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The original question was: Alice knows: $a,b$ and $x$ such that $a^{(x\cdot x)} = b$ Bob knows: $a,b$ and DrLecter referenced this paper (fixed the link), which covers the question. Now, the question was changed to Alice knows: $b$ and $x$ such that $x^x=b$; Bob knows: $b$. The given structure was: ... multiplicative group $G$ ...


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Below is one possibility, but for a large set of values not a really efficient one as the work and the size is linear in the number of values. Say we are working in a cyclic multiplicative group $G$ of prime order $p$ with generators $g$ and $h$ (such that the discrete logarithm between $g$ and $h$ is unknown to the users) and I assume that the values on ...


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Your approach makes getting information other than count of cards in possession of each player at least as hard as breaking the PRFness of HMAC. To make it information-theoretically impossible "for all the ... each player", $\;\;$ if different card_values have different lengths then use $\;\;$ SHA256(commonly_agreed_public_salt:card_value) $\;\;$ ...



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