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Unformatted text preview: 6.045J/18.400J: Automata, Computability and Complexity Nancy Lynch Homework 11 Due: April 30, 2007 Elena Grigorescu Readings: Sections 7.4, 7.5, GareyJohnson (optional) (Some of the solutions are borrowed from Susan Hohenberger and Vinod Vaikuntanathan) Problem 1 : (Sipser 7.39) In the proof of the CookLevin theorem, a window is defined to be a 2 by 3 rectangle of cells. Show why the proof would have failed if we had used 2 by 2 windows instead. Solution 1 : Generally speaking, trouble can arise when checking left moves of the TM head. In particular, we can show that there exist two rows of configurations, such that the top row is a legal configuration, and the bottom row is not a configuration that could be produced from the top row by the transition function, but at all 2 × 2 windows are legal. Consider the following 2 × 3 window, where the top row is part of a legal configuration (and obviously the bottom row does not legally follow – the regular TM cannot be in two states at once). q 1 q 2 q 3 However, the 2 × 2 window scheme cannot catch this problem. It only checks the following two windows: q 1 q 2 , which could be a valid transition; for example, the result of δ ( q 1 , 1) → ( q 2 , 1 ,L ). The problem is that this window can’t see whether the head is looking at a 1 or a 0, so it assumes it is valid. q 1 q 3 , which could be a valid transition; for example, the result of δ ( q 1 , 0) → ( q 3 , ,R ). If both of these checks pass, then obviously we have a problem. Problem 2 : (Sipser 7.36) Show that, if P=NP, a polynomial time algorithm exists that, given a Boolean formula φ , actually produces a satisfying assignment for φ if it is satisfiable.if it is satisfiable....
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 Spring '07
 NancyLynch

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