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CS106B,Lecture11ExhaustiveSearchandBacktracking

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Plan for Today • Newrecursiveproblem-solvingtechniques

–  ExhaustiveSearch–  Backtracking

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Exhaustive search • exhaustivesearch:Exploringeverypossiblecombinationfromasetofchoicesorvalues.–  oftenimplementedrecursively–  SometimescalledrecursiveenumerationApplications:–  producingallpermutationsofasetofvalues–  enumeratingallpossiblenames,passwords,etc.–  combinatoricsandlogicprogramming

• Oftenthesearchspaceconsistsofmanydecisions,eachofwhichhasseveralavailablechoices.–  Example:Whenenumeratingall5-letterstrings,eachofthe5lettersisadecision,andeachofthosedecisionshas26possiblechoices.

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Exhaustive search Ageneralpseudo-codealgorithmforexhaustivesearch:

Explore(decisions):

–  iftherearenomoredecisionstomake:Stop.

–  else,let'shandleonedecisionourselves,andtherestbyrecursion.foreachavailablechoiceCforthisdecision:• ChooseCbymodifyingparameters.• ExploretheremainingdecisionsthatcouldfollowC.• Un-chooseCbyreturningparameterstooriginalstate(ifnecessary).

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Exhaustive Search Model Choosing1.  Wegenerallyiterateoverdecisions.Whatareweiteratingoverhere?The

iterationwillbedonebyrecursion.2.  Whatarethechoicesforeachdecision?Doweneedaforloop?

Exploring3.  Howcanwerepresentthatchoice?Howshouldwemodifytheparameters?

a)  Doweneedtouseawrapperduetoextraparameters?

Un-Choosing4.  Howdoweun-modifytheparametersfromstep3?Doweneedtoexplicitly

un-modify,oraretheycopied?Aretheyun-modifiedatthesamelevelastheyweremodified?

BaseCase5.  Whatshouldwedointhebasecasewhenwe'reoutofdecisions?

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Exercise: printAllBinary • WritearecursivefunctionprintAllBinarythatacceptsanintegernumberofdigitsandprintsallbinarynumbersthathaveexactlythatmanydigits,inascendingorder,oneperline.

– printAllBinary(2); printAllBinary(3);

00 00001 00110 01011 011 100 101 110 111

printBinary

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printAllBinary Choosing1.  Wegenerallyiterateoverdecisions.Whatareweiteratingoverhere?The

iterationwillbedonebyrecursion.2.  Whatarethechoicesforeachdecision?Doweneedaforloop?

Exploring3.  Howcanwerepresentthatchoice?Howshouldwemodifytheparametersand

storeourpreviouschoices?a)  Doweneedtouseawrapperduetoextraparameters?

Un-Choosing4.  Howdoweun-modifytheparametersfromstep3?Doweneedtoexplicitly

un-modify,oraretheycopied?Aretheyun-modifiedatthesamelevelastheyweremodified?

BaseCase5.  Whatshouldwedointhebasecasewhenwe'reoutofdecisions?

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printAllBinary Choosing1.  Wegenerallyiterateoverdecisions.Whatareweiteratingoverhere?Weare

iteratingovercharactersinthebinarystring2.  Whatarethechoicesforeachdecision?Doweneedaforloop?Choose0or1

Exploring3.  Howcanwerepresentthatchoice?Howshouldwemodifytheparametersand

storeourpreviouschoices?Buildupastringthatwewilleventuallyprint.Addthe0or1toit.Stringtracksourpreviouschoicesa)  Doweneedtouseawrapperduetoextraparameters?Yes

Un-Choosing4.  Howdoweun-modifytheparametersfromstep3?Doweneedtoexplicitly

un-modify,oraretheycopied?Aretheyun-modifiedatthesamelevelastheyweremodified?Ifnewstringsforeachcall,wedon'tneedtoun-choose

BaseCase5.  Whatshouldwedointhebasecasewhenwe'reoutofdecisions?Printthe

string

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printAllBinary solution voidprintAllBinary(intnumDigits){printAllBinaryHelper(numDigits,"");}voidprintAllBinaryHelper(intdigits,stringsoFar){if(digits==0){cout<<soFar<<endl;}else{printAllBinaryHelper(digits-1,soFar+"0");printAllBinaryHelper(digits-1,soFar+"1");}}

