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# Circuits 1 EE T220

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This 94 page Class Notes was uploaded by Maudie Quitzon on Saturday September 12, 2015. The Class Notes belongs to EE T220 at West Virginia University taught by Staff in Fall. Since its upload, it has received 21 views. For similar materials see /class/202860/ee-t220-west-virginia-university in Electrical Engineering at West Virginia University.

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Date Created: 09/12/15

Oumne The class NP Xiaofeng Gu1 1Departmem ur Mathemaucs Westvvgwa Unmarsz NP completen ess 0 Problems in NP 0 Problems in NP 0 Variants of Satisfiability G 3SAT O 2SAT O MAX2AT 0 NAESAT 0 Problems in NP 0 Variants of Satisfiability G 3SAT 0 2SAT O MAX28AT 0 NAESAT e GraphTheoretic Problems 0 INDEPENDENT SET 0 MAXCUT Class NP NP is the class of languages decided by nondeterministic Turing machines in polynomial time NP is the class of languages decided by nondeterministic Turing machines in polynomial time definiti39 Let R g Zquot X 2 be a binary relation on strings R is called polynomially decidable if the language xy xy e R is decided by a deterministic Turing machine in polynomial time NP is the class of languages decided by nondeterministic Turing machines in polynomial time ljefinitio39n Let R g Zquot X 2 be a binary relation on strings R is called polynomially decidable if the language xy xy e R is decided by a deterministic Turing machine in polynomial time R is polynomial balanced if xy e R implies lyl g lxlquot for some k 2 1 That is the length ofthe second component is always bounded by a polynomial in the length of the first l Clas NP cotd Proposition Let L g Xquot be a language L 6 NF ifand only if there is a polynomial decidable and polynomial balanced relation R such that L X By x y e R Let L g Xquot be a language L 6 NF ifand only if there is a polynomial decidable and polynomial balanced relation R such that L X By x y e R i If part Suppose that such an R exists we need to show L is decided by a nondeterministic Turing machine M in polynomial time Let L g Xquot be a language L 6 NF ifand only if there is a polynomial decidable and polynomial balanced relation R such that L X By x y e R i If part Suppose that such an R exists we need to show L is decided by a nondeterministic Turing machine M in polynomial time We construct M as follows On input x M guesses a y oflength at most lxlquot and then verify whether x e R This can be done in polynomial time because R is polynomial decidable If xy e R M accepts otherwise it rejects Problems In NF Vanams or sausnabnm GraphTheorem Problems i l i If part Suppose that such an R exists we need to show L is decided by a nondeterministic Turing machine M in polynomial time We construct M as follows On input x M guesses a y oflength at most lxlquot and then verify whether x e R This can be done in polynomial time because R is polynomial decidable If xy e R M accepts otherwise it rejects ii Only if part Suppose that L 6 NF that is there is a nondeterministic Turing machine N that decides L in time lxlquot for some k Problems In NF Vanams of sausnabnm GraphTheorem Problems l i If part Suppose that such an R exists we need to show L is decided by a nondeterministic Turing machine M in polynomial time We construct M as follows On input x M guesses a y oflength at most lxlquot and then verify whether x e R This can be done in polynomial time because R is polynomial decidable lf xy e R M accepts otherwise it rejects Only if part Suppose that L 6 NF that is there is a nondeterministic Turing machine N that decides L in time lxlquot for some k Define a relation R as follows xy e R if and only ify encodes an accepting computation of N on input x Clearly R is polynomial decidable and polynomial bounded Problems In NF Vanams of sausnabnm GraphTheorem Problems l i If part Suppose that such an R exists