一、DES

1 概述

DES算法全称为Data Encryption Standard，即数据加密算法，它是IBM公司于1975年研究成功并公开发表的。DES算法的入口参数有三个：Key、Data、Mode。其中Key为8个字节共64位，是DES算法的工作密钥；Data也为8个字节64位，是要被加密或被解密的数据；Mode为DES的工作方式,有两种：加密或解密。

2 算法原理

DES算法把64位的明文输入块变为64位的密文输出块，它所使用的密钥也是64位，其算法主要分为两步：

（1）初始置换

其功能是把输入的64位数据块按位重新组合,并把输出分为L0、R0两部分，每部分各长32位，其置换规则为将输入的第58位换到第一位，第50位换到第2位……依此类推,最后一位是原来的第7位。L0、R0则是换位输出后的两部分，L0是输出的左32位，R0是右32位，例：设置换前的输入值为D1D2D3……D64，则经过初始置换后的结果为:L0=D58D50……D8；R0=D57D49……D7。

（2）逆置换

经过16次迭代运算后，得到L16、R16,将此作为输入，进行逆置换，逆置换正好是初始置换的逆运算，由此即得到密文输出。

二、两个密钥的三重DES

由于DES密钥只有56bit，易于遭受穷举时攻击。作为一种替代加密方案，Tuchman提出使用两个密钥的三重DES加密方法，并在1985年成为美国的一个商用加密标准。该方法使用两个密钥，执行三次DES算法，如图2所示。加密的过程是加密-解密-加密，解密的过程是解密-加密-解密。

采用两个密钥进行三重加密的好处有：①两个密钥合起来有效密钥长度有112bit，可以满足商业应用的需要，若采用总长为168bit的三个密钥，会产生不必要的开销。②加密时采用加密-解密-加密，而不是加密-加密-加密的形式，这样有效的实现了与现有DES系统的向后兼容问题。因为当K1=K2时，三重DES的效果就和原来的DES一样，有助于逐渐推广三重DES。③三重DES具有足够的安全性，目前还没有关于攻破三重DES的报道。

什么是三重DES

三重DES是为了增加DES的强度，将DES重复3次所得到的一种密码算法，通常缩写为3DES。

二三重DSE加密

明文经过三次DES处理才能变成最后的密文，由于DES密钥长度实质是56位，因此3DES的密钥长度就是56*3=168比特。

注意：三重DES的三次DES加密是（加密->解密->加密），而不是（加密->加密->加密）,为什么这样设计呢？看看下面这张图就明白了。

如果三次DES加密的密钥都相同，前两次DES，相当于还原成了密文，这样看来，真正起作用的只是最后一次。这样做的好处是：三重DES对DES具备向下兼容性。

还有一种密码叫DES-EDE2,下面我们用图来描述这一种密码。

当然，只有三个密钥互不相等才是正宗的DES-EDE3。

三三重DES解密

三重DES的解密过程和加密过程正好相反，是以密钥3、密钥2、密钥1的顺序执行解密->加密->解密的操作。

四三重DES现状

尽管三重DES目前还被银行等机构使用，但其处理速度不高，除了特别重视向下兼容性的情况外，很少被用于新的用途。

pydes:

