Score:0

Faking Pedersen Commitment

in flag

Today, I found a website for Pedersen commitment scheme; however, the generators g and h are not independent and therefore a prover can open a commitment c into many ways. I computed the commitment c for a message m and a randomness r (assuming that I know s):

c  = g^m * h^r 

   = g^m * (g^s)^r 

   = g^m * (g^(s * r))

   = g^(m + s * r) 

Say, I have committed to the message m, randomness r, and the commitment is c (c = g ^ (m + s * r). Now I want to fake it, i.e., open it for some other message m', not equal to m, and therefore I need to calculate a new randomness r': r' = (m - m' + s * r) * s^(-1). I took the original python code, posted on the website, and did a minor modification, returning s. Moreover, I wrote another function fake_message to compute an arbitrary message for opening it to the same commitment c, but when I am running my modified code, the last line,

r2 = print(vv.open(param, c, m2, r2))

is returning false. My question is: what is wrong with my thinking, and the Python implementation? (I am using pycryptodome 3.10.1)

from Crypto import Random
from Crypto.Util import number
import sys

def generate(param):
      p = param[0]
      q = param[1]
      g = param[2]
      h = param[3]
      s = param[4]
      return p,q,g,h,s

class verifier:
    def setup(self, security):
        p = number.getPrime(2 * security, Random.new().read)
        q = 2*p + 1 # hmm, no testing if q is prime or not.

        g = number.getRandomRange(1, q-1)
        s = number.getRandomRange(1, q-1)
        print("Secret value:\t",s)
        h = pow(g,s,q)
        
        param = (p,q,g,h,s)
        print("p=",p)
        print("q=",q)
        print("g=",g)
        print("h=",h)
        print("s=",s)

        return param

    def open(self, param, c, m, r):
        p,q,g,h,s = generate(param)
        res = (pow(g,m,q) * pow(h,r,q)) % q

        return (c == res)

        
class prover: 
    def commit(self, param, m):
        p,q,g,h,s= generate(param)
        
        r = number.getRandomRange(1, q-1)
        c = (pow(g,m,q) * pow(h,r,q)) % q
        return c, r

    # I am going to open it to a random arbitrary message m2
    def fake_message(self, param, c, m1, r1):
        p,q,g,h,s = generate(param)

        #get a random message
        m2 = number.getRandomRange(1, q-1)
        r2 = ((m1 - m2 + s * r1) * number.inverse(s, q))%q
        return (m2, r2)




security = 80
m = 2

vv = verifier()
pp = prover()

param = vv.setup(security)

c, r = pp.commit(param, m)
print(vv.open(param, c, m, r))
m2, r2 = pp.fake_message(param, c, m, r)
print(vv.open(param, c, m2, r2))
Daniel S avatar
ru flag
The problem is the expression `number.inverse(s,q)` and also `%q`. You need to compute the inverse of $s$ modulo the order of the multiplicative group of $q$ not modulo $q$ (likewise the final reduction should not be modulo $q$). If $q$ is prime the order is $q-1$, but as you correctly note this code does not check whether $q$ is prime. To fix add a check that $q$ is prime and then change to `((m1 - m2 + s * r1) * number.inverse(s, q-1))%(q-1)`
in flag
Thanks @DanielS.
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