Beispiel #1
0
class ecprng:
    # Curve P-256; source: https://safecurves.cr.yp.to/
    p = 2**256 - 2**224 + 2**192 + 2**96 - 1
    a = p - 3
    b = 41058363725152142129326129780047268409114441015993725554835256314039467401291
    ec = ecc.CurveFp(p, a, b)

    _Px = 115113149114637566422228202471255745041343462839792246702200996638778690567225
    _Py = 88701990415124583444630570378020746694390711248186320283617457322869078545663
    Point_P = ecc.Point(ec, _Px, _Py)

    _Qx = 75498749949015782244392151836890161743686522667385613237212787867797557116642
    _Qy = 19586975827802643945708711597046872561784179836880328844627665993398229124361
    Point_Q = ecc.Point(ec, _Qx, _Qy)

    def __init__(self, seed):
        self.seed = seed
        if self.seed:
            assert len(long_to_bytes(self.seed)) == 32

    def update_seed(self, intermediate_state_S_1):
        self.seed = (intermediate_state_S_1 * ecprng.Point_P).x()
        assert len(long_to_bytes(self.seed)) == 32

    def ec_generate(self):
        intermediate_state_S_1 = (self.seed * ecprng.Point_P).x()
        self.update_seed(intermediate_state_S_1)
        r_1 = long_to_bytes(
            (intermediate_state_S_1 * ecprng.Point_Q).x())[-30:]
        r_2 = long_to_bytes((self.seed * ecprng.Point_Q).x())[-30:][:2]
        assert len(r_1 + r_2) == 32
        return bytes_to_long(r_1 + r_2)
Beispiel #2
0
def baby_step_giant_step(curve, G, H, order):

    m = int(math.ceil(gmpy2.sqrt(order)))
    L = {}

    # Baby steps
    for j in range(0, m):
        P_tmp = curve.mul(j, G)
        L[str(P_tmp)] = j

    mG = curve.mul(m, G)

    # Giant steps
    for i in range(0, m):
        P_tmp = curve.mul(i, mG)
        if not P_tmp.isInf():
            P_tmp = ecc.Point(P_tmp.x, (-P_tmp.y) % curve.p)

        P = curve.add(H, P_tmp)

        index = str(P)

        if index in L:
            return (L[index] + i * m) % curve.p

    return None
Beispiel #3
0
def bits_to_point(p):
    (x_size, ) = struct.unpack('!H', p[:2])
    if x_size == 0:
        return ecc.PointInf()

    x = bits_to_int(p[2:x_size + 2])
    y = bits_to_int(p[x_size + 4:])

    return ecc.Point(x, y)
Beispiel #4
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def dechiffrement_Alice(curve):
    '''
    Alice reçoit C1 et C2
    elle calcule daC1 grâce à sa clé privée
    elle calcule l'inverse de daC1
    elle déchiffre C2 qui correspond à M
    '''
    C1, C2 = chiffrement_Bob(curve)
    f = open("cle.txt", "r")
    cle = int(f.read())
    daC1 = ecc.Curve.mul(curve, cle, C1)
    daC1_inv = ecc.Point(daC1.x, -daC1.y)
    M = ecc.Curve.add(curve, C2, daC1_inv)
    return M
Beispiel #5
0
def find_next_e(e):
    r = long_to_bytes(e)[:-2]
    for i in trange(133, 140):
        for j in range(256):
            x = bytes_to_long(chr(i) + chr(j) + r)
            y = find_point(ec, x)
            if test_Point(x, y):
                R = ecc.Point(ec, x, y)
                r_2 = long_to_bytes(
                    (((R * inverse(d, order)).x()) * Point_Q).x())[-30:][:2]
                if long_to_bytes(e)[-2:] == r_2:
                    print "finally"
                    return R
    return R
Beispiel #6
0
    blue = '\033[34m'


# New urandom seed for each session (Not really relevant for the challenge)
prng_obj = prng(16793527392756720769,
                2358102439659339126076356431940385122127543421625845446663,
                False)

p = 2**256 - 2**224 + 2**192 + 2**96 - 1
a = p - 3
b = 41058363725152142129326129780047268409114441015993725554835256314039467401291
ec = ecc.CurveFp(p, a, b)

