Castle means something else, obviously.

]]>Split X => X1, X2. Generate A1, A2, B1, B2, C1, C2.

Combine unmatched halves: A1+B2 => AB, B1+C2 => BC, C1+A2 => CA.

Split AB => AB1, AB2. Generate D1, D2, E1, E2, F1, F2.

Repeat for BC, CA.

You now have 9 parts instead of only 3, and you need four of nine to recover X.

Problems I see:

1. Step 3 can’t just break AB the same way that A and B were broken, otherwise AB1 would be the same as A1 and AB2 would be the same as B2. A more complicated hex mixing would need to take place.

2. Also the BIP39 checksums would be broken by combining unmatched halves, unless more correction is done in step 2.

Yes, you’re correct: if P and Q are conjectures, and Q is stronger than P, then it’s entirely possible that Q has been resolved and P has not (because Q has been proved false!).

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