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No. Yes, by choosing an authenticated encryption scheme with a known $\:I\hspace{.03 in}V\hspace{.04 in}||\hspace{.04 in}C\hspace{.04 in}||\hspace{.04 in}tag\:$ and $k_{\hspace{.02 in}0}$ and $k_1$ such that decrypting $\:I\hspace{.03 in}V\hspace{.04 in}||\hspace{.04 in}C\hspace{.04 in}||\hspace{.04 in}tag\:$ with $k_{\hspace{.02 in}0}$ and $k_1$ yields ...


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They both symmetrical encrypt their keys by itself in an algorithm (or aes with enough iterations) that it takes minutes, even hours to complete (this gives ek1). Then they will do the same thing again (encrypt ek1 by itself) (this gives ek2) and send ek2 to the other person when they both say they are done. If they don't align, both parties then send ek1 to ...


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If its possible that either of the party can get hold of the other's cipher text during transmission and decode it, then they could use that itself for determining if they can decode each-other's messages. Since either of the party can use a random-key thinking the other person will be using the real one, the effectiveness of this technique would be the ...


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I'm new here so I'm not sure about the best way to hold this discussion. So, I am adding a different answer to relate to why my proof sketch showing the impossibility of the problem in this question, versus Ricky's proof above that the protocol in this paper (page 16) is impossible. The answer is very connected to technical details to how you define and ...


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This cannot be done. It is provably impossible. In order to explain this in technical terms, what you are looking for is a FAIR protocol to compute equality of long random strings (I added the latter since it adds a constraint and so in theory could make it easier). In any case, if I had such a protocol, then I could toss a fair unbiased coin. Here is the ...



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