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Suppose that we have a multi-secret sharing scheme as it is described in the literature

Let there be $I$ agents and say that $S$ is the space of the (uniform) random variables $s=(s_1,s_2,\cdots,s_I)\in S$ such that the share $s_1$ is known to $P_1$, $s_2$ is known to $P_2$ and so on. Could someone propose an appropriate multi secret sharing scheme? Every agent $i$ wants to share $s_i$ in such a way that it is not easy computable from a small group of players (I do not know if an $(|I|-1,|I|-1)$-threshold scheme can be applied)

Could anyone provide explicitly a proof? Maybe it seems easy, but I am a bit confused where to start or how to do the maths. I would appreciate it if it is convenient for him/her who could give a proof to use $+$ or $\otimes$ and $mod$ schemes from group theory instead of polynomial explanation because it seems more simple for my understanding.

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  • $\begingroup$ @moderators I have totally re-edited my question, is there a choice where I can re-post it? $\endgroup$
    – Hunger Learn
    Jan 10, 2022 at 8:59
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    $\begingroup$ @JAAAY I almost totally re-edited my question $\endgroup$
    – Hunger Learn
    Jan 10, 2022 at 9:10
  • $\begingroup$ I would recommend you to read arxiv.org/pdf/1806.07197.pdf but an answer has to be given here also. $\endgroup$
    – João Víctor Melo
    Jan 10, 2022 at 13:17
  • $\begingroup$ @JoãoVíctorMelo seriously, the way that they write, all of those who are doing research in cryptography is quite non-efficient. For example, we know in general that $x\underbrace{\to}_{f} y$, which means that $f(x)=y$, but instead they are using an inverse arrow, $x_i\rightarrow inv(u_i)$ totally a mess. Especially when instead of inv you have a function like $VSS_{put}(s)=...$, well this function $VSS_{put}$ must have some properties what are they? None clarifies them. And the latter, they all say that $P_i$ shares $s_i$ among $N-\{i\}$ or $(n-1)$ agents.... $\endgroup$
    – Hunger Learn
    Jan 10, 2022 at 15:07
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    $\begingroup$ Yes, in science sometimes to show some things you have to hide others. $\endgroup$
    – João Víctor Melo
    Jan 10, 2022 at 15:39

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