david wong

Hey! I'm David, a security consultant at Cryptography Services, the crypto team of NCC Group . This is my blog about cryptography and security and other related topics that I find interesting.

How to fairly compare two objects?

posted July 2015

Alex asked me if I knew a way of comparing two sets of data: two players want to compare their guesses on some game, without giving away their guesses. You could think of Zero-Knowledge protocols, but this is usually a one-way proof. This is actually the Socialist Millionaire Problem and it is solved by doing a multi-party computation of a function (a comparison function in our case) on two inputs (the two guesses in our case).

In cryptography, the socialist millionaire problem is one in which two millionaires want to determine if their wealth is equal without disclosing any information about their riches to each other. It is a variant of the Millionaire's Problem whereby two millionaires wish to compare their riches to determine who has the most wealth without disclosing any information about their riches to each other.

But how to make it fair? What if one party stops the protocol at one point, for example when he knows if the guesses are the same or not, so that the other party doesn't learn anything.

This seems like a difficult problem to solve, but an interesting problem that crypto should be able to solve.

Alex found this paper: A fair and efficient solution to the socialist millionaires’ problem, where they explain what they call a "fair" protocol. And the solution is quite elegant! I haven't read the whole thing but the idea is basically to compare bit by bit (I guess under the surface they must use garbled circuits) so that if one party stops the protocol early, he only has one bit of advantage over the other one.

Finally, the fairness of the fair version of our protocol is straightforward. Both Alice and Bob are unable to compute the result of the comparison before the beginning of step 4. Moreover, during the fourth step, Bob's advantage over Alice is at most one bit. So, if Bob decides to abort the protocol and tries to search the remaining bits by exhaustive research, Alice needs no more than twice as much time compared to Bob to compute the same result.

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