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The formula you are looking for is Lagrange Basis Polynomials. Essentially, each share consists of two values, an x coordinate and an y coordinate. The x coordinate might, depending on your specific needs, be implicitly determined by context, such as a preexisting identifier for the entity holding the share. The only requirement is that it is non-zero and ...

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I don't think the approach you sketched helps very much. If the server is compromised, the attacker can pretty easily modify the server-side software to log and record all the cryptographic keys, and then you haven't gained anything. Therefore, I don't think the approach you sketch is likely to be a great way to spend your limited software development ...

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Suppose $s_0, s_1, s_2, \ldots, s_{k-1}$ are elements from the finite field you are working in, where $s_0$ is the secret to be shared, and the $s_i, i > 0$ are randomly chosen nonzero elements of the field. Then, the polynomial used to construct the shares is $$S(x) = s_0 + s_1x + s_2x^2+ \cdots + s_{k-1}x^{k-1}$$ and the shares themselves are $y_i = ... 1 The easy way to reconstruct the secret is using Neville's algorithm. Basically, let the array y contain n shares (assumed to be represented as the type gf_t here), let the array x contain the points at which the polynomial was evaluated to generate those shares, and let sub(), mul() and inv() be functions/macros that perform the appropriate finite field ... 1 All too often I describe some process saying that "he" did something to "his" message, and it makes sense in my own head, but no one else can figure out which of the several people involved that those pronouns refer to. Or worse -- sometimes I say that "the message is encrypted", but I don't say who does it and with which key. That's why all the good ... 1 The best answer is to use verifiable secret sharing (VSS), as I describe here: http://crypto.stackexchange.com/a/6618/351 VSS gives the best parameters and best solution to this problem. If you have a$k$-out-of-$n$secret sharing scheme, VSS can detect any cheater and enable you to reconstruct the secret as long as you have at least$k\$ good shares (even ...

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