Keygen for KeyGenMe[1_2014] (esso_x)

# we assume P1 = P3 and P2 = P4
# in F1 function: a ^ d = 0 infer 0 ^ 0x48EF856C = 0 infer F1 = 0 ^ d = d
# in F2 like F1 function, should we had F2 = d ^ 0xE5824A68
# call general value of P1, P3 is X and general value of P2, P4 is Y,
# from R = R1 ^ R2 == 0x4890882A, to success R must be equal 0x4890882A should R1 ^ R2 = 0x4890882A
# same F1 ^ F2 = 0x4890882A same X ^ Y ^ 0xE5824A68 = 0x4890882A
# infer X ^ Y = 0x4890882A ^ 0xE5824A68 same X ^ Y = 0xAD12C242.
# result X = 0xAD12C242 ^ Y

import random

def F1(a, d):
    return ((a ^ d) & 0x48EF856C) ^ d

def F2(a, d):
    return (((a ^ d) & 0x0B76bF76) ^ d) ^ 0xE5824A68

# P1 = Serial(0x09,8)
# P2 = Serial(0x31,8)
# P3 = Serial(0x39,8)
# P4 = Serial(0x19,8)

P2 = P4 = random.randrange(0,0xFFFFFFFF)
P1 = P3 = 0xAD12C242 ^ P2

R1 = F1(P1,P3)
print "R1 = 0x%08X" % (R1)

R2 = F2(P4,P2)
print "R2 = 0x%08X" % (R2)

R = R1 ^ R2
print "R  = 0x%08X" % (R)

if (R == 0x4890882A):
    print "Done!!"
else:
    print "Oops!!"

# .eg: 000000001234567800000000BF26943A0000000000000000BF26943A12345678