--- begin forwarded text
Status: RO
From: Patrick [EMAIL PROTECTED]
To: 'Digital Bearer Settlement List' [EMAIL PROTECTED]
Subject: Lucrative update
Date: Tue, 18 Feb 2003 13:05:40 -0600
Sender: [EMAIL PROTECTED]
Lucrative release 4 is out.
I know many people are used to seeing releases
At 1:09 PM +1100 2/18/03, Greg Rose wrote:
At 02:06 PM 2/17/2003 +0100, Ralf-Philipp Weinmann wrote:
For each AES-128 plaintext/ciphertext (c,p) pair there
exists exactly one key k such that c=AES-128-Encrypt(p, k).
I'd be very surprised if this were true, and if it was, it might
have bad
... We can ask what is the
probability of a collision between f and g, i.e. that there exists
some value, x, in S such that f(x) = g(x)?
But then you didn't answer your own question. You gave the expected
number of collisions, but not the probability that at least one
exists.
That
At 5:45 PM -0600 2/18/03, Matt Crawford wrote:
... We can ask what is the
probability of a collision between f and g, i.e. that there exists
some value, x, in S such that f(x) = g(x)?
But then you didn't answer your own question. You gave the expected
number of collisions, but not the
Matt Crawford wrote:
But here's the more interesting question. If S = Z/2^128 and F is the
set of all bijections S-S, what is the probability that a set G of
2^128 randomly chosen members of F contains no two functions f1, f2
such that there exists x in S such that f1(x) = f2(x)?
Vanishingly
Ed Gerck [EMAIL PROTECTED] wrote:
For each AES-128 plaintext/ciphertext (c,p) pair with length
equal to or larger than the unicity distance, there exists exactly
one key k such that c=AES-128-Encrypt(p, k).
Excuse my naivete in the math for this, but is it relevant that the unicity
distance
The relevant aspect is that the plaintext and key statistics are the
determining factors as to whether the assertion is correct or not.
In your case, for example, with random keys and ASCII text in English,
one expects that a 128-bit ciphertext segment would NOT satisfy the
requirement for a
