Python中如何实现DES加密

发布时间:2020-07-17 17:06:31 作者:小猪
来源:亿速云 阅读:178

这篇文章主要讲解了Python中如何实现DES加密,内容清晰明了,对此有兴趣的小伙伴可以学习一下,相信大家阅读完之后会有帮助。

加密流程

​首先说一下置换的意思,比如说有5678这个字符串,置换表为2143,置换表中的数表示的是位置,所以字符串变成6587。所有的置换表在程序中。(S盒置换不一样,会另外说明)

密钥部分

Python中如何实现DES加密

明文部分

Python中如何实现DES加密

​ 以一组为例子来说明,一组明文8个字节,64位。有16轮迭代,要运行16次feistel函数。注意在16轮迭代前要把明文进行初始置换,迭代后把左右两边数据合并成64位再进行逆初始运算。

把64位明文左右对半分成两份。

右边的先进行部分进行扩展置换,32位变成48位。

再和对应轮数的子密钥进行异或运算。

再进行S盒运算,48位变成32位。S盒运算具体操作方法是,把48位数据分成8份,每份就有6位数据,比如010110,把头和尾结合位00,变成十进制就是0,中间四位的十进制为11,所以(x,y)为(0,11)

[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]

在上面这个表中表示的就是12,在把12变成2进制就是1100,所以6位就变成了4位。总共有8份数据,也有8个表。每份对应的运算的表都不一样。

再进行P盒运算。

最后和左边的32位进行异或运算。

解密

解密部分除了在feistel函数中调用子密钥的顺序相反外,其他都一样。加密调用的顺序是1-16,解密是16-1。

代码

#################################辅助函数######################################

# 十进制转成二进制
def INT_BIN(NUM):
  i = bin(NUM)[2:]
  if len(i) != 8:
    i = ((8 - len(i)) * '0') + i

  return i

# 置换函数
def Replace(ARR,change):
  ARR1 = []
  for i in ARR:
    a = ''
    for j in change:
      a += i[j-1]
    ARR1.append(a)
  return ARR1

# 异或运算
def XOR(a,b):
  c=""
  for i,j in zip(a,b):
    if i==j:
      c+='0'
    else:
      c+='1'
  return [c]

# 二进制转字符
def ASCII(A):
  text = ''
  for i in A:
    for j in range(8):
      b = i[j*8:(j+1)*8]
      text += chr(int(b,2))
  return text

##############################################################################

#################################密钥生成######################################
# 先PC1置换、将56位密钥对半分L0和R0、分别对L0和R0进行左循环移位,
# (当轮数为第1、2、9、16轮时,移动1位,其余时候移动两位)L0,R0移动1位
# 后得到L1,R1,L1+R0进行PC2置换得到密钥K1,L1和R0继续进行下一轮,直到生成16个子密钥

# PC-1置换表
PC1 = [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,
    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]

# PC-2置换表
PC2 = [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]

# 生成子密钥函数
def GenerateSubkey(Key):
  # 字符串转二进制
  K = ""
  i_byte = bytes(Key, encoding='utf-8')
  for i_bin in i_byte:
    K += INT_BIN(i_bin)

  # PC1置换
  ReplacePc1 = Replace([K],PC1)

  # 生成16组子密钥
  Lmi = []
  Rmi = []
  Lmi.append(ReplacePc1[0][:28])
  Rmi.append(ReplacePc1[0][28:])
  for i in range(1,17):
    if i in (1, 2, 9, 16):
      Lmi.append(Lmi[i-1][1:]+Lmi[i-1][:1])
      Rmi.append(Rmi[i-1][1:]+Rmi[i-1][:1])
    else:
      Lmi.append(Lmi[i-1][2:]+Lmi[i-1][:2])
      Rmi.append(Rmi[i-1][2:]+Rmi[i-1][:2])
  del Lmi[0]
  del Rmi[0]
  del ReplacePc1[0]
  for i in range(16):
    ReplacePc1.append(Lmi[i]+Rmi[i])

