Understanding Unsolvable Problem

Jonathan Irvin GUNAWAN Undergraduate Student School of Computing, National University of Singapore Computing 1, 13 Computing Drive, Singapore 117417 e-mail: [email protected]

Abstract. In recent IOIs, there are several problems that seem unsolvable, until we realise that there is a special case to the problem that makes it tractable. In IOI 2014, the problem ‘Friend’ appears to be a standard NP-hard Maximum Independent Set problem. However, the graph is gen- erated in a very special way, hence there is a way to solve the problem in polynomial time. There were several contestants who didn’t identify the special case in this problem, and hence were stuck at the problem. In this paper, we will study a well-known technique called reduction to show that a problem we are currently tackling is intractable. In addition, we introduce techniques to identify special cases such that contestants will be prepared to tackle these problems. Keywords: special case, unsolvable, NP-hard.

1. Introduction

The problem ‘Friend’ in IOI 2014 required contestants to find a set of vertices with maximum total weight, such that no two vertices in the set are sharing a common edge. This is a classical Weighted Maximum Independent Set problem. We can show that Weight- ed Maximum Independent Set problem is NP-hard by reduction from 3-SAT (Cormen et al., 2009). Since the formulation of NP-completeness 4 decades ago, no one has been able to propose a solution to any NP-hard problem in polynomial time. Clearly, it is not expected that a high school student can solve the problem in 5 hours. None of the Indo- nesian IOI 2014 team solved this problem during the contest. After returning from the competition, I asked the Indonesian team about this problem. None of the team members were aware of the fact that Maximum Independent Set is an NP-hard problem, and thus were stuck trying to solve a general Maximum Independent Set problem. A similar problem also occurred in IOI 2008. The problem ‘Island’ required contestants to find a longest path in a graph with 1,000,000 vertices. The longest path problem is a classic NP-hard problem which can be reduced from the Hamiltonian path problem. If a contestant is not aware that the longest path problem is difficult to solve, the contestant may spend a lot of his/her time just to tackle the general longest path problem, without realising that there is a special case to the given graph. 88 J.I. Gunawan

Generally, some contestants spend too much thinking time trying to solve something that is believed to be unsolvable. If only they realise that their attempt is intractable, they may try a different approach and find a special case of this problem. In section 2 of this paper, we will introduce a classic reduction technique often used in theoretical computer science research. In the context of competitive programming, we may find out that a problem which we are attempting is unlikely to be solvable. After realizing that a problem is intractable, we are going to discuss how to proceed to solve the problem in section 3. Finally, in section 4 we will take a look at some common special cases in competitive programming that can be used to solve these kind of problems.

Fig. 1. The result of Indonesian team in IOI 2014, taken from http://stats.ioinformatics.org. The red squared column highlights the ‘Friend’ problem.

Fig. 2. IOI 2014 tasks statistics, taken from http://stats.ioinformatics.org. ‘Friend’ problem is the second least accepted problem in IOI 2014. It may be because some contestants (at least all the Indonesians) were stuck at trying to solve a general case of Maximum Independent Set.

Fig. 3. The result of Indonesian team in IOI 2008, taken from http://stats.ioinformatics.org. The red squared column highlights the ‘Island’ problem. Understanding Unsolvable Problem 89

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It is 1We say that problem X is reducible to problem Y good to know as many NP-hard problem as possible. This is necessary so that if we encounter a new problem X, we Jonathan Irvin GUNAWANcan use any of the NP-hard problems that we know, reduce it to problem X, and thus prove that X is also NP-hard. 1We say that problem X is reducible to problem Y JonathanJonathan Irvin GUNAWAN Irvin GUNAWAN Jonathan Irvin GUNAWAN Jonathan Irvin GUNAWAN Jonathan Irvin GUNAWAN Jonathan Irvin GUNAWAN Jonathan Irvin GUNAWAN

~~Jonathan Irvin GUNAWAN 90 J.I. Gunawan~~

From the above theorem, we conclude that a MIN-VERTEX-COVER can be easily constructed if we have a MAX-INDEPENDENT-SET. In the beginning of this section, we assume we know that MIN-VERTEX-COVER is a NP-hard problem. It is good to know as many NP-hard problem as possible. This is necessary so that if we encounter a new problem X , we can use any of the NP-hard problems that we know, reduce it to problem X , and thus prove that X is also NP- hard.

