-È ü Í ê ó�ɯá� ��-ó ö �1�;ø�Æ ð Ñ,îAÉ´áK��� Í ÿ�� ö �1�;ø�Æ ð Ñ,îAÉb. Theactionsare:â > ÷ Í Ñ,ôGÆ ACTION:-¦Ñ`Æ�î0È�ï^É�ÈPRECOND:â ÷ Æ þ Ñ,ô � ø�ïYÈ�îAÉ�ÈEFFECT:â ÷ Æ þ Ñ«ô � øï^È�ïYÉ´á�JGÆ â ÷ Æ þ Ñ,ô � øï^È�î0É�É�Éâ > ÷ Í Ñ,ôGÆ ACTION:� ���`ó�Æ��WÈ�îAÈ�ïYÉ�ÈPRECOND:â ÷ Æ þ Ñ,ô � ø�ïYÈ�îAɯá � ���`ó ö �1�;ø�Æ��WÉ�ÈEFFECT:â ÷ Æ��WÈ�ï^É®á â ÷ Æ þ Ñ«ô � øï^È�ïYÉ´á�J â ÷ Æ��WÈ�îAÉYá�J â ÷ Æ þ Ñ,ô � ø�ïYÈ�îAÉ�Éâ > ÷ Í Ñ,ôGÆ ACTION:��� Í ÿ��1b : Æ��WÉ�ÈPRECOND:â ÷ Æ þ Ñ,ô � ø�ïYÈ�îAɯá â ÷ Æ��WÈ�îAÉ®á���� Í ÿ�� ö �E�;ø-Æ���É�ÈEFFECT: û ôGÆ þ Ñ,ô � ø�ïYÈ���ÉYá¤J ü ø Í ê ó ÷ Æ þ Ñ,ô � ø�ïYÈ ü Í ê ó�É�Éâ > ÷ Í Ñ,ôGÆ ACTION:-2ù ö � : Æ��WÉ�ÈPRECOND: ü ø Í ê ó ÷ Æ þ Ñ,ô � øï^È�ó�ÉAá ü ø Í ê ó ÷ Æ��WÈ�ó�Éá â ÷ Æ þ Ñ«ô � øï^È�î0É®á â ÷ Æ���È�î0É�ÈEFFECT: üpö è ø-Æ þ Ñ,ô � øï^È��WÉ�Éâ > ÷ Í Ñ,ôGÆ ACTION:��� Í ÿ�� P Ñ��²ôGÆ��WÉ�ÈPRECOND: û ôGÆ þ Ñ,ô � øï^È��WÉ´á ü ø Í ê ó ÷ Æ þ Ñ,ô � øï^È ü Í ê ó�É�ÈEFFECT:J û ôGÆ þ Ñ,ô � ø�ïYÈ��WÉYá�J ü ø Í ê ó ÷ Æ þ Ñ,ô � ø�ïYÈ ü Í ê ó�Éá ü ø Í ê ó ÷ Æ þ Ñ,ô � ø�ïYÈ�dGÑ��MÉâ > ÷ Í Ñ,ôGÆ ACTION:b?ôH-¸ù ö � : Æ��WÉ�È PRECOND: üKö è ø�Æ þ Ñ,ô � ø�ïYÈ���É�ÈEFFECT:J üKö è ø�Æ þ Ñ«ô � øï^È��WÉ�Éc. In situationcalculus,thegoalis astate�suchthat:üKö è ø�Æ þ Ñ«ô � øï^È ðnö ô ö ô ö �`È��-ÉYáFÆ Û î â ÷ Æ ð Ñ,îAÈ�î0È�� M É´á â ÷ Æ ð Ñ«î0È�îAÈ��`É�ÉIn STRIPS, wecanonly talk aboutthegoalstate;thereis nowayof representingthefactthattheremustbesomerelation(suchasequalityof locationof anobject)betweentwostateswithin theplan.Sothereis no way to representthis goal.d. Actually, wedid includethe� ���`ó ö �1��ø precondition.This is anexampleof thequalifi-cationproblem.11.5 Only positive literalsarerepresentedin astate.Sonotmentioninga literal is thesameashaving it benegative.11.6 Goalsandpreconditionscanonly be positive literals. Soa negative effect canonlymakeit harderto achieveagoal(or apreconditionto anactionthatachievesthegoal).There-fore,eliminatingall negative effectsonly makesaproblemeasier.11.7a. It is feasibleto usebidirectionalsearch,becauseit is possibleto invert the actions.However, most of thosewho have tried have concludedthat biderectionalsearchishttp://librosysolucionarios.net71generallynot efficient, becausethe forward andbackward searchestendto misseachother. This is dueto thelargestatespace.A few planners,suchasPRODIGY (Fink andBlythe,1998)have usedbidirectionalsearch.b. Again, this is feasiblebut not popular. PRODIGY is in fact (in part) a partial-orderplanner:in theforwarddirectionit keepsa total-orderplan(equivalentto a state-basedplanner),andin thebackwarddirectionit maintainsatree-structuredpartial-orderplan.c. An actionâcanbe addedif all the preconditionsofâhave beenachieved by otherstepsin theplan.Whenâis added,orderingconstraintsandcausallinks arealsoaddedto make surethatâappearsafterall theactionsthatenabledit andthata preconditionis notdisestablishedbeforeâcanbeexecuted.Thealgorithmdoessearchforward,butit is not thesameasforwardstate-spacesearchbecauseit canexploreactionsin parallelwhenthey don’t conflict. For example,ifâhasthreepreconditionsthatcanbesatisfiedby thenon-conflictingactionsð ,�, andP, thenthesolutionplancanberepresentedasa singlepartial-orderplan,while a state-spaceplannerwould have to considerall�=Gpermutationsof ð ,�, andP.d. Yes, this is onepossibleway of implementinga bidirectionalsearchin the spaceofpartial-orderplans.11.8 Thedrawing is actuallyrathercomplex, anddoesn’t fit well on this page.Somekeythingsto watchout for: (1) Both� ���and��u t Ò actionsarepossibleat levelâ M ; theplanescanstill fly whenempty. (2) Negative effectsappearin� Î, andaremutex with their positivecounterparts.11.9a. Literalsarepersistent,so if it doesnot appearin thefinal level, it never will andneverdid, andthuscannotbeachieved.b. In a serialplanninggraph,only oneactioncanoccurpertime step.Thelevel cost(thelevel at which a literal first appears)thusrepresentstheminimumnumberof actionsinaplanthatmight possiblyachieve theliteral.11.10 A forwardstate-spaceplannermaintainsa partialplanthatis a strict linearsequenceof actions;the plan refinementoperatoris to addan applicableactionto the endof the se-quence,updatingliteralsaccordingto theaction’s effects.A backwardstate-spaceplannermaintainsa partialplanthat is a reversedsequenceofactions;therefinementoperatoris to addanactionto thebeginningof thesequenceaslongastheaction’s effectsarecompatiblewith thestateat thebeginningof thesequence.11.11 Theinitial stateis:û ôGÆ ð È ý°ö �1�;ø�É®á û ôGÆ��?È â É®á û ôGÆ â È ýMö �1�;ø�É´áK���;ø ö ù~Æ ð ÉYá����;ø ö ù~Æ��°ÉThegoalis:û ôGÆ â È ð É´á û ôGÆ ð È��¦ÉFirstwe’ll explainwhy it is ananomalyfor anoninterleavedplanner. Therearetwo subgoals;supposewe decideto work on û ôGÆ â È ð É first. We canclear�off ofâandthenmoveâhttp://librosysolucionarios.net72 Chapter 11. Planningon to ð . But thenthereis no way to achieve û ôGÆ ð È��¦É without undoingthework we havedone.Similarly, if we work on thesubgoalû ôGÆ ð È��°É first we canimmediatelyachieve it inonestep,but thenwe have to undoit to getâon ð .Now we’ll show how thingswork out with aninterleaved plannersuchasPOP. Sinceû ôGÆ â È ð É isn’t true in the initial state,thereis only oneway to achieve it:þ Ñ è ø�Æ â È�îAÈ ð É ,for someî. Similarly, we alsoneedaþ Ñ è ø-Æ ð È�î A È��¦É step,for someî A. Now let’s lookat theþ Ñ è ø�Æ â È�îAÈ ð É step. We needto achieve its precondition����ø ö ù£Æ â É . We could dothat eitherwithþ Ñ è ø-Æ���È â È�ïYÉor withþ Ñ è ø ý Ñ ý°ö �1�;ø�Æ��WÈ â É . Let’s assumewe choosethelatter. Now if webind�to�, thenall of thepreconditionsfor thestepþ Ñ è ø ý Ñ ý°ö �1��ø-Æ��?È â Éaretrue in the initial state,andwe canaddcausallinks to them. We thennoticethat thereis a threat: theþ Ñ è ø�Æ ð È�î�A�È��¦É stepthreatensthe���;ø ö ù£Æ��¦É condition that is requiredbytheþ Ñ è ø ý Ñ ý°ö �1��ø step. We can resolve the threatby orderingþ Ñ è ø�Æ ð È�î�A�È��°É after theþ Ñ è ø ý Ñ ý°ö �1�;ø step.Finally, noticethatall thepreconditionsforþ Ñ è ø�Æ ð È�î A È��¦É aretrueinthe initial state.Thus,we have a completeplanwith all thepreconditionssatisfied.It turnsout thereis awell-orderingof thethreesteps:þ Ñ è ø ý Ñ ý°ö �1�;ø�Æ��²È â Éþ Ñ è ø-Æ ð È ýMö �1�;ø�È��°Éþ Ñ è ø-Æ â È ý°ö �1�;ø�È ð É11.12 Theactionswe needarethefour from page346:â > ÷ Í Ñ«ôGÆ ACTION: / Í ê ó ÷ � ó�Ñ�ø�È PRECOND: / Í ê ó ÷ � Ñ�> � û ôGÈ EFFECT: / Í ê ó ÷ � ó�Ñ�ø û ôGÉâ > ÷ Í Ñ«ôGÆ ACTION: / Í ê ó ÷ � Ñ�> � È EFFECT: / Í ê ó ÷ � Ñ�> � û ôGÉâ > ÷ Í Ñ«ôGÆ ACTION:dGø �P÷ � ó�Ñ�ø�È PRECOND:d�ø �P÷ � Ñ�> � û ôGÈ EFFECT:d�ø �P÷ � ó�Ñ�ø û ôGÉâ > ÷ Í Ñ«ôGÆ ACTION:dGø �P÷ � Ñ�> � È EFFECT:dGø �P÷ � Ñ�> � û ôGÉOnesolution found by GRAPHPLAN is to executeRightSock andLeftSock in the first timestep,andthenRightShoeandLeftShoein thesecond.Now weaddthefollowing two actions(neitherof whichhaspreconditions):â > ÷ Í Ñ«ôGÆ ACTION: üpö�÷ È EFFECT: üKö{÷�û ôGÉâ > ÷ Í Ñ«ôGÆ ACTION:�¦Ñ ö�÷ È EFFECT:�¦Ñ ö{÷�û ôGÉThepartial-orderplan is shown in FigureS11.1.We saw on page348that thereare6 total-orderplansfor the shoes/socksproblem. Eachof theseplanshasfour steps,andthusfivearrow links. Thenext step,Hat couldgoatany oneof thesefive locations,giving us5¥{¦4"ä�#Çtotal-orderplans,eachwith five stepsandsix links. Thenthefinal step,Coat, cango inany oneof these6 positions,giving us�#Ç�{}5­ä�Ê=7#Çtotal-orderplans.11.13 Theactionsarequitesimilarto themonkey andbanannasproblem—youshouldprob-http://librosysolucionarios.net73Start§Finish¨Left©Sockª RightSockLeft©Shoeª RightShoeHat«CoatFigureS11.1 Partial-orderplanincludinga hatandcoat,for Exercise11.1.ablyassignonly oneof thesetwo problems.Theactionsare:â > ÷ Í Ñ«ôGÆ ACTION:-­Ñ�Æ�îAÈ�ïYÉ�ÈPRECOND:â ÷ Æ � ó ö � ø�ïYÈ�îAɯá�m�ôGÆ�î0È�ù~ÉYá¤m�ôGÆ�ïYÈ�ù~É�ÈEFFECT:â ÷ Æ � ó ö � ø�ïYÈ�ï^ɯá�JJÆ â ÷ Æ � ó ö � øï^È�î0É�É�Éâ > ÷ Í Ñ«ôGÆ ACTION:� ���`ó�Æ��WÈ�î0È�ï^É�ÈPRECOND:â ÷ Æ � ó ö � øï^È�î0É®á � ��-ó ö �1�;ø�Æ��WÉ�ÈEFFECT:â ÷ Æ��WÈ�ïYÉ´á â ÷ Æ � ó ö � ø�ïYÈ�ï^ɯá�J â ÷ Æ��WÈ�îAÉ®áKJ â ÷ Æ � ó ö � øï^È�î0É�Éâ > ÷ Í Ñ«ôGÆ ACTION:��� Í ÿ��1b : Æ���É�ÈPRECOND:â ÷ Æ � ó ö � ø�ïYÈ�îAÉAá â ÷ Æ��WÈ�îAɯáK��� Í ÿ�� ö �E�;ø-Æ���É�ÈEFFECT: û ôGÆ � ó ö � ø�ïYÈ��WÉ®áKJ û ôGÆ � ó ö � ø�ïYÈ õ �;Ñ�Ñ«ù£É�Éâ > ÷ Í Ñ«ôGÆ ACTION:��� Í ÿ�� P ÑW�µôGÆ��WÉ�ÈPRECOND: û ôGÆ � ó ö � ø�ïYÈ���É�ÈEFFECT: û ôGÆ � ó ö � ø�ïYÈ õ �;Ñ�Ñ«ù£É´á¤J û ôGÆ � ó ö � øï^È��WÉ�Éâ > ÷ Í Ñ«ôGÆ ACTION: ý ��ù�ô û ôGÆ���É�È PRECOND: û ôGÆ � ó ö � ø�ïYÈ��WÉ®á â ÷ Æ � ó ö � ø�ïYÈ�îAɯá â ÷ Æ��;È�îAÉ�ÈEFFECT: ý ��ù�ôGø ã û ôGÆ��;É�Éâ > ÷ Í Ñ«ôGÆ ACTION: ý ��ù�ô û���� Æ���É�È PRECOND: û ôGÆ � ó ö � ø�ïYÈ��WÉ®á â ÷ Æ � ó ö � ø�ïYÈ�îAÉ®á â ÷ Æ��;È�îAÉ�ÈEFFECT:J ý ��ù�ôGø ã û ôGÆ��;É�ÉTheinitial stateis:m{ôGÆ � � Í ÷ >Wó Î È / Ñ`Ñ,ÿ Î ÉYá�m{ôGÆ P Ñ�Ñ,ù Î È / Ñ�Ñ,ÿ Î ÉYá�m�ôGÆ P Ñ�Ñ«ù Î È��¦Ñ,ù�ù Í�ã Ñ,ù£Ém{ôGÆ � � Í ÷ >Wó Î È / Ñ`Ñ,ÿ Ï ÉYá�m{ôGÆ P Ñ�Ñ,ù Ï È / Ñ�Ñ,ÿ Ï ÉYá�m�ôGÆ P Ñ�Ñ«ù Ï È��¦Ñ,ù�ù Í�ã Ñ,ù£Ém{ôGÆ � � Í ÷ >Wó Î È / Ñ`Ñ,ÿ Ð ÉYá�m{ôGÆ P Ñ�Ñ,ù Ð È / Ñ�Ñ,ÿ Ð ÉYá�m�ôGÆ P Ñ�Ñ«ù Ð È��¦Ñ,ù�ù Í�ã Ñ,ù£Ém{ôGÆ � � Í ÷ >Wó Î È / Ñ`Ñ,ÿ�¬4ÉYá�m{ôGÆ P Ñ�Ñ,ù*¬#È / Ñ�Ñ,ÿ�¬4ÉYá�m�ôGÆ P Ñ�Ñ«ù�¬4È��¦Ñ,ù�ù Í�ã Ñ,ù£Ém{ôGÆ � ó ö � ø�ïYÈ / Ñ�Ñ,ÿ Ð ÉYá â ÷ Æ � ó ö � øï^È`0g­AÉm{ôGÆ ð Ñ«î Î È / Ñ`Ñ,ÿ Î ÉYá�m{ôGÆ ð Ñ«î Ï È / Ñ`Ñ,ÿ Î ÉYá�m{ôGÆ ð Ñ«î Ð È / Ñ`Ñ,ÿ Î ÉYá�m{ôGÆ ð Ñ«îl¬#È / Ñ`Ñ,ÿ Î É��� Í ÿ�� ö �1�;ø�Æ ð Ñ,î Î É´á���� Í ÿ�� ö �1�;ø�Æ ð Ñ,î Ï É´áK��� Í ÿ�� ö �1�;ø�Æ ð Ñ,î Ð ÉYá���� Í ÿ�� ö �1�;ø�Æ ð Ñ,î ¬ É� ���`ó ö �E�;ø-Æ ð Ñ,î Î ÉYá � ���`ó ö �1�;ø�Æ ð Ñ,î Ï É´á � ��-ó ö �1�;ø�Æ ð Ñ,î Ð É®á � ���`ó ö �1��ø-Æ ð Ñ«î ¬ Éâ ÷ Æ ð Ñ,î Î È`0 Î ÉYá â ÷ Æ ð Ñ,î Ï È`0 Ï ÉYá â ÷ Æ ð Ñ,î Ð È`0 Ð É´á â ÷ Æ ð Ñ«î ¬ È`0 ¬ Éý ��ù�ô�� ã û ôGÆ � � Í ÷ >Wó Î ÉYá ý ��ù`ôGø ãû ôGÆ � � Í ÷ >Wó ¬ Éhttp://librosysolucionarios.net74 Chapter 11. PlanningA planto achieve thegoalis:-¦Ñ`Æ'0g­¯È P Ñ�Ñ,ù Ð É-¦Ñ`Æ P Ñ�Ñ,ù Ð È P Ñ`Ñ,ù Î É-¦Ñ`Æ P Ñ�Ñ,ù Î È`0 Ï É� ���`ó�Æ ð Ñ,î Ï È`0 Ï È P Ñ`Ñ,ù Î É� ���`ó�Æ ð Ñ,î Ï È P Ñ`Ñ,ù Î È P Ñ�Ñ«ù Ï É� ���`ó�Æ ð Ñ,î Ï È P Ñ`Ñ,ù Ï È � � Í ÷ >�ó Ï É11.14 GRAPHPLAN is a propositionalalgorithm,so,justaswe couldsolve certainFOL bytranslatingtheminto propositionallogic, wecansolvecertainsituationcalculusproblemsbytranslatinginto propositionalform. Thetrick is how exactly to do that.TheFinishactionin POPplanninghasasits preconditionsthegoalstate.Wecancreatea Finish actionfor GRAPHPLAN, andgive it theeffect Done. In this casetherewould be afinite numberof instantiationsof theFinishaction,andwe would reasonwith them.11.15 (Figure(11.1)is a little hardto find—it is onpage403.)a. Thepointof thisexerciseis to considerwhathappenswhenaplannercomesupwith animpossibleaction,suchasflying a planefrom someplacewhereit is not. For example,suppose� Îis at ��¡andwe give it theactionof flying from Bangaloreto Brisbane.By (11.1),� Îwasat �¡anddid not fly away, soit is still there.b. Yes,theplanwill still work, becausethefluentshold from thesituationbeforeaninap-plicableactionto thestateafterward,sotheplancancontinuefrom thatstate.c. It dependson thedetailsof how theaxiomsarewritten. Our axiomswereof theformAction is possibleçRule. This tells usnothingaboutthecasewheretheactionisnot possible.We would needto reformulatetheaxioms,or addadditionalonesto saywhathappenswhentheactionis notpossible.11.16 A preconditionaxiom is of theformõ ��ïYÆ � Î È* �¡ È*¢ � ß É M ç â ÷ Æ � Î È* �¡ É M ËThereare û Æ ý {®� � �x{¯� â � Ï Éof theseaxioms,whereý is thenumberof time steps,� � �isthenumberof planesand� â �is thenumberof airports.More generally, if thereareôactionschemataof maximumarity�, with� û � objects,thenthereare û Æ�ôk{ ý {9� û � °_É axioms.With symbol-splitting,we don’t have to describeeachspecificflight, we needonly saythatfor aplaneto fly anywhere, it mustbeat thestartairport.Thatis,õ ��ï Î Æ � Î É M á õ � ï Ï ÆW �¡ É M ç â ÷ Æ � Î È* �¡ É MMoregenerally, if thereareôactionschemataof maximumarity�, with� û � objects,andeachpreconditionaxiomdependson just two of thearguments,thenthereare û Æ�ô9{ ý {®� û � Ï Éaxioms,for aspeedupof û Æ*� û � °�± Ï É .An action exclusionaxiom is of theformJGÆ õ � ï^Æ � Ï È* �¡ È*¢ � ß É M á õ ��ïYÆ � Ï È* �¡ È*�®Ã Þ É M É�ËWith thenotationusedabove, thereare û Æ ý {C� � �v{9� â � Ð Éaxiomsfor õ ��ï . More generally,therecanbeup to û Æ�ôk{ ý {9� û � Ï ° É axioms.http://librosysolucionarios.net75With symbol-splitting,we wouldn’t gain anything for the õ � ï axioms,but we wouldgainin caseswherethereis anothervariablethatis not relevantto theexclusion.11.17a. Yes,this will find a plan whenever the normalSATPLAN finds a plan no longerthaný�² Ý«Ü .b. No.c. Thereis no simpleandclearway to induceWALKSAT to find shortsolutions,becauseit hasnonotionof thelengthof aplan—thefactthattheproblemis aplanningproblemis partof theencoding,not partof WALKSAT. But if we arewilling to do someratherbrutalsurgeryon WALKSAT, we canachieve shortersolutionsby identifying thevari-ablesthat representactionsand(1) tendingto randomlyinitialize the actionvariables(particularlythelaterones)to false,and(1) preferingto randomlyflip anearlieractionvariableratherthana laterone.http://librosysolucionarios.netSolutionsfor Chapter12PlanningandActing in theRealWorld12.1a.Å+Ø¾Ö t Ù�Õ'uv&¯Æ ã É is eligible to beaneffectbecausetheactiondoeshavetheeffectof movingthe clock byã. It is possiblethat the durationdependson the actionoutcome,so ifdisjunctive or conditionaleffectsareusedtheremustbe a way to associatedurationswith outcomes,which is most easily doneby putting the durationinto the outcomeexpression.b. TheSTRIPS modelassumesthatactionsaretimepointscharacterizedonly by theirpre-conditionsandeffects. Evenif anactionoccupiesa resource,thathasno effect on theoutcomestate(asexplainedon page420). Therefore,we mustextendtheSTRIPS for-malism.Wecoulddothisby treatingaRESOURCE: effectdifferentlyfrom othereffects,but thedifferenceis sufficiently largethatit makesmoresenseto treatit separately.12.2 The basicideahereis to recordthe initial resourcelevel in the preconditionandthechangein resourcelevel in theeffect of eachaction.a. Let� >�ù£ø����-Æ��`Édenotethe fact that thereare�screws. We needto add� >�ù~ø��h�`Æ�Ê#Ç#Ç#Éto the initial state,andadda fourth argumentto the q ô ê Í ôGø predicateindicatingthenumberof screwsrequired—i.e.,q ô ê Í ôGø�Æ q Î È�� Î È��#Ç#È`3�Ç#É and q ô ê Í ôGø-Æ q Ï È�� Ï È�5#Ç#È�4#Ç#É .Weadd� >�ù£ø��h�`Æ�� M É to thepreconditionofâ+ã�ã q ô ê Í ôGø andadd�asafourthargumentof the q ô ê Í ôGø literal. Thenadd� >�ù£ø����-Æ�� M ^9�`É to theeffectofâµã_ã q ô ê Í ôGø .b. A simplesolutionis to saythat any actionthat consumesa resourceis potentially inconflictwith any causallink protectingthesameresource.c. The plannercankeeptrack of the resourcerequirementsof actionsaddedto the planandbacktrackwhenever thetotalusageexceedstheinitial amount.12.3 Thereis awide rangeof possibleanswersto thisquestion.Theimportantpoint is thatstudentsunderstandwhatconstitutesacorrectimplementationof anaction:asmentionedonpage424, it mustbea consistentplanwhereall thepreconditionsandeffectsareaccountedfor. Sothefirst thing we needis to decideon thepreconditionsandeffectsof thehigh-levelactions. For GetPermit, assumethe preconditionis owning land,andthe effect is having apermit for thatpieceof land. For HireBuilder, thepreconditionis having theability to pay,andtheeffect is having asignedcontractin hand.76http://librosysolucionarios.net77Onepossibledecompositionfor GetPermit is the three-stepsequenceGetPermitForm,FillOutForm, andGetFormApproved. Thereis a causallink with the conditionHaveFormbetweenthefirst two, andonewith theconditionHaveCompletedForm betweenthelasttwo.Finally, theGetFormApprovedstephastheeffectHavePermit. This is avalid decomposition.ForHireBuilder, supposewechoosethethree-stepsequenceInterviewBuilders, Choose-Builder, andSignContract. ThislaststephasthepreconditionAbleToPayandtheeffectHave-ContractInHand. Therearealsocausallinks betweenthesubsteps,but they don’t affect thecorrectnessof thedecomposition.12.4 Considertheproblemof building two adjacentwalls of thehouse.Mostly thesesub-plansareindependent,but they mustsharethestepof puttingupacommonpostat thecornerof the two walls. If that stepwasnot shared,we would endup with an extra post,andtwounattachedwalls.Notethattasksareoftendecomposedspecificallysoasto minimizetheamountof stepsharing. For example,onecould decomposethe housebuilding task into subtaskssuchas“walls” and“floors.” However, realcontractorsdon’t do it thatway. Insteadthey have“roughwalls” and“roughfloors” steps,followedby a “finishing” step.12.5 In the HTN view, thespaceof possibledecompositionsmay constrainthe allowablesolutions,eliminatingsomepossiblesequencesof primitive actions. For example,the de-compositionof the LAToNYRoundTrip actioncanstipulatethat theagentshouldgo to NewYork. In asimpleSTRIPSformulationwherethestartandgoalstatesarethesame,theemptyplan is a solution. We cangetaroundthis problemby rethinkingthegoaldescription.Thegoalstateis notâ ÷ Æ�d â É , butâ ÷ Æ�d â É0á¦p Í � Í ÷ ø ã Æ�@S³nÉ . We addp Í � Í ÷ ø ã Æ�ï^É asaneffect ofõ � ïYÆ�îAÈ�ïYÉ . Then,thesolutionmustbeatrip thatincludesNew York. Thereremainstheprob-lem of preventingtheSTRIPSplan from includingotherstopson its itinerary; fixing this ismuchmoredifficult becausenegatedgoalsarenotallowed.12.6 Supposewe have a STRIPS actiondescriptionfor ö with precondition:andeffectX.The“action” tobedecomposedisâ >�ó Í ø è ø�Æ�X�É. Thedecompositionhastwosteps:â >Wó Í ø è ø�Æ : Éand ö . This can be extendedin the obvious way for conjunctive effectsand precondi-tions.12.7 Weneedoneaction,â �F� Í ê ô , whichassignsthevaluein thesourceregister(or variableif you prefer, but the term “register” makes it clearerthat we aredealingwith a physicallocation)�«ùto thedestinationregisterã ù:â > ÷ Í Ñ«ôGÆ ACTION:â �F� Í ê ôGÆ ã ù#È��«ù~É�ÈPRECOND: / ø ê Í � ÷ øù£Æ ã ù£É´á / ø ê Í � ÷ øù~Æ��«ù£ÉYá�p ö �+�Aø-Æ ã ù#È ã�è ɯá�p ö �)�0ø�Æ��«ù�È�� è É�ÈEFFECT:p ö �+�Aø�Æ ã ù�È�� è ɯá�J�p ö �+�0ø�Æ ã ù#È ã�è É�ÉNow supposewestartin aninitial statewith / ø ê Í � ÷ øù£Æ / Î É�á / ø ê Í � ÷ ø�ù£Æ / Ï É�á�p ö �+�Aø-Æ / Î È�p Î É�áp ö �+�Aø-Æ / Ï È�p Ï É andwe have thegoalp ö �+�Aø�Æ / Î È�p Ï É´á�p ö �+�0ø�Æ / Ï È�p Î É . Unfortunately, thereis noway to solve thisasis. Weeitherneedto addanexplicit / ø ê Í � ÷ ø�ù£Æ / Ð É conditionto theinitial state,or we needa way to createnew registers.Thatcouldbedonewith anactionforhttp://librosysolucionarios.net78 Chapter 12. PlanningandActing in theRealWorldallocatinganew register:â > ÷ Í Ñ«ôGÆ ACTION:â �2�;Ñ�> ö{÷ ø�Æ�ù£É�ÈEFFECT: / ø ê Í � ÷ ø�ù£Æ�ù£É�ÉThenthefollowing sequenceof stepsconstituesavalid plan:â �2�;ÑF> ö�÷ ø-Æ / Ð Éâ �F� Í ê ôGÆ / Ð È / Î Éâ �F� Í ê ôGÆ / Î È / Ï Éâ �F� Í ê ôGÆ / Ï È / Î É12.8 For thefirst case,whereoneinstanceof actionschemaö is in theplan,thereformula-tion is correct,in thesensethatasolutionfor theoriginaldisjunctive formulationis asolutionfor thenew formulationandviceversa. For thesecondcase,wheremorethanoneinstanceof theactionschemamayoccur, thereformulationis incorrect.It assumesthattheoutcomesof the instancesaregovernedby a singlehiddenvariable,so that if, for example,�is theoutcomeof oneinstanceit mustalsobetheoutcomeof theother. It is possiblethatasolutionfor thereformulatedcasewill fail in theoriginal formulation.12.9 With unboundedindeterminacy, thesetof possibleeffectsfor eachactionis unknownor too large to be enumerated.Hence,the spaceof possibleactionssequencesrequiredtohandleall theseeventualitiesis far too largeto consider.12.10 Using the seconddefinition of���;ø ö ù in the chapter—namely, that thereis a clearspacefor ablock—theonly changeis thatthedestinationremainsclearif it is thetable:â > ÷ Í Ñ«ôGÆ þ Ñ è ø-Æ���È�î0È�ï^É�ÈPRECOND: û ôGÆ��WÈ�îAÉ®á����;ø ö ù£Æ��WÉYá����;ø ö ù£Æ�ï^É�ÈEFFECT: û ôGÆ��WÈ�ïYÉ´áK���;ø ö ù£Æ�îAÉ®áKJ û ôGÆ��WÈ�îAÉ®áFÆ whenï[8ä ý°ö �1�;ø*´OJ�����ø ö ù£Æ�ïYÉ�É�É12.11 Let���;ø ö ô ü betrueiff therobot’s currentsquareis cleanand���;ø ö ô û betrueiff theothersquareis clean.Then� �> �is characterizedbyâ > ÷ Í Ñ«ôGÆ � ��> � È PRECOND:ÈEFFECT:���;ø ö ô ü ÉUnfortunately, moving affectsthesenew literals! Ford�ø �P÷ we haveâ > ÷ Í Ñ«ôGÆ�d�ø �P÷ È PRECOND:â ÷1/ ÈEFFECT:â ÷ dpá�J â ÷1/ á when���;ø ö ô ü ´Q���;ø ö ô û á when����ø ö ô û ´Q���;ø ö ô üáwhenJH���;ø ö ô û ´QJ����;ø ö ô ü áwhenJH���;ø ö ô ü ´QJ����;ø ö ô û Éwith thedualfor / Í ê ó ÷ .12.12 Herewe borrow from thelastdescriptionof thedGø �P÷ onpage433:â > ÷ Í Ñ«ôGÆ � ��> � È PRECOND:ÈEFFECT:Æwhenâ ÷ df´Q���;ø ö ôHd � Æ when���;ø ö ôHdf´QJ����;ø ö ôHdGÉ�ÉáFÆwhenâ ÷�/ ´Q����ø ö ô / � Æ when���;ø ö ô / ´oJ�����ø ö ô / É�É�É12.13 Themainthing to noticehereis thatthevacuumcleanermovesrepeatedlyoverdirtyareas—presumably, until they areclean.Also, eachforwardmoveis typically short,followedhttp://librosysolucionarios.net79by an immediatereversingover the samearea. This is explainedin termsof a disjunctiveoutcome:theareamaybefully cleanedor not, thereversingenablestheagentto check,andtherepetitionensurescompletion(unlessthedirt is ingrained).Thus,wehaveastrongcyclicplanwith sensingactions.12.14a. “Lather. Rinse.Repeat.”This is an unconditionalplan, if taken literally, involving an infinite loop. If the pre-conditionofd ö{÷ ó�øù isJH���;ø ö ô , andthegoalis����ø ö ô , thenexecutionmonitoringwillcauseexecutionto terminateonce���;ø ö ô is achieved becauseat thatpoint thecorrectrepairis theemptyplan.b. “Apply shampooto scalpand let it remainfor several minutes. Rinseand repeatifnecessary.”This is aconditionalplanwhere“if necessary”presumablytestsJ����;ø ö ô .c. “Seeadoctorif problemspersist.”Thisis alsoaconditionalstep,althoughit is notspecifiedherewhatproblemsaretested.12.17 First,weneedto decideif thepreconditionis satisfied.Therearethreecases:a. If it is known to be unsatisfied,the new belief stateis identical to the old (sinceweassumenothinghappens).b. If it is known to besatisfied,theunconditionaleffects(which areall knowledgepropo-sitions)areaddedanddeletedfrom thebelief statein theusualSTRIPS fashion.Eachconditionaleffect whoseconditionis known to betrueis handledin thesameway. Foreachsettingof the unknown conditions,we createa belief statewith the appropriateadditionsanddeletions.c. If the statusof the preconditionis unknown, eachnew belief stateis effectively thedisjunctionof theunchangedbelief statefrom (a) with oneof thebelief statesobtainedfrom (b). To enforcethe“list of knowledgepropositions”representation,wekeepthosepropositionsthatareidenticalin eachof thetwobeliefstatesbeingdisjoinedanddiscardthosethat differ. This resultsin a weaker belief statethan if we were to retain thedisjunction;on theotherhand,retainingthedisjunctionsover many stepscouldleadtoexponentiallylargerepresentations.12.18 For / Í ê ó ÷ wehave theobviousdualversionof Equation12.2:â > ÷ Í Ñ«ôGÆ / Í ê ó ÷ È PRECOND:â ÷ dGÈEFFECT: I Æ â ÷�/ ÉYá�J I Æ â ÷ d�ÉYá when���;ø ö ôHdH´OJ I Æ����;ø ö ôHd�É�áwhen���;ø ö ô / ´ I Æ����;ø ö ô / ÉYá whenJH���;ø ö ô / ´ I Æ�J����;ø ö ô / É�ÉWith� �> �, dirt is sometimesdepositedwhenthesquareis clean.With automaticdirt sensing,http://librosysolucionarios.net80 Chapter 12. PlanningandActing in theRealWorldthis is alwaysdetected,sowehave adisjunctive conditionaleffect:â > ÷ Í Ñ«ôGÆ � ��> � È PRECOND:ÈEFFECT:whenâ ÷ dIá¤J����;ø ö ôHdH´ I Æ����;ø ö ôHdGÉáwhenâ ÷ dIá¤���;ø ö ôHdH´ I Æ����;ø ö ôHdGÉ � J I Æ�����ø ö ôHd�ÉYáwhenâ ÷�/ á�J�����ø ö ô / ´ I Æ����;ø ö ô / Éáwhenâ ÷�/ á�����ø ö ô / ´ I Æ����;ø ö ô / É � J I Æ����;ø ö ô / É12.19 Thecontinuousplanningagentdescribedin Section12.6hasat leastoneof thelistedabilities,namelytheability to acceptnew goalsasit goesalong.A new goalis simplyaddedasanextra openpreconditionin the õ Í ô Í �`ó step,andtheplannerwill find a way to satisfyit, if possible,along with the other remaininggoals. Becausethe datastructuresbuilt bythecontinuousplanningagentasit workson theplan remainlargely in placeastheplan isexecuted,thecostof replanningis usuallyrelatively smallunlessthefailure is catastrophic.Thereis no specifictime boundthat is guaranteed,andin generalno suchboundis possiblebecausechangingeven a singlestatevariablemight requirecompletelyreconstructingtheplanfrom scratch.12.20 Let ý bethepropositionthat thepatientis dehydratedand�bethesideeffect. Wehave â > ÷ Í Ñ«ôGÆ P ù Í ô � È PRECOND:ÈEFFECT:J ý Éâ > ÷ Í Ñ«ôGÆ þ ø ã{Í > ö{÷ ø�È PRECOND:ÈEFFECT:J P áwhen ý ´ � Éandtheinitial stateisJ � á�Æ ý � P É�á�Æ�J ý � J P É . Thesolutionplaniså P ù Í ô � È þ ø ã{Í > ö{÷ ø æ .Therearetwo possibleworlds,onewhereý holdsandonewherePholds.In thefirst,P ù Í ô �causesJ ý andþ ø ã�Í > ö�÷ ø hasno effect; in thesecond,P ù Í ô �hasno effect andþ ø ã�Í > ö�÷ øcausesJ P. In bothcases,thefinal stateisJ � á¤J ý á�J P .12.21 Onesolutionplaniså ý ø*� ÷ È if ÔAØO� Ù�ؾֵ�x�lÖµuo¶5Ù%y thenå P ù Í ô � È þ ø ã�Í > ö�÷ ø æ'æ .http://librosysolucionarios.netSolutionsfor Chapter13Uncertainty13.1 The“first principles”neededherearethedefinitionof conditionalprobability,� Æ'0·� ³nÉJä� Æ'0 ᤳMÉ ì � Æ�³nÉ , andthedefinitionsof thelogical connectives. It is not enoughto saythatif ð á â is “given” thenâmustbetrue! Fromthedefinitionof conditionalprobability, andthefactthatâ á â Ú âandthatconjunctionis commutative andassociative, we have� Æ â � ð á â É5ä� Æ â áFÆ ð á â É�É� Æ ð á â Éä � Æ ð á â É� Æ ð á â Éä�Ê13.2 Themainaxiomisaxiom3:� Æ ö � �Wɲä � Æ ö É Y � Æ��WÉ^ � Æ ö á}�WÉ . For thediscreterandomvariable0, let ö betheevent that0 äjî Î, and�betheevent that0hasany othervalue.Thenwe have� Æ'0 ä¹î Î � 0 ä�Ñ ÷ ó�øù£ÉJä � Æ'0 ä¹î Î É Y � Æ'0 ä�Ñ ÷ ó�øù£É Y Çwherewe know that� Æ'0 äÀî Î á\0>äbÑ ÷ ó�øù£É is 0 becausea variablecannottake on twodistinctvalues.If wenow breakdown thecaseof0 ä�Ñ ÷ ó�øù�� , weeventuallyget� Æ'0 ä¹î Î��¸�¸�¸v� 0 ä¹îl¹�ÉJä � Æ'0 ä¹î Î É Y ¸�¸�¸ Y � Æ'0 ä¹î�¹�ÉAËBut the left-handside is equivalent to� Æ ÷ ù��Aø�É , which is 1 by axiom 2, so the sumof theright-handsidemustalsobe1.13.3 Probablytheeasiestway to keeptrack of what’s goingon is to look at theprobabil-ities of the atomic events. A probability assignmentto a set of propositionsis consistentwith the axiomsof probability if the probabilitiesareconsistentwith an assignmentto theatomiceventsthatsumsto 1 andhasall probabilitiesbetween0 and1 inclusive. We call theprobabilitiesof theatomicevents ö , � , > , andã, asfollows:ð J ðâa bJ âc dWethenhave thefollowing equations:� Æ â ÉJä ö�Y �lä�Ç#Ë 3� Æ ð ÉGä öºY >lä�Ç#Ë��� Æ âk� ð É5ä öºY � Y >2ä�Ç#Ë�4� Æ ý ù��Aø�ÉGä öºY � Y > Y ã ä�Ê81http://librosysolucionarios.net82 Chapter 13. UncertaintyFrom these,it is straightforward to infer that ö ä Ç#Ë��,��ä Ç#Ë��,>�ä Ç#Ë'Ê, andã ä Ç#Ë�4.Therefore,� Æ â á ð Élä ö äÀÇ#Ë�� . Thustheprobabilitiesgivenareconsistentwith a rationalassignment,andtheprobability� Æ â á ð É is exactlydetermined.