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  modulo arithmetic V xor         


Author: Antony Clements
Date: Jan 26, 2008 23:23

I have been doing some reading on using modulo arithmetic as opposed to
using the xor function, i know that the xor function is modulo 2 arithmetic,
the question is, would breaking the xor function down into N steps decrease
the memory footprint of an algorithm, if so the reduction in time should
vary depending on the size of input 1 and input 2 and how many of them there
are, as part of a stream cipher for example.

the further question arises of how to break it down into steps, i know how
to break down modulo exponentiation but not any of the other forms of modulo
arithmetic.

links to reference material or explenations would be most welcome.
no comments
  What's up with Skype in Germany?         


Author: Peter Pearson
Date: Jan 26, 2008 22:03

Would any German-speaking sci.crypt participant be so obliging
as to visit
http://wikileaks.org/wiki/Skype_and_the_Bavarian_trojan_in_the_middle
and describe the leaked memo? The phrase "trojan in the middle"
seems confused. I wouldn't expect Skype to be vulnerable
to a man-in-the-middle attack. Does this attack depend on
infecting the victim's computer with malware?

--
To email me, substitute nowhere->spamcop, invalid->net.
11 Comments
  Re: Factors mod N         


Author: Mike Fleagle
Date: Jan 26, 2008 21:37

"JSH" gmail.com> wrote in message
news:dd1d2b2f-7000-42d2-9cf8-3c2dfb0fcf36@n20g2000hsh.googlegroups.com...
> The recent research I've discussed with much frequency recently can be
> further generalized, so that you can calculate factors modulo N, where
> N can be a prime number or a composite. I will make one variable
> change from previous expositions going from 'n' to 'm' since I'm now
> using capital 'N' instead of 'p':
>
> With z^2 = y^2 + mT, where T is the target composite to be factored, m
> is a non-zero integer of your choice, and N is a prime or composite of
> your choice, you may find the following to be true:

so we gotta picka "m",
picka "N",
and picka "y" or picka "z".
> z = (2a)^{-1}(1 + 2a^2)k mod N

Now we gotta picka "k"
and picka "a"

We are up to guessing 5, *FIVE* numbers, JSH
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6 Comments
  Re: Help wanted by encryption ignoramus         


Author: Peter Pearson
Date: Jan 26, 2008 21:33

On Sat, 26 Jan 2008 16:53:27 -0800 (PST), BobKellock wrote:

[snip]
> I don't imagine that the "encryption" algortihm is very
> complex because it'a not being used for security
> purposes. On the other hand it's not super simple as, for
> instance, swapping the position of two bytes in the source
> generates a vastly different result.
Show full article (0.85Kb)
1 Comment
  Re: Help needed by encyption ignoramus         


Author: Mike Fleagle
Date: Jan 26, 2008 21:27

it is probably a generic CRC, many just use the common one or two.

"BobKellock" chainganger.co.uk> wrote in message
news:73402a23-2122-4e43-9c5c-d4bda7349445@m34g2000hsb.googlegroups.com...
> I'm writing a utility to diagnose problems in users' data files of an...
Show full article (1.51Kb)
no comments
  Re: Factors mod N         


Author: amzoti
Date: Jan 26, 2008 20:14

On Jan 26, 7:47 am, JSH gmail.com> wrote:
> The recent research I've discussed with much frequency recently
> James Harris

Your head is so far up your ass that you can't tell research from
nonsense!
no comments