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*To*: [email protected]*Subject*: Truly Stealthy PGP*From*: Hal <[email protected]>*Date*: Sat, 5 Mar 1994 07:33:34 -0800*Sender*: [email protected]

Eric points out the difficulty of making a "stealth PGP" which is 100% indistinguishable from a string of random bits. The problem is that we have to encode the RSA encrypted number, m, which is less than n, the RSA modulus. PGP first puts out two bytes of bit length, then m. This obviously won't do, since the bit length is generally much less than 2^16 and so these two bytes are a dead giveaway. However, we could leave these two bytes off and just output m as raw bits, padded to the length of n. The recipient knows n so he would be able to extract m. The problem here, as Eric points out, is that m is less than n, so the high bits of m will look non-random. If the high two bytes of n are, say, 0x0C12, then m's high two bytes will never be bigger than this. This will allow the opponent to do much better than 50% on guessing which files have embedded messages. This was discussed some time back on the pgp developers' list, and at that time the suggestion was made to add a multiple of n to m so that it covered a fuller range of values. The recipient would then just take the exponent mod n and try that. Mathematically, call L the next multiple of 256 above n. (0x10000... in the example above.) We want to choose k so that M = m + k*n is randomly distributed between 0 and L-1 if m is randomly distributed between 0 and n-1. This may not be possible in this form. Perhaps there is another deterministic and reversible transformation would accomplish it, though. In that case we would have M = f(m,n) such that f can be reversed given M and n (we can recover m). As a trivial example of this problem, given n=2 and L=3, try to come up with a way to turn a random 0/1 value into a random 0/1/2 value which is both reversible and produces each of 0/1/2 with 33% probability. Seems pretty tough! Hal

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