(This latterfactcanbeseenalsofrom axiom3 onpage422.)If� Æ â9� ð É+äªÇ#Ë�6, then� Æ â á ð ɵä ö äbÇ. Thus,even thoughthebetoutlinedinFigure13.3losesifâand ð arebothtrue,theagentbelievesthis to beimpossiblesothebetis still rational.13.4 ? (?,?) arguesroughlyasfollows: Supposewe presentanagentwith a choice:eitherdefinitelyreceive monetarypayoff: {cÿ, or choosea lottery thatpaysÿif event q occursand0 if q doesnot occur. Thenumber:for which theagentis indifferentbetweenthetwochoices(assuminglinearutility of money), is definedto betheagent’s degreeof belief in q .A setof degreesof beliefspecifiedfor somecollectionof eventswill eitherbecoherent,which is definedto meanthat thereis no setof betsbasedon thesestatedbeliefsthat willguaranteethat theagentwill losemoney, or incoherent, which is definedto meanthat thereis suchasetof bets.DeFinetti showedthatcoherentbeliefssatisfytheaxiomsof probabilitytheory.Axiom 1:Ǥ» : »¬Ê, for any q andÿ. If an agentspecifies: i Ê, thenthe agentis offering to paymorethanÿto entera lottery in which thebiggestprize isÿ. If anagentspecifies: ]³Ç, thenthe agentis offering to pay eitherÿor 0 in exchangefor a negativeamount.Eitherway, theagentis guaranteedto losemoney if theopponentacceptstherightoffer.Axiom 2:: ä�Êwhen q isÙ�Ö,Øx�and: ä Çwhen q isz t � " � . Supposetheagentassigns: Aasthedegreeof belief in aknown trueevent.Thentheagentis indifferentbetweenapayoffof: A ÿandoneofÿ. This is only coherentwhen: A ähÊ. Similarly, a degreeof belief of: Afor aknown falseeventmeanstheagentis indifferentbetween: A ÿand0. Only: A^ä�Çmakesthiscoherent.Axiom 3: Giventwo mutuallyexclusive, exhaustive events, q Î and q Ï , andrespectivedegreesof belief:�Îand:_Ïandpayoffsÿ Îandÿ Ï, it mustbe that thedegreeof belief forthecombinedevent q Î� q Ï equals:�Î Y :_Ï . Theideais thatthebeliefs:�Îand:�Ïconstituteanagreementto pay:�Î ÿ Î Y :_Ï ÿ Ï in orderto entera lottery in which theprize isÿ¤¼whenq ¼ occurs.So thenetgain is definedas ê ¼ ä�ÿ ¼ ^ : Î ÿ Î Y : Ï ÿ Ï . To avoid thepossibilityof theÿ¤¼amountsbeingchosento guaranteethatevery ê ¼ is negative,wehave to assurethatthedeterminantof thematrix relatingÿ ¼to ê ¼ is zero,sothatthelinearsystemcanbesolved.This requiresthat:�Î Y :_Ï ä³Ê. The resultextendsto the casewithômutually exclusive,exhaustive eventsratherthantwo.13.5 This is a classiccombinatoricsquestionthat could appearin a basictext on discretemathematics.The point hereis to refer to the relevant axiomsof probability: principally,axiom 3 on page422. The questionalso helpsstudentsto graspthe conceptof the jointprobabilitydistribution asthedistribution over all possiblestatesof theworld.a. Thereare ½Ï½ =Æ�4=��{}4#Ê�{�4#Çh{B3�;h{B3�7#É ì Æ�ÊD{}��{���{\3�{�4#É = 2,598,960possiblefive-cardhands.http://librosysolucionarios.net83b. By thefair-dealingassumption,eachof theseis equallylikely. By axioms2 and3, eachhandthereforeoccurswith probability1/2,598,960.c. Therearefour handsthatareroyal straightflushes(onein eachsuit). By axiom3, sincetheeventsaremutuallyexclusive, theprobabilityof aroyal straightflushis just thesumof theprobabilitiesof theatomicevents,i.e.,4/2,598,960= 1/649,740.d. Again, we examinethe atomiceventsthat are“four of a kind” events. Thereare13possible“kinds” andfor each,thefifth cardcanbeoneof 48 possibleothercards.Thetotalprobabilityis thereforeÆ�Ê=��{B3�7#É ì �#È�4=;=7#È�;=5#Ç�ä�Ê ì 3�È�Ê=5=4 .Thesequestionscaneasilybeaugmentedby morecomplicatedones,e.g.,what is theproba-bility of gettinga full housegiventhatyoualreadyhave two pairs?Whatis theprobabilityofgettingaflushgiventhatyouhave threecardsof thesamesuit?Or youcouldassignaprojectof producingapoker-playingagent,andhave a tournamentamongthem.13.6 The main point of this exerciseis to understandthevariousnotationsof bold versusnon-boldP, anduppercaseversuslowercasevariablenames. The rest is easy, involving asmallmatterof addition.a. Thisasksfor theprobabilitythatToothache is true.� Æ«Ù%u�u�Ù%y t ×*yo�~É5ä�Ç#Ë'Ê#Ç=7 Y Ç#Ë'Ç#Ê=� Y Ç#Ë'Ç#Ê=5 Y Ç#Ë'Ç=5E3cä�Ç#Ë��b. Thisasksfor thevectorof probabilityvaluesfor therandomvariableCavity. It hastwovalues,which we list in theorder ¾ Ù�Ö�Øx��È z t � " �O¿ . First addupÇ#Ë'Ê#Ç=7 Y Ç#Ë'Ç#Ê=� Y Ç#Ë'Ç=6=� YÇ#Ë'Ç#Ç=7¦ä�Ç#Ë��. Thenwe havePÆ`Ô t �-Õ%Ù%��ÉJä ¾ Ç#Ë��#È�Ç#Ë�7=¿JËc. Thisasksfor thevectorof probabilityvaluesfor Toothache, giventhatCavity is true.PÆFÀ�u�u4Ù%y t ×�yv����× t ��Õ�Ù%��ÉJä ¾ Æ�Ë'Ê#Ç=7 Y Ë'Ç#Ê=�#É ì Ç#Ë��#È�Æ�Ç#Ë'Ç=6=� Y Ç#Ë'Ç#Ç=7#É ì Ç#Ë��=¿¸ä ¾ Ç#Ë�5#È�Ç#Ë 3�¿d. This asksfor thevectorof probabilityvaluesfor Cavity, giventhateitherToothacheorCatch is true.Firstcompute� Æ«Ù%u�u4Ù%y t ×�yv� � × t Ù�×*y�ÉJä�Ç#Ë'Ê#Ç=7 Y Ç#Ë'Ç#Ê=� Y Ç#Ë'Ç#Ê=5 Y Ç#Ë'Ç=5E3 YÇ#Ë'Ç=6=� Y Ç#Ë'ÊE3=3cä�Ç#Ë 3�Ê=5 . ThenPÆ`Ô t �-Õ%Ù%����Ù%u�u4Ù%y t ×*yv� � × t Ù�×�y�ÉJä¾ Æ�Ç#Ë'Ê#Ç=7 Y Ç#Ë'Ç#Ê=� Y Ç#Ë'Ç=6=�#É ì Ç#Ë 3�Ê=5#È�Æ�Ç#Ë'Ç4Ê=5 Y Ç#Ë'Ç=5E3 Y Ç#Ë'ÊE3=3�É ì Ç#Ë 3�Ê=5=¿2ä¾ Ç#Ë 3�5#Ê=4#È�Ç#Ë�4=�=7E3�¿13.7 Independenceis symmetric(that is, ö and�areindependentiff�and ö areindepen-dent)so� Æ ö � ��ÉGä � Æ ö É is thesameas� Æ��v� ö ÉJä � Æ��WÉ. Soweneedonly prove that� Æ ö � �WÉJä� Æ ö É is equivalentto� Æ ö á��WÉJä � Æ ö É � Æ��WÉ . Theproductrule,� Æ ö á��WÉJä � Æ ö � �WÉ � Æ��WÉ , canbe usedto rewrite� Æ ö á��WÉ"ä � Æ ö É � Æ��WÉ as� Æ ö � �WÉ � Æ��Wɦä � Æ ö É � Æ��WÉ , which simplifiesto� Æ ö � �WÉJä � Æ ö É13.8 Wearegiventhefollowing information:� Æ ÷ ø�� ÷ � ã�Í �-ø ö �`ø�É}ä Ç#Ë�;=;� Æ�J ÷ ø�� ÷ � J ã{Í �-ø ö �`ø�ÉLä Ç#Ë�;=;� Æ ã�Í �-ø ö �`ø�É}ä Ç#Ë'Ç#Ç#Ç#Êhttp://librosysolucionarios.net84 Chapter 13. Uncertaintyandtheobservation ÷ ø*� ÷ . What thepatientis concernedaboutis� Æ ã{Í �-ø ö �`øO� ÷ ø*� ÷ É . Roughlyspeaking,thereasonit is agoodthingthatthediseaseis rareis that� Æ ã{Í �`ø ö �-øx� ÷ ø�� ÷ É ispropor-tionalto� Æ ã�Í �`ø ö �-ø-É , soalowerprior forã{Í �`ø ö �-ø will meanalowervaluefor� Æ ã�Í �`ø ö �`øO� ÷ ø*� ÷ É .Roughlyspeaking,if 10,000peopletakethetest,weexpect1 to actuallyhavethedisease,andmostlikely testpositive, while the restdo not have thedisease,but 1% of them(about100people)will testpositiveanyway, so� Æ ã{Í �-ø ö �`øO� ÷ ø*� ÷ É will beabout1 in 100.Moreprecisely,usingthenormalizationequationfrom page428:� Æ ã�Í �`ø ö �-øx� ÷ ø�� ÷ Éä Á�ÂÄÃ�Å`Æ`Ã,Ç È ¼ ÆµÅ Ý Æ`ÅEÉ)Á�ÂÄÈ ¼ Æ`Å Ý ÆµÅ1ÉÁ�ÂÄÃ�Å`Æ`Ã,Ç È ¼ ÆµÅ Ý Æ`ÅEÉ)Á�ÂÄÈ ¼ Æ`Å Ý ÆµÅ1ÉËÊxÁ�Â)Ã�ŵÆ`Ã,ÇÍÌ�ȼ Æ`Å Ý Æ`ÅEÉ)Á�ÂËÌ�È ¼ Æ`Å Ý Æ`ÅEÉä M�Î Ï�Ï*ÐOM�Î M�M�M ÎM�Î Ï�Ï*ÐOM�Î M�M�M Î Ê M�Î M Î ÐOM�Î Ï�Ï�Ï�Ïä�Ë'Ç#Ç=;=7#ÇE3Themoral is thatwhenthediseaseis muchrarerthanthetestaccuracy, a positive testresultdoesnotmeanthediseaseis likely. A falsepositive readingremainsmuchmorelikely.Hereis analternative exercisealongthesamelines: A doctorsaysthatan infantwhopredominantlyturnstheheadto the right while lying on thebackwill be right-handed,andonewho turnsto the left will be left-handed.Isabellapredominantlyturnedherheadto theleft. Giventhat90%of thepopulationis right-handed,whatis Isabella’s probabilityof beingright-handedif thetestis 90%accurate?If it is 80%accurate?Thereasoningis thesame,andtheansweris 50%right-handedif thetestis 90%accu-rate,69%right-handedif thetestis 80%accurate.13.9 Thebasicaxiomto usehereis thedefinitionof conditionalprobability:a. WehavePÆ â È ð � q ÉGä PÆ â È ð È q ÉPÆ q ÉandPÆ â � ð È q É PÆ ð � q ÉGä PÆ â È ð È q ÉPÆ ð È q ÉPÆ ð È q ÉPÆ q Éä PÆ â È ð È q ÉPÆ q ÉhencePÆ â È ð � q ÉGä PÆ â � ð È q É PÆ ð � q Éb. Thederivationhereis thesameasthederivationof thesimpleversionof Bayes’Ruleon page426. First we write down the dual form of the conditionalizedproductrule,simplyby switchingâand ð in theabove derivation:PÆ â È ð � q ÉGä PÆ ð � â È q É PÆ â � q ÉThereforethetwo right-handsidesareequal:PÆ ð � â È q É PÆ â � q É�ä PÆ â � ð È q É PÆ ð � q ÉDividing throughby PÆ ð � q É wegetPÆ â � ð È q ÉGä PÆ ð � â È q É PÆ â � q ÉPÆ ð � q Éhttp://librosysolucionarios.net8513.10 Thekey to this exerciseis rigorousandfrequentapplicationof thedefinitionof con-ditional probability, PÆ'0·� ³ÁÉ�äPÆ'0KÈ�³nÉ ì PÆ�³nÉ . The original statementthat we aregivenis:PÆ â È ð � �¦ÉGä PÆ â � �°ÉPÆ ð � �¦ÉWe start by applying the definition of conditionalprobability to two of the terms in thisstatement:PÆ â È ð � �¦ÉGä PÆ â È ð È��°ÉPÆ��¦É and PÆ ð � �¦ÉGä PÆ ð È��¦ÉPÆ��°ÉNow wesubstitutetheright handsideof thesedefinitionsfor theleft handsidesin theoriginalstatementto get:PÆ â È ð È��¦ÉPÆ��¦É äPÆ â � �°É PÆ ð È��¦ÉPÆ��°ÉNow we needthedefinitiononcemore:PÆ â È ð È��¦ÉGä PÆ â � ð È��°É PÆ ð È��¦ÉWesubstitutethis right handsidefor PÆ â È ð È��°É to get:PÆ â � ð È��¦É PÆ ð È��°ÉPÆ��°É äPÆ â � �¦É PÆ ð È��°ÉPÆ��¦ÉFinally, we cancelthePÆ ð È��¦É andPÆ��¦És to get:PÆ â � ð È��¦ÉGä PÆ â � �°ÉThe secondpart of the exercisefollows from by a similar derivation, or by noticing thatâand ð areinterchangeablein theoriginal statement(becausemultiplication is commutativeandâ È ð meansthesameas ð È â ).In Chapter14, we will seethat in termsof Bayesiannetworks, theoriginal statementmeansthat�is the lone parentofâandalsothe lone parentof ð . The conclusionis thatknowing thevaluesof ð and�is thesameasknowing just thevalueof�in termsof tellingyousomethingaboutthevalueofâ.13.11a. Thereareôwaysto pick a coin, and2 outcomesfor eachflip (althoughwith thefakecoin, the resultsof the flip areindistinguishable), so thereare��ôtotal atomicevents.Of those,only 2 pick thefake coin,and� Y Æ�ôS^OÊ#É resultin heads.Sotheprobabilityof a fake coin givenheads,� Æ z tvÑ ���Íyv� t Ò " É , is� ì Æ�� Y ô¤^OÊ#ÉJäZ� ì Æ�ô Y Ê#É .b. Now thereare��°�ôatomicevents,of which��°pick thefakecoin,and��° Y Æ�ô�^cÊ#É resultin heads.Sotheprobabilityof a fake coin givena run of�heads,� Æ z tvÑ ���Íyv� t Ò " ° É , is��° ì Æ���° Y Æ�ô¤^OÊ#É�É . Notethisapproaches1 as�increases,asexpected.If� ä¹ô�ä�Ê=�,for example,than� Æ z toÑ ���Íyv� t Ò " Î M ÉJä�Ç#Ë�;=;=6=� .c. Thereare two kinds of error to consider. Case1: A fair coin might turn up heads�timesin a row. Theprobabilityof this isÊ ì � ° , andtheprobabilityof a fair coin beingchosenisÆ�ôk^�Ê#É ì ô . Case2: The fake coin is chosen,in which casethe procedurehttp://librosysolucionarios.net86 Chapter 13. Uncertaintyalwaysmakesan error. The probability of drawing the fake coin isÊ ì ô . So the totalprobabilityof erroriså Æ�ô¤^OÊ#É ì � ° Y Ê æ ì ô13.12 The importantpoint hereis that althoughthereareoften many possibleroutesbywhich answerscanbecalculatedin suchproblems,it is usuallybetterto stick to systematic“standard”routessuchas Bayes’ Rule plus normalization. Chapter14 describesgeneral-purpose,systematicalgorithmsthatmake heavy useof normalization.We could guessthat� Æ � � J þ É�Ò�Ç#Ë'Ç=4, or we couldcalculateit from theinformationalreadygiven(althoughtheideahereis to assumethat� Æ � Éis notknown):� Æ � � J þ ÉJä � Æ�J þ � � É � Æ � É� Æ�J þ É ä Æ�Ê�^ � Æ þ � � É�É � Æ � ÉÊ�^ � Æ�J þ É ä Ç#Ë�;=;=;=7h{�Ç#Ë'Ç=4Ç#Ë�;=;=;=;=7 ä�Ç#Ë'ÇE3�;=;=;#ÊNormalizationproceedsasfollows:� Æ þ � � ÉfÓ � Æ � � þ É � Æ þ ÉGä�Ç#Ë�4 ì 4#Ç#È�Ç#Ç#Çnä�Ç#Ë'Ç#Ç#Ç#Ç#Ê� Æ�J þ � � É�Ó � Æ � � J þ É � Æ�J þ ÉJä�Ç#Ë'ÇE3�;=;=;#ÊD{LÇ#Ë�;=;=;=;=7°ä�Ç#Ë'ÇE3�;=;=;� Æ þ � � É�ä M�Î M�M�M�M ÎM�Î M�M�M�M Î Ê M�Î M ¬ Ï�Ï�Ï ä�Ç#Ë'Ç#Ç#Ç=�� Æ�J þ � � ÉJä M�Î M�M�M�M ÎM�Î M�M�M�M Î Ê M�Î M ¬ Ï�Ï�Ï ä�Ç#Ë�;=;=;=713.13 The questionwould have beenslightly moreconsistentif we hadasked aboutthecalculationof PÆ ü � q Î È q Ï É insteadof� Æ ü � q Î È q Ï É . Showing thatagivensetof informationis sufficient is relatively easy: find an expressionfor PÆ ü � q Î È q Ï É in termsof the giveninformation. Showing insufficiencycanbe doneby showing that the informationprovideddoesnot containenoughindependentnumbers.a. Bayes’RulegivesPÆ ü � q Î È q Ï ÉGä PÆ q Î È q Ï � ü É PÆ ü ÉPÆ q Î È q Ï ÉHencethe informationin (ii) is sufficient—in fact, we don’t needPÆ q Î È q Ï É becausewe canusenormalization. Intuitively, the information in (iii) is insufficient becausePÆ q Î � ü É andPÆ q Ï � ü É provide no informationaboutcorrelationsbetweenq Î and q Ïthat might be inducedby ü . Mathematically, supposeü hasÿpossiblevaluesandq Î and q Ï haveô Îandô Ïpossiblerespectively. PÆ ü � q Î È q Ï É containsÆ�ÿÔ^�Ê#É�ô Î ô Ïindependentnumbers,whereasthe information in (iii) containsÆ�ÿÕ^�Ê#É Y ÿ�Æ�ô Î ^Ê#É Y ÿ�Æ�ô Ï ^ Ê#É numbers—clearlyinsufficient for largeÿ,ô Î, andô Ï. Similarly, theinformationin (i) containsÆ�ô Î ô Ï ^}Ê#É Y ÿ Y ÿ�Æ�ô Î ^}Ê#É Y ÿ�Æ�ô Ï ^}Ê#É numbers—againinsufficient.b. If q Î and q Ï areconditionallyindependentgiven ü , thenPÆ q Î È q Ï � ü ÉGä PÆ q Î � ü É PÆ q Ï � ü É�ËUsingnormalization,(i), (ii), and(iii) areeachsufficient for thecalculation.13.14 Whendealingwith joint entries,it is usuallyeasiestto geteverythinginto theform ofprobabilitiesof conjunctions,sincethesecanbeexpressedassumsof joint entries.Beginningwith theconditionalindependenceconstraintPÆ'0KÈ�³�� � É5äPÆ'0·� � ÉPÆ�³�� � Éhttp://librosysolucionarios.net87we canrewrite it usingthedefinitionof conditionalprobabilityoneachtermto obtainPÆ'0KÈ�³AÈ � ÉPÆ � É ä PÆ'0KÈ � ÉPÆ � É PÆ�³AÈ � ÉPÆ � ÉHencewe canwrite anexpressionfor joint entries:PÆ'0KÈ�³AÈ � ÉGä PÂÄÖ × É PÂ)Ø × ÉPÂ × É ä Ù PÂ)Ö ÚF × É Û PÂ Ü Ø × ÉÛ�Ü Ù PÂ Ü ÚF × ÉThis gives us 8 equationsconstrainingthe 8 joint entries,but several of the equationsareredundant.13.15 Therelevantaspectof theworld canbedescribedby two randomvariables:ð meansthetaxi wasblue,andd ð meansthetaxi looked blue. The informationon thereliability ofcolor identificationcanbewrittenas� Æ�d ð � ð É5ä�Ç#Ë�6=4 � Æ�J�d ð � J ð ÉAä�Ç#Ë�6=4Weneedto know theprobabilitythatthetaxi wasblue,giventhatit lookedblue:� Æ ð � d ð ÉTÓ � Æ�d ð � ð É � Æ ð É�Ó�Ç#Ë�6=4 � Æ ð É� Æ�J ð � d ð ÉTÓ � Æ�d ð � J ð É � Æ�J ð É�Ó�Ç#Ë��=4#Æ�ʺ^ � Æ ð É�ÉThuswe cannotdecidetheprobabilitywithout someinformationabouttheprior probabilityof bluetaxis,� Æ ð É . For example,if we knew thatall taxiswereblue, i.e.,� Æ ð ɲä±Ê, thenobviously� Æ ð � d ð É5ä�Ê . Ontheotherhand,if weadoptLaplace’sPrincipleof Indifference,which statesthatpropositionscanbedeemedequallylikely in theabsenceof any differenti-ating information,thenwehave� Æ ð ɵä Ç#Ë�4and� Æ ð � d ð ɵäbÇ#Ë�6=4. Usuallywe will havesomedifferentiatinginformation,sothisprincipledoesnotapply.Given that 9 out of 10 taxis are green,and assumingthe taxi in questionis drawnrandomlyfromthetaxi population, wehave� Æ ð ÉGä�Ç#Ë'Ê . Hence� Æ ð � d ð ÉTÓ�Ç#Ë�6=4h{�Ç#Ë'Ê�Ó�Ç#Ë'Ç=6=4� Æ�J ð � d ð ÉTÓ�Ç#Ë��=4h{LÇ#Ë�;¥Ó�Ç#Ë��=�=4� Æ ð � d ð É5ä M�Î M*Ý ½M�Î M*Ý ½ Ê M�ÎÏ�Ͻä�Ç#Ë��=4� Æ�J ð � d ð É5ä M�Î Ï�Ï ½M�Î M*Ý ½ Ê M�ÎÏ�Ͻä�Ç#Ë�6=413.16 Thisquestionis extremelytricky. It is avariantof the“Monty Hall” problem,namedafterthehostof thegameshow “Let’sMakeaDeal” in whichcontestantsareaskedto choosebetweentheprize they have alreadywon andanunknown prizebehinda door. Severaldis-tinguishedprofessorsof statisticshave very publically got thewronganswer. Certainly, allsuchquestionscanbesettledby repeatedtrials!Let õ Ü = “x will befreed”, q Ü = “x will beexecuted”. If the informationprovidedbytheguardis expressedas õlÞ thenwe get:� Æ qHß � õ�Þ ÉGä� Æ õlÞ � qHß É ¸ � Æ qHß É� Æ õ Þ Éä Ê ¸ Ê ì �� ì �ä Ê�This would be quite a shocktoâ—his chancesof executionhave increased!On the otherhand,if theinformationprovidedby theguardis expressedas õ AÞ = “The guardsaidthat õlÞ ”thenwe get:� Æ qHß � õ AÞ ÉGä� Æ õ AÞ � q ß É ¸ � Æ q ß É� Æ õ AÞ Éä Ê ì � ¸ Ê ì �Ê ì �ä Ê�http://librosysolucionarios.net88 Chapter 13. UncertaintyThusthekey thingthatis missedby thenaiveapproachis thattheguardhasachoiceof whomto inform in thecasewhereA will beexecuted.Onecanproducevariantsof thequestionthatreinforcetheintuitionsbehindthecorrectapproach.For example: Supposetherenow a thousandprisonerson deathrow, all but oneof whom will be pardoned.PrisonerA finds a printout with the last pagetorn off. It givesthe namesof 998 pardonees,not including A’s name. What is the probability that A willbeexecuted?Now supposeA is left-handed,andtheprogramwasprinting thenamesof allright-handedpardonees.What is the probability now? Clearly, in the first caseit is quitereasonableforâto get worried, whereasin the secondcaseit is not—thenamesof right-handedprisonersto bepardonedshouldnot influenceA’s chances.It is this secondcasethatappliesin the original story, becausethe guardis precludedfrom giving informationaboutâ.13.17 Wecanapplythedefinitionof conditionalindependenceasfollows:PÆ`Ô t Ø " ���eÉJä PÆeÈPÔ t Ø " �_É ì PÆ eÉJä é PÆeÈPÔ t Ø " �_ɯËNow, divide theeffect variablesinto thosewith evidence,E, andthosewithout evidence,Y.WehavePÆ`Ô t Ø " ���eÉLä éyPÆyÈeÈPÔ t Ø " �PÉä éyPÆ`Ô t Ø " �_É PÆ y ��Ô t Ø " �_É à PÆ�ø à ��Ô t Ø " �_Éä éPÆ`Ô t Ø " �_É à PÆ�ø à ��Ô t Ø " �PÉyPÆ`Ô t Ø " �PÉ PÆ y ��Ô t Ø " �_Éä éPÆ`Ô t Ø " �_É à PÆ�ø à ��Ô t Ø " �PÉwherethe last line follows becausethe summationover y is 1. Therefore,the algorithmcomputesthe productof the conditionalprobabilititesof the evidencevariablesgiven eachvalueof thecause,multiplieseachby theprior probabilityof thecause,andnormalizestheresult.13.18 This questionis essentiallypreviewing materialin Chapter23 (page842), but stu-dentsshouldhave little difficulty in figuring out how to estimatea conditionalprobabilityfrom completedata.a. Themodelconsistsof theprior probabilityPÆ�� ö{÷ ø ê Ñ«ù�ïYÉ andtheconditionalprobabil-ities PÆ�U�Ñ,ù ã ¼�� � ö�÷ ø ê Ñ,ù�ï^É . For eachcategory>, PÆ�� ö�÷ ø ê Ñ,ù`ï"ä[>WÉ is estimatedasthefractionof all documentsthatareof category>. Similarly, PÆ�U�Ñ,ù ã ¼Pä ÷ ù��0øO� � ö�÷ ø ê Ñ,ù�ïwä�>�Éis estimatedasthefractionof documentsof category>thatcontainwordÍ.b. Seetheanswerfor 13.17.Here,every evidencevariableis observed,sincewe cantellif any givenwordappearsin agivendocumentor not.http://librosysolucionarios.net89c. Theindependenceassumptionis clearlyviolatedin practice.For example,thewordpair“artificial intelligence”occursmore frequentlyin any given documentcategory thanwouldbesuggestedby multiplying theprobabilitiesof “artificial” and“intelligence”.13.19 This probability model is alsoappropriatefor Minesweeper(Ex. 7.11). If the totalnumberof pits is fixed,thenthevariables� ¼ à and� ° á areno longerindependent.In general,� Æ � ¼ à äoÙ�Ö,Øx��� � ° á äoÙ�Ö,Øx�_ÉT] � Æ � ¼ à äºÙ�Ö�Øx�l� � ° á ä z t � " �_Ébecauselearningthat� ° á äoÙ�Ö,Øx� makesit lesslikely thatthereis amineatå Í È`r æ(astherearenow fewer to spreadaround). The joint distribution placesequalprobability on all assign-mentsto� Î� Ï Ë�Ë�Ë � ¬ ¬thathave exactly 3 pits,andzeroon all otherassignments.Sincethereare15squares,theprobabilityof each3-pit assignmentisÊ ìνРä}Ê ì 3�4=4 .To calculatetheprobabilitiesof pits inå Ê#È�� æandå �#È�� æ, we startfrom Figure13.7. Wehavetoconsidertheprobabilitiesof completeassignments,sincetheprobabilityof the“other”region assignmentdoesnot cancelout. We cancountthetotal numberof 3-pit assignmentsthatareconsistentwith eachpartialassignmentin 13.7(a)and13.7(b).In 13.7(a),therearethreepartialassignmentswith� Î� Ð ä ÷ ù��0ø :� Thefirst fixesall threepits,socorrespondsto 1 completeassignment.� The secondleaves 1 pit in the remaining10 squares,so correspondsto 10 completeassignments.� Thethird alsocorrespondsto 10completeassignments.Hence,thereare21completeassignmentswith� Î� Ð ä ÷ ù��Aø .In 13.7(b),therearetwo partialassignmentswith� Î� Ð ä �Yö �2�`ø :� Thefirst leaves1 pit in theremaining10squares,socorrespondsto 10completeassign-ments.� Thesecondleaves2 pits in theremaining10squares,socorrespondstoÎ MÏ ä�3�4com-pleteassignments.Hence,thereare55completeassignmentswith� Î� Ð ä �Yö �2�-ø . Normalizing,we obtainPÆ � Î� Ð É5ä é ¾ �#Ê#È�4=4=¿¸ä ¾ Ç#Ë��=6=5#È�Ç#Ë�6=�E3�¿lËWith� Ï� Ï ä ÷ ùF�0ø , thereare four partial assignmentswith a total ofÎ MÏ Y � ¸ Î MÎ YÎ MM ä�5=5completeassignments.With� Ï� Ï ä �Yö �2�`ø , thereis only one partial assignmentwithÎ MÎ ä}Ê#Çcompleteassignments.HencePÆ � Ï� Ï É5ä é ¾ 5=5#È�Ê#Ç=¿¸ä ¾ Ç#Ë�7=5=7#È�Ç#Ë'Ê=�=�=¿lËhttp://librosysolucionarios.netSolutionsfor Chapter14ProbabilisticReasoning14.1 Adding variablesto an existing net can be donein two ways. Formally speaking,oneshouldinsertthevariablesinto thevariableorderingandrerunthenetwork constructionprocessfrom thepoint wherethefirst new variableappears.Informally speaking,oneneverreallybuildsanetwork by astrictordering.Instead,oneaskswhatvariablesaredirectcausesor influenceson whatotherones,andbuilds local parent/childgraphsthatway. It is usuallyeasyto identify wherein suchastructurethenew variablegoes,but onemustbeverycarefulto checkfor possibleinduceddependenciesdownstream.a.mo>�ï�U�ø ö{÷ ó�øù is not causedby any of the car-relatedvariables,so needsno parents.It directly affects the batteryand the startermotor.� ÷�ö ù ÷ ø�ù þ Ñ ÷ Ñ,ù is an additionalpreconditionfor� ÷�ö ù ÷ � . Thenew network is shown in FigureS14.1.b. Reasonableprobabilitiesmayvary a lot dependingon thekind of carandperhapsthepersonalexperienceof theassessor. Thefollowing valuesindicatethegeneralorderofRadioBatteryIgnition GasâStartsãMovesIcyWeather12ä2ä2ä2ä8å12äStarterMotorãFigure S14.1 Car network amended to include æ!ç2è�é?ê1ë�ìWíQêµî andï ìWë�îEìWêµîvð?ñ`ìWñ`îvé?ñ`îvò1ó2ô!õ (ï ð?é ).90http://librosysolucionarios.net91magnitudeandrelative valuesthatmakesense:� A reasonableprior for IcyWeathermight be0.05(perhapsdependingon locationandseason).� � Æ ðnö�÷�÷ ø�ù�ïH� mo>�ï�U�ø ö{÷ ó�øù~ÉGä�Ç#Ë�;=4 , � Æ ðnö�÷�÷ ø�ù�ïH� J�mo>�ï�U�ø ö�÷ ó�ø�ù£É5ä Ç#Ë�;=;=6 .� � Æ � ÷�ö ù ÷ ø�ù þ Ñ ÷ Ñ,ùö� mo>�ï�UÀø ö{÷ ó�øù£É5ä�Ç#Ë�;=7 , � Æ ðnö�÷�÷ ø�ù�ïH� J�mo>�ï�U�ø ö�÷ ó�ø�ù£É5ä�Ç#Ë�;=;=; .� � Æ /"ö ã�Í Ñ�� ðnö�÷�÷ ø�ù�ïYÉJä�Ç#Ë�;=;=;=; , � Æ /"ö ã{Í Ñ�� J ðnö{÷�÷ ø�ù�ï^É5ä�Ç#Ë'Ç=4 .� � Æ�m ê ô Í ÷ Í Ñ,ô�� ðMö�÷�÷ øù`ïYÉJä�Ç#Ë�;=;=7 , � Æ�m ê ô Í ÷ Í Ñ«ô�� J ðMö�÷�÷ øù`ïYÉ5ä�Ç#Ë'Ç#Ê .� � Æ�- ö �`ÉJä�Ç#Ë�;=;=4 .� � Æ � ÷�ö ù ÷ �!� m ê ô Í ÷ Í Ñ,ôGÈ � ÷�ö ù ÷ ø�ù þ Ñ ÷ Ñ,ù�È�- ö �`ÉJä�Ç#Ë�;=;=;=;, otherentries0.0.� � Æ þ Ñ è ø*�!� � ÷�ö ù ÷ �-ÉGä�Ç#Ë�;=;=7 .c. With 8 Booleanvariables,thejoint has�1÷�^OÊ= 255independententries.d. Giventhetopologyshown in FigureS14.1,thetotalnumberof independentCPTentriesis 1+2+2+2+2+1+8+2=20.e. The CPT for� ÷�ö ù ÷ � describesa setof nearlynecessaryconditionsthat are togetheralmostsufficient. That is, all theentriesarenearlyzeroexceptfor theentrywhereallthe conditionsaretrue. That entry will be not quite 1 (becausethereis alwayssomeotherpossiblefault thatwedidn’t think of), but asweaddmoreconditionsit getscloserto 1. If we addad�ø ö � nodeasanextra parent,thentheprobability is exactly 1 whenall parentsare true. We canrelatenoisy-AND to noisy-ORusingde Morgan’s rule:â á ðùø JJÆ�J â9� J ð É . That is, noisy-AND is thesameasnoisy-ORexceptthat thepolaritiesof theparentandchild variablesarereversed.In thenoisy-ORcase,we have� Æ�³oä ÷ ùF�0øO� î Î È�Ë�Ë�Ë{È�î ° ÉJä�Ê�^ ú ¼%û Ü�ü�ý Ã�þ ÿ�Å �X�¼whereX�¼is theprobability that thepresenceof theÍth parentfails to causethechild tobetrue. In thenoisy-ANDcase,wecanwrite� Æ�³oä ÷ ùF�0øO� î Î È�Ë�Ë�Ë{È�î ° ÉJä ú ¼%û Ü�üEý��{Ý á ƵŠ�ù�¼whereù�¼is theprobability that theabsenceof theÍth parentfails to causethechild tobefalse(e.g.,it is magicallybypassedby someothermechanism).14.2 This questionexercisesmany aspectsof thestudent’s understandingof Bayesiannet-worksanduncertainty.a. A suitablenetwork is shown in FigureS14.2.Thekey aspectsare:thefailurenodesareparentsof thesensornodes,andthetemperaturenodeis a parentof boththegaugeandthegaugefailure node. It is exactly this kind of correlationthat makesit difficult forhumansto understandwhatis happeningin complex systemswith unreliablesensors.b. No matterwhichwaythestudentdrawsthenetwork, it shouldnotbeapolytreebecauseof thefactthatthetemperatureinfluencesthegaugein two ways.c. TheCPTfor-is shown below. Thewordingof thequestionis a little tricky becauseîandïaredefinedin termsof “incorrect” ratherthan“correct.”http://librosysolucionarios.net92 Chapter 14. ProbabilisticReasoningT G A�FG FA�FigureS14.2 A Bayesiannetwork for thenuclearalarmproblem.ý ä[@KÑ«ù�ÿ ö � ý ä ü Í ê óõ�� J õ�� õ�� J õ��-}ä[@KÑ,ù`ÿ ö � Ê�^�ï Ê�^�î ï î-}ä ü Í ê ó ï î Ê�^�ï Ê�^Lîd. TheCPTforâis asfollows: -}ä�@KÑ,ù`ÿ ö � -}ä ü Í ê óõ�ß J õ�ß õ�ß J õlßâ0 0 0 1J â1 1 1 0e. This partactuallyasksthestudentto do somethingusuallydoneby Bayesiannetworkalgorithms. The greatthing is that doing the calculationwithout a Bayesiannetworkmakesit easyto seethenatureof thecalculationsthatthealgorithmsaresystematizing.It illustratesthemagnitudeof theachievementinvolvedin creatingcompleteandcorrectalgorithms.Abbreviating ý ä ü Í ê ó and-}ä ü Í ê ó by ý and-, theprobabilityof interesthereis� Æ ý � â È�J õ � È�J õ�ß É . Becausethe alarm’s behavior is deterministic,we canreasonthat if the alarm is working and sounds,-must be ü Í ê ó . Becauseõlß andâared-separatedfrom ý , we needonly calculate� Æ ý � J õ � È�-­É .Thereareseveralwaysto go aboutdoingthis. The“opportunistic”way is to noticethattheCPTentriesgiveus� Æ�-\� ý È�J õ�� É , whichsuggestsusingthegeneralizedBayes’Ruleto switch-andý withJ õ � asbackground:� Æ ý � J õ�� È�-­ÉfÓ � Æ�-B� ý È�J õ�� É � Æ ý � J õ�� ÉWethenuseBayes’Ruleagainon thelastterm:� Æ ý � J õ � È�-­ÉfÓ � Æ�-B� ý È�J õ � É � Æ�J õ � � ý É � Æ ý ÉA similar relationshipholdsforJ ý :� Æ�J ý � J õ � È�-­ÉHÓ � Æ�-B� J ý È�J õ � É � Æ�J õ � � J ý É � Æ�J ý ÉNormalizing,weobtain� Æ ý � J õ � È�-­É�ä Á� � Ç Ì���ÉÄÁ�Â.Ì���Ç �ÉÄÁ�Â��ÉÁ� � Ç Ì� � ÉÄÁ�ÂËÌ� � Ç �É)Á�Â��É.ÊxÁ� � ÇÍÌ� Ì� � É)Á�ÂËÌ� � ÇÍÌ�öÉÄÁ�ÂËÌ��Éhttp://librosysolucionarios.net93The“systematic”way to do it is to revert to joint entries(noticingthatthesubgraphof ý ,-, and õ � is completelyconnectedsono lossof efficiency is entailed).Wehave� Æ ý � J õ � È�-­É�ä� Æ ý È�J õ � È�-­É� Æ�-­È�J õ � Éä � Æ ý È�J õ � È�-­É� Æ ý È�-­È�J õ � É Y � Æ ý È�-­È�J õ � ÉNow we usethe chain rule formula (Equation15.1 on page439) to rewrite the jointentriesasCPTentries:� Æ ý � J õ�� È�-­É�ä Á�Â��É)Á�ÂËÌ���Ç �ÉÄÁ� � Ç Ì��!ÉÁ�Â��ÉÄÁ�Â.Ì���Ç öÉÄÁ� � Ç Ì���É.ÊxÁ�ÂËÌ��ÉÄÁ�Â.Ì���ÇÍÌ��ÉÄÁ� � ÇÍÌ� Ì��!Éwhich of courseis the sameas the expressionarrived at above. Letting� Æ ý Ékä :,� Æ õ�� � ý ÉGä ê , and� Æ õ�� � J ý É5ä�ó , we get� Æ ý � J õ � È�-­É�ä: Æ�Ê�^ ê É�Æ�Ê�^LîAÉ: Æ�Ê�^ ê É�Æ�ʺ^�îAÉ Y Æ�Ê�^ : É�Æ�Ê�^Oó�É�î14.3a. Although (i) in somesensedepictsthe “flow of information” during calculation,it isclearly incorrectasa network, sinceit saysthatgiven themeasurementsþ Îandþ Ï,the numberof starsis independentof the focus. (ii) correctly representsthe causalstructure:eachmeasurementis influencedby theactualnumberof starsandthefocus,andthe two telescopesare independentof eachother. (iii) shows a correctbut morecomplicatednetwork—theoneobtainedby orderingthenodesþ Î,þ Ï,@, õ Î , õ Ï . Ifyou orderþ Ïbeforeþ Îyou would get thesamenetwork exceptwith thearrow fromþ Îtoþ Ïreversed.b. (ii) requiresfewerparametersandis thereforebetterthan(iii).c. To computePÆ þ Î � @KÉ, we will needto conditionon õ Î (thatis, considerbothpossiblecasesfor õ Î , weightedby theirprobabilities).PÆ þ Î � @pÉ}äPÆ þ Î � @kÈ õ Î É PÆ õ Î � @KÉ Y PÆ þ Î � @kÈ�J õ Î É PÆ�J õ Î � @KÉäPÆ þ Î � @kÈ õ Î É PÆ õ Î É Y PÆ þ Î � @kÈ�J õ Î É PÆ�J õ Î ÉLet � betheprobability that thetelescopeis out of focus. Theexercisestatesthat thiswill causean “undercountof threeor morestars,” but if@= 3 or lessthe countwillbe 0 if the telescopeis out of focus. If it is in focus, thenwe will assumethereis aprobabilityoføof countingonetwo few, andøof countingonetoo many. TherestofthetimeÆ�Ê�^C�#ø�É, thecountwill beaccurate.Thenthetableis asfollows:@Àä}Ê @fä�� @fä��þ Î ä�Çf + e(1-f) f fþ Î ä�Ê(1-2e)(1-f) e(1-f) 0.0þ Î äZ�e(1-f) (1-2e)(1-f) e(1-f)þ Î äZ�0.0 e(1-f) (1-2e)(1-f)þ Î äa30.0 0.0 e(1-f)Notice that eachcolumnhasto addup to 1. Reasonablevaluesforøand � might be0.05and0.002.http://librosysolucionarios.net94 Chapter 14. ProbabilisticReasoningd. This questioncausesa surprisingamountof difficulty, so it is importantto make surestudentsunderstandthe reasoningbehindan answer. Oneapproachusesthe fact thatit is easyto reasonin the forward direction,that is, try eachpossiblenumberof stars@andseewhethermeasurementsþ Î ä}Êandþ Ï ä[�arepossible.(This is a sort ofmentalsimulationof thephysicalprocess.)An alternative approachis to enumeratethepossiblefocusstatesanddeducethevalueof@for each.Eitherway, thesolutionsare@ äZ�, 4, orR�5.e. Wecannotcalculatethemostlikely numberof starswithoutknowing theprior distribu-tion� Æ�@pÉ. Let thepriorsbe:�Ï,: ¬, and:����. Theposteriorfor@Àä[�is:_Ï ø Ï Æ�Ê�^ � É Ï ;for@fä�3it is at most: ¬ ø � (at most,becausewith@fäK3the out-of-focustelescopecould measure0 insteadof 1); for@ R�5it is at most:���� � Ï . If we assumethat thepriorsareroughlycomparable,then@Àä��is mostlikely becausewe aretold that � ismuchsmallerthanø.For follow-up or alternatequestions,it is easyto comeup with endlessvariationson thesamethemeinvolving sensors,failure nodes,hiddenstate. One can also add in complexmechanisms,asfor the� ÷�ö ù ÷ � variablein exercise14.1.14.5 Thisexerciseis a little tricky andwill appealto moremathematicalyorientedstudents.a. Thebasicideais to multiply thetwo densities,matchtheresultto thestandardform foramultivariateGaussian,andhenceidentify theentriesin theinversecovariancematrix.Let’s begin by looking at themultivariateGaussian.Frompage982in AppendixA wehave� ÆxÉ5ä ÊÆ����GÉ ¹ ���K� ø ±���  x ±�� É����� �  x ±�� É Èwhere� is themeanvectorand�is thecovariancematrix. In our case,x isÆ�î Î î Ï É! andlet the(asyet)unknown � beÆ�ÿ Î ÿ Ï É . Supposetheinversecovariancematrix is� ± Î ä > ãã øThen,if we multiply out theexponent,weobtain^ ÎÏ Æx^ � É � ± Î Æ x ^ � É ä^ ÎÏ ¸ >�Æ�î Î ^Kÿ Î É Ï Y � ã Æ�î Î ^�ÿ Î É�Æ�î Ï ^�ÿ Ï É Y9� Æ�î Ï ^�ÿ Ï É ÏLookingat thedistributionsthemselves,we have� Æ�î Î É5ä Ê" Î�# ��� ø ± ÂÜ � ±�$ � É �&%  Ï!' �� Éand � Æ�î Ï � î Î ÉJä Ê" Ï(# ��� ø ± ÂÜ � ± Â Ý«Ü � Ê�)�É.É � %  Ï!' �� Éhence� Æ�î Î È�î Ï ÉJä Ê" Î " Ï Æ����GÉ ø ± Â' �� Â Ü � ±  Ý,Ü � Ê�)�ÉËÉ � Ê ' �� Â Ü � ± Â Ý«Ü � Ê�)�ÉËÉ � É %  Ï!' �� ' �� Éhttp://librosysolucionarios.net95Wecanobtainequationsfor>,ã, andøby picking out thecoefficientsofî Ï Î,î Î î Ï, andî ÏÏ: >oä Æ " ÏÏ YOö Ï " Ï Î É ì " Ï Î " ÏÏ� ã ä ^¥� ö_ì " ÏÏøDä Ê ì " ÏÏWecanchecktheseby comparingthenormalizingconstants.