Terms
In a secure sharing, we say every sharing as a part of the original message, and we can denote the $i$-th part by $P_i$.
Probability Model
To protect our data, we have to prove the secure sharing scheme is safe. A naive definition of sharing safe can be:
Given any message distribution, the likelihood of a part is independent of any other (K - 1) parts.
The above text can transferred to, for any given message distribution, and some parts $P_{i_j}$ of any message extracted from this distribution,
$$P(P_{i_0}|P_{i_1}, P_{i_2}, \ldots, P_{i_{K-1}}) = P(P_{i_0}).$$
A easiest way to achieve this is using a weaker secure shared transformation T, which is not part of message, and based on the transformation, we have, any (K - 1) parts are valid for any message. That is, for any message M, and any (K - 1) parts $P_0, P_1, \ldots, P_{K-1}$, we have a transformation $T$, such that $S_i(M|T) = P_i$, where $S_i$ is a $i$-th sharing part under transformation $T$. If we can get the $T$ from $K - 1$ parts, the secure is still not guaranteed.
But it's easy to prove that a revertible transformation can be secured shared in a sense that:
Given any $K - 1$ rows of a revertible transformation matrix, the space for the last row is isomorphic to $F^{K-1}$, where $F$ is the under field space.
That is, we lost one random dimension. If this is not acceptable, we can have a chain of transformations, and this chain will converge to the real random secure model.
Practice
In practice, only one transformation, plus a random accumulated random vector, give quite high entropy of every parts, which is verified by gzip.If you have different views of this secure sharing model, please kindly let me know, so that i'm not in a risk I do not know.
4 comments:
Great post, makes me think of erasure coding.
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