DES原理 这里不予以复述， 有很多优秀的博客 原理可以参考这篇博客 https://www.cnblogs.com/songwenlong/p/5944139.html
DES 函数 传入参数为
- text（明文 或者 密文）
- key （解密的key）
- flag （是加密还是解密过程）
# DES 算法实现 flag是标志位 当为-1时， 是DES解密， flag默认为0
def DES (text, key, flag = "0"):
# 初始字段
# IP置换
InitKeyCode = IP(text)
# 产生子密钥 集合
subkeylist = createSubkey(key)
# 获得Ln 和 Rn
Ln = InitKeyCode[0:32]
Rn = InitKeyCode[32:]
# 如果是解密的过程 把子密钥数字逆过来 就变成解密过程了
if (flag == "-1") :
subkeylist = subkeylist[::-1]
for subkey in subkeylist:
while len(Rn) < 32:
Rn = "0" + Rn
while len(Ln) < 32:
Ln = "0" + Ln
# 对右边进行E-扩展
Rn_expand = E_expend(Rn)
# 压缩后的密钥与扩展分组异或以后得到48位的数据，将这个数据送入S盒
S_Input = int(Rn_expand, base=2) ^ int(subkey, base=2)
# 进行S盒替代
S_sub_str = S_sub(S_Input)
#P盒置换 并且
# 左、右半部分交换，接着开始另一轮
(Ln, Rn) = P(Ln, S_sub_str, Rn)
#进行下一轮轮置换
# 最后一轮之后 左、右两半部分并未进行交换
# 而是两部分合并形成一个分组做为末置换的输入。
# 所以要重新置换 一次
(Ln, Rn) = (Rn, Ln)
# 末置换得到密文
re_text = IP_inverse(Ln, Rn)
return re_textDES有各种置换矩阵的定义， 所以提前定义好， 但是这里虽然说是矩阵 但是使用数组来表示的
# S盒 的置换矩阵
S_MATRIX = [(14, 4, 13, 1, 2, 15, 11, 8, 3, 10, 6, 12, 5, 9, 0, 7,
0, 15, 7, 4, 14, 2, 13, 1, 10, 6, 12, 11, 9, 5, 3, 8,
4, 1, 14, 8, 13, 6, 2, 11, 15, 12, 9, 7, 3, 10, 5, 0,
15, 12, 8, 2, 4, 9, 1, 7, 5, 11, 3, 14, 10, 0, 6, 13),
(15, 1, 8, 14, 6, 11, 3, 4, 9, 7, 2, 13, 12, 0, 5, 10,
3, 13, 4, 7, 15, 2, 8, 14, 12, 0, 1, 10, 6, 9, 11, 5,
0, 14, 7, 11, 10, 4, 13, 1, 5, 8, 12, 6, 9, 3, 2, 15,
13, 8, 10, 1, 3, 15, 4, 2, 11, 6, 7, 12, 0, 5, 14, 9),
(10, 0, 9, 14, 6, 3, 15, 5, 1, 13, 12, 7, 11, 4, 2, 8,
13, 7, 0, 9, 3, 4, 6, 10, 2, 8, 5, 14, 12, 11, 15, 1,
13, 6, 4, 9, 8, 15, 3, 0, 11, 1, 2, 12, 5, 10, 14, 7,
1, 10, 13, 0, 6, 9, 8, 7, 4, 15, 14, 3, 11, 5, 2, 12),
(7, 13, 14, 3, 0, 6, 9, 10, 1, 2, 8, 5, 11, 12, 4, 15,
13, 8, 11, 5, 6, 15, 0, 3, 4, 7, 2, 12, 1, 10, 14, 9,
10, 6, 9, 0, 12, 11, 7, 13, 15, 1, 3, 14, 5, 2, 8, 4,
3, 15, 0, 6, 10, 1, 13, 8, 9, 4, 5, 11, 12, 7, 2, 14),
(2, 12, 4, 1, 7, 10, 11, 6, 8, 5, 3, 15, 13, 0, 14, 9,
14, 11, 2, 12, 4, 7, 13, 1, 5, 0, 15, 10, 3, 9, 8, 6,
4, 2, 1, 11, 10, 13, 7, 8, 15, 9, 12, 5, 6, 3, 0, 14,
11, 8, 12, 7, 1, 14, 2, 13, 6, 15, 0, 9, 10, 4, 5, 3),
(12, 1, 10, 15, 9, 2, 6, 8, 0, 13, 3, 4, 14, 7, 5, 11,
10, 15, 4, 2, 7, 12, 9, 5, 6, 1, 13, 14, 0, 11, 3, 8,
9, 14, 15, 5, 2, 8, 12, 3, 7, 0, 4, 10, 1, 13, 11, 6,
4, 3, 2, 12, 9, 5, 15, 10, 11, 14, 1, 7, 6, 0, 8, 13),
(4, 11, 2, 14, 15, 0, 8, 13, 3, 12, 9, 7, 5, 10, 6, 1,
13, 0, 11, 7, 4, 9, 1, 10, 14, 3, 5, 12, 2, 15, 8, 6,
