密码学DES加密
成长路上的必经之路——debug
-------------------------------------------------------------4.18更新------------------------------------------------------------
本次更新主要增加了明文的长度，之前版本明文长度只有8字节，现在长度为任意，适用性更广。
明文m是64bit序列。
初始密钥K是64 bit序列(含8个奇偶校验bit)。
子密钥K1, K2…K16均是48 bit序列。
轮变换函数f (A,J)：输入A(32 bit序列), J(48 bit序列)，输出32 bit序列。
密文c是64 bit序列。
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子密钥生成：
输入初始密钥，生成16轮子密钥K1, K2…K16。 初始密钥(64bit)经过置换PC-1，去掉了8个奇偶校验位，留下56 bit，接着分成两个28 bit的 分组C0与D0，再分别经过一个循环左移函数LS1，得到C1与D1，连成56 bit数据，然后经过置换PC-2，输出子密钥K1，以此类推产生K2至K16。 注意：置换PC-1、PC-2会在下文中会具体给出；
函数LSi为向左循环移动 1个位置(i=1,2,9,16)2个位置(i=3,4,5,6,7,8,10,11,12,13,14,15)。详见下图：
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轮变换函数f：
f是DES加解密中每一轮的核心运算，输入A(32 bit), J(48 bit)，输出32 bit。
将A做一个扩展运算E，变成48 bit，记为E(A)。计算B=E(A)⊕J，将B分为8组B1…B8，每组Bi为6 bit，通过相应的S盒Si，输出Ci为4 bit，将所有Ci连成C(32 bit)，再通过置换P，得到最后的输出f(A,J)，为32 bit。在加密或解密的第i轮，A = Ri-1，J = Ki。
注意：每个S盒Si是4×16的矩阵，输入b0b1b2b3b4b5 (6 bit)，令L是b0b5对应的十进制数，n是b1b2b3b4对应的十进制数，输出矩阵中第L行n列所对应数的二进制表示。
详见下图：
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加/解密： DES的加密和解密几乎一样，不同之处在于加密时输入是明文，子密钥使
用顺序为K1K2…K16；解密时输入是密文，子密钥使用顺序为K16K15…K1。 以加密为例，输入明文分组64 bit，先进行一次初始置换IP，对置换后的数据X0分成左右两半L0与R0，根据第一个子密钥K1对R0实行轮变换f(R0, K1)，将结果与L0作逐位异或运算，得到的结果成为下一轮的R1，R0则成为下一轮的L1。如此循环16次，最后得到L16与R16。可用下列公式简洁地表示：
最后对64 bit数据R16, L16实行IP的逆置换IP-1，即得密文分组。 注意：第16轮变换后并未交换L16与R16，而直接将R16, L16作为IP-1的输入。 详见下图：
代码如下：
import re
IP_Table = [ # 定义IP置换表
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]
IPInv_Table = [ # 定义IP逆置换表
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]
E_Table = [ # 定义E扩展表
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_Table = [ # 定义P置换表
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]
PC1_Table = [ # 定义PC1置换表
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]
PC2_Table = [ # 定义PC2置换表
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]
LS_Table = [ # 定义左移位数表
1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1]
S_Box = [ # S盒
# 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]]
RoundK = [] # 轮密钥
def str2bits(str): # √
r = ''
for i in str:
t = bin(ord(i))[2:]
for j in range(0, 8 - len(t)):
t = '0' + t # 把输出的b给去掉
r += t
return (r)
def bin2str(bin_str):
res = ""
tmp = re.findall(r'.{8}', bin_str)
for i in tmp:
res += chr(int(i, 2))
return res
def change_PC_1(k):
r = ''
for i in PC1_Table:
r += k[i - 1]
return r
def change_PC_2(k):
r = ''
for i in PC2_Table:
r += k[i - 1]
return r
def move(k, i): # 循环移位
l1 = [1, 2, 9, 16]
if (i in l1):
k = k[1:] + k[0]
else:
k = k[2:] + k[:2]
return k
def set_key(): # k:64bits
k = str2bits(input('请输入密钥：')) # 转为64bits密钥k
k_pc1 = change_PC_1(k) # 56bits√
for i in range(16):
if (i == 0): # 第一次进入
l = k_pc1[:28] # 分成左右两部分
r = k_pc1[28:]
l = move(l, i + 1) # 左移
r = move(r, i + 1)
k_pc2 = change_PC_2(l + r) # 48bits
RoundK.append(k_pc2) # l加入到轮密钥列表中
else:
l = move(l, i + 1)
r = move(r, i + 1)
k_pc2 = change_PC_2(l + r) # 48bits
RoundK.append(k_pc2) # l加入到轮密钥列表中
def E_extend(A): # 32bits扩展为48bits
r = ''
for i in E_Table:
r += A[i - 1]
return r
def xor(str1, str2): # 异或运算
r = ''
for i in range(len(str1)):
t = int(str1[i]) ^ int(str2[i])
if (int(t) == 1):
r += '1'
else:
r += '0'
return r
def s(str, i): # 输入一个6bit，输出x一个4bit
row = int(str[0] + str[5], 2)
column = int(str[1:5], 2)
r = str2bits(chr(S_Box[i][row * 16 + column]))[4:] # 结果转为4bits
return r
def f(r, k):
C = ''
R = ''
r = E_extend(r) # 扩展运算
B = xor(r, k)
for i in range(8): # B48bits,分成8组，每组6bits
C += s(B[i * 6:6 * (i + 1)], i) # C:32bits
for i in P_Table: # P置换
R += C[i - 1]
return R # 返回一个32bits
def change_IP(str): # 初始置换IP
r = ''
for i in IP_Table:
r += str[i - 1]
return r
def change_Inv_IP(str): # 初始逆置换
r = ''
for i in IPInv_Table:
r += str[i - 1]
return r
def encrypt(m):
#print(m)
m = str2bits(m) # 64bits
m = change_IP(m) # √
l = m[:32]
r = m[32:]
for i in range(15):
t = r
r = xor(l, f(t, RoundK[i]))
l = t
l = xor(l, f(r, RoundK[15]))
mm = change_Inv_IP(l + r) # 初始逆置换
return mm
def decrypt(mm):
c = change_IP(mm)
l = c[:32]
r = c[32:]
for i in range(15):
t = r
r = xor(l, f(t, RoundK[i]))
l = t
l = xor(l, f(r, RoundK[15]))
c = change_Inv_IP(l + r)
#print('明文：' + bin2str(c))
return c
x = int(input('请选择：1、加密 2、解密\n'))
if x == 1:
R=''
m=input('请输入明文(输入#结束)：')
set_key()
while(len(m)%8!=0):
m+='0'
for i in range(0,len(m),8):
t=encrypt(m[i:i+8])
R+=t
print('密文：'+R)
else:
mm = input('请输入密文：')
r=''
set_key()
RoundK.reverse()
for i in range(0,len(mm),64):
t=decrypt(mm[i:i+64])
r+=t
r=bin2str(r)
for i in range(len(r)-1,0,-1):
if r[i]=="#":
r=r[0:i]
break
print('明文：'+r)
写在最后：
写本篇文章的目的是在于纪念这一次漫长的debug之路。用舍友成功验收的代码调试，结果给他找出两个BUG（对称加密用相同的方法加解密）。最后发现只是自己解密的过程中多转换了一次而已…