Attachment 10058 Details for Bug 22693 – python 2 and 3 compatible script

python 2 and 3 compatible script

attack-pycrypto.py (text/plain), 1.46 KB, created by Philippe Makowski on 2018-03-19 11:11:39 CET

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Creator: Philippe Makowski

Created: 2018-03-19 11:11:39 CET

Size: 1.46 KB

patch

obsolete

>from Crypto.PublicKey import ElGamal
>from Crypto import Random
>import Crypto.Random.random
>import sys
>
>PYTHON_MAJOR_VER = sys.version_info[0]
>if PYTHON_MAJOR_VER == 3:
>    xrange = range
>
>
>#Legendre for our specific setup with safe primes
>def kronecker(x,p):
>    q = (p-1)/2
>    return pow(int(x),int(q),int(p))
>
>def findQNR(p):
>    r = Crypto.Random.random.randrange(2,p-1)
>    while kronecker(r,p) == 1:
>        r = Crypto.Random.random.randrange(2,p-1)
>    return r
>
>def findQR(p):
>    r = Crypto.Random.random.randrange(2,p-1) 
>    return pow(int(r),2,int(p))
>
>#Oracle; we use a 512 bit prime only for better performance
>key = ElGamal.generate(512, Random.new().read)
>
>wrong = 0
>runs = 1000
>print("Running experiment...")
>for i in xrange(runs):
>    #Adversary
>    plaintexts = dict()
>    plaintexts[0] = findQNR(key.p)
>    plaintexts[1] = findQR(key.p)
>
>    #Oracle
>    challenge_bit = Crypto.Random.random.randrange(0,2)
>    r =  Crypto.Random.random.randrange(1,key.p-1) 
>    challenge = key.encrypt(plaintexts[challenge_bit], r)
>
>    #Adversary
>    output = -1
>    if (kronecker(key.y, key.p) == 1) or (kronecker(challenge[0], key.p) == 1):
>        if kronecker(challenge[1], key.p) == 1:
>            output = 1
>        else:
>            output = 0
>    else:
>        if kronecker(challenge[1], key.p) == 1:
>            output = 0
>        else:
>            output = 1
>
>    if output != challenge_bit:
>        wrong = wrong + 1
>
>print("Number of times adversary was wrong: %s" % wrong)

from Crypto.PublicKey import ElGamal
from Crypto import Random
import Crypto.Random.random
import sys

PYTHON_MAJOR_VER = sys.version_info[0]
if PYTHON_MAJOR_VER == 3:
    xrange = range


#Legendre for our specific setup with safe primes
def kronecker(x,p):
    q = (p-1)/2
    return pow(int(x),int(q),int(p))

def findQNR(p):
    r = Crypto.Random.random.randrange(2,p-1)
    while kronecker(r,p) == 1:
        r = Crypto.Random.random.randrange(2,p-1)
    return r

def findQR(p):
    r = Crypto.Random.random.randrange(2,p-1) 
    return pow(int(r),2,int(p))

#Oracle; we use a 512 bit prime only for better performance
key = ElGamal.generate(512, Random.new().read)

wrong = 0
runs = 1000
print("Running experiment...")
for i in xrange(runs):
    #Adversary
    plaintexts = dict()
    plaintexts[0] = findQNR(key.p)
    plaintexts[1] = findQR(key.p)

    #Oracle
    challenge_bit = Crypto.Random.random.randrange(0,2)
    r =  Crypto.Random.random.randrange(1,key.p-1) 
    challenge = key.encrypt(plaintexts[challenge_bit], r)

    #Adversary
    output = -1
    if (kronecker(key.y, key.p) == 1) or (kronecker(challenge[0], key.p) == 1):
        if kronecker(challenge[1], key.p) == 1:
            output = 1
        else:
            output = 0
    else:
        if kronecker(challenge[1], key.p) == 1:
            output = 0
        else:
            output = 1

    if output != challenge_bit:
        wrong = wrong + 1

print("Number of times adversary was wrong: %s" % wrong)

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Attachments on bug 22693: 10054 | 10058