Ê" Î " Ï Æ����GÉ äÊÆ����GÉ ¹ ���K� äÊÆ����GÉ Ê ì ��� ± Î �ä ÊÆ����GÉ Ê ì Æ�>WøN^ ã Ï Éfrom whichwe obtaintheconstraint>Wøf^ ã Ï ä�Ê ì " Ï Î " ÏÏwhich is easilyconfirmed.Similar calculationsyieldÿ Îandÿ Ï, andpluggingthere-sultsbackshowsthat� Æ�î Î È�î Ï Éis indeedmultivariateGaussian.Thecovariancematrixis�Àä > ãã ø± Î ä Ê>�øN^ ã Ïø ^ ã^ ã > ä " Ï Î ö " Ï Îö " Ï Î " ÏÏ Ydö Ï " Ï Îb. The induction is onô, the numberof variables. The basecaseforô�äLÊis trivial.The inductive stepasksus to show that if any� Æ�î Î È�Ë�Ë�Ë-È�îl¹�Éconstructedwith linear–Gaussianconditionaldensitiesis multivariateGaussian,thenany� Æ�î Î È�Ë�Ë�Ë�È�î�¹�È�î�¹ Ê Î Éconstructedwith linear–Gaussianconditionaldensitiesis also multivariateGaussian.Without lossof generality, we canassumethat0g¹ Ê Î is a leaf variableaddedto a net-work definedin thefirstôvariables.By theproductrulewe have� Æ�î Î È�Ë�Ë�Ë{È�îl¹�È�îl¹ Ê Î ÉLä � Æ�îl¹ Ê Î � î Î È�Ë�Ë�Ë{È�î�¹�É � Æ�î Î È�Ë�Ë�Ë{È�î�¹�Éä � Æ�îl¹ Ê Î � : ö ù~øô ÷ �`Æ'0g¹ Ê Î É�É � Æ�î Î È�Ë�Ë�Ë�È�î�¹�Éwhich,by theinductive hypothesis,is theproductof a linearGaussianwith amultivari-ateGaussian.Extendingtheargumentof part(a), this is in turnamultivariateGaussianof onehigherdimension.14.6a. With multiplecontinuousparents,wemustfind awayto maptheparentvaluevectortoa singlethresholdvalue.Thesimplestway to do this is to take a linearcombinationoftheparentvalues.b. For orderedvaluesï Î ]�ï Ï ] ¸�¸�¸ ]�ï È , considertheBooleanproposition� à definedby³ »aï à . The proposition³ºäNï à is just� à á|JJÆ � à ± Î É forr i Ê. Now we canproposeprobit distributionsfor each� à , with means* à alsoin increasingorder.14.7 This questiondefinitely helpsstudentsget a solid feel for variableelimination. Stu-dentsmayneedsomehelpwith thelastpartif they areto do it properly.http://librosysolucionarios.net96 Chapter 14. ProbabilisticReasoninga. � Æ ð � r`È�ÿ�Éä é � Æ ð É Å� Æ�ø�É Ý � Æ ö � �WÈ�ø-É � Æ'r�� ö É � Æ�ÿ¯� ö Éä é � Æ ð É Å� Æ�ø�É Ë�;�{NË�6�{ Ë�;=4�Ë��=;Ë�;E3ªË'Ç#Ç#Ê Y Ë'Ç=4�{NË'Ç#Ê�{Ë'Ç=4�Ë�6#ÊË'Ç=5�Ë�;=;=;ä é � Æ ð É Å� Æ�ø�É Ë�4=;=7=4=�=4fË'Ê=7=�#Ç=4=4Ë�4=;=�=�=� Ë'Ç#Ç#Ê#Ê=�=;=4ä é � Æ ð É Ë'Ç#Ç=��{Ë�4=;=7=4=�=4Ë'Ê=7=�#Ç=4=4 Y Ë�;=;=7�{Ë�4=;=�=�=�Ë'Ç#Ç#Ê#Ê=�=;=4ä é Ë'Ç#Ç#ÊË�;=;=; { Ë�4=;=�=�E3��=4=;Ë'Ç#Ç#ÊE3�;=�=�=4#Êä é Ë'Ç#Ç#Ç=4=;=�=�E3��=4=;Ë'Ç#Ç#ÊE3�;#Ê=7=4=6=5Ò ¾ Ë��=7E3�È�Ë�6#Ê=5=¿b. Including thenormalizationstep,thereare7 additions,16 multiplications,and2 divi-sions.Theenumerationalgorithmhastwo extramultiplications.c. To computePÆ'0 Î � 0g¹µäºÙ�Ö�ØO�_Éusingenumeration,we have to evaluatetwo completebinarytrees(onefor eachvalueof0 Î), eachof depthôe^��, sothetotalwork is û Æ�� ¹ É .Usingvariableelimination,thefactorsnever grow beyondtwo variables.For example,thefirst stepisPÆ'0 Î � 0g¹+äºÙ�Ö,Øx�_Éä éPÆ'0 Î É~Ë�Ë�Ë Ü(+ � �� Æ�îl¹ ± Ï � î�¹ ± Ð É Ü(+ � �� Æ�îl¹ ± Î � îl¹ ± Ï É � Æ'0g¹µäoÙ�Ö,Øx��� îl¹ ± Î Éä éPÆ'0 Î É~Ë�Ë�Ë Ü + � �� Æ�îl¹ ± Ï � î�¹ ± Ð É Ü + � � f Ö+ � � Æ�î�¹ ± Î È�î�¹ ± Ï É f Ö +_Æ�î�¹ ± Î Éä éPÆ'0 Î É~Ë�Ë�Ë Ü + � �� Æ�îl¹ ± Ï � î�¹ ± Ð É f Ö + � � Ö + Æ�î�¹ ± Ï ÉThelast line is isomorphicto theproblemwithôS^ Êvariablesinsteadofô; theworkdoneon thefirst stepis a constantindependentofô, hence(by inductiononô, if youwantto beformal) thetotal work is û Æ�ôGÉ .d. Herewe canperforman inductionon the numberof nodesin the polytree. The basecaseis trivial. For theinductive hypothesis,assumethatany polytreewithônodescanbe evaluatedin time proportionalto the sizeof the polytree(i.e., the sumof theCPTsizes).Now, considera polytreewithô Y Ê nodes.Any nodeorderingconsistentwiththe topologywill eliminatefirst someleaf nodefrom this polytree. To eliminateanyleaf node,we have to do work proportionalto the sizeof its CPT. Then,becausethenetworkis a polytree, we areleft with independentsubproblems,onefor eachparent.http://librosysolucionarios.net97Eachsubproblemtakestotal work proportionalto thesumof its CPTsizes,sothetotalwork forô Y Ê nodesis proportionalto thesumof CPTsizes.C2C1 C3 C4SB0.50.50.5 0.50.5A,C D EFigureS14.3 A Bayesiannetwork correspondingto a SAT problem.14.8 ConsideraSAT problemsuchasthefollowing:Æ�J âk� ð ÉYáFÆ�J ð � �¦ÉYáFÆ�J�� � P ÉYáIÆ�J�� � J P � q ÉTheideais to encodethisasaBayesnet,suchthatdoinginferencein theBayesnetgivestheanswerto theSAT problem.a. FigureS14.3shows the Bayesnet correspondingto this SAT problem. The generalconstructionmethodis asfollows:� Therootnodescorrespondto thelogicalvariablesof theSAT problem.They haveaprior probabilityof 0.5.� Eachclause� ¼is a node. Its parentsarethevariablesin theclause.TheCPT isdeterministicandimplementsthedisjunctiongivenin theclause.(Negativeliteralsin theclauseareindicatedby negationsymbolson thelinks in thefigure.)� A singlesentencenode�hasall theclausesasparentsandaCPTthatimplementsdeterministicconjunction.It is clearthat� Æ � É i Çif f theSAT problemis satisfiable.Hence,we have reducedSAT to Bayesnetinference.SinceSAT is NP-complete,we have shown thatBayesnetinferenceis NP-hard(evenwithoutevidence).b. Theprior probabilityof eachcompleteassignmentis�v± ¹.� Æ � Éis thereforeI ¸ �v± ¹where I is the numberof satisfyingassignments.Hence,we cancount the numberof satisfyingassignmentsby computing� Æ � É ¸ � ¹. This amountsto a reductionof theproblemof countingsatisfyingassignmentsto Bayesnetinference;sincetheformeris#P-complete,thelatteris #P-hard.14.9a. To calculatethe cumulative distribution of a discretevariable,we start from a vectorrepresentation:of the original distribution and a vector�of the samedimension.http://librosysolucionarios.net98 Chapter 14. ProbabilisticReasoningThen,we loop throughÍ, addingup the: ¼valuesaswe go alongandsetting� ¼to therunningsum,¼à ý ¼ : à . To samplefrom thedistribution,wegeneratearandomnumberùuniformly inå Ç#È�Ê æ, andthenreturnîl¼for the smallestÍsuchthat� ¼¥R�ù. A naiveway to find this is to loop throughÍstartingat 1 until� ¼TROù. This takes û Æ � É time. Amoreefficient solutionis binarysearch:startwith the full rangeå Ê#È �`æ, chooseÍat themidpointof therange.If� ¼]¹ù, settherangefromÍto theupperbound,otherwisesettherangefrom the lower boundtoÍ. After û Æ.- /10 � É iterations,we terminatewhentheboundsareidenticalor differ by 1.b. If we are generating@ 2 �samples,we can afford to preprocessthe cumulativedistribution. Thebasicinsightrequiredis that if theoriginal distribution wereuniform,it would bepossibleto samplein û Æ�Ê#É time by returning 3 � ù(4 . That is, we canindexdirectly into thecorrectpartof therange(analograndomaccess,onemightsay)insteadof searchingfor it. Now, supposewe divide the rangeå Ç#È�Ê æinto�equalpartsandconstructa�-elementvector, eachof whoseentriesis a list of all thoseÍfor which� ¼is in the correspondingpart of the range. TheÍwe want is in the list with index3 � ù54 . We retrieve this list in û Æ�Ê#É time andsearchthroughit in order(asin thenaiveimplementation).Letô à bethenumberof elementsin listr. Thentheexpectedruntimeis givenby°à ý Îô à ¸ Ê ì � ä�Ê ì �¥¸°à ý Îô à ä�Ê ì ��¸ û Æ � ÉJä û Æ�Ê#ÉThevarianceof theruntimecanbereducedbyfurthersubdividing any partof therangewhoselist containsmorethansomesmallconstantnumberof elements.c. Oneway to generateasamplefrom a univariateGaussianis to computethediscretizedcumulative distribution (e.g., integrating by Taylor’s rule) and usethe algorithm de-scribedabove. We cancomputethe tableonceandfor all for the standardGaussian(mean0, variance1) and then scaleeachsampledvalueñto " ñ Y6* . If we had aclosed-form,invertibleexpressionfor thecumulative distribution õ Æ�î0É , we couldsam-ple exactly, simply by returning õ ± Î Æ�ù£É . Unfortunatelythe Gaussiandensityis notexactly integrable.Now, thedensityé î0ø ± Ü �&% Ïis exactly integrable,andtherearecuteschemesfor usingtwo samplesandthisdensityto obtainanexactGaussiansample.Weleave thedetailsto theinterestedinstructor.d. WhenqueryingacontinuousvariableusingMontecarloinference,anexactclosed-formposteriorcannotbe obtained.Instead,onetypically definesdiscreteranges,returninga histogramdistribution simply by countingthe(weighted)numberof samplesin eachrange.14.10 Theseproofsaretricky for thosenotaccustomedto manipulatingprobabilityexpres-sions,andstudentsmayrequiresomehints.a. Thereareseveralwaysto prove this. Probablythesimplestis to work directly from theglobalsemantics.First,we rewrite therequiredprobabilityin termsof thefull joint:� Æ�î�¼1� î Î È�Ë�Ë�Ë�È�îl¼ ± Î È�îl¼ Ê Î È�Ë�Ë�Ë-È�îl¹�ÉLä� Æ�î Î È�Ë�Ë�Ë�È�îl¹�É� Æ�î Î È�Ë�Ë�Ë�È�î�¼ ± Î È�îl¼ Ê Î È�Ë�Ë�Ë�È�îl¹�Éhttp://librosysolucionarios.net99ä � Æ�î Î È�Ë�Ë�Ë�È�î ¹ ÉÜ�ü � Æ�î Î È�Ë�Ë�Ë�È�îl¹�Éä¹à ý Î � Æ�î à � : ö ù£øô ÷ �W0 à ÉÜ�ü ¹à ý Î � Æ�î à � : ö ù~øô ÷ �W0 à ÉNow, all termsin theproductin thedenominatorthatdo not containîl¼canbemovedoutsidethesummation,andthencancelwith thecorrespondingtermsin thenumerator.This just leavesuswith thetermsthatdo mentionî�¼, i.e., thosein which0g¼is a childor aparent.Hence,� Æ�î ¼ � î Î È�Ë�Ë�Ë�È�î ¼ ± Î È�î ¼ Ê Î È�Ë�Ë�Ë-È�î ¹ É is equalto� Æ�î�¼�� : ö ù£øô ÷ �W0g¼¥É Ø5798�:<; ¼ á È�þ Å ¹ ÂÄÖ ü É � Æ�ï à � : ö ù~øô ÷ �`Æ�³ à É�ÉÜ�ü � Æ�îl¼�� : ö ù~øô ÷ �W0g¼�É Ø(7.8�:<; ¼ á È�þ Å ¹ ÂÄÖ ü É � Æ�ï à � : ö ù£ø�ô ÷ �-Æ�³ à É�ÉNow, by reversingtheargumentin part(b), we obtainthedesiredresult.b. This is arelatively straightforwardapplicationof Bayes’rule. Let Yä�³ Î È�Ë�Ë�Ë�È�ï>=bethechildrenof0g¼andlet Z à betheparentsof³ à otherthan0g¼. Thenwe havePÆ'0g¼�� þ ð Æ'0g¼.É�ÉäPÆ'0g¼�� � ö ù£ø�ô ÷ �-Æ'0g¼�É�È Y È Z Î È�Ë�Ë�Ë�È Z =,Éä éPÆ'0g¼�� � ö ù£øô ÷ �`Æ'0g¼�É�È Z Î È�Ë�Ë�Ë�È Z =�É PÆ Y � � ö ù£ø�ô ÷ �-Æ'0g¼�É�È`0g¼�È Z Î È�Ë�Ë�Ë�È Z =�Éä éPÆ'0g¼�� � ö ù£øô ÷ �`Æ'0g¼�É�É PÆ Y � 0g¼.È Z Î È�Ë�Ë�Ë{È Z =�Éä éPÆ'0g¼�� � ö ù£øô ÷ �`Æ'0g¼�É�É Ø57.8�:<; ¼ á È�þ Å ¹ Â)Ö ü É� Æ�³ à � � ö ù£ø�ô ÷ �-Æ�³ à É�Éwherethederivation of the third line from thesecondrelieson the fact thata nodeisindependentof its nondescendantsgivenits children.14.11a. Therearetwo uninstantiatedBooleanvariables(���;Ñ�� ã ïand /"ö Í ô ) andthereforefourpossiblestates.b. First,wecomputethesamplingdistributionfor eachvariable,conditionedonitsMarkovblanket.PÆ��\� ù#È��-É}ä éPÆ��°ÉPÆ��x� �°ÉPÆ�ùö� �¦Éä é ¾ Ç#Ë�4#È�Ç#Ë�4=¿ ¾ Ç#Ë'Ê#È�Ç4Ë�4=¿ ¾ Ç#ËË7#È{Ç#Ë��v¿?ä é ¾ Ç#Ë'ÇE3�È�Ç#Ë'Ç=4=¿2ä ¾ 3 ì ;#È�4 ì ;=¿PÆ���� J^ù#È��-É}ä éPÆ��°ÉPÆ��x� �°ÉPÆ�J^ùö� �¦Éä é ¾ Ç#Ë�4#È�Ç#Ë�4=¿ ¾ Ç#Ë'Ê#È�Ç4Ë�4=¿ ¾ Ç#ËË�#È{Ç#Ë�7v¿?ä é ¾ Ç#Ë'Ç#Ê#È�Ç#Ë��#Ç=¿2ä ¾ Ê ì �#Ê#È��#Ç ì �#Ê=¿PÆ / � >WÈ��`È`�MÉ�ä éPÆ / � >WÉ PÆ'��� �-È / Éä é ¾ Ç#Ë�7#È�Ç#Ë��=¿ ¾ Ç#Ë�;=;#È{Ç#Ë�;4Ç=¿µä é ¾ Ç#Ë�6=;=�#È�Ç#Ë'Ê=7#Ç=¿?ä ¾ �=� ì �=6#È�4 ì �=6=¿PÆ / � J�>WÈ��`È`�MÉ�ä éPÆ / � J�>WÉ P Æ'�K� �`È / Éä é ¾ Ç#Ë��#È�Ç#Ë�7=¿ ¾ Ç#Ë�;=;#È{Ç#Ë�;4Ç=¿µä é ¾ Ç#Ë'Ê=;=7#È�Ç#Ë�6=�#Ç=¿?ä ¾ Ê#Ê ì 4#Ê#È`3�Ç ì 4#Ê=¿Strictly speaking,thetransitionmatrix is only well-definedfor thevariantof MCMC inwhichthevariableto besampledis chosenrandomly. (In thevariantwherethevariablesarechosenin a fixedorder, the transitionprobabilitiesdependon wherewe arein theordering.)Now considerthetransitionmatrix.http://librosysolucionarios.net100 Chapter 14. ProbabilisticReasoning� Entrieson the diagonalcorrespondto self-loops. Suchtransitionscanoccurbysamplingeithervariable.For example,X�Æ�Æ�>WÈ�ù£É�?©Æ�>WÈ�ù£É�ÉGä�Ç#Ë�4 � Æ�>v� ù�È��`É Y Ç#Ë�4 � Æ�ùö� >�È��`È`�MÉlä�Ê=6 ì �=6� Entrieswhereonevariableis changedmustsamplethatvariable.For example,X�Æ�Æ�>WÈ�ù£É�?©Æ�>WÈ�J^ù£É�ÉJä�Ç#Ë�4 � Æ�J^ùö� >WÈ��-È`�nÉ�äZ4 ì 4E3� Entrieswherebothvariableschangecannotoccur. For example,X�Æ�Æ�>WÈ�ù£É�?©Æ�JH>�È�JYù~É�ÉGä�ÇThisgivesusthefollowing transitionmatrix,wherethetransitionis from thestategivenby therow labelto thestategivenby thecolumnlabel:Æ�>WÈ�ù£ÉÆ�>WÈ�J^ù£ÉÆ�J�>WÈ�ù£ÉÆ�J�>WÈ�J^ù£ÉÆ�>WÈ�ù~ÉLÆ�>WÈ�J^ù£É}Æ�JH>�È�ù£ÉLÆ�J�>WÈ�J^ù£ÉÊ=6 ì �=6 4 ì 4E3 4 ì Ê=7 ÇÊ#Ê ì �=6 �=� ì Ê=7=; Ç Ê#Ç ì �#Ê� ì ; Ç·4=; ì Ê=4=� �#Ç ì 4#ÊÇ Ê ì 3���Ê#Ê ì Ê#Ç=� �#Ê#Ç ì �=4=6c. QÏrepresentstheprobabilityof goingfrom eachstateto eachstatein two steps.d. Q¹(asô@?BA) representsthe long-termprobabilityof beingin eachstatestartingineachstate;for ergodic Q theseprobabilitiesare independentof the startingstate,soevery row of Q is the sameandrepresentsthe posteriordistribution over statesgiventheevidence.e. We can producevery large powers of Q with very few matrix multiplications. Forexample,we cangetQÏwith onemultiplication,Q¬with two, andQÏ�Cwith�. Unfor-tunately, in a network withôBooleanvariables,thematrix is of size� ¹ {�� ¹, soeachmultiplicationtakes û Æ�� Ð ¹ É operations.14.12a. Theclassesare ý ø ö ÿ , with instancesâ, ð , and�, andþ ö�÷ >Wó , with instancesâ ð ,ð � , and� â. Eachteamhasaquality<andeachmatchhasa ý ø ö ÿ Î andý ø ö ÿ Ï andan û � ÷ >WÑ,ÿ�ø . Theteamnamesfor eachmatchareof coursefixedin advance.Thepriorover quality couldbeuniform andtheprobabilityof a win for team1 shouldincreasewith<°Æ ý ø ö ÿ Î É�^9<°Æ ý ø ö ÿ Ï É .b. Therandomvariablesareâ Ë�<, ð Ë�< ,�?Ë�<,â ð Ë û � ÷ >WÑ«ÿ�ø , ð �?Ë û � ÷ >�Ñ,ÿ�ø , and� â Ë û � ÷ >WÑ,ÿ�ø .Thenetwork is shown in FigureS14.4.c. Theexact resultwill dependon theprobabilitiesusedin themodel. With any prior onqualitythatis thesameacrossall teams,weexpectthattheposteriorover ð �²Ë û � ÷ >WÑ,ÿ�øwill show that�is morelikely to win than ð .d. Theinferencecostin suchamodelwill be û Æ�� ¹ É becauseall theteamqualitiesbecomecoupled.e. MCMC appearsto do well on this problem, provided the probabilitiesare not tooskewed. Our resultsshow scalingbehavior that is roughly linear in the numberofteams,althoughwe did not investigatevery largeô.http://librosysolucionarios.net101A.QDB.Q C.QAB.Outcome BC.Outcome CA.OutcomeFigureS14.4 A Bayesiannetwork correspondingto thesoccermodelof Exercise14.12.http://librosysolucionarios.netSolutionsfor Chapter15ProbabilisticReasoningoverTime15.1 For eachvariableb à that appearsasa parentof a variable0 Ã,Ê Ï , definean auxiliaryvariablebFE á ÈÃ,Ê Î , suchthatb à is parentofbFE á ÈÃ,Ê Î andb�E á ÈÃ,Ê Î is a parentof0 Ã,Ê Ï . This givesusa first-orderMarkov model. To ensurethat the joint distribution over the original variablesis unchanged,we keeptheCPTfor0 Ã,Ê Ï is unchangedexceptfor thenew parentname,andwe requirethatPÆ�b E á ÈÃ,Ê Î � b Ã É is an identity mapping,i.e., thechild hasthesamevalueastheparentwith probability1. Sincetheparametersin thismodelarefixedandknown, thereis noeffective increasein thenumberof freeparametersin themodel.15.2a. For all ÷ , we thefiltering formulaPÆ / à � � Î û à ÉJä é PÆ'� à � / Ã É GIH � � PÆ / à � / à ± Î É � Æ / à ± Î � � Î û à ± ΠɯËAt thefixed point, we additionallyexpectthat PÆ / à � � Î û à ɰä PÆ / à ± Î � � Î û à ± Î É . Let thefixed-pointprobabilitiesbe ¾�J È?ʺ^ J ¿ . This providesuswith a systemof linearequa-tions:KMLONQPSRTLUWVYXZK\[>] ^>N�[>] _�U`K\[�]Ma�N[>] b�UcLedfKg[>] b>N�[>]ca�U`h`PiRTLkjVYXZK\[>] ^>N�[>] _�UlhmKg[>] noLONIRS[>] noLUOdfKg[>] b>Nk[�]Ma�U`jV P[>] ^�h\[>]noLpdq[>] b�jOdq[>] _�hrRS[>] noLFd�[�]Masj K\[�] ^�N�[�] _�U`hrK\[>] noLONIRS[�] n�LOUOdWK\[>] b>N�[>]ca�UljSolvingthissystem,we find that J Ò Ç#Ë�7=;=�=� .b. Theprobabilityconvergesto ¾ Ç#Ë�4#È�Ç#Ë�4=¿ asillustratedin FigureS15.1.Thisconvergencemakessenseif weconsiderafixed-pointequationfor PÆ / Ï Ê ° � b Î È�b Ï É :PÆ / Ï Ê ° � b Î È�b Ï ÉLä ¾ Ç#Ë�6#È�Ç#Ë��=¿ � Æ�ù Ï Ê °=± Î � b Î È�b Ï É Y ¾ Ç#Ë��#È�Ç#Ë�6=¿ � Æ�JYù Ï Ê °=± Î � b Î È�b Ï ÉPÆ�ù Ï Ê ° � b Î È�b Ï ÉLä Ç#Ë�6 � Æ�ù Ï Ê °=± Î � b Î È�b Ï É Y Ç#Ë��#Æ�Ê�^ � Æ�ù Ï Ê °=± Î � b Î È�b Ï É�Éä Ç#Ë 3 � Æ�ù Ï Ê °=± Î � b Î È�b Ï É Y Ç#Ë��Thatis,� Æ�ù Ï Ê ° � b Î È�b Ï É5ä�Ç#Ë�4 .Noticethatthefixedpoint doesnotdependon theinitial evidence.15.3 ThisexercisedevelopstheIslandalgorithmfor smoothingin DBNs (?).102http://librosysolucionarios.net1030 2 4 6 8 10 12 14 16 18 2000.10.20.30.40.50.60.70.80.91Figure S15.1 A graphof the probability of rain asa function of time, forecastinto thefuture.a. Thechaptershows thatPÆX ° �eÎ û Ã É canbecomputedasPÆX ° �eÎ û à ÉJä é PÆX ° �eÎ û ° É PÆ e° Ê Î û à �X ° É5ä é fÎ û ° b° Ê Î û ÃTheforwardrecursion(Equation15.3)showsthatfÎ û ° canbecomputedfrom fÎ û °=± Î ande° , which canin turn be computedfrom fÎ û °=± Ï ande°�± Î , andso on down to fÎ û M andeÎ. Hence,fÎ û ° canbecomputedfrom fÎ û M andeÎ û ° . Thebackwardrecursion(Equation15.7)shows that b° Ê Î û à canbe computedfrom b° Ê Ï û à ande° Ê Î , which in turn canbecomputedfrom b° Ê Ð û à ande° Ê Ï , andsoon up to b;oÊ Î û à ande; . Hence,b° Ê Î û à canbecomputedfrom b;oÊ Î û à ande° Ê Î û ; . Combiningthesetwo, we find that PÆX ° �eÎ û Ã É canbecomputedfrom fÎ û M , b;QÊ Î û à , andeÎ û ; .b. The reasoningfor the secondhalf is essentiallyidentical: for�betweenóand ÷ ,PÆX ° � eÎ û Ã É canbecomputedfrom fÎ û ; , bÃ,Ê Î û à , ande;oÊ Î û à .c. The algorithmtakes3 arguments:an evidencesequence,an initial forward message,andafinal backwardmessage.Theforwardmessageis propagatedto thehalfwaypointand the backward messageis propagatedbackward. The algorithm then calls itselfrecursively on the two halves of the evidencesequencewith the appropriateforwardandbackwardmessages.Thebasecaseis asequenceof length1 or 2.d. At eachlevel of the recursionthealgorithmtraversestheentiresequence,doing û Æ ÷ Éwork. Thereare û Æ.- /10 Ï ÷ É levels,so the total time is û Æ ÷ - /10 Ï ÷ É . Thealgorithmdoesa depth-firstrecursion,so the total spaceis proportionalto the depthof the stack,i.e., û Æ.- /10 Ï ÷ É . Withôislands,the recursiondepthis û Æ.- /10 ¹ ÷ É , so the total time isû Æ ÷ - /10 ¹ ÷ É but thespaceis û Æ�ôq- /10 ¹ ÷ É .http://librosysolucionarios.net104 Chapter 15. ProbabilisticReasoningoverTime15.4 This is averygoodexercisefor deepeningintuitionsabouttemporalprobabilisticrea-soning.First, noticethattheimpossibilityof thesequenceof mostlikely statescannotcomefrom animpossibleobservationbecausethesmoothedprobabilityat eachtime stepincludestheevidencelikelihoodat that time stepasa factor. Hence,the impossibilityof a sequencemust arisefrom an impossibletransition. Now considersucha transitionfrom0 ° ä Í to0 ° Ê Î ä�r for someÍ,r,�. For0 ° Ê Î ä\r to bethemostlikely stateat time� Y Ê , eventhoughit cannotbereachedfrom themostlikely stateattime�, wecansimplyhaveanô-statesystemwhere,say, thesmoothedprobabilityof0 ° ä Í isÆ�Ê Y Æ�ô¤^OÊ#É.t«É ì ô andtheremainingstateshaveprobabilityÆ�ÊT^ut,É ì ô . Theremainingstatesall transitiondeterministicallyto0 ° Ê Î ä�r .Fromhere,it is asimplematterto work outaspecificmodelthatbehavesasdesired.15.5a. Lookingat thefragmentof themodelcontainingjust� M , X M , andXÎ, we havePÆXÎ É5ä °Æwv ý Î� Æ�� M Éx v� ÆxM É PÆ'0 Î � xM È�� M ÉFrom the propertiesof the Kalmanfilter, we know that the integral givesa Gaussianfor eachdifferent value of� M . Hence,the predictiondistribution is a mixture of�Gaussians,eachweightedby� Æ�� M É .b. Theupdateequationfor theswitchingKalmanfilter isPÆX Ã,Ê Î È � Ã,Ê Î � eÎ û Ã,Ê Î Éä éPÆeÃ,Ê Î �X Ã,Ê Î È � Ã,Ê Î É°Æ H ý Î xH PÆ xà È�� à �eÎ û Ã É PÆ X Ã,Ê Î È � Ã,Ê Î � xà È�� à Éä éPÆeÃ,Ê Î �X Ã,Ê Î É°Æ H ý Î� Æ�� à �eÎ û Ã É PÆ � Ã,Ê Î � � à ÉxH PÆ xà �eÎ û Ã É PÆ X Ã,Ê Î � xà È�� à ÉWe aregiven thatPÆxà �eÎ û Ã É is a mixtureofÿGaussians.EachGaussianis subjectto�differentlinear–Gaussianprojectionsandthenupdatedby a linear-Gaussianobserva-tion, sowe obtainasumof� ÿGaussians.Thus,after ÷ stepswe have� ÃGaussians.c. Eachweight representsthe probability of one of the� Ãsequencesof valuesfor theswitchingvariable.15.6 This is asimpleexercisein algebra.Wehave� Æ�î Î � ñ Î ÉLä é ø ± �� xzy � � Û �`{ �| �y ø ± �� x Û � �~} v { �| �v9� | �Ûä é ø ± �� x | �v � | �Û { x�y � � Û �`{ � � | �y x Û � �~} v { �| �y x | �v � | �Û {ä é ø ±��� x | �v � | �Û { x�y �� � � y � Û � � Û � � { � | �y x Û � � � � } v Û � � } �v {| �y x | �v � | �Û {ä é ø ±��� x | �v � | �Û � | �y { Û � � � � xMx | �v � | �Û { y � � | �y } v { Û � �>�| �y x | �v � | �Û {http://librosysolucionarios.net105012345670 2 4 6 8 10 12Variance of posterior distribution�Number of time stepssx2=0.1, sz2=0.1sx2=0.1, sz2=1.0sx2=0.1, sz2=10.0sx2=1.0, sz2=0.1sx2=1.0, sz2=1.0sx2=1.0, sz2=10.0sx2=10.0, sz2=0.1sx2=10.0, sz2=1.0sx2=10.0, sz2=10.0FigureS15.2 Graphfor Ex. 15.7,showing theposteriorvariance�1�� asa functionof ì forvariousvaluesof �1�� and �1�� .ä é A ø± �� x Û � � x |�v � | �Û { y � � | �y } v| �v � | �Û � | �y { �x | �v9� | �Û { | �y�� x | �v!� | �Û � | �y { Ë15.7a. SeeFigureS15.2.b. Wecanfind afixedpointby solving" Ï ä Æ " Ï Y " ÏÜ É " Ï�" Ï Y " ÏÜ Y " Ï�for " Ï . Usingthequadraticformulaandrequiring " Ï R�Ç , we obtain" Ï ä ^ " ÏÜ Y " ¬Ü Y 3 " ÏÜ " Ï�" Ï�Weomit theproofof convergence,which,presumably, canbedoneby showing thattheupdateis a contraction(i.e., afterupdating,two differentstartingpointsfor " à becomecloser).c. As " ÏÜ ? Ç, we seethat thefixedpoint " Ï ? Çalso. This is because" ÏÜ ä±Çimpliesa deterministicpathfor theobject. Eachobservation suppliesmoreinformationaboutthispath,until its parametersareknown completely.http://librosysolucionarios.net106 Chapter 15. ProbabilisticReasoningoverTimeAs " Ï� ?©Ç, thevarianceupdategives " Ã,Ê Î ? Çimmediately. Thatis, if wehaveanexactobservation of theobject’s state,thentheposterioris a deltafunctionaboutthatobservedvalueregardlessof thetransitionvariance.15.8 Thereis oneclass,ÀAÕ'�h� ". Eachelementof the classhasa" Ù t Ù%� attribute withôpossiblevaluesandanu~� " ��ÖE� t Ù�Õ'uv& attributewhichmaybediscreteor continuous.Theparentof theu~� " ��ÖE� t Ù�Õ.uv& attribute is the" Ù t Ù%� attribute of thesametime step.Eachtime stephas$ Ö`�«Òv�,×W� "W" u�Örelationto anothertime step,andthe parentof the" Ù t Ù%� attribute is the" Ù t Ù%�attributeof the$ Ö`�«Òo�«×W� "W" u4Ötimestep.15.9a. The curve of interestis the onefor q ÆWw t Ù�Ù%�-Ö�� à ��Ë�Ë�Ë�4=4=4=4#Ç#Ç#Ç#Ç#Ç#Ç5Ë�Ë�Ë9É . In the absenceof any usefulsensorinformationfrom the batterymeter, the posteriordistribution forthebatterylevel is thesameastheprojectionwithout evidence.The transitionmodelfor thebatteryincludesasmallprobabilityfor downwardtransitionsin thebatterylevelat eachtime step,but zeroprobability for upward transitions(thereareno rechargingactionsin themodel).Thus,thestationarydistribution towardswhich thebatteryleveltendshasvalue0 with probability 1. The curve for q ÆWw t Ù�Ù%��ÖE� à ��Ë�Ë�Ë*4=4=4=4#Ç#Ç#Ç#Ç#Ç#Ç5Ë�Ë�Ë9Éwill asymptoteto 0.b. SeeFigureS15.3. The CPT forw��¦��Ù%��Ö Îhasa probability of transientfailure (i.e.,reporting0) thatincreaseswith temperature.c. The agentcanobviously calculatethe posteriordistribution overÀ���� $ à by filteringthe observation sequencein the usualway. This posteriorcan be informative if theeffect of temperatureon transientfailureis non-negligible andtransientfailuresoccurmorefrequentlythando majorchangesin temperature.Essentially, the temperatureisestimatedfrom thefrequency of “blips” in thesequenceof batterymeterreadings.15.10 Theprocessworksexactlyason page507.Westartwith thefull expression:PÆ / Ð � � Î È`� Ï È`� Ð ÉGä é þ � þ �� Æ�ù Î É � Æ'� Î � ù Î É � Æ�ù Ï � ù Î É � Æ'� Ï � ù Ï ÉPÆ / Ð � ù Ï É PÆ'� Ð � / Ð ÉWhichever orderwe pushin thesummations,thevariableeliminationprocessnever createsfactorscontainingmorethantwo variables,which is thesamesizeastheCPTsin theoriginalnetwork. In fact, given an HMM sequenceof arbitrary length,we caneliminatethe statevariablesin any order.15.11 Themodelis shown in FigureS15.4.Thekey pointto noteis thatany phonevariationatonepoint (here,[aa]vs.[ey] in thefourthphone)resultsin variationatthreepointsbecauseof theeffectson theprecedingandsucceedingphones.15.12 The [C1,C2] and[C6,C7] mustcomefrom theonsetandend,respectively. TheC3couldcomefrom 2 sources,theonsetor themid, andthe[C4, C4] combinationcouldcomefrom 3 sources:mid-mid, mid-end,or end-end.You can’t go directly from onsetto end,sotheonly possiblepaths(alongwith theirpathandoutputprobabilities)areasfollows:http://librosysolucionarios.net1071BatteryBattery0�1BMeter0�BMBroken 1BMBroken0�Temp 1TempFigureS15.3 Modificationof Figure15.13(a)to includetheeffectof externaltemperatureon thebatterymeter.t(_,ow) ow(t,m)m(ow.ey)m(ow,aa)ey(m,t)aa(m,t)t(ey,ow)ow(t,_)t(aa,ow)FigureS15.4 Transitiondiagramfor thetriphonepronunciationmodelin Ex. 15.11.C1 C2 C3 C4 C4 C6 C7OOOMMEE(* .3 .3 .7 .9 .1 .4 .6 .5 .2 .3 .7 .7 .5 .4) = 4.0e-6OOOMEEE(* .3 .3 .7 .1 .4 .4 .6 .5 .2 .3 .7 .1 .5 .4) = 2.5e-7OOMMMEE(* .3 .7 .9 .9 .1 .4 .6 .5 .2 .2 .7 .7 .5 .4) = 8.0e-6OOMMEEE(* .3 .7 .9 .1 .4 .4 .6 .5 .2 .2 .7 .1 .5 .4) = 5.1e-7OOMEEEE(* .3 .7 .1 .4 .4 .4 .6 .5 .2 .2 .1 .1 .5 .4) = 3.2e-8Sothemostprobablepathis OOMMMEE (that is, onset-onset-mid-mid-mid-end-end), withaprobabilityof �1�c�����1�~� � . Thetotalprobabilityof thesequenceis thesumof thefivepaths,or �1�c�����1� ��� .http://librosysolucionarios.netSolutionsfor Chapter16MakingSimpleDecisions16.1 It is interestingto createa histogramof accuracy on this taskfor thestudentsin theclass.It is alsointerestingto recordhow many timeseachstudentcomeswithin, say, 10%oftheright answer. Thenyou geta profile of eachstudent:this oneis anaccurateguesserbutoverly cautiousaboutbounds,etc.16.2 Theexpectedmonetaryvalueof thelottery � is������ �¢¡¤£ �¥ � ��¦1�1��§ �¨ �1�1�1�1�1� �©¦1�1�1�1�1�1�1��£ª¦1�1�c«1�Although ¦1�1�c«1�T¬­¦1� , it is not necessarilyirrationalto buy theticket. First we will considerjust theutilities of themonetaryoutcomes,ignoringtheutility of actuallyplayingthelotterygame.Using ® �.¯k°(±�² ¡ to representtheutility to theagentof having ³ dollarsmorethanthecurrentstate,andassumingthatutility is linear for smallvaluesof money (i.e., ® �.¯k°(±�² ¡µ´³ � ® �.¯k°(±<¶ ¡<·¸® �.¯° ¡.¡ for ·¹�1�¹º»³¼º½�1� ), theutility of thelottery is:® � ��¡u£ �¥ � ®�.¯°1±<¶¿¾ ¡À§ �¨1Á �1�1� Á �1�1� ®�.¯k°(±<¶! ¾.¾.¾s ¾.¾.¾ ¡´ �¥ ® �.¯°1±<¶ ¡À§ �¨1Á �1�1� Á �1�1� ®�.¯°1±<¶! ¾.¾.¾s ¾.¾.¾ ¡This is morethan ® �.¯°1±<¶ ¡ when ® �.¯°1±<¶! ¾.¾.¾s ¾.¾.¾ ¡¤Ã�� ÁoÄ �1� Á �1�1�1® � ¦1�1¡ . Thus,for a purchaseto be rational(whenonly money is considered),theagentmustbe quite risk-seeking.Thiswouldbeunusualfor low-incomeindividuals,for whomthepriceof a ticket is non-trivial. Itis possiblethatsomebuyersdo not internalizethemagnitudeof thevery low probabilityofwinning—toimagineaneventis to assignit a“non-trivial” probability, in effect. Apparently,thesebuyersare betterat internalizingthe large magnitudeof the prize. Suchbuyersareclearlyactingirrationally.Somepeoplemayfeeltheircurrentsituationis intolerable,thatis, ® �.¯° ¡�´½® �.¯°1Å<¶ ¡Q´ÆSÇ . Thereforethesituationof having onedollar moreor lesswould beequallyintolerable,andit wouldberationalto gambleonahigh payoff, evenif onethathaslow probability.Gamblersalso derive pleasurefrom the excitementof the lottery and the temporarypossessionof at leastanon-zerochanceof wealth.Soweshouldaddto theutility of playingthelottery theterm È to representthethrill of participation.Seenthis way, thelottery is justanotherform of entertainment,andbuying a lottery ticket is no moreirrational thanbuying108http://librosysolucionarios.net109a movie ticket. Either way, you payyour money, you geta small thrill È , and(mostlikely)youwalk awayempty-handed.(Notethatit couldbearguedthatdoingthis kind of decision-theoreticcomputationdecreasesthevalueof È . It is not clearif this is a goodthing or a badthing.)16.3a. Theprobabilitythatthefirst headsappearson the ³ th tossis¨ � ² , so�É�­�� ��¡Q£Ê²¤ËF¶ ¨ �²ÉÌ ¨ ² £Ê²QË�¶ �É£½Íb. Typicalanswersrangebetween$4and$100.c. Assumeinitial wealth(afterpaying Î to play thegame)of ¦ �.Ï ·¸Î�¡ ; then® � �¢¡¤£Ê²QË�¶ ¨ �² Ì �.ÐÒÑ Ó1Ô�Õ~�.Ï ·¸ÎQ§ ¨ ² ¡À§¸Ö�¡AssumeÏ ·»Î×£½¦1� for simplicity. Then® � �¢¡©£Ê²QË�¶ ¨ �² Ì �.Ð¢Ñ Ó1Ô Õ � ¨ ² ¡<§¸Ö�¡£Ê²QË�¶ ¨ �²�Ì Ð ³f§ØÖ£ ¨ Ð §¸Öd. ThemaximumamountÎ is givenby thesolutionofÐÒÑ Ó1Ô�Õ¤Ï §¸ÖףʲQËF¶ ¨ �² Ì �.ÐÒÑ Ó1Ô�Õ~�.