OK37 partition residue — Alexander’s ask

Ask. Partition the Östergård–Kaikkonen $[37,25]_2$ radius-3 seed $H=[I_{12}\mid M]$ (DM 178 Table 2; $M$ hex B16, not BI6) into $p\le 17$ blocks with a $(3,\ell)$ property so Construction QM$_4^3$ at $m=4$ could give $\ell_2(24,3)\le 607$ versus the table’s 618.

Floor. Cadical SAT shows the minimum $(3,0)$-partition size is $p=18$. Every $p\le 17$ is UNSAT. Explicit 18-block labels: partition_H_OK37_p18_ell0.txt (block sizes: 15 singletons + $7+7+8$).

Why $\ell\ge 1$ is impossible. The seed has no dependent triples, so syndrome $0$ has no weight-1/2/3 representation; every $(3,\ell)$ cover with $\ell\ge 1$ fails at the trivial partition already.

Why 607 fails. QM$_4^3$ at $m=4$ needs $p\le 17$ on this seed. The certified minimum is $p=18$, so the $n=607$ lift does not apply. No dent.

Replay.

cd problems/covering/compute
python3 -c "
from ok37_seed import ok37_columns, is_partition_3ell
labels=[int(x) for line in open('partition_H_OK37_p18_ell0.txt')
        if not line.startswith('#') for x in line.split()]
print(is_partition_3ell(ok37_columns(), labels, ell=0), len(set(labels)))
"
# → True 18

Campaign code and SAT residue: q12/. Summary JSON: ok37_p_le17_results.json.