S_Box的构造
实现思路
- 根据拓展欧几里算法得出对应位置的逆元
- 进行仿射变化
- 将得出的答案进行输出
问题：设计除法函数的时候，会出现溢出的情况。
下面贴出代码:
import sys
def max_index(num):
for i in range(8):
if not (num>>(i+1)):
return i
#进行乘法运算
def gf_multiply(a, b):
if(b > a):
a ,b = b ,a
res = 0
if(b & 0x01):
res = a
for i in range(8):
if(b & (0x01 << i)):
temp = a
for j in range(0,i):
#判断最高位是否为1
if not(temp & 0x80):
temp <<= 1
else:
temp <<= 1
temp = temp ^ 0x1B
res ^= temp
return res
#进行除法晕眩
def gf_divide(m,b):
m_msb = max_index(m)
b_msb = max_index(b)
#递归的终点，当除数大于被除数
if( m_msb < b_msb ):
global r
r = m
return 0
#获取商的值
d = m_msb - b_msb
tem = b
tem = tem << d
#求解得到下一次的被除数
m = m ^ tem
#将除法得到的每一轮结果进行相加(其中r为余数)
return ( 1 << d ) | gf_divide( m, b),r
#求逆元，辗转相除，ax+by=1
def get_inverse(num):
if (num == 0):
return 0
a = num
b = 0x11B
w0 = 0
x = 1
q,r = gf_divide(b,a)
value = w0 ^ gf_multiply(q,x)
while(1):
if(r == 0 | r ==1 ):
break
b = a
a = r
q,r = gf_divide(b, a)
w0 = x
x = value
value = w0 ^ gf_multiply(q,x)
return x
#仿射变化
def map(num):
c = 0x63
temp = 0x0
res = 0x0
for i in range(0,8):
temp = temp ^ ((num >> i) & 0x01) ^ ((num >> ((i + 4) % 8)) & 0x01)
temp = temp ^ ((num >> ((i + 5) % 8)) & 0x01) ^ ((num >> ((i + 6) % 8)) & 0x01)
temp = temp ^ ((num >> ((i + 7) % 8)) & 0x01) ^ ((c >> i) & 0x01)
res = res | (temp << i)
return res
if __name__=="__main__":
sbox = [[0 for i in range(16)] for j in range(16)]
for i in range(0, 16):
for j in range(0, 16):
tem = get_inverse( (i << 4) + j)
sbox[i][j] = map(tem)
print (sbox)
