Score:1

Can BDOZ and SPDZ implement secure 2-party computation?

aw flag

Considering that P1 has value x, P2 has value y and they want to compute x+y without telling the other what he has,So P1 and P2 secret-share their values to each other. Now P1 gets x1 and y1, P2 gets x2 and y2 and they locally compute xi+yi. If P1 want to compute x+y = (x1+y1)+(x2+y2), P2 need to send x2+y2 to P1, but P1 can get x2 by computing x-x1, thus P1 can get y2 and then get the value y, which is P2's private value. This is no doubt a violation of secure. So my question is, whether BDOZ and SPDZ can implement secure two-way computing or not?

Score:1
us flag

Considering that P1 has value x, P2 has value y and they want to compute x+y without telling the other what he has . . . thus P1 can . . . get the value y, which is P2's private value. This is no doubt a violation of secure.

If you know $x$ and I have a secret value $y$, and you somehow learn $x+y$, then of course you can deduce $y$. It doesn't matter how you learned $x+y$ --- you could learn it from a magic fairy or some cryptographic protocol. The fact that you learn my input is not the cryptography's fault.

When we formalize security of 2PC, we don't say "a corrupt party learns nothing about the other party's input." Instead, we say "a corrupt party learns nothing beyond what can be inferred from the output." In your example, $y$ can be inferred from $x+y$ (and $x$), so a 2PC protocol doesn't need to hide it. It is impossible to hide $y$ and still let the parties learn $x+y$.

To answer your main question, yes SPDZ and BDOZ work for any number of parties, even 2 parties.

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