1996-11-17 - Re: PGP3.0 & ElGamal

Header Data

From: Steve Reid <steve@edmweb.com>
To: “Mark M.” <markm@voicenet.com>
Message Hash: c5a963865cc94b67ddef1b185dd4f7c54d465ae5b0f8add76b150356f8c2b843
Message ID: <Pine.BSF.3.91.961116193436.193B-100000@bitbucket.edmweb.com>
Reply To: N/A
UTC Datetime: 1996-11-17 04:26:12 UTC
Raw Date: Sat, 16 Nov 1996 20:26:12 -0800 (PST)

Raw message

From: Steve Reid <steve@edmweb.com>
Date: Sat, 16 Nov 1996 20:26:12 -0800 (PST)
To: "Mark M." <markm@voicenet.com>
Subject: Re: PGP3.0 & ElGamal
Message-ID: <Pine.BSF.3.91.961116193436.193B-100000@bitbucket.edmweb.com>
MIME-Version: 1.0
Content-Type: text/plain


>> Yes, but I believe 3DES has an effective key length of only 112 bits.
>> Of course, even this is more than sufficient for a long time to come. 
> 3DES can have an effective key length of 168 bits if 3 keys are used
> instead of two.  There are no security problems that I know of from 
> using 3 keys. 

The Meet In The Middle attack (MITM, not to be confused with Man In The
Middle) is a time-memory tradeoff that works against any multiple
encryption. 

Triple encryption:
ciphertext = encrypt(k3, decrypt(k2, encrypt(k1, plaintext)))
plaintext  = decrypt(k3, encrypt(k2, decrypt(k1, ciphertext)))

An MITM attack, as I understand it, works by decrypting from one end and 
encrypting from the other...

Step 1- decrypt(x3, ciphertext) for every possible x3, and store all the
results. This requires 2^56 operations and 2^56*8 bytes (550 petabytes?)
of memory when done against 3DES. Quite a lot, but it might be doable. 

Step 2- decrypt(x2, encrypt(x1, plaintext)) for every possible x1,x2. 
This require 2^112 steps with 3DES. If the result you get can be found in
the table built by step 1 then you've figured all three keys. 





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