7.24sipserfall09sol - CPSC 421 Fall 2009 Homework 5:...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Page1 / 2

7.24sipserfall09sol - CPSC 421 Fall 2009 Homework 5:...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online