Ï ·ØÎQ§ ¨ ² ¡À§ØÖ�¡For oursimplecase,we haveÐÒÑ Ó1Ô�Õ ÎQ§»Ö×£ ¨ Ð §¸Öor ÎZ£u¦�Ù .16.4 This is aninterestingexerciseto do in class.Choose� ¶ £�¦1�1�1� , � Õ = $100,$1000,$10000,$1000000.Ask for ashow of handsof thosepreferringthelotteryatdifferentvaluesof Ú . Studentswill almostalwaysdisplayrisk aversion,but theremaybeawide spreadin itsonset.A curve canbeplottedfor theclassby finding thesmallestÚ yielding a majority votefor thelottery.16.5 Theprogramitself is pretty trivial. But notethat therearesomestudiesshowing youget betteranswersif you asksubjectsto move a slider to indicatea proportion,ratherthanaskingfor a probabilitynumber. So having a graphicaluserinterfaceis an advantage.Themainpoint of theexerciseis to examinethedata,exposeinconsistentbehavior on thepartofthesubjects,andseehow peoplevary in their choices.16.6 The protocol would be to ask a seriesof questionsof the form “which would youprefer” involving amonetarygain(or loss)versusanincrease(or decrease)in a risk of death.http://librosysolucionarios.net110 Chapter 16. Making SimpleDecisionsFor example,“would you pay$100for ahelmetthatwould eliminatecompletelytheone-in-a-million chanceof deathfrom abicycle accident.”16.7 Thecompleteproof is givenby Keeney andRaiffa (1976).16.8 This exercisecanbesolvedusingan influencediagrampackagesuchasIDEAL. Thespecificvaluesarenotespeciallyimportant.Noticehow thetediumof encodingall theentriesin the utility tablecriesout for a systemthat allows the additive, multiplicative, andotherformssanctionedby MAUT.Oneof thekey aspectsof thefully explicit representationin Figure16.5is its amenabil-ity to change.By doing this exerciseaswell asExercise16.9,studentswill augmenttheirappreciationof the flexibility affordedby declarative representations,which canotherwiseseemtedious.a. For this part,onecouldusesymbolicvalues(high, medium,low) for all thevariablesandnot worry too much aboutthe exact probability values,or one could useactualnumericalrangesandtry to assessthe probabilitiesbasedon someknowledgeof thedomain.Evenwith three-valuedvariables,thecostCPThas54 entries.b. Thispartalmostcertainlyshouldbedoneusingasoftwarepackage.c. If eachaircraftgenerateshalf asmuchnoise,weneedto adjusttheentriesin the ÛWÜwÝwÞ5ßCPT.d. If thenoiseattributebecomesthreetimesmoreimportant,theutility tableentriesmustall be altered. If an appropriate(e.g., additive) representationis available, then onewouldonly needto adjusttheappropriateconstantsto reflectthechange.e. This part shouldbe doneusing a software package.Somepackagesmay offer VPIcalculationalready. Alternatively, onecaninvoke thedecision-makingpackagerepeat-edly to do all thewhat-if calculationsof bestactionsandtheir utilities, asrequiredinthe VPI formula. Finally,onecanwrite general-purposeVPI codeasan add-onto adecision-makingpackage.16.9 Theinformationassociatedwith theutility nodein Figure16.6is anaction-valuetable,andcanbeconstructedsimply by averagingout the àáß Ð ÈoâkÞ , ÛãÜ9Ý�Þsß , and ä�Ü(Þ!È nodesin Fig-ure16.5.As explainedin thetext , modificationsto aircraftnoiselevelsor to theimportanceof noisedo not resultin simplechangesto theaction-valuetable. Probablytheeasiestwayto do it is to gobackto theoriginal tablein Figure16.5.Theexercisethereforeillustratesthetradeoffs involvedin usingcompiledrepresentations.16.10 The answerto this exercisedependson the probability valueschosenby the stu-dent.16.11 Thisquestionis asimpleexercisein sequentialdecisionmaking,andhelpsin makingthe transitionto Chapter17. It alsoemphasizesthe point that the valueof information iscomputedby examiningtheconditionalplanformedby determiningthebestactionfor eachpossibleoutcomeof the test. It may be usefulto introduce“decisiontrees”(asthe term isusedin thedecisionanalysisliterature)to organizetheinformationin thisquestion.(SeePearl(1988),Chapter6.) Eachpartof thequestionanalyzessomeaspectof thetree. Incidentally,http://librosysolucionarios.net111TestBuyUOutcomeQualityFigureS16.1 A decisionnetwork for thecar-buyingproblem.the questionassumesthat utility andmonetaryvaluecoincide,and ignoresthe transactioncostsinvolvedin buying andselling.a. Thedecisionnetwork is shown in FigureS16.1.b. Theexpectednetgainin dollarsiså¹�.æ ± ¡ � ¨ �1�1�F·Ø� ¥ �1�1¡I§ å¹�.æ � ¡ � ¨ �1�1�p· ¨1¨ �1�1¡�£½�1�c«� ¥ �1��§¸�1�c����· ¨ �1�É£ ¨1ç �c. Thequestioncouldprobablyhavebeenstatedbetter:Bayes’theoremisusedto computeå¹�.æ ±�è åÉÐ Þ5Þs¡ , etc.,whereasconditionalizationis sufficient to computeå¹�.å¹Ð ÞsÞ5¡ .å¹�.å¹Ð Þ5Þs¡Q£ åÉ�.å¹Ð Þ5Þ è æ ± ¡ å¹�.æ ± ¡À§ åÉ�.å¹Ð Þ5Þ è æ � ¡ å¹�.æ � ¡£ª�1�c�����1�c«µ§¸�1�c� ¥ ���1�c�¹£½�1� Ä1Ä1¥UsingBayes’theorem:å¹�.æ ± è å¹Ð ÞsÞ5¡�£å¹�.å¹Ð Þ5Þ è æ ± ¡ å¹�.æ ± ¡å¹�.åÉÐ Þ5Þs¡ £ �1�c�����1�c«�1� Ä1Ä1¥ ´½�1�c��Ù ¨ �å¹�.æ � è å¹Ð ÞsÞ5¡�´é�µ·»�1�c��Ù ¨ ��£½�1�c� ¥ « çå¹�.æ ± è ê å¹Ð ÞsÞ5¡�£å¹� ê åÉÐ Þ5Þ è æ ± ¡ å¹�.æ ± ¡å¹� ê å¹Ð Þ5Þ5¡ £ �1� ¨ �©�1�c«�1�c�1� ¥ ´½�1� Ùk�1« çå¹�.æ � è ê å¹Ð ÞsÞ5¡�´é�µ·»�1� Ùk�1« ç £½�1� ¥ � ¨ �d. If thecarpassesthetest,theexpectedvalueof buying iså¹�.æ ± è å¹Ð ÞsÞ5¡ � ¨ �1�1��·¸� ¥ �1�1¡i§ å¹�.æ � è å¹Ð ÞsÞ5¡ � ¨ �1�1��· ¨1¨ �1�1¡£ë�1�c��Ù ¨ �T� ¥ �1�µ§¸�1�c� ¥ « ç ��· ¨ �1�¹£ª�1«1�1� ç1¨Thusbuying is thebestdecisiongivenapass.If thecarfails thetest,theexpectedvalueof buying iså¹�.æ ± è ê å¹Ð ÞsÞ5¡ � ¨ �1�1��·Ø� ¥ �1�1¡I§ å¹�.æ � è ê å¹Ð Þ5Þs¡ � ¨ �1�1�F· ¨1¨ �1�1¡£ë�1� Ùk�1« ç � ¥ �1�µ§¸�1� ¥ � ¨ �T��· ¨ �1�¹£ ç1¨ � ¥ �Buying is againthebestdecision.http://librosysolucionarios.net112 Chapter 16. Making SimpleDecisionse. Sincetheactionis thesamefor bothoutcomesof thetest,thetestit*elf is worthless(ifit is theonly possibletest)andtheoptimalplanis simply to buy thecarwithout thetest.(This is a trivial conditionalplan.) For the testto be worthwhile, it would needto bemorediscriminatingin orderto reducetheprobabilityå¹�.æ ±×è ê å¹Ð ÞsÞ5¡ . The testwouldalsobeworthwhileif themarketvalueof thecarwereless,or if thecostof repairsweremore.An interestingadditionalexerciseis to prove thegeneralpropositionthatif ì is thebestactionfor all theoutcomesof a testthenit mustbe thebestactionin theabsenceof thetestoutcome.16.12 Intuitively, the value of information is nonnegative becausein the worst caseonecouldsimplyignoretheinformationandactasif it wasnotavailable.A formalproofthereforebegins by showing that this policy resultsin thesameexpectedutility. The formula for thevalueof informationis�å¹í.î��.�¢ï ¡¤£ ° åÉ�.��ï £©ß ï ° è � ¡ � ® � ì�ð!ñmò è � Á �¢ï £¼ß ï ° ¡ · � ® � ì è � ¡If theagentdoesì giventheinformation��ï, its expectedutility is° åÉ�.� ï £¼ß ï ° è � ¡ � ® � ì è � Á � ï £©ß ï ° ¡Q£ � ® � ì è � Á � ï £uß ï ° ¡wherethe equality holds becausethe LHS is just the conditionalizationof the RHS withrespectto��ï. By definition,� ® � ì�ð ñrò è � Á ��ï £uß ï ° ¡¤ó � ® � ì è � Á ��ï £uß ï ° ¡hence��å¹í.î×�.��ï ¡�ó½� .16.13 This is relatively straightforward in theAIMA2e coderelease.We needto addnodetypesfor actionnodesandutility nodes;we needto beableto run standardBayesnet infer-enceonthenetwork givenfixedactions,in ordertocomputetheposteriorexpectedutility; andwe needto write an“outer loop” thatcantry all possibleactionsto find thebest.Giventhis,addingVPI calculationis straightforward,asdescribedin theanswerto Exercise16.8.http://librosysolucionarios.netSolutionsfor Chapter17MakingComplex Decisions17.1 Thisquestionhelpsto bring homethedifferencebetweendeterministicandstochasticenvironments.Here,evena shortsequencespreadstheagentall over theplace.Theeasiestway to answerthequestionsystematicallyis to draw a treeshowing thestatesreachedaftereachstepandthe transitionprobabilities. Thenthe probability of reachingeachleaf is theproductof theprobabilitiesalongthepath,becausethetransitionprobabilitiesareMarkovian.If the samestateappearsat morethanoneleaf, the probabilitiesof the leavesaresummedbecausetheeventscorrespondingto the two pathsaredisjoint. (It is alwaysa goodideatoensurethat studentsjustify thesekinds of probabilisticcalculations,especiallysincesome-timesthe “naive” approachgetsthe wronganswer.) Thestatesandprobabilitiesare: (3,1),0.01;(3,2),0.08;(3,3),0.09;(4,2),0.18;(4,3),0.64.17.2 Stationarityrequirestheagentto have identicalpreferencesbetweenthesequencepairô Þ ¾ Á Þ ¶ Á Þ Õ Á �o�o��õ Á ô Þ ¾ Á Þ9ö ¶ Á ÞwöÕ~Á �o�o� õ andbetweenthesequencepairô Þ ¶ Á Þ Õ Á �o�o��õ Á ô Þwö ¶ Á Þ9öÕ~Á �o�o� õ . If theutility of a sequenceis its maximumreward,wecaneasilyviolatestationarity. For example,ô Ù Á � Á � Á � Á � Á �o�o�cõI÷ ô Ù Á � Á � Á � Á � Á �o�o�Mõbut ô � Á � Á � Á � Á �o�o�øõiù ô � Á � Á � Á � Á �o�o�øõZ�Wecanstill define ®�ú � Þs¡ astheexpectedmaximumrewardobtainedby executingû startingin Þ . The agent’s preferencesseempeculiar, nonetheless.For example,if the currentstateÞ hasreward ü max, the agentwill be indifferent amongall actions,but oncethe action isexecutedandthe agentis no longer in Þ , it will suddenlystart to careaboutwhat happensnext.17.3 A finite searchproblem(seeChapter3) is definedby an initial state Þ ¾ , a successorfunction¯¤� Þs¡ thatreturnsasetof action–statepairs,astepcostfunction Î � Þ Á Ð Á Þ ö ¡ , andagoaltest.An optimalsolutionis a least-costpathfrom Þ ¾ to any goalstate.To constructthe correspondingMDP, define ü � Þ Á Ð Á Þ9öý¡O£þ·ÿÎ � Þ Á Ð Á Þwö&¡ unless Þ is agoalstate,in which caseü � Þ Á Ð Á Þ ö ¡O£u� (seeEx. 17.5for how to obtaintheeffect of a three-argumentreward function); define � � Þ Á Ð Á Þ9öý¡O£u� if�.Ð Á Þwöý¡�� ¯¤� Þ5¡ ; and ��£¼� . An optimalsolutionto this MDP is a policy thatfollows theleast-costpathfrom eachstateto its nearestgoalstate.17.4113http://librosysolucionarios.net114 Chapter 17. Making Complex Decisionsa. Intuitively, the agentwantsto get to state3 assoonaspossible,becauseit will pay acostfor eachtimestepit spendsin states1 and2. However, theonly actionthatreachesstate3 (action Ö ) succeedswith low probability, sotheagentshouldminimizethecostit incurswhile trying to reachthe terminalstate. This suggeststhat the agentshoulddefinitelytry action Ö in state1; in state2, it mightbebetterto try actionÐto getto state1 (which is thebetterplaceto wait for admissionto state3), ratherthanaimingdirectlyfor state3. Thedecisionin state2 involvesanumericaltradeoff.b. Theapplicationof policy iterationproceedsin alternatingstepsof valuedeterminationandpolicy update.� Initialization: ®����¿·É� Á · ¨1Á � , å ���¿Ö Á Ö� .� Valuedetermination:Æ ¶ £ë·¹�µ§¸�1�c� Æ� §¸�1� ç Æ ¶Æ Õ £ë· ¨ §¸�1�c� Æ � §¸�1� ç Æ ÕÆ� £ë�Thatis, Æ ¶ £ª·¹�1� andÆ Õ £½· ¨ � .Policy update: In state1,ï � � � Á Ð Á�� ¡ Æ ï £½�1�c���©· ¨ �µ§¸�1� ¨ �©·¹�1�¹£½·¹�1�whileï � � � Á Ö Á�� ¡ Æ ï £½�1�c���������1� ç �©·¹�1�¹£½· çsoaction Ö is still preferredfor state1.In state2,ï � � � Á Ð Á�� ¡ Æ ï £½�1�c���©·¹�1�µ§¸�1� ¨ �©· ¨ �¹£½·¹� ¨whileï � � � Á Ö Á�� ¡ Æ ï £½�1�c���������1� ç �©· ¨ �¹£½·¹�1�soactionÐis preferredfor state1. Weset �������������������� �"! Ñ #%$ andproceed.� Valuedetermination:Æ ¶ £ë·¹�µ§¸�1�c� Æ� §¸�1� ç Æ ¶Æ Õ £ë· ¨ §¸�1�c� Æ ¶ §¸�1� ¨ Æ ÕÆ� £ë�OncemoreÆ ¶ £½·É�1� ; now, Æ Õ £ª·¹� ¥ . Policy update: In state1,ï � � � Á Ð Á�� ¡ Æ ï £½�1�c���©·¹� ¥ §¸�1� ¨ �©·¹�1�¹£½·¹��Ùwhileï � � � Á Ö Á�� ¡ Æ ï £½�1�c���������1� ç �©·¹�1�¹£½· çhttp://librosysolucionarios.net115soaction Ö is still preferredfor state1.In state2,ï � � � Á Ð Á�� ¡ Æ ï £½�1�c���©·¹�1�µ§¸�1� ¨ �©·¹� ¥ £½·¹�1�whileï � � � Á Ö Á�� ¡ Æ ï £½�1�c���������1� ç �©·¹� ¥ £½·¹�1�1� ¥soactionÐis still preferredfor state1. �&�'�(�����'�(���� remains)+*�, $ , andwe termi-nate.Notethattheresultingpolicy matchesour intuition: whenin state2, try to moveto state1, andwhenin state1, try to move to state3.c. An initial policy with actionÐin bothstatesleadsto anunsolvableproblem.TheinitialvaluedeterminationproblemhastheformÆ ¶ £é·É�ק¸�1� ¨ Æ ¶ §Ø�1�c� Æ ÕÆ Õ £é· ¨ §¸�1�c� Æ ¶ §Ø�1� ¨ Æ ÕÆ-� £é�andthefirst two equationsareinconsistent.If we wereto try to solve themiteratively,wewouldfind thevaluestendingto ·¹Í .Discountingleadsto well-definedsolutionsby boundingthepenalty(expecteddis-countedcost)anagentcanincur at eitherstate.However, thechoiceof discountfactorwill affect the policy that results. For � small, the cost incurredin the distantfutureplays a negligible role in the value computation,because� ² is near0. As a result,anagentcouldchooseaction Ö in state2 becausethediscountedshort-termcostof re-mainingin thenon-terminalstates(states1 and2) outweighsthediscountedlong-termcostof action Ö failing repeatedlyandleaving theagentin state2. An additionalexer-cisecouldaskthestudentto determinethevalueof � at which theagentis indifferentbetweenthetwo choices.17.5 This is a deceptively simpleexercisethat teststhestudent’s understandingof thefor-maldefinitionof MDPs.Somestudentsmayneedahint or anexampleto getstarted.a. Thekey hereis to getthemaxandsummationin theright place.For ü � Þ Á Ð ¡ we have® � Þ5¡Q£/.0!213 ô ü � Þ Á Ð ¡<§4� 5�67� � Þ Á Ð Á Þ ö ¡.® � Þ ö ¡andfor ü � Þ Á Ð Á Þwöý¡ we have® � Þ5¡Q£/.0!213 5�67� � Þ Á Ð Á Þ ö ¡ ô ü � Þ Á Ð Á Þ ö ¡I§8�i® � Þ ö ¡¿õZ�b. Therearea variety of solutionshere. One is to createa “pre-state” Ú�9ß � Þ Á Ð Á Þ ö ¡ forevery Þ , Ð , Þ9ö , suchthatexecutingÐin Þ leadsnot to Þ9ö but to Ú�9ß � Þ Á Ð Á Þ9ö&¡ . In this stateis encodedthefactthattheagentcamefrom Þ anddidÐto gethere.Fromthepre-state,http://librosysolucionarios.net116 Chapter 17. Making Complex Decisionsthereis just oneaction Ö thatalwaysleadsto Þ ö . Let thenew MDP have transition� ö ,reward ü�ö , anddiscount��ö . Then� ö � Þ Á Ð Á Ú�9ß � Þ Á Ð Á Þ ö ¡.¡¤£:� � Þ Á Ð Á Þ ö ¡��ö � Ú�9ß � Þ Á Ð Á Þwöý¡ Á Ö Á Þ9öý¡Q£½�ü ö � Þ Á Ð ¡¤£½�ü ö � Ú�9ß � Þ Á Ð Á Þ ö ¡ Á Ö�¡�£:� �<;= ü � Þ Á Ð Á Þ ö ¡��öZ£:� ;=c. In keepingwith theideaof part(b), we cancreatestatesÚ�Ü(Þ!È � Þ Á Ð ¡ for every Þ , Ð , suchthat� ö � Þ Á Ð Á Ú�Ü(Þ9È � Þ Á Ð Á Þ ö ¡.¡¤£½���ö � Ú�Ü(Þ!È � Þ Á Ð Á Þwö¿¡ Á Ö Á Þwöý¡¤£:� � Þ Á Ð Á Þwö&¡ü�ö � Þs¡Q£½�ü ö � Ú�Ü(Þ!È � Þ Á Ð Á Þ ö ¡.¡¤£>� �<;= ü � Þ Á Ð ¡� ö £:� ;=17.6 Theframework for thisproblemis in "uncertainty/d omai ns/ 4x 3- mdp. lis p" .Thereis still somesynthesisfor thestudentto do for answerb. For c. someexperimentalde-signis necessary.17.7 Thiscanbedonefairly simplyby:� Callpolicy-iteration (from "uncertainty/a lg or ith ms/d p. li sp" )ontheMarkov DecisionProcessesrepresentingthe4x3 grid, with valuesfor thestepcostrangingfrom, say, 0.0to 1.0 in incrementsof 0.02.� For any two adjacentpoliciesthatdiffer, run binarysearchon thestepcostto pinpointthethresholdvalue.� Convince yourselfthat you haven’t missedany policies,eitherby usingtoo coarseanincrementin stepsize(0.02),or by stoppingtoosoon(1.0).Oneuseful observation in this context is that the expectedtotal reward of any fixedpolicy is linearin 9 , theper-steprewardfor theemptystates.Imaginedrawing thetotalrewardof a policy asa functionof 9 —a straightline. Now draw all thestraightlinescorrespondingto all possiblepolicies.Therewardof theoptimalpolicy asa functionof 9 is just themaxofall thesestraightlines. Thereforeit is a piecewise linear, convex functionof 9 . Hencethereis avery efficient way to find all theoptimalpolicy regions:� Forany two consecutivevaluesof 9 thathavedifferentoptimalpolicies,find theoptimalpolicy for themidpoint. Oncetwo consecutive valuesof 9 give thesamepolicy, thentheinterval betweenthetwo pointsmustbecoveredby thatpolicy.� Repeatthisuntil two pointsareknown for eachdistinctoptimalpolicy.� Suppose� 9 3 ¶ Á�? 3 ¶ ¡ and� 9 3 Õ Á�? 3 Õ ¡ arepointsfor policyÐ, and� 9(@ ¶ Á�? @ ¶ ¡ and� 9(@ Õ Á�? @ Õ ¡arethenext two points,for policy Ö . Clearly, we candraw straightlines throughthesepairsof pointsandfind their intersection.Thisdoesnotmean,however, thatthereis nootheroptimal policy for the interveningregion. We candeterminethis by calculatinghttp://librosysolucionarios.net117theoptimal policy for the intersectionpoint. If we geta differentpolicy, we continuetheprocess.The policiesandboundariesderived from this procedureareshown in FigureS17.1. Thefigure shows that therearenine distinct optimal policies! Notice that as 9 becomesmorenegative, the agentbecomesmorewilling to dive straightinto the –1 terminalstateratherthanfacethecostof thedetourto the+1 state.The somewhat ugly codeis asfollows. Notice that becausethe lines for neighboringpoliciesarevery nearlyparallel,numericalinstability is aseriousproblem.(defun policy-surface (mdp r1 r2 &aux prev (unchanged nil))"returns points on the piecewise-linear surfacedefined by the value of the optimal policy of mdp as afunction of r"(setq rvplist(list (cons r1 (r-policy mdp r1)) (cons r2 (r-policy mdp r2))))(do ()(unchanged rvplist)(setq unchanged t)(setq prev nil)(dolist (rvp rvplist)(let* ((rest (cdr (member rvp rvplist :test #’eq)))(next (first rest))(next-but-one (second rest)))(dprint (list (first prev) (first rvp)’* (first next) (first next-but-one)))(when next(unless (or (= (first rvp) (first next))(policy-equal (third rvp) (third next) mdp))(dprint "Adding new point(s)")(setq unchanged nil)(if (and prev next-but-one(policy-equal (third prev) (third rvp) mdp)(policy-equal (third next) (third next-but-one) mdp))(let* ((intxy (policy-vertex prev rvp next next-but-one))(int (cons (xy-x intxy) (r-policy mdp (xy-x intxy)))))(dprint (list "Found intersection" intxy))(cond ((or (< (first int) (first rvp))(> (first int) (first next)))(dprint "Intersection out of range!")(let ((int-r (/ (+ (first rvp) (first next)) 2)))(setq int (cons int-r (r-policy mdp int-r))))(push int (cdr (member rvp rvplist :test #’eq))))((or (policy-equal (third rvp) (third int) mdp)(policy-equal (third next) (third int) mdp))(dprint "Found policy boundary")(push (list (first int) (second int) (third next))(cdr (member rvp rvplist :test #’eq)))(push (list (first int) (second int) (third rvp))(cdr (member rvp rvplist :test #’eq))))(t (dprint "Found new policy!")(push int (cdr (member rvp rvplist :test #’eq))))))http://librosysolucionarios.net118 Chapter 17. Making Complex Decisions(let* ((int-r (/ (+ (first rvp) (first next)) 2))(int (cons int-r (r-policy mdp int-r))))(dprint (list "Adding split point" (list int-r (second int))))(push int (cdr (member rvp rvplist :test #’eq))))))))(setq prev rvp))))(defun r-policy (mdp r &aux U)(set-rewards mdp r)(setq U (value-iteration mdp(copy-hash-table (mdp-rewards mdp) #’identity):epsilon 0.0000000001))(list (gethash ’(1 1) U) (optimal-policy U (mdp-model mdp) (mdp-rewards mdp))))(defun set-rewards (mdp r &aux (rewards (mdp-rewards mdp))(terminals (mdp-terminal-states mdp)))(maphash #’(lambda (state reward)(unless (memberstate terminals :test #’equal)(setf (gethash state rewards) r)))rewards))(defun policy-equal (p1 p2 mdp &aux (match t)(terminals (mdp-terminal-states mdp)))(maphash #’(lambda (state action)(unless (member state terminals :test #’equal)(unless (eq (caar (gethash state p1)) (caar (gethash state p2)))(setq match nil))))p1)match)(defun policy-vertex (rvp1 rvp2 rvp3 rvp4)(let ((a (make-xy :x (first rvp1) :y (second rvp1)))(b (make-xy :x (first rvp2) :y (second rvp2)))(c (make-xy :x (first rvp3) :y (second rvp3)))(d (make-xy :x (first rvp4) :y (second rvp4))))(intersection-point (make-line :xy1 a :xy2 b)(make-line :xy1 c :xy2 d))))(defun intersection-point (l1 l2);;; l1 is line ab; l2 is line cd;;; assume the lines cross at alpha a + (1-alpha) b,;;; also known as beta c + (1-beta) d;;; returns the intersection point unless they’re parallel(let* ((a (line-xy1 l1))(b (line-xy2 l1))(c (line-xy1 l2))(d (line-xy2 l2))(xa (xy-x a)) (ya (xy-y a))(xb (xy-x b)) (yb (xy-y b))(xc (xy-x c)) (yc (xy-y c))(xd (xy-x d)) (yd (xy-y d))(q (- (* (- xa xb) (- yc yd))(* (- ya yb) (- xc xd)))))http://librosysolucionarios.net119(unless (zerop q)(let ((alpha (/ (- (* (- xd xb) (- yc yd))(* (- yd yb) (- xc xd)))q)))(make-xy :x (float (+ (* alpha xa) (* (- 1 alpha) xb))):y (float (+ (* alpha ya) (* (- 1 alpha) yb))))))))− 1+ 1− 1+ 1− 1+ 1− 1+ 1− 1+ 1− 1+ 1− 1+ 1− 1+ 1− 1+ 1r = [−1.6284 : −1.3702] r = [−1.3702 : −0.7083]r = [−0.7083 : −0.4278] r = [−0.4278 : −0.0850] r = [−0.0850 : −0.0480]r = [−0.0480 : −0.0274] r = [−0.0274 : −0.0218] r = [−0.0218 : 0.0000]r = [− : −1.6284]8Figure S17.1 Optimalpoliciesfor differentvaluesof thecostof a stepin the ACBED envi-ronment,andtheboundariesof theregionswith constantoptimalpolicy.17.8a. For ®GF we have® F � Þs¡O£¼ü � Þs¡<§H.I!213 3åÉ� Þ ö è Ð Á Þ5¡.®GJ � Þ ö ¡andfor ®GJ wehave®GJ � Þs¡O£uü � Þ5¡<§H.0K L3 3å¹� Þ ö è Ð Á Þs¡.® F � Þ ö ¡i�http://librosysolucionarios.net120 Chapter 17. Making Complex Decisions(1,4)(1,3)(1,2)(2,4)(2,3)(2,1)(3,4)(3,2)(3,1)(4,3)(4,2)−1 −1+1+1FigureS17.2 State-spacegraphfor thegameon page190(for Ex. 17.8).b. To dovalueiteration,wesimply turneachequationfrom part(a) into aBellmanupdateandapply themin alternation,applyingeachto all statessimultaneously. Theprocessterminateswhen the utility vector for one player is the sameas the previous utilityvectorfor thesameplayer (i.e., two stepsearlier). (Notethat typically ® F and ®MJ arenot thesamein equilibrium.)c. Thestatespaceis shown in FigureS17.2.d. Wemarktheterminalstatevaluesin boldandinitialize othervaluesto 0. Valueiterationproceedsasfollows:(1,4) (2,4) (3,4) (1,3) (2,3) (4,3) (1,2) (3,2) (4,2) (2,1) (3,1)® F 0 0 0 0 0 +1 0 0 +1 –1 –1®MJ 0 0 0 0 –1 +1 0 –1 +1 –1 –1® F 0 0 0 –1 +1 +1 –1 +1 +1 –1 –1® J –1 +1 +1 –1 –1 +1 –1 –1 +1 –1 –1® F +1 +1 +1 –1 +1 +1 –1 +1 +1 –1 –1®MJ –1 +1 +1 –1 –1 +1 –1 –1 +1 –1 –1andtheoptimalpolicy for eachplayeris asfollows:(1,4) (2,4) (3,4) (1,3) (2,3) (4,3) (1,2) (3,2) (4,2) (2,1) (3,1)ûONF (2,4) (3,4) (2,4) (2,3) (4,3) (3,2) (4,2)û NJ (1,3) (2,3) (3,2) (1,2) (2,1) (1,3) (3,1)17.9 This questionis simplea matterof examiningthedefinitions. In a dominantstrategyequilibriumô Þ ¶ Á �o�o� Á Þ ² õ , it is thecasethat for every player Ý , ÞQP is optimal for every combi-nationÈ � P by theotherplayers:R Ý R È � P R Þ ö P ô ÞQP Á È � P\õ ù÷ ô Þ ö P Á È � Pgõ<�In a Nashequilibrium,we simply requirethat Þ2P is optimal for theparticularcurrentcombi-nation Þ � P by theotherplayers:R Ý R Þ ö P ô Þ2P Á Þ � Pgõ ù÷ ô Þ ö P Á Þ � P\õ<�Therefore,dominantstrategy equilibriumis aspecialcaseof Nashequilibrium.Theconversedoesnothold,aswecanshow simplyby pointingto theCD/DVD game,whereneitherof theNashequilibriais adominantstrategy equilibrium.http://librosysolucionarios.net12117.10 In the following table, the rows are labelledby A’s move andthe columnsby B’smove,andthetableentrieslist thepayoffs to A andB respectively.R P S F WR 0,0 –1,1 1,-1 –1,1 1,-1P 1,-1 0,0 –1,1 –1,1 1,-1S –1,1 1,-1 0,0 –1,1 1,-1F 1,-1 1,-1 1,-1 0,0 –1,1W –1,1 –1,1 –1,1 1,-1 0,0SupposeS choosesamixedstrategyô 9UTkü�V�ÚWT å V<ÞXT ¯ VZY[T'\EV�]^T'_@õ , where9¹§ãÚ�§¸ÞÒ§Y§8]»£u� . Thepayoff to A of B’spossiblepureresponsesareasfollows:ü`TT§�ÚÉ·ØÞ¢§HY·8]å TT·a9 §¸ÞÒ§HY·8]¯ TT§a9 ·�Ú�§bY�·4]\cTT·a9 ·�Ú�·ØÞ¢§8]_ TT§a9 §�Ú�§ØÞ¢·HYIt is easyto seethatnooptionis dominatedoverthewholeregion. Solvingfor theintersectionof thehyperplanes,wefind 9×£ ÚF£uÞi£©�d ç and Ye£e]»£u�d1� . By symmetry, we will find thesamesolutionwhen f choosesamixedstrategy first.17.11 Thepayoff matrix for three-fingerMorra is asfollows:O: one O: two O: threeE: one� £ ¨1Áhg £ · ¨ � £ ·¸� Áhg £u� � £�Ù Áhg £ ·ãÙE: two� £ ·¸� Áhg £©� � £�Ù Áhg £ ·�Ù � £ · ¥1Áhg £ ¥E: three� £ Ù Áhg £ ·ãÙ � £ · ¥1Áhg £ ¥ � £ Ä1Áhg £ · ÄSuppose�choosesamixedstrategyô Ú ¶ TOÜw³�ß�V�Ú Õ T(È�]�ÜiV�Ú � T1È�â�9Oßsß.õ , whereÚ ¶ §�Ú Õ §�Ú � £¼� .Thepayoff to E of O’spossiblepureresponsesareasfollows:Üw³�ßHT ¨ Ú ¶ ·¸��Ú Õ §�Ù1Ú �È�]�ÜjT�·¹��Ú ¶ §©Ù1Ú Õ · ¥ Ú �Èoâ9ßsßbTeÙ1Ú ¶ · ¥ Ú Õ § Ä Ú �It is easyto seethatnooptionis dominatedoverthewholeregion. Solvingfor theintersectionof thehyperplanes,wefind Ú ¶ £u�d�Ù . Ú Õ £u�d ¨ , Ú � £u�d�Ù . Theexpectedvalueis 0.17.12 Everygameis eitherawin for oneside(anda lossfor theother)or a tie. With 2 for awin, 1 for atie,and0 for aloss,2 pointsareawardedfor everygame,sothisis aconstant-sumgame.If 1 point is awardedfor a lossin overtime,thenfor somegames3 pointsareawardedin all. Therefore,thegameis no longerconstant-sum.Supposewe assumethat teamA hasprobability 9 of winning in regular time andteamB hasprobability Þ of winning in regulartime (assumingnormalplay). Furthermore,assumehttp://librosysolucionarios.net122 Chapter 17. Making Complex DecisionsteamB hasa probabilityæof winning in overtime(whichoccursif thereis a tie afterregulartime). Onceovertimeis reached(by any means),theexpectedutilities areasfollows:®7kF £ë�µ§�Ú®7kJ £ë�µ§ æIn normalplay, the expectedutilities arederived from the probability of winning plus theprobabilityof tying timestheexpectedutility of overtimeplay:® F £ ¨ 9ɧ � �µ·89¹·¸Þ5¡ � �µ§�Ú�¡® J £ ¨ Þ�§ � �µ·l9¹·¸Þ5¡ � �µ§ æ ¡HenceA hasanincentive to agreeif ® kF ý®MF , or��§�Úáà ¨ 9¹§ � �µ·49 ·»Þ5¡ � �µ§�Ú�¡ or 9(ÚÉ·89¹§¸Þ9Úɧ¸Þpý� or Úáà 9 ·ØÞ9 §ØÞandB hasanincentive to agreeif ® kJ ý® J , or��§ æ à ¨ Þ¢§ � �µ·89¹·¸Þ5¡ � �µ§ æ ¡ or Þ æ ·ØÞ¢§89 æ §89Tý� oræ à ޢ·899 §ØÞWhenbothof theseinequalitieshold, thereis an incentive to tie in regulationplay. For anyvaluesof 9 and Þ , therewill bevaluesof Ú andæsuchthatbothinequalitieshold.For an in-depthstatisticalanalysisof the actualeffectsof the rule changeanda moresophisticatedtreatmentof the utility functions,see“Overtime! RulesandIncentives in theNationalHockey League”by StephenT. EastonandDuaneW. Rockerbie,availableathttp://people.u le th .c a/˜ ro ck er bi e/O VERTIME. PDF.17.13 We apply iteratedstrict dominanceto find thepurestrategy. First, Pol: do nothingdominatesPol: contract, so we drop thePol: contract row. Next, Fed: contract dominatesFed: do nothingandFed: expandon theremainingrows,sowe dropthosecolumns.Finally,Pol: expanddominatesPol: do nothingon theoneremainingcolumn.Hencetheonly Nashequilibriumis a dominantstrategy equilibriumwith Pol: expandandFed: contract. This isnot Paretooptimal: it is worsefor bothplayersthanthefour strategy profilesin thetop rightquadrant.http://librosysolucionarios.netSolutionsfor Chapter18Learningfrom Observations18.1 The aim hereis to couchlanguagelearningin the framework of the chapter, not tosolve the problem! This is a very interestingtopic for classdiscussion,raising issuesofnaturevs. nurture,theindeterminacy of meaningandreference,andsoon. BasicreferencesincludeChomsky (1957)andQuine(1960).The first stepis to appreciatethe variety of knowledgethatgoesunderthe heading“language.” The infantmustlearnto recognizeandproducespeech,learnvocabulary, learngrammar, learnthesemanticandpragmaticinterpretationof aspeechact,andlearnstrategiesfor disambiguation,amongotherthings.Theperformanceelementsfor this (in humans)andtheir associatedlearningmechanismsareobviously very complex andasyet little is knownaboutthem.A naivemodelof thelearningenvironmentconsidersjusttheexchangeof speechsounds.In reality, the physicalcontext of eachutteranceis crucial: a child mustseethe context inwhich “watermelon”is utteredin order to learn to associate“watermelon”with watermel-ons. Thus,the environmentconsistsnot just of otherhumansbut alsothe physicalobjectsandeventsaboutwhich discoursetakesplace.Auditory sensorsdetectspeechsounds,whileother senses(primarily visual) provide information on the physicalcontext. The relevanteffectorsarethe speechorgansandthe motor capacitiesthat allow the infant to respondtospeechor thatelicit verbalfeedback.Theperformancestandardcouldsimplybetheinfant’sgeneralutility function,howeverthatis realized,sothattheinfantperformsreinforcementlearningto performandrespondtospeechactsso asto improve its well-being—forexample,by obtainingfood andattention.However, humans’built-in capacityfor mimicry suggeststhattheproductionof soundssim-ilar to thoseproducedby otherhumansis a goalin itself. Thechild (onceheor shelearnstodifferentiatesoundsandlearnaboutpointingor othermeansof indicatingsalientobjects)isalsoexposedto examplesof supervisedlearning:anadultsays“shoe”or “belly button” whileindicatingtheappropriateobject. Sosentencesproducedby adultsprovide labelledpositiveexamples,andtheresponseof adultsto theinfant’s speechactsprovidesfurtherclassificationfeedback.Mostly, it seemsthatadultsdonotcorrectthechild’s speech,sotherearevery few neg-ativeclassificationsof thechild’s attemptedsentences.This is significantbecauseearlyworkon languagelearning(suchasthework of Gold, 1967)concentratedjust on identifying thesetof stringsthataregrammatical,assumingaparticulargrammaticalformalism.If thereare123http://librosysolucionarios.net124 Chapter 18. Learningfrom Observationsonly positive examples,thenthereis nothingto rule out thegrammar̄nm _­ÜQ9'o N . Sometheorists(notablyChomsky andFodor)usedwhatthey call the“povertyof thestimulus”argu-mentto saythatthebasicuniversalgrammarof languagesmustbeinnate,becauseotherwise(given the lack of negative examples)therewould beno way thata child could learna lan-guage(undertheassumptionsof languagelearningaslearninga setof grammaticalstrings).Criticshavecalledthisthe“povertyof theimagination”argument—Ican’t think of alearningmechanismthatwould