Skip to content
Related Articles

Related Articles

GATE | GATE-CS-2003 | Question 54
  • Difficulty Level : Expert
  • Last Updated : 01 Jul, 2019

Define languages L0 and L1 as follows :

L0 = {< M, w, 0 > | M halts on w}
L1 = {< M, w, 1 > | M does not halts on w} 

Here < M, w, i > is a triplet, whose first component. M is an encoding of a Turing Machine, second component, w, is a string, and third component, i, is a bit. Let L = L0 ∪ L1. Which of the following is true ?
(A) L is recursively enumerable, but L’ is not
(B) L’ is recursively enumerable, but L is not
(C) Both L and L’ are recursive
(D) Neither L nor L’ is recursively enumerable

Answer: (D)

Explanation: Since Halting problem of Turing Machines is undecidable. So, L = L0 ∪ L1 is undecidable even not semi-decidable. That is not recursive enumerable , therefore, its complement (L’) is also not recursive enumerable.

Option (D) is correct.

Quiz of this Question


My Personal Notes arrow_drop_up
Recommended Articles
Page :