we need to show L is decided by a nondeterministic Turing machine M in polynomial time We construct M as follows On input x M guesses a y oflength at most lxlquot and then verify whether xy e R This can be done in polynomial time because R is polynomial decidable lf xy e R M accepts otherwise it rejects Only if part Suppose that L 6 NF that is there is a nondeterministic Turing machine N that decides L in time lxlquot for some k Define a relation R as follows xy e R if and only ify encodes an accepting computation of N on input x Clearly R is polynomial decidable and polynomial bounded Now we show L X By xy e R Problems In NF Vanams of sausnabnm GraphTheorem Problems l i If part Suppose that such an R exists we need to show L is decided by a nondeterministic Turing machine M in polynomial time We construct M as follows On input x M guesses a y oflength at most lxlquot and then verify whether x e R This can be done in polynomial time because R is polynomial decidable lf xy e R M accepts otherwise it rejects Only if part Suppose that L 6 NF that is there is a nondeterministic Turing machine N that decides L in time lxlquot for some k Define a relation R as follows xy e R if and only ify encodes an accepting computation of N on input x Clearly R is polynomial decidable and polynomial bounded Now we show L X By xy e R Since N decides L Vx e L there must be a y such that xy e R and hence L g X By xy e R Problems In NF Vanams of sausnabnm GraphTheorem Problems 39l i If part Suppose that such an R exists we need to show L is decided by a nondeterministic Turing machine M in polynomial time We construct M as follows On input x M guesses a y oflength at most lxlquot and then verify whether x e R This can be done in polynomial time because R is polynomial decidable lf xy e R M accepts otherwise it rejects Only if part Suppose that L 6 NF that is there is a nondeterministic Turing machine N that decides L in time lxlquot for some k Define a relation R as follows xy e R if and only ify encodes an accepting computation of N on input x Clearly R is polynomial decidable and polynomial bounded Now we show L X By xy e R Since N decides L Vx e L there must be a y such that xy e R and hence L g X By xy e R Conversely Vx e X By xy e R it must be the cast that N accepts X It means x e L and hence x 3yxy e R g L Thus L X By xy e R l What does the proposition tel What does the Note i Any yes instance x of the problem in NP has at least one polynomial certi cate y of its being a yes instance What does the Note i Any yes instance x of the problem in NP has at least one polynomial certi cate y of its being a yes instance ii We may not know how to discover this certificate in polynomial time but we are t sure it exists if the instance is a yes Ins ance What does the i Any yes instance x of the problem in NP has at least one polynomial certi cate y of its being a yes instance ii We may not know how to discover this certificate in polynomial time but we are 39 tance sure it exists if the instance is a yes Ins iii Naturally no instance may not have such certificate Problems In NF Vanams of sausnabnm mphTheorems Problems i Any yes instance x of the problem in NP has at least one polynomial certificate y of its being a yes instance ii We may not know how to discover this certificate in polynomial time but we are sure it exists if the instance is a yes instance iii Naturally no instance may not have such certificate SAT The certificate isjust an assignment that satisfies the Boolean expression HAMILTON PATH the certificate is precisely a Hamilton path in the graph Fro emsIuNP Variants of Sans ablllty mTh 39 5 Outline o Variants of Satisfiability G 3SAT a In HP e I Frab39 IL Cook s Theorem SAT is NP complete QEfi iti Hi kSAT where k 2 1 is an integer is the special case of SAT in which the formula is in CNF and all clauses have k literals Problems In NP Variants of Sans ablllty mT em max 5 Proposition 3 SAT is