work, so it mustbe innate. Indeed,if we go to probabilisticcontextfreegrammars,thenit is possibleto learna languagewithoutnegative examples.18.2 Learningtennisis muchsimpler thanlearningto speak. The requisiteskills canbedividedinto movement,playingstrokes,andstrategy. Theenvironmentconsistsof thecourt,ball, opponent,andone’s own body. Therelevant sensorsarethevisualsystemandpropri-oception(the senseof forceson andpositionof one’s own body parts). The effectorsarethe musclesinvolved in moving to the ball andhitting the stroke. The learningprocessin-volvesboth supervisedlearningandreinforcementlearning. Supervisedlearningoccursinacquiringthepredictive transitionmodels,e.g.,wheretheopponentwill hit theball, wheretheball will land,andwhat trajectorytheball will have afterone’s own stroke (e.g.,if I hita half-volley this way, it goesinto the net, but if I hit it that way, it clearsthe net). Rein-forcementlearningoccurswhenpointsarewon andlost—this is particularlyimportantforstrategic aspectsof play suchasshotplacementandpositioning(e.g.,in 60% of the pointswhereI hit a lob in responseto a cross-courtshot, I endup losing the point). In the earlystages,reinforcementalsooccurswhenashotsucceedsin clearingthenetandlandingin theopponent’s court. Achieving this smallsuccessis itself a sequentialprocessinvolving manymotorcontrolcommands,andthereis no teacheravailableto tell the learner’s motorcortexwhichmotorcontrolcommandsto issue.18.3 This is a deceptively simplequestion,designedto point out the plethoraof “excep-tions” in real-world situationsand the way in which decisiontreescapturea hierarchyofexceptions.Onepossibletreeis shown in FigureS18.1.Onecan,of course,imaginemanymoreexceptions.Thequalificationproblem,definedoriginally for actionmodels,appliesafortiori to condition–actionrules.18.4 In standarddecisiontrees,attribute testsdivide examplesaccordingto the attributevalue. Thereforeany examplereachingthe secondtestalreadyhasa known value for theattribute andthesecondtestis redundant.In somedecisiontreesystems,however, all testsareBooleanevenif theattributesaremultivaluedor continuous.In this case,additionaltestsof theattributecanbeusedto checkdifferentvaluesor subdivide therangefurther(e.g.,firstcheckif p ý� , andthenif it is, checkif q©Ã½�1� ).18.5 The algorithmmay not returnthe “correct” tree,but it will returna treethat is logi-cally equivalent,assumingthat themethodfor generatingexampleseventuallygeneratesallpossiblecombinationsof input attributes.This is truebecauseany two decisiontreedefinedon thesamesetof attributesthatagreeon all possibleexamplesare,by definition, logicallyequivalent.Theactuallyform of thetreemaydiffer becausetherearemany differentwaystorepresentthesamefunction. (For example,with two attributes S and f wecanhaveonetreehttp://librosysolucionarios.net125No YesNo YesNo YesNorNorYessNorYessNo YesFrontOfQueue?IntersectionBlocked?CarAheadMoving?No YesNorYessNo YesCrossTraffic?Pedestrians?NorTurning?No RightLeftOncomingTraffic?tCyclist?uNo YesNoYessFigure S18.1 A decisiontreefor decidingwhetherto move forwardat a traffic intersec-tion, givena greenlight.with S at theroot andanotherwith f at theroot.) Theroot attributeof theoriginal treemaynot in fact be theonethatwill be chosenby the informationgaingeuristicwhenappliedtothetrainingexamples.18.6 This is aneasyalgorithmto implement.Themainpoint is to have somethingto testotherlearningalgorithmsagainst,andto learnthebasicsof what a learningalgorithm is intermsof inputsandoutputsgiventheframework providedby thecoderepository.18.7 If we leave out anexampleof oneclass,thenthemajority of theremainingexamplesareof theotherclass,sothemajority classifierwill alwayspredictthewronganswer.18.8 This questionbringsa little bit of mathematicsto bearon theanalysisof thelearningproblem,preparingthe groundfor Chapter20. Error minimization is a basictechniqueinboth statisticsandneuralnets. The main thing is to seethat the error on a given trainingsetcanbewritten asa mathematicalexpressionandviewed asa functionof thehypothesischosen.Here,thehypothesisin questionis asinglenumberìl� ô � Á ��õ returnedat theleaf.a. If ì is returned,theabsoluteerroris� £»Ú � �µ·¸ìi¡À§�³�ì㣽ì � ³W·�Ú�¡<§�Ú £ ³ when ì�£½�£ Ú when ì�£½�This is minimizedby settingì�£½� if Úáû³ì�£½� if Úᬻ³http://librosysolucionarios.net126 Chapter 18. Learningfrom ObservationsThatis, ì is themajority value.b. First calculatethesumof squarederrors,andits derivative:� £»Ú � �µ·¸ìi¡ Õ §ã³�ì Õv îv%w £ ¨ ì³f· ¨ Ú � ��·»ìi¡�£ ¨ ì � Ú�§�³�¡<· ¨ ÚThefactthat thesecondderivative,v = îv%w = £ ¨ � Ú�§u³�¡ , is greaterthanzeromeansthat�is minimized(notmaximized)wherev îv%w £½� , i.e.,when ìW£ xx ±�² .18.9 Supposethatwe draw y examples.Eachexamplehas³ input featuresplusits classi-fication,sothereare¨ ²�±<¶distinctinput/outputexamplesto choosefrom. For eachexample,thereis exactlyonecontradictoryexample,namelytheexamplewith thesameinput featuresbut theoppositeclassification.Thus,theprobabilityof findingno contradictionisnumberof sequencesof mnon-contradictoryexamplesnumberof sequencesof mexamples£ ¨ ²�±<¶ � � ¨ ²�±<¶ ·Ø�1¡O�o�o� � ¨ ²�±<¶ ·8y §¸�1¡¨hz7{ ²�±<¶+|£ ¨ ²�±<¶�}� ¨ ²�±<¶ ·8y¼¡} ¨ z7{ ²�±<¶+|For ³ª£ �1� , with 2048possibleexamples,a contradictionbecomeslikely with probabilityý�1� ¥ after54 exampleshave beendrawn.18.10 This resultemphasizesthefactthatany statisticalfluctuationscausedby therandomsamplingprocesswill resultin anapparentinformationgain.Theeasypart is showing that thegain is zerowheneachsubsethasthesameratio ofpositive examples.Thegainis definedasí ÚÚ�§ã³ Á³Ú�§�³ ·~P Ë<¶Ú�P�§�³GPÚ�§�³í Ú�PÚ�P�§�³GP Á³GPÚ�P�§�³GPSinceÚ £ Ú P and ³ £ ³ P , if Ú P d � Ú P §©³ P ¡Ò£@Ú ï � Ú ï §�³ ï ¡ for all Ý , � thenwe musthaveÚ�P�d � Ú�P§W³GPl¡Q£ Ú-d � Ú¹§W³�¡ for all Ý , andalso³GP�d � Ú�P§W³GP`¡Q£ ³�d � Ú¹§W³�¡ . Fromthis,weobtain� Ð Ým³ £ í ÚÚ�§©³ Á ³Ú�§�³ ·~P Ë<¶Ú P §�³ PÚ�§�³í ÚÚ�§�³ Á ³Úɧ�³£ í ÚÚ�§©³ Á ³Ú�§�³ �µ·~P Ë<¶ Ú�P�§�³GPÚ�§�³ £½�18.11 This is a fairly small, straightforward programmingexercise.The only hardpart istheactual�Õcomputation;you might want to provide your studentswith a library functionto do this.18.12 This is anotherstraightforward programmingexercise.The follow-up exerciseis torun teststo seeif themodifiedalgorithmactuallydoesbetter.http://librosysolucionarios.net12718.13 Let theprior probabilitiesof eachattributevaluebeå¹� ? ¶ ¡ Á �o�o� Á å¹� ? ² ¡ . (Theseprob-abilities areestimatedby the empirical fractionsamongthe examplesat the currentnode.)Frompage540,theintrinsic informationcontentof theattribute isí��.å¹� ? ¶ ¡ Á �o�o� Á åÉ� ? ² ¡.¡¤£²P Ë<¶ ·åÉ� ? P ¡ Ñ Ó1Ô ? PGiven this formula and the empirical estimatesofå¹� ? P ¡ , the modification to the codeisstraightforward.18.1418.15 Note: this is the only exerciseto cover the materialin section18.6. Although thebasicideasof computationallearningtheoryarebothimportantandelegant,it is not easytofind goodexercisesthataresuitablefor anAI classasopposedto a theoryclass.If you areteachinga graduateclass,or an undergraduateclasswith a strongemphasison learning,itmight bea goodideato usesomeof theexercisesfrom KearnsandVazirani(1994).a. If eachtest is an arbitrary conjunctionof literals, then a decisionlist can representan arbitrary DNF (disjunctive normal form) formula directly. The DNF expressionä ¶�� ä Õ � ÌoÌoÌ � ä ² , where ä�P is a conjunctionof literals, can be representedby adecisionlist in which ä�P is the Ý th testandreturns�<9 Æ ß if successful.Thatis:ä ¶ m ��9 Æ ß�Vä Õ m ��9 Æ ß�V�o�o�ä ² m ��9 Æ ß�V��9 Æ ß m \ Ð�� ÞsßSinceany BooleanfunctioncanbewrittenasaDNF formula,thenany Booleanfunctioncanberepresentedby adecisionlist.b. A decisiontreeof depthÏcanbetranslatedinto adecisionlist whosetestshaveatmostÏliteralssimplyby encodingeachpathasatest.Thetestreturnsthecorrespondingleafvalueif it succeeds.SincethedecisiontreehasdepthÏ, no pathcontainsmorethanÏliterals.http://librosysolucionarios.netSolutionsfor Chapter19Knowledgein Learning19.1 In CNF, thepremisesareasfollows:ê Û Ð È`Ý�Ü9³ Ð�� ÝmÈ�� � q Á ³�¡ � ê Û Ð ÈlÝwÜw³ Ð�� ÝmÈ�� � � Á ³�¡ � ê � Ð ³Z� Æ Ð ��ß � q Á � ¡ � � Ð ³Z� Æ Ð ��ß � � Á � ¡Û Ð È`ÝwÜw³ Ð�� ÝmÈ�� � \¹ß%95³ Ð ³�o>Ü Á f79 Ð�� Ý � ¡� Ð ³Z� Æ Ð ��ß � \¹ß%9(³ Ð ³�o�Ü Á å Ü�9(È Æ � Æ ß�Þ5ß�¡We canprove thedesiredconclusiondirectly ratherthanby refutation.Resolve thefirst twopremiseswith �2q�d\Éß+9(³ Ð ³�o�Üi� to obtainê Û Ð È`Ý�Ü9³ Ð�� ÝmÈ�� � � Á f79 Ð�� Ý � ¡ � ê � Ð ³Z� Æ Ð ��ß � \Éß+9(³ Ð ³�o�Ü Á � ¡ � � Ð ³Z� Æ Ð ��ß � � Á � ¡Resolve thiswith � Ð ³Z� Æ Ð ��ß � \Éß+9(³ Ð ³�o�Ü Á å ÜQ95È Æ � Æ ß�Þ5ß�¡ to obtainê Û Ð È`Ý�Ü9³ Ð�� ÝmÈ�� � � Á f79 Ð�� Ý � ¡ � � Ð ³Z� Æ Ð ��ß � � Á å ÜQ9(È Æ � Æ ßsÞsßs¡which is thedesiredconclusionÛ Ð ÈlÝwÜw³ Ð�� ÝrÈ�� � � Á f79 Ð�� Ý � ¡�� � Ð ³Z� Æ Ð ��ß � � Á å Ü�9(È Æ � Æ ß�Þ5ß�¡ .19.2 This questionis tricky in places. It is importantto seethe distinction betweenthesharedandunsharedvariableson theLHS andRHSof thedetermination.Thesharedvari-ableswill be instantiatedto theobjectsto becomparedin ananalogicalinference,while theunsharedvariablesareinstantiatedwith theobjects’observedandinferredproperties.a. Herewearetalkingaboutthezip codesandstatesof houses(or addressesor towns). Iftwo houses/addresses/towns have thesamezip code,they arein thesamestate:� ÝmÚ�ä�Ü�o�ß � q Á � ¡¤ù ¯ È Ð È�ß � q Á Þ5¡The determinationis true becausethe US PostalServicenever draws zipcodebound-ariesacrossstatelines(perhapsfor somereasonhaving to dowith interstatecommerce).b. Heretheobjectsbeingreasonedaboutarecoins,anddesign,denomination,andmassarepropertiesof coins.SowehaveäÉÜwÝm³ � Î�¡�� � àáßsÞ!Ý��>³ � Î Á o�¡M�fàáß!³�Ü�y Ýr³ Ð ÈlÝwÜw³ � Î Á Ð ¡¤ù ��Ð ÞsÞ � Î Á y¼¡.¡This is (very nearlyexactly) truebecausecoinsof agivendenominationanddesignarestampedfrom thesameoriginal die usingthesamematerial;sizeandshapedeterminevolume;andvolumeandmaterialdeterminemass.128http://librosysolucionarios.net129c. Herewe have to be careful. The objectsbeingreasonedaboutarenot programsbutruns of a given program. (This determinationis also one often forgottenby noviceprogrammers.)Wecanusesituationcalculusto referto theruns:R Ú í ³<Ú Æ È � Ú Á Ý Á Þ5¡Qù g Æ ÈlÚ Æ È � Ú Á Ü Á Þs¡HeretheR Ú capturesthe Ú variableso that it doesnot participatein thedeterminationasoneof the sharedor unsharedvariables.The situationis the sharedvariable. Thedeterminationexpandsout to thefollowing Hornclause:í ³<Ú Æ È � Ú Á Ý Á Þ ¶ ¡G� í ³<Ú Æ È � Ú Á Ý Á Þ Õ ¡�� g Æ ÈlÚ Æ È � Ú Á Ü Á Þ ¶ ¡�� g Æ È`Ú Æ È � Ú Á Ü Á Þ Õ ¡That is, if Ú hasthesameinput in two differentsituationsit will have thesameoutputin thosesituations.This is generallytruebecausecomputersoperateon programsandinputsdeterministically;however, it is importantthat“input” includetheentirestateofthecomputer’s memory, file system,andsoon. Noticethatthe“naive” choiceí ³<Ú Æ È � Ú Á Ýw¡¤ù g Æ ÈlÚ Æ È � Ú Á Ü(¡expandsout toí ³<Ú Æ È � Ú ¶ Á Ýw¡�� í ³<Ú Æ È � Ú Õ Á Ýw¡�� g Æ ÈlÚ Æ È � Ú ¶ Á Ü(¡�� g Æ ÈlÚ Æ È � Ú Õ Á Ü(¡whichsaysthatif any two programshave thesameinput they producethesameoutput!d. Here the objectsbeing reasonedarepeoplein specifictime intervals. (The intervalscouldbethesamein eachcase,or differentbut of thesamekind suchasdays,weeks,etc.Wewill stick to thesameinterval for simplicity. As above,weneedto quantifytheinterval to “precapture”thevariable.)Wewill useä � Ý�y Ð Èoß � q Á Î Á Ýw¡ to meanthatpersonq experiencesclimate Î in interval Ý , andwe will assumefor thesake of variety thataperson’s metabolismis constant.R Ý ä � Ý�y Ð Èoß � q Á Î Á Ýw¡G� àÉÝwß!È � q Á o Á Ýw¡�� � qSß+9Î`ÝwÞ5ß � q Á ß Á Ýw¡�� � ß!È Ð Ö�Ü � ÝwÞ�y � q Á y¼¡ù � Ð Ýr³ � q Á ] Á Ý�¡While the determinationsseemsplausible,it leaves out suchfactorsaswater intake,clothing,disease,etc.Thequalificationproblemariseswith determinationsjustaswithimplications.e. Let f Ð�� oo³�ßsÞsÞ � q Á Ö�¡ meanthatpersonq hasbaldnessÖ (whichmightbe f Ð�� o , å¹Ð 95ÈlÝ Ð�� ,or � Ð Ý�9�� , say).A first stabat thedeterminationmight be� Ü9Èoâkß%9 � y Á qi¡���\ Ð Èoâkß%9 � � Á y¼¡G��f Ð�� o�³�ß�Þ5Þ � � Á Ö�¡�ù/f Ð�� oo³�ßsÞsÞ � q Á Ö�¡but thiswouldonly allow aninferencewhentwo peoplehave thesamemotherandma-ternalgrandfatherbecausethe y and� aretheunsharedvariablesontheLHS.Also, theRHShasno unsharedvariable.Noticethat thedeterminationdoesnot sayspecificallythatbaldnessis inheritedwithout modification;it allows, for example,for a hypothet-ical world in which the maternalgrandchildrenof a bald man are all hairy, or viceversa.This might not seemparticularlynatural,but considerotherdeterminationssuchas“Whetheror not I file a tax returndetermineswhetheror not my spousemustfile atax return.”http://librosysolucionarios.net130 Chapter 19. Knowledgein LearningThebaldnessof thematernalgrandfatheris therelevantvaluefor prediction,sothatshouldbetheunsharedvariableon theLHS. Themotherandmaternalgrandfatheraredesignatedby skolemfunctions:� Ü9Èoâkß%9 �.��� qS¡ Á qi¡G��\ Ð Èoâkß%9 � \ �.��� qS¡.¡ Á ��� qS¡.¡���f Ð�� oo³�ßsÞsÞ � \ �.�­� qi¡.¡ Á Ö ¶ ¡ù�f Ð�� oo³�ßsÞsÞ � q Á Ö Õ ¡If weuse\ Ð È�â�ß+9 and� ÜwÈ�â�ß+9 asfunctionsymbols,thenthemeaningbecomesclearer:f Ð�� o�³�ß�Þ5Þ � \ Ð Èoâkß+9 �.� ÜwÈ�â�ß+9 � qi¡.¡ Á Ö ¶ ¡Qù/f Ð�� oo³�ß�Þ5Þ � q Á Ö Õ ¡Justto check,thisexpandsintof Ð�� o�³�ß�Þ5Þ � \ Ð Èoâkß+9 �.� ÜwÈ�â�ß+9 � qi¡.¡ Á Ö ¶ ¡Z��f Ð�� oo³�ß�Þ5Þ � \ Ð Èoâkß%9 �.� ÜwÈoâkß+9 � �<¡.¡ Á Ö ¶ ¡��f Ð�� oo³�ßsÞsÞ � q Á Ö Õ ¡�� f Ð�� o�³�ß�Þ5Þ � � Á Ö Õ ¡whichhastheintendedmeaning.19.3 Becauseof the qualificationproblem, it is not usually possiblein most real-worldapplicationsto list on the LHS of a determinationall the relevant factorsthat determinetheRHS.Determinationswill usuallythereforebe true to an extent—thatis, if two objectsagreeon the LHS thereis someprobability (preferablygreaterthanthe prior) that the twoobjectswill agreeon the RHS. An appropriatedefinition for probabilisticdeterminationssimply includesthis conditionalprobability of matchingon the RHS given a matchon theLHS. For example,we could define Û Ð È`Ý�Ü9³ Ð�� ÝmÈ�� � q Á ³�¡Tù � Ð ³Z� Æ Ð ��ß � q Á � ¡ � �1� ç �1¡ to meanthat if two peoplehave the samenationality, thenthereis a 90% chancethat they have thesamelanguage.19.4 This exercisetestthestudent’s understandingof resolutionandunification,aswell asstressingthenondeterminismof theinverseresolutionprocess.It shouldhelpa lot in makingtheinverseresolutionoperationlessmysteriousandmoreamenableto mathematicalanalysis.It is helpful first to draw out theresolution“V” whendoingtheseproblems,andthento do acarefulcaseanalysis.a. Thereis no possiblevaluefor ä Õ here.Theresolutionstepwouldhave to resolve awayboththeå¹� q Á �<¡ on theLHS of ä ¶ andthe � � q Á �Z¡ on theright, which is not possible.(Resolutioncan remove morethanoneliteral from a clause,but only if thoseliteralsareredundant—i.e.,onesubsumestheother.)b. Without lossof generality, let ä ¶ containthenegative (LHS) literal to beresolvedaway.The LHS of ä ¶ thereforecontainsoneliteral�, while the LHS of ä Õ mustbe empty.TheRHSof ä Õ mustcontain� ö suchthat�and� ö unify with someunifier � . Now wehave a choice:å¹� S Á f�¡ on the RHSof ä could comefrom theRHS of ä ¶ or of ä Õ .Thusthetwo basicsolutiontemplatesareä ¶ £ � � \ Ð�� Þ5ß ; ä Õ £>��9 Æ ß�� � ö � å¹� S Á f¡+� � ¶ä ¶ £ � � å¹� S Á f¡+� � ¶ ; ä Õ £>��9 Æ ß�� � öWithin thesetemplates,thechoiceof�is entirelyunconstrained.Suppose�is � � q Á �<¡and� ö is � � S Á f¡ . Thenå¹� S Á f¡+�>� ¶ could beå¹� q Á �<¡ (orå¹� S Á �Z¡ orå¹� q Á f¡ ) andhttp://librosysolucionarios.net131thesolutionsareä ¶ £/� � q Á �Z¡�� \ Ð�� Þ5ß ; ä Õ £:��9 Æ ß�� � � S Á f¡ � å¹� q Á �Z¡ä ¶ £/� � q Á �Z¡�� å¹� q Á �Z¡ ; ä Õ £:��9 Æ ß�� � � S Á f¡c. As before,let ä ¶ containthenegative (LHS) literal to beresolvedaway, with� ö on theRHSof ä Õ . Wenow have four possibletemplatesbecauseeachof thetwo literalsin äcouldhave comefrom either ä ¶ or ä Õ :ä ¶ £ � � \ Ð�� Þ5ß ; ä Õ £ å¹� q Á �<¡+� � ¶ � � ö � å¹� q Á Y � �Z¡.¡+� � ¶ä ¶ £ � � å¹� q Á Y � �<¡.¡+� � ¶ ; ä Õ £ å¹� q Á �<¡+� � ¶ � � öä ¶ £ � � å¹� q Á �<¡+� � ¶ � \ Ð�� Þ5ß ; ä Õ £>��9 Æ ß�� � ö � åÉ� q Á Y � �<¡.¡+� � ¶ä ¶ £ � � å¹� q Á �Z¡+� � ¶ � å¹� q Á Y � �<¡.¡+� � ¶ ; ä Õ £>��9 Æ ß�� � öAgain,we have a fairly freechoicefor�. However, since ä containsq and � , � cannotbind thosevariables(elsethey would not appearin ä ). Thus, if�is � � q Á �Z¡ , then� ömustbe � � q Á �<¡ alsoand � will beempty.19.5 Wewill assumethatPrologis thelogic programminglanguage.It is certainlytruethatany solutionreturnedby thecall to ü ß�Þ5Ü � ? ß will beacorrectinverseresolvent.Unfortunately,it is quite possiblethat the call will fail to returnbecauseof Prolog’s depth-firstsearch.Iftheclausesin ü ßsÞsÜ � ? ß and ®µ³<ÝQY�� areinfelicitously arranged,theproof treemight go downthe branchcorrespondingto indefinitely nestedfunction symbolsin the solutionandneverreturn. This canbe alleviatedby redesigningthe Prolog inferenceengineso that it worksusingbreadth-firstsearchor iterative deepening,althoughthe infinitely deepbrancheswillstill bea problem.Notethatany cutsusedin thePrologprogramwill alsobea problemfortheinverseresolution.19.6 Thisexercisegivessomeideaof theratherlargebranchingfactorfacingtop-down ILPsystems.a. It is important to note that position is significant—å¹� S Á f¡ is very different fromå¹� f Á SF¡ ! The first argumentposition can containone of the five existing variablesor anew variable.For eachof thesesix choices,thesecondpositioncancontainoneofthefive existing variablesor a new variable,exceptthat the literal with two new vari-ablesis disallowed. Hencethereare35 choices. With negatedliterals too, the totalbranchingfactoris 70.b. This seemsto be quite a tricky combinatorialproblem. The easiestway to solve itseemsto be to start by including the multiple possibilitiesthat are equivalent underrenamingof thenew variablesaswell asthosethat containonly new variables.Thentheseredundantor illegal choicescanberemoved later. Now, we canuseup to 9�·6�new variables. If we use º Ý new variables,we canwrite� ³ § Ýw¡¡ literals, so usingexactly ÝTà � variableswe canwrite� ³¼§ Ýw¡ · � ³u§ Ý¢· �1¡ literals. Eachof theseis functionally isomorphicunderany renamingof thenew variables.With Ý variables,http://librosysolucionarios.net132 Chapter 19. Knowledgein Learningthereareare Ý renamings.Hencethe total numberof distinct literals (including thoseillegal oneswith no old variables)is³ § �¶P Ë<¶� ³f§�Ýw¡ · � ³f§�Ýi·Ø�1¡ Ý }Now wejustsubtractoff thenumberof distinctall-new literals.With º»Ý new variables,thenumberof (notnecessarilydistinct)all-new literalsis Ý� , sothenumberwith exactlyÝFà � is Ý · � ݤ·6�1¡ . Eachof thesehasÝ } equivalentliterals in theset.This givesusthefinal total for distinct,legal literals:³ § �¶P Ë<¶� ³f§�Ýw¡ · � ³f§�Ýi·Ø�1¡ Ý } · �¶P Ë<¶Ý · � ÝS·¸�1¡ Ý }which candoubtlessbesimplified. Onecancheckthat for 9 £ ¨and ³¸£ ¥this gives35.c. If a literal containsonly new variables,theneithera subsequentliteral in the clausebody connectsoneor moreof thosevariablesto oneor moreof the “old” variables,or it doesn’t. If it does,thenthesameclausewill begeneratedwith thosetwo literalsreversed,suchthat the restrictionis not violated. If it doesn’t, thenthe literal is eitheralwaystrue (if thepredicateis satisfiable)or alwaysfalse(if it is unsatisfiable),inde-pendentof the“input” variablesin thehead.Thus,theliteral wouldeitherberedundantor would rendertheclausebodyequivalentto \ Ð�� Þ5ß .19.7 FOIL is availableon thewebat http://www-2.cs.cmu.edu/afs/cs/project/ai-repository-/ai/areas/learning/systems/foil/0.html (andpossiblyotherplaces).It is worthwhileto experi-mentwith it.http://librosysolucionarios.netSolutionsfor Chapter20StatisticalLearningMethods20.1 Thecodefor this exerciseis a straightforward implementationof Equations20.1and20.2. FigureS20.1shows the resultsfor datasequencesgeneratedfrom â � and â�¢ . (Plots00.20.40.60.810 20 40 60 80 100Posterior probability of hypothesisNumber of samples in dP(h1 | d)P(h2 | d)P(h3 | d)P(h4 | d)P(h5 | d)0.40.50.60.70.80.910 20 40 60 80 100Probability that next candy is lime£Number of samples in d(a) (b)00.20.40.60.810 20 40 60 80 100Posterior probability of hypothesisNumber of samples in dP(h1 | d)P(h2 | d)P(h3 | d)P(h4 | d)P(h5 | d)0.40.50.60.70.80.910 20 40 60 80 100Probability that next candy is lime£Number of samples in d(c) (d)Figure S20.1 Graphsfor Ex. 20.1. (a) Posteriorprobabilities ¤<¥¡¦§�¨ª©«%¬Q­�­�­�¬�©h®7¯ over asamplesequenceof length100generatedfrom ¦�° (50%cherry+ 50% lime). (b) Bayesianprediction ¤<¥¡©h®M±M«�²e³µ´·¶X¸�¨¹©«+¬Q­�­Q­�¬�©h®X¯ given the data in (a). (c) Posteriorprobabilities¤<¥¡¦§�¨ª©i«�¬�­Q­�­%¬Q©�®º¯ over a samplesequenceof length 100 generatedfrom ¦(» (25% cherry+ 75%lime). (d) Bayesianprediction ¤<¥¡©h®M±M«'²[³µ´·¶X¸�¨¹©«+¬Q­�­Q­�¬�©h®X¯ giventhedatain(c).133http://librosysolucionarios.net134 Chapter 20. StatisticalLearningMethodsfor â ¶ and â Õ areessentiallyidenticalto thosefor â � and â�¢ .) Resultsobtainedby studentsmayvary becausethedatasequencesaregeneratedrandomlyfrom thespecifiedcandydis-tribution. In (a), the samplesvery closelyreflect the true probabilitiesandthe hypothesesotherthan â � areeffectively ruledout very quickly. In (c), theearlysampleproportionsaresomewherebetween50/50and25/75;furthermore,â � hasahigherprior than â ¢ . As aresult,â � and â�¢ vie for supremacy. Between50 and60 samples,apreponderanceof limesensuresthedefeatof â � andthepredictionquickly convergesto 0.75.20.2 Typical plotsareshown in FigureS20.2.BecausebothMAP andML chooseexactlyonehypothesisfor predictions,thepredictionprobabilitiesareall 0.0,0.25,0.5,0.75,or 1.0.For smalldatasetstheML predictionin particularshows very largevariance.20.3 This is a nontrivial sequentialdecisionproblem,but can be solved using the toolsdevelopedin the book. It leadsinto generalissuesof statisticaldecisiontheory, stoppingrules,etc.Here,we sketchthe“straightforward” solution.00.20.40.60.810 20 40 60 80 100Probability that next candy is lime£Number of samples in d00.20.40.60.810 20 40 60 80 100Probability that next candy is lime£Number of samples in d(a) (b)00.20.40.60.810 20 40 60 80 100Probability that next candy is lime£Number of samples in d00.20.40.60.810 20 40 60 80 100Probability that next candy is lime£Number of samples in d(c) (d)FigureS20.2 Graphsfor Ex.20.2.(a)PredictionfromtheMAP hypothesisgivenasamplesequenceof length100generatedfrom ¦ ° (50%cherry+ 50%lime). (b) Predictionfrom theML hypothesisgiventhedatain (a). (c) Predictionfrom theMAP hypothesisgivendatafrom¦(» (25%cherry+ 75%lime). (d) Predictionfrom theML hypothesisgiventhedatain (c).http://librosysolucionarios.net135We canthink of this problemasa simplified form of POMDP(seeChapter17). The“belief states”aredefinedby thenumbersof cherryandlime candiesobserved sofar in thesamplingprocess.Let thesebe ä and � , andlet ® � ä Á ��¡ betheutility of thecorrespondingbelief state.In any givenstate,therearetwo possibledecisions:Þ5ß �¼� and Þ Ð y Ú � ß . Thereis asimpleBellmanequationrelating � and ® for thesamplingcase:� � ä Á � Á Þ Ð y Ú � ß�¡�£ å¹� Î�âkß+9i9i� è ä Á ��¡.® � ä@§¸� Á �¢¡À§ å¹�+� Ý�y¼ß è ä Á ��¡.® � ä Á �u§¸�1¡Let the posteriorprobability of each â P beå¹� â P è ä Á �¢¡ , the sizeof the bagbe Û , and thefraction of cherriesin a bagof type Ý be YhP . Thenthevalueobtainedby selling is given bythevalueof thesampledcandies(whichAnn getsto keep)plusthepricepaidby Bob(whichequalstheexpectedutility of theremainingcandiesfor Bob):� � ä Á � Á Þ5ß �¼� ¡�£ªäÉνFf§¸��¾�FW§ På¹� â P è ä Á ��¡ ô � Y P Û ·ØäÉ¡.Î J § �.� �F·HY P ¡.Û ·¸�¢¡¿¾ J õandof coursewe have® � ä Á ��¡¢£/.0!21Z�� � ä Á � Á Þsß �¼� ¡ Á � � ä Á � Á Þ Ð y Ú � ß�¡+�É�Thuswecansetupadynamicprogramto compute� giventheobviousboundaryconditionsfor thecasewhereä §��á£©Û . Thesolutionof thisdynamicprogramgivestheoptimalpolicyfor Ann. It will have thepropertythat if sheshouldsell at� ä Á ��¡ , thensheshouldalsosellat� ä Á ��§ Ï ¡ for all positiveÏ. Thus,theproblemis to determine,for eachä , thethresholdvalueof � at or above which sheshouldsell. A minor complicationis that the formula forå¹� â�P è ä Á ��¡ shouldtake into accountthenon-replacementof candiesandthefinitenessof Û ,otherwiseoddthingswill happenwhen äا¸� is closeto Û .20.4 The Bayesianapproachwould be to take both drugs. The maximumlikelihoodap-proachwould be to take the anti-f drug. In the casewherethereare two versionsof f ,theBayesianstill recommendstakingbothdrugs,while themaximumlikelihoodapproachisnow to take theanti-S drug,sinceit hasa40%chanceof beingcorrect,versus30%for eachof the f cases.This is of courseacaricature,andyouwouldbehard-pressedto find adoctor,evena rabidmaximum-likelihoodadvocatewho would prescribelike this. But you canfindoneswho doresearchlike this.20.5 Boostednaive Bayeslearningis discussedby ? (?). The applicationof boostingtonaiveBayesis straightforward. ThenaiveBayeslearnerusesmaximum-likelihoodparameterestimationbasedon counts,so usinga weightedtraining setsimply meansaddingweightsratherthancounting.EachnaiveBayesmodelis treatedasadeterministicclassifierthatpicksthemostlikely classfor eachexample.20.6 Wehave� £½·ay �.Ñ Ó1Ô�À § Ñ Ó1Ô�Á ¨ û¢¡À· ï� � ï · � � ¶ q ï §H� Õ ¡.¡ Õ¨ À Õhencetheequationsfor thederivativesat theoptimumare � � ¶�£ · ï qï�� � ï · � � ¶ q ï §H� Õ ¡.¡À Õ £½�http://librosysolucionarios.net136 Chapter 20. StatisticalLearningMethods � � Õ £ · ï� � ï · � � ¶ q ï §>� Õ ¡.¡À Õ £½� � À £ · y À § ï� � ï · � � ¶ q ï §H� Õ ¡.¡ ÕÀ � £ª�andthesolutionscanbecomputedas� ¶ £ yï q ï � ï · ï � ï ï q ïy ï q Õ ï · ï q ïÕ� Õ £ �y ï � � ï ·H� ¶ q ï ¡À Õ £ �y ï � � ï · � � ¶ q ï §H� Õ ¡.¡Õ20.7 Therearea coupleof waysto solve this problem. Here,we show the indicatorvari-able methoddescribedon page743. Assumewe have a child variable à with parentsp ¶ Á �o�o� Á p ° and let the rangeof eachvariablebe �1� Á �� . Let the noisy-ORparametersbeæ P�£ å¹� à £u� è pIP�£u� Á p � PO£u�1¡ . Thenoisy-ORmodelthenassertsthatåÉ� à £¼� è q ¶ Á �o�o� Á q ° ¡¤£ª�µ·°P Ë�¶æ�ÄiÅP �Assumewe have y complete-datasampleswith values� ï for à and qÆP ï for eachpIP . Theconditionallog likelihoodforå¹� à è p ¶ Á �o�o� Á p ° ¡ is givenby�Y£ ï Ñ Ó1Ô �µ· Pæ ÄiÅ ïPÇ ñPæ ÄiÅ ïP¶ � Ç ñ£ ï � ïÀÑ Ó1Ô �µ· Pæ ÄiÅ ïP § � �µ·8� ï ¡ P qPïÀÑ Ó1Ô æ PThegradientwith respectto eachnoisy-ORparameteris � æ P £ ï · � ï qÆP ï P æÄiÅ ïPæ P �µ· P æ ÄiÅïP § � �µ·8� ï ¡¿qÆP ïæ P£ ï qPï ��·l� ï · P æ ÄiÅïPæ P �µ· P æ ÄiÅïP20.8a. By integratingover the rangeô � Á ��õ , show that the normalizationconstantfor the dis-tribution È $ )�! ô Ð Á Örõ is given by ì £ÊÉ �.Ð §�Ö�¡+dÉ �.Ð ¡+É � Ö�¡ where É � qi¡ is the Gammafunction, definedby É � q᧠�1¡O£�q Ì É � qi¡ and É � �1¡O£¼� . (For integer q , É � qq§»�1¡O£[q } .)Ë�Ì¡Í+Í�ÌÎÐÏ�ѼҼÓ�Ô Õ+Ñhttp://librosysolucionarios.net137We will solve this for positive integerÐand Ö by inductionoverÐ. Let ì �.Ð Á Ö�¡ bethenormalizationconstant.For thebasecases,we haveì � � Á Ö�¡¢£½�d¶¾ � ¾ � �µ·H�~¡ @ � ¶ oÖ�É£½·¹�d ô � Ö� �µ·b�~¡ @ õ ¶¾ £½Öand É � �µ§¸Ö�¡É � �1¡+É � Ö�¡ £ Ö Ì É � Ö�¡� Ì É � Ö�¡ £½Ö��For theinductive step,we assumefor all Ö thatì �.Ð ·Ø� Á ÖQ§¸�1¡¤£ É �.Ð §¸Ö�¡É �.Ð ·¸�1¡+É � ÖQ§¸�1¡ £Ð ·¸�ÖÌ É �.Ð §¸Ö�¡É �.Ð ¡+É � Ö�¡Now weevaluateì �.Ð Á Ö�¡ usingintegrationby parts.Wehave�d1ì �.Ð Á Ö�¡¼£¶¾ � 3 � ¶ � ��·H�~¡ @ � ¶ oÖ�£ ô � 3 � ¶ Ì � Ö� �µ·b�~¡ @ õ ¾ ¶ § Ð ·Ø�Ö¶¾ � 3 � Õ � �µ·H�~¡ @ oÖ�£ �ק Ð ·»�Ö �ì �.Ð ·¸� Á ÖQ§Ø�1¡Henceì �.Ð Á Ö�¡�£ ÖÐ ·¸� ì�.Ð ·¸� Á ÖQ§¸�1¡¤£ ÖÐ ·¸�Ð ·¸�ÖÌ É �.Ð §ØÖ�¡É �.Ð ¡+É � Ö�¡ £É �.Ð §¸Ö�¡É �.Ð ¡+É � Ö�¡asrequired.b. Themeanis givenby thefollowing integral:× �.Ð Á Ö�¡¼£ëì �.Ð Á Ö�¡¶¾ � Ì � 3 � ¶ � �µ·H�>¡ @ � ¶ o��£ëì �.Ð Á Ö�¡¶¾ � 3 � �µ·H�~¡ @ � ¶ o��£ëì �.Ð Á Ö�¡+d1ì �.Ð §Ø� Á Ö�¡¢£ É �.Ð §¸Ö�¡É �.Ð ¡+É � Ö�¡Ì É �.Ð §¸�1¡+É � Ö�¡É �.Ð §¸ÖS§¸�1¡£ É�.Ð §ØÖ�¡É �.Ð ¡+É � Ö�¡Ì Ð É �.Ð ¡+É � Ö�¡�.Ð §¸Ö�¡+É �.Ð §¸ÖS§¸�1¡ £ÐÐ §¸Ö �c. Themodeis foundby solvingfor oØÈ $ )�! ô Ð Á Ö`õ � �~¡+do���£u� :ooÖ�� ì �.Ð Á Ö�¡+� 3 � ¶ � �µ·b�~¡ @ � ¶ ¡£ ì �.Ð Á Ö�¡ ô �.Ð ·Ø�1¡+� 3 � Õ � �µ·>�>¡ @ � ¶ · � ÖS·¸�1¡+� 3 � ¶ � ��·H�~¡ @ � Õ õI£½�� �.