  # PC2置换
  return Replace(ReplacePc1,PC2)

###########################################################################

#################################明文处理###################################
# 明文填充,采用PKCS #5规则,如果刚好满足每组有8个字节,则再添加一组,每个字节为
# 000010000,如果最后一组没有8个字节,则把这一组填充成8个字节,填充的字节为少掉的
# 字节的数目,比如有7个字节,则填充00000001

# 对明文进行填充,分组
def InitPlaintext(Plaintext):
  DecimalList = []
  BytesList = []
  BinList = []
  
  # 字符串转成10机制
  i_byte = bytes(Plaintext, encoding='utf-8')
  for i_bin in i_byte:
    DecimalList.append(i_bin)

  # 刚好满足分组
  if len(DecimalList) % 8 == 0:
    for i in range(8):
      DecimalList.append(8)
    for i in range(int(len(DecimalList)/8)):
      BytesList.append(DecimalList[i*8:(i+1)*8])
  # 不满足分组
  else:
    INT = 8 - len(DecimalList) % 8
    for i in range(INT):
      DecimalList.append(INT)
    for i in range(int(len(DecimalList)/8)):
      BytesList.append(DecimalList[i*8:(i+1)*8])
  
  # 10进制转2进制
  for i in BytesList:
    TMP = ''
    for j in i:
      TMP += INT_BIN(j)
    BinList.append(TMP)
  
  return BinList

###########################################################################

################################feistel函数################################
#ip初始置换表
IPINIT = [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]

#扩展E置换表
EExten = [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]

#P盒置换表
PBOX = [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]

#逆初始置换表
P1 = [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]

#8个s盒
S_1 = [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]
 
S_2 = [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]
 
S_3 = [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]
 
S_4 = [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]
 
S_5 = [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]
 
S_6 = [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]
 
S_7 = [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]
 
S_8 = [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]

S_9 = []
S_9.append(S_1)
S_9.append(S_2)
S_9.append(S_3)
S_9.append(S_4)
S_9.append(S_5)
S_9.append(S_6)
S_9.append(S_7)
S_9.append(S_8)

# S盒置换
def S(R):
  s = ''
  for i in range(8):
    a = R[i*6:(i+1)*6]
    x = int(a[0]+a[-1],2)
    y = int(a[1:5],2)
    s += INT_BIN(S_9[i][x*15+y])[4:]
  return[s]


# feistel函数
def feistel(L, R, K):
  # 扩展置换
  Expand = Replace(R,EExten)
  # 异或运算
  Expand = XOR(Expand[0],K)
  # S盒运算
  Expand = S(Expand[0])
  # P盒
  Expand = Replace(Expand,PBOX)
  # 异或运算
  Expand = XOR(L[0],Expand[0])

  return Expand[0]

###########################################################################

################################加、解密函数################################
# 加密
def Encrypt(PlanText,Key):
  # 初始置换
  IP1 = Replace(InitPlaintext(PlanText),IPINIT)
  # 生成子密钥
  SubkeyList = GenerateSubkey(Key)

  # 16轮迭代
  Ciphertext = []
  for i in IP1:
    L = i[:32]
    R = i[32:]
    for k in SubkeyList:
      TMP = feistel([L],[R],k)
      L = R
      R = TMP
    # 逆初始置换
    Ciphertext.append(Replace([R+L],P1)[0])
  return Ciphertext,SubkeyList

# 解密
def Decrypt(Ciphertext,Key):
  # 初始置换
  IP1 = Replace(Ciphertext,IPINIT) 

  # 16轮迭代
  PlanText = []
  for i in IP1:
    L = i[:32]
    R = i[32:]
    for k in Key[::-1]:
      TMP = feistel([L],[R],k)
      L = R
      R = TMP
    # 逆初始置换
    PlanText.append(Replace([R+L],P1)[0])
  return PlanText

###########################################################################

if __name__ == "__main__":
  miwen,miyao = Encrypt('computer','networks')
  print(miwen)
  print(ASCII(Decrypt(miwen,miyao)))

看完上述内容,是不是对Python中如何实现DES加密有进一步的了解,如果还想学习更多内容,欢迎关注亿速云行业资讯频道。

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