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Unformatted text preview: CS 360 Introduction to the Theory of Computing Spring 2008 Assignment 3 Solutions 1. [6 points] Prove that the language A = { x # y : x is a prefix of y } is not contextfree. Solution. It is assumed that the alphabet = { , 1 , # } and that for each x # y A we have x,y { , 1 } * . Suppose toward a contradiction that A is contextfree and let n be the pumping length of A . Consider the string n 10 n #0 n 10 n A . The pumping lemma for contextfree languages tells us that there exist strings a,b,c,d,e with: n 10 n #0 n 10 n = abcde  bcd  n bd negationslash = For every i we have ab i cd i e A We show that any strings a,b,c,d,e satisfying the first three conditions cannot also satisfy the fourth, which yields the desired contradiction. For each such a,b,c,d,e , let =  bd  . Exactly one of the following holds: a contains # . Choose i = 0 and write ab cd e = x # y . Then  x  = 2 n + 1 and  y  = 2 n + 1 . In particular,  x  >  y  , implying that x is not a prefix of y and so ab cd e negationslash A . e contains # . Choose i = 2 and write ab 2 cd 2 e = x # y . Then  x  = 2 n + 1 + and  y  = 2 n + 1 . Once again,  x  >  y  , implying that x is not a prefix of y and so ab 2 cd 2 e negationslash A . bd contains # . Choose i = 2 . Then ab 2 cd 2 e contains two occurrences of # and so ab 2 cd 2 e negationslash A . c contains # . Choose i = 2 and write ab 2 cd 2 e = x # y . As  bcd  n , it holds that x = 0 n 10 n + p and y = 0 n + q 10 n for some p,q not both zero. If p > then x has n + p > n occurrences of after its 1 , but y has only n occurrences of after its 1 . In particular, x is not a prefix of y and so ab 2 cd 2 e negationslash A . If q > then y has n + q > n occurrences of before its 1 , but x has only n occurrences of before its 1 . In particular, x is not a prefix of y and so ab 2 cd 2 e negationslash A . 2. [4 points] Consider the following Turing machine, whose input alphabet is = { 1 } ....
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 Spring '08
 JohnWatrous

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