Python AES-CBC implementation using AES-ECB
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Python AES-CBC implementation using AES-ECB
$begingroup$
I am looking for ways to make this code more "pythonic" and any issues with my implementation. This was designed to solve Crytopal's Challenge 10. This challenge involves recreating the AES-CBC cipher using a library-provided AES-ECB function.
from Cryptodome.Cipher import AES
def decrypt_aes_ecb(data, key):
cipher = AES.new(key, AES.MODE_ECB)
return cipher.decrypt(data)
def encrypt_aes_ecb(data, key):
cipher = AES.new(key, AES.MODE_ECB)
return cipher.encrypt(data)
def pkcs7(val, block_size=16):
remaining = block_size - len(val) % block_size
if remaining == block_size:
remaining = 16
ret = val + chr(remaining).encode() * remaining
return ret
def unpkcs7(val, block_size=16):
pad_amount = val[-1]
if pad_amount == 0:
raise Exception
for i in range(len(val) - 1, len(val) - (pad_amount + 1), -1):
if val[i] != pad_amount:
raise Exception
return val[:-pad_amount]
def decrypt_aes_cbc(data, key, iv = b'x00' * 16, pad=True):
prev_chunk = iv
decrypted = []
for i in range(0, len(data), 16):
chunk = data[i : i + 16]
decrypted += xor(decrypt_aes_ecb(chunk, key), prev_chunk)
prev_chunk = chunk
if pad:
return unpkcs7(bytes(decrypted))
return bytes(decrypted)
def encrypt_aes_cbc(data, key, iv = b'x00' * 16, pad=True):
if pad:
padded = pkcs7(data)
else:
padded = data
prev_chunk = iv
encrypted = []
for i in range(0, len(padded), 16):
chunk = padded[i : i + 16]
encrypted_block = encrypt_aes_ecb(xor(chunk, prev_chunk), key)
encrypted += encrypted_block
prev_chunk = encrypted_block
return bytes(encrypted)
python python-3.x programming-challenge cryptography
asked 2 mins ago
InfuzionInfuzion
101
New contributor
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am looking for ways to make this code more "pythonic" and any issues with my implementation. This was designed to solve Crytopal's Challenge 10. This challenge involves recreating the AES-CBC cipher using a library-provided AES-ECB function.
from Cryptodome.Cipher import AES
def decrypt_aes_ecb(data, key):
cipher = AES.new(key, AES.MODE_ECB)
return cipher.decrypt(data)
def encrypt_aes_ecb(data, key):
cipher = AES.new(key, AES.MODE_ECB)
return cipher.encrypt(data)
def pkcs7(val, block_size=16):
remaining = block_size - len(val) % block_size
if remaining == block_size:
remaining = 16
ret = val + chr(remaining).encode() * remaining
return ret
def unpkcs7(val, block_size=16):
pad_amount = val[-1]
if pad_amount == 0:
raise Exception
for i in range(len(val) - 1, len(val) - (pad_amount + 1), -1):
if val[i] != pad_amount:
raise Exception
return val[:-pad_amount]
def decrypt_aes_cbc(data, key, iv = b'x00' * 16, pad=True):
prev_chunk = iv
decrypted = []
for i in range(0, len(data), 16):
chunk = data[i : i + 16]
decrypted += xor(decrypt_aes_ecb(chunk, key), prev_chunk)
prev_chunk = chunk
if pad:
return unpkcs7(bytes(decrypted))
return bytes(decrypted)
def encrypt_aes_cbc(data, key, iv = b'x00' * 16, pad=True):
if pad:
padded = pkcs7(data)
else:
padded = data
prev_chunk = iv
encrypted = []
for i in range(0, len(padded), 16):
chunk = padded[i : i + 16]
encrypted_block = encrypt_aes_ecb(xor(chunk, prev_chunk), key)
encrypted += encrypted_block
prev_chunk = encrypted_block
return bytes(encrypted)
python python-3.x programming-challenge cryptography
asked 2 mins ago
InfuzionInfuzion
101
New contributor
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am looking for ways to make this code more "pythonic" and any issues with my implementation. This was designed to solve Crytopal's Challenge 10. This challenge involves recreating the AES-CBC cipher using a library-provided AES-ECB function.
