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Unformatted text preview: 1 The Satisfiability Problem Cook’s Theorem: An NPComplete Problem Restricted SAT: CSAT, 3SAT 2 Boolean Expressions ◆ Boolean, or propositionallogic expressions are built from variables and constants using the operators AND, OR, and NOT. ◗ Constants are true and false, represented by 1 and 0, respectively. ◗ We’ll use concatenation (juxtaposition) for AND, + for OR,  for NOT, unlike the text . 3 Example : Boolean expression ◆ (x+y)(x + y) is true only when variables x and y have opposite truth values. ◆ Note : parentheses can be used at will, and are needed to modify the precedence order NOT (highest), AND, OR. 4 The Satisfiability Problem ( SAT ) ◆ Study of boolean functions generally is concerned with the set of truth assignments (assignments of 0 or 1 to each of the variables) that make the function true. ◆ NPcompleteness needs only a simpler question (SAT): does there exist a truth assignment making the function true? 5 Example : SAT ◆ (x+y)(x + y) is satisfiable. ◆ There are, in fact, two satisfying truth assignments: 1. x=0; y=1. 2. x=1; y=0. ◆ x(x) is not satisfiable. 6 SAT as a Language/Problem ◆ An instance of SAT is a boolean function. ◆ Must be coded in a finite alphabet. ◆ Use special symbols (, ), +,  as themselves. ◆ Represent the ith variable by symbol x followed by integer i in binary. 7 Example : Encoding for SAT ◆ (x+y)(x + y) would be encoded by the string (x1+x10)(x1+x10) 8 SAT is in NP ◆ There is a multitape NTM that can decide if a Boolean formula of length n is satisfiable. ◆ The NTM takes O(n 2 ) time along any path. ◆ Use nondeterminism to guess a truth assignment on a second tape. ◆ Replace all variables by guessed truth values. ◆ Evaluate the formula for this assignment. ◆ Accept if true. 9 Cook’s Theorem ◆ SAT is NPcomplete. ◗ Really a stronger result: formulas may be in conjunctive normal form (CSAT) – later. ◆ To prove, we must show how to construct a polytime reduction from each language L in NP to SAT. ◆ Start by assuming the most resticted possible form of NTM for L (next slide). 10 Assumptions About NTM for L 1. One tape only. 2. Head never moves left of the initial position. 3. States and tape symbols are disjoint. ◆ Key Points : States can be named arbitrarily, and the constructions manytapestoone and twowayinfinitetapetoone at most square the time. 11 More About the NTM M for L ◆ Let p(n) be a polynomial time bound for M....
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 Spring '08
 Motwani,R
 CSAT, Boolean formula

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