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I'll keep the reasoning abstract. If I start from a mathematical formulation of a problem $A$ known to be $NP$-hard, I add a set of constraints which creates a problem $A'$.

However, I do know that there exists some instance of the problem for which the new set of constraint is empty, which brings, for those instance, back to the $NP$-hard formulation.

Is this enough to state that $A'$ is $NP$-hard?

Daniele Cuomo
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    To prove that a problem $A$ is NP-hard, we typically pick a problem $B$ already known to be NP-hard, and reduce $B$ to $A$. Try using this proof technique in your case. – Yuval Filmus Jul 18 '21 at 10:43
  • Thank you, I edited my question. So, can I say that $A'$ is NP-hard because $A$ is $NP$-hard and i can reduce $A$ to $A'$ by simply noticing that $A$ is a case of $A'$ with a set of constraint being empty? – Daniele Cuomo Jul 18 '21 at 10:58
  • If you manage to prove that a problem is NP-hard, then it is NP-hard. – Yuval Filmus Jul 18 '21 at 11:08

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No (unless $P=NP$), here is a counter-example:

Consider $A:=SAT$. It is well known that this is an NP-complete problem (and hence also an NP-hard problem).

Now, we will add the following constraint: "every $\phi$ with length bigger than $100$ has to have at most $2$ variables in each clause".

For formulas with length less than $100$ the constraint doesn't apply, hence the condition you stated holds.

However, the resulting language is in $P$, since we can reduce it to $2SAT$ which is known to be solved in polynomial time. To do the reduction, simply "brute force" the answer for formulas $\phi$ with length less than $100$, and otherwise keep the formula as-is.

nir shahar
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  • Thanks for your answer. I feel like we are talking about different kinds of constraint. The one of your examples seems to me that puts a constraint over the existing instances. Instead the constraint I meant reduces only the space of feasible solutions, not the solution space itself. – Daniele Cuomo Jul 18 '21 at 15:15
  • @DanieleCuomo What do you mean by "space of the feasible solutions", and how is that different from the "solution space"? Can you please elaborate a bit more on the differences? – nir shahar Jul 18 '21 at 15:41
  • For example in the SAT problem, I can find an assignment which does not satisfy all the clauses. In that case the solution is unfeasible, but it is still part of the solution space. – Daniele Cuomo Jul 18 '21 at 16:11
  • How would you define it for an arbitrary language? – nir shahar Jul 18 '21 at 16:25
  • I would partition the co-domain of the decision variables in two parts. A set of feasible solutions and a set of unfeasible ones. – Daniele Cuomo Jul 19 '21 at 15:07
  • So you are suggesting to make this into a promise problem? – nir shahar Jul 19 '21 at 16:28
  • Apologies for using a dictionary more from Operation Research field instead of Computer Science. Anyway I think that the answer is no, the NO instances are what I call the unfeasible solutions, while the YES instances are the feasible ones. – Daniele Cuomo Jul 20 '21 at 09:33
  • I am not familiar with the definition of a problem by language, but I try. My scenario I think begins with a language $L$, which I presume is a set of YES instances $L_{y}$ and NO instances $L_{n}$. Starting from $L$, I create an $L'$ such that $L'{y}\subset L{y}$ and $L_{n}\subset L'_{n}$. – Daniele Cuomo Jul 20 '21 at 09:45