For problems X and Y, Y is NP-complete and X reduces to Y in polynomial time. Which of the following is TRUE?

**(A)** If X can be solved in polynomial time, then so can Y

**(B)** X is NP-complete

**(C)** X is NP-hard

**(D)** X is in NP, but not necessarily NP-complete

**Answer:** **(D)** **Explanation:**

In order to solve these type of questions in GATE, we will give 2 important theorems. Proofs of these is beyond the scope of this explanation. For Proofs please refer to Introduction To Algorithms by Thomas Cormen.

**Theorem – 1
**When a given Hard Problem (NPC, NPH and Undecidable Problems) is reduced to an unknown problem in polynomial time, then unknown problem also becomes Hard.

*Case – 1 * When NPC(NP-Complete) problem is reduced to unknown problem, unknown problem becomes NPH(NP-Hard).

*Case – 2* When NPH(NP-Hard) problem is reduced to unknown problem, unknown problem becomes NPH(NP-Hard).

*Case – 3* When undecidable problem is reduced to unknown problem, unknown problem becomes also becomes undecidable.

Remember that Hard problems needs to be converted for this theorem but not the other way.

**Theorem – 2**

When an unknown problem is reduced to an Easy problem(P or NP) in polynomial time, then unknown problem also becomes easy.

*Case – 1 * When an unknown problem is reduced to a P type problem, unknown problem also becomes P.

*Case – 2* When an unknown problem is reduced to a NP type problem, unknown problem also becomes NP.

Remember that unknown problems needs to be converted for this theorem to work but not the other way.

In the given question, X which is unknown problem is reduced to NPC problem in polynomial time so Theorem – 1 will not work. But all NPC problems are also NP, so we can say that X is getting reduced to a known NP problem so that **Theorem – 2 **is applicable and X is also NP. In order to make it NPC, we have to prove it NPH first which is not the case as Y is not getting reduced to X.

This solution is contributed by **Pranjul Ahuja**.

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