1992-12-04 - No Subject

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From: Karl L. Barrus <barrus@tree.egr.uh.edu>
To: cypherpunks@toad.com
Message Hash: 2f431cfa3d49b13abd23485eb4f3929d313692b83118ce227aa1a80b444ff7f6
Message ID: <9212041802.AA13345@tree.egr.uh.edu>
Reply To: N/A
UTC Datetime: 1992-12-04 18:02:46 UTC
Raw Date: Fri, 4 Dec 92 10:02:46 PST

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From: Karl L. Barrus <barrus@tree.egr.uh.edu>
Date: Fri, 4 Dec 92 10:02:46 PST
To: cypherpunks@toad.com
Subject: No Subject
Message-ID: <9212041802.AA13345@tree.egr.uh.edu>
MIME-Version: 1.0
Content-Type: text/plain


FutureNerd Steve Witham writes:

>I should be able to verify with my own eyes how cypher technology works.  
>Otherwise I'm trusting my security to somebody's black box.

This is important, which is why its nice to have source code.
However, be prepared to invest a little :-) time in the study of PGP's
code; I recently printed it out and must have sucked a toner cartridge
nearly dry as it took more than 500 pages!  Check out 'wc -l *.[ch]'
which results in 22720 lines for me.

>        The signature algorithm (MD5?)
>                128 bits?
>                Is that based on RSA?

Get the md5.doc from rsa.com - available via anonymous ftp.  It
explains the algorithm (which is largely bitwise operations on blocks
of the input stream) and even includes source code.

>        A cryptostrong pseudorandom # generator?
>                Is this based on RSA?

Somebody posted a cryptographically strong prng to sci.crypt recently,
but I haven't had a chance to look it over.  I have my own rng - a ten
sided die, which I haven't had occasion to use, yet.  

>        A binary<-->ascii translator

Phil Karn's DES package includes source code for uuencode, which
performs this translation.  I'm sure it is somewhere in the PGP code
as well.  There is also RFC 1113.

>What's the formula for RSA again?
>        out = in * something ^ somethingelse mod yetanother??
>        I know it can't be, because the key is only one number.

Close!  That looks like the formula for solving ax mod n = d for x
given a, n, d.

RSA: pick two primes p and q.  Then n=pq.  As Hal Finney said, the two
primes should be "good" in the sense that p-1 and q-1 have no small
prime factors.  Obviously, you can't avoid having 2 divide p-1 and
q-1, but that should be about it.  Also, p and q should differ in
length by a few digits otherwise an enemy may begin to try factoring n
by starting near sqrt(n).

Pick a decryption exponent d.  d and phi(n) should be relatively
prime, where phi(x) = Euler totient function = # of numbers less than
x which are relatively prime to x = x-1 for x prime.  For n, phi(n) =
(p-1)(q-1).  Calculate the inverse of d mod phi(n) and call it e, your
encryption exponent (e = d^(-1) mod phi(n)).  Publish n and e as your
public info, and keep d secret as your decryption key.

Somebody encrypts a message by calculating M^e mod n.  You decrypt the
ciphertext by calculating C^d mod n.

For example: p=7, q=11 => n=77 and phi(n)=60.
I pick d=13, gcd(13,60) = 1 so it is acceptable.
Then e = 13^(-1) mod 60 = 37.  Check: 13 37 mod 60 = 1.
A message M=20 is encrypted as 20^37 mod 77 = 48.
I decrypted the ciphertext 48 like this: 48^13 mod 77 = 20, and I got
back the original message.

/-----------------------------------\
| Karl L. Barrus                    |
| barrus@tree.egr.uh.edu (NeXTMail) |
| elee9sf@menudo.uh.edu             |
\-----------------------------------/





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