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How is sum of two share of (2,n) shamir scheme equal to the secret?

I quote the specific part of the paper you are mentioning : Picks $X, r$ randomly Splits $X$ into $n$ numbers using $(2, n)$-secret sharing scheme [19], $\mathcal{X}_{s}=\left\{x_{0}, x_{1}, x_{2}, \...
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How is sum of two share of (2,n) shamir scheme equal to the secret?

I can't answer the question in the text of the this question, but have an answer to the question in the title. It is not the sum of two shares but instead a weighted sum of two shares that is equal ...
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