16.410 lecture 04: soundness and completeness of …...• conjecture: for any n, the algorithm...

7/2/11

1

1

Soundness and Completeness of State Space Search

Sertac Karaman 16.410‐13 Sept 20th, 2010

2

Assignments •  Remember:

Problem Set #2: Uninformed search Out last Wednesday, Due this Wednesday, September 22nd

•  Reading: –  Today: Proofs and InducNon: Lecture 2 and 3 Notes of 6.042. –  Wednesday: [AIMA] Ch. 6.1; 24.3‐5 Constraint SaNsfacNon.

•  To learn more: Constraint Processing, by Rina Dechter –  Chapter 2: Constraint Networks –  Chapter 3: Consistency Enforcing and PropagaNon

9/20/1

AutonomousSystems:

• PlancomplexsequencesofacVons

• ScheduleVghtresources• Monitoranddiagnosebehavior

• Repairorreconfigurehardware. formulateasstatespacesearch.

BrianWilliams,Fall10 3

FormalizingGraphSearch


C

S

B

G A

D

Input:AsearchproblemSP=<g,S,G>where• graphg=<V,E>,• startvertexSinV,and• goalvertexGinV.

Output:AsimplepathP=<S,v2,…G>ingfromStoG.(i.e.,<vi,v >∈E,andv ifi≠j).i+1 i≠vj

0

2

9/20/10

3


S

B

G

A

<S>

<B,S><A,S>

<G,B,S><A,B,S>

GraphSearchisaKindofStateSpaceSearch

S

BA

GB

GraphSearchisaKindOfTreeSearch


SoluVon:DepthFirstSearch(DFS)S

D

B A

C G

C G

D

C G

S

D

B A

C G

C G

D

C G

SoluVon:BreadthFirstSearch(BFS)

9/20/10

4


SoluVon:DepthFirstSearch(DFS)S

D

B A

C G

C G

D

C G

S

D

B A

C G

C G

D

C G

SoluVon:BreadthFirstSearch(BFS)

Depth-first:

Add path extensions to front of Q

Pick first element of Q

Breadth-first:

Add path extensions to back of Q

Pick first element of Q


TheWorstofTheWorst

Search Method

Worst Time

Worst Space

Shortest Path?

Guaranteed to find path?

Depth-first ~bm b*m No Yes for finite graph

Breadth-first ~bm bm Yes unit lngth Yes

Worst case time is proportional to number of nodes visited Worst case space is proportional to maximal length of Q

S

D

B A

C G

C G

D

C G

Whichisbeler,depth‐firstorbreadth‐first?

C

S B

G A

D

• Assumed=mintheworstcase,andcallbothm.

• Best‐firstcan’texpandtolevelm+1,justm.

9/20/10

5

ElementsofAlgorithmDesignDescripVon:(lastMonday)

–  Problemstatement.–  Stylizedpseudocode,sufficienttoanalyzeandimplementthe

algorithm.–  ImplementaVon(lastWednesday).

Analysis:(lastWednesday)•  Performance:

–  Timecomplexity:•  howlongdoesittaketofindasoluVon?

–  Spacecomplexity:•  howmuchmemorydoesitneedtoperformsearch?

•  Correctness:(today)–  Soundness:

•  whenasoluVonisreturned,isitguaranteedtobecorrect?–  Completeness:

•  isthealgorithmguaranteedtofindasoluVonwhenthereisone?


Outline

•  Review

•  ProoftechniquesandtheaxiomaVcmethod

•  Proofsofsoundnessandcompletenessofsearchalgorithms

•  LimitsofaxiomaVcmethod


9/20/10

6

EnvelopegameProbabili(es do not work!

•  Iputanamount$Nand$2Nintotwodifferentenvelopes(you do not know N).

•  Iopenoneofthem,ithas$X.

•  Wouldyoupicktheopenoneortheother?

•  Reasoning 1:(I pick one at random) –  seeinginsideanenvelopedoesnotmaler…

•  Reasoning 2: (I pick the second one) –  IfIgetthisenvelope,Iget$X.–  IfIgettheotherenvelope,Iget,onaverage:

(1/2)X/2+(1/2)2X=(5/4)X

UnexpectedhangingparadoxInduc(on does not work!

•  Ajudgetellsacriminalthat“the criminal will be hanged on a weekday at noon next week, he will not know when he will be hanged, it will be a total surprise”.