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A tree of calls • printAllBinary(2);

–  Thiskindofdiagramiscalledacalltreeordecisiontree.–  Thinkofeachcallasachoiceordecisionmadebythealgorithm:

• ShouldIchoose0asthenextdigit?• ShouldIchoose1asthenextdigit?

digits soFar

2 ""

1 "0"

0 "00" 0 "01"

1 "1"

0 "10" 0 "11"

0 1

0 1 0 1

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The base case voidprintAllBinaryHelper(intdigits,stringsoFar){if(digits==0){cout<<soFar<<endl;}else{printAllBinaryHelper(digits-1,soFar+"0");printAllBinaryHelper(digits-1,soFar+"1");}}

–  Thebasecaseiswherethecodestopsafterdoingitswork.• pAB(3)->pAB(2)->pAB(1)->pAB(0)

–  Eachcallshouldkeeptrackoftheworkithasdone.

–  Basecaseshouldprinttheresultoftheworkdonebypriorcalls.• Workiskepttrackofinsomevariable(s)-inthiscase,stringsoFar.

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Exercise: printDecimal • WritearecursivefunctionprintDecimalthatacceptsanintegernumberofdigitsandprintsallbase-10numbersthathaveexactlythatmanydigits,inascendingorder,oneperline.

– printDecimal(2); printDecimal(3);

00 00001 00102 002.. ...98 99799 998 999

– Userecursion.*

printDecimal

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printDecimal Choosing1.  Wegenerallyiterateoverdecisions.Whatareweiteratingoverhere?The

iterationwillbedonebyrecursion.2.  Whatarethechoicesforeachdecision?Doweneedaforloop?

Exploring3.  Howcanwerepresentthatchoice?Howshouldwemodifytheparameters

storeourpreviouschoices?a)  Doweneedtouseawrapperduetoextraparameters?

Un-Choosing4.  Howdoweun-modifytheparametersfromstep3?Doweneedtoexplicitly

un-modify,oraretheycopied?Aretheyun-modifiedatthesamelevelastheyweremodified?

BaseCase5.  Whatshouldwedointhebasecasewhenwe'reoutofdecisions?

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printDecimal solution voidprintDecimal(intdigits){printDecimalHelper(digits,"");}voidprintDecimalHelper(intdigits,stringsoFar){if(digits==0){cout<<soFar<<endl;}else{for(inti=0;i<10;i++){printDecimalHelper(digits-1,soFar+integerToString(i));}}}

– Observation:Whenthesetofdigitchoicesavailableislarge,usingalooptoenumeratethemavoidsredundancy.(Thisisokay!)

– Note:Loopoverchoices,notdecisions

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Announcements

• Homework3dueonWednesdayat5PM• Shreyawillbeguest-lecturingonMonday

– Myofficehourswillbecancelledthatday(stillavailableviaemail)• MidtermReviewSessiononTuesday,July24,from7-9PMinGatesB01

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Backtracking • backtracking:Findingsolution(s)bytryingallpossiblepathsandthenabandoningthemiftheyarenotsuitable.

–  a"bruteforce"algorithmictechnique(triesallpaths)–  oftenimplementedrecursively–  Couldinvolvelookingforonesolution

• Ifanyofthepathsfoundasolution,asolutionexists!Ifnonefindasolution,nosolutionexists

–  Couldinvolvefindingallsolutions–  Idea:it'sexhaustivesearchwithconditionsApplications:–  parsinglanguages–  games:anagrams,crosswords,wordjumbles,8queens,sudoku–  combinatoricsandlogicprogramming–  escapingfromamaze

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Backtracking: One Solution Ageneralpseudo-codealgorithmforbacktrackingproblems

searchingforonesolutionBacktrack(decisions):

–  iftherearenomoredecisionstomake:• ifourcurrentsolutionisvalid,returntrue• else,returnfalse

–  else,let'shandleonedecisionourselves,andtherestbyrecursion.foreachavailablevalidchoiceCforthisdecision:• ChooseCbymodifyingparameters.• ExploretheremainingdecisionsthatcouldfollowC.Ifanyofthemfindasolution,returntrue

