hw3sol - Problem Set 3 Solutions CS373 - Spring 2011 Due:...

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Unformatted text preview: Problem Set 3 Solutions CS373 - Spring 2011 Due: Thursday Mar 17 at 2:00 PM in class (151 Everitt Lab) 1. Mirrors [ Category : Proof, Points : 18] Let A = { w { a,b } * | w = w R } . (a) Determine the set of equivalence classes of A on A . Using this, conclude that A is not regular. (4 Points) [SOLUTION] Every string has its own equivalence class. We will show that two strings w and x , w 6 = x , have distinguishing strings. If | w | = | x | , then ww R A but xw R 6 A since w R 6 = x R . Thus w R is distinguishing. Otherwise, wlog let | w | > | x | and write w = w 1 w 2 where | w 1 | = | x | . ww R = w 1 w 2 w R 2 w R 1 A , so consider xw R = xw R 2 w R 1 . w 2 must be a palindrome for xw R to be in A , so w R is distinguishing if w 2 6 = w R 2 . If w 2 is a palindrome, then w 2 is not, where is di erent than the rst symbol of w 2 . ww R = w 1 w 2 w R 2 w R 1 A , but xw R = xw R 2 w R 1 6 A since w R 2 6 = w 2 . Hence, w R is distinguishing. (b) Prove that A cannot be generated by any CFG in which S aSa | bSb are the only production rules that have terminals on the right side. (6 Points) [SOLUTION] Let n a ( w ) be the number of a 's in w , and let n b ( w ) be the number of b 's in w . If S aSa | bSb are the only production rules that have a terminal on the right side, then n a ( w ) and n b ( w ) must both be even. The string aaab is in A , but the number of a 's is not even. Therefore, aaab cannot be generated by a grammar where S aSa | bSb are the only production rules that have a terminal on the right side. Therefore, such a grammar cannot generate A . (c) A mirror grammar is a CFG ( V, ,R,S ) where the only production rules are of the form X aY b X a X for X,Y V , a,b . Prove that every regular language has a mirror grammar, but not all CFLs with mirror grammars are regular. Hint : Recall that a rule of the form X aY can be seen as a transition from X to Y in a nite automaton. For mirror grammars, think of Y as being more than just one state. (8 Points) [SOLUTION] To show that not all mirror grammars generate regular languages, here is a gram- mar for { n 1 n | n } : S S 1 | 1 We can construct a mirror grammar G = ( V, ,R,S ) for a regular language by considering its DFA M = ( Q, ,,q ,F ) . On input w , a rule X aY b will correspond to reading a symbol each from the start and end of w simultaneously. Reading backwards in a DFA requires tracking a subset of possible states. Thus each variable will correspond to a pair ( q,P ) where q is the current state while reading forwards and P is the subset of states reading backwards. The grammar will stop when q and P meet in the middle. Formally, V = Q Pow ( Q ) S = ( q ,F ) R contains ( q,P ) a ( q ,P ) b if ( q,a ) = q and P = { p Q | ( p,b ) P } ....
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This note was uploaded on 04/28/2011 for the course CS 373 taught by Professor Viswanathan,m during the Spring '08 term at University of Illinois, Urbana Champaign.

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hw3sol - Problem Set 3 Solutions CS373 - Spring 2011 Due:...

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