一个人炫耀什么，说明内心缺少什么。
算法分析
1.DES是一个对称密码体制，加密解密使用同一秘钥，有效密钥长度为56比特。
2.DES是一个分组密码算法，明文分组和密文分组长度均为64比特。
3.DES使用Feistel结果，具有加密相似特性，加解密算法相同，只是解密子密钥与加密子密钥的使用顺序相反。
4.DES由初始置换，16轮迭代，逆初始置换组成。
5.具体过程如下：
- 64位明文经过初始置换（IP）而重新排列，并将其分为左右两个分组L0和R0，各为32位。
- 在秘钥的参与下，对左右两个分组进行16轮相同轮函数的迭代。最后一轮输出为64位，且其左半部分和右半部分不进行交换。
- 最后的与输出再通过初始逆置换产生64位密文。
算法实现
import re
# Permutation and translation tables for DES
# 压缩置换矩阵 从64位里选56位
pc1 = [56, 48, 40, 32, 24, 16, 8,
0, 57, 49, 41, 33, 25, 17,
9, 1, 58, 50, 42, 34, 26,
18, 10, 2, 59, 51, 43, 35,
62, 54, 46, 38, 30, 22, 14,
6, 61, 53, 45, 37, 29, 21,
13, 5, 60, 52, 44, 36, 28,
20, 12, 4, 27, 19, 11, 3
]
# number left rotations of pc1
shifttimes = [
1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1
]
# permuted choice key (table 2)
# 压缩置换矩阵 从56位里选48位
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]
# The (in)famous S-boxes
# S盒 的置换矩阵
sbox = [
# S1
[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],
# S2
[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],
# S3
[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],
# S4
[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],
# S5
[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],
# S6
[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],
# S7
[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],
# S8
[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],
]
# 32-bit permutation function P used on the output of the S-boxes
# P置换的置换矩阵
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]
# initial permutation IP
# IP置换的 置换矩阵
ip = [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]
# Expansion table for turning 32 bit blocks into 48 bits
# E扩展置换矩阵
e_expansion = [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]
# final permutation IP^-1
# IP逆置换矩阵
ipreverse = [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]
# IP置换
def IP(text):
#如果长度不是64位 就退出
assert len(text) == 64
t = ""
#通过循环 进行IP置换
for i in ip :
t += text[i - 1]
return t
# 秘钥移位
def shift(s, times):
s = s[times:] + s[0:times]
return s
# 生成子秘钥
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 = pc1[0:28]
Rlist = pc1[28:56]
# 初始生成 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
# 定义返回的subkey
subkey = []
# 轮函数生成 48位密钥开始轮置换
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, shifttimes[i])
R0 = shift(R0, shifttimes[i])
#合并左右部分
merge = 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 pc2:
tempkey += merge[i - 1]
assert len(tempkey) == 48
# 加入生成子密钥
subkey.append(tempkey)
return subkey
# E扩展
def E_expand(Ri):
expand = ""
for i in e_expansion:
expand += Ri[i - 1]
assert len(expand) == 48
return expand
# S盒代换
def S_replace(s):
# 从第二位开始的子串,去掉0X
s = bin(s)[2:]
while len(s) < 48:
s = "0" + s
index = 0
replace = ""
for S in sbox:
# 输入的高低两位做为行数row
row = int(s[index] + s[index + 5], base=2)
# 中间四位做为列数L
col = int(s[index + 1:index + 5], base=2)
# 得到result的单个四位输出
out = bin(S[row * 16 + col])[2:]
# 尾数为0，要把0补全
while len(out) < 4:
out = "0" + out
# 合并单个输出
replace += out
# index + 6 进入下一个六位输入
index += 6
assert len(replace) == 32
return replace
# P 盒置换
def P(Ri):
t = ""
for i in p:
t += Ri[i - 1]
