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    Carmelics

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    LoyalLoyalJusticeJustice
    Made withinDC&Austin
    Statements
    321,452
    Perspectives
    108,905
    Topics
    42
    Home/Original/inverse
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    Inverse View

    It is not the case that If P equals NP, factoring a natural number would be no more difficult than verifying that a given factorization is correct

    ?Set your confidence on the premises below to see your aggregate.

    Reasons For

    2 perspectives
    Reason for 1 of 2
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    • 1.P=NP would equalize worst-case complexity classes, but factoring's average-case hardness could persist independently of class membership.
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    • 2.Levin's average-case complexity theory shows that worst-case class collapse does not entail uniform tractability across problem instances.
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    • 3.The claim conflates decision-problem complexity with the search problem of producing a factorization, which require separate complexity-theoretic treatment.
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    Reason for 2 of 2
    ?
    • 1.Integer factorization is not NP-complete, so P=NP need not imply factoring is in P unless factoring reduces to an NP-complete problem.
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    • 2.Factoring occupies an intermediate status under standard conjectures, placing it in a region where P=NP may not collapse its complexity to P.
      ?

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    Reasons Against

    1 perspective
    Reason against
    ?
    • 1.Verifying a factorization is a polynomial-time task
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    • 2.Integer factorization is not known to be in P but is in NP
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    • 3.If P equals NP, finding and verifying solutions to NP problems are equally difficult
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