Ð ·¸�1¡ � �F·H�~¡S£ � ÖQ·¸�1¡+�� �¹£Ð ·¸�Ð §¸ÖS· ¨d. È $ )�! ô Ù Á Ù õ>£ ì � Ù Á Ù ¡+�ØÚ � ¶ � � ·j�~¡¡Ú � ¶ tendsto very large valuescloseto �¤£©� and �Q£¼� ,i.e., it expressestheprior belief that thedistribution characterizedby � is nearlydeter-ministic (eitherpositively or negatively). After updatingwith a positive examplewehttp://librosysolucionarios.net138 Chapter 20.StatisticalLearningMethodsobtainthedistribution È $ )�! ô �F§ Ù Á Ù õ , which hasnearlyall its massnear �Q£u� (andtheconversefor a negative example),i.e., we have learnedthat thedistribution character-ized by � is deterministicin the positive sense.If we seea “counterexample”, e.g.,apositiveandanegativeexample,weobtainÈ $ )�! ô �À§ Ù Á �À§ Ù õ , whichis closeto uniform,i.e., thehypothesisof near-determinismis abandoned.20.9 Considerthemaximum-likelihoodparametervaluesfor theCPTof nodeà in theorig-inal network, wherean extra parentp °1±<¶ will be addedto à . If we settheparametersforå¹� � è q ¶ Á �o�o� Á q ° Á q °1±<¶ ¡ in thenew network to beidenticaltoå¹� � è q ¶ Á �o�o� Á q ° ¡ in theoriginalnetowrk, regardlessof thevalue q °1±<¶ , thenthe likelihoodof the datais unchanged.Maxi-mizing thelikelihoodby alteringtheparameterscanthenonly increasethelikelihood.20.10a. With threeattributes,thereareseven parametersin the modelandthe empiricaldatagive frequenciesfor¨ � £u� classes,which supply7 independentnumberssincethe 8frequencieshave to sumto the total samplesize. Thus,theproblemis neitherunder-nor over-constrained.With two attributes,therearefive parametersin the modelandtheempiricaldatagivefrequenciesfor¨ Õ £ Ù classes,whichsupply3 independentnum-bers.Thus,theproblemis severelyunderconstrained.Therewill bea two-dimensionalsurfaceof equallygoodML solutionsandtheoriginal parameterscannotberecovered.b. The calculationis sketchedandpartly completedon pages729 and730. Completingit by handis tedious;studentsshouldencouragedto implementthemethod,ideally incombinationwith abayesnetpackage,andtraceits calculations.c. Thequestionshouldalsoassume�Q£u�1� ¥ to simplify theanalysis.With everyparameteridenticaland �Q£u�1� ¥ , thenew parametervaluesfor bag1 will bethesameasthoseforbag2, by symmetry. Intuitively, if we assumeinitially that thebagsareidentical,thenit is asif we hadjustonebag.Likelihoodis maximizedunderthis constraintby settingtheproportionsof candytypeswithin eachbagto theobservedproportions.(Seeproofbelow.)d. Webegin by writing out � for thedatain thetable:� £ ¨ «1� Ñ Ó1ÔÀ� ���2Û ¶ �QÜ ¶ �2Ý ¶ § � �µ·H�~¡+�2Û Õ �2Ü Õ �2Ý Õ ¡§ ç � Ñ Ó1ÔZ� ���2Û ¶ �QÜ ¶ � ��·H�QÝ ¶ ¡À§ � �µ·H�>¡+�QÛ Õ �QÜ Õ � �µ·H�2Ý Õ ¡.¡À§ ÌoÌoÌNow wecandifferentiatewith respectto eachparameter. For example, � �QÝ ¶ £ ¨ «1� ���2Û ¶ �2Ü ¶���2Û ¶ �2Ü ¶ �2Ý ¶ § � ��·>�~¡+�QÛ Õ �2Ü Õ �QÝ Õ· ç � �Ö�QÛ ¶ �QÜ ¶��� Û ¶ � Ü ¶ � �µ·H� Ý ¶ ¡<§ � �µ·H�~¡+� Û Õ � Ü Õ � ��·>� Ý Õ ¡ §ÌoÌoÌNow if �QÛ ¶ £Þ�2Û Õ , �2Ü ¶ £Þ�2Ü Õ , and �2Ý ¶ £��QÝ Õ , thedenominatorssimplify, everythingcancels,andwehave � �QÝ ¶ £��¨ «1��QÝ ¶ ·ç �� �µ·H�2Ý ¶ ¡ §ÌoÌoÌ £�� ¥1¥ ��QÝ ¶ ·Ù ¥ �� �µ·H�2Ý ¶ ¡http://librosysolucionarios.net139First,notethat ��d  �2Ý Õ will havethesamevalueexceptthat � and� �·W�~¡ arereversed,sotheparametersfor bags1 and2 will move in lock stepif ��£©�1� ¥ . Second,notethat �Od  �2Ý ¶ £¼� when �2Ý ¶ £ ¥1¥ �d � Ù ¥ �§ ¥1¥ �1¡ , i.e., exactly the observed proportionofcandieswith holes. Finally, we cancalculatesecondderivativesandevlauatethematthefixedpoint. For example,we obtainÂ Õ � � ÕÝ ¶ £ Û�� Õ �2Ý ¶ � ��·H�QÝ ¶ ¡ � ¨ �QÝ ¶ ·¸�1¡which is negative (indicating the fixed point is a maximum)only when � Ý ¶ ¬þ�1� ¥ .Thus,in generalthefixedpoint is asaddlepoint assomeof thesecondderivativesmaybepositiveandsomenegative. Nonetheless,EM canreachit by moving alongtheridgeleadingto it, aslongasthesymmetryis unbroken.20.11 XOR (in factany Booleanfunction)is easiestto constructusingstep-functionunits.BecauseXOR is not linearlyseparable,wewill needahiddenlayer. It turnsout that justonehiddennodesuffices. To designthenetwork, we canthink of theXOR functionasOR withtheAND case(both inputson) ruled out. Thusthehiddenlayer computesAND, while theoutputlayercomputesORbut weightstheoutputof thehiddennodenegatively. Thenetworkshown in FigureS20.3doesthetrick.t = 0.5 t = 0.2W = 0.3W = 0.3W = 0.3W = 0.3W =− 0.6FigureS20.3 A network of step-functionneuronsthatcomputestheXOR function.20.12 Theexamplesmapfromô q ¶ Á q Õ õ toô q ¶ Á q ¶ Á q Õ õ coordinatesasfollows:ô ·¹� Á ·¹��õ (negative) mapstoô ·¹� Á §¹��õô ·¹� Á §¹��õ (positive) mapstoô ·É� Á ·¹��õô §¹� Á ·¹��õ (positive) mapstoô §É� Á ·¹��õô §¹� Á §¹��õ (negative) mapstoô §¹� Á §¹��õThus,thepositive exampleshave q ¶ q Õ £ ·@� andthenegative exampleshave q ¶ q Õ £ § � .The maximummargin separatoris the line q ¶ q Õ £u� , with a margin of 1. The separatorcorrespondsto the q ¶ £u� and q Õ £u� axesin theoriginalspace—thiscanbethoughtof asthelimit of ahyperbolicseparatorwith two branches.20.13 The perceptronadjuststhe separatinghyperplanedefinedby the weightsso as tominimizethetotal error. Thequestionassumesthat theperceptronis trainedto convergence(if possible)on theaccumulateddatasetaftereachnew examplearrives.Therearetwo phases.With few examples,thedatamayremainlinearly separableandthehyperplanewill separatethepositiveandnegativeexamples,althoughit will notrepresenthttp://librosysolucionarios.net140 Chapter 20. StatisticalLearningMethodstheXOR function.Oncethedatabecomenon-separable,thehyperplanewill movearoundtofind a local minimum-errorconfiguration.With parity data,positive andnegative examplesaredistributeduniformly throughouttheinputspaceandin thelimit of largesamplesizestheminimumerror is obtainedby a weightconfigurationthatoutputsapproximately0.5 for anyinput. However, in this region theerrorsurfaceis basicallyflat, andany smallfluctuationinthe local balanceof positive andnegative examplesdueto thesamplingprocessmaycausetheminimum-errorplaneto movearounddrastically.Hereis somecodeto try outagiventrainingset:(setq *examples*’(((T . 0) (I1 . 1) (I2 . 0) (I3 . 1) (I4 . 0))((T . 1) (I1 . 1) (I2 . 0) (I3 . 1) (I4 . 1))((T . 0) (I1 . 1) (I2 . 0) (I3 . 0) (I4 . 1))((T . 0) (I1 . 0) (I2 . 1) (I3 . 1) (I4 . 0))((T . 0) (I1 . 0) (I2 . 0) (I3 . 1) (I4 . 1))((T . 1) (I1 . 1) (I2 . 0) (I3 . 0) (I4 . 0))((T . 0) (I1 . 1) (I2 . 1) (I3 . 0) (I4 . 0))((T . 1) (I1 . 0) (I2 . 1) (I3 . 1) (I4 . 1))((T . 0) (I1 . 0) (I2 . 1) (I3 . 1) (I4 . 0))))(deftest ex20.13((setq problem(make-learning-problem:attributes ’((I1 0 1) (I2 0 1) (I3 0 1) (I4 0 1)):goals ’((T 0 1)):examples (subseq *examples* 0 <n>)))) ;;;; vary <n> as needed;; Normally we’d call PERCEPTRON-LEARNINGhere, but we need added control;; to set all the weights to 0.1, so we build the perceptron ourselves:((setq net (list (list (make-unit :parents (iota 5):children nil:weights ’(-1.0 0.1 0.1 0.1 0.1):g #’(lambda (i) (step-function 0 i)))))));; Now we call NN-LEARNING with the perceptron-update method,;; but we also make it print out the weights, and set *debugging* to t;; so that we can see the epoch number and errors.((let ((*debugging* t))(nn-learningproblem net#’(lambda (net inputs predicted target &rest args)(format t ‘‘˜&˜A Weights = ˜{˜4,1F ˜}˜%’’(if (equal target predicted) ‘‘YES’’ ‘‘NO ‘‘)(unit-weights (first (first net))))(apply #’perceptron-update net inputs predicted target args))))))Up to the first 5 examples,the network convergesto zeroerror, andconvergesto non-zeroerrorthereafter.20.14 Accordingto ?(?), thenumberof linearlyseparableBooleanfunctionswith ³ inputsisÞ ² £ ¨ ²P ËQ¾¨ ² ·¸�Ýhttp://librosysolucionarios.net141For ³©ó ¨wehaveÞ ² º ¨ � ³W§¸�1¡ ¨ ² ·¸�³ £ ¨ � ³ã§»�1¡ Ì � ¨ ² ·¸�1¡ }³ } � ¨ ² ·�³ ·¸�1¡ } º¨ � ³f§Ø�1¡ � ¨ ² ¡ ²³ } º ¨ ² =sothefractionof representablefunctionsvanishesas³ increases.20.15 Theseexamplesweregeneratedfrom a kind of majority or voting function, whereeachinput hasa differentnumberof votes:10 forí ¶, 4 forí Õtoí ¢ , 2 forí � , and1 forí½ß. Ifyouassignthisproblem,it is probablyagoodideato tell thestudentsthis. FigureS??showsa perceptronanda feed-forward netwith logical nodesthat representthis function. Our in-tuition was that the function shouldhave beeneasyto learnwith a perceptron,but harderwith otherrepresentationssuchasa decisiontree. However, it turnsout that therearenotenoughexamplesfor evenaperceptronto learn.In retrospect,thatshouldnotbetoosurpris-ing, asthereareover �1� ¶¿à differentBooleanfunctionsof six inputs,andonly 14 examples.Runningthe following code(which makesuseof the codein learning/nn.lis p andlearning/percep tr on.l isp ) we find that theperceptronquickly convergesto learnall 14 trainingexamples,but it performsatchancelevel on thefive-exampletestsetwemadeup (getting2 or 3 out of 5 right on mostruns). (Thestudentwho did not know what theun-derlyingfunctionwaswouldhave to keepout someof theexamplesto useasa testset.)Theweightsvary widely from run to run, althoughin every run theweight forí ¶is thehighest(asit shouldbefor thespecifiedfunction),andtheweightsforí � andí½ßareusuallylow, butsometimesí½ßis higherthanothernodes.This maybea resultof the fact that theexampleswerechosento representsomeborderlinecaseswhereí½ßcaststhe decidingvote. So thisservesasa lesson:if youare“clever” in choosingexamples,but rely on a learningalgorithmthatassumesexamplesarechosenat random,you will run into trouble.Hereis thecodeweused:(defun test-nn (net problem &optional(examples (learning-problem-examples problem)))(let ((correct 0))(for-each example in examples do(if (eql (cdr (first example))(first (nn-output net (rest example)(learning-problem-attributes problem)nil)))(incf correct)))(values correct ’out-of (length examples))))(deftest ex20.15((setq examples’(((T . 1) (I1 . 1) (I2 . 0) (I3 . 1) (I4 . 0) (I5 . 0) (I6 . 0))((T . 1) (I1 . 1) (I2 . 0) (I3 . 1) (I4 . 1) (I5 . 0) (I6 . 0))((T . 1) (I1 . 1) (I2 . 0) (I3 . 1) (I4 . 0) (I5 . 1) (I6 . 0))((T . 1) (I1 . 1) (I2 . 1) (I3 . 0) (I4 . 0) (I5 . 1) (I6 . 1))((T . 1) (I1 . 1) (I2 . 1) (I3 . 1) (I4 . 1) (I5 . 0) (I6 . 0))((T . 1) (I1 . 1) (I2 . 0) (I3 . 0) (I4 . 0) (I5 . 1) (I6 . 1))((T . 0) (I1 . 1) (I2 . 0) (I3 . 0) (I4 . 0) (I5 . 1) (I6 . 0))((T . 1) (I1 . 0) (I2 . 1) (I3 . 1) (I4 . 1) (I5 . 0) (I6 . 1))((T . 0) (I1 . 0) (I2 . 1) (I3 . 1) (I4 . 0) (I5 . 1) (I6 . 1))((T . 0) (I1 . 0) (I2 . 0) (I3 . 0) (I4 . 1) (I5 . 1) (I6 . 0))http://librosysolucionarios.net142 Chapter 20. StatisticalLearningMethods((T . 0) (I1 . 0) (I2 . 1) (I3 . 0) (I4 . 1) (I5 . 0) (I6 . 1))((T . 0) (I1 . 0) (I2 . 0) (I3 . 0) (I4 . 1) (I5 . 0) (I6 . 1))((T . 0) (I1 . 0) (I2 . 1) (I3 . 1) (I4 . 0) (I5 . 1) (I6 . 1))((T . 0) (I1 . 0) (I2 . 1) (I3 . 1) (I4 . 1) (I5 . 0) (I6 . 0)))))((setq problem(make-learning-problem:attributes ’((I1 0 1) (I2 0 1) (I3 0 1) (I4 0 1) (I5 0 1) (I6 0 1)):goals ’((T 0 1)):examples examples)))((setq net (perceptron-learning problem)))((setq weights (unit-weights (first (first net)))))((test-nn net problem)(= * 14))((test-nn net problem’(((T . 1) (I1 . 1) (I2 . 0) (I3 . 0) (I4 . 1) (I5 . 0) (I6 . 0))((T . 0) (I1 . 0) (I2 . 0) (I3 . 1) (I4 . 1) (I5 . 1) (I6 . 1))((T . 1) (I1 . 1) (I2 . 1) (I3 . 0) (I4 . 0) (I5 . 1) (I6 . 0))((T . 0) (I1 . 1) (I2 . 0) (I3 . 0) (I4 . 0) (I5 . 0) (I6 . 0))((T . 1) (I1 . 0) (I2 . 1) (I3 . 1) (I4 . 1) (I5 . 1) (I6 . 1))))(= * 5)))20.16 The probability Ú outputby the perceptronis � � ï _ ï�Ð5ï ¡ , where� is the sigmoidfunction.Since�>ö1£e� � �µ·4��¡ , wehave Ú-d  _ ï £:� ö � ï _ ï Ð ï ¡ Ð ï £ Ú � �µ·�Ú�¡ Ð ïFor adatumwith actualvalue� , thelog likelihoodis� £:� Ñ Ó1Ô Ú�§ � �µ·8�<¡ Ñ Ó1Ô�� �µ·�Ú�¡sothegradientof thelog likelihoodwith respectto eachweightis � _ ï £ � ÚÌ Â Ú _ ï¹· �µ·8��µ·�ÚÌ Â Ú _ ï£ ��Ú � ��·�Ú�¡ Ð5ïÚ · � �µ·8�<¡¿Ú � �µ·©Ú�¡ Ðsï�µ·�Ú £ � �T·�Ú�¡ Ð5ï £�áOâQâp� Ð5ï �20.17 Thisexercisereinforcesthestudent’s understandingof neuralnetworksasmathemat-ical functionsthat canbeanalyzedat a level of abstractionabove their implementationasanetwork of computingelements.For simplicity, we will assumethat theactivation functionis thesamelinear functionat eachnode: � � qi¡ £ Î�qW§�o . (Theargumentis thesame(onlymessier)if we allow different νP and o�P for eachnode.)a. Theoutputsof thehiddenlayerare� ï £>� ° _ °~ ï í.° £½Î ° _ °~ ï í.° §HoThefinal outputsareg PS£:� ï _ ï  P�� ï £½Î ï _ ï  P Î ° _ °~ ï í.° §Ho §Hohttp://librosysolucionarios.net143Now we just have to seethatthis is linearin theinputs:g PS£½Î Õ ° í.° ï _ °~ ï _ ï  P�§Ho �µ§¸Î ï _ ï  PThuswecancomputethesamefunctionasthetwo-layernetwork usingjustaone-layerperceptronthat hasweights _ °~ P £ ï _ °~ ï _ ï  P andan activation function � � qi¡T£Î Õ q�§Ho �µ§ØÎ ï _ ï  P .b. Theabove reductioncanbeusedstraightforwardly to reducean ³ -layernetwork to an� ³�·6�1¡ -layer network. By induction,the ³ -layernetwork canbereducedto a single-layernetwork. Thus,linearactivationfunctionrestrictneuralnetworksto representonlylinearly functions.20.18 Theimplementationof neuralnetworkscanbeapproachedin severaldifferentways.Thesimplestis probablyto storetheweightsin an ³¼�T³ array. Everythingcanbecalculatedasif all the nodeswerein eachlayer, with the zeroweightsensuringthat only appropriatechangesaremadeaseachlayeris processed.Particularly for sparsenetworks, it can be more efficient to usea pointer-basedim-plementation,with eachnoderepresentedby a datastructurethat containspointersto itssuccessors(for evaluation)andits predecessors(for backpropagation).Weights _ ï  P areat-tachedto nodeÝ . In bothtypesof implementation,it is convenientto storethesummedinputÝm³GPᣠï _ ï  P Ðsï and the value �>ö � Ým³GPl¡ . The coderepositorycontainsan implementationof the pointer-basedvariety. Seethe file learning/algori th ms/nn .l is p, and thefunctionnn-learning in thatfile.20.19 Thisquestionis especiallyimportantfor studentswhoarenotexpectedto implementor useaneuralnetwork system.Togetherwith 20.15and20.17,it givesthestudentaconcrete(if slender)graspof what the network actuallydoes. Many othersimilar questionscanbedevised.Intuitively, the datasuggestthat a probabilisticpredictionå¹� g Æ È`Ú Æ È<£u�1¡W£ �1�c� isappropriate.Thenetwork will adjustit* weightsto minimizetheerrorfunction.Theerroris� £ �¨ P� � P · Ð P ¡ Õ £ � ¨ ô �1� � ��· Ð ¶ ¡Õ § ¨ � � �µ· Ð ¶ ¡ Õ õI£ ¥ � g Õ ¶ ·»�1� g ¶ § ¥ �Thederivative of theerrorwith respectto thesingleoutputÐ ¶is � Р¶£½�1�1� Ð ¶ ·¸�1�Settingthe derivative to zero,we find that indeedÐ ¶ £ �1�c� . The studentshouldspot theconnectionto Ex. 18.8.20.20 Theapplicationof cross-validationis straightforward—themethodologyis thesameasthatfor any parameterselectionproblem.With 10-foldcross-validation,for example,eachsizeof hiddenlayer is evaluatedby training on 90% subsetsand testingon the remaining10%. Thebest-performingsizeis thenchosen,trainedon all thetrainingdata,andtheresultis returnedas the system’s hypothesis.The purposeof this exerciseis to have the studenthttp://librosysolucionarios.net144 Chapter 20. StatisticalLearningMethodsunderstandhow to designtheexperiment,runsomecodeto seeresults,andanalyzetheresult.The higher-orderpurposeis to causethe studentto questionresultsthat make unwarrantedassumptionsabouttherepresentationusedin a learningproblem,whetherthatrepresentationis thenumberof hiddennodesin aneuralnet,or any otherrepresentationalchoice.20.21 Themainpurposeof thisexerciseis to make concretethenotionof thecapacityof afunctionclass(in this case,linearhalfspaces).It canbehardto internalizethis concept,buttheexamplesreallyhelp.a. Threepointsin generalpositionon a planeform a triangle. Any subsetof the pointscan be separatedfrom the rest by a line, as can be seenfrom the two examplesinFigureS20.4(a).b. FigureS20.4(b)shows two caseswherethepositive andnegative examplescannotbeseparatedby a line.c. Fourpointsin generalpositionon a planeform a tetrahedron.Any subsetof thepointscanbe separatedfrom the restby a plane,ascanbe seenfrom the two examplesinFigureS20.4(c).d. FigureS20.4(d)shows a casewherea negative point is insidethe tetrahedronformedby four positive points;clearlynoplanecanseparatethetwo sets.(a) (b)(c) (d)FigureS20.4 IllustrativeexamplesforEx. 20.21.http://librosysolucionarios.netSolutionsfor Chapter21ReinforcementLearning21.1 Thecoderepositoryshows anexampleof this, implementedin thepassive Ù��f� envi-ronment.Theagentsarefoundunderlisp/learning/ agents /pa ss iv e* .l isp andthe environmentis in lisp/learning/do main s/ 4x 3-p as si ve -mdp. li sp . (TheMDP is convertedto a full-blown environmentusing the function mdp->environmen twhichcanbefoundin lisp/uncertain ty /e nvi ro nments /mdp. li sp .)21.2 Considera world with two states,̄¾and¯ ¶, with two actionsin eachstate:staystillor move to theotherstate.Assumethemoveactionis non-deterministic—itsometimesfails,leaving theagentin thesamestate.Furthermore,assumetheagentstartsin¯ ¾andthat¯ ¶is aterminalstate.If theagenttriesseveralmoveactionsandthey all fail, theagentmayconcludethat � �.¯ ¾ Á�ã�ä�å � Á ¯ ¶ ¡ is 0, andthusmay choosea policy with û �.¯ ¾ ¡�£çæ'è¡��é , which is animproperpolicy. If we wait until theagentreaches̄¶beforeupdating,we won’t fall victimto this problem.21.3 This questionessentiallyasksfor a reimplementationof a generalschemefor asyn-chronousdynamicprogrammingof which the prioritized sweepingalgorithm is an exam-ple(MooreandAtkeson,1993).For a., thereis codefor apriority queuein boththeLisp andPythoncoderepositories.Somostof thework is theexperimentationcalledfor in b.21.4 Whenthereareno terminalstatesthereareno sequences,so we needto definese-quences.We cando that in severalways. First, if rewardsaresparse,we cantreatany statewith a rewardastheendof asequence.We canuseequation(21.2);theonly problemis thatwe don’t know theexact totl rewardof thestateat theendof thesequence,becauseit is nota terminalstate. We canestimateit using the current ® � Þs¡ estimate.Anotheroption is toarbitrarily limit sequencesto ³ states,andthenconsiderthenext ³ states,etc.A variationonthis is to useasliding window of states,sothatthefirst sequenceis states�Q�o�o�!³ , thesecondsequenceis¨ �o�o�.³f§¸� , etc.21.5 The idea hereis to calculatethe reward that the agentwill actually obtain using agivenestimateof ® anda givenestimatedmodel�. This is distinct from thetrueutility ofthestatesvisited.First,wecomputethepolicy for theagentby calculating,for eachstate,theactionwith thehighestestimatedutility:åÉ� Ýw¡�£/!2* Ô .0!213 ï � 3P ï ® � � ¡145http://librosysolucionarios.net146 Chapter 21. ReinforcementLearningThentheexpectedvaluescanbe found by applyingvaluedeterminationwith policyåandcanthenbecomparedto theoptimalvalues.21.6 Theconversionof thevacuumworld problemspecificationinto anenvironmentsuit-ablefor reinforcementlearningcanbedoneby merging elementsof mdp->environmen tfrom lisp/uncertain ty /en vi ro nments/ mdp. li sp with elementsof the corre-spondingfunctionproblem->enviro nment from lisp/search/age nt s. lis p. Thepoint hereis twofold: first, thatthereinforcementlearningalgorithmsin Chapter20 arelim-ited to accessibleenvironments;second,that thedirt makesa hugedifferenceto thesizeofthe statespace.Without dirt, the statespaceisg � ³�¡ with ³ locations. With dirt, the statespaceisg � ¨ ² ¡ becauseeachlocationcanbe cleanor dirty. For this reason,input general-ization is clearly necessaryfor ³ above about10. This illustratesthe misleadingnatureof“navigation” domainsin which the statespaceis proportionalto the “physical size” of theenvironment. “Real-world” environmentstypically have somecombinatorialstructurethatresultsin exponentialgrowth.21.7 Codenot shown. Several reinforcementlearningagentsare given in the directorylisp/learning/a gent s .21.8 This utility estimationfunctionis similar to equation(21.9),but addsa termto repre-sentEuclideandistanceon agrid. Usingequation(21.10),theupdateequationsarethesamefor � ¾ through � Õ , andthenew parameter� � canbecalculatedby taking thederivative withrespectto � � :� ¾ � � ¾ §Øì � Æ ï�� Þ5¡<·Êê®Gë � Þ5¡.¡ Á� ¶ � � ¶ §Øì � Æ ï�� Þ5¡<· ê®Gë � Þ5¡.¡¿q Á� Õ � � Õ §Øì � Æ ï � Þ5¡<· ê®Gë � Þ5¡.¡¿� Á� � � � � §Øì � Æ ï�� Þ5¡<· ê®Gë � Þ5¡.¡ � q ·8qÆì1¡ Õ § � ��·8�Öì(¡ Õ �21.9 Possiblefeaturesinclude:� Distanceto thenearest§¹� terminalstate.� Distanceto thenearest·¹� terminalstate.� Numberof adjacent§É� terminalstates.� Numberof adjacent·É� terminalstates.� Numberof adjacentobstacles.� Numberof obstaclesthatintersectwith apathto thenearest§É� terminalstate.21.10 This is a relatively time-consumingexercise. Codenot shown to computethree-dimensionalplots.Theutility functionsare:a. ® � q Á �<¡¤£½�×·4� �.� �1��·4qi¡À§ � �1�F·8�Z¡.¡ is thetrueutility, andis linear.b. Sameasin a,exceptthat ® � �1� Á �1¡�£ª·¹� .c. Theexactutility dependson theexactplacementof theobstacles.Thebestapproxima-tion is thesameasin a. Thefeaturesin exercise21.9might improve theapproximation.http://librosysolucionarios.net147d. Theoptimal policy is to headstraightfor thegoal from any point on the right sideofthewall, andto headfor (5, 10) first (andthenfor thegoal) from any point on the leftof thewall. Thus,theexactutility functionis:® � q Á �Z¡¼£ë�µ·8� �.� �1�F·lqi¡À§ � �1�µ·8�<¡.¡ (if qãó ¥)£ë�µ·8� �.� ¥ ·8qi¡À§ � �1�F·8�<¡.¡À· ¥ � (if q©¬ ¥)Unfortunately, this is not linear in q and � , asstated.Fortunately, we canrestatetheoptimal policy as“headstraightup to row 10 first, thenheadright until column10.”Thisgivesusthesameexactutility asin a,andthesamelinearapproximation.e. ® � q Á �<¡¤£½��·l� � è ¥ ·�q è § è ¥ ·�� è ¡ is thetrueutility. This is alsonot linearin q and� ,becauseof theabsolutevaluesigns.All canbefixedby introducingthefeaturesè ¥ ·�q èandè ¥ ·4� è .21.11 Themodificationinvolvescombiningelementsof theenvironmentconverterfor games(game->environme nt in lisp/search/gam es .li sp ) with elementsof thefunctionmdp->environmen t . Therewardsignalis just theutility of winning/drawing/losingandoccursonly at theendof thegame.Theevaluationfunctionusedby eachagentis theutilityfunction it learnsthroughthe TD process.It is importantto keepthe TD learningprocess(which is entirely independentof the fact that a gameis being played)distinct from thegame-playingalgorithm.Usingtheevaluationfunctionwith a deepsearchis probablybetterbecauseit will helptheagentsto focusonrelevantportionsof thesearchspaceby improvingthequality of play. Thereis, however, a tradeoff: thedeeperthesearch,themorecomputertime is usedin playingeachtraininggame.21.12 Codenot shown.21.13 Reinforcementlearningas a general“setting” can be appliedto almostany agentin any environment. The only requirementis that therebe a distinguishedreward signal.Thus,given the signalsof pain, pleasure,hunger, andso on, we canmaphumanlearningdirectly into reinforcementlearning—althoughthis saysnothingabouthow the “program”is implemented.What this view missesout, however, is the importanceof other forms oflearningthat occurin humans.Theseinclude“speeduplearning” (Chapter21); supervisedlearningfrom otherhumans,wherethe teacher’s feedbackis takenasa distinguishedinput;andtheprocessof learningtheworld model,whichis “supervised”by theenvironment.21.14 DNA doesnot, asfar aswe know, sensetheenvironmentor build modelsof it. Therewardsignalis thedeathandreproductionof theDNA sequence,but evolution simplymod-ifies theorganismratherthanlearninga ® or � function. The really interestingproblemisdecidingwhatit is that is doingtheevolutionarylearning.Clearly, it is not theindividual (ortheindividual’sDNA) thatis learning,becausetheindividual’sDNA getstotally intermingledwithin afew generations.Perhapsyoucouldsaythespeciesis learning,but if soit is learningto produceindividualswho survive to reproducebetter;it is not learninganything to do withthespeciesasa wholeratherthanindividuals. In TheSelfishGene, RichardDawkins (1976)proposesthatthegeneis theunit thatlearnsto succeedasan“individual” becausethegeneispreservedwith small,accumulatedmutationsover many generations.http://librosysolucionarios.netSolutionsfor Chapter22Communication22.1 No answerrequired;just readthepassage.æ m íXîp� æ'�&ï¿ð�����è¡ñ å � Á �'�&òóï��(â Á+ô ��â2õ ä �i¡0ö îe� �'�&òóï��(â Á+ô ��â2õ ä �i¡ è �o�o�íXîp� ����õQ� Á �'�&òóï���â Á+ô �(âQõ ä �i¡ m î â ä � ä �&� � ����õQ� Á �'�&òCïQ��â Á+ô �(âQõ ä �i¡íXîp� ����õQ� Á �'�&òóï���â Á�÷ ��ñ¿â��<¡ m í ��òó� � �'�&òóï���â�¡ è í ä �&� � �'�&òCïQ��â�¡ è �o�o�ö îp� �'�&òóï���â Á+ô �(âQõ ä �i¡ m ö îe� �'�&òóï��(â Á+ô ��â2õ ä �i¡ íXîe��ø ï¼ðØ�Q��è¡ñ å � Á Á ¡ è �o�o�îOîùm î â�� ô�ä õ�ñ¿è¡ñ ä � í7îp��ø ï¼ðØ�Q��è¡ñ å � Á Á ¡î â ä � ä �&� � æ���ï¼ðØ�Q��è¡ñ å � Á æ'ñ¿�'���&úû��â Á�ü ñ¿â2õ�è�¡ m Iî â ä � ä �&� � æ���ï¼ðØ�Q��è¡ñ å � Á æ'ñ¿�'���&úû��â Á æ'��� ä �'�I¡ m youî â ä � ä �&� � æ���ï¼ðØ�Q��è¡ñ å � Á æ'ñ¿�'���&úû��â Á�÷ ��ñ"â��I¡ m heèsheèitî â ä � ä �&� � æ���ï¼ðØ�Q��è¡ñ å � Á î úû�&â���ú Á�ü ñ¿âQõQè�¡ m weî â ä � ä �&� � æ���ï¼ðØ�Q��è¡ñ å � Á î úû�&â���ú Á æ'�Q� ä ���À¡ m youî â ä � ä �&� � æ���ï¼ðØ�Q��è¡ñ å � Á î úû�&â���ú Á�÷ ��ñ¿â��À¡ m theyî â ä � ä �&� ��ø ï¼ðØ�Q��è¡ñ å � Á æ'ñ¿�'���&úû��â Á�ü ñ¿â2õ�è�¡ m meî â ä � ä �&� ��ø ï¼ðØ�Q��è¡ñ å � Á æ'ñ¿�'���&úû��â Á æ'��� ä �'�<¡ m youî â ä � ä �&� ��ø ï¼ðØ�Q��è¡ñ å � Á æ'ñ¿�'���&úû��â Á�÷ ��ñ"â��<¡ m himèherèitî â ä � ä �&� ��ø ï¼ðØ�Q��è¡ñ å � Á î úû�&â���ú Á�ü ñ¿âQõQè�¡ m usî â ä � ä �&� ��ø ï¼ðØ�Q��è¡ñ å � Á î úû�&â���ú Á æ'�Q� ä ���<¡ m youî â ä � ä �&� ��ø ï¼ðØ�Q��è¡ñ å � Á î úû�&â���ú Á�÷ ��ñ¿â��<¡ m themö��(âQï � æ'ñ¿�'���&úû��â Á�ü ñ¿â2õ�è�¡ m smellö��(âQï � æ'ñ¿�'���&úû��â Á æ'��� ä �'�<¡ m smellö��(âQï � æ'ñ¿�'���&úû��â Á�÷ ��ñ"â��<¡ m smellsö��(âQï ��î úû�&â���ú Á ¡ m smellFigure S22.1 A partial DCG for ý « , modified to handlesubject–verb number/personagreementasin Ex. 22.2.22.2 SeeFigureS22.1for apartialDCG.Weincludebothpersonandnumberannotational-thoughEnglishreallyonly differentiatesthethird personsingularfor verbagreement(exceptfor theverbbe).22.3 SeeFigureS22.2148http://librosysolucionarios.net149íXîp� ����õQ� Á �'�&òóï���â Á�÷ ��ñ¿â��<¡ m í ��òó� � �'�&òóï���â�¡íXîp� ����õQ� Á î úû�&â���ú Á�÷ ��ñ¿â��<¡ m í ä �&� ��î úû��â���ú`¡íXîp� ����õQ� Á �'�&òóï���â Á�÷ ��ñ¿â��<¡ m þ âQè¡ñ¿��úû� � �'�&òCïQ��â>¡ í ä ��� � �'�&òóï��(â~¡þ â2è¡ñ¿��úû� � æ'ñ¿�'�(��úû��â�¡ m aèanètheþ â2è¡ñ¿��úû� ��î úª�&â���ú`¡ m theèsomeèmanyFigure S22.2 A partial DCG for ý « , modified to handlearticle–nounagreementas inEx. 22.3.22.4 Thepurposeof thisexerciseis togetthestudentthinkingaboutthepropertiesof naturallanguage.Thereis awide varietyof acceptableanswers.Hereareours:� Grammar and Syntax Java: formally definedin a referencebook. Grammaticalityiscrucial; ungrammaticalprogramsarenot accepted.English: unknown, never formallydefined,constantlychanging.Most communicationis madewith “ungrammatical”ut-terances.Thereis anotionof gradedacceptability:someutterancesarejudgedslightlyungrammaticalor a little odd,while othersareclearlyright or wrong.� SemanticsJava: thesemanticsof a programis formally definedby thelanguagespec-ification. More pragmatically, onecansay that the meaningof a particularprogramis the JVM codeemittedby the compiler. English: no formal semantics,meaningiscontext dependent.� Pragmatics and Context-DependenceJava: somesmall partsof a programare leftundefinedin the languagespecification,andaredependenton thecomputeron whichtheprogramis run. English:almosteverythingaboutanutteranceis dependenton thesituationof use.� Compositionality Java: almostall compositional.Themeaningof “A + B” is clearlyderived from themeaningof “A” andthemeaningof “B” in isolation. English: somecompositionalparts,but many non-compositionaldependencies.� Lexical Ambiguity Java: asymbolsuchas“Avg” canbelocally ambiguousasit mightrefer to a variable,a class,or a function. The ambiguitycanbe resolved simply bycheckingthe declaration;declarationsthereforefulfill in a very exact way the roleplayedby backgroundknowledgeandgrammaticalcontext in English. English:muchlexical ambiguity.� Syntactic Ambiguity Java: thesyntaxof the languageresolvesambiguity. For exam-ple, in “if (X) if (Y) A; elseB;” onemight think it is ambiguouswhetherthe “else”belongsto thefirst or second“if, ” but thelanguageis specifiedsothatit alwaysbelongsto thesecond.English:muchsyntacticambiguity.� ReferenceJava: thereis apronoun“this” to referto theobjecton which amethodwasinvoked. Otherthanthat,thereareno pronounsor othermeansof indexical reference;no “it,” no “that.” (Comparethis to stack-basedlanguagessuchasForth, wherethestackpointeroperatesasa sortof implicit “it.”) Thereis referenceby name,however.Note thatambiguitiesaredeterminedby scope—iftherearetwo or moredeclarationshttp://librosysolucionarios.net150 Chapter 22. Communicationof thevariable“X”, thenauseof X refersto theonein theinnermostscopesurroundingtheuse.English:many techniquesfor reference.