3sat - CS 601/IITB Randomizd algorithm for 3 SAT Abhiram...

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Unformatted text preview: CS 601/IITB Randomizd algorithm for 3 SAT Abhiram Ranade Input: CNF formula. We will consider 3-CNF for simplicity. n variables. Output: A satisfying assignment. The obvious algorithm is to try out all 2 n assignments. For (2 n ) time. Since the problem is extremely important, exponential time algorithms have also been looked at, and the goal is minimize the exponent, i.e. minimize q where time = O ( q n ). Actually, we dont mind polynomial factors say n 2 q n . So let us use ~ O ( f ) to denote f 2 o ( n ) . So we are looking for algorithms with time ~ O ( q n ) for as small q as possible. We will see a randomized ~ O ((4 = 3) n ) algorithm. This is nearly the best algorithm known today { some minor variations of it improve the time in a fairly minor manner. It beats the best deterministic algorithms which are better than ~ O (2 n ). 1 The Algorithm The algorithm starts with a random initial state, and tries to improve it in the sense of going closer to a satisfying assignment. 1. Start with a random assignment to all variables. 2. For i = 1 :: 3 n If the current assignment is satisfying, then stop. Else let c be a clause that is false. Randomly pick any variable in c and negate it. 3. End for. Suppose s £ is any satisfying assignment, and s i the assignment constructed by the algo- rithm after i iterations. Let d i denote the Hamming distance between s i and d i . If clause c is not satisfying, then s i   1 ; s £ must di er in one of the variables in c . If we negate that variable, then d i = d i   1   1 and we will make progrss. Thus we have least 1/3 chance of making progress, and at most 2/3 chance of making regress. But it turns out that this gamble is good enough!...
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This note was uploaded on 01/04/2010 for the course CSE CS601 taught by Professor Prof.ranade during the Summer '09 term at IIT Bombay.

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3sat - CS 601/IITB Randomizd algorithm for 3 SAT Abhiram...

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