This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 TMs 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....
View Full
Document
 Spring '10
 mr.

Click to edit the document details