_Px = 53881495764268889303293517690095107010093794097958309592680107528631746121613
_Py = 69534606358473748292927094386662082099432383517498778127513290350658945146669
Point_P = ecc.Point(ec, _Px, _Py)

iteration = True
counter = 0

while iteration == True and counter < 10:
    print colors.blue + "Choose one between authentication protocols listed below:" + colors.reset
    print colors.orange + "[1] Asynchronous SchnorrID" + colors.reset
    print colors.orange + "[2] Synchronous SchnorrID" + colors.reset
    choice = int(raw_input("Enter your choice: "))
    print ""

    if choice == 1:
        print "Here are the coordinates of the base point P: ", _Px, _Py
        _Qx, _Qy = map(
            int,
Beispiel #7
0
                    return R
    return R


st = lambda x: str(x).strip('L')

if __name__ == "__main__":

    p = 2**256 - 2**224 + 2**192 + 2**96 - 1
    a = p - 3
    b = 41058363725152142129326129780047268409114441015993725554835256314039467401291
    ec = ecc.CurveFp(p, a, b)

    _Px = 115113149114637566422228202471255745041343462839792246702200996638778690567225
    _Py = 88701990415124583444630570378020746694390711248186320283617457322869078545663
    Point_P = ecc.Point(ec, _Px, _Py)

    _Qx = 75498749949015782244392151836890161743686522667385613237212787867797557116642
    _Qy = 19586975827802643945708711597046872561784179836880328844627665993398229124361
    Point_Q = ecc.Point(ec, _Qx, _Qy)
    d = 1735
    x = 53881495764268889303293517690095107010093794097958309592680107528631746121613
    y = 69534606358473748292927094386662082099432383517498778127513290350658945146669

    P = ecc.Point(ec, x, y)
    Q = 123 * P

    order = 115792089210356248762697446949407573529996955224135760342422259061068512044369

    #io = process('./encrypt.py')
    io = remote('34.74.30.191', 3333)
Beispiel #8
0
        (gcd, x0, x1) = xgcd(ni, tmp)

        x += x_prime * x1 * tmp

    return x % N


# (A, B, N)

A = 0
B = 0
N = 0

X = 0
Y = 0

curve = ecc.Curve(A, B, N)
G = ecc.Point(X, Y)

sent = [
    0x00, 0x30, 0x00, 0x16, 0x0d, 0x6c, 0x24, 0xb0, 0x5a, 0xf7, 0xff, 0x4f,
    0xa6, 0x28, 0xeb, 0xce, 0xfd, 0x43, 0xdd, 0xad, 0x1a, 0x57, 0xac, 0xb9,
    0xa4, 0x65, 0x00, 0x16, 0x0a, 0x00, 0x63, 0x5f, 0x98, 0x88, 0x1c, 0x47,
    0x07, 0x50, 0x48, 0x3e, 0xa0, 0x59, 0x77, 0xc1, 0x93, 0x28, 0x9a, 0xeb,
    0x50, 0x64
]

H = bits_to_point("".join(map(chr, sent)).encode())

x = pohlig_hellman(curve, G, H)
print("x = %s" % x)
Beispiel #9
0
import os.path
import sys
import ecc
sys.path.append(os.path.join(os.path.dirname(__file__), '..'))
''' Création de la courbe elliptiques '''
A = 109454571331697278617670725030735128145969349647868738157201323556196022393856
B = 107744541122042688792155207242782455150382764043089114141096634497567301547839
''' Ordre du point P '''
l = 109454571331697278617670725030735128146004546811402412653072203207726079563233
''' Ordre de la courbe '''
N = 109454571331697278617670725030735128145969349647868738157201323556196022393859
'''n pour la multiplication '''
n = 2
''' Point de la courbe elliptique '''
P = ecc.Point(
    82638672503301278923015998535776227331280144783487139112686874194432446389503,
    43992510890276411535679659957604584722077886330284298232193264058442323471611
)
Q = ecc.Point(
    100597391921786027039183722380481804805320476080319934670061678997404767442782,
    80123073214026054915454326239165515159448240266403681526048086449062769463365
)
'''Point M du message'''
M = ecc.Point(
    100597391921786027039183722380481804805320476080319934670061678997404767442782,
    80123073214026054915454326239165515159448240266403681526048086449062769463365
)