~~3. How to Proceed~~

Suppose we already know that a problem is unsolvable (i.e. any known algorithm will not solve this problem in time). In competition, it is impossible to complain that “This 33 How How to to proceed proceedis unsolvable, can you eliminate this problem?” to the judges, since the judges believe they have a solution. Such a request is absurd when there are already several contestants SupposeSuppose we we already already know know that that a a problem problem is is unsolvable unsolvable (i.e. (i.e.who any any knownhave known solved algorithm algorithm that problem. will will not not solve solveAlso, this thisin a problem problemmajor competition in in (e.g. ACM International time).time). In In competition, competition, it it is is impossible impossible to to complain complain that that "This "ThisCollegiate is is unsolvable, unsolvable, Programming can can you you Contest eliminate eliminate World this this Finals, problem?" problem?" IOI), it is unlikely that the judges have toto the the judges, judges, since since the the judges judges believe believe they they have have a a solution. solution.incorrect Such Such a a solution. request request is is absurd absurd when when there there are are already already severalseveral contestants contestants who who have have solved solved that that problem. problem. Also, Also, in in a a major major competition competition (e.g. (e.g. ACM ACM International International Collegiate Collegiate ProgrammingProgramming Contest Contest World World Finals, Finals, IOI), IOI), it it is is unlikely unlikely that that the the3.1. judges judges Approximation have have incorrect incorrect solution. solution.3 How to proceed Suppose we already know that a problem is unsolvable (i.e. any known algorithm will not solve this problem in 3.13.1 Approximation Approximation time).In Inreal competition, life, when we it iscannot impossible find the to complainoptimal solution, that "This we iscan unsolvable, try to find can theyou solution eliminate this problem?" InIn real real life, life, when when we we cannot cannot find find the the optimal optimal solution, solution, we weto can can thethat try tryjudges, is to to close find find since theto the the the solution solution optimal judges that that believesolution. is is close close they More to to havethe the specifically, optimal optimal a solution. we Such try to a find request a solution is absurd thatwhen is there are already solution.solution. More More specifically, specifically, we we try try to to find find a a solution solution that thatseveral is is not not contestants larger larger than than who αα (where have(where(where solved α>α> that11)) timestimes times problem. the the optimal optimal optimalAlso, in solution a major for competition a minimisation (e.g. ACM prob- International Collegiate solutionsolution for for a a minimisation minimisation problem. problem. The The most most common commonProgramming approximation approximationlem. The Contestmost algorithmalgorithm common World approximation IFinals, I found found IOI), in in textbooks textbooks it algorithm is unlikely is is the theI that found the in judges textbooks have is incorrect the 2-approx solution.- 2-approximation2-approximation MIN-VERTEX-COVER MIN-VERTEX-COVER problem, problem, which which means meansimation that that the the MIN-VERTEX-COVER algorithm algorithm will will not not choose choose problem, more more than than which twice twice means that the algorithm will not thethe number number of of vertices vertices than than the the optimal optimal solution. solution. However, However, approximationchoose approximation more than problems problems twice the rarely rarely number occur occur of3.1 in invertices competitive competitive Approximation than the optimal solution. However, ap- programmingprogramming (especially (especially IOI). IOI). One One of of the the reason reason is is because becauseproximation to to create create this this problems kind kind of of rarely problem, problem, occur the the in judges judgescompetitive have have toprogramming to (especially IOI). One knowknow the the optimal optimal solution solution in in order order to to verify verify that that the the contestant’s contestant’sIn realof life, thesolution when solutionreason we is isis cannot because indeed indeed findα αto-approximation.-approximation. thecreate optimal this kind solution, However, However,of problem, we can the try tojudges find have the solution to know that the is close to the optimal generatinggenerating the the optimal optimal solution solution is is impossible impossible (or (or takes takes a asolution. long longoptimal time). time). More Sincesolution Since specifically, this this in approach approachorder we to try verify is is to not not find that really really athe solution suitablecontestant’s suitable that for for issolution not larger is indeed than α-approxima(where α>- 1) times the optimal competitivecompetitive programming, programming, I I will will not not discuss discuss this this approach approachsolution in in detail.tion. detail. for However, a minimisation generating problem. the optimal The most solution common is impossible approximation (or takes algorithm a long time). I found in textbooks is the 2-approximationSince this approach MIN-VERTEX-COVER is not really suitable problem, for whichcompetitive means programming, that the algorithm I will will not not dis choose- more than twice the numbercuss this of approach vertices thanin detail. the optimal solution. However, approximation problems rarely occur in competitive 3.23.2 Pruning Pruning programming (especially IOI). One of the reason is because to create this kind of problem, the judges have to ThisThis approach approach is is useful useful in in some some competitive competitive programming programmingknow problems. problems. the optimal In In ACM ACM solution International International in order Collegiate Collegiate to verify Programming Programming that the contestant’s solution is indeed α-approximation. However, 3.2. Pruning ContestContest (ICPC) (ICPC) World World Finals Finals 2010, 2010, problem problem I I (Robots (Robotsgenerating on on Ice) Ice) required required the optimal the the contestant contestant solution is to to impossible count count the the (or number number takes aof of long time). Since this approach is not really suitable for HamiltonianHamiltonian Paths Paths with with constraints constraints ( (ACMACM ICPC ICPC World Worldcompetitive Finals Finals 2010 2010 programming, problem problem statement statement I willn.d.), notn.d.), discuss which which this is is known known approach to to in detail. bebe a a NP-hard NP-hard problem. problem. While While finding finding all all possible possible paths paths is is impossible, impossible,This approach the the solutionis solution useful for forin thissome this problem problemcompetitive is is to to programming prune prune problems. 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IOI. approachconstraints Usually, Usually, is useful(ACM IOI IOI problems problems inICPC some World require requirecompetitive Finals deep deep programming2010 analysis analysis problem from from problems. statement, In n.d.), ACM which International is known Collegiate Programming thethe contestant. contestant. It It is is rare rare that that we we can can get get Accepted Accepted by by only onlyContest "hacking" "hacking"to be (ICPC) a NP-hard a a complete complete World problem. Finals search search 2010, While algorithm. algorithm. problem finding Therefore, Therefore, I (Robotsall possible we we on Ice)paths requiredis impossible, the contestant the solution to count the number of willwill not not discuss discuss this this technique technique in in detail. detail. Hamiltonian Paths with constraints (ACM ICPC World Finals 2010 problem statement n.d.), which is known to be a NP-hard problem. While finding all possible paths is impossible, the solution for this problem is to prune the exponential algorithm we use to find all possible paths (ACM ICPC World Finals 2010 Solutions n.d.). If at 3.33.3 Finding Finding Small Small Constraints Constraints some point we know that it is impossible to visit the rest of the unvisited points, then we can prune the path and SupposeSuppose there there is is an an NP-hard NP-hard problem problem with with large large NN thatthatbacktrack is is impossible impossible immediately. to to solve solve However, exponentially exponentially it is not (e.g. (e.g. veryN> suitableN>5050). for). IOI. Usually, IOI problems require deep analysis from SometimesSometimes we we should should also also look look for for other other small small constraints constraintsthe that that contestant. may may help. help. It Foris For rare example, example, that we a a can SUBSET-SUM SUBSET-SUM get Accepted problem problem by only "hacking" a complete search algorithm. Therefore, we isis considered considered an an NP-hard NP-hard problem, problem, and and will will not not be be solvable solvablewill with withnot discussNN == 100 100 this.. When When technique we we are are in given detail.given the the constraint constraint that that allall the the elements elements inside inside the the array array are are small small (e.g. (e.g. [0[0,,100]100]),), this this problem problem can can be be solved solved using usingOO((NXNX22))DynamicDynamic Programming,Programming, where whereXX isis the the upper upper bound bound of of the the elements elements inside inside the the array. array. Even Even though though3.3 the the running running Finding time time of of Small the the Constraints algorithmalgorithm is is exponential exponential to to the the size size of of the the input, input, the the algorithm algorithm is is still still fast fast enough enough for for the the given given constraint. constraint. Another Another wayway to to apply apply this this technique technique is is when when the the value value of ofNN isis notSuppose not too too large large there (e.g. (e.g. is anN50). algorithmalgorithm for forNN == 36 36isis unlikely unlikely to to run run in in one one second, second, weSometimes we can, can, for forwe instance, instance, should use usealso a a lookOO(2(2N/ forN/22) other)MeetMeet small In In The The constraints Middle Middle that may help. For example, a SUBSET-SUM problem algorithmalgorithm for for solving solving a a problem problem like like SUBSET-SUM. SUBSET-SUM. While WhileisO consideredO(2(2N/N/22))isis anstill still NP-hard exponential exponential problem, to toNN, and, it it is is will much much not faster faster be solvable than than with N = 100. When we are given the constraint that 2 OO(2(2NN)),, and and the the range range of ofNN thatthat it it can can solve solve is is twice twice the theall range range the elements of ofNN usingusing inside a aOO the(2(2NN) array)solution.solution. are small (e.g. [0, 100]), this problem can be solved using O(NX ) Dynamic Programming, where X is the upper bound of the elements inside the array. Even though the running time of the algorithm is exponential to the size of the input, the algorithm is still fast enough for the given constraint. Another 3.43.4 Finding Finding Special Special Cases Cases way to apply this technique is when the value of N is not too large (e.g. N<40). For example, while O(2N ) ThisThis is is the the most most suitable suitable approach approach in in IOI, IOI, and and thus thus is isalgorithm the the main main forfocus focusN of= of 36 this thisis paper. paper.unlikely To To to solve runsolve in IOI IOI one 2008 2008 second, Island Island we can, for instance, use a O(2N/2) Meet In The Middle andand IOI IOI 2014 2014 Friend, Friend, we we need need to to use use this this approach. approach.algorithm We We must must for find find solving a a special special a problem constraint constraint like SUBSET-SUM. in in the the problem problem While such suchO(2N/2) is still exponential to N, it is much faster than thatthat this this constraint constraint allows allows the the problem problem to to be be solvable solvable in inO(2 polynomial polynomialN ), and the time. time. range We We of can canN checkthat check it whether whethercan solve an an is additional additionaltwice the range of N using a O(2N ) solution.

