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Feldman Verifiable secret sharing verify

xk flag

I recently started to learn about Shamir secret sharing and Feldman's VSS Scheme. I know the concepts But I can't figure out how it works. mostly because many of modulates are with "p" and some of them are with "q". I can't find a good sample that helps me understand it better.

here's the example I am working with:

assume that I want to work with $q = 59$ and my prime for Shamir secret sharing, and $secret = 22$. I have a 2 of 3 threshold SSS. and my polynomial is : $f(x) = 25x+22$. also assume that the index of participant is 12.

$f(12) = 25*12+22 = 322 = 27 \pmod{59}$

until now, this was just Shamir secret sharing. now I need to add versifications. I use $p = 709$ and we know that : $q | p - 1$. and $g = 37$ is a generator of q

now commitments:

$c_0 = g^{secret} = 37^{22} = 106 \pmod{709}$

$c_1 = g^{a_1} = 37^{25} = 670 \pmod{709}$

now, participant 1 wants to verify his share:

left side : $g^{f(12)} = 37 ^ {27} = 493 \pmod{709}$

right side : $c_0*c_1^x = 106 * 670^{12} = 196 \pmod{709}$

why are they not equal? am I doing modulates wrong?


I added reference to Wikipedia page of Feldman VSS scheme. should I add anymore explain about Shamir and Feldman scheme to clearify?

Daniel S avatar
ru flag
The problem is that 37 is not of order $q$ in the multiplicative group mod 709. You can check that $37^{59}\equiv 522\pmod{709}$. Try using $g=20$ instead.
kodlu avatar
sa flag
while @DanielS has given a nice hint, can you *please* clearly define what the Feldman verification is, so this question is useful to others?
user109261 avatar
xk flag
thanks for your reply @DanielS. how can I find that generator? generator of q modulus p when p and q are both primes and q | p - 1?
Daniel S avatar
ru flag
You pick a number and raise it to the power $(p-1)/q$ modulo $p$. If the answer is not 1 then the answer will work as a generator of a subgroup of order $q$. In this case we might pick 5 and then $5^{(p-1)/q}\equiv 20\pmod{709}$.
user109261 avatar
xk flag
thank you. that was very helpful. @DanielS
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