from PIL import Image
from numpy import *
from BitVector import BitVector as bv
im = Image.open('FullMapEE.bmp')
L = list(im.getdata())
for l, v in enumerate(L):
if L[l] == 255:
L[l] = 0
else:
L[l] = 1
#add 5 more rows
for i in range(850):
L.append(0)
for K in range(16):
for i in range(170):
# read entire columns
buf = bv(size=8)
for j in range(8):
# read entire block
buf[7 - j] = int(L[i + 170 * j + 170 * 8 * K])
print "0x" + buf.get_hex_string_from_bitvector(), ",",
print
#! /usr/bin/env python
from BitVector import BitVector as bv
from numpy import poly1d as P
n = 4
m = bv(bitlist=[1, 0, 0, 1, 1])
print "computing GF(2^n)"
print "n = {0}".format(n)
print "modulus m = "
print P(list(m))
for i in range(0, 2 ** n):
a1 = bv(bitlist=[0, 0, 1, 0])
a = a1
if i == 0:
# 0 case.
a = bv(bitlist=[0, 0, 0, 0])
else:
for j in range(1, i):
a = a.gf_multiply_modular(a1, m, n)
print "============================================"
print "g^{0}".format(i)
print "Binary: {0}".format(a)
print "Hex: {0}".format(a.getHexStringFromBitVector())
print "Polynomial:"
print P(list(a))
#! /usr/bin/env python #perform the MixCols step in the AES cipher. from BitVector import BitVector as bv n = 8 #for GF(2^8) #modulus for AES #x^8 + x^4 + x^3 + x + 1 m = bv(bitlist = [1,0,0,0,1,1,0,1,1]) col00 = bv(intVal=0x7C,size = 8) col01 = bv(intVal=0xF2,size = 8) col02 = bv(intVal=0x2B,size = 8) col03 = bv(intVal=0xAB,size = 8) bv2 = bv(intVal=2, size = 8) bv3 = bv(intVal=3, size = 8) newcol00 = col00.gf_multiply_modular(bv2,m,n) ^ col01.gf_multiply_modular(bv3,m,n) ^ col02 ^ col03 newcol01 = col01.gf_multiply_modular(bv2,m,n) ^ col02.gf_multiply_modular(bv3,m,n) ^ col00 ^ col03 newcol02 = col02.gf_multiply_modular(bv2,m,n) ^ col03.gf_multiply_modular(bv3,m,n) ^ col00 ^ col01 newcol03 = col03.gf_multiply_modular(bv2,m,n) ^ col00.gf_multiply_modular(bv3,m,n) ^ col02 ^ col01 print newcol00.getHexStringFromBitVector() print newcol01.getHexStringFromBitVector() print newcol02.getHexStringFromBitVector() print newcol03.getHexStringFromBitVector()
#! /usr/bin/env python
#effecient modular exponentiation algorithm
#from page 283, Stallings crypto book 5th ed.
#computes a^b mod n
# inputs: a, b, n
from BitVector import BitVector as bv
k = 10
a = 5
bval = 596
b = bv(intVal=bval, size=k)
n = 1234
print "performing the naive calculation:"
print "result:{0}".format(a**bval % n)
print "Performing fast modular exponentiation:"
print "a:{0}, b:{1}, n:{2}".format(a,bval,n)
c = 0
f = 1
for i in reversed(xrange(k)):
j = k- 1 - i #to fix the wierd indexing in BitVector
print "i:{3}, f:{0}, c:{1}, bi:{2}".format(f,c,b[j],i)
c = 2 * c
f = (f * f) % n
if (b[j] == 1):
c = c + 1