3.4 Finding Special Cases JonathanJonathan Irvin Irvin GUNAWAN GUNAWAN This is the most suitable approach in IOI, and thus is the main focus of this paper. To solve IOI 2008 Island and IOI 2014 Friend, we need to use this approach. We must ﬁnd a special constraint in the problem such that this constraint allows the problem to be solvable in polynomial time. We can check whether an additional

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Suppose we already know that a problem is unsolvable (i.e. any known algorithm will not solve this problem in time). In competition, it is impossible to complain that "This is unsolvable, can you eliminate this problem?" to the judges, since the judges believe they have a solution. Such a request is absurd when there are3 already How to proceed several contestants who have solved that problem. Also, in a major competition (e.g. ACM International Collegiate Programming Contest World Finals, IOI), it is unlikely that theSuppose judges have we already incorrect know solution. that a problem is unsolvable (i.e. any known algorithm will not solve this problem in time). In competition, it is impossible to complain that "This is unsolvable, can you eliminate this problem?" 3.1 Approximationto the judges, since the judges believe they have a solution. Such a request is absurd when there are already several contestants who have solved that problem. 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A and B will have the same sum,of as equalf for(a) every,fsum.(a +0 1),fk (a +−| 2)−,− ...f(|b) |forb− somea |2 a>)=SS} f.0.(a By By+2+3.. lemma As lemma As −kand3and) 2,− 2, SS  isareisare an an connected connected independent independent set. set. Therefore, Therefore, SS isis not not a a { } || −−JonathanJonathan|| || Irvin−− Irvin≤|| GUNAWAN GUNAWAN≤ ∈∈ −− || || || ||∈∈≤ ≤ byby an an edge, edge,MAX-INDEPENDENT-SET. we we note note that that SS maximumismaximumis not notb an ana INDEPENDENT-SET.independentJonathan INDEPENDENT-SET.independent2 f(Jonathan Irvina +3 set. set. GUNAWANk This)+ This Irvinf is( isa This This a aGUNAWAN+1+3 contradiction. contradiction. is is a ak contradiction.)= contradiction.f(a Therefore Therefore+2+3 Therefore Thereforek)VV SS isisVV the the minimumS minimumS isis a a vertex vertex cover. cover. Proof. Pick any consecutiveB f subsequencea k , of bF kwitha 2 length multiples. of three, whichb wea will2 denote by F  = −− Jonathan Irvin GUNAWAN = ( +2+3InIn the the) beginning beginning0 f( ofa of−+3 this this−kJonathanJonathan)+ section, section,f(a +1+3 we weIrvin Irvin assume assume GUNAWAN GUNAWANk)= we wefIn( knowa know the+2+3f(a beginning that+3 thatkk MIN-VERTEX-COVER))+ MIN-VERTEX-COVERf of(a this+1+3 section,k)=−− wef( assume isa is+2+3 a a NP-hard NP-hard wek) know problem. problem. that MIN-VERTEX-COVER It It is is is a NP-hard problem. It is Jonathan Irvin GUNAWAN VERTEX-COVER.VERTEX-COVER.A = f({a +3Jonathank), 0 k Irvin≤− GUNAWAN≤− 3 f(}a +1+3k), 0 k − − f(a),f(a + 1),f(a + 2){, ...f(b) for some≤ ≤a> SS.. By By lemma lemma 2, 2,SS isis an an independent independent set. set. Therefore, Therefore,SS isis not not a a || −− || || −− || JonathanJonathan Irvin| Irvin| || GUNAWAN GUNAWAN|| || maximummaximum independent independent set. set. This This is is a a contradiction. contradiction. Therefore ThereforeVV SS isis the the minimum minimum vertex vertex cover. cover. −− FromFrom the the above above theorem, theorem, we we conclude conclude that that a a MIN-VERTEX-COVER MIN-VERTEX-COVER can can be be easily easily constructed constructed if if we we have have a a MAX-INDEPENDENT-SET.MAX-INDEPENDENT-SET.

InIn the the beginning beginning of of this this section, section, we we assume assume we we know know that that MIN-VERTEX-COVER MIN-VERTEX-COVER is is a a NP-hard NP-hard problem. problem. It It is is goodgood to to know know as as many many NP-hard NP-hard problem problem as as possible. possible. This This is is necessary necessary so so that that if if we we encounter encounter a a new new problem problem X, X, we we cancan use use any any of of the the NP-hard NP-hard problems problems that that we we know, know, reduce reduce it it to to problem problem X, X, and and thus thus prove prove that that X X is is also also NP-hard. NP-hard.

~~11WeWe say say that that problem problem X X is is reducible reducible to to problem problem Y Y~~

