381hw9solns - CS381, Homework #9 Solutions Question 1...

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CS381, Homework #9 Solutions Question 1 (6.3.2) Convert the following CFG to a PDA S aAA A aS | bS | a The PDA P = ( Q, Σ , Γ , δ, q 0 , Z 0 , F ) is defined as Q = { q } Σ = { a, b } Γ = { a, b, S, A } q 0 = q Z 0 = S F = {} And the transition function is defined as: δ ( q, ±, S ) = { ( q, aAA ) } δ ( q, ±, I ) = { ( q, aS ) , ( q, bS ) , ( q, a ) } δ ( q, a, a ) = { ( q, ± ) } δ ( q, b, b ) = { ( q, ± ) } 1
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Homework 9 Exercise 6.3.3 Solutions In the following, S is the start symbol, e stands for the empty string, and Z is used in place of Z 0 . 1. S -> [qZq] | [qZp] The following four productions come from rule (1). 2. [qZq] -> 1[qXq][qZq] 3. [qZq] -> 1[qXp][pZq] 4. [qZp] -> 1[qXq][qZp] 5. [qZp] -> 1[qXp][pZp] The following four productions come from rule (2). 6. [qXq] -> 1[qXq][qXq] 7. [qXq] -> 1[qXp][pXq] 8. [qXp] -> 1[qXq][qXp] 9. [qXp] -> 1[qXp][pXp] The following two productions come from rule (3). 10. [qXq] -> 0[pXq] 11. [qXp] -> 0[pXp] The following production comes from rule (4). 12. [qXq] -> e The following production comes from rule (5). 13. [pXp] -> 1 The following two productions come from rule (6). 14. [pZq] -> 0[qZq] 15. [pZp] -> 0[qZp]
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Exercise 7.2.1(b) We will use L to denote the language {a n b n c i | i n}. For any constant n > 0, take a string to be z = a n b n c n . Clearly z ε L. Now the string will be decomposed into z = uvwxy, with vwx ε and |vwx| n. We then have several cases to consider:
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381hw9solns - CS381, Homework #9 Solutions Question 1...

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