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UGC-NET | NTA UGC NET 2019 June – II | Question 22

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  • Last Updated : 03 Dec, 2021

Which of the following problems is/are decidable problem(s) (recursively enumerable) on a Turing machine M?
(a) G is a CFG with L(G)=∅
(b) There exist two TMs M​1​ and M2​ such that L(M) ⊆{L(M​1​)UL(M​2​)}= language of all TMs
(c) M is a TM that accepts w using a most 2​|w|​ cells of tape

(A) a and b only
(B) a only
(C) a, b and c
(D) c only

Answer: (C)

Option C: a, b and c

a): L(G)=∅ is a emptiness problem and emptiness problem for context free languages is decidable.

(b): Recursive: The reason is we can assume L(M2)=∑* and hence any TM (M1) will be definitely subset of L(M2). So we only need to see that M1 is a valid Turing machine and no need to check whether it is subset of L(M2) as it is obviously subset of L(M2)=∑*. Checking validity of Turing machine is always a halting process, hence it is recursive.

(c): L={(M,w)|M is a TM that accepts w using at most 2|w| squares of its tape}.
Recursive: Let m be he number of states in M, and k be the size of the alphabet that M uses, and r=|w|. If M uses at most 2r squares of its tape, then there are at most Alpha=mK2r 2r configurations(why?). If M runs on w for more than α steps, and does not use more than 2r squares of its tape, then M must be in the one configuration at least twice(pigeonhole principle), in which case M would enter an infinite loop on input w. Based on this, we design a machine M* that decides L that works as follows:
M* runs M on w for at most α+1 steps.
If M accepts w which steps with using at most 2r squares, M* halts and accepts.
If M rejects w within α steps with using at most 2r squares, M* halts and rejects.
If M goes beyond 2r squares, M* halts and rejects.
If M does not halt and does not go beyond 2r squares, M* rejects.

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