~~JonathanJonathan Irvin Irvin GUNAWAN GUNAWAN 3 How to proceed~~

Suppose we already know that a problem is unsolvable (i.e. any known algorithm will not solve this problem in time). In competition, it is impossible3 How to complain to that proceed "This is unsolvable, can you eliminate this problem?" Supposeto the we judges,already sinceknow the that judges a problem believe is unsolvable they have (i.e.a solution. any known Such algorithm a request will is absurd not solve when this there problem are inalready time).several In competition, contestants it who is impossible have solved to that complain problem. that Also, "This in a is major unsolvable, competition can you (e.g. eliminate ACM International this problem?" Collegiate to theProgramming judges, since Contest the judges World believe Finals, they IOI), have it is a unlikely solution. that Such the ajudges request have is incorrectabsurd when solution. there are already several contestants who have solved that problem. Also, in a major competition (e.g. ACM International Collegiate Programming Contest World Finals, IOI), it is unlikely3.1 Approximation that the judges have incorrect solution. In real life, when we cannot find the optimal solution, we can try to find the solution that is close to the optimal solution. More specifically, we try to3.1 find a Approximation solution that is not larger than α (where α>1) times the optimal In realsolution life, when for we a minimisation cannot find the problem. optimal The solution, most we common can try approximation to find the solution algorithm that Iis found close to in the textbooks optimal is the solution.2-approximation More specifically, MIN-VERTEX-COVER we try to find a solution problem, that which is notmeans larger that than the algorithmα (where willα> not1) timeschoose the more optimal than twice solutionthe for number a minimisation of vertices problem. than the optimal The most solution. common However, approximation approximation algorithm problems I found rarely in textbooks occur in competitive is the 2-approximationprogramming MIN-VERTEX-COVER (especially IOI). One problem, of the reason which means is because that to the create algorithm this willkind not of chooseproblem, more the than judges twice have to α the numberknow the of vertices optimal than solution the in optimal order tosolution. verify that However, the contestant’s approximation solution problems is indeed rarely occur-approximation. in competitive However, programminggenerating (especially the optimal IOI). solution One of is the impossible reason is (or because takes a to long create time). this Since kind this of problem, approach the is not judges really have suitable to for knowcompetitive the optimal programming, solution in order I will to not verify discuss that this the contestant’sapproach in detail. solution is indeed α-approximation. However, generating the optimal solution is impossible (or takes a long time). Since this approach is not really suitable for competitive programming, I will not discuss this approach3.2 in Pruning detail. N N/2 This approach is usefulFigure in some 4: Comparisoncompetitive programmingof O(2 ) and problems.O(2 ) SUBSET-SUM In ACM International algorithm Collegiate running Programming time for various input sizes. This Contest (ICPC) Worldexperiment Finals 2010, is run problem 1003.2 times I (Robots Pruning for each on value Ice) required of N on the a MacBook contestant Pro to (Retina, count the 13-inch, number Early of 2015) N N/N N/2 This approachHamiltonianFigure is 4: useful Paths ComparisonFigure in with some 4: constraints of competitive ComparisonO(2 ) and (ACM programming ofOO(2(2 ICPC2))and WorldSUBSET-SUM problems.O(2 Finals) InSUBSET-SUM 2010 ACM algorithm problem International running statement algorithm timeCollegiaten.d.), running for various Programmingwhich time input is for known various sizes. to This input sizes. 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If whileN f2(1) (modis odd. 3). Therefore,F (N) canbecause there be partitioned isF no( wayN) intois to even partition two while multisetsFf((1)N). ofis odd. equal Therefore, sum. there is no way to partition F (N). haveN 1 0(havemodN3). By1 lemma0(mod 3, 3)F .(N By) lemmacan be≡ 3, partitionedF (N) can into be two partitioned multisets into of equal two multisets sum.Theorem Therefore, of equal 3. the sum.If N Therefore,1 (mod the 3),F(N) cannot be partitioned into two multisets of equal sum. Theorem 4. If N 2 (mod− 3).≡F (N) can be− partitionedTheorem≡ 4. intoTheoremIf twoN multisets2 4.(modIf N of 3). equalF2( (modN sum.) can 3) be. F partitioned(N) can be into partitioned two multisets into two of equal multisets sum.≡ of equal sum. sum of all elements in F (N) is even. However, the sum of≡ all elements in F (N)=F (N)+f(1), which is odd ≡ sum of all elementsProof. in F (AssignN)≡is even.f(1) to However,A and f the(2) sumto B of. all Since elementsf(1) = infF(2)(N,)= we areF (N now)+ tryingf(1), whichto partition is oddF (N)=F (N) Theorem 4. If N 2 (mod 3). F (N) can be partitionedTheorem into 4. twoIf multisetsN Proof.2 (mod ofSuppose equal 3). F sum.F(N(N) can)= beF ( partitionedN) f(1). into Note two that multisetsF (N) ofcontains equal sum.N 1 elements. Since N 1(mod 3), we because F (N) is even while f(1) is odd. Therefore,≡ there is no way to partition F (N). ≡ − Proof. Assign f(1) to A and f(2) to becauseB. SinceFProof.f((1)N) =isAssign evenf(1)(2),Proof. while weff(1)(2) aref.toAssign(1)F nowA(isNandtrying odd.)fwill(1)f(2) Therefore,to have totoA partitionNandB. Sincef there2(2)elements.F (tof isN(1) no)=B. = way SinceFf((2)N to),Nf partition we(1) are2( = fmod now(2)F (N, trying3) we)., we are to obtain now partition tryingN 2F to( partitionN0()=mod−F (FN3).)(N By)= lemmaF (N) − ≡ − − −≡ haveN 1 0(mod−3).≡ By lemma 3, F−(N) can be− partitioned into two multisets of equal sum. Therefore, the f(1) f(2). F (N) will have N 2 elements.Proof. Sincef(1) NAssignf3,(2)2(F. f(modFf(1)N(1)()Ncan3))tofwill,(2)A webe.and have partitionedobtainF (fNN(2))Nwillto2 intoelements.B2 have. Since two0(N mod multisets Sincef2(1)elements.3)Proof.. =N Byf of(2) lemma equal2(Assign, Since wemod sum, areNf3)(1) now, which we2(to tryingobtainmodA impliesand− 3) toN≡f,(2) we partition our2 obtainto theorem.B0(.FmodN Since(N)=23)f.(1) ByF0(( =N lemmamod)f(2)3),. we By are lemma now trying to partition F (N)=F (N) − Theorem 4.− If NTheorem2 (mod 4. 3)If.−NF≡(N2) can (mod be− 3) partitioned. F (N) can into− be− two partitioned≡ multisets− into of equal two≡ multisets sum. of≡sum equal of sum. all elements− ≡ in F (N−) is≡ even. However,− the sum of all elements in F (N)=F (N)+−f(1), which is odd 3, F (N) can be partitioned into two multisets≡ f of3,(1) equalF (Nf(2) sum,)≡can. F3, which be(FN partitioned()Nwill implies) can have be our intoN partitioned theorem. two2 elements. multisets into Sincetwo of equal multisetsfN(1) sum,2(f(2) ofmod which equal. F3)(, impliesN sum, we) will obtain which our haveN theorem. impliesN 2 2 elements. our0(mod theorem.3) Since. ByN lemma2(mod 3), we obtain N 2 0(mod 3). By lemma − − ≡− because F (N−−) is≡ even while f(1) is odd.≡ Therefore, there is no way− to≡ partition F (N). 3, F (N) canTherefore, be partitioned solving into this two problem multisets is reduced of equal3, toF sum,( checkingN) whichcan be whether implies partitioned ourN theorem.1 into (mod two 3) multisets. 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We will look at some4 common Some examples example of special of cases special that may occur cases4 in competitive Some example programming of problems. special cases We will look at some common examples of specialWe cases will look thatWe atmay some will occur look common in at competitive some examples common programming of special examples cases problems. of special that may cases occur that inTherefore, may competitive occur solving in programming competitive this problem programming problems. is reduced problems. to checking whether N 1 (mod 3). 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Planar graph is a graph that can be drawn on a flat surface without having two edges crossing each other (West We will look at someWe will common look at examples some common of special examples cases that of special mayby occur cases4.1 the construction in that Planarcompetitive may occur of graphs programmingthe in function. competitive problems. programming problems.4.1 Planar graphs Planar graph is a graph that can be drawn on aPlanar flat surface graphet al. withoutPlanar is 2001). a graph graph having There that is aretwo a can graph many edges beProof. drawn that graph crossingIf canN on problems beis a each flatdivisible drawn other surface which on by (West aare withoutthree, flat easy surface then to having solveF without(N two if) theis edges ahaving graph prefix crossing is two of planar.F edgeswith each4 We crossing length other willSome provide(West multiples each example other several of (West three. By of lemma special 3, F (N) cases et al. 2001). 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Consider the problem The bruteIn force simple solution general for ofthis graph, colouring problemThe the Maximum numberis a to graph run Clique ofDFS such edges from problemthat can every no be tworequires ‘left’ up adjacent tovertex us a toquadratic which vertices find the orderhave maximum the with same set respect of colour. vertices to the Note in number a that graph, if of there such isthat a clique every pairof of In simple generalgives graph, us a runningvertices the number time (i.e. of ofO( edgesV 2)).. 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Consider the problem holds for V graph3 by usinggiven theinthat this Euler’s has problemO( formula.E) isin always its Therefore, runningof a colouring planar time the graph. can a number graph beWe changed first suchof edges relabel that into is no OtheO( two(V V‘right’)).. adjacent Naturally, Computing vertices vertices any shortest algorithm have path the in same a general colour. graph Note using that if there is a clique of ≥ Jonathan Irvin GUNAWAN 2 that has O(Ein) in ascending its running Bellman-Fordorder time according can be algorithm changedto the sizey-coordinate. takes intok inOO a((VEV graph,) .The) Computing(Halim theplanar set & properties Halim of shortest the 2013), vertices pathensures but in inside itathat generalonly thefor takes clique graphO( mustV using) in be a coloured planar graph. with k Countingcolours. Therefore, the entire Bellman-Ford algorithm takestheO number(VE) (Halim of connected & Halim components 2013), but using it only DFS takes in aO general(V 2) in graph a planar takes graph.O(V Counting+ E), but it only takes O(V ) in a the number of connected componentsplanar graph. using Therefore, DFS in if a the general problem graph requires takes O us(V to+ computeE), butthe it only number takes ofO( connectedV ) in a components in a planar 2 vertex v is visitable from vertex u if there Jonathanis a path from Irvin u to v GUNAWAN planar graph. Therefore, ifgraph, the problem even though requires the us constraint to compute states the thatnumberV of100 connected, 000,E components100, 0002, in the a planarstandard DFS solution still runs ≤ ≤ graph, even though the constraintunder one states second that (inV competitive100, 000,E programming,100, 0002, we the assume standard that DFS 1 million solution operations still runs can be done in 1 second ≤ ≤ under one second (in competitive(Halim & programming, Halim 2013)). we assume that 1 million operations can be done in 1 second (Halim & Halim 2013)). 4.1.2 Maximum Clique problem 4.1.2 Maximum Clique problem The Maximum Clique problem requires us to find the maximum set of vertices in a graph, such that every pair of The Maximum Clique problemvertices requires in the us set to is find directly the maximum connected set by of an vertices edge. By in reductiona graph, such from that vertex every cover, pair the of maximum clique problem vertices in the set is directlyon connected general graph by an is edge. NP-hard. By reduction However, from it is vertex easy to cover, solve the this maximum problem clique in planar problem graph. Consider the problem on general graph is NP-hard.of colouring However, a graph it is easy such to that solve no this two problem adjacent in vertices planar graph.have the Consider same colour. the problem Note that if there is a clique of of colouring a graph suchsize thatk noin atwo graph, adjacent the set vertices of the have vertices the insidesame colour. the clique Note must that be if coloured there is witha cliquek colours. of Therefore, the entire size k in a graph, the set of the vertices inside the clique must be coloured with k colours. Therefore, the entire Jonathan Irvin GUNAWAN Jonathan Irvin GUNAWAN