1, 4, 11, 13, 12, 3, 7, 14, 10, 15, 6, 8, 0, 5, 9, 2,
6, 11, 13, 8, 1, 4, 10, 7, 9, 5, 0, 15, 14, 2, 3, 12),
(13, 2, 8, 4, 6, 15, 11, 1, 10, 9, 3, 14, 5, 0, 12, 7,
1, 15, 13, 8, 10, 3, 7, 4, 12, 5, 6, 11, 0, 14, 9, 2,
7, 11, 4, 1, 9, 12, 14, 2, 0, 6, 10, 13, 15, 3, 5, 8,
2, 1, 14, 7, 4, 10, 8, 13, 15, 12, 9, 0, 3, 5, 6, 11)]
# P置换的置换矩阵
P_MATRIX = [16, 7, 20, 21, 29, 12, 28, 17, 1, 15, 23, 26, 5, 18, 31, 10,
2, 8, 24, 14, 32, 27, 3, 9, 19, 13, 30, 6, 22, 11, 4, 25]
# IP置换的 置换矩阵
IP_MATRIX = [58, 50, 42, 34, 26, 18, 10, 2,
60, 52, 44, 36, 28, 20, 12, 4,
62, 54, 46, 38, 30, 22, 14, 6,
64, 56, 48, 40, 32, 24, 16, 8,
57, 49, 41, 33, 25, 17, 9, 1,
59, 51, 43, 35, 27, 19, 11, 3,
61, 53, 45, 37, 29, 21, 13, 5,
63, 55, 47, 39, 31, 23, 15, 7]
# 压缩置换矩阵 从56位里选48位
COMPRESS_MATRIXS = [14, 17, 11, 24, 1, 5,
3, 28, 15, 6, 21, 10,
23, 19, 12, 4, 26, 8,
16, 7, 27, 20, 13, 2,
41, 52, 31, 37, 47, 55,
30, 40, 51, 45, 33, 48,
44, 49, 39, 56, 34, 53,
46, 42, 50, 36, 29, 32]
# E扩展置换矩阵
E_MATRIX = [32, 1, 2, 3, 4, 5,
4, 5, 6, 7, 8, 9,
8, 9, 10, 11, 12, 13,
12, 13, 14, 15, 16, 17,
16, 17, 18, 19, 20, 21,
20, 21, 22, 23, 24, 25,
24, 25, 26, 27, 28, 29,
28, 29, 30, 31, 32, 1]
# IP逆置换矩阵
IP_INVERSE_MATRIX = [40, 8, 48, 16, 56, 24, 64, 32, 39, 7, 47, 15, 55, 23, 63, 31,
38, 6, 46, 14, 54, 22, 62, 30, 37, 5, 45, 13, 53, 21, 61, 29,
36, 4, 44, 12, 52, 20, 60, 28, 35, 3, 43, 11, 51, 19, 59, 27,
34, 2, 42, 10, 50, 18, 58, 26, 33, 1, 41, 9, 49, 17, 57, 25]对于置换 我们的明文是用一个长度为64的字符01串构成的 所以置换 只需要新的字符串 然后按照置换矩阵的下表来把原字符串的单个字符加入
def IP(Mingwen):
#如果长度不是64 就退出
assert len(Mingwen) == 64
ret = ""
#通过循环 进行IP置换
for i in IP_MATRIX:
ret = ret + Mingwen[i - 1]
return ret对输入的str 按照shift_count的大小 左移shift_count位 这个函数主要用于子密钥生成
def shift(str, shift_count):
try:
if len(str) > 28:
raise NameError
except TypeError:
pass
str = str[shift_count:] + str[0:shift_count]
return str此函数用来生成子密钥 生成16组子密钥 用于和扩展分组 异或
def createSubkey(key):
# 如果key长度不是64 就退出
assert len(key) == 64
#DES的密钥由64位减至56位，每个字节的第8位作为奇偶校验位
#把56位 变成 2个28位
Llist = [57, 49, 41, 33, 25, 17, 9,
1, 58, 50, 42, 34, 26, 18,
10, 2, 59, 51, 43, 35, 27,
19, 11, 3, 60, 52, 44, 36]
Rlist = [63, 55, 47, 39, 31, 23, 15,
7, 62, 54, 46, 38, 30, 22,
14, 6, 61, 53, 45, 37, 29,
21, 13, 5, 28, 20, 12, 4]
# 初试生成 左右两组28位密钥
L0 = ""
R0 = ""
for i in Llist:
L0 += key[i - 1]
for i in Rlist:
R0 += key[i - 1]
assert len(L0) == 28
assert len(R0) == 28
#轮函数生成 48位密钥
#定义轮数
Movetimes = [1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1]