N Pcomplete t is easy to see that 3SAT 6 NF We can construct a nondeterministic Turing machine to guess a truth assignment for the variables and check in polynomial time whether the assignment satisfies all the threeliteral clauses First easy to see that 3SAT 6 NF We can construct a nondeterministic Turing machine to guess a truth assignment for the variables and check in polynomial time whether the assignment satisfies all the threeliteral clauses Then we can reduce SAT to 3SAT Suppose c is a kliteral clause in the input CNF expression lfk 1 c xthen c xxx 39 It It First Is easy to see that 3SAT 6 NF We can construct a nondeterministic Turing machine to guess a truth assignment for the variables and check in polynomial time whether the assignment satisfies all the threeliteral clauses Then we can reduce SAT to 3SAT Suppose c is a kliteral clause in the input CNF expression lfk 1 c xthen c xxx 39 lfk 2 c xy then c xyy It First Is easy to see that 3SAT 6 NF We can construct a nondeterministic Turing machine to guess a truth assignment for the variables and check in polynomial time whether the assignment satisfies all the threeliteral clauses Then we can reduce SAT to 3SAT Suppose c is a kliteral clause in the input CNF expression lfk 1 c xthen c xxx 39 lfk 2 c xy then c xyy lfk 3 c xyz It First Is easy to see that 3SAT 6 NF We can construct a nondeterministic Turing machine to guess a truth assignment for the variables and check in polynomial time whether the assignment satisfies all the threeliteral clauses Then we can reduce SAT to 3SAT Suppose c is a kliteral clause in the input CNF expression lfk 1 c xthen c xxx 39 lfk 2 c xy then c xyy lfk 3 c xyz lfk 4 c X1X2X3X4 rewrite as x1X2u A X3X4U First tIs easy to see that 3SAT 6 NF We can construct a nondeterministic Turing machine to guess a truth assignment for the variables and check in polynomial time whether the assignment satisfies all the threeliteral clauses Then we can reduce SAT to 3SAT Suppose c is a kliteral clause in the input CNF expression lfk 1 c xthen c xxx 39 lfk 2 c xy then c xyy lfk 3 c xyz lfk 4 c X1X2X3X4 rewrite as x1X2u A X3X4U When k 2 4 c X1X2X3X4 xk rewri e as X1 2411 A X3 111112 A X4 u 2 Us A 39 39 39 A M71300 Wis Note In analyzing the complexity of a problem we are trying to de ne the precise boundary between the polynomial and NPcomplete cases In analyzing the complexity of a problem we are trying to de ne the precise boundary between the polynomial and NPcomplete cases For SAT we already know that 3SAT is NPcomplete how about 2SAT Fro emsIuNP Variants of Sans ablllty mTh 39 5 Outline o Variants of Satisfiability 0 2SAT ZSAT and Graph G Let as be an instance of 2SAT We define a graph G as follows ZSAT and Graph G i Let as be an instance of 2SAT We define a graph G as follows a The vertices of G are the variables of lt15 and their negations ZSAT and Graph G i Let as be an instance of 2SAT We define a graph G as follows a The vertices of G are the variables of lt15 and their negations b There is an edge 0 9 if and only ifthere is a clause u V 9 or 19 V ain as ZSAT and Graph G Definition Let as be an instance of 2SAT We define a graph G as follows a The vertices of G are the variables of lt15 and their negations b There is an edge 0 9 if and only ifthere is a clause u V 9 or 19 V ain as c G has a weird symmetry If 0419 is an edge then so is ea Frnblems m NP Variants of Sans ablllty mphTheoretic Problems Let as be an instance of 2SAT We define a graph G as follows a The vertices of G are the variables of lt15 and their negations b There is an edge 0419 if and only ifthere is a clause u V 9 or 19 V ain as c G has a weird symmetry If 0419 is an edge then so is ux 1501 VX2AX1 V xs A xi VX2 A X2 VX3 6015 7 3 ZSAT and Graph G as is unsatisfiable if and only if there is a variable x such that there are paths from x to from x a 39 Frablems m NP us as is unsatisfiable if and only if there is a variable x such that