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    If P = NP, then finding a satisfying valuation for a prop... — Carmelics
    Home/Truth & Knowledge
    HistoryEditSee Inverse

    If P = NP, then finding a satisfying valuation for a propositional formula would be no harder than constructing its truth table

    Proof of definition segmentsTruth & Knowledge
    ?Rate how convincing each reason is below to see the overall strength.
    1 reason for
    2 reasons against

    Reasons For

    1 perspective
    Reason for
    ?
    • 1.P is the class of problems decidable efficiently
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    • 2.NP is the class of problems verifiable efficiently given a certificate
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    • 3.If P = NP, deciding and verifying coincide up to a polynomial factor for all NP problems
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    Reasons Against

    2 perspectives
    Reason against 1 of 2
    ?
    • 1.Hartmanis and Stearns (1965) grounded complexity in resource-bounded computation where 'no harder than' requires a precise reduction type, but the claim leaves the reduction unspecified.
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    • 2.Without specifying whether the reduction is many-one, Turing, or Cook, the comparative difficulty relation invoked is semantically underdetermined and cannot bear the logical weight the argument assigns it.
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    • 3.Cook's original 1971 formalization treats SAT's hardness relative to NP-completeness, not relative to truth table enumeration, making the chosen baseline philosophically unmotivated.
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    Reason against 2 of 2
    ?
    • 1.Constructing a truth table is exponential in the number of variables, not polynomial, so it cannot serve as the baseline for 'no harder than'.
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    • 2.The claim conflates asymptotic complexity classes with concrete algorithmic equivalence: P=NP would mean SAT is solvable in polynomial time, making it strictly easier than truth table construction, not equally hard.
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    Topics

    Truth & KnowledgeProof of definition segments

    Related

    Constructing a truth table is exponential in the number of variables, not polyno...Cook's original 1971 formalization treats SAT's hardness relative to NP-complete...Hartmanis and Stearns (1965) grounded complexity in resource-bounded computation...If P = NP, deciding and verifying coincide up to a polynomial factor for all NP ...
    +4 moreShow less
    NP is the class of problems verifiable efficiently given a certificateP is the class of problems decidable efficientlyThe claim conflates asymptotic complexity classes with concrete algorithmic equi...Without specifying whether the reduction is many-one, Turing, or Cook, the compa...

    Similar

    If P equals NP, finding a satisfying valuation for a propositional for...98%If P = NP, then finding a satisfying valuation for a propositional for...94%Finding a satisfying valuation for a propositional formula is an NP pr...90%A formula is in SAT if and only if a satisfying valuation exists83%

    Source

    AI-extracted1/3 agreementValid
    SEP: computational-complexity
    View source passageHide passage
    Among these are Grover’s algorithm (Grover 1996) for searching an unsorted database (which runs in time \(O(n^{1/2})\), whereas the best possible classical algorithm is \(O(n)\)) and Shor’s algorithm (Shor 1999) for integer factorization (which runs in \(O(\log_2(n)^3)\), whereas the best known classical algorithm is \(O(2^{\log_2(\log_2(n))^{1/3})}\)). Since it can be shown that quantum models can simulate models such as the classical Turing machine, \(\textbf{BQP}\) contains \(\textbf{P}\) and
    Extraction notes

    Validity: Extracted via Max plan + API grounding/validity checks

    Details

    Type
    claim
    Perspectives
    3 (1 for, 2 against)
    Edits
    1 edit