• a_non_monotonic_function@lemmy.world
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    3 days ago

    No, we don’t use it as an adjective. It is a set. That’s literally all it is.

    And predicating anything on the assumption that P equals NP is rather absurd. Of course, we can’t rule it out as a possibility. But virtually nobody in the discipline believes that to be the case. And honestly, why would it be? If there were polynomial time solutions to that many problems of that degree of importance, surely we would have discovered something by now.

    • MangoCats@feddit.it
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      2 days ago

      virtually nobody in the discipline believes that to be the case.

      While I agree that NP is likely not within P, that “No True Scotsman” argument is exceptionally weak for such a well disciplined field.

    • solrize@lemmy.ml
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      2 days ago

      No, we don’t use it as an adjective.

      From the first sentence of https://en.wikipedia.org/wiki/NP-hardness : “In computational complexity theory, a computational problem H is called NP-hard if…”.

      Of course, we can’t rule it out as a possibility.

      Exactly, we don’t know for certain. Of course it’s very unlikely, but in math when we say we know something for certain, it means there’s a theorem to that effect. There reason to think that factoring is not NP-hard but we don’t know for certain. If you still claim otherwise, can you cite a theorem?

      Anyway, yes, you’re confused, and at this point you’re spouting misinformation. You might consider reading a book or taking a class.

      If there were polynomial time solutions to that many problems of that degree of importance, surely we would have discovered something by now.

      It’s still an open problem, there’s a $1 million Clay prize waiting for you to claim it if you have a proof either way.