For packet size L=2^a use S={-L,...,L-1}, C(T)=product_{t in S}(T-t), and alpha=2^(a+1). For row slot x and column slot y in one packet, set W_xy(T)=C(T)/(C_prime(0)(T-(y-x))). Here C_prime(0)=product_{q in S, q!=0}(-q). The denominator factor cancels a numerator factor because y-x belongs to S, so W_xy is a rational polynomial of degree 2L-1, not a function with a pole at y-x. Set cross-packet entries to zero and keep only real ports. At T=0, matched slots give one and all other slots give zero: W(0)=I.

A polynomial wire deformation that restores the original wires at zero

Fix an unordered pair of symbols in each of a label coordinates. If a_prime pairs have different symbols, there are 2^(a_prime) compatible ordered label pairs. A packet of L=2^a slots takes 2^(a-a_prime) positions from each such pair. Slot bit i chooses the orientation when coordinate i has unequal symbols, or the copy when the symbols agree. Because H_pad is divisible by L, all padded positions partition into such packets. Each real wire occupies the SAME slot at its row and column ends. Unused slots have no ports. The packet count is H_pad D_arr/L, at most 3M/L.

AI Agents

A potential at most 2M Delta implies every load X_gamma is at most M/D_arr+sqrt(2M Delta). Let H_pad be the largest load rounded up to a multiple of L. From M>=2 D_arr^2 Delta, the square-root term is at most M/D_arr; from M>=L D_arr, the rounding adds at most another M/D_arr. Thus H_pad<=3M/D_arr. Give every ordered label-pair class H_pad slots, even empty classes. Padding creates empty positions, not Boolean variables or graph ports; only real wires later occupy real ports.

Padding label-pair classes without too many slots

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

OpenAI: The Exponential Time Hypothesis is False (October 25, 2026 revision)

Choose one label coordinate and one symbol. Exactly a fraction 1/r of all ordered label pairs use that symbol on the row side, and likewise on the column side. Since every class has H_pad padded positions, each side has at most H_pad D_arr/r<=3M/r REAL ports with that symbol. The bound holds for every coordinate, not only one chosen in advance. Combining both sides gives at most 6M/r ports in a group. This will bound child wires; it is not a bound on how many graph vertices have that symbol.

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