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2.4-2.9

# 2.4-2.9 - Solution to CS243 Assignment3 1 Text(Sipser...

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Solution to CS243 Assignment3 1. Text (Sipser, second edition) Chapter 2 (p.128) 2.4 2.4b S 0 | 1 | 0R0 | 1R1 R 0R | 1R | ε 2.4c S 0T | 1T T 0S | 1S | ε 2.4e S ε | 0 | 1 | 0S0 | 1S1 2.4f (V, , Φ , S), where V={S}, and = {0,1} 2. Text (Sipser, second edition) Chapter 2 (p.129) 2.6bd 2.6b S R | T | W W ba | XW | WX (out of order) X b | a R a | aR | aRb (in order, but more a) T b | Tb | aTb (in order, but more b) 2.6d S R#Y | Y#R | Y Y ε | a | b | aYa | bYb | a#R#a | b#R#b (x i = x j R occurs in Y) R R#X | X (any number of strings) X aX | bX | ε (any string) 3. Text (Sipser, second edition) Chapter 2 (p.129) 2.9 S AR | TC A aA | ε C cC | ε R bRc | ε T aTb | ε This CFG is ambiguous. It has two derivation for abc: abc abc | | AR TC / \ / \ aR bRc aTb cC | | | | ε ε ε ε

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4. Text (Sipser, second edition) Chapter 2 (p.130) 2.20 Proof idea: Let N B be the NFA for the language B. We can construct the following PDA to recognize A\B. This PDA non-deterministically enters N B , and move within N B based on what is on the stack top: (1) if it’s a symbol a, then move to q k = δ B (q i ,a); (2) if it’s a variable V, then replace the stack top with string y, based on rules specified in the CFG of language A. It can be seen that if the PDA end up in the accept state for input string w, there must exist a string x in B so that wx is in A. Thus this PDA recognize the language A\B, i.e., A\B is CFL.
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