Covert adversaries have the property that they may deviate arbitrarily from the protocol specification in an attempt to cheat, but do not wish to be caught doing so. Does this model includes also the case in which adversaries do not cheat in case it does not maximize their utility function, e.g., they lost an escrow?
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$\begingroup$ I think you meant to say "unselfish" in the title. Is that so? $\endgroup$– PatriotFeb 6, 2021 at 2:34
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$\begingroup$ I meant selfish, or rational, that is the way these kind of adversaries are usually called in literature. $\endgroup$– LorenzoMar 30, 2021 at 21:42
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There is a close connection between covert adversaries and selfish ones. In particular, if you know the utility function of the selfish adversary, then you can compute the deterrent factor that you need for covert adversary. However, it is important to note that the model of covert adversaries considers corrupt and honest parties, where here corrupt would mean selfish. It does not cover the scenario of all parties being selfish (and so all parties may deviate from the specified protocol if it is in their interest).
In order to see how to model this scenario, a good starting point is Rationality and Adversarial Behavior in Multi-Party Computation by Lysyanskaya and Triandopoulos (from CRYPTO 2006).
answered May 10, 2020 at 5:43
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$\begingroup$ Thanks for the answer. So, in case I have parties who may deviate arbitrarily from the protocol but they do not because it is not rational (given their utility function and an eventual punishment mechanism of the protocol), they are still considered just malicious? $\endgroup$– LorenzoMay 10, 2020 at 5:58
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1$\begingroup$ It's not that easy. I added a reference to the answer for you to look at. $\endgroup$ May 10, 2020 at 7:47