•  Criminal’sreasoning:–  HecannotbehangedonaFriday(byThursdayawernoon,he

willknow–itwon’tbeasurprise).–  Then,hecannotbehangedonThursdayeither.–  Then,hecannotbehangedatall…Sohefeelssafe.

(HewashangedonWed.atnoon–itwasatotalsurprise…)

•  Whatwentwrongwithcriminal’sdeducVon?

9/20/10

7

TheaxiomaVcmethod

•  InventedbyEuclidaround300BC(inAlexandria,Egypt).

•  5axioms ofgeometrymenVonedinhisworkElements.

•  StarVngfromtheseaxioms,Euclidestablishedmany“proposi2ons”byproviding“proofs”.

EuclidofAlexandria

OldestsurvivingcopyoftheElements onapapyrusfoundinOxyrhynchus, Egypt. EuclidstatuteinOxford

TheaxiomaVcmethod

•  AdefiniVonofa“proof”:•  Anysequenceoflogical deduc(onsfromaxioms andpreviouslyprovenproposi(ons/statements thatconcludeswiththeproposiVoninquesVon.

•  Therearemanytypesof“proposi2ons”:•  Theorem:Importantresults,mainresults•  Lemma:apreliminaryproposiVonforproving

laterresults•  Corollary:Aneasy(butimportant)conclusion,

anawerthoughtofatheorem.

EuclidofAlexandria

EuclidstatuteinOxford

Images are in the public domain.

9/20/10

8

TheaxiomaVcmethod

•  Euclid’saxiom‐proofapproachisnowfundamentaltomathemaVcs!

•  Amazingly,essenVallyallmathemaVcscanbederivedfromjustahandfulofaxioms…

•  Howtoevenstartaproof?•  Therearemany“templates”(outlines,ortechniques)

•  Thedetailsdiffer…

EuclidofAlexandria

EuclidstatuteinOxford

Provinganimplica2on

•  SeveralmathemaVcalclaimsareoftheform:•  “If P, then Q”,orequivalently“P implies Q”.

• QuadraJcs:•  Ifand,2 b

"ax + bx + c = 0 a = 0 thenx =

! ±

•  InequaliJes:•  If,then

• Goldbach’sConjecture:•  Ifn>2,thennisasumoftwoprimes.

b2 ! 4ac

2a

0 ! x ! 2 !x3 + 4x + 1 > 0

!

9/20/10

9

ProvingimplicaVons:Simplestprooftechnique

•  Toprove“PimpliesQ”,–  “AssumeP”andshowthatQlogicallyfollows.

•  Theorem:–  If,then0 ! x ! 2 !x3 + 4x + 1 > 0

•  Proof:–  Assume(P)0 ! x ! 2–  Then,,,andarx 2! x x + 2 eallnon‐negaVve–  Then,isnon‐negx(2! x)(2 + x) aVve–  Then,isnon‐negx(2! x)(2 + x) + 1 aVve–  Then,mulVplyingoutthelewsidegives

•  (Q!x3 + 4x + 1 > 0 )

ProofbyContradicVon

•  ToprovethatastatementPisTrue.– Assumethatitisnot.–  Showthatsomeabsurd(clearlyfalse)statementfollows.

•  Formalized:InordertoproveastatementPisTrue – AssumethatPisFalse,– DeducealogicalcontradicVon(negaVonofapreviousstatementoranaxiom).

9/20/10

10

ProofbyContradicVon

•  Theorem:isanirr!

aVonalnumber.•  Proof:

•  AssumethatisnotirraVonal.•  Then,isaraVonalnumberandcanbewrilenas=a/bwherea andb areintegersandfrac(on is in lowest terms

•  Then,squaringbothsidesandrearranginggives2=a2/b2•  Then,a mustbeeven•  Then,a2mustbeamulVpleof4•  Then,2b2mustalsobeamulVpleof4•  Then,bisalsoeven•  Then,thefracVonisnotinthelowestterms(sincea andbarebotheven)

2!

2!2

!2

ProofbyInducVon

PickparameterNtodefineproblemsize.•  NumberofedgesinthesoluVon.•  Numberofnodesingraph.

BaseStep:N=0(orsmallN)•  ProvepropertyforN=0

InducJonStep:•  AssumepropertyholdsforN•  ProvepropertyforN+1

•  Conclusion:propertyholdsforallproblemsizesN.20

9/20/10

11

ProofbyInducVonformalized

•  LetP(i)beastatementwithparameteri.

