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"Ralph" == Donald ``Ralph'' Wood writes to Perry Metzger:

Ralph> I do not object to criticism, when I am wrong, but I do Ralph> object to using just highly subjective opinions to attack me, Ralph> or anyone for that matter. The same Ralph of IPG who offered an unbreakable system or they would sell the company for $1? [Repost from March 19, 1996] Return-Path: owner-cypherpunks@toad.com Received: from relay3.UU.NET (relay3.UU.NET [192.48.96.8]) by deanna.miranova.com (8.7.3/8.6.9) with ESMTP id TAA01777 for <steve@miranova.com>; Tue, 19 Mar 1996 19:52:42 -0800 Received: from toad.com by relay3.UU.NET with SMTP id QQahuk09205; Tue, 19 Mar 1996 22:31:17 -0500 (EST) Received: by toad.com id AA29880; Tue, 19 Mar 96 09:32:18 PST Received: from pangaea.hypereality.co.uk by toad.com id AA29874; Tue, 19 Mar 96 09:32:09 PST Received: (from remail@localhost) by pangaea.hypereality.co.uk (8.6.9/8.6.9) id RAA17262 for cypherpunks@toad.com; Tue, 19 Mar 1996 17:32:47 GMT Hypereality Systems : <WWW: http://www.hypereality.co.uk/> Date: Tue, 19 Mar 1996 17:32:47 GMT Message-Id: <199603191732.RAA17262@pangaea.hypereality.co.uk> To: cypherpunks@toad.com From: cpunk@remail.ecafe.org (ECafe Anonymous Remailer) Subject: IPG cracked with known plaintext Remailed-By: ECafe Anonymous Remailer Complaints-To: complaints@remail.ecafe.org X-Www: http://www.ecafe.org/~remail/ X-Notice: The contents of this message are neither appoved or X-Notice: condoned by ecafe.org or our host Hypereality Systems. X-Notice: We bear no liability for misuse of this system. X-Warn: *** This message was remailed through an anonymous remailer *** X-Warn: *** Replying to it will not send your reply to the sender *** Sender: owner-cypherpunks@toad.com Precedence: bulk Lines: 77 Xref: deanna.miranova.com cypherpunks:199 This information is preliminary and is based on an attempt to understand the IPG algorithm information. That description is not clear in some areas, however, hence this analysis is tentative at this time. First let us describe the IPG system in more conventional C: a[0] to a[63] are initialized to random 8-bit values. (The description is unclear and almost makes it sound like they are initialized to a random 8-bit value anded with 0x3500, which would of course be zero. The attack below will assume that this bizarre step is not done, but will still apply even if it is.) b[0] to b[63] are initialized to random primes selected from some pool. c[0] to c[63] are also initialized to random primes selected from a different pool. d is initialized to a random 8 bit value. The algorithm is: for ( ; ; ) { for (i=0; i<63; i++) { a[i] = (a[i] + b[i]) % c[i]; d = (d + a[i]) & 255; *data++ ^= d; /* xor with data */ } } Note first that with a known plaintext attack, the value of d can be calculated for each iteration, simply by xor'ing the plaintext and ciphertext. So we can easily recover a series of d values under this assumption. Known plaintext is a plausible cryptographic assumption in many contexts. Note second that we can assume that b[i] is less than c[i]. It appears from the description that this will be true, although it is a little unclear. If b[i] is greater than c[i] then simply do b[i] = b[i] % c[i] before beginning the loop. This will produce the same results since (a + (b mod c)) mod c is equal to (a + b) mod c. Note third that when a[i] and b[i], both less than c[i], are added mod c[i], the result will be equal to one of two things: a[i]+b[i], or a[i]+b[i]-c[i]. The reason is that the sum a[i]+b[i] must be less than 2*c[i] so the "mod" operation will be at most a single subtraction of c[i]. In general, half the time it will be necessary to subtract c[i], and half the time it will not. Now, as mentioned above, with known plaintext we can deduce the series of d values. Since each d differs from its predecessor by adding a[i], this allows us to calculate the low 8 bits of a[i] simply by taking the difference between successive d's. Every 64 bytes, i repeats. We know the low byte of a[i] from the previous iteration, and we know it for this iteration. Half of the time (on average) a[i] will change simply by adding b[i], in which case the low 8 bits will change by exactly the low 8 bits of b[i]. So if we take the difference between a[i] values spaced 64 bytes apart, half of the time these values will be a constant which is equal to the low byte of b[i]. The other half the time, the low 8 bits will change by adding b[i] and subtracting c[i]. So the low 8 bits of (b[i]-c[i]) is the other possible constant value which will be seen when you take the difference of a[i] every 64 bytes. So with a few multiples of 64 bytes of known plaintext, you will quickly find all the possible b[i] and b[i]-c[i] low bytes. By itself this should significantly narrow down the possibilities for b[i] and c[i], in many cases to a single prime. Even without this the algorithm can now be run forward or backward with only two possible known changes to a[i] at each step, and the entire message can be easily deduced. So this algorithm is easily broken with known plaintext.