Figure 5: An instance of problem ‘Traffic‘. In this example, the expected output is {4, 4, 0, 2}, since the top ‘left’ Figure 5: AnFigure instance 5: An of problem instance ‘Traffic‘. of problem In this‘Traffic‘. example, In this the example, expected the output expected is {4, output 4, 0, 2}, is {4, since 4, 0, the 2}, top since ‘left’vertex the top can ‘left’ visit 4 ‘right’ vertices, the second top ‘left’ vertex can visit 4 ‘right’ vertices, the third top ‘left’ vertex Figure 5: An instancevertex of problem can visitvertex ‘Traffic‘. 4 ‘right’ can visit In vertices, this 4 ‘right’ example, the vertices, second the expected top the ‘left’second output vertex top is ‘left’ can {4, visit vertex4, 0, 4 2}, ‘right’ can since visit vertices, the 4 ‘right’ top the ‘left’ vertices, third top the ‘left’ third vertex topcannot ‘left’ vertex visit any ‘right’ vertex, and the bottom ‘left’ vertex can visit 2 ‘right’ vertices vertex can visit 4 ‘right’cannot vertices, visitcannot any the ‘right’ second visit vertex, any top ‘right’ ‘left’ and vertex vertex,the bottom can and visit the ‘left’ 4 bottom ‘right’vertex ‘left’ vertices,can visit vertex the 2 ‘right’ can third visit vertices top 2 ‘left’ ‘right’ vertex vertices cannot visit any ‘right’ vertex, and the bottom ‘left’ vertex can visit 2 ‘right’ vertices The bruteforce solution for this problem is to run DFS from every ‘left’ vertex which gives us a running time The bruteforceThe solution bruteforce for solution this problem for this is to problem run DFS is to from run every DFS ‘left’ from vertex every ‘left’ which vertex gives which us a running gives us time a running time 2vertex v is visitable from vertex u if there is a path from u to v The bruteforce solution for this problem is to run DFS from every ‘left’ vertex which gives us a running time 2vertex v is visitable2vertex fromv is visitable vertex u fromif there vertex is au pathif there from isu ato pathv from u to v 2vertex v is visitable from vertex u if there is a path from u to v Jonathan Irvin GUNAWAN Jonathan IrvinJonathan GUNAWAN Irvin GUNAWAN Jonathan Irvin GUNAWAN Understanding Unsolvable Problem 95