return t
# Ri_P(P置换后的输出)和Li（上一轮次的L）异或
def requirexor(Ri_P , Li, Ri): # Ri上一轮次的R
# P盒置换的结果与最初的64位分组左半部分L0异或
RI = int(Ri_P, base=2) ^ int(Li, base=2)
RI = bin(RI)[2:] # RI这一轮次的R
while len(RI) < 32:
RI = "0" + RI
assert len(RI) == 32
# 这一轮次的L即为上一轮次的R，即LI=Ri
LI = Ri
return (LI, RI)
# IP逆置换
def IP_reverse(L16, R16):
t = L16 + R16
res = ""
for i in ipreverse:
res += t[i - 1]
assert len(res) == 64
return res
# 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"):
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 把子密钥次序反过来 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_expand(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 = S_replace(S_Input)
# P盒置换
tem = P(S_sub)
# 获得下一轮的Ln和Rn
(Ln, Rn) = requirexor(tem, Ln, 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
# 所以要置换 一次
(Ln, Rn) = (Rn, Ln)
# IP逆置换得到密文
re_text = IP_reverse(Ln, Rn)
return re_text
# 字符串转换为二进制字符串
def strtobin(s):
res = []
for c in s:
tem = bin(ord(c)).replace('b', '')
# 转为字符串时，后7位中，如果存在前面为0，会自动去掉，需要加回来，使之满足8位
if len(tem) < 8:
tem = "0" + tem
res.append(tem)
return ''.join(res)
# 二进制转字符串
def bintostr(s):
tem = ""
for i in s:
tem += str(chr(int(i, base=2)))
return tem
# 将明文字符串分割为指定长度字符串并存于列表中
def cut_text(text, lenth):
tem = re.findall('.{' + str(lenth) + '}', text)
tem.append(text[(len(tem) * lenth):])
# 由于分割后，末尾出现一个空字符，故去掉
result = [i for i in tem if i != '']
return result
if __name__ == "__main__":
# 秘钥，长度必须为64位
key = "asdfghjk"
#明文
plaintext = "ifnottothesunforsmilingwarmisstillinthesuntherebutwewilllaughmoreconfidentcalmifturnedtofoundhisownshadowappropriateescapethesunwillbethroughtheheartwarmeachplacebehindthecornerifanoutstretchedpalmcannotfallbutterflythenclenchedwavingarmsgivenpowerificanthavebrightsmileitwillfacetothesunshineandsunshinesmiletogetherinfullbloom"
print("明文为： \n" + plaintext)
# 加密
key = strtobin(key)
# 当明文长度不为64位的倍数时，填充空格满足条件
while len(plaintext) % 8 != 0:
plaintext += " "
# 将明文分割成每组8字节，即64位
mlist = cut_text(plaintext, 8)
ciphertext = ""
for t in mlist:
mtext = strtobin(t)
# 对每组明文分别加密
ciphertext += DES(mtext, key)
# 将结果转换成16进制表示，并去掉头部0X
print("加密后的密文为： \n" + hex(int(ciphertext, base=2))[2:].upper())
#解密
ctext = ""
# 加密得到的密文为二进制，故直接分割为64位每组
clist = cut_text(ciphertext, 64)
for c in clist:
ctext += DES(c, key, "-1")
# 将二进制字符串按每字节分割
result = cut_text(ctext, 8)
# 转换为相应字符并合并成字符串
ans = bintostr(result)
print("解密后的明文为： \n" + ans.rstrip())
加密与解密
当key = "asdfghjk"，时，加密与解密如图所示：
安全性分析
如今，DES的安全性已经无法满足需求（三重DES安全性能达到要求，但存在许多缺陷）故存在许多可行攻击。
1.固有的安全问题：互补性和弱密钥问题。
- 互补性： 对明文和密钥逐位取补，则加密后的密文同样为原密文的补。故互补性使得DES在选择明文攻击下所需的工作量减半。
- 弱密钥：初始密钥会生成16个相同的子密钥，这样的弱密钥有
0x0101010101010101
0xFEFEFEFEFEFEFEFE
0xE0E0E0E0F1F1F1F1
0x1F1F1F1F0E0E0E0E
- 半弱密钥：即用K2加密明文，可以用K1解密，这种半弱密钥有：
0x011F011F010E010E:0x1F011F010E010E01
0x01E001E001F101F1:0xE001E001F101F101
0x01FE01FE01FE01FE:0xFE01FE01FE01FE01
0x1FE01FE00EF10EF1:0xE01FE01FF10EF10E
0x1FFE1FFE0EFE0EFE:0xFE1FFE1FFE0EFE0E
0xE0FEE0FEF1FEF1FE:0xFEE0FEE0FEF1FEF1
- 还有四分之一弱密钥，八分之一弱密钥等。这些秘钥的安全性很差，故一般生成秘钥后，都x需要进行弱密码检查。
2.穷举攻击：如今，穷举时间已经减少到不足24小时。
3.差分分析：这种方法对破译16轮的DES不能提供一种实用的方法，但对破译轮数较低的DES是很成功的。
4.线性分析：用这种方法破译DES比差分分析方法更有效。可用2^47个已知明文破译8－轮DES。