Exercises 10 - CPSC 413 F18.pdf

2sat input a φ in which each clause has exactly 2

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2SAT Input: A φ in which each clause has exactly 2 distinct literals Output: “Yes”, if there exists an assignment that satisfies φ . “No” otherwise. 1. Give a satisfying assignment to φ 2 . 2. Give an example of a 2SAT formula that is not satisfiable.
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3. Formulas This exercise contains no questions, but instead some terminology we will be using while discussing NP and reducibility. A formula is an expression like the following, φ 1 = ( x 1 x 4 ) ( x 1 x 2 x 3 x 4 ) ( x 1 x 2 x 3 ) ( x 2 x 3 x 4 ) ( x 4 ) ( x 1 x 3 x 4 ) . This is a formula over the 4 variables x 1 , x 2 , x 3 , x 4 . The notation x 1 denotes the logical negation of x 1 . The 8 expressions x 1 , x 1 , x 2 , x 2 , x 3 , x 3 , x 4 , x 4 are called literals . A literal is either a variable x i or the negation of a variable x i . The formula φ 1 contains 6 clauses . The first clause is ( x 1 x 4 ) and the last is ( x 1 x 3 x 4 ). We assign either true or false to each of the four variables. Let us consider the assignment A 1 = ( x 1 , x 2 , x 3 , x 4 ) = (0 , 1 , 1 , 0), where 1 is shorthand for true and 0 shorthand for false . We can evaluate formula φ 1 on assignment A 1 : φ 1 ( A 1 ) = (0 1) (0 0 0 1) (1 1 0) (0 1 0) 1 (1 0 1) = 1 1 1 1 1 1 = 1 . Here a b is the logical or of a and b , and a b is the logical and of a and b . We conclude that formula φ 1 evaluates to true on the assignment A 1 . We say that a formula φ is satisfiable if there exists an assignment A for which φ ( A ) evaluates to true . Formula φ 1 is satisfiable since φ 1 ( A 1 ) evaluates to true . 4. One-in-Three SAT gadget Consider the following formula φ g in the three Boolean variables x 1 , x 2 , x 3 , φ g = ( x 1 x 2 x 3 ) ( x 1 x 2 x 3 ) ( x 1 x 2 x 3 ) ( x 1 x 2 x 3 ) ( x 1 x 2 x 3 ) .
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  • Fall '13
  • GeoffCruttwell
  • Dynamic Programming, Quantification, vertex cover, independent set, NP-complete

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