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Unformatted text preview: ECS 120: Theory of Computation Homework 8 Problem 1. Sipser, Exercise 3.7. The description is not a legitimate Turing machine because the reject condition in step 3 has to wait for potentially all settings of integers to the k variables to happen and be evaluated. This is of course not a legitimate condition because there are infinitely many such settings (ie the number of conditions is not bounded by a constant), so the TM’s finite control cannot be designed. Problem 2. Sipser, Problem 3.16, b) and c). 3.16, b) Solution 1 Let M 1 and M 2 be TMs recognizing L 1 and L 2 , respectively. Construct M that recognizes L = L 1 ◦ L 2 as follows. On input x , M tries in sequence each of the finitely many bipartitions of x , x = x 1 x 2 (how many are there?). For each partition, M checks if x 1 is accepted by M 1 and if x 2 is accepted by M 2 , by alternating between the computations of M 1 and M 2 , step by step. If both machines accept, M accepts x . If no good partitioning is found among all possible partitionings, M rejects. Solution 2 One can use a Nondeterministic Turing machine M which given a word x “guesses” the correct partitioning of x into two parts, and then just verifies that M 1 and M 2 accept x 1 and x 2 respectively.respectively....
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This note was uploaded on 04/29/2010 for the course ECS 222 taught by Professor Mr. during the Spring '10 term at UC Davis.
 Spring '10
 mr.

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