Wicher Malten

Many thanks, this was very helpful! It seems to me that (2) would not work since ordinary triples are also needed for this protocol, but (1) sounds perfect. That option...

A `.reveal()` in `shamir` takes 2 rounds but in `mal-shamir` it takes 1 round, is that correct? ``` x = sint(5) start_timer(1) x.reveal() stop_timer(1) ```

Oh, that sounds like the "king node" approach of Damgård-Nielsen? I'm curious why it would be limited to the opening part of the protocol, and to [dishonest-majority](https://mp-spdz.readthedocs.io/en/v0.3.8/readme.html?highlight=star#finding-the-most-efficient-protocol) protocols - is...

Inside a large circuit, I now have an array `arr` of secret-shared `sint` values which are used in multiplications and additions later in the circuit. Depending on some random values...

Made some changes, but I still don't understand why for the following code the round complexity with `shamir` is larger than 7? (The line `y[i] = cint(0)` is probably incorrect,...

Is there a straightforward way around that?

Many thanks for those suggestions, but I'm not entirely sure I follow so here's a simplified version of the problem: ``` y = [sint.get_random() for i in range(size)] x_clear =...

That's unfortunate, but many thanks again for your effort. Following [this](https://github.com/rdragos/MP-SPDZ/blob/rabbit/Programs/Source/mcomp.mpc ) code, I'd now like to separately benchmark the online and offline phase, running a protocol `f(x,y,prep_material)` thousands or...

By which you mean a `MultiArray` or `Tensor`? For an f requiring 2 edabits, I just tried something like this: ``` edabit_values = sint.Tensor([n,2,1]) edabit_bits = sint.Tensor([n,2,bit_length]) def preprocessing(): edabit0,...

Not sure I follow - each instance of this f requires 2 edabits, so that's $2 \cdot bitlength$ sbits and 2 sints; are you saying I should produce $2 \cdot...