Does hardness of a special case imply hardness of a general case?

Question

I am wondering in general if it is correct to claim that if a special case of a problem is NP-Hard, the general case of that problem is NP- Hard too?

For example: Min Set-Cover is NP-Hard does it imply without having to prove that Min Set-k-Cover where each element needs to be covered k times is NP-Hard too?

score 5 · Answer 1 · answered Mar 14 '13 at 06:24

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If frafl's answer doesn't give it away immediately, perhaps think in the reverse direction, if the special case is NP-hard, what would be implied by the general case being in P?

answered Mar 14 '13 at 06:24

Luke Mathieson

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score 3 · Answer 2 · answered Mar 14 '13 at 06:19

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To show that a general problem $G=\{(w,k)\mid \text{Property}\}$ is NP-complete you need a polynomial time computable many-one reduction function $f$ from e.g. $S_l =\{w\mid (w,l) \in G\}$ to $G$. How could $f$ look like?

answered Mar 14 '13 at 06:19

frafl

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score 2 · Answer 3 · edited Mar 18 '13 at 20:50

A problem is a set of instances. We consider a problem "hard" if there are instances that require lots of resources to solve (time, space, brain cells, whatever). If a subset of the problem is hard, then the problem is hard. If a subset is easy, that doesn't mean the problem is easy (there might be hard instances elsewhere). Think of the hardness as a "worst possible case" measure.

[Yes, I know this is horribly non-rigorous.]

Does hardness of a special case imply hardness of a general case?

3 Answers3