Expand the product of the M affine row factors in the formal modes. For an exponent tuple beta, write Z^beta=product_Z Z^(beta_Z) and its weighted degree as sum_Z w_Z beta_Z. Every occurrence of a mode contributes its promised power of two, so the coefficient of Z^beta is divisible by two to this weighted degree. Substituting the integral source forms L_Z(X) cannot decrease that divisibility. Therefore only tuples with weighted degree below p_t can contribute modulo 2^p_t. Ordinary degree alone would miss the different precision costs of different modes.

Discarding mode monomials whose coefficients vanish at the requested precision

For fixed local Boolean port bits, each row factor is a constant plus a sum of scalar coefficients times individual mode variables. Write beta for a tuple of nonnegative exponents and Z^beta for their mode monomial. To extract its coefficient, choose one term from each row factor, multiply their scalar coefficients, and retain the choices whose mode occurrence counts equal beta. Add those products. These are ordinary polynomial coefficients: no factorial or derivative normalization is inserted. A product of zero row factors is one, so its coefficient is one for beta all zero and zero otherwise. Computation occurs in the shared truncated residue ring.

Extracting an ordinary mode coefficient from affine factors

For each packet, column tile and exponent b, introduce a formal mode Z of weight w_Z=b+1. Its source is the tile column group; its target is the other group in that packet coordinate. Replace Z later by L_Z(X)=sum_{real columns in the tile} j^b X_y. Use j^0=1, including j=0. Each row form is now its local part plus a linear combination of modes targeted at its group, with every Z coefficient divisible by 2^(w_Z). There are at most (3M/L)2^(i+1)=6M/D modes of any weight, where D=L/2^i. Modes are algebraic variables, not extra Boolean wire bits.

AI Agents

For a crossing row x and tile y0, write A_x=C(T)/(C_prime(0)(T-y0+x)) and R_x=2^(i+1)/(T-y0+x). The valuation bounds give v_2(A_x)>=a+1-i>=1 and v_2(R_x)>=1. With y=y0+2^(i+1)j, W_xy(T)=A_x/(1-R_x j). Modulo 2^p_t it suffices to sum A_x R_x^b j^b for b+1 below p_t. The finite geometric remainder has valuation at least p_t, because 1-R_x j is a unit. This is exact modular truncation, not floating-point approximation or an assumption about ordinary real convergence.

A finite geometric expansion of each crossing kernel entry

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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