非线性组合序列的快速相关攻击算法A和算法B
自己做密码算法，写的很垃圾，需要拿走import math
from scipy.special import comb
from lfsr import LFSR
#计算1的位数
def bit_count(val):
cnt = 0
while val:
cnt += 1
val = val & (val - 1)
return cnt
#汉明距离变换
hamming = {15 : [[]] * 16, 16 : [[]] * 17}
for i in range(1, 1 << 16):
cnt = bit_count(i)
hamming.append(i)
if i < (1 << 15):
hamming.append(i)
class MS:
def __init__(self, mask, n, z, p):
self.mask = mask #抽头掩码
self.n = n #级数
self.z = list(z) #密钥流序列
self.p = p #相关概率
self.len_z = len(z) #密钥流长度
self.t = 0 #抽头数
for i in range(n):
if (1 << i) & self.mask:
self.t += 1
self.m = round(self.M()) #平均每一位方程数
self.s_init = self.S(self.t) #每一位正确的后验概率s
@staticmethod
def bit_stream_to_int(a):
return int(''.join(map(str, a)), 2)
def M(self):
return math.log2(self.len_z / (2 * self.n)) * (self.t + 1)
def S(self, t):
if t == 1:
return self.p
return self.p * self.S(t - 1) + (1 - self.p) * (1 - self.S(t - 1))
#生成校验等式
def get_eq(self):
tap =
for i in range(self.n):
if (self.mask >> i) & 1:
tap.append(self.n - i - 1)
tap.reverse()
eqs =
while True:
if (tap[-1] << 1) >= self.len_z:
break
tmp = tap.copy()
for i, val in enumerate(tmp):
tmp = val << 1
eqs.append(tmp)
tap = tmp
return eqs
#算法A得到loc位正确的概率
def calc_eq(self, eqs, loc):
shift_eqs = []
for eq in eqs:
for pos in eq:
offset = loc - pos
if eq + offset < 0 or eq[-1] + offset >= self.len_z:
continue
shift_eqs.append()
m = len(shift_eqs)
if m == 0:
return 0, 0, 0
h = 0
for eq in shift_eqs:
# print(eq)
xor_sum = 0
for i in eq:
xor_sum ^= self.z
if xor_sum == 0:
h += 1
p1 = comb(m, h) * pow(self.s_init, h) * pow(1 - self.s_init, m - h)
p0 = comb(m, h) * pow(self.s_init, m - h) * pow(1 - self.s_init, h)
return m, h, p1 / (p1 + p0)
#生成矩阵
def gen_linear_eq(self):
length = max(self.len_z, self.n)
tap = []
for i in range(self.n):
if (self.mask >> i) & 1:
tap.append(i + 1)
eqs = []
for i in range(self.n):
eqs.append(1 << i)
for i in range(self.n, length):
res = 0
for j in tap:
res ^= eqs
eqs.append(res)
return eqs
#根据生成矩阵解方程
@staticmethod
def solve(assume, n):
eq_len = len(assume)
mat = []
for i in range(eq_len):
mat.append( * n)
b = * eq_len
for i in range(eq_len):
b = assume
for j in range(n):
mat = (assume >> j) & 1
for i in range(n):
tmp = -1
for j in range(i, eq_len):
if mat:
tmp = j
break
if tmp == -1:
return []
mat, mat = mat, mat
b, b = b, b
for j in range(eq_len):
if not mat or i == j:
continue
b ^= b
for k in range(i, n):
mat ^= mat
if not any(mat):
return []
# print(b[:n])
return b[:n]
#计算初始状态并验证
def get_init_stat(self, locs, linear_eq):
assume = [(linear_eq], x) for x in locs]
b = []
idx = self.n
# print("----- try solve equations -----")
while not b:
b = MS.solve(assume[:idx], self.n)
idx += 1
# print("----- solve success -----")
stat = MS.bit_stream_to_int(b)
# print("----- genrate original LFSR -----")
l = LFSR(stat, self.mask, self.n)
for i in range(self.n):
l.step_back()
init_stat = l.init
# print("init:", init_stat)
# print("----- genrate original LFSR finished -----")
same_cnt = 0
for i in range(self.len_z):
same_cnt += int(self.z == l.next())
rate = same_cnt / self.len_z
if abs(rate - self.p) < 0.05:
return init_stat
else:
#hamming变换
for i in range(self.n + 1):
for filp in hamming:
change = init_stat ^ filp
l.init = change
cnt = 0
for i in range(self.len_z):
cnt += int(self.z == l.next())
rate = cnt / self.len_z
if abs(rate - self.p) < 0.05:
return change
return init_stat
#算法A攻击函数
def crackA(self):
eqs = self.get_eq()
# print("----- select candidates -----")
candidates = []
for i in range(self.len_z):
m, h, p_star = self.calc_eq(eqs, i)
if p_star > 0.5:
candidates.append((p_star, i, m, h))
candidates.sort(reverse=True)
# candidates = candidates[:2*self.n]
# print(candidates[:5])
# print("----- select candidates finished -----")
linear_eq = self.gen_linear_eq()
locs = [(cand, self.z]) for cand in candidates]
return self.get_init_stat(locs, linear_eq)
#计算s_init
@staticmethod
def var_S(var_p, t):
assert(t == len(var_p))
if t == 1:
return var_p
s = MS.var_S(var_p[:-1], t - 1)
return var_p[-1] * s + (1 - var_p[-1]) * (1 - s)
def Q(self, h):
res = 0
for i in range(h + 1):
res += comb(self.m, i) * (self.p_false(i) + self.p_true(i))
return res
def I(self, h):
res = 0
for i in range(h + 1):
res += comb(self.m, i) * (self.p_false(i) - self.p_true(i))
return res
def p_true(self, h):
return self.p * pow(self.s_init, h) * pow(1 - self.s_init, self.m - h)
def p_false(self, h):
return (1 - self.p) * pow(1 - self.s_init, h) * pow(self.s_init, self.m - h)
def p_update(self, h):
t = self.p_true(h)
f = self.p_false(h)
return t / (t + f)
def calc_h_max(self):
h_max = 0
I_max = 0
for i in range(self.m + 1):
newI = self.I(i)
if newI > I_max:
I_max = newI
h_max = i
return h_max
#计算迭代后验概率的4个部分
def parcheck(self, eqs, arr_p):
poly = self.t + 1
s = [ for i in range(self.len_z)]
for eq in eqs:
curtaps = eq.copy()
offset = self.len_z - eq[-1]
# print('\n' , offset , '\n')
for i in range(offset):
# print(i , ' ')
xor_sum = 0
for tap in curtaps:
xor_sum ^= self.z
for j in range(poly):
var_p = ] for k in range(poly) if k != j]
cur_s = MS.var_S(var_p, self.t)
cur_bit = curtaps
if xor_sum == 0:
s *= cur_s
s *= 1 - cur_s
else:
s *= cur_s
s *= 1 - cur_s
for j in range(poly):
curtaps += 1
return s
#计算迭代后验概率
@staticmethod
def var_p_update(p, arr_s):
t = p * arr_s * arr_s
div = t + (1 - p) * arr_s * arr_s
#问题出在这里
if div == 0:
return 1
return t / (t + (1 - p) * arr_s * arr_s)
#检验是否满足校验方程
@staticmethod
def check(eq, z):
offset = len(z) - eq[-1]
for i in range((offset)):
xor_sum = 0
for tap in eq:
xor_sum ^= z
if xor_sum:
return False
for j, val in enumerate(eq):
eq = val + 1
return True
#算法B攻击函数
def crackB(self):
prim_z = self.z.copy()
eqs = self.get_eq()
h_max = self.calc_h_max()
p_thr = (self.p_update(h_max) + self.p_update(h_max + 1)) / 2
N_thr = self.Q(h_max) * self.len_z
# print("------------------------------------------")
# print("Round\tIteration\tN_w")
# print("------------------------------------------")
for r in range(1, 1000):
arr_p =
for iter in range(1, 6):
arr_ss = self.parcheck(eqs, arr_p)
for i in range(self.len_z):
arr_p = MS.var_p_update(arr_p, arr_ss)
N_w = 0
for i in range(self.len_z):
if arr_p < p_thr:
N_w += 1
# print(r, '\t', iter, '\t\t', N_w)
if N_w >= N_thr:
break
cnt_filp_bit = 0
for pos in range(self.len_z):
if arr_p < p_thr:
self.z ^= 1
cnt_filp_bit += 1
# print("------------------------------------------")
if MS.check(eqs[:], self.z) or cnt_filp_bit == 0:
stat = MS.bit_stream_to_int(self.z[:self.n])
l = LFSR(stat, self.mask, self.n)
for i in range(self.n):
l.step_back()
self.z = prim_z
return l.init
return list() 容我先理解一下 感谢分享！
只会C、C++，还好里面的语法规则看得懂。 谢谢分享~~~ 学习学习 这个世界最难得不是看不懂，而是都是专业术语。。。确没有解释 时间复杂度有点高了 先容我学学密码学…感谢大佬分享 让我先学学密码学这里太难懂了