20060413, 14:18  #1 
Sep 2005
177_{8} Posts 
New Algorithm  Tester(s) required
I've just posted the following assertion over at M+2:
http://groups.google.com/group/Merse...9949e8e2560b6b Anybody feel like helping me test it? J 
20060413, 16:10  #2  
Nov 2003
2^{2}·5·373 Posts 
Quote:
via naiive multiplication methods and n (log p loglogp logloglog p) via FFT techniques. We only test numbers that are 1 mod 2n. From 2^x to 2^(x+1) there are 2^x/(2n) = 2^(x1)/n such numbers. Each takes the amount of time given above. Now multiply. I do not know where you get the exponent "13". 

20060413, 17:33  #3 
Sep 2005
127 Posts 
Thanks for your response, and analysis.
However, 1) I fear you have misunderstood what exactly the algorithm does. (and therefore how effective it is) 2) The result quoted is based an empirical one. J 
20060413, 18:01  #4  
Nov 2003
1110100100100_{2} Posts 
Quote:
Please correct the misunderstanding. State the algorithm exactly. My reading of what you wrote is that the algorithm's purpose was to find small factors of much larger numbers of the form 2^n+1. 2) If one is looking to analyse an algorithm one starts by determining its complexity as a function of the input size. Then, and only then does one try to find explicit constants in place of the implied constants given by the O() estimates. 

20060413, 18:15  #5 
Sep 2005
127 Posts 
(thanks again for your response)
1) The source code is there  its complexity is only akin to say, rho method 2) My justification for claim of polynomial (ie logarithmic) speed: a) [the second part of your original analysis] The number of steps required will tend to a constant (sic) b) [the first part ditto] Each step can be performed in polynomial/logarithmic time using 'Russian Peasant' method of exponentiation 
20060413, 18:30  #6  
Nov 2003
2^{2}·5·373 Posts 
Quote:
State the objective(s) of the computation. State the method. Saying that "it" is akin to Pollard Rho says nothing unless you tell us what "it" is. WHAT ARE YOU TRYING TO DO?????? 

20060413, 18:57  #7  
∂^{2}ω=0
Sep 2002
República de California
26376_{8} Posts 
Quote:
James (a.k.a. bearnol), if you can't formally justify your asymptoticcomplecity claims in a way most reasonable folks can understand, then please be so kind as to at least provide some nontrivial example factorizations, their runtimes, and basic notes about your implementation. I don't want to hear any guff along the lines of "well, I haven't actually *coded* it yet..." or "My implementation uses offtheshelfpackage X, which is, like, totally suboptimal and disguises the true breakthrough speed of the algorithm"  those are the kinds of dodges used by cranks, rogues and scoundrels, i.e. the kinds of people I'm sure you wouldn't want to be seen as keeping company with. 

20060413, 19:06  #8  
Nov 2003
2^{2}·5·373 Posts 
Quote:
I am *trying* to be helpful. But the o.p. fails to answer direct questions! 

20060413, 19:19  #9 
Sep 2005
1111111_{2} Posts 
WE is an algorithm for factoring (general) integers.
WEP is a faster version that is specific to M+2 integers only. [there should be sufficient documentation on my website(s): http://bearnol.isageek.com/Mersenn...neplustwo.html http://www.bearnol.pwp.blueyonder.co.uk/Math/index.html ] 
20060413, 19:34  #10 
∂^{2}ω=0
Sep 2002
República de California
2·13·443 Posts 
James  I repeat, please provide some example factorizations and timings. Why should folks who have many demands on their time (that probably includes just about all the ones qualified to judge the veracity of claims such as yours) be bothered to read reams of online documentation unless you provide at least a glimmer of an idea that there may be something to your claim?
To put it bluntly: show us some data. Otherwise stop wasting our time. 
20060413, 20:35  #11  
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across
2^{4}×641 Posts 
Quote:
James' track record so far strongly suggests that anyone who goes any further into his pronouncements deserves all the grief they may get. That's not to say he hasn't got something. It's only to say that his reputation precedes him. If (and only if, IMO) he starts to publish rigorous mathematical analysis of his ideas, such analysis to use the language conventional in mathematics, will he be worth taking seriously. All the above IMAO. YMMV. Paul 

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