from Cryptodome.Cipher import AES
def decrypt_aes_ecb(data, key):
cipher = AES.new(key, AES.MODE_ECB)
return cipher.decrypt(data)
def encrypt_aes_ecb(data, key):
cipher = AES.new(key, AES.MODE_ECB)
return cipher.encrypt(data)
def pkcs7(val, block_size=16):
remaining = block_size - len(val) % block_size
if remaining == block_size:
remaining = 16
ret = val + chr(remaining).encode() * remaining
return ret
def unpkcs7(val, block_size=16):
pad_amount = val[-1]
if pad_amount == 0:
raise Exception
for i in range(len(val) - 1, len(val) - (pad_amount + 1), -1):
if val[i] != pad_amount:
raise Exception
return val[:-pad_amount]
def decrypt_aes_cbc(data, key, iv = b'x00' * 16, pad=True):
prev_chunk = iv
decrypted = []
for i in range(0, len(data), 16):
chunk = data[i : i + 16]
decrypted += xor(decrypt_aes_ecb(chunk, key), prev_chunk)
prev_chunk = chunk
if pad:
return unpkcs7(bytes(decrypted))
return bytes(decrypted)
def encrypt_aes_cbc(data, key, iv = b'x00' * 16, pad=True):
if pad:
padded = pkcs7(data)
else:
padded = data
prev_chunk = iv
encrypted = []
for i in range(0, len(padded), 16):
chunk = padded[i : i + 16]
encrypted_block = encrypt_aes_ecb(xor(chunk, prev_chunk), key)
encrypted += encrypted_block
prev_chunk = encrypted_block
return bytes(encrypted)
python python-3.x programming-challenge cryptography
asked 2 mins ago
InfuzionInfuzion
101
New contributor
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am looking for ways to make this code more "pythonic" and any issues with my implementation. This was designed to solve Crytopal's Challenge 10. This challenge involves recreating the AES-CBC cipher using a library-provided AES-ECB function.
from Cryptodome.Cipher import AES
def decrypt_aes_ecb(data, key):
cipher = AES.new(key, AES.MODE_ECB)
return cipher.decrypt(data)
def encrypt_aes_ecb(data, key):
cipher = AES.new(key, AES.MODE_ECB)
return cipher.encrypt(data)
def pkcs7(val, block_size=16):
remaining = block_size - len(val) % block_size
if remaining == block_size:
remaining = 16
ret = val + chr(remaining).encode() * remaining
return ret
def unpkcs7(val, block_size=16):
pad_amount = val[-1]
if pad_amount == 0:
raise Exception
for i in range(len(val) - 1, len(val) - (pad_amount + 1), -1):
if val[i] != pad_amount:
raise Exception
return val[:-pad_amount]
def decrypt_aes_cbc(data, key, iv = b'x00' * 16, pad=True):
prev_chunk = iv
decrypted = []
for i in range(0, len(data), 16):
chunk = data[i : i + 16]
decrypted += xor(decrypt_aes_ecb(chunk, key), prev_chunk)
prev_chunk = chunk
if pad:
return unpkcs7(bytes(decrypted))
return bytes(decrypted)
def encrypt_aes_cbc(data, key, iv = b'x00' * 16, pad=True):
if pad:
padded = pkcs7(data)
else:
padded = data
prev_chunk = iv
encrypted = []
for i in range(0, len(padded), 16):
chunk = padded[i : i + 16]
encrypted_block = encrypt_aes_ecb(xor(chunk, prev_chunk), key)
encrypted += encrypted_block
prev_chunk = encrypted_block
return bytes(encrypted)
python python-3.x programming-challenge cryptography
python python-3.x programming-challenge cryptography
asked 2 mins ago
InfuzionInfuzion
101
New contributor
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 mins ago
InfuzionInfuzion
101
New contributor
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 mins ago
InfuzionInfuzion
101
New contributor
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 mins ago
InfuzionInfuzion
101
asked 2 mins ago
InfuzionInfuzion
101
101
New contributor
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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