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    BHP is NP-complete — Carmelics
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    Connected to 3 discussions

    Supports→BHP is in P only if P = NP
    Supports→BHP is in P only if P equals NP
    Supports→The Boolean Halting Problem (BHP) is in P only if P equals NP

    BHP is NP-complete

    All sources support it
    Overall Strength:30%
    2 reasons for
    2 reasons against

    Reasons For

    2 perspectives
    Reason for 1 of 2
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    • 1.BHP is in NP
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    • 2.BHP is hard for NP
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    Reason for 2 of 2
    ?
    • 1.BHP is in NP
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    • 2.BHP is NP-hard
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    Reasons Against

    2 perspectives
    Reason against 1 of 2
    ?
    • 1.NP-completeness proofs require a precise formal definition of BHP, yet 'Bounded Halting Problem' admits multiple non-equivalent formulations in the literature.
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    • 2.If BHP is defined over Turing machines with bounded steps, the reduction from SAT presupposes a fixed encoding scheme whose choice affects the complexity classification.
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    • 3.Complexity classifications are encoding-relative, so claiming BHP is NP-complete without specifying the encoding is a category error, not a theorem (cf. Papadimitriou 1994).
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    Reason against 2 of 2
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    • 1.NP-hardness is established via polynomial-time many-one reductions, which themselves presuppose P ≠ NP is not provably false—a claim independent of ZFC under some interpretations.
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    • 2.If NP-completeness proofs rely on unresolved foundational assumptions about computational complexity classes, the claim 'BHP is NP-complete' is conditionally established at best, not categorically so.
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    Topics

    All sources support it

    Connections

    1 linked claim · 4 topics

    Truth & Knowledge11 linkedProof of definition segments3 linkedPhilosophy of Language3 linkedModality & Possibility1 linked
    BHP is in P only if P = NP

    Related

    BHP is NP-hardBHP is hard for NPBHP is in NPBHP is in P only if P = NP
    +12 moreShow less
    BHP is in P only if P equals NPComplexity classifications are encoding-relative, so claiming BHP is NP-complete...If BHP is defined over Turing machines with bounded steps, the reduction from SA...If NP-completeness proofs rely on unresolved foundational assumptions about comp...

    Similar

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

    Source

    AI-extracted
    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

    Details

    Type
    premise
    Perspectives
    4 (2 for, 2 against)
    Edits
    1 edit
    If a problem is NP-complete and in P, then P = NP
    If any NP-complete problem is in P, then every problem in NP reduces to it and i...
    NP is closed under polynomial-time many-one reducibility
    NP is closed under polynomial-time many-one reductions
    NP is closed under polynomial-time reductions
    NP-completeness proofs require a precise formal definition of BHP, yet 'Bounded ...
    NP-hardness is established via polynomial-time many-one reductions, which themse...
    The Boolean Halting Problem (BHP) is in P only if P equals NP