i thought about an idea. probably its obvious
when you see Euclid's proof that there are infinity prime numbers
. so instead of search only big numbers.
search for every number, dont miss even 1 . and you can create easly a new numebr
P=1*2*3*5*7*11*13 +1 will be new prime .. and so on .
but you must have the all primes between 1 and your largest known prime .
well this is just an idea
bye
eko
when you see Euclid's proof that there are infinity prime numbers
. so instead of search only big numbers.
search for every number, dont miss even 1 . and you can create easly a new numebr
P=1*2*3*5*7*11*13 +1 will be new prime .. and so on .
but you must have the all primes between 1 and your largest known prime .
well this is just an idea
bye
eko
f0dder, http://developer.intel.com/design/pentium4/manuals/index2.htm there should be a Request a hardcopy of this document on the download pages of manuals (click on that link of the docs you want).
Maverick, I've done the same algo on 680x0, x86, but not MMX.
Hutch, has a version on his web page - not by me.
I just discovered that the algo that we indipendently invented was invented even before us, thousands years ago, by Erastotene:
http://www.math.utah.edu/~alfeld/Eratosthenes.html
"The Sieve of Eratosthenes"
I gave only a quick look but it really seems the same algo.
I bet he didn't do his first version on a 680x0 or a 65xx. ;)
PS: changing topic, you posted this URL some time ago:
http://www.rsasecurity.com/rsalabs/challenges/factoring/numbers.html
Does it mean that if one finds the two numbers that once factored give those big numbers one wins the prize?
When I have some free time I may look into this.. I've a very weird idea to test.
PS: changing topic, you posted this URL some time ago:
http://www.rsasecurity.com/rsalabs/challenges/factoring/numbers.html
Does it mean that if one finds the two numbers that once factored give those big numbers one wins the prize?
You can order hardcopy of AMD docs too. Look under Support -> Literature Order Center.
I found it here
I found it here