With the payload confirmed verbatim in RAM, the disc bytes can be read directly.
The grid dimensions are the first two u16 of the payload: level 0 is 11 x 33 =
363 cells, level 1 24 x 16 = 384, level 2 1 x 53 = 53. Level 2 being one cell wide
is worth noting -- the grids are not all similar in shape.
Two readings disproved, both by measuring the whole payload instead of a sample.
An ascending u32 offset table. A probe from payload +1024 reads f4ae0000 40af0000
c4b2000020b20000, which is three ascending values before the fourth goes down
(0xb2c4 to 0xb220). Read as a table, that was a three-value impression. Scanning
all three levels for maximal ascending u32 runs of 8 or more entries finds zero,
on every level.
A global alternating-zero stride. The head of level 0 shows an obvious (value, 0)
pattern from about i16[37]:
[32..] [1808, 1872, 2198, 0, 34, 86, 25848, 1, 25824, 1,
-13028, 0, -12284, 0, -12228, 0, -12716, 0, ...]
Measured over the whole payload, both alignments score far below what that sample
suggests: even-slot aligned 18% second-half zero, odd-slot aligned 13%. The
sample again.
What the global numbers do say is that the payload is heavily zero-bearing -- 18%
against a chance rate of about 0.0015% -- which is what a sparse or variable-length
encoding looks like, and consistent with the variable-length reading established
earlier. The 18-against-13 split is a real alignment preference but a weak one.
The cell layout is still undecoded. The useful position now is that the bytes can
be read as they are and that they are sparse and variable-length rather than
fixed-stride, so the next attempt should look for a terminator-delimited walk
through the sparse region rather than for a record size.
ReaperOfVeriod
338af9b7bd