Blum-Goldwasser Probalistic Encryption

With public key encryption, Alice could have two possible messages (a ‘0’ or a ‘1’) that she sends to Bob. If Eve knows the possible…

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Blum-Goldwasser Probabilistic Encryption

With public key encryption, Alice could have two possible messages (a ‘0’ or a ‘1’) that she sends to Bob. If Eve knows the possible messages (a ‘0’ or a ‘1’), she will then cipher each one with Bob’s public key and then matches the results against the cipher message that Alice sends. Eve can thus determine what Alice has sent to Bob. In order to overcome this the Blum-Goldwasser method is a public key algorithm that uses a probabilistic public-key encryption scheme [here]:

The encryption method uses the Blum-Blum-Shub (BBS) technique to generate the keystream [here]. Initially we create two prime numbers (p and q), and then calculate N:

N = pq

The public key is N, and the private key is p and q, and where p and q:

p (mod 4) = 3

q (mod 4) = 3

For example we can select p= 239, q= 179, as both will give us 3 when we do a (mod 4) operation:

>>> p=239
>>> q=179
>>> p%4
3
>>> q%4
3

The basic method, as defined by Wikipedia, is:

The code is as follows [here]:

import numpy as np
import binascii
import Crypto.Util.number
from Crypto import Random
import sys
import random
def xgcd(a, b):
"""return (g, x, y) such that a*x + b*y = g = gcd(a, b)"""
x0, x1, y0, y1 = 0, 1, 1, 0
while a != 0:
(q, a), b = divmod(b, a), a
y0, y1 = y1, y0 - q * y1
x0, x1 = x1, x0 - q * x1
return b, x0, y0
# Method from https://stackoverflow.com/questions/7396849/convert-binary-to-ascii-and-vice-versa
def to_bits(text, encoding='utf-8', errors='surrogatepass'):
bits = bin(int(binascii.hexlify(text.encode(encoding, errors)), 16))[2:]
return bits.zfill(8 * ((len(bits) + 7) // 8))
def from_bits(bits, encoding='utf-8', errors='surrogatepass'):
n = int(bits, 2)
return int2bytes(n).decode(encoding, errors)
def int2bytes(i):
hex_string = '%x' % i
n = len(hex_string)
return binascii.unhexlify(hex_string.zfill(n + (n & 1)))
# Derived from https://github.com/coenvalk/blum-goldwasser-probabilistic-encryption/blob/master/blumgoldwasser.py
def BGW_enc(p, q, x, m):
n = p * q

    # assert p%4 == 3 and q%4 == 4
    k = int(np.log2(n))
h = int(np.log2(k))
    t = len(m) / h
    xi = x
c = ''
for i in range(t):
mi = m[i*h:(i + 1)*h]
xi = (xi ** 2) % n
xi_bin = bin(xi)
pi = xi_bin[-h:]
        mi_int = int(mi, 2)
pi_int = int(pi, 2)
        ci = pi_int ^ mi_int
ci_bin = format(ci, '0' + str(h) + 'b')
c += ci_bin
    xt = (xi ** 2) % n
return c, xt

def BGW_dec(p, q, a, b, xt, c):
assert a*p + b*q == 1
n=p*q
k = int(np.log2(n))
h = int(np.log2(k))
    t = len(c) / h
    d1 = (((p + 1) / 4)**(t + 1)) % (p - 1)
d2 = (((q + 1) / 4)**(t + 1)) % (q - 1)
    u = pow(xt,d1,p)
v = pow(xt,d2, q)
    x0 = (v*a*p + u*b*q) % n
    xi = x0
m = ''
for i in range(t):
        ci = c[i*h:(i + 1)*h]
xi = (xi**2) % n
xi_bin = bin(xi)
pi = xi_bin[-h:]
ci_int = int(ci, 2)
pi_int = int(pi, 2)
        mi = pi_int ^ ci_int
mi_bin = format(mi, '0' + str(h) + 'b')
m += mi_bin
    return m
p = 499
q = 547
bits=10
msg='Hello'
if (len(sys.argv)>1):
bits=int(sys.argv[1])
if (len(sys.argv)>2):
msg=(sys.argv[2])
while True:
p=Crypto.Util.number.getPrime(bits, randfunc=Crypto.Random.get_random_bytes)
if (p%4==3): break
while True:
q=Crypto.Util.number.getPrime(bits, randfunc=Crypto.Random.get_random_bytes)
if (q%4==3): break
m=to_bits(msg)
a=1
b=1
_,a,b=xgcd(p,q)
r= random.getrandbits(bits)
x0 = (a*p*r + b*q+r) % (p*q)

c, xt = BGW_enc(p, q, x0, m)
print ("Message: %s" % msg)
print (" %s" % m)
print ("\nNo of bits in prime is %d" % bits)
print ("p= %d" % p)
print ("q= %d" % q)
print ("a= %d" % a)
print ("b= %d" % b)
print ("r= %d" % r)
print ("x0= %d" % x0)
print ("ap+bq: %d" % (a*p+b*q))
print "\nCiphertext:", c
d = BGW_dec(p, q, a, b, xt, c)

print ("Decrypted: %s" % from_bits(d))

A sample run is [here]:

Message: Hello
0100100001100101011011000110110001101111
No of bits in prime is 16
p= 65447
q= 34231
a= -7482
b= 14305
r= 37115
x0= 1908144401
ap+bq: 1
Ciphertext: 0011100101011111010000110011100011010010
Decrypted: Hello