•  ProofbyinducVonstatesthefollowingimplicaVon:•  “P(0)isTrue”(1)and“P(i)impliesP(i+1)”(2)•  (1)and(2)implies“P(i)isTrueforalli”.

•  InducVonisoneofthecore principles ofmathemaVcs.

•  Itisgenerallytakenasanaxiom,ortheaxiomsaredesignedsothatinducVonprinciplecanbeproven.

21

AninducVonexample

•  Theorem:fn(n + 1)1 + 2 + · · · + n = oralln.

•  Proof:– Base case:P(0)isTrue.

•  Because,0=0.–  InducVonstep:P(n)impliesP(n+1)

•  Assumethatthehypothesisholdsforn. •  Forn+1:

2

n(n + 1)1 + 2 + · · · + n + (n + 1) = + n + 1

2(n + 1)(n + 2)

=2

9/20/10

12

AfaultyinducVon

•  Theorem[!]:Allhorsesarethesamecolor.•  Proof[!]:

–  Base case:P(1)isTrue.•  because,thereisonlyonehorse.

–  Induc(on step:P(i)impliesP(i+1).•  AssumethatP(i)isTrue.•  Bytheinduc(on hypothesisfirstihorsesarethesamecolor,andthelastihorsesarealsothesamecolor.

h1, h2, . . . , hi,hi+1 h1, h2, . . . , hi, hi+1

•  Soallthei+1horsesmustbethesamecolor.•  Hence,P(i+1)isalsoTrue.

•  Whatwentwronghere?

ProofbyInvarianceA common technique in algorithm analysis •  Showthatacertainpropertyholdsthroughoutinanalgorithm.

•  AssumethatthepropertyholdsiniVally.•  Showthatinanystepthatthealgorithmtakes,thepropertysVllholds.

•  Then,propertyholdsforever.

•  ItisasimpleapplicaVonofinducVon.Why?

9/20/10

13

ProvingstatementsaboutalgorithmsHandle with care!

•  Correctnessofsimplestalgorithmsmaybeveryhardtoprove…

•  Collatzconjecture:•  Algorithm(HalfOrTriplePlusOne‐HOTPO):

• Givenanintegern. 1.  Ifniseven,thenn=n/22.  Ifnisodd,thenn=3n+13.  Ifn=1,thenterminate,elsegotostep1.

•  Conjecture:Foranyn,thealgorithmalwaysterminates(withn=1).

ProvingstatementsaboutalgorithmsHandle with care!

Collatzconjecture:•  Firstproposedin1937.

•  Itisnot knownwhethertheconjectureistrueorfalse.

First1000numbers

PaulErdős(1913‐1996) ‐famousnumbertheorist–

“Mathema(cs is not yet ready for such problems”, 1985.

First100numbers

Images are in the public domain. Images by Keenan Pepper and Jon McLoone.

http://en.wikipedia.org/wiki/File:Collatz-graph-all-30-no27.svg

http://en.wikipedia.org/wiki/File:Collatz1000mathematica.png

9/20/10

14

SoundnessandCompletenessofSearchAlgorithms

•  Today:

•  provestatementsaboutthesearchalgorithmswehavestudiedintheclass.

•  studywhetherthealgorithmreturnsacorrectsoluVon.

•  studywhetherthealgorithmreturnsasoluVonatallwhenoneexists.

SoundnessandCompleteness

Givenaproblem PR,analgorithm thataYemptstosolvethisproblemmayhavethefollowingproperJes:

Soundness: •  ThesoluVonreturnedbythealgorithmiscorrect.

Completeness:•  ThealgorithmalwaysreturnsasoluVon,ifoneexists.•  IfthereisnosoluVon,thealgorithmreportsfailure.

Also,OpJmality:•  ThealgorithmreturnstheopVmalsoluVon,ifitreturnsone.

28

9/20/10

15

29

SomeOtherNoVonsofSoundnessandCompleteness

ProbabilisJcCompleteness:•  ThealgorithmreturnsasoluVon,ifoneexists,withprobability

approachingtooneasthenumberofiteraVonsincreases.•  IfthereisnosoluVon,itmayrunforforever.

ProbabilisJcSoundness:•  Theprobabilitythatthe“soluVon”reportedsolvesthe

problemapproachesone,asthenumberofiteraVonsincreases.

