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Author: walid.sabawalid.saba
Date: Jun 1, 2008 11:13
Dear all,
Data compression is not in any way my area of specialization and my
two questions might very well reflect this ignorance. However, I would
really appreciate your feedback on the following:
1- can we assume that we can test data compression algorithms on a
sequence of binary digits (a large sequence of 1’s and 0’s) regardless
of the original data type (text, image, video, audio, etc); I am here
looking for a simple ‘yes’ or ‘no’
2- if the answer to (1) is ‘yes’, then what is the best (lossless)
compression rate out there? I am here looking for answer like ‘70%%’,
etc. If this rate depends on the type of the original data, then the
answer could be (text,70%%), (image,85%%), etc.
I thank you in advance for your feedback.
Regards,
Walid Saba
http://walid.saba.googlepages.com
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6 Comments |
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Author: kkkhalid3kkkhalid3
Date: Jun 1, 2008 08:40
1) If you are looking for the truth to soothe and please your self,
just click on this link.
2) If you are looking for the secrt of the family happiness, just
click on this link
3) If you are looking for solving any social problem you face, just
click on this link.
http://www.lecalame.org/english/html/index.htm
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no comments
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Author: Jeffrey WaltonJeffrey Walton
Date: Jun 1, 2008 06:09
Hi All,
This is probably a mistake on my part, but what is the number
following the letter on an EC curve? For example, P-112 or K-571. I
was under the impression it was the size of the prime p or the group
size in 2^m.
~JW
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1 Comment |
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Author: JSHJSH
Date: May 31, 2008 09:31
I'm hoping to shorten the arguing phase with the remarkably simple
solution to the factoring problem I've outlined in a previous post by
pointing out a few things:
Given a target composite T, and
z^2 = y^2 + nT
where n is an integer chosen for z to have 3 as a factor--so if T mod
3 = 1 then n=5 will work--then z exists as
z = (f_1 + f_2)/2
when f_1*f_2 = nT, and of course you want integer factors.
The method I have then allows you to find z using a variable I call k,
where
z = 3k/2
and k^2 = 2^{-1} (nT) modulo p
where p is an odd prime less than k as k is an even positive integer.
May seem complicated all in a rush like that but it's incredibly
simple.
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11 Comments |
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Author: JSHJSH
Date: May 31, 2008 09:08
Wow. Years of research and then the answer turns out to be incredibly
simple, but that's how it can be with mathematics.
What I like about this method is that there isn't really any room for
people to argue with me about whether or not it solves the factoring
problem (I've posted it in another thread), and it has to be
incredibly fast, as easily shown by mathematical proof.
So no dumb arguments about me having to demonstrate as I haven't
gotten around to testing it yet, at all, as I've just relied on
mathematical proof and it's so simple.
It boggles the mind how simple it is.
Needless to say this research result gives FULL VALIDATION to my
research across the board and will usher in a new era in mathematics
as a discipline around the world.
That brings forward the correct prime counting function, the correct
information about randomness and primes, so that the Goldbach
Conjecture is handled and the Twin Primes conjecture is handled, and
the Riemann Hypothesis can probably be quickly handled.
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13 Comments |
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Author: JSHJSH
Date: May 30, 2008 23:15
If you have a target composite T, and use
z^2 = y^2 + nT
where n is used to force z to be divisible by 3, then there will be an
integer k such that
z = 3k/2.
But now find an integer x and odd prime p such that
x^2 = y^2 mod p
where also
2x = k.
So given k as defined before, you find an x that equals one half of
it, and consider a prime p where
(k/2)^2 = y2 mod p.
Then it's trivial to show that
k^2 = 2^{-1}(nT) mod p
and, if f_1*f_2 = nT, then
f_1 = k mod p
and
f_2 = 2k mod p.
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47 Comments |
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Author: magmag
Date: May 30, 2008 14:15
I read this requirement in the FIPS 140-2 Derived test requirements:
AS05.20: (Levels 3 and 4) The tamper response and zeroization
circuitry shall remain operational when plaintext secret and private
cryptographic keys or CSPs are contained within the cryptographic
module.
Does this imply that if your plaintext secret/private key/CSP's are
stored in nonvolatile memory, that you must use battery-backed logic
that can perform zeroization in the event of power loss in order to be
FIPS 140-2 level 3 or 4 compliant? Is there any literature to support
this claim? Thanks.
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no comments
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Author: hamadahamada
Date: May 30, 2008 09:21
Besides parents' rights, a great emphasis is also laid on the rights
of other relatives. In Islamic terms, 'Silah-Rahimi' is used to denote
good treatment towards the relatives'.
In the Qur'an, where the Muslims are enjoined to show kindness to
parents, they are also required to treat the other relatives with love
and sympathy and to pay due regard to their rights as well.
Allah has declared, "I am Allah, I am Ar-Rahman (The Merciful), I have
created the bond of kinship and named it Rahim,-which I have derived
from the root of my name of Rahman. Thus, whoever, shall join it tie.
Rahim, I shall join him, and whoever will break it, I shall break
him."
The Almighty has designed the system of birth in such a way, that
whoever is born, is tied to the bonds of kinship-and these bonds carry
certain claims and rights. Thus, whoever fulfils these claims, by
being kind to his relatives and treating them well, Allah will "join
him" i.e. He will make him His own and bestow His favor and mercy on
him. And whosoever will violate these claims, Allah will "break him"
i.e. He will have nothing to do with him.
Fulfilling The Rights Of Relatives
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Author: Kim SchulzKim Schulz
Date: May 30, 2008 02:02
Hi
I have a setup where two systems needs to transfer some data securely
over an insecure connection.
Normally I would use a station-to-station(STS) setup with AES
encryption of the data. In this case, however,
one of the systems does not have the ability to store a private key
over time (used in STS signing to identify the system towards the
other system).
Does some protocol exist that does not need one system to store a
private key but still gives the ability to have the secret key K
generated at runtime (as STS does it)?
kim
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6 Comments |
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Author: puggidpuggid
Date: May 29, 2008 12:21
I was wondering if someone tell me how I go about calculating
collision rates for a "key" that is 2 characters (lowercase alpha
only) long?
I realise it's 26x26 but if I generate two random numbers between 1-26
is that all that's involved? So each time I generate this key, I have
a 1/676 chance of a collision?
Thanks for the help
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5 Comments |
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