Consider three decision problems P1, P2 and P3. It is known that P1 is decidable and P2 is undecidable. Which one of the following is TRUE?**(A)** P3 is decidable if P1 is reducible to P3**(B)** P3 is undecidable if P3 is reducible to P2**(C)** P3 is undecidable if P2 is reducible to P3**(D)** P3 is decidable if P3 is reducible to P2’s complement**Answer:** **(C)****Explanation:** **Background:** In computational complexity theory, a decision problem has only two possible outputs yes or no.

A decision problem is said to be decidable if there exists an effective method or algorithm that returns a correct yes/no answer to that problem.

A decision problem is said to be undecidable if there does not exist a single algorithm that always lead to a correct yes/no solution.

In terms of reducibility: A ≤_{p} B denotes A is a decision problem that is reducible to B in polynomial time p. This simply means that A’s instance can be transformed into B’s instance and following the solution of B we can get a solution for the problem A.

So here we can draw some conclusions:

1. If A ≤_{p}B and B is decidable then A is also decidable. This is because if there exists a specific algorithm for solving B and we can also reduce A to B then we can have a solution of A as well. Hence A is decidable. However the reverse is not true i.e. if A ≤_{p}B and A is decidable then B is also decidable because A can have an algorithm existing for its correct solution but might be the case that B does not. 2. If A ≤_{p}B and A is undecidable then B is also undecidable. This is because if A is undecidable even when it can be reduced to B that simply reflects even B cannot provide an algorithm by which we can solve B and hence A. So decision problem B is also undecidable.

However the reverse is not true here as well i.e. if A ≤_{p} B and B is undecidable then A is also undecidable because there might exist an algorithm for A that can provide a solution to A.

Using the above stated conclusions we can say that option 1, 2 and 4 are false and option 3 is true.

Option 1: P1 ≤_{p}P3 and given P1 is decidable gives no conclusion for P3. Option 2: P3 ≤_{p}P2 and given P2 is undecidable gives no conclusion for P3. Option 3: P2 ≤_{p}P3 and given P2 is undecidable gives conclusion for P3 to be undecidable. Option 4: P3 ≤_{p}P2’s complement and given P2 is undecidable therefore P2’s complement is also undecidable gives no conclusion for P3.

This explanation is contributed by **Yashika Arora.**

**Visit the following articles to learn more:**

undecidability-and-reducibility

Wikipedia: Reduction_(Complexity)

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