AsymptoJcOpJmality:•  Thealgorithmdoesnotnecessarilyreturnanop(mal soluVon,

butthecostofthesoluVonreportedapproachestheopVmalasthenumberofiteraVonsincreases.


Problem:StateSpaceSearch

C

S

B

G A

D

Input:AsearchproblemS=<g,S,G>where• graphg=<V,E>,• startvertexSinV,and• goalvertexGinV.

Output:AsimplepathP=<S,v2,…G>ingfromStoG.

9/20/10

16

PseudoCodeForSimpleSearchLet g be a Graph G be the Goal vertex of g. S be the start vertex of g Q be a list of simple partial paths in GR,

1. Initialize Q with partial path (S) as only entry; set Visited = ( ); 2. If Q is empty, fail. Else, pick some partial path N from Q; 3. If head(N) = G, return N; (goal reached!) 4. Else

a) Remove N from Q; b) Find all children of head(N) (its neighbors in g) not in Visited

and create a one-step extension of N to each child; c) Add to Q all the extended paths; d) Add children of head(N) to Visited; e) Go to step 2. 31

SoundnessandCompletenessTheorems

Wewouldliketoprovethefollowingtwotheorems:

Theorem1(Soundness):Simplesearchalgorithmissound.

Theorem2(Completeness):Simplesearchalgorithmiscomplete.

Wewilluseablendofprooftechniquesforprovingthem.

32

    

  

   

9/20/10



Letusprove3lemmasbeforeprovingthistheorem.

33

Alemmatowardstheproof

•  Lemma1:If<v1,v2,…,vk>isapathinthequeueatanygivenVme,thenvk=S.

•  Proof:(byinvariance)– Base case:IniVally,thereisonly<S>inthequeue.Hence,theinvariant holds.

–  Induc(on step:Let’scheckthattheinvariant conVnues toholdineverystepofthealgorithm.

17

9/20/10

18

PseudoCodeForSimpleSearchInvariant:If<v1,v2,…,vk>isapathinthequeueatanygivenJme,thenvk=S.



and create a one-step extension of N to each child; c) Add to Q all the extended paths; d) Add children of head(N) to Visited; e) Go to step 2.

35





Before this line: assumethatinvariantholds.AIer this line: showthatinvariantissJlltrue.

Inthiscasenonewpathisaddedtothequeue. 36

    

  

   

    

  

   

9/20/10

19












Inthiscasenonewpathisaddedtothequeue(apathisremoved) 38

    

  

   

    

  

   

9/20/10

20






Severalpathsadded,eachsaJsfytheinvariantsinceNsaJsfiesit. 39







    

  

   

    

  

   

9/20/10

21







Anotherlemmatowardstheproof

•  DefiniJon:Apath<v0,v1,…,vk>isvalidiff(vi!1, vi) ! E oralli ! {1, 2, . . . , k}

•  Lemma2:If<v1,v2,…,vk>isapathinthequeueatanygivenVme,thenitisvalid.

•  Proof:(byinvariance)–  Base case:IniVallythereisonlyonepath<S>,whichisvalid.Hence,theinvariant holds.


    

  

   

9/20/10

22

PseudoCodeForSimpleSearchInvariant: If<v1,v2,…,vk>isapathinthequeueatanygivenVme,thenitisvalid




43







    

  

   

    

  

   

9/20/10

23













    

  

   

    

  

   

9/20/10

24












Notethatvalidityholdsforallnewlyaddedpath(fromLine4.b) 48

    

  

   

    

  

   

9/20/10

25













    

  

   

    

  

   

9/20/10

26

Yetanotherlemmatowardstheproof

•  Lemma3:If<v1,v2,…,vk>isapathinthequeueatanygivenVme,thenitisasimplepath(containsnocycles).

•  Proof:(byinvariance)– Base case:IniVally,thereisonly<S>inthequeue.Hence,theinvariant holds.


PseudoCodeForSimpleSearchInvariant:If<v1,v2,…,vk>isapathinthequeueatanygivenVme,thenitisasimplepath.




