I am currently studying the concept of NP - complete, and I came across a question that I'm not sure about my own answer (currently, it's a "Yes."). The question is: assuming Problem A requires finding a a maximum number of people. I was able to show that Problem A's answer is equivalent to the cardinality of the maximum independent set of an undirected graph $G$. Now, as we know that finding a maximum independent set of an undirected graph is NP-hard, does it imply Problem A must also be NP-hard, which means A is NP-complete?
2026-03-26 06:20:20.1774506020
Equivalence of NP-complete problems
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(In all cases below, by "reduction" I mean "polynomial-time reduction.")
To prove NP-hardness you must show a reduction from an NP-hard problem to your problem. You have a reduction from your problem to an NP-hard problem (indeed a parsimonious reduction from your problem to a #P-complete problem) but that is insufficient to show NP-hardness or #P-completeness. If you don't have a proof of NP-hardness, you don't have a proof of NP-completeness either.