20140413, 16:27  #1 
Dec 2011
After milion nines:)
2^{2}·5·71 Posts 
How to calculate FFT lengths of candidates
I get this data from first megabit drive
554 candidates (144k: 49.8%, 160k: 50.2%) Can someone tell me how to calculate K weight of candidate? What formula is used ( I sow that AVX uses some different approach) 
20140414, 08:26  #2 
Feb 2003
2^{2}×3^{2}×53 Posts 
In principle the procedure is quite simple:
For each candidate in the input file the FFT length is obtained and then the percentages are computed. In practise I'm using a small tool (see attachment) based on my "llrtools". 
20140414, 13:27  #3 
Feb 2003
2^{2}×3^{2}×53 Posts 
The attached file contains a binary for Windows machines.
Note that this is a command line tool. You'll need to run it from your DOS prompt: Code:
fft_percentage.exe input.txt Last fiddled with by Thomas11 on 20140414 at 13:28 
20140417, 11:18  #4 
Feb 2003
2^{2}×3^{2}×53 Posts 
Due to the recent interest (and also since the original LLR tools didn't contain the parameters for AVX machines), I'm posting here a Javascript version of the FFT length calculator  just a proof of concept, so don't expect any sophisticated piece of software...
It's a simple HTML file which you should be able to open in your favourite (Javascript enabled) web browser. Usage is quite simple: Enter the values for k, nmin and nmax, select the proper cpu type (avx, sse2, or x87) and hit the "Get FFT lengths" button. Note that zeropadded FFTs use a different algorithm which is not yet implemented. Thus, for K>1000000 the points where the FFT lengths are changing are slightly different. Feel free to further modify the code. 
20140417, 14:24  #5 
Dec 2011
After milion nines:)
2^{2}·5·71 Posts 
I dont wont to be rude but since I ask this for candidates on base 10 this is pretty unusable for me :(

20140417, 18:03  #6  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·5·23·41 Posts 
Quote:
Quote:
Must be the language barrier. Anyway, just run this Code:
# ./sllr64 d q"2*10^10590021" Base factorized as : 2*5 Base prime factor(s) taken : 5 Starting N+1 prime test of 2*10^10590021 Using AMD K8 FFT length 224K, Pass1=896, Pass2=256, a = 2 (In fact, if you set up llr.ini properly and wait for a few seconds, you will also get the time per iteration, ... and then you kill it.) On a different CPU, the FFT kernel and size may be different. Repeat for random samples from your future work files. Plot. Scratch your forehead. Analyze. Plan ahead. Keep it simple. 

20140417, 19:05  #7 
Dec 2011
After milion nines:)
2^{2}·5·71 Posts 
Batalov, I truly love yours answers, and in fact that is way I found FFT length , start LLR , look at ,and stop it. But I was thinking if there is another ( faster way)....

20140417, 20:50  #8 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·5·23·41 Posts 
You can of course,
 understand the LLR source (and GWNUM library that it calls)  get the appropriate parts of the code and  repackage them in a new program or a script that will "predict" what particular FFT size will be used. You will have to update this accessory program every time the LLR+GWNUM changes, when there's a relevant change. There is no magic single formula!  there are decision paths (ifelseifelseifelse...) that the GWNUM library uses when it is called to initialize the FFT control structures for a number. Find and read the PrimeGrid topic about that: they initially considered this to be too tedious, but then surprisingly this is what they exactly did: prerun each number for a split second to know exactly what FFT size will be used and based their credit system on that knowledge. 
20140417, 20:51  #9 
Feb 2003
2^{2}×3^{2}×53 Posts 
There is an undocumented feature in LLR. Just add the following line to your llr.ini file:
Code:
SetupOnly=1 However, for some forms the keyword is just ignored. 
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