Two results, one of them a correction.
1. "(6,7) is just the constant 4096/0 pair, not a texcoord" -- wrong. It looked
like an artifact because a constant is trivially a pair of multiples of 4096.
But partitioned by i16[3]'s high byte it is a genuine fixed-point pair in its
own right:
i16[3] high byte ZERO (n = 89,341) divisor rate by slot pair
pair /1 /4 /16 /256
( 0, 1) 100.0% 11.5% 4.6% 3.3% baseline
( 2, 3) 100.0% 11.0% 4.5% 3.3% baseline
( 4, 5) 100.0% 92.9% 92.0% 91.8% the texcoords
( 6, 7) 100.0% 89.5% 87.3% 86.5% a SECOND fixed-point pair
If (6,7) were a constant it would score 100%, not 89.5%. The constant is the
common case WITHIN it, not the whole of it -- which is exactly the distinction
the constant-artifact lesson is about, and I applied it backwards last time.
So the ordinary element carries TWO fixed-point pairs, (4,5) and (6,7), with
(0,1) and (2,3) at baseline. What (6,7)'s role is remains open -- second UV
set, normal, or a zero-heavy fixed-point pair all fit.
2. The element LAYOUT differs between the three cases. Scoring all four slot
pairs within the non-zero high-byte group, no pair stands out:
i16[3] high byte NON-ZERO (n = 6,271)
pair /1 /4 /16 /256
( 0, 1) 100.0% 4.6% 3.5% 0.7%
( 2, 3) 100.0% 2.0% 0.6% 0.0%
( 4, 5) 100.0% 1.9% 0.3% 0.1%
( 6, 7) 100.0% 7.5% 5.7% 4.5%
Best is 7.5% against a 2-5% baseline. So for those 6,271 vertices the 16 bytes
are not laid out as position/texcoord/normal at all -- a different element
shape, selected by the high byte.
This also explains why the global slot-pair score looked ambiguous to begin with:
pooling the two layouts averages a 92% pair and a 7% pair into one number that
describes neither. Fourth time this session that partitioning produced the answer
rather than merely bounding it.
ReaperOfVeriod
922248a4ec