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Crafting station queue: Jobs wait for every station · case 04

Parallel stations behave like one serial station.

Member previewVariant 4 · 3 implementations · 8 checks per implementation

Case contract

jobs = [[name, duration], ...] in queue order. Effective duration = max(1, floor(duration*(100-speed)/100)). Each job starts on the station that frees earliest (lowest index on ties) at that station free time. Return names of jobs finished by now (finish <= now) ordered by finish time then queue position.

Why this case matters

Game economies leak or destroy currency when one crafting or pricing rule is off by one boundary, rounding stage or state update; the defect is observable in exact integer outcomes.

One recorded failure

Sample boundary fixture

This sample comes from the broken implementation of a controlled reproducer.

Boundary fixtureActualExpectedOutcome
two stations tie #1["a"]["a", "b", "c"]Failed

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