NP-Completeness of 3SAT, 1-IN-3SAT and MAX 2SAT

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1 NP-Completeness of 3SAT, 1-IN-3SAT and MAX 2SAT 3SAT The 3SAT problem is the following. INSTANCE : Given a boolean expression E in conjunctive normal form (CNF) that is the conjunction of clauses, each of which is the disjunction of three distinct literals. QUESTION : Is E satisfiable? The CSAT problem is the following. INSTANCE : Given a boolean expression E in CNF. QUESTION : Is E satisfiable; that is, is there a truth assignment to the variables of E so that each clause in E becomes true? The Cook-Levin theorem asserts that SATISFIABILITY is NP-complete. Although 3-CNF expressions are a subset of the CNF expressions, they are complex enough in the sense that testing for satisfiability turns out to be NP-complete. Theorem : 3SAT is NP-complete. Proof : Evidently 3SAT is in NP, since SAT is in NP. To determine whether a boolean expression E in CNF is satisfiable, nondeterministically guess values for all the variables and then evaluate the expression. If E turns out to be true, then accept. This can be carried out in nondeterministic polynomial time. Thus 3SAT is in NP. To prove NP-completeness, we shall reduce CSAT to 3SAT. Let a given CNF expression be E = e 1 e 2... e k where each e i is a disjunction of literals. We replace each clause e i (as shown below) to create a new expression F such that i) F is a clause in 3SAT form. ii) time taken to construct F is linear in the length of E. iii) a truth assignment satisfies E if and only if it can be extended to a satisfying truth assignment for F. The construction of F is as follows. 1. If an e i is a single literal, say (x) or (x) we introduce two new variables u and v. We replace (x) by the conjunction of four clauses as (x+u+v)(x+u+v)(x+u+v)(x+u+v). Since u and v appear in all combinations, the only way to satisfy all four clauses is to make x true. Scribe : R. Karthik

2 2. Suppose an e i is the disjunction of two literals, (x + y). We introduce a new variable z and replace e i by the conjunction of two clauses (x + y + z)(x + y + z). As in case 1, the only way to satisfy both clauses is to satisfy (x + y). 3. If an e i is the disjunction of three literals it is already in the form required for 3-CNF, so we take e i as such to construct F. 4. Suppose e i = (x 1 +x x m ) for some m 4. We introduce new variables y 1, y 2,..., y m 3 and replace e i by the conjunction of clauses (x 1 + x 2 + y 1 )(x 3 + y 1 + y 2 )(x 4 + y 2 + y 3 )... (x m 2 + y m 4 + y m 3 )(x m 1 + x m + y m 3 ). (1) If there is a truth assignment τ that satisfies e i then at least one literal (one of the x s) in x 1 + x x m should be true; say τ makes x j true; that is in... (x j 1 + y j 3 + y j 2 )(x j + y j 2 + y j 1 )(x j+1 + y j 1 + y j )... Then in (1) above, if we make y 1, y 2,..., y j 2 true and make y j 1, y j,..., y m 3 false, we satisfy all the clauses of (1). Thus τ can be extended to satisfy these clauses. On the other hand, if τ makes all the x s false we can reason (as follows) that it is not possible to extend τ to make (1) true. The reason is that, since x 1 and x 2 are false, y 1 must be true (otherwise this clause will become false and will collapse e i to false) in the first clause; since x 3 is false and y 1 is false, y 2 must be true to keep the situation alive. By continuing the argument, we will reason that y m 3 is true. But, alas, x m 1 is false, x m is false and y m 3 is false and we see that e i cannot be satisfied. The above argument shows how to reduce an instance E of CSAT to an instance F of 3SAT, such that F is satisfiable if and only if E is satisfiable. The construction evidently requires time that is linear in the length of E, because none of the four cases above expand a clause by more than a factor 32/3 (that is the ratio of symbol counts in case 1) and it is possible to built F in polynomial time. Since CSAT is NP-complete, it follows that 3SAT is also NP-complete. 1-IN-3SAT The 1-IN-3SAT problem is the following. INSTANCE : A collection of clauses C 1,..., C m, m > 1; each C i is a disjunction of exactly three literals. QUESTION : Is there a truth assignment to the variables occurring so that exactly one literal is true in each C i? Let X = {x 1,..., x 5 } be the set of variables. Let the clause set C = {C 1, C 2, C 3 } be the following : C 1 = {x 1, x 2, x 3 }, C 2 = {x 1, x 4, x 5 }, C 3 = {x 2, x 4, x 5 }. With this (X, C) we consider the 1-IN-3SAT problem. We say a clause is correctly satisfied if and only if the clause is satisfied due to exactly one literal in it. To correctly satisfy C 1 we try setting only x 1 = in C 1. This when applied to C 2 requires x 4 = and x 5 = for correct satisfaction of C 2. But x 4 = and x 5 = in C 3 requires x 2 = for correct satisfaction of C 3 this violates the correct satisfaction of C 1. Similarly we get a contradiction if we try setting only x 2 = in C 1. The only solution is x 1 =, x 2 = and x 3 = and exactly one of x 4 or x 5 is. Note that as a 3SAT instance (X, C) admits other solutions as well.

