GATE 1997
December 6, 2023Theory-of-Computation
December 6, 2023GATE 1997
| Question 45 |
Which one of the following is not decidable?
| Given a Turing machine M, a stings s and an integer k, M accepts s within k steps | |
| Equivalence of two given Turing machines | |
| Language accepted by a given finite state machine is not empty | |
| Language generated by a context free grammar is non empty |
Question 45 Explanation:
(A) It is not halting problem. In halting problem number of steps can go upto infinity and that is the only reason why it becomes undecidable.
In (A) the number of steps is restricted to a finite number ‘k’ and simulating a TM for ‘k’ steps is trivially decidable because we just go to step k and output the answer.
(B) Equivalence of two TM’s is undecidable.
For options (C) and (D) we do have well defined algorithms making them decidable.
In (A) the number of steps is restricted to a finite number ‘k’ and simulating a TM for ‘k’ steps is trivially decidable because we just go to step k and output the answer.
(B) Equivalence of two TM’s is undecidable.
For options (C) and (D) we do have well defined algorithms making them decidable.
Correct Answer: B
Question 45 Explanation:
(A) It is not halting problem. In halting problem number of steps can go upto infinity and that is the only reason why it becomes undecidable.
In (A) the number of steps is restricted to a finite number ‘k’ and simulating a TM for ‘k’ steps is trivially decidable because we just go to step k and output the answer.
(B) Equivalence of two TM’s is undecidable.
For options (C) and (D) we do have well defined algorithms making them decidable.
In (A) the number of steps is restricted to a finite number ‘k’ and simulating a TM for ‘k’ steps is trivially decidable because we just go to step k and output the answer.
(B) Equivalence of two TM’s is undecidable.
For options (C) and (D) we do have well defined algorithms making them decidable.