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treadcore · check-in report CI-020

Each Switch Alone, Without Running Out of Tanks

Check-in report CI-020. Paul Richeson, treadcore.org; Claude Code (Claude Opus 5.5) as the pair. 20 October 2026. Companion records: journal 0072; LAB-REPORT-BRAIN.md section 44; CI-001, CI-016.

Source: documentation/checkins/CI-020-2026-10-20-without-running-out.md, built by orrery's tools/report/collect.py.

Abstract

Backlog 29 asked whether the census, with half the seat-time dead, hid the parts that pay late. Every capability off by default was measured alone, at unlimited respawns and at 3, under today's defaults, 12 seeds each. The base cycle and the escort read inside the noise both ways: not confirmed. Three of 24 readings fell outside two errors, where about one is expected by chance. Only smooth, negative in all four readings on record, stands as a finding. bridge_jobs (+1.68) and duck (-2.26) are leads for confirmation. On Paul's choice, smooth becomes a weight with an insignificant default.

Index terms: multiple comparisons, attrition, ablation, census design.

I. Introduction

A. Intent

Backlog 29, from CI-003: a census in which the late parts can act.

B. The prompts

commit. push. continue backlog. i love your analysis

smooth sounds like we should remove it.

ok. maybe not remove it. give it an insignificant coefficient

is there an improved and faster way for you to test ?

II. Decision Log

#CauseDecisionNot taken, and why
D1The defaults changed since LR 24Both passes under today's defaultsReading unlimited against LR 24 alone: two changes at once
D2RESPAWNS_UNLIMITED is 255255The roadmap's 254
D3Paul: remove smoothShown it bundles six behaviours; offered split or removalRemoving unasked-about parts that may be neutral
D4Paul: "an insignificant coefficient"Backlog 40, w_smooth 0.05; blend the numeric parts, threshold the yes-or-no ones at 0.5A switch kept off: Paul's rule makes an either/or a scalar
D5Paul: a faster way to testSix ways, ranked; the frozen binary first (backlog 41)More seeds alone: noise, not CPU, is the cost
D63 of 24 outside two errorsOnly the consistent one (smooth) a finding; two leads to 48 fresh seeds (backlog 39)Acting on single-pass readings
D7Brain edits wait for the runsNo source changed during either pass

III. Methods

RESPAWNS=255 JOBS=3 make switches, then RESPAWNS=3 JOBS=3 make switches: every capability off by default alone as the trial, the defaults (the landfall's three on) for the baselines; team_rotate, bridge_jobs and posture with board on for every team; 12 seeds of 900 s on naval.map, the fair census.

IV. Results

See LR 44's table. Outside two errors: smooth -3.32 ± 1.47 (unlimited), bridge_jobs +1.68 ± 0.70 (3), duck -2.26 ± 1.05 (3). Never acting: board alone, island_site.

V. Discussion

  1. The attrition hypothesis is not confirmed. Given the full 900 s without running out, the late parts still read nothing.
  2. Read a table of tests as a table. At two errors, one in twenty is outside by chance. Consistency across independent passes, or a confirmation on fresh seeds, separates a finding from a draw.
  3. Faster testing (Paul's question): noise costs more than CPU. Freeze the binary so work goes on during a census; measure each part on its own counts; test single behaviours in scenarios (backlog 24); more cores; a cheaper match (finding L2); sequential rules set in advance.

Next: backlog 41, then 40, then 39.