there are paths from x to x and from x a 39 639 i If part Suppose that such a x exists we want to show as is unsatisfiable If as is satisfied by an assignment T we have two cases Frablems m NP us llablllty as is unsatisfiable if and only if there is a variable x such that there are paths from x to x and from x a 39 639 If part Suppose that such a x exists we want to show as is unsatisfiable If as is satisfied by an assignment T we have two cases a Tx true There is a path from x to x and T x true and T x false then there must be an edge 0419 along this path such that TDt true and false Since 0419 is an edge in G u V 9 is a clause in as which is not satisfied by T a contradiction Frablems m NP Variants of sausnabumy PrawnTheorems Problems Vontd as is unsatisfiable if and only if there is a variable x such that there are paths from x to x and from x a 39 Proof If part Suppose that such a x exists we want to show as is unsatisfiable If as is satisfied by an assignment T we have two cases a Tx true There is a path from x to x and T x true and T x false then there must be an edge 0419 along this path such that TDt true and false Since 0419 is an edge in G u V l is a clause in as which is not satisfied by T a contradiction b Tx false Use path x to x and the same argument as a Only if part Suppose that as is unsatisfiable we want to show there is such a variable x Ifthere is no such an x and then by contradiction we prove it we are going to construct a satisfying assignment Problems In NP nsflablllty Problems Vanants u raph fheove Only if part Suppose that as is unsatisfiable we want to show there is such a variable x fthere is no suc x we are going to construct a satisfying assignment and then by contradiction we prove it a For a node a ifthere is a path from a to at then a must be assigned false b fthere s no path from a to at then all nodes that reachable fro true a m a are assigned nd all nodes from which u is reachable are assigned false Repeat the step until all nodes have assignments we can get a satisfying assignment Frablems m NP us Vanants Iablllty mammeme Problems Only if part Suppose that as is unsatisfiable we want to show there is such a variable x fthere is no such an x we are going to construct a satisfying assignment and then by contradiction we prove i a For a node a ifthere is a path from a to at then a must be assigned false b fthere s no path from a to at then all nodes that reachable from a are assigned t nd all nodes from which u is reachable are assigned false Repeat the step until all nodes have assignments we can get a satisfying assignment We have two problems39 i Is the step in b welldefined Yes ii Continue doing the steps we will get an assignment Why is it a satisfying assignment7 Problems In HP Variants of Sans ablllty mT em F abiams 2SAT is in NL and therefore in P NL is closed under complement We can recognize unsatisfiable expressions in NL Guess a variable x and paths from x to x and back in nondeterministic logarithmic space E Problems In NP Variants of Sans ablllty nT V em Pro 5 o Variants of Satisfiability 0 MAX2 SAT Problems In NP sausn my emc Fro MAXkSAT Definition We are given a set of clauses each with two literals in it and an integer K MAX2SAT is the problem whetherthere is an assignment that satisfies at least K of the clauses We are given a set of clauses each with two literals in it and an integer K MAX2SAT is the problem whetherthere is an assignment that satisfies at least K of the clauses Problems In NP Variants of Sans ablllty em Pro 5 MAXZSAT MAX2 SAT is N Pcomplete The iem MAX2 SAT is N Pcomplete s prise Let us consider a small instance first given ten clauses X Y Z W x v y y v z z v x x V aw y V aw z V w How many clauses can be satisfied MAX2SAT is NPcomplete 1556be Let us consider a small instance first given ten clauses X Y Z W x v y y v z z v x x V aw y V aw z V w ow many clauses can be satisfied If an assignment satisfy x V y V 2 then