I wrote
> What's the average number of ones in a randomly-chosen N-bit number?
Duh..
Sorry, I was thinking of the average number of bits that change on
increment. The answer's obviously N/2 for the above.
--
Mike Stay
Cryptographer / Programmer
AccessData Corp.
mailto:[EMAIL PROTECTED]
Bill Frantz writes:
> It seems to me you could use an existing public key infrastructure, e.g.
> PGP, but build a different message format with the stego requirements in
> mind. Off the top of my head (using PGP 2.6):
>
> (size, data)
> (256, key) - RSA encrypted key padded with pseudo-random pa
At 12:07 -0400 1999.06.30, Ron Rivest described the Beer Bottle Cypher,
asking:
>
>The actual security of this cipher seems to be an open question... Can it
>be broken?
>
Have you tried getting an export license for it?
Martin Minow
[EMAIL PROTECTED]
The Beer Bottle Cipher
Ron Rivest
6/30/99
Last week an MIT student hacker broke into the famous Yale University
secret drinking society known as "Skull and Bones". He made a
startling discovery that has implication
--- begin forwarded text
Date: Wed, 30 Jun 1999 10:51:22 +0200 (MESZ)
From: Christof Paar <[EMAIL PROTECTED]>
To: DCSB <[EMAIL PROTECTED]>
Subject: Papers at CHES
Please find below a list of accepted papers and invited presentations at
CHES (Workshop on Cryptographic Hardware and Embedded Syst
At 9:42 AM -0700 6/29/99, Russell Nelson wrote:
>So you've got a chicken-and-egg problem -- you have to have yet
>another set of public keys for your stego crypto algorithm.
It seems to me you could use an existing public key infrastructure, e.g.
PGP, but build a different message format with the
--
> From: David Honig <[EMAIL PROTECTED]>
> To: Jay Holovacs <[EMAIL PROTECTED]>; Russell Nelson <[EMAIL PROTECTED]>;
[EMAIL PROTECTED]
> Subject: Re: Eason/Kawaguchi stego
>
> \begin{nuance}
> Except that encrypted LSBs will be perfectly uniformly distributed
> and normal noise won't.
At 11:44 AM 6/25/99 -0700, bram wrote:
>> > > > > There are 52! bridge hands, so a random hand has
>> > > > > log2(56!) = 226 bits of entropy or 68 decimal digits worth.
>> > No, just 52! / (13!)^4 hands, which is around 2^96.
>> The interesting part is to come up with an algorithm that only uses
