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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
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