it can be extended to satisfy seven of the clauses and no more Frablems m NP us Variants quotability GraphTheme Problems KW l 1 MAX2 SAT is NPcomplete J Poof Let us consider a small instance first given ten clauses X Y Z W x v y y v z z v x x V aw y V aw z V aw How many clauses can be satisfied If an assignment satisfy x V y V 2 then it can be extended to satisfy seven of the clauses and no more Then 3SAT can be reduced to MAX2SAT given any instance as of 3SAT we can construct an instance R of MAX2SAT for each clause C a V 9 V 39y of as we add to R the following ten clauses a my v w w v w w v w w v m l MAXZSAT contd Proof 0 v w v w w w MAXZSAT contd 04 V em V mm W wl If has m clauses then R has 10m Set K 7m We claim that as is satisfiable if and only there are at least K clauses can be satisfied in R D Outline o Variants of Satisfiability 0 NAESAT NAEAT Definition NAESAT A Boolean expression in CNF is said to be NAEsatisfied ifin each clause at least one literal is true and at least one literal is false NAESAT A Boolean expression in CNF is said to be NAEsatisfied ifin each clause at least one literal is true and at least one literal is false NAESAT is NPcomplete J NAESAT A Boolean expression in CNF is said to be NAEsatisfied ifin each clause at least one literal is true and at least one literal is false NAESAT is NPcomplete J In Theorem 82 we have proved CIRCUIT SAT is NPcomplete Now we reduce CIRCUIT SAT to NAESAT as Example 83 on how to reduce CIRCUIT SAT to SAT NAESAT A Boolean expression in CNF is said to be NAEsatisfied ifin each clause at least one literal is true and at least one literal is false NAESAT is NPcomplete J In Theorem 82 we have proved CIRCUIT SAT is NPcomplete Now we reduce CIRCUIT SAT to NAESAT as Example 83 on how to reduce CIRCUIT SAT to SAT We dd to all one or twoliteral clauses the same literal call it 2 We claim that the resulting set of clauses are NAEsatisfiable if and only if the original circuit is satisfiable Problems In NP Variants of sausnabumy GraphTheoretlc Problems MAE ST NAESAT A Boolean expression in CNF is said to be NAEsatisfied ifin each clause at J least one literal is true and at least one literal is false N AESAT is N Pcomplete In Theorem 82 we have proved CIRCUIT SAT is NPcomplete Now we reduce CIRCUIT SAT to NAESAT as Example 83 on how to reduce CIRCUIT SAT to SAT We add to all one or twoliteral clauses the same literal call it 2 We claim that the resulting set of clauses are NAEsatisfiable if and only if the original circuit is satisfiable Suppose that there is an assignment T that NAE satisfies all clauses Then T a so NAEsatisfies all clauses In one ofthese assignments 2 takes the value false This assignment then satisfies all original clauses before the addition of z and therefore there is a satisfying assignment for the circuit I Problems In NP Variants of sausnabumy GraphTheoretlc Problems NAESAT A Boolean expression in CNF is said to be NAEsatisfied ifin each clause at J least one literal is true and at least one literal is false N AESAT is N Pcomplete f In Theorem 82 we have proved CIRCUIT SAT is NPcomplete Now we reduce CIRCUIT SAT to NAESAT as Example 83 on how to reduce CIRCUIT SAT to SAT We add to all one or twoliteral clauses the same literal call it 2 We claim that the resulting set of clauses are NAEsatisfiable if and only if the original circuit is satisfiable Suppose that there is an assignment T that NAE satisfies all clauses Then 7 NAEsatisfies all clauses In one ofthese assignments 2 takes the value false This assignment then satisfies all original clauses before the addition of z and therefore there is a satisfying assignment for the circuit Conversely it there is an assignment that satisfies the circuit Then there is an assignment T that satisfies all clauses We add 2 and set 2 false in T then in no clause all literals are true under T Hence the resulting clauses are NAEsatisfied under T um vam savanabmw