hw3sol - CS 181 Winter 2005 Formal Languages and Automata...

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CS 181 — Winter 2005 Formal Languages and Automata Theory Problem Set #3 Solutions Problem 3.1. (8 points) Give a regular expression for the language ( rb ) * || ( br ) * , where || is the shuffle operator from Homework 2. (First construct an ε -move-free finite automaton that accepts this language; then transform the automaton into an equivalent regular expression. Hint: At each step of the construction, try to find the simplest automaton for the language). Solution In the previous assignment, we found how to construct an NFA for the shuffle of two languages A 1 and A 2 . We will perform that construction here to get a machine for their shuffle. We start by finding machines for A 1 = ( rb ) * and A 2 = ( br ) * . We’d like these machines to be as simple as possible, and it turns out two states suffice for each. Thus, NFAs for A 1 and A 2 are M 1 and M 2 , respectively, as shown below: B r b A D b r C M 1 M 2 Using our construction from the last homework, we build a machine that simulates running both machines and non-deterministically decides which letter to send to which machine. This gives the following NFA: BC r b AC AD BD r b r b r b We now take this machine an find a corresponding regular expression. We start by adding a new initial and final state, then one-by-one removing states BC , AD , then BD .
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AC BC r b AD BD r b r b r b ε ε AC r b AD BD r b rb ε ε rb rb rb AC BD ε ε rb+br rb+br rb+br rb+br AC ε ε ( rb+br ) + [( rb+br )( rb+br )*( rb+br )] The resulting machine gives the regular expression. [( rb br ) ( rb br )( rb br ) * ( rb br )] * = ( rb br ) * . Let us check that this regular expression is reasonable. Notice that this regular expression gives the languages of strings that do not start or end with rr or bb nor contain rrr nor bbb .
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If a word w is in the shuffle of w 1 A 1 and w 2 A 2 , and if it started with rr then none of these letters can be from w 2 since w 2 must start with b . Then they must both be w 1 letters. However, this means w 1 contains rr which cannot be the case. We can use similar logic to note that w cannot start with bb , and cannot contain rrr nor bbb .
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