notes1 - Some notes on the equivalence of npdas and...

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Unformatted text preview: Some notes on the equivalence of npdas and context-free grammars. Theorem. A language is context-free iff there is an npda that accepts it. This theorem is proved in two Lemmas: Lemma 1 . Let G be a context-free grammar. Then there is an npda M such that L ( M ) = L ( G ) . Lemma 2 . Let M be an npda. Then there is a context-free grammar G such that L ( M ) = L ( G ) . For the proof of Lemma 1, we will describe a procedure which, given a context-free grammar G = ( V, T, S, P ), produces an equivalent npda M = ( Q, , , , q , z, F ). We set Q = { q , q 1 , q f } , F = { q f } , = T , = V T { z } . Now for the function: (1) ( q , , z ) = { ( q 1 , Sz ) } . (2) ( q 1 , w ) ( q 1 , , A ) for all w ( V T ) * and all A V such that A w is in P . (3) ( q 1 , a, a ) = { ( q 1 , ) } for each a . (4) ( q 1 , , z ) = { ( q f , ) } . Example. Let G be the grammar given by: S aSb | A | B | A aA | B bB | bb This grammar is my solution to exercise (6b) from section 5.1. L ( G ) = { a n b m : n 6 = m- 1 } . Let us employ the procedure described above. This will give us an npda M with the following function. ( q , , z ) = { ( q 1 , Sz ) } . ( q 1 , , S ) = { ( q 1 , aSb ) , ( q 1 , A ) , ( q 1 , B ) , ( q 1 , ) } . ( q 1 , , A ) = { ( q 1 , aA ) , ( q 1 , ) } . ( q 1 , , B ) = { ( q 1 , bB ) , ( q 1 , bb ) } . ( q 1 , a, a ) = { ( q 1 , ) } ( q 1 , b, b ) = { ( q 1 , ) } ( q 1 , , z ) = { ( q f , ) } . In order to see that the npda M is equivalent to the grammar G , notice that there is a one-to-one correspondence between left-most derivations of sentences from G and sequences of valid moves of M . Example. Here is a typical derivation from G : S asb aaSbb aaAbb aaaAbb aaabb. Here is the corresponding sequence of valid moves of M : ( q , aaabb, z ) ( q 1 , aaabb, Sz ) ( q 1 , aaabb, aSbz ) ( q 1 , aabb, Sbz ) ( q 1 , aabb, aSbbz ) ( q 1 , abb, Sbbz ) ( q 1 , abb, Abbz ) ( q 1 , abb, aAbbz ) ( q 1 , bb, Abbz ) ( q 1 , bb, bbz ) ( q 1 , b, bz ) ( q 1 , , z ) ( q 1 , , ) accept 1 That is all I will say about Lemma 1. Now let us discuss Lemma 2. Definition. Let M = ( Q, , , , q , z, F ) be an npda. We say that M is in standard reduced form iff (a) M has exactly three states: Q = { q , q 1 , q f } . (b) F = { q f } . (c) There is a symbol S with S 6 = z such that the only clause which involves the state q is the clause: ( q , , z ) = { ( q 1 , Sz ) } . (d) The only -clause which involves the state q f is the clause: ( q 1 , , z ) = { ( q f , ) } ....
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This note was uploaded on 07/17/2011 for the course MAD 3512 taught by Professor Staff during the Spring '07 term at University of Florida.

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notes1 - Some notes on the equivalence of npdas and...

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