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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?

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    $\begingroup$ 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. $\endgroup$ Jul 18 at 10:43
  • $\begingroup$ 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? $\endgroup$ Jul 18 at 10:58
  • $\begingroup$ If you manage to prove that a problem is NP-hard, then it is NP-hard. $\endgroup$ Jul 18 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.

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  • $\begingroup$ 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. $\endgroup$ Jul 18 at 15:15
  • $\begingroup$ @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? $\endgroup$
    – nir shahar
    Jul 18 at 15:41
  • $\begingroup$ 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. $\endgroup$ Jul 18 at 16:11
  • $\begingroup$ How would you define it for an arbitrary language? $\endgroup$
    – nir shahar
    Jul 18 at 16:25
  • $\begingroup$ I would partition the co-domain of the decision variables in two parts. A set of feasible solutions and a set of unfeasible ones. $\endgroup$ Jul 19 at 15:07

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