Pycrypto

[luca72]

Hello
I have to make an easy operation but reading the pycrypto doc. a never
see AES example
I have to cript this key 'ea523a664dabaa4476d31226a1e3bab0' with the
AES.
Can you help me for make it with pycrypto

Regards Luca

 

[wittempj]

Hello
I have to make an easy operation but reading the pycrypto doc. a never
see AES example
I have to cript this key 'ea523a664dabaa4476d31226a1e3bab0' with the
AES.
Can you help me for make it with pycrypto

Regards Luca

You can do this as follows:

py> from Crypto.Cipher import AES
py> # key has to be 16, 24 or 32 bytes for AES
py> crypt = AES.new('abcdefghijklmnop', AES.MODE_ECB)
# we're lucky, the string to encrypt is a multiple of 16 in length
py> txt = 'ea523a664dabaa4476d31226a1e3bab0'
py> c = crypt.encrypt(txt)
py> c
'w\x81\xe3\xdd\x066\x9eY\xc7\xce~O\x9e\xfb\xef\xfa\xb5\x8a\xac\x7f\xca\x9fl{\xe5\xfd6\x80\xe3\x81%\xb9'
py> crypt.decrypt(c)
'ea523a664dabaa4476d31226a1e3bab0'

see http://www.amk.ca/python/writing/pycrypt for the docs. if you have
to encrypt data which has not a multiple of length 16 you have to pad
it e.g. with spaces, and then strip the decrypt() result.

 

[Laszlo Nagy]

You can do this as follows:

py> from Crypto.Cipher import AES
py> # key has to be 16, 24 or 32 bytes for AES
py> crypt = AES.new('abcdefghijklmnop', AES.MODE_ECB)
# we're lucky, the string to encrypt is a multiple of 16 in length
py> txt = 'ea523a664dabaa4476d31226a1e3bab0'
py> c = crypt.encrypt(txt)
py> c
'w\x81\xe3\xdd\x066\x9eY\xc7\xce~O\x9e\xfb\xef\xfa\xb5\x8a\xac\x7f\xca\x9fl{\xe5\xfd6\x80\xe3\x81%\xb9'
py> crypt.decrypt(c)
'ea523a664dabaa4476d31226a1e3bab0'

see http://www.amk.ca/python/writing/pycrypt for the docs. if you have
to encrypt data which has not a multiple of length 16 you have to pad
it e.g. with spaces, and then strip the decrypt() result.

Or use CBC mode? I'm not familiar with pycrypto but I know that CBC mode
can crypt/decrypt text with any size.

Laszlo

 

[wittempj]

Or use CBC mode? I'm not familiar with pycrypto but I know that CBC mode
can crypt/decrypt text with any size.

Laszlo

Not in this implementation:
py> from Crypto.Cipher import AES
py> crypt = AES.new('abcdefghijklmnop', AES.MODE_CBC)
py> c = crypt.encrypt('1')
Traceback (most recent call last):
File "<stdin>", line 1, in ?
ValueError: Input strings must be a multiple of 16 in length

 

[luca72]

Thanks

Luca

 

[Laszlo Nagy]

Not in this implementation:
py> from Crypto.Cipher import AES
py> crypt = AES.new('abcdefghijklmnop', AES.MODE_CBC)
py> c = crypt.encrypt('1')
Traceback (most recent call last):
File "<stdin>", line 1, in ?
ValueError: Input strings must be a multiple of 16 in length

This is strange. In theory, any ECB mode cipher can be used to create a
CBC mode cipher.
AFAIK, CBC creates one encrypted block, and uses the one byte from the
plain text to xor it
with the last encrypted byte. Finally it shifts the encrypted block.
This way each input byte will
have a corresponding output byte, and there is no size limit for the
plain text.

Frankly, I could write the CBC mode cipher using the (already existing)
ECB cipher. Why we have this limitation?

Laszlo

 

[luca72]

Excuse me again,
If the string is not a sting but hex number how i have to proced :

look this page:
http://www.cs.eku.edu/faculty/styer/460/Encrypt/JS-AES.html

Regards Luca

 

[Philippe Martin]

Hi,

Look at the bin2ascii module.

