TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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NumberSAT is Sharp-P-complete under parsimonious reductions

Statement

$N u m b e r S A T$ is $# P$ -complete under parsimonious reductions.

Facts & Assumptions

Given: an arbitrary function $f \in # P$ .

[L1]

Such an $f$ is the accepting-path count of a fixed polynomial-time nondeterministic machine, by Sharp-P and Gap-P functions.

[L2]

$N u m b e r S A T \in # P$ , by NumberSAT belongs to Sharp-P.

[L3]

The Cook--Levin construction can preserve accepting paths in a bijection with satisfying assignments, by The Cook--Levin construction can be made parsimonious.

[L4]

Parsimony means exact count equality under a polynomial-time map, by Parsimonious reductions between counting functions.

Proof

technique · direct

1.1

Choose a machine $N$ with $f (x) = # {acc}_{N} (x)$ as supplied by [L1]. Apply [L3] to compute the formula $r (x) = φ_{N, x}$ in polynomial time.

L1L3givenconstruct

2.1

For every input $x$ , the bijection in [L3] gives the exact chain $f (x) = # {acc}_{N} (x) = # sat (φ_{N, x}) = N u m b e r S A T (r (x)) .$ Thus $r$ is a parsimonious reduction by [L4].

L3L4step 1.1

3.1

Since $f$ was arbitrary, step 2.1 proves $# P$ -hardness, and [L2] proves membership.

L2step 2.1∎

Depends on

Used by

Polynomial time with a Sharp-P oracle Definition
NP is contained in P with a Sharp-P oracle Proposition

Dependency tree · one level

5 results within one dependency step of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Arora and Barak, Computational Complexity: A Modern Approach (standard reference, not scraped)
Lance Fortnow, Counting Complexity (standard reference, not scraped)