#-*- coding: utf8 -*-#Initial permut matrix for the datas
# IP initial permut
PI = [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]#Initial permut made on the key
CP_1 = [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]#Permut applied on shifted key to get Ki+1
CP_2 = [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]#Expand matrix to get a 48bits matrix of datas to apply the xor with Ki
E = [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]#SBOX
S_BOX = [[[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],
]
]#Permut made after each SBox substitution for each round
P = [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]#Final permut for datas after the 16 rounds
PI_1 = [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]#Matrix that determine the shift for each round of keys
SHIFT = [1,1,2,2,2,2,2,2,1,2,2,2,2,2,2,1]def string_to_bit_array(text):#Convert a string into a list of bitsarray = list()for char in text:binval = binvalue(char, 8)#Get the char value on one bytearray.extend([int(x) for x in list(binval)]) #Add the bits to the final listreturn arraydef bit_string_to_array(bitText):array = list()for bit in bitText.replace(" ",""):array.append(int(bit))return arraydef bit_array_to_string(array): #Recreate the string from the bit arrayres = ''.join([chr(int(y,2)) for y in [''.join([str(x) for x in _bytes]) for _bytes in  nsplit(array,8)]])return res# 根据ascil编码把字符转成对应的二进制
def binvalue(val, bitsize): #Return the binary value as a string of the given sizebinval = bin(val)[2:] if isinstance(val, int) else bin(ord(val))[2:]if len(binval) > bitsize:raise ("binary value larger than the expected size")while len(binval) < bitsize:binval = "0"+binval #Add as many 0 as needed to get the wanted sizereturn binvaldef nsplit(s, n):#Split a list into sublists of size "n"return [s[k:k+n] for k in range(0, len(s), n)]def showBits(s, n):#Split a list into sublists of size "n"return " ".join(["".join([str(i) for i in s[k:k+n]]) for k in range(0, len(s), n)])ENCRYPT=1
DECRYPT=0class des():def __init__(self):self.password = Noneself.text = Noneself.keys = list()def run(self, key, text, action=ENCRYPT):self.password = keyself.text = text# 产生所有子密钥self.generatekeys() #Generate all the keysresult = list()block = self.textblock = self.permut(block,PI)#Apply the initial permutationg, d = nsplit(block, 32) #g(LEFT), d(RIGHT)tmp = Nonefor i in range(16): #Do the 16 roundsd_e = self.expand(d, E) #Expand d to match Ki size (48bits)if action == ENCRYPT:tmp = self.xor(self.keys[i], d_e)#If encrypt use Kielse:tmp = self.xor(self.keys[15-i], d_e)#If decrypt start by the last keytmp = self.substitute(tmp) #Method that will apply the SBOXestmp = self.permut(tmp, P)tmp = self.xor(g, tmp)   # tmp = f(Rn-1,Kn)g = d   # Ln = Rn-1d = tmp # Rn = Ln-1 + f(Rn-1,Kn)print(f"L{i+1}:{showBits(g, 8)}")print(f"R{i + 1}:{showBits(d, 8)}")result += self.permut(d+g, PI_1) #Do the last permut and append the result to resultfinal_res = "".join([str(i) for i in result])return final_res# if padding and action==DECRYPT:#     return self.removePadding(final_res) #Remove the padding if decrypt and padding is true# else:#     return final_res #Return the final string of data ciphered/deciphereddef substitute(self, d_e):#Substitute bytes using SBOXsubblocks = nsplit(d_e, 6)#Split bit array into sublist of 6 bitsresult = list()for i in range(len(subblocks)): #For all the sublistsblock = subblocks[i]row = int(str(block[0])+str(block[5]),2)#Get the row with the first and last bitcolumn = int(''.join([str(x) for x in block[1:][:-1]]),2) #Column is the 2,3,4,5th bitsval = S_BOX[i][row][column] #Take the value in the SBOX appropriated for the round (i)bin = binvalue(val, 4)#Convert the value to binaryresult += [int(x) for x in bin]#And append it to the resulting listreturn result# 置换def permut(self, block, table):#Permut the given block using the given table (so generic method)return [block[x-1] for x in table]# 扩展def expand(self, block, table):#Do the exact same thing than permut but for more clarity has been renamedreturn [block[x-1] for x in table]def xor(self, t1, t2):#Apply a xor and return the resulting listreturn [x^y for x,y in zip(t1,t2)]def generatekeys(self):#Algorithm that generates all the keysself.keys = []# 把密钥转成这个形式 00110100 11000010.....key = self.passwordkey = self.permut(key, CP_1) #Apply the initial permut on the keyg, d = nsplit(key, 28) #Split it in to (g->LEFT),(d->RIGHT)for i in range(16):#Apply the 16 roundsg, d = self.shift(g, d, SHIFT[i]) #Apply the shift associated with the round (not always 1)tmp = g + d #Merge themself.keys.append(self.permut(tmp, CP_2)) #Apply the permut to get the Kidef shift(self, g, d, n): #Shift a list of the given valuereturn g[n:] + g[:n], d[n:] + d[:n]def addPadding(self):#Add padding to the datas using PKCS5 spec.pad_len = 8 - (len(self.text) % 8)self.text += pad_len * chr(pad_len)def removePadding(self, data):#Remove the padding of the plain text (it assume there is padding)pad_len = ord(data[-1])return data[:-pad_len]def encrypt(self, key, text):return self.run(key, text, ENCRYPT)def decrypt(self, key, text):return self.run(key, text, DECRYPT)if __name__ == '__main__':plain_text = bit_string_to_array("00000001 00100011 01000101 01100111 10001001 10101011 11001101 11101111")key = bit_string_to_array("00010011 00110100 01010111 01111001 10011011 10111100 11011111 11110001")d = des()r = d.encrypt(key,plain_text)print(r)# 测试cyberchefr = bit_string_to_array("01011010 11110101 00010101 10110000 01001011 11101111 01010010 01000001")r2 = d.decrypt(key,r)print("Ciphered: %r" % r)print("Deciphered: ", r2)# 07040502033636363636363636363636363636363636363636363636363636363636363636363636363636363636363636363636363636363636363636363636
#
# 6d6e6f68695c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c5c

