Python：SM4

SM4

Issued by the State Password Administration SM4
SM4.0（ Original name SMS4.0） It is a block cipher standard adopted by the government of the people's Republic of China

Realization

Program operation instructions

According to the document above ：

• The packet length of the algorithm is 128 The bit , The key length is 128 The bit . Both encryption algorithm and key expansion algorithm adopt 32 Wheel nonlinear iterative structure . The decryption algorithm has the same structure as the encryption algorithm , Round key only rk Are used in reverse order , The decryption round key is the reverse order of the encryption round key .
• Finally call the function enc() that will do
• Comments in the program are only used to verify the correctness of the algorithm , No impact on final results

Parameters

• Input ： Hexadecimal plaintext string （ Maximum length is 128bit That is, the maximum length of the string is 16） And hexadecimal encryption key string mk（ The maximum length is also 128bit That is, the maximum length of the string is 16）
• Output ： Hex ciphertext string
• For other functions and constants, please refer to the official documentation above

Code

def sbox(a):
s=['d690e9fecce13db716b614c228fb2c05',
'2b679a762abe04c3aa44132649860699',
'9c4250f491ef987a33540b43edcfac62',
'e4b31ca9c908e89580df94fa758f3fa6',
'4707a7fcf37317ba83593c19e6854fa8',
'686b81b27164da8bf8eb0f4b70569d35',
'1e240e5e6358d1a225227c3b01217887',
'd40046579fd327524c3602e7a0c4c89e',
'eabf8ad240c738b5a3f7f2cef96115a1',
'e0ae5da49b341a55ad933230f58cb1e3',
'1df6e22e8266ca60c02923ab0d534e6f',
'd5db3745defd8e2f03ff6a726d6c5b51',
'8d1baf92bbddbc7f11d95c411f105ad8',
'0ac13188a5cd7bbd2d74d012b8e5b4b0',
'8969974a0c96777e65b9f109c56ec684',
'18f07dec3adc4d2079ee5f3ed7cb3948']
b=''
for i in range(4):
r=int(a[2*i],16)
c=int(a[2*i+1],16)
b=b+s[r][c*2:(c+1)*2]
return int(b,16)
def cl(x,i):#Cycle left
x='{:032b}'.format(x)
i=i%32
x=x[i:]+x[:i]
return int(x,2)
fk=[int('A3B1BAC6',16),int('56AA3350',16),int('677D9197',16),int('B27022DC',16)]
ck=[]
for i in range(32):
cki=[]
for j in range(4):
cki.append('{:02x}'.format(((4*i+j)*7)%256))
ck.append(int(''.join(cki),16))
''' ck=['00070e15','1c232a31','383f464d','545b6269', '70777e85','8c939aa1','a8afb6bd','c4cbd2d9', 'e0e7eef5','fc030a11','181f262d','343b4249', '50575e65','6c737a81','888f969d','a4abb2b9', 'c0c7ced5','dce3eaf1','f8ff060d','141b2229', '30373e45','4c535a61','686f767d','848b9299', 'a0a7aeb5','bcc3cad1','d8dfe6ed','f4fb0209', '10171e25','2c333a41','484f565d','646b7279'] '''
def f(x,rk,i):
return x[i]^t(x[i+1]^x[i+2]^x[i+3]^rk)
def t(x):
return l(tau(x))
def t_(x):#t'
return l_(tau(x))
def tau(a):
a='{:08x}'.format(a)
b=sbox(a)
return b
def l(b):
c=b^cl(b, 2)^cl(b, 10)^cl(b, 18)^cl(b, 24)
return c
def l_(b):#l'
c=b^cl(b, 13)^cl(b, 23)
return c
def enc(x,mk):
if len(x)!=32:
x=(32-len(x))*'0'+x
p=[x[:8],x[8:16],x[16:24],x[24:]]
x=p
for i in range(len(x)):
x[i]=int(x[i],16)
rk=kea(mk, fk, ck)
for i in range(32):
#print('rk['+'{:02d}'.format(i)+'] = '+'{:08x}'.format(rk[i]),end=' ')
x.append(f(x,rk[i],i))
#print('x['+'{:02d}'.format(i)+'] = '+'{:08x}'.format(x[-1]))
#print()
y=x[32:][::-1]
for i in range(len(y)):
y[i]='{:08x}'.format(y[i])
y=''.join(y)
return y
def kea(mk,fk,ck):#Key extension algorithm
if len(mk)!=32:
mk=(32-len(mk))*'0'+mk
p=[mk[:8],mk[8:16],mk[16:24],mk[24:]]
mk=p
for i in range(len(mk)):
mk[i]=int(mk[i],16)
k=[]
for i in range(4):
k.append(mk[i]^fk[i])
for i in range(32):
k.append(k[i]^t_(k[i+1]^k[i+2]^k[i+3]^ck[i]))
rk=k[4:]
return rk
if __name__=='__main__':
p=input('Your plaintext(hex):')
mk=input('Your encryption key(hex):')
print()
y=enc(p, mk)
print('Ciphertext:',end=(' '))
for i in range(4):
print(y[i*8:(i+1)*8],end=' ')
print()

Running effect

Follow the example in the documentation ：
Remove the blank space in the plaintext and encryption key during verification
Process parameters are output in comments

• Example 1
  Encrypt a set of plaintext with a key
  bright —— writing ：01234567 89abcdef fedcba98 76543210
  Encryption key ：01234567 89abcdef fedcba98 76543210
  And output round key rk[i] With each round of output status x[i] ：
  
  Finally get the ciphertext ：681edf34 d206965e 86b3e94f 536e4246
  The process related parameters and results are consistent with the documents

• Example 2
  Use the same encryption key to encrypt a group of plaintext repeatedly 1000000 Time
  bright —— writing ：01234567 89abcdef fedcba98 76543210
  Encryption key ：01234567 89abcdef fedcba98 76543210
  
  Get the ciphertext ：595298c7 c6fd271f 0402f804 c33d3f66