From: Jack Mott <thecrow@iconn.net>
To: cypherpunks@toad.com
Message Hash: 5959533f571bcbdc3774f9411898e675a436fc1f7990e88130d8b199caffd9df
Message ID: <316EF0D7.3B80@iconn.net>
Reply To: N/A
UTC Datetime: 1996-04-13 07:51:56 UTC
Raw Date: Sat, 13 Apr 1996 15:51:56 +0800
From: Jack Mott <thecrow@iconn.net>
Date: Sat, 13 Apr 1996 15:51:56 +0800
To: cypherpunks@toad.com
Subject: Known Plaintext attacks on symmertric algorithms
Message-ID: <316EF0D7.3B80@iconn.net>
MIME-Version: 1.0
Content-Type: text/plain
Now maybe I have this all wrong, but it is my understanding that a known
plaintext attack is when the cracker knows part of the plaintext of an
encrypted file. Then he/she uses that and runs the inverse of the
algorithm to calculate the key.
Whether or not I am right about what known plaintext means, isn't the
entirely possible on all of the symmetric algorithms out there? If I
grab a file that I know is, say, a standard credit card transaction
form, and I know what the first 256 bytes are because they are always
the same, shouldn I always be able to find the entire key that
corresponds with those 256 bytes? (assuming the key is 2048 bits or
less) And then with that key decrypt the whole file?
Maybe I am missing something but it seems that all the symmetric
algorithms are vulnerable to this, and I thought of a fix, but it
involves having two keys (or one thats twice as big)
--
thecrow@iconn.net
"It can't rain all the time"
RSA ENCRYPTION IN 3 LINES OF PERL
---------------------------------------------------------
#!/bin/perl -sp0777i<X+d*lMLa^*lN%0]dsXx++lMlN/dsM0<j]dsj
$/=unpack('H*',$_);$_=`echo 16dio\U$k"SK$/SM$n\EsN0p[lN*1
lK[d2%Sa2/d0$^Ixp"|dc`;s/\W//g;$_=pack('H*',/((..)*)$/)
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1996-04-13 (Sat, 13 Apr 1996 15:51:56 +0800) - Known Plaintext attacks on symmertric algorithms - Jack Mott <thecrow@iconn.net>