GraphTheoretic Problems INDEPENDENT SET MAXOUT e GraphTheoretic Problems 0 INDEPENDENT SET ms IHNP Vana saunabmw GraphTheoretic Problems INDEPENDENT SET MAXOUT INDEPENDENT SET Definition Let G VE be an undirected graph and V g V We say that V is independent if w e V U E 39 w GraphTheoretic Problems ET Let G VE be an undirected graph and V g V We say that V is independent if Vij E V ij g E QEfi iti H5 INDEPENDENT SET problem Given an undirected graph and an integer K is there an independent set V with iV i 2 K Vana 5n GraphTheoretic Problems ms nu NP INDEPENDENT SET snabumy mum INDEPENDENT SET contd In the graph below V V2 V4 is an independent set V1 V2 V4 ms nu NP amaan Wm 53 ugfziunsm SET GraphTheoretic Problems 39 INDEPENDENT SET contd 7 INDEPENDENT SET is NPComplete Frnblems m NP Vanams er sausnabnm GraphTheoretic Problems tel MEWS my 2 INDEPENDENT SET is NPcomplete 1 Reduce 3SAT to INDEPENDENT SET Given an instance as of 3SAT with m clauses We can construct a graph R a For each one of the m clauses we create a separate triangle in the graph b Each node of the triangle corresponds to a literal in the clause c There is an edge between two nodes u and v in different triangles if and only if v u ems IHNP Vana sausnabnmj GraphTheoretic Problems INDEPENDENT SET MAXOUT Proof contd x1 V X2 V X3 A EM V n V EX3AEX1 V X2 V X3 ms IHNP Vana sausnabnmj GraphTheoretic Problems INDEPENDENT SET MAXOUT Proof contd Proof m clauses correspond mtriangles Set K m We claim that as is satisfiable if and only ifthere is an independent set V ofK nodes in graph R ms IHNP Vana sausnabnmj GraphTheoretic Problems INDEPENDENT SET MAXOUT Proof contd Proof m clauses correspond mtriangles Set K m We claim that as is satisfiable if and only ifthere is an independent set V ofK nodes in graph R To see this ifa satisfying assignment exists then we identify a true literal in each clause and pick the node in the triangle ofthis clause labeled by this literal ms IHNP Vana sausnabnmj GraphTheoretic Problems INDEPENDENT SET MAXOUT Proof contd m clauses correspond mtriangles Set K m We claim that as is satisfiable if and only ifthere is an independent set V ofK nodes in graph R To see this ifa satisfying assignment exists then we identify a true literal in each clause and pick the node in the triangle ofthis clause labeled by this literal Conversely if sucha set V exists just set the vertices in V to be true and then we can get a satisfying assignment D ms IHNP Vana sauenabmw GraphTheoretic Problems Application of INDEPEN SET CLIQU INDEPENDENT SET MAXOUT A clique in an undirected graph is a set of pairwise adjacent vertices A que an undirected graph is a set of pairwise adjacent vertices CLIQUE problem Given an undirected graph G and an integer K whetherthere is a set ofK vertices that form a clique by having all possible edges between them 39 v a GraphTheoretic Problems Ewd CLIQUE problem Given an undirected graph G and an integer K whetherthere is a set ofK vertices that form a clique by having all possible edges between them C U Q U E N Pcomplete 39 n wu T GraphTheoretic Problems E w39 A clique in an undirected graph is a set of pairwise adjacent vertices Definition CLIQUE problem Given an undirected graph G and an integer K whetherthere is a set ofK vertices that form a clique by having all possible edges between them C Ll Q U E is N Pcomplete Outline of proof vertex subset C is a clique in a graph G if and only if it is an independent set in G the complement of G l ms IHNP Vana sausnabnm GraphTheoretic Problems INDEPENDENT SET MAXOUT CLIQUE contd C v1 V2 V3 is a clique in the first graph and also it is a independent set in the second graph which is the complement of the first graph V1 V4 Vi V4 V2 V3 V2 V3 ms IHNP Vana sausnabnmj GraphTheoretic Problems Application of INDEPENDEN ET NDEOVER INDEPENDENT SET MAXOUT 7 Definition A node cover of an undirected graph G V E is a set C g V that contains at least one endpoint of every