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: CS 181 Winter 2008 Formal Languages and Automata Theory Problem Set #8 Solutions Problem 8.1. (10 points) Show that P is closed under union, concatenation, and star. Solution Suppose A,B are two languages in P. Then let M A be the polynomial time Turingdecider for A and M B the same for B . The M A has some asymptotic runtime O ( n k ) and M B has runtime O ( n ) for some integers k and . Without loss of generality, we assume that k . We first show that A B P. We can build the following Turing decider M 1 for A B . M 1 first runs M A on its input, if M A accepts, then M 1 accepts. If M A rejects, then M 1 acts as M B on the same input. Clearly, this machine decides A B . Let us now analyze the running time of M 1 . Notice that for every input word, in the worst case we run M A then M B . Then this takes O ( n k + n ) = O ( n max( k, ) ) = O ( n k ). Since k is a constant, it follows that M 1 runs in polynomial time. The existence of M 1 proves that A B P. Next, we show that AB P by building the Turing decider M 2 for it. Notice that given an input string of length n , there are n +1 ways to partition it into two substrings (e.g. 123 can be partitioned as 123 , 1 23 , 12 3 or 123 ). M 2 simply tries each partitioning one at a time. For each partitioning, it runs M A on the left half of the string then M B on the right half. If both machines accept, then M 2 halts and accepts. If one of them rejects, then M 2 tries the next partitioning. If all partitionings fail, then M 2 rejects. Let us now analyze the running time of M 2 . Notice that for each partitioning, we run M A on a string of length at most n and M B on a string of length at most n . Thus, for each partioning M 2 takes time O ( n max( k, ) ) = O ( n k ). However, there are n + 1 partitionings. Thus, in total, M 2 takes O ( n k +1 ) time. This is still polynomial and so AB P. Lastly, we show that A * P by building a Turing decider M 3 for it. M 3 will work by using dynamic programming. Notice that a string s is in A * if s A or if s can be written s = s 1 s 2 where s 1 ,s 2 A * and s 1 ,s 2 are nonempty (and so are proper substrings of s ). Then for input string s , we can first calculate which length 1 substrings are in A * (there are n of them). Then using this information, we can calculate which of the length 2 substrings are in A * (all n 1 of them) by checking to see if they are in A , and otherwise checking to see if we can partition it into substrings in A * . We then continue this for the n 2 length3 substrings, the n 3 length4 substrings, etc. This continues until we can calculate whether the length n substring (namely, s itself) is in A * ....
View
Full
Document
 Spring '08
 griebach

Click to edit the document details