# GATE | GATE-CS-2005 | Question 45

• Last Updated : 28 Jun, 2021

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

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.