� Background KnowledgeJava: noneneededto interpreta program,althougha local“context” is built up asdeclarationsareprocessed.English:muchneededto do disam-biguation.� Understanding Java: understandingaprogrammeanstranslatingit to JVM bytecode.English:understandinganutterancemeans(amongotherthings)respondingto it appro-priately;participatingin adialog(or choosingnotto participate,but having thepotentialability to do so).As a follow-up question,you might want to comparedifferentlanguages,for example:En-glish,Java,Morsecode,theSQL databasequerylanguage,thePostscriptdocumentdescrip-tion language,mathematics,etc.22.5 This exerciseis designedto clarify the relationbetweenquasi-logicalform and thefinal logical form.a. Yes. Without a quasi-logicalform it is hardto write rulesthat produce,for example,two differentscopesfor quantifiers.b. No. It justmakesambiguitiesandabstractionsmoreconcise.c. Yes.Youdon’t needto explicitly representapotentiallyexponentialnumberof disjunc-tions.d. Yesandno. Theform is moreconcise,andsoeasierto manipulate.On theotherhand,the quasi-logicalform doesn’t give you any cluesas to how to disambiguate.But ifyoudo have thoseclues,it is easyto eliminateawholefamily of logical formswithouthaving to explicitly expandthemout.22.6 Assumingthatí Þ is theinterpretationof “is,” andí È is theinterpretationof “it,” thenwe getthefollowing:a. It is awumpus:ÿ ßØßX� í Þ �.í È Á ô ÿ ] _ Æ y Ú Æ Þ � ]¡¿õr¡�� à Æ 9(Ým³Z� � ÛãÜ�] Á ß�¡b. Thewumpusis dead:ÿ ßØßX�©àáß Ð o � ô ÿ } ] _ Æ y Ú Æ Þ � ]¡¿õr¡Z� à Æ 9(Ým³Z� � ÛãÜQ] Á ß�¡c. Thewumpusis in 2,2:ÿ ßØßX� í Þ � ô ÿ } ] _ Æ y Ú Æ Þ � ]¡¿õ Á �Z¡�� í ³ � � Á ô ¨1Áo¨ õr¡G� à Æ 9(Ým³Z� � ÛãÜQ] Á ßs¡We shoulddefinewhatí Þ means—onereasonableaxiom for one senseof “is” would beR q Á � í Þ � q Á �Z¡�� � q £^�<¡ . (This is the “is thesameas” sense.Thereareothers.)Theformulaÿ qÊ_ Æ y Ú Æ Þ � qi¡ is a reasonablesemanticsfor “It is a wumpus.” Theproblemis ifweusethatformula,thenwehavenowhereto gofor “It wasawumpus”—thereis noeventtowhich we canattachthetime information.Similarly, for “It wasn’t a wumpus,” we can’t useê ÿ qc_ Æ y Ú Æ Þ � qi¡ , nor couldwe useÿ q ê _ Æ y Ú Æ Þ � qS¡ . So it is bestto have anexpliciteventfor “is.”http://librosysolucionarios.net15122.7 This is a very difficult exercise—mostreadershave no ideahow to answertheques-tions (except perhapsto rememberthat “too few” is betterthan “too many”). This is thewholepoint of theexercise,aswe will seein exercise23.14.22.8 Thepurposeof this exerciseis to getsomeexperiencewith simplegrammars,andtoseehow context-sensitivegrammarsaremorecomplicatedthancontext-free.Oneapproachtowriting grammarsis to write down thestringsof thelanguagein anorderlyfashion,andthenseehow a progressionfrom onestring to thenext couldbecreatedby recursive applicationof rules.For example:a. The languageÐ ² Ö ² : Thestringsare � , Ð Ö , Ð�Ð Ö�Ö , . . . (where � indicatesthenull string).Eachmemberof this sequencecanbe derived from the previous by wrappinganÐatthestartanda Ö at theend.Thereforeagrammaris:¯ m �¯ m Ð�¯ Öb. Thepalindromelanguage:Let’s assumethealphabetis justÐ, Ö and Î . (In general,thesizeof thegrammarwill beproportionalto thesizeof thealphabet.Thereis no way towrite a context-freegrammarwithout specifyingthealphabet/lexicon.) Thestringsofthelanguageinclude � , a, b, c, aa,bb,cc,aaa,aba,aca,bab,bbb,bcb,. . . . In general,astringcanbeformedby bracketingany previousstringwith two copiesof any memberof thealphabet.Soa grammaris:¯ m �è Ð è Ö è Î è Ð�¯¼Ð è Ö ¯ Ö è Î ¯ Îc. The duplicatelanguage:For the moment,assumethat the alphabetis justÐ Ö . (It isstraightforward to extendto a largeralphabet.)Theduplicatelanguageconsistsof thestrings: � , Ð�Ð , Ö�Ö , Ð�Ð�Ð�Ð , Ð Ö Ð Ö , Ö�Ö�Ö�Ö , Ö Ð Ö Ð , . . .Notethatall stringsareof evenlength.Onestrategy for creatingstringsin this languageis this:� Start with markers for the front and middle of the string: we can usethe non-terminal \ for thefront and�for themiddle. Soat this point we have thestring\ � .� Generateitemsat the front of thestring: generateanÐfollowed by an S , or a Öfollowedby a f . Eventuallywe get,say, \ Ð S Ð SpÖ�f � . Thenwe no longerneedthe \ marker andcandeleteit, leavingÐ S Ð SFÖ�f � .� Move thenon-terminalsS and f down the line until just beforethe�. We endup withÐ�Ð Ö�S7SXf � .� Hop the S s and f s over the�, converting eachto a terminal(Ðor Ö ) aswe go.Thenwedeletethe�, andareleft with theendresult:Ð�Ð Ö Ð�Ð Ö .http://librosysolucionarios.net152 Chapter 22. CommunicationHereis agrammarto implementthisstrategy:¯ m \ � (startingmarkers)\ m \ Ð S (introducesymbols)\ m \½Öaf\ m � (deletethe \ marker)S Ð m Ð S (movenon-terminalsdown to the�)S6Ö m ÖaSf Ð�m Ð ffÿÖ m ÖafS � m �BÐ(hopover�andconvert to terminal)f � m � Ö� m � (deletethe�marker)Hereis a traceof thegrammarderivingÐ�Ð Ö Ð�Ð Ö :�� �� ����� �������� ����������� ����������� ������ ��������������������� ������� ������� ����������22.9 Grammar(A) doesnot work, becausethereis no way for theverb“walked” followedby theadverb “slowly” andtheprepositionalphrase“to the supermarket” to be parsedasaverbphrase.A verbphrasein (A) musthave eithertwo adverbsor bejusta verb. Hereis theparseundergrammar(B):S---NP-+-Pro---Someone||-VP-+-V---walked||-Vmod-+-Adv---slowly||-Vmod---Adv---PP---Prep-+-to||-NP-+-Det---the||-NP---Noun---supermarkethttp://librosysolucionarios.net153Hereis theparseundergrammar(C):S---NP-+-Pro---Someone||-VP-+-V---walked||-Adv-+-Adv---slowly||-Adv---PP---Prep-+-to||-NP-+-Det---the||-NP---Noun---supermarket22.10 Codenot shown.22.11 Codenot shown.22.12 This is the grammarsuggestedby the exercise. Thereareadditionalgrammaticalconstructionsthat could be coveredby moreambitiousgrammars,without addingany newvocabulary.¯ m Û å½�åÛ ånm ÛWÜ Æ ³Û ånm S7o � ß�Î`ÈlÝ ? ß ÛãÜ Æ ³Û ånm Û å Û åª� ß+9Ö (for relative clause)�åcm � ß%9ÖÒÛ åSXo � ß�Î`ÈlÝ ? ß m f Æ YZY Ð�� ÜÛãÜ Æ ³ m Ö Æ YZY Ð�� Ü� ß%9OÖ m Ö Æ YZY Ð�� ÜRunningtheparserfunctionparses from thecoderepositorywith this grammarandwithstringsof theform f Æ YZY Ð�� Ü ² , andthenjust countingthenumberof results,we get:N 1 2 3 4 5 6 7 8 9 10Number of parses 0 0 1 2 3 6 11 22 44 90To countthenumberof sentences,deMarcken implementeda parserlike thepacked forestparserof example22.10,exceptthat therepresentationof a forestis just anintegercountofthenumberof parses(andthereforethecombinationof ³ adjacentforestsis just theproductof theconstituentforests).He thengetsasingleintegerrepresentingtheparsesfor thewhole200-word sentence.22.13 Here’sonewayto draw theparsetreefor thestoryonpage823.Theparsetreeof thestudents’storieswill depnedon their choice.Segment(Evaluat io n)Segment(1) ‘‘A funny thing happened’’Segment(Ground- Fi gure)Segment(Cause)Segment(Enable)Segment(2) ‘‘John went to a fancy restaurant’’http://librosysolucionarios.net154 Chapter 22. CommunicationSegment(3) ‘‘He ordered the duck’’Segment(4) ‘‘The bill came to $50’’Segment(Cause)Segment(Enable)Segment(Explanat io n)Segment(5) ‘‘John got a shock...’’Segment(6) ‘‘He had left his wallet at home’’Segment(7) ‘‘The waiter said it was all right’’Segment(8) ‘‘He was very embarrassed...’ ’22.14 Now wecananswerthedifficult questionsof 22.7:� Thestepsaresortingtheclothesintopiles(e.g.,whitevs. colored);goingto thewashingmachine(optional); taking the clothesout and sorting into piles (e.g., socksversusshirts);puttingthepilesaway in theclosetor bureau.� Theactualrunningof thewashingmachineis neverexplicitly mentioned,sothatis onepossibleanswer. Onecouldalsosaythatdrying theclothesis amissingstep.� Thematerialis clothesandperhapsotherwashables.� Puttingtoomany clothestogethercancausesomecolorsto runontootherclothes.� It is betterto do too few.� Sothey won’t run; sothey get thoroughlycleaned;so they don’t causethemachinetobecomeunbalanced.http://librosysolucionarios.netSolutionsfor Chapter23ProbabilisticLanguageProcessing23.1 Codenot shown. The approachsuggestedherewill work in somecases,for authorswith distinctvocabularies.For moresimilar authors,otherfeaturessuchasbigrams,averageword andsentencelength,partsof speech,andpunctuationmight help. Accuracy will alsodependon how many authorsarebeingdistinguished.Oneinterestingway to make thetaskeasieris to groupauthorsinto maleandfemale,andtry to distinguishthesex of anauthornotpreviously seen.Thiswassuggestedby thework of ShlomoArgamon.23.2 Codenot shown. Thedistribution of wordsshouldfall alonga Zipfian distribution: astraightline on a log-log scale.Thegeneratedlanguageshouldbesimilar to theexamplesinthechapter.23.3 Codenot shown. Therearenow several open-sourceprojectsto do Bayesianspamfiltering, sobewareif youassignthisexercise.23.4 Doing the evaluationis easy, if a bit tedious(requiring150 pageevaluationsfor thecomplete10 documents� 3 engines� 5 queries).Explainingthedifferencesis morediffi-cult. Somethingsto checkarewhetherthegoodresultsin oneengineareeven in theotherenginesat all (by searchingfor uniquephraseson the page);checkwhetherthe resultsarecommerciallysponsored,areproducedby humaneditors,or arealgorithmicallydeterminedby asearchrankingalgorithm;checkwhethereachenginedoesthefeaturesmentionedin thenext exercise.23.5 Onegoodwaytodothisis tofirst findasearchthatyieldsasinglepage(or afew pages)by searchingfor rarewordsor phraseson thepage.Thenmake thesearchmoredifficult byaddingavariantof oneof thewordsonthepage—awordwith differentcase,differentsuffix,differentspelling,or a synonym for oneof thewordson thepage,andseeif thepageis stillreturned.(Make surethat thesearchenginerequiresall termsto matchfor this techniquetowork.)23.6 Computationslikethisaregivenin thebookManagingGigabytes(Wittenetal., 1999).Here’s oneway of doing thecomputation:Assumean averagepageis about10KB (givingusa 10TB corpus),andthat index sizeis linear in thesizeof thecorpus.Bahleet al. (2002)show anindex sizeof about2GB for a 22GBcorpus;soour billion pagecorpuswould haveanindex of about1TB.155http://librosysolucionarios.net156 Chapter 23. ProbabilisticLanguageProcessing23.7 Codenot shown. The simplestapproachis to look for a string of capitalizedwords,followed by “Inc” or “Co.” or “Ltd.” or similar markers. A morecomplex approachis toget a list of company names(e.g. from an online stock service),look for thosenamesasexactmatches,andalsoextractpatternsfrom them.Reportingrecallandprecisionrequiresaclearly-definedcorpus.23.8 Themainpoint of this exerciseis to show thatcurrenttranslationsoftwareis far fromperfect.Themistakesmadeareoftenamusingfor students.23.9 Hereis astartof agrammar:time => hour ":" minute| extendedhour| extendedhour "o’clock"| difference before_after extendedhourhour => 1 | 2 | ... | 24 | "one" | ... | "twelve"extendedhour => hour | "midnight" | "noon"minute => 1 | 2 | ... | 60before-after => "before" | "after" | "to" | "past"difference => minute | "quarter" | "half"23.10a. “I have never seenabetterprogramminglanguage”is easyfor mostpeopleto see.b. “Johnlovesmary” seemsto bepreferedto “Mary lovesJohn”(onGoogle,by amarginof 2240to 499,andbyasimilarmargin onasmallsampleof respondents),but bothareof courseacceptable.c. This oneis quite difficult. The first sentenceof the secondparagraphof Chapter22is “Communicationis the intentionalexchangeof information broughtaboutby theproductionandperceptionof signsdrawn from asharedsystemof conventionalsigns.”However, this cannotbereliably recoveredfrom thestringof wordsgivenhere.Codenotshown for testingtheprobabilitiesof permutations.23.11 In parlimentarydebate,a standardexpressionof approval is “bravo” in French,and“hear, hear” in English. That meansthat in going from Frenchto English, “bravo” wouldoftenhave a fertility of 2, but for Englishto French,thefertility distribution of “hear” wouldbehalf 0 andhalf 1 for this usage.For otherusages,it would have variousvalues,probablycenteredcloselyaround1.http://librosysolucionarios.netSolutionsfor Chapter24Perception24.1 Thesmallspacesbetweenleavesactaspinholecameras.Thatmeansthatthecircularlight spotsyou seeareactuallyimagesof thecircularsun.You cantestthis theorynext timethereis a solareclipse:thecircular light spotswill have a crescentbite takenout of themasthe eclipseprogresses.(Eclipseor not, the light spotsareeasierto seeon a sheetof paperthanon theroughforestfloor.)24.2 Givenlabelsontheoccludingedges(i.e., they areall arrows pointingin theclockwisedirection),therewill beno backtrackingat all if theorderis S7fä�à ; eachchoiceis forcedby the existing labels. With theorder fà SFä , the amountof backtrackingdependson thechoicesmade.Thefinal labellingis shown in FigureS24.1.24.3 Recallthattheimagebrightnessof aLambertiansurface(page743)isgivenbyí>� q Á �<¡¤£Ïn� q Á �Z¡.� s. Herethe light sourcedirections is alongthe q -axis. It is sufficient to considerahorizontalcross-section(in the q –� plane)of thecylinderasshown in FigureS24.2(a).Then,thebrightnessí�� qi¡µ£ Ï��lÓ# � � qS¡ for all thepointson theright half of thecylinder. TheleftA�B�C�D�+ ++−+ ++FigureS24.1 Labellingof theL-shapedobject(Exercise24.2).157http://librosysolucionarios.net158 Chapter 24. Perceptionhalf is in shadow. As qã£:9 �lÓ# � , wecanrewrite thebrightnessfunctionasí>� qS¡¤£ ° Ä whichrevealsthat the isobrightnesscontoursin the lit partof thecylinder mustbeequallyspaced.Theview from the�-axisis shown in FigureS24.2(b).xyxrzillumination�θviewer (a)�(b)�Figure S24.2 (a) Geometryof thesceneasviewedfrom alongthe � -axis. (b) Thescenefrom the � -axis,showing theevenlyspacedisobrightnesscontours.24.4 Welist thefour classesandgive two or threeexamplesof each:a. depth: Betweenthe top of thecomputermonitor andthewall behindit. Betweenthesideof theclock tower andthesky behindit. Betweenthewhite sheetsof paperin theforegroundandthebookandkeyboardbehindthem.b. surfacenormal: At thenearcornerof thepagesof thebookonthedesk.At thesidesofthekeys on thekeyboard.c. reflectance: Betweenthewhite paperandtheblack lineson it. Betweenthe“golden”bridgein thepictureandthebluesky behindit.d. illumination: On thewindowsill, theshadow from thecenterglasspanedivider. On thepaperwith Greektext, the shadow alongthe left from the paperon top of it. On thecomputermonitor, theedgebetweenthewhite window andthebluewindow is causedby differentilluminationby theCRT.24.5 This exerciserequiressomebasicalgebra,andenoughcalculusto know that ��� �"!$#&%� � #(' � # � . Studentswith freshmancalculusasbackgroundshouldbeableto handleit. Notethatall differentiationis with respectto ) . Crucially, thismeansthat �*�,+&!$#-%/.0�*�,+!�12.�)3%54 .Wework thesolutionfor thediscretecase;thecontinuous(integral) caseis similar.���768�"! # %9� :��*�,+&!,�"�,)<;3+!�! # (definitionof 6 )% :(���*�,+&!,�"�,)=;3+!�!$# (derivative of asum)% : �*�,+&!,�>#?�,)@;3+&! ' �A#,�,+!,�"�,)=;3+&! (since ��� �"!$#*%/� �># ' �A#B� )% :C�*�,+&!,� # �,)@;3+&! (since .D�*�,+!�12.�)3%54 )% �E6&�># (definitionof 6 )24.6 Beforeansweringthisexercise,wedraw adiagramof theapparatus(topview), shownin FigureS24.3.Noticethatwe make theapproximationthatthefocal lengthis thedistancefrom the lensto the imageplane;this is valid for objectsthatarefar away. Notice that thishttp://librosysolucionarios.net159512x512F pixels10cm10cm16cm16mObjectG0.5m0.5mFigureS24.3 Top view of thesetupfor stereoviewing (Exercise24.6).questionasksnothingaboutthe H coordinatesof points;we might aswell have a singlelineof 512pixelsin eachcamera.a. Solve this by constructingsimilar triangles:whosehypotenuseis thedottedline fromobject to lens,and whoseheight is 0.5 metersandwidth 16 meters. This is similarto a triangleof width 16cmwhosehypotenuseprojectsonto the imageplane;we cancomputethat its heightmustbe 0.5cm;this is theoffset from the centerof the imageplane.Theothercamerawill haveanoffsetof 0.5cmin theoppositedirection.Thusthetotal disparityis 1.0cm,or, at 512pixels/10cm,a disparityof 51.2pixels,or 51, sincethereareno fractionalpixels. Objectsthatarefartheraway will have smallerdisparity.Writing this asanequation,where . is thedisparityin pixels and I is thedistancetotheobject,we have:.D%5JLKM2N J pixelsN 4 cmK N2Ocm K 42P M mIb. In otherwords,this questionis askinghow muchfurther than16mcouldanobjectbe,andstill occupy thesamepixelsin theimageplane?Rearrangingtheformulaabove byswapping . and I , andpluggingin valuesof 51 and52 pixels for . , we getvaluesofI of 16.06and15.75meters,for a differenceof 31cm(a little over a foot). This is therangeresolutionat16 meters.c. In otherwords,this questionis askinghow far away would anobjectbe to generateadisparityof onepixel? Objectsfartherthanthis arein effect out of range;we can’t saywherethey arelocated.Rearrangingthe formula above by swapping . and I we get51.2meters.24.7 In the3-D case,the two-dimensionalimageprojectedby anarbitrary3-D objectcanvary greatly, dependingon the poseof the object. But with flat 2-D objects,the imageisalwaysthe“same,” exceptfor its orientationandposition. All featurepointswill alwaysbepresentbecausetherecanbe no occlusion,so QR%TS . Supposewe computethe centerofgravity of the imageandmodel featurepoints. For the model this canbe doneoffline; forthe imagethis is UD�,S! . Thenwe cantake thesetwo centerof gravity points,alongwith anhttp://librosysolucionarios.net160 Chapter 24. Perceptionarbitraryimagepoint andoneof the S modelpoints,andtry to verify thetransformationforeachof the S cases.Verificationis UD�,SWV X2Y�S! asbefore,sothewholeprocessis UD�,S[Z\V X2Y�S! .A follow-up exerciseis to look at someof thetricks usedby Olson(1994)to seeif they areapplicablehere.24.8 A, B, C canbe viewed in stereoandhencetheir depthscanbe measured,allowingtheviewer to determinethatB is nearest,A andC areequidistantandslightly furtheraway.Neither D nor E can be seenby both cameras,so stereocannotbe used. Looking at thefigure,it appearsthatthebottleoccludesD from Y andE from X, soD andE mustbefurtherawaythanA, B, C, but their relativedepthscannotbedetermined.Thereis, however, anotherpossibility (noticedby Alex Fabrikant). Rememberthateachcameraseesthecamera’s-eyeview not thebird’s-eye view. X seesDABC andY seesABCE. It is possiblethatD is verycloseto cameraX, so closethat it falls outsidethe field of view of cameraY; similarly, Emight be very closeto Y andbe outsidethe field of view of X. Hence,unlessthe camerashavea180-degreefield of view—probablyimpossible—thereis nowayto determinewhetherD andE arein front of or behindthebottle.24.9a. False.This canbequitedifficult, particularlywhensomepoint areoccludedfrom oneeyebut not theother.b. True. Thegrid createsanapparenttexturewhosedistortiongivesgoodinformationasto surfaceorientation.c. False.It appliesonly to trihedralobjects,excludingmany polyhedrasuchasfour-sidedpyramids.d. True.A bladecanbecomea fold if oneof theincidentsurfacesis warped.e. True.Thedetectabledepthdisparityis inverselyproportionalto ] .f. False.g. False.A disk viewededge-onappearsasastraightline.24.10 Thereare at least two reasons:(1) The leftmost car appears bigger and carsareusuallyroughlysimilar in size,thereforeit is closer. (2) It is assumedthattheroadsurfaceisanapproximatelyhorizontalgroundplane,andthatbothcarsareon it. In thatcase,becausetheleftmostcarappearslower in theimage,it mustbecloser.http://librosysolucionarios.net25 ROBOTICS25.1 To answerthis question,considerall possibilitiesfor the initial samplesbeforeandafterresampling.Thiscanbedonebecausethereareonly finitely many states.ThefollowingC++programcalculatestheresultsfor finite ^ . Theresultfor ^_%/` is simplytheposterior,calculatedusingBayesrule.intmain (int argc , char *arg v[]){// par se command line argu mentif (ar gc != 3){cerr << "Us age: " << arg v[0] << " <number of samples> "<< " <number of stat es>" << endl;exit (0);}int numSample s = atoi( argv [1]) ;int numStates = ato i(arg v[2] );cerr << "number of sampl es: " << numSampl es << endl<< "number of state s: " << numState s << endl;asse rt(n umSamples >= 1);asse rt(n umStates >= 1);// generate cou nterint samples[n umSamples ];for (int i = 0; i < numSampl es; i++ )samples[i ] = 0;// set up pro babi lity tabl esasse rt(n umStates == 4); // pre sentl y defi ned for 4 stat esdoub le cond Prob OfZ[ 4] = {0.8 , 0.4, 0.1, 0.1} ;doub le post erio rPro b[num Stat es];for (int i = 0; i < numState s; i++)post erior Prob [i] = 0.0 ;doub le even tPro b = 1.0 / pow(numState s, numSampl es);//lo op thro ugh all possi bili tiesfor (int done = 0; !done ; ){// compute impo rtanc e weig hts (is pro babil ity dist ribut ion)doub le weig ht[n umSamples ], tota lWei ght = 0.0 ;for (int i = 0; i < numSampl es; i++ )tota lWeig ht += weigh t[i] = cond Prob OfZ[s ampl es[i ]];// nor maliz e themfor (int i = 0; i < numSampl es; i++ )weig ht[i] /= total Weight;// cal culat e cont ribut ion to post erio r proba bili tyfor (int i = 0; i < numSampl es; i++ )post erior Prob [samp les[ i]] += even tPro b * wei ght[ i];// inc remen t coun terfor (int i = 0; i < numSampl es && i != -1;){samples[i ]++;if (sam ples [i] >= numState s)samples[ i++] = 0;elsei = -1;if (i == numSample s)done = 1;}}// pri nt resu ltcout << "Resu lt: ";for (int i = 0; i < numState s; i++)cout << " " << poste rior Prob [i];cout << endl;// cal culat e asymptoti c expe ctati ondoub le tota lWei ght = 0.0 ;for (int i = 0; i < numState s; i++)tota lWeig ht += condP robO fZ[i ];cout << "Unbi ased :";for (int i = 0; i < numState s; i++)cout << " " << condP robO fZ[i ] / tot alWeight;cout << endl;// cal culat e KL div ergen cedoub le kl = 0.0 ;for (int i = 0; i < numState s; i++)kl += poste rior Prob[ i] * (lo g(pos teri orPro b[i] ) -log( condP robO fZ[i] / tot alWei ght) );cout << "KL div erge nce: " << kl << endl ;}(a) (b)FigureS25.1 Codeto calculateanswerto exercise25.1.a. The program(correctly) calculatesthe following posteriordistributions for the fourstates,asafunctionof thenumberof sampleŝ . Notethatfor ^_% N, themeasurementis ignoredentirely! Thecorrectposteriorfor ^a%/` is calculatedusingBayesrule.161http://librosysolucionarios.net162 Chapter 25. Roboticsb cedsampleat f�gih c(dsampleat fkjlh c(dsampleat fkmnh c(dsampleat f�onhbqpsr0.25 0.25 0.25 0.25bqp=t0.368056 0.304167 0.163889 0.163889bqp=u0.430182 0.314463 0.127677 0.127677bqp=v0.466106 0.314147 0.109874 0.109874bqp=w0.488602 0.311471 0.0999636 0.0999636bqp=x0.503652 0.308591 0.0938788 0.0938788bqpsy0.514279 0.306032 0.0898447 0.0898447bqp=z0.522118 0.303872 0.0870047 0.0870047bqp={0.528112 0.30207 0.0849091 0.0849091bqpsr�|0.532829 0.300562 0.0833042 0.0833042bqp~}0.571429 0.285714 0.0714286 0.0714286b. Pluggingthe posteriorfor ^�%a` into the definition of the Kullback Liebler Diver-gencegivesus:b �W�d��c��?c h b �W�&d��c���c hbqpsr |�� u�z�x�u�t�{ bqp~y |>� |�|�z�|�v�{�z�tbqp=t |���r�t�{�u�v�u bqp<z |>� |�|�w�{�u�|�t�vbqp=u |�� |�w�x�uAr�{ bqp<{ |>� |�|�v�w�v�t�|�wbqp=v |�� |�t�{�vAy�w bqp~r�| |>� |�|�u�w�z�x�x�ubqp=w |>� | r�y�wAy�|�w bqp�} |c. Theproof for ^�% Nis trivial, sincethere-weightingignoresthemeasurementproba-bility entirely. Therefore,theprobabilityfor generatingasamplein any of thelocationsin � is givenby theinitial distribution,which is uniform.For ^9%_J , a proof is easilyobtainedby consideringall J���% N2Owaysin whichinitial samplesaregenerated:number samples probability �2���2�B��� weights probabilityof resamplingof sampleset for eachsample for eachsample for eachlocationin �1 0 0 ����� � � � �� �� ���� 0 0 02 0 1 ������ � � �� �� �� � ��¡ 0 03 0 2 ���� ��£¢� � �� � �� 0 �� �?� 04 0 3 ���� ��£¢� � �� � �� 0 0 �� �?�5 1 0 ����� � �� �� �� �� � ��¡ 0 06 1 1 ������ �� �� �� 0 ���� 0 07 1 2 ���� ��£¢�� �� � �0 �� ¢ � ¢ 08 1 3 ���� ��£¢�� �� � �0 �� ¢ 0 � ¢9 2 0 ����� � ��£¢� �� �� 0 �� �?� 010 2 1 ������ ��£¢� � �� 0 �� ¢ � ¢ 011 2 2 ���� ��£¢ ��£¢ �� �� 0 0 ���� 012 2 3 ���� ��£¢ ��£¢ �� �� 0 0 �� � �� �13 3 0 ����� � ��£¢� �� �� 0 0 �� �?�14 3 1 ������ ��£¢� � �� 0 �� ¢ 0 � ¢15 3 2 ���� ��£¢ ��£¢ �� �� 0 0 �� � �� �16 3 3 ���� ��£¢ ��£¢ �� �� 0 0 0 ����sumof all probabilities� �� �?�¥¤ �� � ¢�?�� �$¢�?�� �$¢http://librosysolucionarios.net163A quick checkshouldconvinceyou thatthesenumbersarethesameasabove. Placingthis into the definition of the Kullback Liebler divergencewith the correctposteriordistribution,givesus 42P N J2¦2§�¨©§ .For ^ª%«` we know that thesampleris unbiased.Hence,theprobabilityof gen-eratinga sampleis the sameasthe posteriordistribution calculatedby Bayesfilters.Thosearegivenabove aswell.d. Herearetwo possiblemodifications.First, if the initial robot locationis known withabsolutecertainty, the samplerabove will alwaysbe unbiased.Second,if the sensormeasurement¬ is equallylikely for all states,that is ­"��¬-® ¯l°�!±%²­"��¬-® ¯ Z !±%²­"��¬-® ¯i³2!±%­"��¬"® ¯ � ! , it will alsobeunbiased.An invalid answer, which we frequentlyencounteredin class,pertainsto thealgorithm(insteadof theproblemformulation). For example,replacingparticlefilters by the exact discreteBayesfiler remediesthe problembut isnota legitimateanswerto thisquestion.Neitheris theuseof infinitely many particles.25.2 ImplementingMonteCarlolocalizationrequiresa lot of work but is apremierewaytogaininsightsinto thebasicworkingsof probabilisticalgorithmsin robotics,andtheintricaciesinherentin realdata.We have usedthis exercisein many courses,andstudentsconsistentlyexpressedhaving learneda lot. Westronglyrecommendthisexercise!Theimplementationis notasstraightforwardasit mayappearat first glance.Commonproblemsinclude:´ Thesensormodelmodelstoo little noise,or thewrong typeof noise. For example,asimpleGaussianwill notwork here.´ The motion modelassumestoo little or too muchnoise,or the wrong type of noise.HereaGaussianwill work fine though.´ The implementationmay introduceunnecessarilyhigh variancein the resultingsam-pling set,by samplingtoo often,or by samplingin thewrongway. This problemman-ifests itself by diversity disappearingprematurely, often with the wrong samplessur-viving. While thebasicMCL algorithm,asstatedin thebook,suggeststhatsamplingshouldoccur after eachmotion update,implementationsthat samplelessfrequentlytend to yield superiorresults. Further, drawing samplesindependentlyof eachotheris inferior to so-calledlow variancesamplers.Hereis a versionof low variancesam-pling, in which µ denotestheparticlesand ¶ their importanceweights.Theresultingresampledparticlesresidein theset �*# .function LOW-VARIANCE-WEIGHTED-SAMPLE-WITH-REPLACEMENT( �[·¸¶ ):�*#¹%/º8»]¼% ½¾$¿ ° ¶�À Á�Âà %ÅÄ�Æ�Ç�È(��42ɸ]�!for S~% Nto ^ doÁ¼%/Æ�Ä�Y©ÊÌË Ç"Í ÍÎ ¿ ° ¶ÏÀ Q<ÂÑÐ Ãadd �ÒÀ Á� to � #à %/� à ' Ä�Æ�ÇÈ(��42ɸÓ�!�! modulo ]return �[#http://librosysolucionarios.net164 Chapter 25. RoboticsTheparameterÓ determinesthespeedatwhichwecycle throughthesampleset.Whileeachsample’s probabilityremainsthesameasifit weresampledindependently, there-sultingsamplesaredependent,andthevarianceof thesampleset � # is lower(assumingÓÕÔÖ] ). As a pleasantsideeffect, the low-variancesamplesis alsoeasilyimplementedin UL��^�! time,which is moredifficult for theindependentsampler.´ Samplesarestartedin theoccupiedor unknown partsof themap,or areallowed intothosepartsduringtheforwardsampling(motionprediction)stepof theMCL algorithm.´ Too few samplesare used. A few thousandshoulddo the job, a few hundredwillprobablynot.Thealgorithmcanbespedupby pre-cachingall noise-freemeasurements,for all ) -H - × posesthat the robot might assume.For that, it is convenientto definea grid over thespaceof allposes,with 10centimetersspatialand2 degreesangularresolution.Onemight thencomputethenoise-freemeasurementsfor thecentersof thosegrid cells. Thesensormodelis clearlyjust a functionof thosecorrectmeasurements;andcomputingthosetakesthebulk of time inMCL.25.3 Let Ø betheshoulderandÙ betheelbow angle.Thecoordinatesof theendeffectorarethengivenby thefollowing expression.Here ¬ is theheightand) thehorizontaldisplacementbetweentheendeffectorandtherobot’s base(origin of thecoordinatesystem):Ú� p |>Û$Üx�|>Û$Ü Ý Þ�ßáà�âãåä Þeâæ v�|>Û$Ü Ý Þ�ßçà d â Ý7è hãåä Þ d â Ý7è h æ v�|�Û$ÜNotice that this is only oneway to definethekinematics.Thezero-positionsof theanglesØ andÙ canbeanywhere,andthemotorsmayturn clockwiseor counterclockwise.Herewechosedefinetheseanglesin awaythatthearmpointsstraightupat Ø3%ÅÙs%/4 ; furthermore,increasingØ andÙ makesthecorrespondingjoint rotatecounterclockwise.Inversekinematicsis theproblemof computingØ and Ù from theendeffectorcoordi-nates) and ¬ . For that, we observe that the elbow angle Ù is uniquelydeterminedby theEuclideandistancebetweentheshoulderjoint andtheendeffector. Let uscall this distance. . Theshoulderjoint is locatedO 42Ó�Q above theorigin of thecoordinatesystem;hence,thedistance. is given by .