Pycrypto

**wittempj@hotmail.com** · Jun 16 '06, 11:55 AM

Re: Pycrypto

luca72 wrote:[color=blue]
> Hello
> I have to make an easy operation but reading the pycrypto doc. a never
> see AES example
> I have to cript this key 'ea523a664dabaa 4476d31226a1e3b ab0' with the
> AES.
> Can you help me for make it with pycrypto
>
> Regards Luca[/color]

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('abcdef ghijklmnop', AES.MODE_ECB)
# we're lucky, the string to encrypt is a multiple of 16 in length
py> txt = 'ea523a664dabaa 4476d31226a1e3b ab0'
py> c = crypt.encrypt(t xt)
py> c
'w\x81\xe3\xdd\ x066\x9eY\xc7\x ce~O\x9e\xfb\xe f\xfa\xb5\x8a\x ac\x7f\xca\x9fl {\xe5\xfd6\x80\ xe3\x81%\xb9'
py> crypt.decrypt(c )
'ea523a664dabaa 4476d31226a1e3b ab0'

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** · Jun 16 '06, 12:15 PM

Re: Pycrypto

[color=blue]
> 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('abcdef ghijklmnop', AES.MODE_ECB)
> # we're lucky, the string to encrypt is a multiple of 16 in length
> py> txt = 'ea523a664dabaa 4476d31226a1e3b ab0'
> py> c = crypt.encrypt(t xt)
> py> c
> 'w\x81\xe3\xdd\ x066\x9eY\xc7\x ce~O\x9e\xfb\xe f\xfa\xb5\x8a\x ac\x7f\xca\x9fl {\xe5\xfd6\x80\ xe3\x81%\xb9'
> py> crypt.decrypt(c )
> 'ea523a664dabaa 4476d31226a1e3b ab0'
>
> 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.
>[/color]
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@hotmail.com** · Jun 16 '06, 12:55 PM

Re: Pycrypto

Laszlo Nagy wrote:[color=blue][color=green]
> > 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('abcdef ghijklmnop', AES.MODE_ECB)
> > # we're lucky, the string to encrypt is a multiple of 16 in length
> > py> txt = 'ea523a664dabaa 4476d31226a1e3b ab0'
> > py> c = crypt.encrypt(t xt)
> > py> c
> > 'w\x81\xe3\xdd\ x066\x9eY\xc7\x ce~O\x9e\xfb\xe f\xfa\xb5\x8a\x ac\x7f\xca\x9fl {\xe5\xfd6\x80\ xe3\x81%\xb9'
> > py> crypt.decrypt(c )
> > 'ea523a664dabaa 4476d31226a1e3b ab0'
> >
> > 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.
> >[/color]
> Or use CBC mode? I'm not familiar with pycrypto but I know that CBC mode
> can crypt/decrypt text with any size.
>
> Laszlo[/color]

Not in this implementation:
py> from Crypto.Cipher import AES
py> crypt = AES.new('abcdef ghijklmnop', 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** · Jun 16 '06, 01:25 PM

Re: Pycrypto

Thanks

Luca

**Laszlo Nagy** · Jun 16 '06, 01:35 PM

Re: Pycrypto

[color=blue]
> Not in this implementation:
> py> from Crypto.Cipher import AES
> py> crypt = AES.new('abcdef ghijklmnop', 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
>[/color]
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** · Jun 16 '06, 02:05 PM

Re: Pycrypto

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

look this page:

Page not found - Eastern Kentucky University

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

Regards Luca

**Philippe Martin** · Jun 16 '06, 02:55 PM

Re: Pycrypto

Hi,

Look at the bin2ascii module.

Philippe

luca72 wrote:
[color=blue]
>
> 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/...pt/JS-AES.html
>
> Regards Luca[/color]

**luca72** · Jun 16 '06, 03:15 PM

Re: Pycrypto

Hello again i have solve doing this:

from Crypto.Cipher import AES
stri=(chr(int(' 9b',16))+chr(in t('d3',16))+chr (int('2d',16))+ chr(int('24',16 ))+chr(int('af' ,16))+chr(int(' c9',16))+chr(in t('e9',16))+chr (int('d7',16))+ chr(int('46',16 ))+chr(int('69' ,16))+chr(int(' 71',16))+chr(in t('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(i nt('52',16))+ch r(int('3a',16)) +chr(int('66',1 6))+chr(int('4d ',16))+chr(int( 'ab',16))+chr(i nt('aa',16))+ch r(int('44',16)) +chr(int('76',1 6))+chr(int('d3 ',16))+chr(int( '12',16))+chr(i nt('26',16))+ch r(int('a1',16)) +chr(int('e3',1 6))+chr(int('ba ',16))+chr(int( 'b0',16)))
testo = str(testo)
c = crypt.encrypt(t esto)

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

Regards

Luca

**Marc 'BlackJack' Rintsch** · Jun 16 '06, 04:35 PM

Re: Pycrypto

In <1150470895.257 327.302460@h76g 2000cwa.googleg roups.com>, luca72 wrote:
[color=blue]
> Hello again i have solve doing this:
>
> from Crypto.Cipher import AES
> stri=(chr(int(' 9b',16))+chr(in t('d3',16))+chr (int('2d',16))+ chr(int('24',16 ))+chr(int('af' ,16))+chr(int(' c9',16))+chr(in t('e9',16))+chr (int('d7',16))+ chr(int('46',16 ))+chr(int('69' ,16))+chr(int(' 71',16))+chr(in t('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(i nt('52',16))+ch r(int('3a',16)) +chr(int('66',1 6))+chr(int('4d ',16))+chr(int( 'ab',16))+chr(i nt('aa',16))+ch r(int('44',16)) +chr(int('76',1 6))+chr(int('d3 ',16))+chr(int( '12',16))+chr(i nt('26',16))+ch r(int('a1',16)) +chr(int('e3',1 6))+chr(int('ba ',16))+chr(int( 'b0',16)))
> testo = str(testo)
> c = crypt.encrypt(t esto)
>
> I don't know if this is the best way , but anyway it work[/color]

In [26]:import binascii

In [27]:binascii.unhex lify('ea523a664 dabaa4476d31226 a1e3bab0')
Out[27]:'\xeaR:fM\xab\ xaaDv\xd3\x12&\ xa1\xe3\xba\xb0 '

Ciao,
Marc 'BlackJack' Rintsch

**James Stroud** · Jun 16 '06, 08:55 PM

Re: Pycrypto

Laszlo Nagy wrote:[color=blue]
>[color=green]
>> Not in this implementation:
>> py> from Crypto.Cipher import AES
>> py> crypt = AES.new('abcdef ghijklmnop', 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
>>[/color]
>
> 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
>
>[/color]

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:

Block cipher mode of operation - Wikipedia

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** · Jun 16 '06, 09:55 PM

Re: Pycrypto

[color=blue]
> In [26]:import binascii
>
> In [27]:binascii.unhex lify('ea523a664 dabaa4476d31226 a1e3bab0')
> Out[27]:'\xeaR:fM\xab\ xaaDv\xd3\x12&\ xa1\xe3\xba\xb0 '
>
> Ciao,
> Marc 'BlackJack' Rintsch[/color]

Ciao Marc

Grazie Thanks

**K.S.Sreeram** · Jun 17 '06, 01:45 PM

Re: Pycrypto

wittempj@hotmai l.com wrote:[color=blue]
> ValueError: Input strings must be a multiple of 16 in length[/color]

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** · Jun 26 '06, 08:05 PM

Re: Pycrypto

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