suffix - Computer Science E-207 A Proof by Construction and...

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Unformatted text preview: Computer Science E-207 A Proof by Construction and Mutual Inclusion Let L be a language and define Suffix( L ) = { x : w * s.t. wx L } . Prove that, if L is regular, then so is Suffix( L ) . Proof Idea: Since L is regular, let M = ( Q, ,q ,,F ) be a DFA recognizing L . Assume without loss of generality that M has no states that are unreachable from q . Informally, we change M into an NFA M by creating a new start state q and adding transitions from q to all other states. This allows M to skip the prefix part of any word in L . Note that the reason we dont want any unreachable states here is that this construction might otherwise let us jump to some state that no word could bring M to, and this would be a bad thing! Proof: Formally, let M be as above and define M = ( Q , ,q , ,F ) where Q = Q { q } , ( q, ) = { ( q, ) } for all q Q and , ( q , ) = Q , ( q , ) = for all , and ( q, ) =...
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suffix - Computer Science E-207 A Proof by Construction and...

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