1
$\begingroup$

I am reading Fast and Secure Three-party Computation: The Garbled Circuit Approach And in the introduction section it said that:

the round complexity of these protocols is proportional to the circuit-depth of the computation being performed

What does this exactly means? Why does Yao's garbled circuit does not have this kind of issue?Thanks in advance!

$\endgroup$

1 Answer 1

1
$\begingroup$

When running a secret-sharing based MPC protocol, the parties must interact for each multiplication gate of the circuit. If two multiplication gates are independent of one another, then we can combine the communication for both gates into a single message. But if one multiplication gate feeds into a second multiplication gate then the communication for those gates must come in separate, sequential rounds. So the total number of communication rounds has to be at least the multiplicative depth of the circuit (multiplicative depth = maximum number of multiplication gates along any path from input wire to output wire).

In a garbled circuit MPC protocol, one party (garbler) prepares some cryptographic information for every gate in the circuit and sends it to the other party (evaluator) in one big message. The evaluator can process this information gate by gate, but it doesn't require any interaction to process it. The garbler has already prepared all the information the evaluator will ever need, for the entire circuit. So no matter how deep the circuit is, the garbler just sends a single message.

I don't know if I can answer the question of "why" these protocols are like this. It's just the way these protocols are.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.