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    SAT is NP-complete — Carmelics
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    SAT is NP-complete

    All sources support it
    ?Rate how convincing each reason is below to see the overall strength.
    3 reasons for
    0 reasons against

    Reasons For

    3 perspectives
    Reason for 1 of 3
    ?
    • 1.FACTORIZATION is in NP
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    • 2.Every problem X in NP can be reduced in polynomial time to SAT by encoding the computation of the accepting nondeterministic Turing machine for X as a propositional formula
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    • 3.SAT is therefore NP-hard
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    Reason for 2 of 3
    ?
    • 1.FACTORIZATION is in NP
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    • 2.Every problem X in NP is polynomial-time reducible to SAT
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    • 3.For any NP machine N and input x, a propositional formula φ_{N,x} can be constructed such that φ_{N,x} is satisfiable if and only if N accepts x within p(n) steps
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    Reason for 3 of 3
    ?
    • 1.FACTORIZATION is in NP
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    • 2.Every problem X in NP is polynomial-time reducible to SAT via a reduction through the computation history of the non-deterministic Turing machine accepting X
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    Connections

    3 topics

    Truth & Knowledge7 linkedProof of definition segments3 linkedPhilosophy of Language2 linked

    Related

    Every problem X in NP can be reduced in polynomial time to SAT by encoding the c...Every problem X in NP is polynomial-time reducible to SATEvery problem X in NP is polynomial-time reducible to SAT via a reduction throug...For any NP machine N and input x, a propositional formula φ_{N,x} can be constru...
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    SAT is in NPSAT is therefore NP-hard

    Similar

    BHP is NP-complete100%SAT is NP-complete.98%BHP is NP-complete.98%BHP (the Bounded Halting Problem) is NP-complete.92%

    Source

    AI-extracted1/3 agreementValid
    SEP: computational-complexity
    View source passageHide passage
    We additionally say that a class \(\textbf{C}\) is closed under \(\leq_P\) if \(Y \in \textbf{C}\) and \(X \leq_P Y\) implies \(X \in \textbf{C}\). [16] A problem \(Y\) is said to be hard for a class \(\textbf{C}\) if \(X \leq_P Y\) for all \(X \in \textbf{C}\). Finally \(Y\) is said to be complete for \(\textbf{C}\) if it is both hard for \(\textbf{C}\) and also a member of \(\textbf{C}\). e. so-called NP-complete problems. A canonical example of such a problem is a time-bounded variant of the
    Extraction notes

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

    Details

    Type
    claim
    Perspectives
    3 (3 for, 0 against)
    Edits
    1 edit