Philippe

 

[luca72]

Hello again i have solve doing this:

from Crypto.Cipher import AES
stri=(chr(int('9b',16))+chr(int('d3',16))+chr(int('2d',16))+chr(int('24',16))+chr(int('af',16))+chr(int('c9',16))+chr(int('e9',16))+chr(int('d7',16))+chr(int('46',16))+chr(int('69',16))+chr(int('71',16))+chr(int('32',16))+chr(int('45',16))+chr(int('5f',16))+chr(int('27',16))+chr(int('0b',16)))
luca = str(stri)
crypt = AES.new(luca, AES.MODE_ECB)
testo=(chr(int('ea',16))+chr(int('52',16))+chr(int('3a',16))+chr(int('66',16))+chr(int('4d',16))+chr(int('ab',16))+chr(int('aa',16))+chr(int('44',16))+chr(int('76',16))+chr(int('d3',16))+chr(int('12',16))+chr(int('26',16))+chr(int('a1',16))+chr(int('e3',16))+chr(int('ba',16))+chr(int('b0',16)))
testo = str(testo)
c = crypt.encrypt(testo)

I don't know if this is the best way , but anyway it work

Regards

Luca

 

[Marc 'BlackJack' Rintsch]

Hello again i have solve doing this:

from Crypto.Cipher import AES
stri=(chr(int('9b',16))+chr(int('d3',16))+chr(int('2d',16))+chr(int('24',16))+chr(int('af',16))+chr(int('c9',16))+chr(int('e9',16))+chr(int('d7',16))+chr(int('46',16))+chr(int('69',16))+chr(int('71',16))+chr(int('32',16))+chr(int('45',16))+chr(int('5f',16))+chr(int('27',16))+chr(int('0b',16)))
luca = str(stri)
crypt = AES.new(luca, AES.MODE_ECB)
testo=(chr(int('ea',16))+chr(int('52',16))+chr(int('3a',16))+chr(int('66',16))+chr(int('4d',16))+chr(int('ab',16))+chr(int('aa',16))+chr(int('44',16))+chr(int('76',16))+chr(int('d3',16))+chr(int('12',16))+chr(int('26',16))+chr(int('a1',16))+chr(int('e3',16))+chr(int('ba',16))+chr(int('b0',16)))
testo = str(testo)
c = crypt.encrypt(testo)

I don't know if this is the best way , but anyway it work

In [26]:import binascii

In [27]:binascii.unhexlify('ea523a664dabaa4476d31226a1e3bab0')
Out[27]:'\xeaR:fM\xab\xaaDv\xd3\x12&\xa1\xe3\xba\xb0'

Ciao,
Marc 'BlackJack' Rintsch

 

[James Stroud]

This is strange. In theory, any ECB mode cipher can be used to create a
CBC mode cipher.
AFAIK, CBC creates one encrypted block, and uses the one byte from the
plain text to xor it
with the last encrypted byte. Finally it shifts the encrypted block.
This way each input byte will
have a corresponding output byte, and there is no size limit for the
plain text.

Frankly, I could write the CBC mode cipher using the (already existing)
ECB cipher. Why we have this limitation?

Laszlo

CBC mode is cipher block chaining, so it still works as a block cipher,
which means that it must be that len(text) % block_size == 0. In other
words, CBC does not shift by one byte but by block_size bytes. See:

http://en.wikipedia.org/wiki/Block_cipher_mode_of_operation

James

--
James Stroud
UCLA-DOE Institute for Genomics and Proteomics
Box 951570
Los Angeles, CA 90095

http://www.jamesstroud.com/

 

[luca72]

In [26]:import binascii

In [27]:binascii.unhexlify('ea523a664dabaa4476d31226a1e3bab0')
Out[27]:'\xeaR:fM\xab\xaaDv\xd3\x12&\xa1\xe3\xba\xb0'

Ciao,
Marc 'BlackJack' Rintsch

Ciao Marc

Grazie Thanks

 

[K.S.Sreeram]

ValueError: Input strings must be a multiple of 16 in length

As James Stroud noted, a CBC mode cipher is still a block cipher, and
the input *must* be a multiple of the block size.

OpenSSL provides a standard padding mechanism so that there are no input
size limitations for any cipher.

Have a look at http://tachyon.in/ncrypt/
It provides access to the OpenSSL ciphers (including padding).

Regards
Sreeram


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[luca72]

Hello again

You know if is possible save all the encryption process in a text file
and not only the result?

This will be wery helpful for compare the pycrypto step by step
operation, with the hand made operation and see where hand made make a
mistake