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Date: Fri, 24 May 2024 15:02:39 +0000
To: Andrew Poelstra <apoelstra@wpsoftware.net>
From: "'Rama Gan' via Bitcoin Development Mailing List" <bitcoindev@googlegroups.com>
Cc: "bitcoindev@googlegroups.com" <bitcoindev@googlegroups.com>
Subject: Re: [bitcoindev] Penlock, a paper-computer for secret-splitting BIP39
seed phrases
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> Are you sure? It seems that if two shares have the same value in a given
> position, the line through them should be constant, meaning that every
> other share will have the same constant value.
For the 2-of-M split, the secret is encoded as the difference between two
consecutive shares instead of being a point at a given index. If both the secret
and share A have a header `HEAD`, then share B will start with `====` (zeros)
and share C will be the additive inverse of `HEAD`.
The secret is the "slope" of the line; for the shares headers to be constant,
the solution would be to fill the corresponding spots with zeros on the secret.
So yes it _is_ possible, but then the 2-of-M and the K-of-M cases will behave
differently which could be a source of confusion. I guess it is the
cons of going for a composite scheme.
-- Rama Gan
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