HW6Soln

HW6Soln - Homework 6 Solution CSCI 2670 3.7 The problem is with step 2 The Turing machine has to evaluate the polynomial at every possible

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Homework 6 Solution CSCI 2670 October 11, 2005 3.7 The problem is with step 2. The Turing machine has to evaluate the polynomial at every possible combination of integer values for x 1 , x 2 , …, x k . Since there are an infinite number of possibilities, the machine will never get past this step and accept if the polynomial has some integer root. 3.15 b) Assume we have two decidable languages A and B. We want to show that A ° B is also decidable. Let M 1 and M 2 be two deciders such that L(M 1 ) = A and L(M 2 ) = B. Consider the following Turing machine: C = “On input w 1 Let l = length(w) 2 For each i = 0, 1, 2, …, l 3 Let w l = the leftmost i symbols in w 4 Let w r = the rightmost l-i symbols in w 5 Run M 1 on input w l 6 If it accepts 7 Run M 2 on input w r 8 If it accepts accept 9 Next i 10 reject Claim: C decides A ° B. This is clear since if w is in A ° B, there is some i for which w l is in A and w r is in B. When the loop reaches that i , C will accept. If w is not in A
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This note was uploaded on 02/07/2011 for the course CS 501 taught by Professor Sm during the Spring '11 term at Indiana.

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HW6Soln - Homework 6 Solution CSCI 2670 3.7 The problem is with step 2 The Turing machine has to evaluate the polynomial at every possible

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