claimroutineinferred_dep

Fixed-density counting via tensor contraction

Claim·§D.2 Fixed clause density·inside thmSubexponential deterministic k-CNF model counting

Statement (paper)

Let s=n+ms=n+m. Applying Theorem  over Z/2n+1Z\mathbb Z/2^{n+1}\mathbb Z with P0=n+1P_0=n+1 and ℓ=1\ell=1 to the constructed bipartite network, and multiplying by 2u2^u modulo 2n+12^{n+1} when uu declared variables are unused, yields #F\#F as the least nonnegative residue, in time sOk(1)2Ok(s/log⁡s)s^{O_k(1)}2^{O_k(s/\sqrt{\log s})} and polynomial space. In particular, when m⩽Dnm\le Dn we have n⩽s⩽(1+D)nn\le s\le(1+D)n, so the running time is (n+m)Ok,D(1)2Ok,D(n/log⁡n)(n+m)^{O_{k,D}(1)}2^{O_{k,D}(n/\sqrt{\log n})}.

In plain English

Contracting the clause network over Z/2n+1Z\mathbb Z/2^{n+1}\mathbb Z (with P0=n+1P_0=n+1, ℓ=1\ell=1), and multiplying by 2u2^u for uu unused variables, gives #F\#F in time sOk(1)2Ok(s/log⁡s)s^{O_k(1)}2^{O_k(s/\sqrt{\log s})} and polynomial space, where s=n+ms=n+m. When m⩽Dnm\le Dn this is (n+m)Ok,D(1)2Ok,D(n/log⁡n)(n+m)^{O_{k,D}(1)}2^{O_{k,D}(n/\sqrt{\log n})}.

Why it's here

The clause network from the previous claim has Ok(s)O_k(s) edges and small degree, so the contraction theorem applies. Working modulo 2n+12^{n+1} is enough because #F⩽2n\#F\le 2^n, so the residue is the exact count. Since n⩽s⩽(1+D)nn\le s\le(1+D)n when m⩽Dnm\le Dn, the bound in terms of ss becomes the stated fixed-density bound of the main theorem.
Classified routine:
Direct application of the paper's tensor-contraction theorem to the clause network from the previous claim, plus bookkeeping: #F ≤ 2^n makes the residue mod 2^{n+1} exact, each unused variable contributes a factor of 2, and n ≤ s ≤ (1+D)n converts the bound in s into a bound in n.

Sources

PaperOpen
Section
§D.2 Fixed clause density
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quantitative-reduction.tex:86
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Friction

  • info
    inferred_dep
    outline: "Fixed-density bound via contraction theorem" rests on claim-thm-main-1