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I have been trying to understand the interplay between robustness and privacy of an SMC protocol. It is easy to come up with a protocol that is private but not robust.

Can someone provide an example of a secure multiparty computation protocol that is not private but is robust, under some threshold $t$ corrupt parties?

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    $\begingroup$ What does robustness of an SMC protocol mean? ​ ​ $\endgroup$
    – user991
    Commented Apr 9, 2016 at 15:25
  • $\begingroup$ "A protocol is robust there exists an efficient simulator $S$ such that for every adversary attacking the protocol, $S$ can efficiently compute an allowed influence with the same effect" - Book titled: Secure Multiparty Computation and Secret Sharing $\endgroup$
    – sun
    Commented Apr 9, 2016 at 16:00

1 Answer 1

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Sure.


Parties: ​ more than 1
Input: ​ each party has a secret input
Output: ​ empty string

"Protocol": ​ Each party broadcasts their secret input then outputs the empty string.


That is robust for arbitrary thresholds, but is
not private against even a single semi-honest adversary.

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  • $\begingroup$ Just an interesting observation from your example is that any protocol in the semi-honest model is robust, and hence we can come up with many protocols which are robust but not private. But can we design a protocol that is robust in the malicious model, but not private? Moreover, is it right for me to think that we talk about robustness only in the presence of malicious adversaries? $\endgroup$
    – sun
    Commented Apr 9, 2016 at 16:27
  • $\begingroup$ Yes, see my answer. ​ Not quite. ​ ​ ​ ​ $\endgroup$
    – user991
    Commented Apr 9, 2016 at 16:36

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