Suppose Alcie,Bob,and Charlie each has a secret value $a,b,c$ respectively. they want to compute a value $r\cdot (a+b+c)$ together. $r$ is a random value, and all parties know nothing about it.
I tried to use OLE to address this problem,but it works only in two party setting,i.e. $r\cdot (a+b)$。So is there any possible solutions?


1 Answer 1


So is there any possible solutions?

How about: have the party jointly pick a random $d$ value, and we are done.

In this case, $r = d / (a + b + c)$; as long as $a+b+c \ne 0$, it will always exist (assuming that we're working in a field).

No one without knowledge of $a, b, c, d$ can know the value of $r$, which I believe is what was requested.

Of course, this doesn't work if you're in a ring/field with a nontrivial probability of a random element being noninvertible (e.g. $\mathbb{Z}/{2^n}$ or $GF(2)$)

  • $\begingroup$ thanks for your answer! Another questioin is if $a,b,c$ was extended to vectors $\vec{a},\vec{b},\vec{c}$,they need to compute $r\cdot (\vec{a}+\vec{b}+\vec{c})$。it seems doesn't work $\endgroup$
    – Rui T.
    Apr 16 at 13:49
  • $\begingroup$ @RuiT. if that's the scenario you're interested in, you probably should mention it in your question... $\endgroup$
    – poncho
    Apr 16 at 14:11
  • $\begingroup$ sorry about that!!i thought the former scenario could be easily extended to the latter ,so i simplify the question incorrectly。sorry about wasting your time! $\endgroup$
    – Rui T.
    Apr 16 at 14:44

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