edge Problems IHNP Vanams er sausnabnm 91339 GraphTheoretic Problems 39 quot node cover of an undirected graph G V E is a set C g V that contains at least one endpoint of every edge NODE COVER problem Given a graph and an integer K whether there is a node cover 0 with K or fewer vertices 39 v a GraphTheoretic Problems Ewd A node cover of an undirected graph G V E is a set C g V that contains at least J one endpoint of every edge Definition NODE COVER problem Given a graph and an integer K whether there is a node cover 0 with K or fewer vertices NODE COVER is NPcomplete n u GraphTheoretic Problems E RJ39 Definitin A node cover of an undirected graph G V E is a set C g V that contains at least one endpoint of every edge Definition ODE COVER problem Given a graph and an integer K whether there is a node cover 0 with K or fewer vertices iNODE CSOVER is N Outline of proof Vertex subset C is a node cover of a graph G if and only if V 7 C is an independent set l ms IHNP Vana sausnabnmj GraphTheoretic Problems INDEPENDENT SET MAXOUT NODE COVER contd 7 0 v1 V3 is a node cover in the graph V 7 0 V2 V4 is an independent set v1 V4 V2 V3 F ems Ill NP Vanams m savanabmw GraphTheoretic Problems INDEPENDENT SET MAXCUT Outline e GraphTheoretic Problems 0 MAXCUT HP 39 ammy GraphTheoretic Problems INDEPENDENT SET MAXCUT Definition A cut in an undirected graph G V E is a partition ofvertices into two nonempty sets S and V 7 S And the size ofa cut S V 7 S is the number of edges between S h llllel fallgl dl L1ng mm l GraphTheoretic Problems 39 Definition A cut in an undirected graph G V E is a partition ofvertices into two nonempty sets S and V 7 S And the size ofa cut S V 7 S is the number of edges between S Definitipn MINCUT problem To find a cut with the smallest size in a graph MAXCUT problem To find a cut with the largest size in a graph GraphTheoretic Problems Definitionquot A cut in an undirected graph G V E is a partition ofvertices into two nonempty sets S and V 7 S And the size ofa cut S V 7 S is the number of edges between S Definition MINCUT problem To find a cut with the smallest size in a graph MAXCUT problem To find a cut with the largest size in a graph MINCUTIs in P m NP Vanams m saunabmw GraphTheoretic Problems INDEPENDENT SET MAXCUT MAX C U T is N Pcomplete W immermaw L1ng mm GraphTheoretic Problems MAX C U T is N Pcomplete We reduce NAE3SAT to MAXCUT Given m clauses with three literals each Ci C2 V V V Cm and the variables are xi n T en we construct a graph G Vertex set xmxb 39 Edge Each clause 0 gtXn corresponds to a triangle in G n multiple edges between x and x where n is the number of occurrences of x or a D ems IHNP Vana sausnabnmj GraphTheoretic Problems INDEPENDENT SET MAXCUT Proof contd x1 V X2 A x1 V EX3 A xw V n VX3 E x1 V X2 V X2 A x1 V EX3 V EX3 A EM V n V X3 ms IHNP Vana sausnabnmj GraphTheoretic Problems INDEPENDENT SET MAXCUT Proof contd Proof K 5m We claim that there is an assignment NAEsatisfying m clauses if and only if there is a cut S V 7 S with at least K edges in the graph ms IHNP Vana sausnabnmj GraphTheoretic Problems INDEPENDENT SET MAXCUT Proof contd Proof Let K 5m We claim that there is an assignment NAEsatisfying m clauses if and only if there is a cut S V 7 S with at least K edges in the graph To see this ifthere is an assignment NAE satisfying all clauses it is easy to get a cut of size 5m true literals form a set ms IHNP Vana sausnabnmj GraphTheoretic Problems INDEPENDENT SET MAXCUT Proof contd Let K 5m We claim that there is an assignment NAEsatisfying m clauses if and only if there is a cut S V 7 S with at least K edges in the graph To see this ifthere is an assignment NAE satisfying all clauses it is easy to get a cut of size 5m true literals form a set 8 Conversely if there is a such a out then set the literals in S true and literals in V 7 S false and we can get a NAEsatisfying assignment D

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