52

    

  

   

9/20/10

27













    

  

   

    

  

   

9/20/10

28













    

  

   

    

  

   

9/20/10

29





WewouldliketoshowthateachnewlyaddedpathissimpleassumingNissimple.Proof: (by contradic2on) Assume one path is not simple. Then, a children of head(N) appears in N. But, this is contradicts Line 4.b 57







    

  

   

    

  

   

9/20/10

30









Proof:bycontradicVon…

60

    

  

   

9/20/10

31

ProofofSoundness

Assumethatthesearchalgorithmisnotsound:Letthereturnedpathbe < v0, v1, . . . , vk >

Then,oneofthefollowingmustbeTrue:•  1.ReturnedpathdoesnotstartwithS:

vk = S

•  2.ReturnedpathdoesnotcontainGathead:v0 = G

•  3.SometransiVoninthereturnedpathisnotvalid:(vi!1, vi) !/ E forsome i ! {1, 2, . . . , vk}

•  4.Returnedpathisnotsimple:fvi = vj orsomewithi, j ! {0, 1, . . . , k} i = j

61

ProofofSoundness

•  1.ReturnedpathdoesnotstartwithS: vk = S

•  But,thiscontradictsLemma1!

•  Lemma1:If<v1,v2,…,vk>isapathinthequeueatanygivenVme,thenvk=S.

62

!

!

!

!

9/20/10

32

ProofofSoundness


•  Butclearly,thereturnedpathhasthepropertythatHead(N)=G

•  Recallthepseudocode:

63





64

!

    

  

   

9/20/10

33

ProofofSoundness

•  3.SometransiVoninthereturnedpathisnotvalid:f(vi!1, vi) !/ E orsome i ! {1, 2, . . . , vk}

•  ContradictsLemma2!

•  Lemma2:If<v1,v2,…,vk>isapathinthequeueatanygivenVme,thenitisvalid.

65

ProofofSoundness


•  ContradictsLemma3!

•  Lemma3:If<v1,v2,…,vk>isapathinthequeueatanygivenVme,thenitisasimplepath(containsnocycles).

66

!

9/20/10

34

ProofofSoundness

Assumethatthesearchalgorithmisnotsound:Letthereturnedpathbe < v0, v1, . . . , vk >

Then,oneofthefollowingmustbeTrue:•  1.ReturnedpathdoesnotstartwithS:

vk = S


•  3.SometransiVoninthereturnedpathisnotvalid:(vi!1, vi) !/ E forsome i ! {1, 2, . . . , vk}


67

ProofofSoundness

Assumethatthesearchalgorithmisnotsound:

WereachacontradicVoninallcases.

Hence,thesimplesearchalgorithmissound.

!

!

!

9/20/10

35

ProofofCompleteness


Needtoprove:

•  IfthereisapathtoreachfromStoG,thenthealgorithmreturnsonepaththatdoesso.

69





    

  

   

9/20/10

36

Acommontechniqueinanalysisofalgorithms

•  Let’sslightlymodifythealgorithm

•  Wewillanalyzethemodifiedalgorithm.

•  Then,“project”ourresultstotheoriginalalgorithm.


1. Initialize Q with partial path (S) as only entry; set Visited = ( ); 2. If Q is empty, fail. Else, pick some partial path N from Q; 3. // If head(N) = G, return N; (goal reached!) 4. Else



    

  

   

9/20/10

37

ProofofCompleteness

•  Themodifiedalgorithmterminateswhenthequeueisempty.

•  Letusproveafewlemmasregardingthebehaviorofthemodifiedalgorithm

ProofofCompleteness

•  Lemma1:Apaththatistakenoutofthequeueisnotplacedintothequeueagainatalaterstep.

•  Proof:(using logical deduc(on) •  Another way to state this:“Ifp=<v0,v1,…,vk>isapath

thatistakenoutofthequeue,thenp=<v0,v1,…,vk>isnotplacedintothequeueatalaterstep.”

•  Assumethatp=<v0,v1,…,vk>istakenoutofthequeue.

•  Then,pmustbeplacedintothequeueatanearlierstep.

•  Then,v0mustbeinthevisitedlistatthisstep.•  Then,p=<v0,v1,…,vk>cannotplacedintothequeue

atalaterstep,sincev0isinthevisitedlist.

9/20/10

38

ProofofCompleteness

•  DefiniJon:Avertexvisreachable fromS,ifthereexistsapath<v0,v1,…,vk>thatstartsfromSandendsatv,i.e.,vk=Sandv0=v.

•  Lemma2:IfavertexvisreachablefromS,thenvisplacedintothevisitedlistawerafinitenumberofsteps.

