Security and Cryptography in Python - Block Cipher(1)
Security and Cryptography in Python - Block Cipher(1)
DES
https://en.wikipedia.org/wiki/Data_Encryption_Standard
GOST
https://en.wikipedia.org/wiki/GOST_(block_cipher)
pyDes
https://github.com/twhiteman/pyDes/blob/master/pyDes.py
DES
ECB mode
from pyDes import *
message = "0123456701234567"
key = "DESCRYPT"
iv = bytes([0]*8)
k = des(key, ECB, iv, pad=None, padmode=PAD_PKCS5)
cipher = k.encrypt(message)
print("Length of plain text:", len(message ))
print("Length of cipher text:", len(cipher))
print("Encrypted:", cipher[0:8])
print("Encrypted:", cipher[8:16])
print("Encrypted:", cipher[16:])
message = k.decrypt(cipher)
print("Decrypted:", message)
CBC Mode
from pyDes import *
message = "0123456701234567"
key = "DESCRYPT"
iv = bytes([0]*8)
k = des(key, CBC, iv, pad=None, padmode=PAD_PKCS5)
cipher = k.encrypt(message)
print("Length of plain text:", len(message ))
print("Length of cipher text:", len(cipher))
print("Encrypted:", cipher[0:8])
print("Encrypted:", cipher[8:16])
print("Encrypted:", cipher[16:])
message = k.decrypt(cipher)
print("Decrypted:", message)
DES and modifying the cipher texts - ECB Mode
from pyDes import *
def modify(cipher):
mod = [0]*len(cipher)
mod[10] = ord(' ') ^ ord('1')
mod[11] = ord(' ') ^ ord('0')
mod[12] = ord('1') ^ ord('0')
return bytes([mod[i] ^ cipher[i] for i in range(len(cipher))])
message = "Give Bob: 10$"
key = "DESCRYPT"
iv = bytes([0]*8)
k = des(key, ECB, iv, pad=None, padmode=PAD_PKCS5)
# Alice sending the encrypted message
cipher = k.encrypt(message)
print("Length of plain text:", len(message ))
print("Length of cipher text:", len(cipher))
print("Encrypted:", cipher)
# Bob modifying the cipher text
cipher = modify(cipher)
# this is the bank decrypting the message
message = k.decrypt(cipher)
print("Decrypted:", message)
Running Result:
Change the modified block and run it again:
Change the modified block again and run it again:
DES and modifying the cipher texts - CBC Mode
from pyDes import *
def modify(cipher):
mod = [0]*len(cipher)
mod[9] = 1
return bytes([mod[i] ^ cipher[i] for i in range(len(cipher))])
message = "Give Bob: 10$ and send them to him"
key = "DESCRYPT"
iv = bytes([0]*8)
k = des(key, CBC, iv, pad=None, padmode=PAD_PKCS5)
# Alice sending the encrypted message
cipher = k.encrypt(message)
print("Length of plain text:", len(message ))
print("Length of cipher text:", len(cipher))
print("Encrypted:", cipher)
# Bob modifying the cipher text
cipher = modify(cipher)
# this is the bank decrypting the message
message = k.decrypt(cipher)
print("Decrypted:", message)
Running Result:
相信未来 - 该面对的绝不逃避,该执著的永不怨悔,该舍弃的不再留念,该珍惜的好好把握。