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?

Question

If L and L are recursively enumerable, then L is:

Answer 1

NP-completeness needs two things: the problem is in NP, and every NP problem can be reduced into it. Proving the second part directly for every NP problem one by one isn't feasible.
Instead, since 3-SAT is already known to be NP-complete, everything in NP already reduces to 3-SAT. So if we can further show 3-SAT reduces to $\chi$, we get a chain: any NP problem to 3-SAT to $\chi$, which means any NP problem reduces to $\chi$ too.
Reducing $\chi$ the other way, into 3-SAT, would only show $\chi$ is easy, not hard, so that direction doesn't help prove NP-completeness.
So the correct answer is the 3-SAT problem can be reduced to $\chi$ in polynomial time.

Answer 2

It helps to think of reductions as transferring "hardness" in one specific direction: if problem A reduces to problem B, then B is at least as hard as A. With that lens:

$\chi$ reduces to 3-SAT: This transfers hardness from $\chi$ onto 3-SAT, meaning 3-SAT is at least as hard as $\chi$, the opposite of what we want to show about $\chi$.
3-SAT reduces to $\chi$: This transfers hardness from 3-SAT onto $\chi$, meaning $\chi$ is at least as hard as 3-SAT. Since 3-SAT is already the hardest kind of problem in NP, this makes $\chi$ at least that hard too, which is what NP-completeness requires.
$\chi$ reduces to any other NP problem: Again transfers hardness away from $\chi$ onto something else, showing $\chi$ is easy relative to that other problem, not that it's hard.
Some NP problem reduces to $\chi$: This does transfer hardness onto $\chi$, but only from one arbitrary, possibly weak, NP problem, not from a problem already known to be as hard as all of NP, so it's not strong enough.

Only pulling hardness from the already-hardest known problem, 3-SAT, into $\chi$ is strong enough to establish NP-completeness.

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

Answer 3

It helps to think of reductions as transferring "hardness" in one specific direction: if problem A reduces to problem B, then B is at least as hard as A. With that lens:

$\chi$ reduces to 3-SAT: This transfers hardness from $\chi$ onto 3-SAT, meaning 3-SAT is at least as hard as $\chi$, the opposite of what we want to show about $\chi$.
3-SAT reduces to $\chi$: This transfers hardness from 3-SAT onto $\chi$, meaning $\chi$ is at least as hard as 3-SAT. Since 3-SAT is already the hardest kind of problem in NP, this makes $\chi$ at least that hard too, which is what NP-completeness requires.
$\chi$ reduces to any other NP problem: Again transfers hardness away from $\chi$ onto something else, showing $\chi$ is easy relative to that other problem, not that it's hard.
Some NP problem reduces to $\chi$: This does transfer hardness onto $\chi$, but only from one arbitrary, possibly weak, NP problem, not from a problem already known to be as hard as all of NP, so it's not strong enough.

Only pulling hardness from the already-hardest known problem, 3-SAT, into $\chi$ is strong enough to establish NP-completeness.

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

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?

Show Hint

The Correct Option is B

Approach Solution - 1

Approach Solution -2

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