They answered ask #2 -- the gap is deterministic, not a load -- and held me to
re-stating the choice against four measurements (0, 6, 4, 6) rather than the three
it was made on. That is my own standard for check-all's stale allowance turned
around.
The arithmetic moved against 0: it costs 2.0x the best fit on four measurements
against 1.7x on three, and is now the minority outcome, wrong on 3 of 4 where a
uniform 6 is wrong on 2 of 4.
Still 0, because the reason was never the fit -- it adds no black the game does
not have, and is measured-correct for a real transition rather than for an average
of them. Picking 6 because it appears twice is choosing a mode from four samples
when their finding is that a rule exists and nobody has found it.
But one leg is gone and is not being kept quietly: part of the case for 0 was that
the quantity might be machine-dependent and unauthorable. It is not. That removes
an excuse without supplying a value.
Added a tripwire: if the gap is non-zero in most transitions and no rule emerges,
omitting a real quantity stops being caution and becomes a different invention.
The port is wrong by 4-6 units on three of four known transitions and the why now
says so.
Also records as settled: the outgoing ramp is the declared final ramp, my {8,10,10}
against their measured multiset, two independent routes.
Co-Authored-By: Claude Opus 5 (1M context) <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_01N7FiFFFwbvG2uxdcEh8HyF
82ae10a002