#定义返回的subKey
retkey = []
#开始轮置换
for i in range(0, 16):
#获取左半边 和 右半边 shift函数用来左移生成轮数
L0 = shift(L0, Movetimes[i])
R0 = shift(R0, Movetimes[i])
#合并左右部分
mergedKey = L0 + R0
tempkey = ""
# 压缩置换矩阵 从56位里选48位
#选出48位子密钥
for i in COMPRESS_MATRIXS:
tempkey += mergedKey[i - 1]
assert len(tempkey) == 48
#加入生成子密钥
retkey.append(tempkey)
return retkey用于把右边的32位扩展成48位
def E_expend(Rn):
retRn = ""
for i in E_MATRIX:
retRn += Rn[i - 1]
assert len(retRn) == 48
return retRn压缩后的密钥与扩展分组异或以后得到48位的数据，将这个数据送入S盒，进行替代运算。替代由8个不同的S盒完成，每个S盒有6位输入4位输出。48位输入分为8个6位的分组，一个分组对应一个S盒，对应的S盒对各组进行代替操作。
# S盒替代运算
def S_sub(S_Input):
#从第二位开始的子串 去掉0X
S_Input = bin(S_Input)[2:]
while len(S_Input) < 48:
S_Input = "0" + S_Input
index = 0
retstr = ""
for Slist in S_MATRIX:
# 输入的高低两位做为行数row
row = int(S_Input[index] + S_Input[index + 5], base=2)
# 中间四位做为列数L
col = int(S_Input[index + 1:index + 5], base=2)
# 得到 result的 单个四位输出
ret_single = bin(Slist[row * 16 + col])[2:]
while len(ret_single) < 4:
ret_single = "0" + ret_single
# 合并单个输出
retstr += ret_single
# index + 6 进入下一个六位输入
index += 6
assert len(retstr) == 32
return retstrS盒代替运算的32位输出按照P盒进行置换
最后，P盒置换的结果与最初的64位分组左半部分L0异或，然后左、右半部分交换，接着开始另一轮。
def P(Ln, S_sub_str, oldRn):
# P 盒置换
tmp = ""
for i in P_MATRIX:
tmp += S_sub_str[i - 1]
# P盒置换的结果与最初的64位分组左半部分L0异或
LnNew = int(tmp, base=2) ^ int(Ln, base=2)
LnNew = bin(LnNew)[2:]
while len(LnNew) < 32:
LnNew = "0" + LnNew
assert len(LnNew) == 32
# 左、右半部分交换，接着开始另一轮
(Ln, Rn) = (oldRn, LnNew)
return (Ln, Rn)末置换是初始置换的逆过程，DES最后一轮后，左、右两半部分并未进行交换，而是两部分合并形成一个分组做为末置换的输入 置换后得到密文或者解密的明文
def IP_inverse(L16, R16):
tmp = L16 + R16
retstr = ""
for i in IP_INVERSE_MATRIX:
retstr += tmp[i - 1]
assert len(retstr) == 64
return retstr用来测试效果
if __name__ == "__main__":
key = "0001001000110100010101100111100010010001001000110100010101100111"
Mingwen = "1001100001110110010101000011001000010001010001110010010110000011"
#打印明文的16进制形式
print("明文的16进制形式: " + hex(int(Mingwen, base=2)).upper())
ciphertext = DES(Mingwen, key)
#打印加密后的密文
print("加密后的密文: " + hex(int(ciphertext, base=2)).upper())
falseKey = "1001001000110100010101100111100010010001001000110100010101100110"
decode_ciphertext = DES(ciphertext, key, "-1")
#打印解密后的明文 看是否相同
print("解密后的明文: " + hex(int(decode_ciphertext, base=2)).upper())
decode_ciphertext = DES(ciphertext, falseKey, "-1")
# 打印给定错误的key 解密后的明文 看是否不同
print("给定错误的key 解密后的明文:" + hex(int(decode_ciphertext, base=2)).upper())