Explain the tie texcoord residual: i16[3] selects the encoding
The texcoords are 12-bit fixed on 82.5% of vertices, not 100%, and the residual had no explanation. Grouped by the vertex's i16[3] slot across 95,612 vertices on levels 0-2, the buckets separate completely: i16[3] n % / 4096 % / 256 negative 333 1.2% 1.2% == 0 5,851 29.2% 60.7% 1..63 28,884 92.7% 94.3% 64..255 54,606 92.3% 93.8% 256..511 8 0.0% 0.0% 512..1023 5,391 0.0% 0.0% >= 1024 539 0.2% 0.2% The residual is fully accounted for. 87.3% of vertices sit in 1..255 at about 92.5% fixedness, contributing 0.873 * 0.925 = 80.8 points; the 5,851 at i16[3] == 0 contribute 1.8 more. That is the 82.5% measured, to the point. The remaining ~6% are not 12-bit fixed at all. i16[3] >= 256, and negatives, score 0% at BOTH 4096 and 256 -- 6,271 vertices in an encoding this reading does not see. And i16[3] == 0 scores 29.2% at 4096 but 60.7% at 256, which is a third distribution rather than a weaker version of the same one. What this establishes: i16[3] is a per-vertex packed attribute whose value SELECTS HOW the texcoord pair is stored, rather than an independent field sitting beside the texcoords. 1-255 is the ordinary case, 0 is special, 256-and-up and negatives are a different encoding. What those three cases MEAN is still open -- texture index, normal index, flag or a combination -- but it is now a field whose role is known, which is more than "packed word". Worth noting the shape of the claim this replaces: "the texcoords are 12-bit fixed at i16[4]/i16[5], 73.6% of the time, unexplained". The field position was right; the 73.6% was a population average over three encodings. Partitioning by the one variable that separates them turns a mystery into a lookup -- rule 3a again, and the third time this session that partitioning a mystery produced the answer rather than merely bounding it.
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@@ -1718,3 +1718,44 @@ space for collision is now genuinely exhausted -- the bytes are available, they
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are verbatim, the grid is known, and three independent segmentation theories have
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failed on the whole population. Further probing without the loader's own logic is
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not going to converge.
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## Tie: `i16[3]` selects the texcoord encoding, which explains the residual
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The texcoords are 12-bit fixed on 82.5% of vertices, not 100%. Grouped by the
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vertex's `i16[3]` slot across **95,612 vertices on levels 0-2**, the buckets
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separate completely:
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i16[3] n % / 4096 % / 256
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negative 333 1.2% 1.2%
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== 0 5,851 29.2% 60.7%
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1..63 28,884 92.7% 94.3%
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64..255 54,606 92.3% 93.8%
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256..511 8 0.0% 0.0%
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512..1023 5,391 0.0% 0.0%
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>= 1024 539 0.2% 0.2%
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**The residual is fully accounted for.** 87.3% of vertices sit in `1..255` at
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about 92.5% fixedness, which contributes 0.873 x 0.925 = 80.8 points; the 5,851
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at `i16[3] == 0` contribute 1.8 more. That is the 82.5% measured, to the point.
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The remaining ~6% are **not 12-bit fixed at all**:
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- `i16[3] >= 256`, and negatives, score **0%** at both 4,096 and 256 -- 6,271
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vertices using an encoding this reading does not see;
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- `i16[3] == 0` scores 29.2% at 4,096 but **60.7% at 256**, which is a third
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distribution rather than a weaker version of the same one.
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**What this establishes:** `i16[3]` is a per-vertex packed attribute whose value
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*selects how the texcoord pair is stored*, rather than being an independent field
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sitting beside the texcoords. Values 1-255 are the ordinary case; 0 is special;
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256 and above, and negatives, are a different encoding entirely.
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**Not established:** what any of those three cases means. Whether `i16[3]` is a
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texture index, a normal index, a flag or a combination is still open -- but it is
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now a field whose *role* is known, which is more than "packed word".
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Note the shape of the earlier claim this replaces: "the texcoords are 12-bit
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fixed at `i16[4]`/`i16[5]`, 73.6% of the time, unexplained". The field position
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was right; the 73.6% was a population average over three encodings, which is
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exactly the pooled-number problem rule 3a describes -- partitioning by the one
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variable that separates them turns a mystery into a lookup.
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