Re: A Last Word on Ciphertext Expansion Ratio - Promise.

[Dennis Lee Bieber <wlfraed@ix.netcom.com>]

On Sat, 6 Dec 2014 07:45:59 -0800 (PST), Austin Obyrne
<austin.obyrne@hotmail.com> declaimed the following:


>
>Appendix. - The palindrome "able was I ere I saw elba" is encrypted here with this new ciphertext as a demonstration just to show the difference.
>
>New.
>
>322   693  -583  484  883  -817  444  719  -593  87 149  -14  -448 -902  1044  356 686  -648        373   645  -532  467  912  -831  -401 -966 1091  -231 -520  670 -440 -925 1052  87  166        -14    301   612  -535 114   122   13 -446 -963 1046  -291 -549  610 -443 -927 1049  439        934  -859  356  636 -549  391  725  -613  -441 -939  1051  67 121  -34 415  697 -622        502    921  -799  354  612 -551
>

	Danger, Will Robinson, Danger

	The above indicates signed integer output... But that could be 16, 32,
or 64 bit integers internally.

	99.9% of the recognized encryption routines would take your 25-byte
(presuming ASCII) input and give back a 25-byte output. If the output is
then rendered for human reading it may expand to 50 bytes of hex digits, or
75 bytes if the hex bytes are space separated. Nowhere would it expand to
over 200 bytes.

>>> from Crypto.Cipher import DES3
>>> from Crypto import Random
>>> from Crypto.Util import Counter
>>> key = "SubKey01SubKey02SubKey03"
>>> len(key)
24
>>> len(key) * 8
192
>>> nonce = Random.new().read(DES3.block_size/2)
>>> nonce
'\xe2\xf7\\u'
>>> len(nonce)
4
>>> ctr = Counter.new(DES3.block_size*8/2, prefix=nonce)
>>> cipher = DES3.new(key, DES3.MODE_CTR, counter=ctr)
>>> plaintext = "Able was I ere I saw Elba"
>>> msg = nonce + cipher.encrypt(plaintext)
>>> len(msg)
29
>>> len(plaintext)
25
>>> len(plaintext) + len(nonce)
29
>>> msg
'\xe2\xf7\\u\xd3\xa3y\xe2\xd4\xdeqL\xfbJ\x02u\xbf\x9fQ%\xc6\x1f\x07\x82\x1a\xbd\x9c\x04\xad'
>>> " ".join("%2.2X" % ord(c) for c in msg)
'E2 F7 5C 75 D3 A3 79 E2 D4 DE 71 4C FB 4A 02 75 BF 9F 51 25 C6 1F 07 82 1A
BD 9C 04 AD'
>>> len(" ".join("%2.2X" % ord(c) for c in msg))
86
>>> " ".join("%d" % ord(c) for c in msg)
'226 247 92 117 211 163 121 226 212 222 113 76 251 74 2 117 191 159 81 37
198 31 7 130 26 189 156 4 173'
>>> print msg
?u?y?qL???Q%O\a??\x04

	The only reason the encrypted message is longer than the plaintext is
because it has been salted with a 4-byte random value, and that value is
provided so the decryption can set itself up for the same "randomness". The
reason for such a salting is so that two users, say, using the same key,
and the same plain text, would get different encrypted forms. You can't
look at the result and say "I know what user A sent... What user B sent is
identical, therefore I know what B sent" without even trying to decrypt the
message.
-- 
	Wulfraed                 Dennis Lee Bieber         AF6VN
    wlfraed@ix.netcom.com    HTTP://wlfraed.home.netcom.com/