20081111, 07:02  #23  
May 2007
Kansas; USA
2^{2}·2,539 Posts 
Quote:
CAREFUL!! This is a little tricky to code! You only want to delete k's divisible by 3^q if and only if k / 3^q is still remaining and you sometimes have to check several kvalues remaining for a k if it is divisible by 9 or 27 or 81, etc. To do otherwise would not be good. Here is some "pseudocode" so to speak: Code:
read file of kvalues remaining do until end of file m = kvalue / 3 remainder r do while r = 0 if m = any kvalue in file delete current kvalue endif m = m / 3 remainder r enddo read file of kvalues remaining enddo Gary Last fiddled with by gd_barnes on 20081111 at 07:02 

20081111, 10:38  #24 
Mar 2006
Germany
2826_{10} Posts 
i've uploaded a new version of the low n scripts (the one with PFGW).
i tuned up the modular filtering (see example for base 3) with all filtered k's written i an own file and better logging! hope Gary can verify the filtering with the "3^q"thing in the last post. if all is ok, i think there're no more changes to do and this step is complete! 
20081111, 10:55  #25  
May 2007
Kansas; USA
2^{2}·2,539 Posts 
Quote:
Did you automatically remove ALL k / 3^q ? That is the impression that I got in your PM. If so, that is incorrect. See my pseudocode in the prior post here. You have to check if k / 3 ^q reduces to a k that is remaining. If it does (it does ~90% of the time), THEN you can remove it. Gary 

20081111, 11:03  #26  
Mar 2006
Germany
2·3^{2}·157 Posts 
Quote:
have to change again! lol made those changes last night when i understand it that way. but this is something else to consider! ATTENTION: don't use the modular reduction yet! it's not ok yet for this purpose!!!! PS: to filter all k's like k==1 mod 7 for Riesel Base 15 it is ok and you can use this. Last fiddled with by kar_bon on 20081111 at 11:34 Reason: PS 

20081111, 11:37  #27  
May 2007
Kansas; USA
2^{2}·2,539 Posts 
Quote:
Karsten, I'm not sure how you are going to handle this situation: Let's say you are testing k=250M253M. You come across, perhaps k=252000000 reamining. You divide by 3 and now you have to check if k=84000000 is remaining. If that's not remaining, you have to divide by 3 again and see if k=28000000 is remaining. If so, you can remove k=252000000, if not, k=252000000 remains because 28000000 is not divisible by 3. One problem: You have to have a file of k's remaining to compare to and that file will contain many more k's than are in the file that you are processing. In other words, I think you're going to have to read against a completely separate file PLUS your current file. Perhaps the best thing to do would be to put all k's remaining into a separate 'BIG' file before starting plus ALL k's from your current krange. Let me be VERY specific on that: To start with for the range of k=250M253M, the file would contain something like 5379555, 8013488, 12345678, 22345019, 250000000, 250000002, 250000004, 250000006, 250000008, etc. That is it only contains previous remaining k's < 250M and then ALL k's in your current krange. As you process your current krange, you remove k's from this other 'BIG' file plus your current file. You'll then be able to take your current file and do a lookup on that 'BIG' file of k's remaining for all k / 3^q for each k. Please note that this situation is mostly unique to bases 3, 7, and 15. On all other bases, we always sieve the entire file at one time. There shouldn't be any use for this 'BIG' file of k's remaining on all (or most) other bases. Even on Sierp base 31, I tested all k's at once to begin with before splitting it up on to 4 cores. For a conjecture of k>6.3M, it was slow going for a little while. You've chosen the most difficult base (by far) to make your automated process work on. One good thing: If you can get it to work for base 3, it should work for any base! Now, we'll just have to tell it how to remove algebraic factors! I won't even go there. Gary Last fiddled with by gd_barnes on 20081113 at 15:48 

20081111, 11:46  #28 
Mar 2006
Germany
5412_{8} Posts 
i've downloaded the pages from KEP's Riesel Base 3 effort with all remaining k's so far (i think 1866 to k=250M).
i see what i can do. yes, my intention for these scripts was to search much easier such high k's for the conjectures found here. you're right: if these work fine for bases like 3,7 and 15 it will work for others. i'm running this for my Riesel base 35 with only about 1300 k's left. :) 
20081111, 20:48  #29  
Jan 2006
Hungary
2^{2}·67 Posts 
Quote:
any k can be written as 3^a*b, with a >= 0 and b not divisible by 3. 1) if there is no 0 < n <= a with b*3^n is prime you do not have to test this k. 2) if not 1) then test this k. You can do this because we are working our way up so when b < k we can assume that b has already been tested. As we do not accept k*3^n1 with negative n we have to exclude those with the small loop in 1) This method will have some doubles but then there is no need to keep everything as reference. Cheers, Willem. 

20081112, 00:37  #30  
May 2007
Kansas; USA
2^{2}·2,539 Posts 
Quote:
I cannot imagine that Karsten wants to get into an additional "primality testing loop" to see if each k that is divisible by 3 should remain. I'll be curious to see what he comes up with. 

20081112, 00:45  #31  
May 2007
Kansas; USA
2^{2}×2,539 Posts 
Quote:
Where are you getting 1866 k's remaining to k=250M on Riesel base 3? Per my web page here, there are 977 k's remaining to k=250M. That was up to date as of 12 hours ago. I need to check the Mini drive for Riesel base 3 and see if there were any more primes today. But I know he is also sieving k=250M500M that he isn't done with yet. Perhaps he has some k's remaining in that range. Gary Last fiddled with by gd_barnes on 20081112 at 00:47 

20081112, 09:17  #32  
Mar 2006
Germany
2·3^{2}·157 Posts 
Quote:
my fault! 

20081113, 10:16  #33 
Jan 2005
479 Posts 
[quote=kar_bon;148625]you only have this in do.bat:
Code:
@echo off call doranges 3 1 1000 Hmm... This indeed looks way better :) I get the same stats as you do now. I'll see what I can do for starters to 1k, and worry about deleting the mod 3 values later :> 
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