Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

GATE | GATE-CS-2016 (Set 1) | Question 53

  • Difficulty Level : Basic
  • Last Updated : 01 Oct, 2021
Consider the transition diagram of a PDA given below with input alphabet ∑ = {a, b}and stack alphabet Γ = {X, Z}. Z is the initial stack symbol. Let L denote the language accepted by the PDA.
gt11

(A) A
(B) B
(C) C
(D) D


Answer: (D)

Explanation:  

In the given PDA we can number the states as q1, q2 and q3 from left to right.

Attention reader! Don’t stop learning now.  Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.

Learn all GATE CS concepts with Free Live Classes on our youtube channel.

The transitions of the PDA are as follows:



  1. (q1, a, Z) ->(q1, XZ )

  2. (q1, a, X) ->(q1, XX )

  3. (q1, b, X) ->(q2, € )

  4. (q2, b, X) –>(q2, € )

  5. (q2, €, Z) ->(q3, Z)

As initial state is the final state hence null string is always accepted by the PDA.
The first two transitions show that state remains q1 (final state) on ‘a’ input alphabet and with every ‘a’ we push X onto the stack. Hence a
n is always accepted for n≥0.
Transitions 3 and 4 shows that for input alphabet ‘b’ and stack symbol X (i.e. ‘a’ occurred in the string) we can pop X from the stack. Transition 5 shows that we can move to the final state (q3) only when the string is empty and stack symbol is Z. This is possible when we have popped all X from the stack i.e. ‘b’ occurred exactly the same times as ‘a’. Hence a
nbn is always accepted for n≥0. The language accepted by the PDA is { an | n≥0 } U { anbn | n≥0 } and is a Deterministic CFL.
(a
nbn | n≥0 is not regular but is accepted by a PDA). Hence option (D).

This solution is contributed by Yashika Arora.



Quiz of this Question

My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!