Is $v=\ (r_i)^{-1}\cdot z $, a uniformly random value of a field?
Solution 1:
No, since the restriction $r_i > \frac{q}{2}$ means $v$ can only assume $2p + 1 - (p + 1) = p$ different values.
Is $v=\ (r_i)^{-1}\cdot z $, a uniformly random value of a field?
Solution 1:
No, since the restriction $r_i > \frac{q}{2}$ means $v$ can only assume $2p + 1 - (p + 1) = p$ different values.