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Unformatted text preview: CPSC 421 Fall 2009 Homework 5: Solutions 7.21 of Sipser: If f ( x 1 , . . ., x n ) is a formula function, then the function g defined on one more variable, g ( x 1 , . . . , x n , x n +1 ) = f ( x 1 , . . . , x n ) , has twice as many satisfying assignments as f . Thus f has at least one satisfying assignment iff g has two or more satisfying assignments. Thus 3SAT ≤ DOUBLE − SAT. 7.24 of Sipser: (a) If one variable in a clause is assigned to true and the other to false, then in the negation the values are interchanged and remain unequal. Therefore this holds for a 3cnf-formula in which each clause contains unequal truth values. (b) Let y 1 ∨ y 2 ∨ y 3 be true. Then at least one of y 1 , y 2 , y 3 is true. Say y 1 is true. Then setting z i to false and b to false, we have both clauses y 1 ∨ y 2 ∨ z i and z i ∨ y 3 ∨ b have unequal values (regardless of y 2 and y 3 ). Similarly for y 2 . Finally say that y 1 , y 2 are false and y 3 is true. Then taking z i to be true we have the...
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- Spring '10