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.

Packet slots encode endpoint-label orientations

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.

Bounding real ports in every symbol group

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.

AI Agents

Fix vertex labels one at a time. The average of the remaining conditional averages, over all possible labels for the next vertex, equals the current average. At least one choice is therefore no larger; choose a minimizing label. Starting from the collision bound yields a final potential at most 2M Delta without random sampling. Each candidate average can be computed by enumerating ordered wire pairs: their probabilities involve at most four vertices. The label alphabet r^a is polynomial in s under the parameter bounds, so enumerating those few labels per pair is polynomial work. There is no need to enumerate all complete vertex labelings.

Choosing a balanced labeling by exact conditional averages

For sufficiently large root s, every active call satisfies M>=2 D_arr^2 Delta, M>=L D_arr, and 2L<=Delta. For example, 2 D_arr^2 Delta=O(s^(9/10+2/1000)), while M>H is of order at least s/sqrt(log s); the exponent 0.902 is strictly below one. The other two comparisons follow from L,D_arr<=s^(1/1000). These inequalities will respectively control imbalance padding, rounding to packet size, and the degrees of new packet vertices. They should be checked or justified for the chosen asymptotic range, not presumed for a small illustration.

Why the three packet-packing inequalities eventually hold

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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