2.1 Solution. E T F a. E E + T T + T F + T a + T a + F a + a

Transcription

1 . Solution E T F a E E + T T + T F + T a + T a + F a + a

2 E E + T E + T + T T + T + T F + T + T a + T + T a + F + T a + a + T a + a + F a + a + a

3 E T F ( E) ( T ) ( F) (( E)) (( T )) (( F)) (( a))

4 . Solution m n a) Language A={a b c n m, n 0} has the same number of b and c, while language n n B={a b c m m, n 0} has the same number of a and b. n n Hence, A B={a b c n n 0}. Then we have the exact same language given in Example.36 at page 6 of the textbook. So we can apply the proof in this example to prove the claim that the class of context-free languages is not closed under intersection. b) In order to show that the class of context-free languages is not closed under complementation, we first need to show that the class of context-free languages is closed under union operation. Let G = ( V, Σ, R, S) generate L and G = ( V, Σ, R, S) generate L, so that the two grammars have no variables in common. We can form a grammar G=(V, Σ,R,S) such that L(G)=L L. We set V = V V { S}, Σ = Σ Σ, and R = R R { S S } { S S }. This construction proves that context-free languages are closed under union. Then we prove the claim in the problem.(b) by Suppose that the class of context-free languages is closed under complementation. Then because A and B are contextfree, A and B are also context-free. According to DeMorgan s law, A B = A B. Because context-free languages are closed under union, A B is context free, and according to our assumption, the complement of A B is also context-free, which implies that A B is contextfree. However, we already showed in (a), that context-free languages are not closed under intersection. Hence, we get a.5 Solution We can simply give a counterexample. Suppose the context-free language A, and the corresponding grammar is G = {{ S},{(,)},{ S ( S), S ε}, S}. Then if we add S SS and get the new grammar the language generated by grammar does not generate ' G will have (()()), which is not contained in * A in this case. * A. Hence, the new ' G,.6 Proof Assume that context-free languages L and L are generated by CFGs G = ( V, Σ, R, S) and G = ( V, Σ, R, S). We need to make sure that V V =. To achieve this, we could simply rename variables whenever necessary. If we can construct a new CFG, denoted as G=(V, Σ,R,S), * which can generate the languages L L, L L, and L, respectively, then the class of context-free languages is closed under the regular operations.

5 (a) To prove that the class of context-free languages is closed under union, let G=(V, Σ,R,S), where V = V V { S}, Σ = Σ Σ, and R = R R { S S} { S S}. (b) To prove that the class of context-free languages is closed under concatenation, let G=(V, Σ,R,S), where V = V V { S}, Σ = Σ Σ, and R = R R { S SS}. (c) To prove that the class of context-free languages is closed under star, let G=(V, Σ,R,S), where V = V { S}, Σ = Σ, and R = R R { S SS} { S ε}..3 Proof We assume that B is a CFL and obtain a Let p be the pumping length of B that p p p p is guaranteed to exist by the pumping lemma. Select the string s = 0 0. Clearly s is a member of B and of length at least p. We can show that no matter how we divide s into uvxyz, one of the three conditions of the lemma is violated. According to condition 3 of the pumping lemma, vxy following ways: p, we can only place vxy in the (a) vxy completely falls in the first 0 p If we pump v and y, then the new string is no longer a palindrome, and the number of 0s will be greater than the number of s, which is a (b) vxy falls between 0 p and p In this case, v will only contain 0s while y will only contain s. So if we pump s, the new string is no longer palindrome, which is a (c) vxy completely falls in the first p Similar to (a), after pumping s, the number of s will be greater than the number of 0s, which is a (d) vxy falls between p and p After pumping s, the number of s will be greater than the number of 0s, which is a (e) vxy completely falls in the second p

6 This case is same with (c). (f) vxy falls between p and 0 p Similar to (b), v will only contain s while y will only contain 0s. So if we pump s, the new string is no longer palindrome, which is a (g) vxy completely falls in the second 0 p This case is same with (a). Hence, B is not context free..3 Proof We prove the claim by Suppose C is context free. Let p be the pumping length. Select a string s = 3 4 p p p p, which is a member of C with length at least p. No matter how we divide s into uvxyz, there will always be a According to condition 3 of the pumping lemma, vxy following ways: p, we can only place vxy in the (a) vxy completely falls in p After pumping s, the number of s and the number of s are no longer the same, which is a (b) vxy falls between p and 3 p After pumping s, not only will the number of s and s, but also the number of 3s and 4s will become different, which is a (c) vxy falls completely in 3 p After pumping s, the number of 3s will be greater than the number of 4s, which is a (d) vxy falls between 3 p and p Similar to (b), we can obtain a contradiction after pumping s. (e) vxy falls completely in p

7 After pumping s, the number of s and the number of s are no longer the same, which is a (f) vxy falls between p and 4 p Similar to (b), we can obtain a contradiction after pumping s. (g) vxy falls completely in 4 p After pumping s, the number of 4s will be greater than the number of 3s, which is a Hence, C is not context free..44 Proof Because A and B is regular, we can construct DFA and DFA to accept A and B correspondingly. Also, we could construct NFA using DFA and DFA to accept A B = { xy x A and y B}. In order to accept A B = { xy x A and y B and x = y }, we need to keep track of the number characters accepted by DFA and the number of characters accepted by DFA, and if they are equal, the whole string should be accepted. We can achieve this by modifying NFA into a PDA. The state diagram of PDA is the following: x, ε x y, x ε ε, ε $ ε, ε ε ε,$ ε NOTE: There may be more than two states in either DFA or DFA. Let s look at the part DFA. The transition means that at the same time DFA accepting x, we push x on to the stack; similarly, as for DFA, at the same time it accepting y, we pop what we have on the stack.

8 Because this PDA can accept A B = { xy x A and y B and x = y }, A B = { xy x A and y B and x = y } is context free.