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Two-stone handicap uses the wrong diagonal · case 03

The first two stones go to upper-left and lower-right.

Member previewVariant 3 · 3 implementations · 8 checks per implementation

Case contract

Input [size, count]. Star line e = 2 (0-based) below 13x13 else 3; f = size-1-e; mid = size//2. Counts outside 2..9 give []. Even boards allow at most 4. Order: [e,f],[f,e],[f,f],[e,e]; counts 6-9 add [mid,e],[mid,f]; 8-9 add [e,mid],[f,mid]; odd counts from 5 add the centre last.

Why this case matters

Go servers and scoring tools compute this value automatically; a wrong answer changes a game result.

One recorded failure

Sample boundary fixture

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

Boundary fixtureActualExpectedOutcome
handicap placement case 0[[2, 2], [4, 4], [2, 4], [4, 2], [3, 2], [3, 4], [2, 3], [4, 3]][[2, 4], [4, 2], [4, 4], [2, 2], [3, 2], [3, 4], [2, 3], [4, 3]]Failed

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