Initial rows number at most 3n; at most 3n further rows are ever included, so every candidate has at most 6n equations. A row mentioning at most s clauses expands to at most 2^k s monomials of degree at most k on the ORIGINAL n variables. Initial row length is at most D0 and added-list length is P. If B is bounded by a fixed power of log n for fixed k, both P and D0 are polylogarithmic, so one fixed sparsity constant c covers all queries, including discarded candidates. At most 3n(2m)^P+1 counts suffice. There is no circular SAT oracle: the queried primitive was derived for sparse polynomial systems independently of this reduction.

Why every candidate query fits the sparse common-zero counter

There are (2m)^P ordered candidate lists. For each candidate row h, use the sparse common-zero counter to compute N(A;h). All satisfying assignments pass, so N(A;h)=#F+D(A;h) with the SAME unknown #F across candidates. A row minimizing N therefore minimizes false positives D. The average surviving D is at most 3D(A)/4, so a minimizing row achieves that bound. Repeat for exactly 3n rounds, breaking ties deterministically. At most (3/4)^(3n)2^n=(27/32)^n<1 false positives remain; their integer count is zero. The procedure never needs to know #F to select a row or decide when to stop.

AI Agents

A candidate is an ordered list of P independent pairs (e_j,c_j), each a uniform clause index and independent uniform bit. Repeated indices and zero coefficients remain part of the list; combine them modulo two only to form the resulting row. A remaining false positive violates at least m/P clauses, so the chance that no selected index hits its support is at most (1-1/P)^P<=exp(-1)<1/2. Given a hit, choose one hit position and fix all other coefficients. Its free uniform bit makes the row parity uniform, even with repeated indices. Hence each false positive is rejected with probability at least 1/4. Lists, not distinct resulting rows, carry the uniform distribution.

Why a short candidate parity row rejects many remaining false positives

Fix k and c. The manuscript counter accepts ANY system of at most c n binary polynomial equations, each with degree at most k and at most (log n)^c monomials. It does not require that its zeros already equal the satisfying assignments of a SAT formula. High-incidence branching leaves polylogarithmic scope and incidence bounds. On each branch, use ell=n+1 from the original variable count and the lifted-degree/bit analysis with the checked layout. Since n_prime>=n/2, m and ell remain O(n_prime). The resulting stated time is 2^{O_{k,c}(rho(n))}, rho(n)=n(log log n)^2/sqrt(log n), and polynomial space. Failed finite checks fall back to exhaustive counting. This is an asymptotic manuscript result, not a demonstrated practical solver.

The sparse common-zero counter used by the SAT reduction

OpenAI. The Exponential Time Hypothesis is False. Manuscript dated October 17, 2026, revised October 25, 2026. This reference pins commit 42907c9e999cc00d6abe7a50bfc1023250edd1b9. The introduction defines the exact-counting task used by this learning path; citing it does not establish that its main theorem is correct.

https://github.com/openai/math/blob/42907c9e999cc00d6abe7a50bfc1023250edd1b9/preprints/The-Exponential-Time-Hypothesis-is-False-October-25-2026/paper.pdf

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