3 Theorem : 1-IN-3SAT is NP-complete. Proof : It can be argued that 1-IN-3SAT is in NP. To prove that the problem is NP-hard, the following reduction shows 3SAT is polynomially reducible to 1-IN-3SAT, where 3SAT is NP-hard. That is, we need to show that any instance of 3SAT can be efficiently transformed to an instance of 1-IN-3SAT so that 3SAT is satsifiable if and only if 1-IN-3SAT is satisfiable. We consider any C i = {x i1, x i2, x i3 } in 3SAT. In the corresponding 1-IN-3SAT instance, we will produce the following three clauses with the addition of four new variables a i, b i, c i, d i : C i1 = {x i1, a i, b i }, C i2 = {x i2, b i, c i }, C i3 = {x i3, c i, d i } Hence if the 3SAT instance has n variables and m clauses, in the 1-IN-3SAT instance, there will be n + 4m variables and 3m clauses. This transformation can be done efficiently. Assume that the 3SAT instance is not satisfiable. Then in any C i, x i1 =, x i2 = and x i3 =. We try to extend this truth assignment to a satisfying truth assignment of 1-IN- 3SAT. In the 1-IN-3SAT instance, in C i2 since x i2 = exactly one of b i or c i should be so that C i2 gets correctly satisfied; if b i =, C i1 will not get satisfied; if c i =, C i3 will not get satisfied. In other words, if the 3SAT instance is not satisfied then no truth assignment of variables can make the 1-IN-3SAT instance satisfied correctly. Now assume that the 3SAT instance is satisfiable. Every C i is in any solution. We consider the following cases that make C i = in the 3SAT instance: 1. x i2 = (we have 4 cases) : In the 1-IN-3SAT instance we can set b i =, c i = in C i2 ; we also set a i = x i1 and d i = x i3 so that C i1 and C i3 are correctly satisfied. 2. x i2 =, x i1 =, x i3 = : In the 1-IN-3SAT instance we can set a i =, b i =, c i =, d i = so that C i1, C i2 and C i3 are correctly satisfied. 3. Only x i1 = : In the 1-IN-3SAT instance we set a i =, b i =, c i =, d i =. This correctly satisfies C i1, C i2 and C i3. 4. Only x i3 = : In the 1-IN-3SAT instance we set a i =, b i =, c i =, d i =. This correctly satisfies C i1, C i2 and C i3. In summary, a satisfying truth assignment to the variables in 3SAT can be extended to a correctly satisfying truth assignment of the variables in 1-IN-3SAT. We conclude that 1-IN-3SAT is also NP-complete. MAX 2SAT The following is the maximum satisfiability with at most two literals per clause (MAX 2SAT) problem. INSTANCE : Given a set C 1,..., C p of clauses, each being a disjunction of at most two literals; and an integer k where 1 k < p. QUESTION : Is there a truth assignment to the variables so that k or more clauses are satisfied? Note that if k = p, the above problem reduces to the 2SAT problem which can be solved

4 efficiently in polynomial time. The following theorem relies on reducing 3SAT to MAX 2SAT. The proof rewritten slightly essentially follows [2]. Theorem : MAX 2SAT is NP-complete. Proof : First we note that if a truth setting of variable makes at least k clauses true, then by substituting these truth values we can efficiently check the truth or falsity clause by clause, keeping a counter to hold the number of clauses found to be true so far. Hence MAX 2SAT is in NP. The following is a reduction from 3SAT to MAX 2SAT. That is, given an instance of 3SAT we construct an instance of MAX 2SAT so that a satisfying truth assignment of 3SAT can be extended to a satisfying truth assignment of MAX 2SAT. Note that a satisfying truth assignment is one that makes at least k clauses true in the MAX 2SAT instance. Let S be the instance of 3SAT where the clauses are C 1,..., C m where C i = {x i, y i, z i }, where each x i, y i, and z i represents either a variable or its negation and 1 i m. From S we build an instance S of MAX 2SAT as follows. Each C i in S corresponds to a clause group C i in S where C i = {(w i ), (x i ), (y i ), (z i ), (x i + y i ), (y i + z i ), (x i + z i ), (w i + x i ), (w i + y i ), (w i + z i )} where w i is a new variable and 1 i m. We set k = 7m. Each clause in S as constructed above has at most two literals. It can be seen that the clauses in S can be efficiently generated from the clauses in S in polynomial time. We now argue that a satisfying truth assignment of S exists if and only if it can be extended to a satisfying truth assignment for S appropriately. Assume that S is satisfiable. Then in a typical clause C i = {x i, y i, z i } either one or two or all the three variables are true. a) Let x i = T, y i = F, z i = F. With this truth assignment in S, if w i = T, six clauses of C i become true; if w i = F, seven clauses of C i become true. b) Let x i = T, y i = T, z i = F. This truth assignment in S together with w i = T or w i = F, makes seven clauses of C i true. c) Let x i = T, y i = T, z i = T. With this truth assignment in S, if w i = T, seven clauses of C i become true; if w i = F, six clauses of C i become true. In summary, a satisfying truth assignment of S can be extended to a satisfying truth assignment of S where exactly seven clauses in each clause group get satisfied. Moreover no setting of w i causes more than seven of the ten clauses to be true in each clause group in S. Now assume that S is not satisfiable. Then in at least one clause C i = {x i, y i, z i } in S, we have x i = F, y i = F, z i = F. With this truth assignment in S, if w i = T, four clauses of C i become true; if w i = F, six clauses of C i become true. That is, if S is not satisfiable, no truth setting can make at least seven clauses true in each clause group C i. Since 3SAT is NP-complete, we conclude that MAX 2SAT is also NP-complete.

5 References [1] J. E. Hopcroft, R. Motwani, and J. D. Ullman. Introduction to Automata Theory, Languages, and Computation, third edition. Pearson Addison-Wesley, [2] M. R. Garey, D. S. Johnson, and L. Stockmeyer. Some simplified NP-complete graph problems. Theo. Comp. Sci., 1(3): , [3] T. J. Schaefer. The complexity of satisfiability problems, in Proceedings of the 10th Annual ACM Symposium on Theory of Computing, pages , 1978.