# GATE2001-1.5

8k views

Which of the following statements is true?

1. If a language is context free it can always be accepted by a deterministic push-down automaton
2. The union of two context free languages is context free
3. The intersection of two context free languages is a context free
4. The complement of a context free language is a context free

edited

(A) is wrong as a language can be context free even if it is being accepted by non-deterministic PDA for ex- $\{ ww^r: w \in \Sigma^*$ and $w^r$  is reverse of $w\}$

(C) and (D) not always true as Context free languages are not closed under Complement or Intersection.

edited
0

I think example of A is wrong ... it will be { WWr  :  W ∈(a,b)and W is reverse} ... correct me if i am wrong ...

Edit : { WWr  :  W ∈(a,b)and W is reverse}  ... both r npda .... My mistake ...

0

Answer is only (B) because Deterministic push down automata can recognize only deterministic context free languages and cannot recognize non- deterministic context free languages.

Non Deterministic push down automata can recognize all the context free languages(both deterministic and non- deterministic).

You can also refer similar GATE question https://gateoverflow.in/3650/gate2004-it-9?show=4336#a4336.

0

The statement in your answer should have been this - "(A) is wrong as a language can be context-free even if it is being not accepted by deterministic PDA for e.g., ...". We don't care whether a CFL is accepted by a non-deterministic PDA or not. Here we concentrate on Deterministic PDA. The key concept here is NDPA ≠ DPDA, i.e., to say NDPA is more powerful than DPDA.

Ans: B not closed under intersection and complement

## Related questions

1
2.3k views
Construct DFA's for the following languages: $L=\left\{w \mid w \in \{a,b\}^*, \text{ w has baab as a substring } \right\}$ $L=\left\{w \mid w \in \{a,b\}^*, \text{ w has an odd number of a's and an odd number of b's } \right\}$
Given an arbitrary non-deterministic finite automaton (NFA) with $N$ states, the maximum number of states in an equivalent minimized DFA at least $N^2$ $2^N$ $2N$ $N!$
Consider the following two statements: $S_1: \left\{ 0^{2n} \mid n \geq 1 \right\}$ is a regular language $S_2: \left\{0^m1^n0^{m+n} \mid m \geq 1 \text{ and } n \geq 1 \right\}$ is a regular language Which of the following statement is correct? Only $S_1$ is correct Only $S_2$ is correct Both $S_1$ and $S_2$ are correct None of $S_1$ and $S_2$ is correct
Let a decision problem $X$ be defined as follows: $X$: Given a Turing machine $M$ over $\Sigma$ and any word $w \in \Sigma$, does $M$ loop forever on $w$? You may assume that the halting problem of Turing machine is undecidable but partially decidable. Show that $X$ is undecidable Show that $X$ is not even partially decidable