Fig. 5. An instance of problem ‘Traffic’. In this example, the expected output is {4, 4, 0, 2}, since the top ‘left’ vertex can visit 4 ‘right’ vertices, the second top ‘left’ vertexof O( canV 2 )visit. It 4 is very difficult to find a solution (if any) faster than O(V 2) for this problem. However, there is an ‘right’ vertices, the third top ‘left’ vertex cannot visit any ‘right’ vertex, and the bottom ‘left’ vertex can visit 2 ‘right’ vertices. important constraint on the problem. The graph given in this problem is always a planar graph. We first relabel the ‘right’ vertices in ascending order according to the y-coordinate. The planar properties ensures that for every ‘left’ vertex, the sequence of ‘right’ visitable vertices from that ‘left’ vertex is a contiguous sequence, assuming that ofofOO((VV22)).. It It is is very very difficult difficult to to find find a a solution solution (if (if any) any)of faster fasterO(V 2 than) than. ItO isOof(( veryVVO22)()V difficultforfor2) this. this It is problem. to problem. very find difficult aevery solutionHowever, However, ‘left’ to find(if vertex, there there any) a solution is is fasterthe an an sequence than (if any)O (ofV faster 2‘right’) for than this visitableO problem.(V 2 )verticesfor However, this from problem. therethat However,‘left’ is an vertex there is is an we have removed all ‘right’ vertices which are not visitable from any ‘left’ vertex. With this property, there is a importantimportant constraint constraint on on the the problem. problem. The The graph graph given givenimportant in in this this problem problem constraintimportant is is always always on the constraint a a problem. planar planar graph. ona graph. Thecontiguous the graph Weproblem. We first first givensequence, relabel relabel The in graphthis assuming problem given that in is this alwayswe problemhave a planarremoved is always graph. all a‘right’ We planar first vertices graph. relabel which We first are relabel not visitable from any ‘left’ vertex. With this property, there is a O(V log V ) solutionsolution (CEOI 2011 Solutions n.d.). thethe ‘right’ ‘right’ vertices vertices in in ascending ascending order order according according to to the thetheyy-coordinate.-coordinate. ‘right’ verticesthe The The ‘right’in planar planar ascending vertices properties properties order in ascending ensures according ensures that that order to for thefor according every everyy-coordinate. to the They-coordinate. planar properties The planar ensures properties that for ensures every that for every (CEOI 2011 Solutions, n.d.). ‘left’‘left’ vertex, vertex, the the sequence sequence of of ‘right’ ‘right’ visitable visitable vertices vertices from from‘left’ that that vertex, ‘left’ ‘left’ the vertex vertex‘left’ sequence is vertex, is a a contiguous contiguous of the ‘right’ sequence visitable sequence, sequence, of ‘right’ vertices assuming assuming visitable from that that that vertices ‘left’ from vertex that is a ‘left’ contiguous vertex is sequence, a contiguous assuming sequence, that assuming that wewe have have removed removed all all ‘right’ ‘right’ vertices vertices which which are are not not visitable visitablewe have from from removed any anywe ‘left’ ‘left’ all have ‘right’ vertex. vertex. removed vertices With With all this this ‘right’ which property, property, verticesare not there there visitable which is is a area from not any visitable ‘left’ from vertex. any With ‘left’ this vertex. property, With there this property, is a there is a 4.2 Bipartite Graph OO((VV loglogVV))solutionsolution ( (CEOICEOI 2011 2011 Solutions Solutionsn.d.).n.d.). O(V log V ) solutionO(V log (CEOIV ) solution 2011 Solutions (4.2.CEOI Bipartite 2011n.d.). Solutions Graph n.d.). Bipartite graph is a graph in which the vertices can be partitioned into two disjoint sets U and V such that all edges connect a vertex from U and a vertex from V . Some problems have a bipartite graph as an input although 4.24.2 Bipartite Bipartite Graph Graph 4.2 Bipartite4.2 Graph Bipartite Graph the problem statement does not explicitly state that the given input graph must be bipartite. The problem that we Bipartite graph is a graph in which the vertices can be partitioned into two disjoint sets used as an introduction for this paper, IOI 2014 Friend is a very good example. The construction of the graph BipartiteBipartite graph graph is is a a graph graph in in which which the the vertices vertices can can be beBipartite partitioned partitioned graph into intoBipartite is a two two graph disjoint disjoint graph in which is sets sets a graph theUU andand vertices in VV which suchsuchsuch can the that thatthat be vertices partitioned allall edges can connect intobe partitioned two a vertex disjoint into from sets two U disjointandand Va vertexsuch sets U that fromand all V .such Some that all in subtask 5 of this problem implicitly ensures that the final graph will always be bipartite. There are several edgesedges connect connect a a vertex vertex from fromUU andand a a vertex vertex from fromVV.. Someedges Some connectproblems problems aedges have have vertex aconnect a bipartitefrom bipartiteU aand vertex graph graph aproblems vertex fromas as an anU inputfrom input haveandV although aalthougha. vertex Somebipartite from problems graphV . Someas have an aproblemsinput bipartite although have graph athe bipartite as problem an input graph statement although as an does input not although graph problems that are NP-hard for general graph but solvable in polynomial time if the graph is bipartite. Since thethe problem problem statement statement does does not not explicitly explicitly state state that that the thethe given given problem input input statement graph graphthe problem must must does be be statement not bipartite. bipartite. explicitlyexplicitly does The The state problemnot problem state that explicitly thatthe that giventhe statewe we given input that input the graph given graph must input must be bipartite.graph be bipartite. must The be The problem bipartite. problem that The that we problem we used that we bipartite graph and bipartite matching was recently included in IOI 2015 syllabus (Forisek 2015), we can expect usedused as as an an introduction introduction for for this this paper, paper, IOI IOI 2014 2014 Friend Friendused is is asa a very very an introduction good goodused example. example. as an for introduction thisThe The paper, construction constructionas an forIOI introduction this 2014 of of paper, the Friendthe graph graphfor IOI is this 2014a very paper, Friend good IOI isexample.2014 a very Friend good The is a constructionexample. very good The example. of construction the graph The construc of thegraph that this type of problem may be conceived in the near future of IOI. We will take a look at several examples. inin subtask subtask 5 5 of of this this problem problem implicitly implicitly ensures ensures that that thein the subtask final final graph graph 5 ofin will willthis subtask always always problem 5 be be of implicitly bipartite. thisbipartite. problemtion ensures of There There the implicitly graph arethat are several several the in ensures subtask final graph that 5 of the willthis final alwaysproblem graph be implicitly willbipartite. always ensures There be bipartite. arethat several the Therefinal aregraph several graphgraph problems problems that that are are NP-hard NP-hard for for general general graph graph but butgraph solvable solvable problems in in polynomial polynomialgraph that are problems time NP-hardtime if if that the the for graph aregraphwill general NP-hard always is is bipartite. bipartite. graph be for butbipartite. general Since Since solvable graph There in butpolynomialare solvableseveral timegraph in polynomial if problems the graph time that is bipartite. ifare the NP-hard graph Since isfor bipartite. general Since 4.2.1 Vertex Cover and Independent Set (and maximum matching) bipartitebipartite graph graph and and bipartite bipartite matching matching was was recently recently included includedbipartite in in graph IOI IOI 2015bipartite 2015 and bipartitesyllabus syllabus graph (Forisek (Forisekmatching and bipartitegraph 2015), 2015), was but recently matching we we solvable can can expectincluded expect wasin polynomial recently in IOI included 2015 time syllabusif the in IOIgraph (Forisek 2015 is bipartite. syllabus 2015), Since (Forisek we can bipartite expect2015), graph we can and expect bipartite matching was recently included in IOI 2015 syllabus (Forišek, 2015), we can thatthat this this type type of of problem problem may may be be conceived conceived in in the the near nearthat future future this of typeof IOI. IOI. ofthat We We problem this will will taketype take may aof a be look look problem conceived at at several several may in be examples. theexamples. conceived near future in the of IOI. near We future will of take IOI. a Welook will at several take a look examples.As we at severaldiscussed examples. in an earlier section, both of the MIN-VERTEX-COVER and MAX-INDEPENDENT-SET problems expect that this type of problem may be conceived in the near future of IOI. We will are NP-hard. However, both of these problems are solvable in polynomial time on bipartite graph. By Konig’s take a look at several examples. 4.2.14.2.1 Vertex Vertex Cover Cover and and Independent Independent Set Set (and (and4.2.1 maximum maximum Vertex4.2.1 matching) matching) Cover Vertex and Independent Cover and Independent Set (and maximum Set (and matching) maximum matching)theorem, the size of the minimum vertex cover in bipartite graph is equal to the size of the maximum bipartite matching (Bondy & Murty 1976), and the size of the maximum independent set in bipartite graph is equal to the AsAs we we discussed discussed in in an an earlier earlier section, section, both both of of the the MIN-VERTEX-COVER MIN-VERTEX-COVERAs we discussedAs in andwe andan discussed earlier MAX-INDEPENDENT-SET MAX-INDEPENDENT-SET section, in an4.2.1. bothearlier of Vertex section, the MIN-VERTEX-COVER Cover problems problems both and of the Independent MIN-VERTEX-COVER and Set MAX-INDEPENDENT-SET (and Maximum and MAX-INDEPENDENT-SET Matching) problems problems number of the vertices minus the size of the maximum bipartite matching. Therefore, both problems are equivalent areare NP-hard. NP-hard. However, However, both both of of these these problems problems are are solvable solvableare NP-hard. in in polynomial polynomial However,are NP-hard. time time both on on However, of bipartite bipartite theseAs problems bothwe graph. graph. discussed of theseBy By are Konig’s Konig’s solvable in problems an earlier in are polynomial section, solvable both time in polynomialof on the bipartite MIN-VERTEX-COVER time graph. on bipartite By Konig’s graph. and MAX- By Konig’s to finding the size of the maximum bipartite matching, which can be solved in O(V 3) time. Finding the size of the theorem,theorem, the the size size of of the the minimum minimum vertex vertex cover cover in in bipartite bipartitetheorem, graph graph the is issizetheorem, equal equal of the to to the minimumthe size size of of vertex the theINDEPENDENTSET maximumminimum maximum cover in vertex bipartite bipartite cover problems graph in bipartite is areequal NP-hard. graphto the is size However, equal of the to the maximum both size of of these bipartite the problemsmaximum are bipartite maximum matching in bipartite graph using Hopcroft-Karp algorithm or Maximum-Flow algorithm is much simpler matchingmatching (Bondy (Bondy & & Murty Murty 1976), 1976), and and the the size size of of the thematching maximum maximum (Bondy independent independentmatching & Murty set set (Bondy in 1976),in bipartite bipartite & and Murtysolvable graph thegraph 1976), size is isin ofequal equalpolynomial and the to tomaximumthe the the size time of independent theon maximumbipartite set graph. independent in bipartite By Konig’s setgraph in theorem, bipartite is equal tothe graph the size is of equal the to the to solve, as compared to using Edmonds Blossom algorithm on general graph. This is actually the solution of the numbernumber of of the the vertices vertices minus minus the the size size of of the the maximum maximumnumber bipartite bipartite of matching. matching. the verticesnumber Therefore, Therefore, minus of the the vertices both both size problems problemsminimum of minus the maximum the are arevertex size equivalent equivalent of cover bipartite the maximumin bipartite matching. bipartite graph Therefore, is matching. equal both to the Therefore, problems size of the are both maximum equivalent problems bipartite are equivalent 5th subtask of IOI 2014 Friend. toto finding finding the the size size of of the the maximum maximum bipartite bipartite matching, matching,to which which finding can can the be be size solved solvedto finding of the in inOO maximum the((VV33 size))time.time. of bipartite thematching Finding Finding maximum matching, the the(Bondy size size bipartite of of whichand the the Murty, matching, can be 1976), solved which and in can Othe( beV size3 solved) time. of the in Finding Omaximum(V 3) thetime. sizeindependent Finding of the the set size in of the maximummaximum matching matching in in bipartite bipartite graph graph using using Hopcroft-Karp Hopcroft-Karpmaximum algorithm algorithm matchingmaximum or or Maximum-Flow Maximum-Flow in bipartite matching graph algorithm algorithm in bipartiteusing isHopcroft-Karp is much muchgraph simpler simpler using algorithmHopcroft-Karp or Maximum-Flow algorithm or Maximum-Flow algorithm is much algorithm simpler is much simpler toto solve, solve, as as compared compared to to using using Edmonds Edmonds Blossom Blossom algorithm algorithmto solve, on on as general general comparedto graph. solve, graph. to as This using This compared is is Edmonds actually actually to using the Blossom the solution solution Edmonds algorithm of of theBlossom the on general algorithm graph. on generalThis is actually graph. This the solution is actually of the solution of the 55thth subtasksubtask of of IOI IOI 2014 2014 Friend. Friend. 5th subtask of IOI5th 2014subtask Friend. of IOI 2014 Friend. 4.3 Directed Acyclic Graph A directed acyclic graph is a directed graph that does not contain any cycle. Similar to planar and bipartite graphs, there are several graph problems that are much easier to solve if the graph is a directed acyclic graph. 4.34.3 Directed Directed Acyclic Acyclic Graph Graph 4.3 Directed4.3 Acyclic Directed Graph Acyclic Graph 4.3.1 Minimum Path Cover AA directed directed acyclic acyclic graph graph is is a a directed directed graph graph that that does does notA not directed contain contain acyclic any anyA cycle. cycle. directed graph Similar Similar is acyclic a directed to to planar planargraph graph andis and a thatdirectedbipartite bipartite does graph graphs, graphs, not contain that does any not cycle. contain Similar any to cycle. planar Similar and bipartite to planar graphs, and bipartite graphs, therethere are are several several graph graph problems problems that that are are much much easier easierthere to to solve solve are if severalif the thethere graph graph graph are is is problems a aseveral directed directed graph that acyclic acyclic are problems much graph. graph. easierthat are to much solve if easier the graph to solve is a if directed the graph acyclic is a directed graph.MIN-PATH-COVER acyclic graph. is a problem that requries us to find the minimum number of vertex-disjoint paths needed to cover all of the vertices in a graph. By a simple reduction from Hamiltonian Path, this problem is NP-hard. 4.3.14.3.1 Minimum Minimum Path Path Cover Cover 4.3.1 Minimum4.3.1 Path Minimum Cover Path Cover A graph has a Hamiltonian Path if and only if we only need one path to cover all of the vertices. However, this problem can be solved in polynomial time for a directed acyclic graph. For a graph G =(V,E), we create a new MIN-PATH-COVERMIN-PATH-COVER is is a a problem problem that that requries requries us us to toMIN-PATH-COVER find find the the minimum minimumMIN-PATH-COVER number number is a problem of of vertex-disjoint vertex-disjoint that is a requries problem paths paths us that to needed needed find requries the minimum us to find number the minimum of vertex-disjoint number of vertex-disjointpaths needed paths needed graph G =(V V ,E), where V = V = V and E = (u, v) V V :(u, v) E . Then it can be toto cover cover all all of of the the vertices vertices in in a a graph. graph. By By a a simple simple reduction reductionto cover all from from of Hamiltoniantothe Hamiltonian cover vertices all of in Path, Path, thea graph. vertices this this problem problem By in a a simple graph. is is NP-hard. 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4.44.4 Miscellaneous Miscellaneous 4.4 Miscellaneous4.4 Miscellaneous We will use Google Code Jam 2008 Round 1A ‘Milkshake’ problem for this example. This problem requires the contestant to ﬁnd a solution with the minimum number of true variables that satisfy a CNF-SAT problem with up 4.4.14.4.1 Special Special case case of of CNF-SAT CNF-SAT problem problem 4.4.1 Special4.4.1 case Special of CNF-SAT case of problem CNF-SAT problem