ProofofCompleteness•  Lemma2:IfavertexvisreachablefromS,thenvisplacedintothevisitedlist

awerafinitenumberofsteps.•  Proof:(by contradic(on)

•  AssumevisreachablefromS,but itisneverplacedonthevisitedlist.•  SincevisreachablefromS,thereexistsapaththatisoftheform<v0,v1,…,

vk>,wherev0=vandvk=S.•  Letvibethefirstnode(starVngfromvk)inthechainthatisneveraddedto

thevisitedlist.•  (1)Notethatviwasnotinthevisitedlistbeforethisstep.•  (2)Notealsothat(vi+1,vi)isinE.•  Sincevi+1wasinthevisitedlist,thequeueincludedapath<vi+1,…,vk>(not

necessarilythesameasabove),wherevk=S.•  Thispathmusthavebeenpoppedfromthequeue,sincethereareonly

finitelymanydifferentparValpathsandnopathisaddedtwice(byLemma1)andviwasnotinthevisitedlist(seestatement1above).

•  Sinceitispoppedfromthequeue,then<vi+1,vi,…,vk>mustbeplacedintoqueue(seestatement2above)andviplacedintothevisitedlist

•  Redstatementscontradict!

9/20/10

39

ProofofCompleteness

•  Lemma2:IfavertexvisreachablefromS,thenvisplacedintothevisitedlistawerafinitenumberofsteps.

•  Corollary:Inthemodifiedalgorithm,Gisplacedintothevisitedqueue.

•  “Project” back to the original algorithm: •  Thisisexactlywhentheoriginalalgorithmterminates

ProofofCompleteness


•  Proof:FollowsfromLemma2evaluatedintheoriginalalgorithm.

78

9/20/10

40





SummarizeCompletenessandSoundness

•  Hence,wehaveproventwotheorems:



•  Soundnessandcompletenessisarequirementformostalgorithms,althoughwewilltheirrelaxaVonsquiteowen

    

  

   

9/20/10

41

BacktotheAxiomaVcMethodDoes it really work?

•  EssenVallyallofwhatweknowinmathemaVcstodaycanbederivedfromahandfulofaxiomscalledtheZarmelo‐FrankelsettheorywiththeaxiomofChoice(ZFC).

•  TheseaxiomsweremadeupbyZarmelo(theydidnotexista priori,unlikephysicalphenomena).

•  Wedonotknowwhethertheseaxiomsarelogically consistent!–  Soundscrazy!But,happenedbefore…

Around 1900, B. Russell discovered that the axioms of that (me were logically inconsistent, i.e., one could prove a contradic(on.


•  ZFCaxiomsgivesonewhatshe/hewants:–  Theorem:5+5=10.

•  However,absurdstatementscanalsobedriven:–  Theorem(Banach‐Tarski):Aballcanbecutintoafinitenumberofpiecesandthenthepiecescanberearrangedtobuildtwoballsofthesamesizeoftheoriginal.

Clearly, this contradicts our geometric intui(on! Image by Benjamin D. Esham, in the public domain.

http://en.wikipedia.org/wiki/File:Banach-Tarski_Paradox.svg

9/20/10

42



Onthefundamentallimitsofmathema2cs •  Godelshowedin1930thattherearesomeproposiVonsthat

aretrue,butdonotlogicallyfollowfromtheaxioms.

•  Theaxiomsarenotenough!

•  But,Godelalsoshowedthatsimplyaddingmoreaxiomsdoesnoteliminatethisproblem.Anysetofaxiomsthatis

notcontradictorywillhavethesameproblem!

•  Godel’sresultsaredirectlyrelatedtocomputaVon.TheseresultswerelaterusedbyAlanTuringin1950stoinventarevoluVonaryidea:computer…

Images of Godel, Turing, and Einstein removed due to copyright restrictions.

9/20/10

Whatyoushouldknow

•  ThedefiniVonsofaproposiVon,proof,theorem,lemma,andcorollary.

•  ProoftechniquessuchasproofbycontradicVon,inducVon,invarianceproofs.

•  NoVonsofsoundnessandcompleteness.•  Proving soundness and completeness of search algorithms.

43

MIT OpenCourseWare http://ocw.mit.edu

16.410 / 16.413 Principles of Autonomy and Decision Making Fall 2010

For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

http://ocw.mit.edu

http://ocw.mit.edu/terms

16.410 lecture 04: soundness and completeness of …...• conjecture: for any n, the algorithm...

Documents