# Variations on Shamir secret sharing

Let's say you have six shares (let's call them A, B, C, D, E and F),
where a minimum of three shares are required to reconstitute the
secret. Is there a way to further divide the shares, such that for each
share, there are ten possibilities for that share? For example, ten As,
ten Bs, etc. Each is interchangeable with the other shares in it's own
letter, but you couldn't reconstitute the secret from multiple shares
within a letter.

My second question is if it's possible to require one of the shares.
Using the above example, let's say you need A, but any two other shares
would work.

Lastly, a combination of the above approaches, where A is required and
B-F are divided into multiple possible shares.

Any insight into these issues would be great. Anyone know of this type
of stuff being implemented or discussed in a paper? If someone can
suggest irc channels and the like where this type of stuff is
discussed, I'd appreciate that too.

Thanks,

Philip

.

## Relevant Pages

• Re: Variations on Shamir secret sharing
... Is there a way to further divide the shares, such that for each share, there are ten possibilities for that share? ... Each is interchangeable with the other shares in it's own letter, but you couldn't reconstitute the secret from multiple shares within a letter. ...
(sci.crypt)
• Re: Cantor Confusion
... Suppose I have an edge that is divided in countably many shares. ... We need not divide anything by infinite numbers, ... axiom holds and with geometry where it does not hold. ...
(sci.math)
• Re: Cantor Confusion
... > An edge is considered to be a set of shares. ... If you divide n edges into m equal shares each, ... We need not divide anything by infinite numbers, ... Every level of the tree has a finite number n. ...
(sci.math)