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IOI 2015 Since syllabus (Forisek 2015), we can expect bipartite graph and bipartitebipartite matching graph was and recently bipartitebipartite included matching graph in IOI and was 2015 bipartite recently syllabusbipartite4.2.1 matchingincluded (Forisek Vertex graph in was 2015),IOI and recently Cover 2015 bipartite we syllabus can included and matching expect Independent (Forisek inthatthat IOI was this this 2015),2015 recently type type syllabus Set we of of canincluded problem problem(and (Forisek expect maximum mayin may IOI 2015), be be 2015 conceived conceived matching) we syllabus can expectin in (Forisekthe the near near 2015), future future ofwe of IOI. IOI. can We We expect will will take take a a look look at at several several examples. examples. that this type of problem maythat be this conceived type of problem in the near may future be conceived of IOI. 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Acyclic Graph 4.3 Directed Acyclic4.3 GraphDirected AcyclicA4.3 directed Directed acyclic Graph graph Acyclic is a4.3 directed Directed Graph graph that does Acyclic not contain Graph any cycle. Similar to planar and bipartite graphs, Athere directed are several acyclic graph graph problems is a directed that graph are much that doeseasier not to contain any cycle. Similar to planar and bipartite graphs, A directed acyclic graph is aA directed directed graph acyclic that graph doesA is directed not a directed contain acyclic graph any graph cycle. that is does SimilarA a directed not tosolve contain planar graphacyclic if the any and that graph graph cycle. bipartite does is4.3.1is Similara not a directeddirected graphs, contain Minimum totherethere planar graphacyclic any are are cycle. thatand Pathgraph. several several bipartite does Similar Cover graph graphnot graphs, to contain problems problemsplanar any and that thatcycle. bipartite are are Similar much much graphs, to easier easier planar to to and solve solve bipartite if if the the graph graph graphs, is is a a directed directed acyclic acyclic graph. graph. there are several graph problemsthere are that several are much graph easier problems to solve that if are the much graph easier is a directed to solve acyclic if the graph graph. is a directed acyclic graph. there are several graph problemsthere are that several are much graph easierthere problems to are solve several thatMIN-PATH-COVER if theare graph muchgraph problems easier is a directed to is that solve a problem are acyclic if muchthe graph. graphthat easier requries is toa directed solve usto if the acyclic find graph the graph. minimum is a directed number acyclic of graph. vertex-disjoint paths needed 4.3.1. Minimum Path Cover to cover all of the vertices in a graph. By a simple reduction from Hamiltonian Path, this4.3.1 problem Minimum is NP-hard. Path Cover MIN-PATH-COVER4.3.1 Minimum is a Path problem Cover4.3.1 that requiresMinimum us Path to find Cover the minimum number of 4.3.1 Minimum PathA graph Cover4.3.1 has aMinimum Hamiltonian Path4.3.1 Path Cover if and Minimum only if Pathwe only Cover need one path to cover all of the vertices. However, this vertex-disjoint paths needed MIN-PATH-COVER to cover all of the vertices is a problem in a graph. that requries By a simple us to refind- the minimum number of vertex-disjoint paths needed G V,E MIN-PATH-COVER is a problemMIN-PATH-COVER that requries usisMIN-PATH-COVER a to problem find the thatproblem minimum requries can is number a be us problemMIN-PATH-COVER solvedto find ofduction vertex-disjointthat in the polynomial requriesminimum from isHamiltonian us time paths anumber to problem findfor needed a ofthe directed Path,that vertex-disjointtoto minimum cover cover requries this acyclic problem all all number usof of graph. paths theto the is find vertices ofvertices NP-hard. needed For vertex-disjoint the a graphminimum in in A a a graph graph. graph.=( numberpaths has By By a aneeded a) Hamiltonian, simpleof simple we vertex-disjoint create reduction reduction a new from from paths Hamiltonian Hamiltonian needed Path, Path, this this problem problem is is NP-hard. 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Then Then it it can can be be out ∪ in out in { ∈ out × in ∈ } graph G =(Vout Vin,E)graph, whereGV=(outV=outVin V=inVgraph,Eand), whereGE =(= VV(outoutu, v=) VVininV,Eout=graphV),, where whereVandinG:(E u,V=(=out vV) out(=u,EV vin)V.=in ThenV,EVoutand) it, where canVEin = be:(shownVshownu,out(u, v)= v) by byVEinV Konig’s Konig’s ..out= Then ThenV and Vit Theoremitin Theorem can canE:(:( beu,= be vshown)( that thatu,E v )byG.. Konig’s Then ThenVhasout a it itmatching VTheorem can canin :( be beu,v that of) size E .m Thenifif and and it can only only be if if there there exist exist V m vertex-disjoint ∪ ∪ { ∪ ∈ × {∈ ∪} ∈ × { ∈ ∈ } × { ∈ }∈ × ∈ } || |−|− shown by Konig’s Theoremshown that G by has Konig’s a matching Theoremshown of that size by Konig’sGm ifhas and a Theorem matching only ifshown there that of size byG exist Konig’shashasmV if a and matching Theoremm onlyvertex-disjoint ifof of4.4 there thatsize size G existm Miscellaneouspaths hasififif and andandV a that matching only onlym cover vertex-disjoint if ifif there therethere all of of size exist exist them verticesVif andm only in vertex-disjointvertex-disjointG. if. there exist pathsV m vertex-disjoint | |− | |− | |− | |− paths that cover all of the verticespaths that in G cover. all of thepaths vertices that cover in G. all of thepaths vertices thatthat in coverGcover. allall ofof thethe vertices in in G .. 4.4.1 Special case of CNF-SAT problem 4.4 Miscellaneous We will use Google Code Jam4.4 2008 Round Miscellaneous 1A ‘Milkshake’4.4 problem Miscellaneous for this example. This problem requires the 4.4 Miscellaneous4.4 Miscellaneous4.4. Miscellaneous4.4 Miscellaneous contestant to find a solution with the minimum number of true variables that satisfy4.4.1 a CNF-SAT Special problem case with of CNF-SAT up problem 4.4.1 Special case of4.4.1 CNF-SAT Special problem case4.4.1 of CNF-SAT Special problem case of4.4.1 CNF-SAT Special problem case of CNF-SAT problem 4.4.1. Special Case of CNF-SATWe willProblem use Google Code Jam 2008 Round 1A ‘Milkshake’ problem for this example. This problem requires the We will use Google Code JamWe will2008 use Round Google 1A Code ‘Milkshake’We Jam will 2008use problemJonathan Google Round for Code 1A Irvin this ‘Milkshake’ JamWe GUNAWANexample. will 2008We use ThisRoundproblem will Google problemuse 1A forGoogle Code ‘Milkshake’ this requires Jam example.Code 2008 theJam problemcontestantcontestant Round This2008 forproblem Round 1A this to to ‘Milkshake’ find find example. 1A requires a‘Milkshake’ a solution solution problemthe This with with problem problem for the the this minimum minimum requires for example. this the numberexample. number This of ofproblem true true variables variables requires that that the satisfy satisfy a a CNF-SAT CNF-SAT problem problem with with up up contestant to find a solutioncontestant with the minimum to find a solution numbercontestant withof true the to variables find minimum a solution that numbercontestant satisfy with ofThis the a true CNF-SAT to minimum problem findvariables a problemsolution numberrequires that satisfy withof the true theupcontestant a CNF-SATvariables minimum to that problemfind number satisfy a solution withof a true CNF-SAT up variableswith the problem thatminimum satisfy with number up a CNF-SAT of problem with up true variables that satisfy a JonathanCNF-SAT Irvin problem GUNAWAN with up to 2,000 variables (Google Jonathan Irvin GUNAWANJonathan Irvin GUNAWANJonathan Irvin GUNAWANJonathanCode Irvin Jam GUNAWAN 2008 Round 1A, ‘Milkshake’ problem, n.d.). The CNF-SAT is a satisfiability problem given in a conjunctive normal form (i.e. conjunction of disjunction of literals) which was proven to be NP-hard (Cook, 1971). Therefore, it is unlikely that there is an algorithm to solve a CNF-SAT problem with 2,000 variables in less than 8 minutes3. However, there is a special property in this problem, in which at most one unnegated literal exists in each clause. Therefore, all clauses can be converted into Horn clauses. With this property, a linear time algorithm exists. (Google Code Jam 2008 Round 1A, ‘Milkshake’ solution, n.d.).

