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This is actually a fairly trivial case of secret sharing: In the first case, we select a random value $R$, we give $X$ the value $R \oplus M$, and we give both Y and Z the value $R$. Obviously, $X$ alone, nor $Y$ and $Z$ together cannot reconstruct $R$. In the second case, we select random values $R_1$, $R_2$ and $R_3$, we give: $R_1 \oplus M$ to $X$ ...


1

As far as I know, you can not do multiplication with (m,m) shamir secret sharing. The typical method to do multiplication on shamir secret shares increases the degree of the sharing polynomial, which is why the parties run an additional protocol to reduce the degree. That is why the degree of the sharing polynomial must be less than $m/2$ if there are $m$ ...


0

No one is stopping you from running the polynomial setup yourself. Knowledge of the polynomial is equivalent to knowing the actual secret. And knowledge of the complete polynomial is required to evaluate the polynomial at any point. But this doesn't have to be a third party; it might be the case that your particular context uses a third party for this task, ...


6

We simply have to trust this party because this scheme requires a trusted setup. We can use verifiable secret sharing, that allows the parties to check whether the shares they have obtained are consistent. If we do not want to have a trusted dealer who sets up the shares at all, we have to switch to dealer-free secret sharing schemes. See for instance, ...


0

It depends on the situation. The third party can be a computer in a datacenter with armed guards at the door. Or it can be an employee who doesn't want to screw up and be fired. Or other things. Of course, none of those solutions are 100% foolproof, and it would be better to have a protocol which doesn't use a trusted third party, but if no such protocol is ...



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