• Un-chooseCbyreturningparameterstooriginalstate(ifnecessary).–  Ifnosolutionswerefound,returnfalse

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Backtracking: All Solutions Ageneralpseudo-codealgorithmforbacktrackingproblems

searchingforallsolutionsBacktrack(decisions):

–  iftherearenomoredecisionstomake:• ifourcurrentsolutionisvalid,addittoourlistoffoundsolutions• else,donothingorreturn

–  else,let'shandleonedecisionourselves,andtherestbyrecursion.foreachavailablevalidchoiceCforthisdecision:• ChooseCbymodifyingparameters.• ExploretheremainingdecisionsthatcouldfollowC.Keeptrackofwhichsolutionstherecursivecallsfind.

• Un-chooseCbyreturningparameterstooriginalstate(ifnecessary).–  Returnthelistofsolutionsfoundbyallthehelperrecursivecalls.

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Backtracking Model Choosing1.  Wegenerallyiterateoverdecisions.Whatareweiteratingoverhere?Whatarethe

choicesforeachdecision?Doweneedaforloop?Exploring3.  Howcanwerepresentthatchoice?Howshouldwemodifytheparametersandstore

ourpreviouschoices(avoidingarms-lengthrecursion)?a)  Doweneedtouseawrapperduetoextraparameters?

4.  Howshouldwerestrictourchoicestobevalid?5.  Howshouldweusethereturnvalueoftherecursivecalls?Arewelookingforall

solutionsorjustone?Un-choosing6.  Howdoweun-modifytheparametersfromstep3?Doweneedtoexplicitlyun-modify,

oraretheycopied?Aretheyun-modifiedatthesamelevelastheyweremodified?BaseCase7.  Whatshouldwedointhebasecasewhenwe'reoutofdecisions(usuallyreturntrue)?8.  Isthereacaseforwhentherearen'tanyvalidchoicesleftora"bad"stateisreached

(usuallyreturnfalse)?9.  Arethebasecasesorderedproperly?Areweavoidingarms-lengthrecursion?

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Exercise: Dice roll sum • WriteafunctiondiceSumthatacceptstwointegerparameters:anumberofdicetoroll,andadesiredsumofalldievalues.Outputallcombinationsofdievaluesthatadduptoexactlythatsum.

diceSum(2,7); diceSum(3,7); {1,1,5}

{1,2,4}{1,3,3}{1,4,2}{1,5,1}{2,1,4}{2,2,3}{2,3,2}{2,4,1}{3,1,3}{3,2,2}{3,3,1}{4,1,2}{4,2,1}{5,1,1}

{1,6}{2,5}{3,4}{4,3}{5,2}{6,1}

diceSum

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Dice Roll Sum Choosing1.  Wegenerallyiterateoverdecisions.Whatareweiteratingoverhere?Whatarethe

choicesforeachdecision?Doweneedaforloop?Exploring3.  Howcanwerepresentthatchoice?Howshouldwemodifytheparametersandstore

ourpreviouschoices(avoidingarms-lengthrecursion)?a)  Doweneedtouseawrapperduetoextraparameters?

4.  Howshouldwerestrictourchoicestobevalid?5.  Howshouldweusethereturnvalueoftherecursivecalls?Arewelookingforall

solutionsorjustone?Un-choosing6.  Howdoweun-modifytheparametersfromstep3?Doweneedtoexplicitlyun-modify,

oraretheycopied?Aretheyun-modifiedatthesamelevelastheyweremodified?BaseCase7.  Whatshouldwedointhebasecasewhenwe'reoutofdecisions(usuallyreturntrue)?8.  Isthereacaseforwhentherearen'tanyvalidchoicesleftora"bad"stateisreached

(usuallyreturnfalse)?9.  Arethebasecasesorderedproperly?Areweavoidingarms-lengthrecursion?

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Easier: Dice rolls • Suggestion:Firstjustoutputallpossiblecombinationsofvaluesthatcouldappearonthedice.

