Equality test

This functionality, denoted as $[a=b]$ returns a shared bit corresponding to the equality property. We consider only the particular case where the RHS is zero, because general equality is implemented trivially as $[a-b=0]$.

To obtain $[x=0]$ we do the following:

1. $[r \in_R \mathbb{F}_p]_B$
2. Compute $[r] = \underset{i=0}{\overset{l-1}\sum} 2^i[r_i]$
3. Open $[x+r]$, denote as $b_i$ its bit decomposition.
4. Set $[q_i] = b_i ? [r_i] : (1-[r_i])$
5. $[x=0] = \underset{i=0}{\overset{l-1}\vee}[q_i]$

It is important to note that because $r$ is uniformly distributed in $\mathbb{F}_p$, revealing $r+x$ doesn't leak any information about $x$.