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Suppose we have multiple parties ${P_i}$ each holding a secret share $(y_i,k_i)$ of the secret $(y,k)$. How can we calculate the inverse value $(y+k)^{-1}$ using multiparty computation such that these secret shares are never revealed?

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First, to simplify notation, note that computing $(y+k)^{-1}$ is equivalent to computing $z^{-1}$, where $z=y+k$, which can be locally computed by the parties. Furthermore, I assume that $z \in \mathbb{F}_p$ is a field element (e.g., integers mod $p$).

Now, to compute $z^{-1} \in \mathbb{F}_p$, each party can (locally) sample a random share of a value $r \in \mathbb{F}_p \setminus \{0\}$. Then, the parties reveal $zr$ to compute $(zr)^{-1} \in \mathbb{F}_p$, in the clear. Finally, the parties use their shares of $r$ to engage in a MPC protocol for multiplication to compute $(zr)^{-1}\cdot r = z^{-1} \in \mathbb{F}_p$. A similar protocol was described in Algorithm for computing modular inverse in MPC under a different context.

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  • $\begingroup$ You seem to have made quite a few assumptions about the secret sharing scheme $\endgroup$
    – integrator
    Commented Jun 6, 2023 at 11:28
  • $\begingroup$ My answer would probably have worked better as a comment. The main assumption is that @KanchanBisht is using an additive secret sharing scheme over a field, which appears reasonable based on how the question was formulated. $\endgroup$ Commented Jun 6, 2023 at 12:19
  • $\begingroup$ To clarify, I think your answer is good. I was just (clumsily) pointing out it would be better to explicitly mention you were assuming additivity (especially since you swept under the rug where that property is used: "which can be locally computed by the parties"). $\endgroup$
    – integrator
    Commented Jun 6, 2023 at 17:42

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