Question

To prove that a problem $\chi$ in NP is NP-complete, it would be sufficient to show which one of the following to be true?

• A. The problem $\chi$ can be reduced to the 3-SAT problem in polynomial time
• B. The 3-SAT problem can be reduced to $\chi$ in polynomial time
• C. The problem $\chi$ can be reduced to any other problem in NP in polynomial time
• D. Some problem in NP can be reduced to $\chi$ in polynomial time

Hint:

For NP-completeness, always reduce a known NP-complete problem to the new problem.

Answer 1

The Correct Option is B

Step 1: Definition of NP-completeness.
A problem is NP-complete if it satisfies two conditions: it is in NP, and every problem in NP can be reduced to it in polynomial time.
Step 2: Role of known NP-complete problems.
3-SAT is a well-known NP-complete problem. Reducing 3-SAT to another problem proves that the target problem is at least as hard as 3-SAT.
Step 3: Interpreting the reduction direction.
To show $\chi$ is NP-complete, a known NP-complete problem must reduce to $\chi$, not the other way around.
Step 4: Final conclusion.
Thus, showing that 3-SAT can be reduced to $\chi$ in polynomial time is sufficient to prove NP-completeness.

Answer 2

To prove a problem \(\chi\) already known to be in NP is NP-complete, we must additionally show that every problem in NP can be reduced to it in polynomial time. Since directly reducing every NP problem individually is impractical, we instead rely on a known NP-complete problem, since a polynomial-time reduction from a known NP-complete problem to \(\chi\) transitively shows every NP problem reduces to \(\chi\) too. Let's check each option:

1. \(\chi\) can be reduced to 3-SAT in polynomial time: This shows \(\chi\) is no harder than 3-SAT, but that only proves \(\chi\) is in NP (or at most as hard as an NP-complete problem), it says nothing about every NP problem reducing to \(\chi\), so it can't establish NP-completeness of \(\chi\).
2. 3-SAT can be reduced to \(\chi\) in polynomial time: Since 3-SAT is NP-complete, every problem in NP already reduces to 3-SAT in polynomial time. Chaining that with a reduction from 3-SAT to \(\chi\) means every NP problem also reduces to \(\chi\), which is exactly the condition needed (combined with \(\chi \in NP\)) to call \(\chi\) NP-complete.
3. \(\chi\) can be reduced to any other problem in NP in polynomial time: This again shows \(\chi\) is easy relative to other problems, the wrong direction, it doesn't demonstrate that hard problems reduce into \(\chi\).
4. Some problem in NP can be reduced to \(\chi\) in polynomial time: This is too weak, a reduction from just one arbitrary (possibly easy) NP problem doesn't establish that ALL NP problems reduce to \(\chi\), only a reduction from an NP-complete problem carries that guarantee.

Only reducing a known NP-complete problem into \(\chi\) transfers the "as hard as everything in NP" property onto \(\chi\).

Therefore, the correct answer is The 3-SAT problem can be reduced to \(\chi\) in polynomial time.