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 Non-deterministic 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