l = [1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0]
def showBits(s, n):#Split a list into sublists of size "n"return " ".join(["".join([str(i) for i in s[k:k+n]]) for k in range(0, len(s), n)])print(showBits(l, 8))

CBC模式调用：

from pyDes import des, CBC, PAD_PKCS5
import binascii# 秘钥
KEY = 'Codeoooo'def des_encrypt(s):"""DES 加密:param s: 原始字符串:return: 加密后字符串，16进制"""secret_key = KEY  # 密码iv = secret_key  # 偏移# secret_key:加密密钥，CBC:加密模式，iv:偏移, padmode:填充des_obj = des(secret_key, CBC, iv, pad=None, padmode=PAD_PKCS5)# 返回为字节secret_bytes = des_obj.encrypt(s, padmode=PAD_PKCS5)# 返回为16进制return bytes.decode(binascii.b2a_hex(secret_bytes))def des_descrypt(s):"""DES 解密:param s: 加密后的字符串，16进制:return:  解密后的字符串"""secret_key = KEYiv = secret_keydes_obj = des(secret_key, CBC, iv, pad=None, padmode=PAD_PKCS5)decrypt_str = des_obj.decrypt(binascii.a2b_hex(s), padmode=PAD_PKCS5)return bytes.decode(decrypt_str)print(des_encrypt('codeooo'))
print(des_descrypt('4ff67fd9b661f01d'))

ECB模式调用：

from pyDes import des, ECB, PAD_PKCS5
import base64DES_SECRET_KEY = '12345678'
data = 'codeooo'
des_obj = des(DES_SECRET_KEY, ECB, DES_SECRET_KEY, padmode=PAD_PKCS5)
secret_bytes = str(base64.b64encode(des_obj.encrypt(data)), encoding = 'utf-8')
print(secret_bytes)

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DES对称加密及三重DES（3DES）

一、DES

1 概述

2 算法原理

二、两个密钥的三重DES

CBC模式调用：

ECB模式调用：

DES对称加密及三重DES（3DES）相关推荐

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DES对称加密及三重DES（3DES）

一、DES

1 概述

2 算法原理

二、 两个密钥的三重DES

CBC模式调用：

ECB模式调用：

DES对称加密及三重DES（3DES）相关推荐

最新文章

热门文章

二、两个密钥的三重DES