• Thisisjustexhaustivesearch!•  Ingeneral,startingwithexhaustivesearchandthenaddingconditionsisnotabadidea

diceSum(2,7); diceSum(3,7);

{1,1}{1,2}{1,3}{1,4}{1,5}{1,6}{2,1}{2,2}{2,3}{2,4}{2,5}{2,6}

{3,1}{3,2}{3,3}{3,4}{3,5}{3,6}{4,1}{4,2}{4,3}{4,4}{4,5}{4,6}

{5,1}{5,2}{5,3}{5,4}{5,5}{5,6}{6,1}{6,2}{6,3}{6,4}{6,5}{6,6}

{1,1,1}{1,1,2}{1,1,3}{1,1,4}{1,1,5}{1,1,6}{1,2,1}{1,2,2}...{6,6,4}{6,6,5}{6,6,6}

diceRoll

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A decision tree chosen available

- 4dice

1 3dice

1,1 2dice

1,1,1 1die

1,1,1,1

1,2 2dice 1,3 2dice 1,4 2dice

2 3dice

1,1,2 1die 1,1,3 1die

1,1,1,2 1,1,3,1 1,1,3,2

1,4,1 1die ... ... ...

...

... ... ... ...

valueforfirstdie?

valueforseconddie?

valueforthirddie?

diceSum(4,7);

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Initial solution voiddiceSum(intdice,intdesiredSum){Vector<int>chosen;diceSumHelper(dice,desiredSum,chosen);}voiddiceSumHelper(intdice,intdesiredSum,Vector<int>&chosen){if(dice==0){if(sumAll(chosen)==desiredSum){cout<<chosen<<endl;//basecase}}else{for(inti=1;i<=6;i++){chosen.add(i);//choosediceSumHelper(dice-1,desiredSum,chosen);//explorechosen.remove(chosen.size()-1);//un-choose}}}intsumAll(constVector<int>&v){//addsthevaluesingivenvectorintsum=0;for(intk:v){sum+=k;}returnsum;}

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Wasteful decision tree chosen available desiredsum

- 3dice 5

1 2dice

1,1 1die

1,1,1

1,2 1die 1,3 1die 1,4 1die

6 2dice

...

2 2dice 3 2dice 4 2dice 5 2dice

1,5 1die 1,6 1die

1,1,2 1,1,3 1,1,4 1,1,5 1,1,6

1,6,1 1,6,2

...

diceSum(3,5);

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Optimizations • Weneednotvisiteverybranchofthedecisiontree.

–  Somebranchesareclearlynotgoingtoleadtosuccess.– Wecanpreemptivelystop,orprune,thesebranches.

•  Inefficienciesinourdicesumalgorithm:–  Sometimesthecurrentsumisalreadytoohigh.

• (Evenrolling1forallremainingdicewouldexceedthedesiredsum.)

–  Sometimesthecurrentsumisalreadytoolow.• (Evenrolling6forallremainingdicewouldexceedthedesiredsum.)

–  Thecodemustre-computethesummanytimes.• (1+1+1=...,1+1+2=...,1+1+3=...,1+1+4=...,...)

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diceSum solution voiddiceSum(intdice,intdesiredSum){Vector<int>chosen;diceSumHelper(dice,0,desiredSum,chosen);}voiddiceSumHelper(intdice,intsum,intdesiredSum,Vector<int>&chosen){if(dice==0){if(sum==desiredSum){cout<<chosen<<endl;//solutionfoundbasecase}}elseif(sum+1*dice>desiredSum//invalidstatebasecase||sum+6*dice<desiredSum){return;}else{for(inti=1;i<=6;i++){chosen.add(i);//choosediceSumHelper(dice-1,sum+i,desiredSum,chosen);//explorechosen.remove(chosen.size()-1);//un-choose}}}

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A Twist • HowwouldyoumodifydiceSumsothatitprintsonlyuniquecombinationsofdice,ignoringorder?–  (e.g.don'tprintboth{1,1,5}and{1,5,1})

diceSum2(2,7); diceSum2(3,7);

{1,1,5}{1,2,4}{1,3,3}{1,4,2}{1,5,1}{2,1,4}{2,2,3}{2,3,2}{2,4,1}{3,1,3}{3,2,2}{3,3,1}{4,1,2}{4,2,1}{5,1,1}

{1,6}{2,5}{3,4}{4,3}{5,2}{6,1}