Context free languages and Push-down automata

(D) is false. L1 is regular, so its complement would also be regular. L1 is a regular language of the form 0^* 1^* 0^*. L2 on the other hand is a CFL as it can be derived from the following CFG L2 = { 0^p 1^q 0^r | p,q,r>0   And p notEqualTo r } S -> AC|CA C -> 0C0|B A -> 0A|0 B -> 1B|epsilon If coming up with a CFG for L2 is difficult, one can intuitively see that by reducing it to a simpler problem. L2 is very similar to a known CFL L3 = { a^m b^l | m notEqualTo n } (A) L2 is context free, which is true [CORRECT] (B) L1 intersection L2 is context free, which is again true because L1 is a regular language and L2 is a CFL. RL union CFL is always a CFL. Hence [CORRECT] (C) Complement of L2 is recursive, which is true due to the fact that complement of a CFL is CSL for sure (Context sensitive language), which in turn (CSL) is a subset of recursive languages. Hence [CORRECT] (D) Complement of L1 is context free but not regular, which is false due to closure laws of regular languages. Complement of a RL is always a RL. Hence [INCORRECT]   This solution is contributed by Vineet Purswani .

NDPDA can handle languages or grammars with ambiguity, but DPDA cannot handle languages with ambiguity and any context-free grammar.

1. P ∩ Q would be Q, due to the given fact that Q ⊆ P, hence context free but not regular. 2. P − Q = P ∩ Q might not even be a context free language, due to the closure properties of context free languages. 3. Σ∗ − P is equivalently complement of P, hence regular. Refer to closure laws of regular languages. 4. Σ∗ − Q is equivalently complement of Q, hence it might not even be a context free language. Refer to closure laws of CFLs. Reference: http://quiz.geeksforgeeks.org/theory-of-computation-closure-properties-of-context-free-languages/ See http://www.geeksforgeeks.org/automata-theory-set-4/ This solution is contributed by Vineet Purswani.

L1 is regular. Its DFA is given as

L2 is not regular, can be proved using pumping lemma (refer to Ullman). But L2 is CFL.

        S → AB
        A → 0A|ε
        B → 1B|ε

L3 is not CFL, can be proved using pumping lemma (refer to Ullman). But L3 is Recursive. Every regular language is also a CFL. So PDA can be used to recognized L1 and L2. As a CFL and Regular language is algo a Recursive language. Hence, Turing machine can be used to recognise L1, L2 and L3. L2 is not regular, can be proved using pumping lemma (refer to Ullman). But L2 is CFL.

        S → AB
        A → 0A|ε
        B → 1B|ε

L3 is not CFL, can be proved using pumping lemma (refer to Ullman). But L3 is Recursive. Every regular language is also a CFL. So PDA can be used to recognised L1 and L2. As a CFL and Regular language is algo a Recursive language. Hence, turing machine can be used to recognise L1, L2 and L3. Source: http://clweb.csa.iisc.ernet.in/rahulsharma/gate2011key.html

  All these languages have valid CFGs that can derive them. Hence, all of them are CFLs. Intuitively, (A) & (B) are well known CFLs and CFGs for (C) & (D) could be made by little modifications in A & B’s CFGs.     This solution is contributed by vineet purswani.

The possible palindrome generated by above grammar can be of odd length only as there is no rule for S -> For example generated palindromes are aba, aaa, bab, ababa, aaaaa, ..

The language L1 accept strings {c, abc, abcab, aabbcab, aabbcaabb, …} and L2 accept strings {a, b, c, ab, abc, aabc, aabbc, … }. Intersection of these two languages is which is context free, but not regular.

Let us first design a deterministic pushdown automata for the given language.

• For each occurrence of ‘0’ , we PUSH X in the stack.
• When ‘2’ appears, no stack operation is performed. But, state of the automata is changed.
• For each occurrence of ‘1’ , we POP X from the stack.
• If at the end Z₀ is on the stack top then input string is accepted

We also design a Turing machine for the given language.

• When ‘0’ appears in the input string , we replace it with X .Then, traverse to the rightmost corner and replace ‘1’ with Y.
• We go back to the leftmost ‘0’ and repeat the above process.
• While traversing rightwards from the beginning of the input string, if after X, ‘2’ appears and after ‘2’, Y appears then we reach the HALT state. Thus, the given language is recursive. Every recursive language is a CFL. Thus, option (B) is the answer. Please comment below if you find anything wrong in the above post.

Given below production rules.

S --> aB        S --> bA
B --> b         A --> a
B --> bS        A --> aS
B --> aBB       A --> bAA

We can derive aabbab using below sequence

S  -> aB      [Using S --> aB] 
   -> aaBB    [Using B --> aBB]
   -> aabB    [Using B --> b]
   -> aabbS   [Using B --> bS]
   -> aabbaB  [Using S --> aB]
   -> aabbab  [Using B --> b]

When it asks about the no of derivations tree, we should consider either left most derivation(LMD) or right most derivations(RMD), but not both. Here two left most derivations are possible for the correct string of the previous question "aabbab" from the given grammar. LMD-1 S -> aB [Using S --> aB] -> aaBB [Using B --> aBB] -> aabB [Using B --> b] -> aabbS [Using B --> bS] -> aabbaB [Using S --> aB] -> aabbab [Using B --> b] LMD-2 S -> aB [Using S --> aB] -> aaBB [Using B --> aBB] -> aabSB [Using B --> bS] -> aabbAB [Using S --> bA] -> aabbaB [Using A --> a] -> aabbab [Using B --> b] The Derivation tress are shown below :