Reductions MCQ Quiz - Objective Question with Answer for Reductions - Download Free PDF
Last updated on Mar 17, 2025
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Reductions Question 1:
Let A be recursively enumerable language which is not recursive and A’ be its complement, also B be recursive language and B’ be its complement. Let X and Y be two languages such that A’ reduces to X and Y reduces to B’ and the reduction is many to one. Which of the following is/are true?
I. X can be recursively enumerable language
II. X is not recursively enumerable language
III. Y is recursive language
IV. Y may or may not be recursive language
Answer (Detailed Solution Below)
Reductions Question 1 Detailed Solution
- If B is recursive language (REC) then B’ is also REC and recursive language is decidable. Y reduces to B Since B is decidable then Y is also decidable and decidable languages are recursive.
- A is recursively enumerable language (REL) but not REC and hence A’ is also not REL. Since A’ reduces to X then X is also not REL.
Reductions Question 2:
Which of the following are correct?
Answer (Detailed Solution Below)
Reductions Question 2 Detailed Solution
The correct answer is option 1 and option 4.
Key Points
A language L1 is polynomial-time reducible to L2 (L1 ≤P L2) if there is a polynomial-time computable function f from languages to languages such that for all x, x ∈ L1 iff f(x) ∈ L2. If L1 is polynomial-time reducible to L2, then an algorithm to solve L2 can be used to obtain an algorithm to solve L1.
Option 1: If L1 ≤m L2 and L2 ≤m L3 then L1 ≤m L3
True, L1 is polynomial-time reducible to L2 and L2 is polynomial-time reducible to L3. Then L1 is polynomial-time reducible to L3 is acceptable.
Option 2: If L1 ≤m L2 then L2 ≤m L1
False, L1 is polynomial-time reducible to L2 then we can't say L2 is polynomial-time reducible to L1.
False, L1 is polynomial-time reducible to L2 then we can't say complement of L2 is polynomial-time reducible to complement of L1.
True, L1 is polynomial-time reducible to complement of L1 then L1 is decidable. A language is decidable when L and L complement both decidable.
Reductions Question 3:
Consider two decision problems 𝑄1, 𝑄2 such that 𝑄1 reduces in polynomial time to 3 - SAT and 3 - SAT reduces in polynomial time to 𝑄2. Then which one of the following is consistent with the above statement?
Answer (Detailed Solution Below)
Reductions Question 3 Detailed Solution
I) If a is reducible to B then we can say that B is at least as difficult as A. (that is by reducing a problem can be complicated or transformed).
II) All NP complete problems are also NP hard
III) 3 SAT is NP – complete.
Given,