é% ) Z ' ��¬Ò; O 42Ó�Q~! Z . An alternative way to calculate. is byrecovering it from theelbow angleÙ andthe two connectedjoints (eachof which is ¨©42Ó�Qlong): .D%/J-꣨©42Ó¡Qëê�ì�X2íÒî Z . Thereadercaneasilyderive thisfrom basictrigonomy, exploitingthefactthatboththeelbow andtheshoulderareof equallength.Equatingthesetwo differentderivationsof . with eachothergivesusÚ j Ý d �ï x�|>Û$Ü h j pðz�|>Û?Ü æ ãiä Þ è t (25.1)orè pòñ&t æió�ô ãiãåä ÞÚ j Ý d �ï x�|�Û$Ü h jz�|>Û$Ü (25.2)In most cases,Ù can assumetwo symmetricconfigurations,one pointing down and onepointingup. Wewill discussexceptionsbelow.To recover theangle Ø , we notethat theanglebetweentheshoulder(thebase)andtheendeffectoris givenby Æ�Älì�õ�Æ�Ç0J2�,)·¸¬�; O 42Ó�Q~! . Here Æ�Ä�ì�õ�Æ�ÇEJ is thecommongeneralizationhttp://librosysolucionarios.net165of thearcustangensto all four quadrants(checkit out—it is a function in C). Theangle Øis now obtainedby adding î Z , againexploiting that theshoulderandtheelbow areof equallength:â p ó�ô ãiö ó à t d Ú � �\ï x�|>Û$Ü hAï è t (25.3)Of course,the actualvalueof Ø dependson the actualchoiceof the valueof Ù . With theexceptionof singularities,Ù cantake on exactly two values.Theinversekinematicsis uniqueif Ù assumesasinglevalue;asaconsequence,sodoesalpha.For this to bethecase,we needtható�ô ãåãiä ÞÚ j Ý d �ï x�|>Û?Ü h jz�|>Û$Ü pð|(25.4)This is the caseexactly when the argumentof the Æ�Ä�ì¸ì�X2í is 1, that is, when the distance.W%Ö÷242Ó¡Q andthearmis fully stretched.Theendpoints )&·¸¬ thenlie on a circle definedby) Z ' ��¬C; O 42Ó¡Q~! Z %Ö÷242Ó�Q . If thedistance.WøÖ÷242Ó¡Q , thereis no solutionto theinversekinematicproblem:thepoint is simply too faraway to bereachableby therobotarm.Unfortunately, configurationslike thesearenumericallyunstable,asthequotientmaybeslightly largerthanone(dueto truncationerrors).Suchpointsarecommonlycalledsingu-larities, andthey cancausemajorproblemsfor robotmotionplanningalgorithms.A secondsingularityoccurswhentherobot is “folded up,” that is, Ù/% N ÷24úù . Heretheendeffector’sposition is identicalwith that of the robot elbow, regardlessof the angle Ø : )5%_42Ó�Q and¬û% O 42Ó�Q . This is an importantsingularity, as thereare infinitely many solutionsto theinversekinematics.As long as Ùü% N ÷24úù , the valueof Ø canbe arbitrary. Thus,this sim-ple robotarmgivesusanexamplewheretheinversekinematicscanyield zero,one,two, orinfinitely many solutions.25.4 Codenot shown.25.5a. Theconfigurationsof therobotsareshown by theblackdotsin thefollowing figures.FigureS25.2 Configurationof therobots.http://librosysolucionarios.net166 Chapter 25. Roboticsb. Theabove figureanswersalsothesecondpartof this exercise:it shows theconfigura-tionspaceof therobotarmconstrainedby theself-collisionconstraintandtheconstraintimposedby theobstacle.c. Thethreeworkspaceobstaclesareshown in thefollowing diagrams:FigureS25.3 Workspaceobstacles.d. This questionis a greatmind teaserthat illustratesthedifficulty of robotmotionplan-ning! Unfortunately, for anarbitraryrobot,aplanarobstaclecandecomposetheworkspaceinto anynumberof disconnectedsubspaces.To see,imaginea 1-DOFrigid robot thatmoveson a horizontalrod,andpossesses^ upward-pointingfingers,like a giantfork.A singleplanarobstacleprotrudingvertically into oneof the free-spacesbetweenthefingerscouldeffectively separatetheconfigurationspaceinto ^ ' Ndisjointsubspaces.A secondDOF will not changethis.More interestingis the robot arm usedas an examplethroughoutthis book. Byslightly extendingtheverticalobstaclesprotrudinginto the robot’s workspacewe candecomposethe configurationspaceinto five disjoint regions. The following figuresshow the configurationspacealongwith representative configurationsfor eachof thefiveregions.Is five themaximumfor any planarobjectthatprotrudesinto theworkspaceof thisparticularrobotarm? We honestlydo not know; but we offer a $1 reward for thefirstpersonwho presentsto usa solutionthatdecomposestheconfigurationspaceinto six,seven,eight,nine,or tendisjointregions.For therewardto beclaimed,all theseregionsmustbe clearly disjoint, andthey mustbe a two-dimensionalmanifold in the robot’sconfigurationspace.For non-planarobjects,theconfigurationspaceis easilydecomposedinto any num-ber of regions. A circular object may force the elbow to be just aboutmaximallybent;theresultingworkspacewould thenbea very narrow pipethatleave theshoulderlargelyunconstrained,but confinestheelbow to anarrow range.Thispipeis theneasilychoppedinto piecesby smalldentsin thecircularobject;thenumberof suchdentscanbeincreasedwithout bounds.http://librosysolucionarios.net16725.6 A simpledeliberatecontrollermight work asfollows: Initialize therobot’s mapwithan empty map, in which all statesare assumedto be navigable, or free. Then iteratethefollowing loop: Find the shortestpathfrom the currentpositionto the goal positionin themapusingA*; executethefirst stepof this path;sense;andmodify themapin accordancewith thesensedobstacles.If therobotreachesthegoal,declaresuccess.Therobotdeclaresfailure whenA* fails to find a pathto the goal. It is easyto seethat this approachis bothcompleteandcorrect. The robot alwaysfind a pathto a goal if oneexists. If no suchpathexists,theapproachdetectsthis throughfailureof thepathplanner. Whenit declaresfailure,it is indeedcorrectin thatno pathexists.A commonreactive algorithm,whichhasthesamecorrectnessandcompletenessprop-erty asthe deliberateapproach,is known asthe BUG algorithm. The BUG algorithmdis-tinguishestwo modes,theboundary-following andthego-to-goalmode.Therobotstartsingo-to-goalmode.In this mode,therobotalwaysadvancesto theadjacentgrid cell closesttothegoal. If this is impossiblebecausethecell is blockedby anobstacle,therobotswitchestotheboundary-following mode. In this mode,therobot follows theboundaryof theobstacleuntil it reachesa point on theboundarythat is a local minimumto thestraight-linedistanceto thegoal. If sucha point is reached,therobot returnsto thego-to-goalmode.If therobotreachesthegoal,itdeclaressuccess.It declaresfailurewhenthesamepoint is reachedtwice,which canonly occurin theboundary-following mode. It is easyto seethat theBUG algo-rithm is correctandcomplete.If a pathto the goal exists, the robot will find it. Whentherobot declaresfailure, no pathto the goal may exist. If no suchpathexists, the robot willFigureS25.4 Configurationspacefor eachof thefive regions.http://librosysolucionarios.net168 Chapter 25. Roboticsultimatelyreachthesamelocationtwice anddetectit* failure.Bothalgorithmscancopewith continuousstatespacesprovidesthatthey canaccuratelyperceive obstacles,planpathsaroundthem(deliberative algorithm)or follow their boundary(reactive algorithm).Noisein motioncancausefailuresfor bothalgorithms,especiallyif therobothasto move througha narrow openingto reachthegoal.Similarly, noisein perceptiondestroys both completenessandcorrectness:In both casesthe robot may erroneouslycon-cludea goalcannotbereached,just becauseits perceptionwasnoise.However, a deliberatealgorithmmight build a probabilisticmap,accommodatingthe uncertaintythat arisesfromthenoisysensors.Neitheralgorithmasstatedcancopewith unknown goal locations;how-ever, thedeliberatealgorithmis easilyconvertedinto anexploration algorithmby which therobotalwaysmovesto thenearestunexploredlocation.Suchanalgorithmwouldbecompleteandcorrect(in the noise-freecase).In particular, it would be guaranteedto find andreachthegoalwhenreachable.TheBUG algorithm,however, wouldnotbeapplicable.A commonreactive techniquefor findingagoalwhoselocationis unknown is randommotion;thisalgo-rithm will with probabilityonefind agoalif it is reachable;however, it is unableto determinewhento give up,andit maybehighly inefficient. Moving obstacleswill causeproblemsforboththedeliberateandthereactive approach;in fact, it is easyto designanadversarialcasewheretheobstaclealwaysmovesinto therobot’sway. For slow-moving obstacles,acommondeliberatetechniqueis to attacha timer to obstaclesin thegrid, anderasethemafteracertainnumberof timesteps.Suchanapproachoftenhasa goodchanceof succeeding.25.7 Thereareanumberof waysto extendthesingle-leg AFSM in Figure25.22(b)intoasetof AFSMs for controllinga hexapod. A straightforward extension—thoughnot necessarilythemostefficientone—isshown in thefollowing diagram.Herethesetof legsis dividedintotwo, namedA andB, andlegsareassignedto thesesetsin alternatingsequence.Thetoplevelcontroller, shown ontheleft, goesthroughsix stages.Eachstagelifts asetof legs,pushestheonesstill on thegroundbackwards,andthenlowersthelegsthathave previously beenlifted.The samesequenceis then repeatedfor the other set of legs. The correspondingsingle-leg controller is essentiallythe sameas in Figure 25.22(b),but with addedwait-stepsforsynchronizationwith thecoordinatingAFSM. The low-level AFSM is replicatedsix times,oncefor eachleg.For showing that this controlleris stable,we show thatat leastoneleg groupis on thegroundat all times. If this conditionis fulfilled, therobot’s centerof gravity will alwaysbeabove theimaginarytriangledefinedby thethreelegson theground.Theconditionis easilyprovenby analyzingthetop level AFSM. Whenonegroupof legsin ¯ � (or on theway to ¯ �from ¯i³ ), theotheris eitherin ¯ Z or ¯n° , bothof whichareon theground.However, thisproofonly establishesthat therobotdoesnot fall over whenon flat ground;it makesno assertionsabouttherobot’s performanceon non-flatterrain.Our resultis alsorestrictedto staticstabil-ity, that is, it ignoresall dynamiceffectssuchasinertia. For a fast-moving hexapod,askingthatit* centerof gravity beenclosedin thetriangleof supportmaybeinsufficient.25.8 Wehaveusedthisexercisein classto greateffect. Thestudentsgetaclearerpictureofwhy it is hardto do robotics.Theonly drawbackis thatit is a lot of fun to play, andthusthestudentswant to spenda lot of time on it, andtheoneswho arejust observingfeel like theyhttp://librosysolucionarios.net169(a) (b)FigureS25.5 Controllerfor ahexapodrobot.aremissingout. If youhave laboratoryor TA sections,youcando theexercisethere.Bearin mind thatbeingtheBrain is averystressfuljob. It cantakeanhourjust to stackthreeboxes. Choosesomeonewho is not likely to panicor becrushedby studentderision.Help theBrain out by suggestingusefulstrategiessuchasdefininga mutuallyagreedHand-centriccoordinatesystemso that commandsareunambiguous.Almost certainly, the Brainwill startby issuingabsolutecommandssuchas“Move theLeft Hand12 inchespositive ydirection”or “MovetheLeft Handto (24,36).” Suchactionswill neverwork. Themostuseful“invention” thatstudentswill suggestis theguardedmotiondiscussedin Section25.5—thatis, macro-operatorssuchas“Move theLeft Handin thepositive y directionuntil theeyessaytheredandgreenboxesarelevel.” ThisgetstheBrainoutof theloop,soto speak,andspeedsthingsup enormously.We have alsouseda relatedexerciseto show why roboticsin particularandalgorithmdesignin generalis difficult. The instructorusesaspropsa doll, a table,a diaperandsomesafetypins, andasksthe classto comeup with an algorithmfor putting the diaperon thebaby. The instructorthenfollows the algorithm,but interpretingit in the leastcooperativeway possible:putting thediaperon thedoll’s headunlesstold otherwise,droppingthedollon thefloor if possible,andsoon.http://librosysolucionarios.netSolutionsfor Chapter26PhilosophicalFoundations26.1 Wewill take thedisabilities(seepage949)oneata time. Notethatthisexercisemightbebetterasaclassdiscussionratherthanwrittenwork.a. bekind: Certainlythereareprogramsthatarepoliteandhelpful,but to bekind requiresanintentionalstate,sothisoneis problematic.b. resourceful: Resourcefulmeans“clever at finding waysof doing things.” Many pro-gramsmeetthis criteriato somedegree:a compilercanbeclever makinganoptimiza-tion that the programmermight not ever have thoughtof; a databaseprogrammightcleverly createanindex to make retrievals faster;a checkersor backgammonprogramlearnsto play aswell asany human. Onecould arguewhetherthe machinesare“re-ally” clever or just seemto be,but mostpeoplewould agreethis requirementhasbeenachieved.c. beautiful: Its not clearif Turingmeantto bebeautifulor to createbeauty, nor is it clearwhetherhe meantphysicalor inner beauty. Certainlythe many industrialartifactsintheNew York Museumof ModernArt, for example,areevidencethat a machinecanbe beautiful. Therearealsoprogramsthat have createdart. Thebestknown of theseis chronicledin Aaron’s code:Meta-art,artificial intelligence, andthework of HaroldCohen(McCorduck,1991).d. friendlyThisappearsto fall underthesamecategory askind.e. haveinitiative Interestingly, thereis now aseriousdebatewhethersoftwareshouldtakeinitiative. The whole field of softwareagentssaysthat it should;critics suchasBenSchneidermansaythat to achieve predictability, softwareshouldonly be an assistant,not an autonomousagent. Notice that the debateover whethersoftwareshouldhaveinitiative presupposesthatit hasinitiative.f. havea senseof humorWe know of no majoreffort to producehumorousworks. How-ever, this seemsto beachievablein principle. All it would take is someonelike HaroldCohenwhois willing to spendalongtimetuningahumor-producingmachine.Wenotethathumoroustext is probablyeasierto producethanothermedia.g. tell right fromwrongThereis considerableresearchin applyingAI to legal reasoning,andtherearenow toolsthatassistthelawyer in decidingacaseanddoingresearch.Onecouldarguewhetherfollowing legalprecedentsis thesameastelling right from wrong,andin any casethishasaproblematicconsciousaspectto it.170http://librosysolucionarios.net171h. make mistakesAt this stage,every computeruseris familiar with softwarethatmakesmistakes! It is interestingto think back to what the world was like in Turing’s day,whensomepeoplethoughtit would be difficult or impossiblefor a machineto makemistakes.i. fall in loveThis is oneof thecasesthatclearlyrequiresconsciousness.Notethatwhilesomepeopleclaimthattheirpetslovethem,andsomeclaimthatpetsarenotconscious,I don’t know of anybodywhomakesbothclaims.j . enjoystrawberriesandcreamTherearetwo partsto this. First, therehasbeenlittle tonowork ontasteperceptionin AI (althoughtherehasbeenrelatedwork in thefoodandperfumeindustries;seehttp://198.80.36.88/popmech/tech/U045O.htmlfor onesuchar-tificial nose),sowe’re nowhereneara breakthroughon this. Second,the “enjoy” partclearlyrequiresconsciousness.k. make someonefall in lovewith it This criteria is actuallynot too hardto achieve; ma-chinessuchasdolls andteddybearshave beendoing it to childrenfor centuries.Ma-chinesthat talk andhave moresophisticatedbehaviors just have a larger advantageinachieving this.l. learn fromexperiencePartVI shows thatthis hasbeenachievedmany timesin AI.m. usewordsproperlyNo programuseswordsperfectly, but therehave beenmany naturallanguageprogramsthatusewordsproperlyandeffectively within alimited domain(seeChapters22-23).n. be thesubjectof its own thoughtTheproblematicword hereis “thought.” Many pro-gramscanprocessthemselves,aswhena compilercompilesitself. Perhapsclosertohumanself-examinationis the casewherea programhasan imperfectrepresentationof itself. Oneanecdoteof this involvesDougLenat’s Eurisko program.It usedto runfor long periodsof time, andperiodicallyneededto gatherinformation from outsidesources.It “knew” that if a personwereavailable,it could type out a questionat theconsole,andwait for a reply. Lateonenight it saw thatno personwasloggedon,soitcouldn’t askthequestionit neededto know. But it knew thatEurisko itself wasup andrunning,anddecidedit would modify the representationof Eurisko so that it inheritsfrom “Person,” andthenproceededto askitself thequestion!o. haveas much diversity of behavioras manClearly, no machinehasachieved this, al-thoughthereis noprincipledreasonwhy onecouldnot.p. do somethingreally new This seemsto be just an extensionof the idea of learningfrom experience:if you learnenough,you cando somethingreally new. “Really” issubjective, andsomewould saythat no machinehasachieved this yet. On the otherhand,professionalbackgammonplayersseemunanimousin their belief thatTDGam-mon(Tesauro,1992),anentirelyself-taughtbackgammonprogram,hasrevolutionizedtheopeningtheoryof thegamewith its discoveries.26.2 No. Searle’s Chineseroom thesissaysthat thereare somecaseswhererunning aprogramthatgeneratestheright outputfor theChineseroomdoesnotcausetrueunderstand-ing/consciousness. The negationof this thesisis thereforethat all programswith the righthttp://librosysolucionarios.net172 Chapter 26. PhilosophicalFoundationsoutputdo causetrueunderstanding/consciousness. So if you wereto disprove Searle’s the-sis, thenyou would have a proof of machineconsciousness.However, what this questionisgettingat is theargumentbehindthethesis.If youshow thattheargumentis faulty, thenyoumayhave provednothingmore: it might bethat thethesisis true(by someotherargument),or it might befalse.26.3 Yes,thisisalegitimateobjection.Remember, thepointof restoringthebraintonormal(page957) is to be able to ask “What was it like during the operation?” and be sureofgettinga“human”answer, notamechanicalone.But theskepticcanpointout thatit will notdo to replaceeachelectronicdevice with the correspondingneuronthat hasbeencarefullykept aside,becausethis neuronwill not have beenmodifiedto reflect the experiencesthatoccurredwhile theelectronicdevice wasin theloop. Onecouldfix theargumentby saying,for example,thateachneuronhasasingleactivationenergy thatrepresentsits “memory,” andthatwe setthis level in theelectronicdevice whenwe insertit, andthenwhenwe remove it,we readoff thenew activationenergy, andsomehow settheenergy in theneuronthatwe putbackin. Thedetails,of course,dependon your theoryof what is importantin thefunctionalandconsciousfunctioningof neuronsandthe brain; a theorythat is not well-developedsofar.26.4 This exercisedependson what happensto have beenpublishedlately. The NEWSand MAGS databases,available on many online library catalogsystems,can be searchedfor keywords suchas Penrose,Searle,ChineseRoom, Dreyfus, etc. We found about90reviews of Penrose’s books. Here are someexcerptsfrom a fairly typical one, by AdamSchulman(1995).RogerPenrose,thedistinguishedmathematicalphysicist,hasagainenteredthelists to ridtheworld of a terribledragon.Thenameof this dragonis ”strongartificial intelligence.”StrongAl, asits defenderscall it, is botha widely heldscientificthesisandanongoingtechnologicalprogram.Thethesisholdsthatthehumanmindis nothingbut afancy calcu-latingmachine-”-acomputermadeof meat”–andthatall thinking is merelycomputation;theprogramis to build fasterandmorepowerful computersthatwill eventuallybeabletodo everythingthehumanmind cando andmore.Penrosebelievesthat thethesisis falseandtheprogramunrealizable,andheis confidentthathecanprovetheseassertions.�i�i�In Part I of Shadowsof theMind Penrosemakeshis rigorouscasethathumanconscious-nesscannotbefully understoodin computationalterms.�i�å�How doesPenroseprovethatthereis moreto consciousnessthanmerecomputation?Most peoplewill alreadyfind itinherentlyimplausiblethatthediversefacultiesof humanconsciousness–self-awareness,understanding,willing, imagining,feeling–differ only in complexity from the workingsof, say, anIBM PC.Studentsshouldhave no problemfinding thingsin this andotherarticleswith which to dis-agree.Thecomp.ai Newsnetgroupis alsoagoodsourceof rashopinions.Dubiousclaimsalsoemerge from the interactionbetweenjournalists’desireto writeentertainingandcontroversialarticlesandacademics’desireto achieveprominenceandto beviewed asaheadof the curve. Here’s onetypical result— Is Nature’s Way TheBestWay?,Omni,February1995,p. 62:http://librosysolucionarios.net173Artificial intelligencehasbeenone of the leastsuccessfulresearchareasin computerscience. That’s becausein the past, researcherstried to apply conventionalcomputerprogrammingto abstracthumanproblems,suchas recognizingshapesor speakinginsentences.But researchersatMIT’ sMediaLabandBostonUniversity’sCenterfor Adap-tive Systemsfocuson applyingparadigmsof intelligencecloserto whatnaturedesignedfor humans,which includeevolution,feedback,andadaptation,areusedto producecom-puterprogramsthatcommunicateamongthemselvesandin turnlearnfrom theirmistakes.ProfilesIn Artificial Intelligence, DavidFreedman.This is not an argumentthat AI is impossible,just that it hasbeenunsuccessful.The fulltext of the article is not given, but it is implied that the argumentis that evolution workedfor humans,thereforeit is a betterapproachfor programsthan is “conventionalcomputerprogramming.” This is a commonargument,but onethat ignoresthe fact that (a) therearemany possiblesolutionsto a problem;onethathasworkedin thepastmaynot bethebestinthepresent(b) we don’t have a goodtheoryof evolution, sowe maynot beableto duplicatehumanevolution, (c) naturalevolution takes millions of yearsand for almostall animalsdoesnot resultin intelligence;thereis no guaranteethatartificial evolution will do better(d)artificial evolution (or geneticalgorithms,ALife, neuralnets,etc.) is not theonly approachthatinvolvesfeedback,adaptationandlearning.“Conventional”AI doesthisaswell.26.5 Thisalsomightmake agoodclassdiscussiontopic. Hereareourattempts:intelligence: a measureof theability of anagentto make theright decisions,giventheavailableevidence.Giventhesamecirc*mstances,a moreintelligentagentwill make betterdecisionson average.thinking : creatinginternalrepresentationsin serviceof thegoalof comingto aconclu-sion,makinga decision,or weighingevidence.consciousness: beingawareof one’s own existence,andof one’s currentinternalstate.Herearesomeobjections[with replies]:For intelligence, too much emphasisis put on decision-making. Haven’t you everknown a highly intelligent personwho madebaddecisions?Also no mentionis madeoflearning.Youcan’t beintelligentby usingbrute-forcelook-up,for example,couldyou?[Theemphasison decision-makingis only a liability whenyou areworkingat too coarsea gran-ularity (e.g., “What shouldI do with my life?”) Onceyou look at smaller-grain decisions(e.g.,“ShouldI answera,b, c or noneof theabove?),yougetat thekindsof thingstestedbycurrentIQ tests,while maintainingthe advantagesof the action-orientedapproachcoveredin Chapter1. As to thebrute-forceproblem,think of intelligencein termsof anecologicalniche:anagentonly needsto beasintelligentasis necessaryto besuccessful.If this canbeaccomplishedthroughsomesimplemechanism,fine. For thecomplex environmentsthatwehumansarefacedwith, morecomplex mechanismsareneeded.]For thinking , wehavethesameobjectionsaboutdecision-making,but in general,think-ing is theleastcontroversialof thethreeterms.For consciousness, the weaknessis thedefinition of “aware.” How doesonedemon-strateawareness?Also, it is not one’s true internalstatethat is important,but somekind ofabstractionor representationof someof thefeaturesof it.http://librosysolucionarios.net174 Chapter 26. PhilosophicalFoundations26.6 It ishardtogiveadefinitiveanswerto thisquestion,but it canprovokesomeinterestingessays.Many of thethreatsareactuallyproblemsof computertechnologyor industrialsocietyin general,with somecomponentsthat canbe magnifiedby AI—examplesincludelossofprivacy to surveillance,andtheconcentrationof power andwealthin thehandsof themostpowerful. As discussedin the text, the prospectof robotstaking over the world doesnotappearto beaseriousthreatin theforeseeablefuture.26.7 Biologicalandnucleartechnologiesprovidemushmoreimmediatethreatsof weapons,yieldedeitherby statesor by smallgroups.Nanotechnlogythreatensto producerapidly re-producingthreats,eitherasweaponsor accidently, but thefeasibilityof this technologyis stillquitehypothetical.As discussedin thetext andin thepreviousexercise,computertechnologysuchascentralizeddatabases,network-attachedcameras,andGPS-guidedweaponsseemtoposeamoreseriousportfolio of threatsthanAI technology, at leastasof today.26.8 To decideif AI is impossible,we mustfirst defineit. In this book, we’ve chosenadefinition that makes it easyto show it is possiblein theory—fora given architecture,wejust enumerateall programsandchoosethe best. In practice,this might still be infeasible,but recenthistoryshows steadyprogressat a wide varietyof tasks.Now if we defineAI astheproductionof agentsthatactindistinguishablyform (or at leastasintellgientlyas)humanbeingson any task,thenonewould have to saythatlittle progresshasbeenmade,andsome,suchasMarvin Minsky, bemoanthefactthatfew attemptsareevenbeingmade.Othersthinkit is quite appropriateto addresscomponenttasksratherthan the “whole agent”problem.Our feeling is that AI is neitherimpossiblenor a oomingthreat. But it would be perfectlyconsistentfor someoneto ffel thatAI is mostlikely doomedto failure,but still thattherisksof possiblesuccessaresogreatthatit shouldnotbepersuedfor fearof success.http://librosysolucionarios.netSolutionsfor Chapter27AI: PresentandFutureThereareno exercisesin this chapter. Therearemany topicsthat areworthy of classdis-cussion,or of paperassignmentsfor thosewho like to emphasizesuchthings. Examplesare:´ Whatarethebiggesttheoreticalobstaclesto successfulAI systems?´ Whatarethebiggestpracticalobstacles?How arethesedifferent?´ Whatis theright goalfor rationalagentdesign?Doesthechoiceof agoalmakeall thatmuchdifference?´ Whatdo youpredictthefutureholdsfor AI?175http://librosysolucionarios.nethttp://librosysolucionarios.netBibliographyAndersson,R. L. (1988). A robot ping-pongplayer: Experimentin real-timeintelligent con-trol. MIT Press,Cambridge,Massachusetts.Bahle, D., Williams, H., and Zobel, J. (2002).Efficient phrasequeryingwith an auxiliary in-dex. In Proceedingsof theACM-SIGIRConfer-enceon Research andDevelopmentin Informa-tion Retrieval, pp.215–221.Chomsky, N. (1957).SyntacticStructures. Mou-ton,TheHagueandParis.Cormen,T. H., Leiserson,C. E., andRivest,R.(1990). Introductionto Algorithms. MIT Press,Cambridge,Massachusetts.Dawkins, R. (1976). TheSelfishGene. OxfordUniversityPress,Oxford,UK.Gold, E. M. (1967). Languageidentificationinthelimit. InformationandControl, 10, 447–474.Heinz,E.A. (2000).Scalablesearch in computerchess. Vieweg, Braunschweig,Germany.Held, M. andKarp, R. M. (1970). Thetravelingsalesmanproblemandminimumspanningtrees.OperationsResearch, 18, 1138–1162.Kay, M., Gawron, J. M., andNorvig, P. (1994).Verbmobil: A Translation Systemfor Face-To-FaceDialog. CSLI Press,Stanford,California.Kearns,M. and Vazirani, U. (1994). An In-troduction to ComputationalLearning Theory.MIT Press,Cambridge,Massachusetts.Keeney, R. L. andRaiffa, H. (1976). Decisionswith Multiple Objectives:PreferencesandValueTradeoffs. Wiley, New York.McCorduck, P. (1991). Aaron’s code: Meta-art, artificial intelligence, andtheworkof HaroldCohen. W. H. Freeman,New York.Moore, A. W. andAtkeson,C. G. (1993). Pri-oritized sweeping—reinforcementlearningwithlessdataandlesstime. Machine Learning, 13,103–130.Norvig, P. (1992). Paradigmsof Artificial Intel-ligenceProgramming:CaseStudiesin CommonLisp. MorganKaufmann,SanMateo,California.Pearl,J.(1988).ProbabilisticReasoningin Intel-ligentSystems:Networksof PlausibleInference.MorganKaufmann,SanMateo,California.Quine, W. V. (1960). Word and Object.MIT Press,Cambridge,Massachusetts.Schulman,A. (1995). Shadows of the mind: Asearchfor the missingscienceof consciousness(bookreview). Commentary, 99, 66–68.Smith, D. E., Genesereth,M. R., andGinsberg,M. L. (1986). Controlling recursive inference.Artificial Intelligence, 30(3), 343–389.Tesauro,G. (1992). Practicalissuesin temporaldifferencelearning. Machine Learning, 8(3–4),257–277.Wahlster, W. (2000). Verbmobil: FoundationsofSpeech-to-Speech Translation. SpringerVerlag.Witten, I. H., Moffat,A., andBell, T. C. (1999).Managing Gigabytes: Compressingand Index-ing Documentsand Images (secondedition).MorganKaufmann,SanMateo,California.177http://librosysolucionarios.net
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