~~3 In Google Code Jam, contestants are given 8 minutes to produce the output upon downloading the input. Understanding Unsolvable Problem 97~~

~~5. Conclusion~~

In conclusion, we can use a well-known reduction technique to prove that a problem that we are currently attempting to solve is impossible (or at least it is very hard such that no people has been able to solve it for more than 40 years). In competitive programming (including IOI), understanding this technique is essential so that we will not be stuck at trying to solve an impossible problem, thus prompting us to find another way to solve the problem. To prove that a problem is NP-hard, it is good to know as many NP-hard problems as possible, so that we can reduce from any one of the problems that we know to the new problem. Some of the classic NP-hard problems include 3-SAT, Vertex Cover, Independent Set, and Subset Sum. After realizing that the problem is NP-hard, we must be able to find the special case that makes the problem solvable. We must be able to find a special property that breaks the reduction proof. Having a lot of practice on these kind of problems will help us to familiarize with the possibilities of a special case. Some of the common special cases include planar, bipartite, and directed acyclic graph.

~~References~~

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J.I. Gunawan is an undergraduate student studying Computer Science in National University of Singapore. He participated a lot of programming contests, including IOI 2012 and 2013 and ACM ICPC World Finals 2014 and 2015. He is also the lead of the Scientific Committee of Indonesian Computing Olympiad team (TOKI) training Indonesian teams for IOI in 2015–2016.