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Old 2011-07-12, 03:10   #45
wblipp
 
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Quote:
Originally Posted by Christenson View Post
Would a user be interested in M^3,M^5,M^7,M^11 all together?
[or would they have to ask for M^(3*5*7*11), which is going to take 15*7*11 = 1155 (> 2^10) --> 11 additional steps, as opposed to simply looking at M, M^2, M^3...M^12, which also takes 11 additional steps]

Can you give me an idea of the population of likely requests? I don't have the sophistication/experience to know which case I should be optimising for, beyond the fact that whoever is using this is going to need to search automatically for the interesting Ms.....as there are going to be a lot of them. The corollary of this is that this search will need to happen inside the same processor that generates the Ms, otherwise we'll be getting into huge communications delays.
I'll keep supplying answers, but I think you making a design error. I believe you are designing unneeded complexity that will make the code slower and harder to develop. I believe it is a mistake to handle the powers of M as a separate processing step. The most effective and efficient implementation will be to multiply the exponent by the factor (183925013 by 12) and solve once for the factors of 791^2207100156-1.

In all the cases I know of, the user will be interested in divisors of the algebraic factors of a composite exponent. The only cases it makes sense to do this in a combined fashion are those where the exponent is a large prime times small factors. It's sufficient to work with the largest exponent because the algebraic factors will always divide it.

Stated another way in more explicit detail, every instance I know of where the user in interested in factors of M^3, M^5, M^7, and M^11, the user will also be interested in factors of M^15, M^21, M^33, M^34, M^55, M^77, M^105, M^165, M^231, M^385, and M^1155. He gets all of these automatically from factors of M^1155. You get a simpler and more efficient implementation when you multiply the exponent by 1155 before starting.
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Old 2011-07-13, 02:01   #46
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wblipp:
There's a non-quiche-eating, REAL programmer in me that doesn't like what you are saying..and wants to use negative subscripts to modify the operating system....but realistically, especially given the time it has taken me to do a "simple" change to mfaktc, you are correct, and the program should simply factor to whatever multiple of the exponent is requested.
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Old 2011-07-13, 13:55   #47
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Quote:
Originally Posted by Christenson View Post
There's a non-quiche-eating, REAL programmer in me that doesn't like what you are saying..and wants to use negative subscripts to modify the operating system
I'm SO glad! It's risky business trying to dissuade volunteer programmers from a course that appeals to them. I'd rather get a burnt, half loaf than no bread at all. But I feel I'm channeling Seymour Cray with these recommendations. Your inner programmer is trying to slow down the application for its major application to put in fancy features that nobody wants. I don't want fancy features, I want bare metal speed.

William

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Old 2012-10-10, 00:27   #48
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Quote:
Originally Posted by wblipp View Post
Is there any plan for GPU trial factoring for a^n-1 when a is not 2?

Oddperfect has a few long standing uses for this (a=41 a=127 a=157 and others)

http://home.earthlink.net/~oddperfec...TrialFactoring

William
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Old 2012-10-10, 07:11   #49
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Bump
Wasn't there a CudaLLr program?

Luigi
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Old 2012-10-11, 04:02   #50
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Luigi-
There is. But LLR doesn't have anything to do with factoring!
-Curtis
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Old 2012-10-11, 13:23   #51
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Luigi-
There is. But LLR doesn't have anything to do with factoring!
-Curtis
I didn't pay attention to the term "trial-factoring", and I just had a 25% cutoff of my salary, so maybe I had a momentary laspse of reason...

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