mersenneforum.org Sieving both sides vs one side at a time
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 2015-11-16, 14:53 #1 paul0   Sep 2011 3·19 Posts Sieving both sides vs one side at a time My line siever currently sieves N(a+bθ)*(a+bm), which is the algebraic side and the rational side at the same time. However, it looks like most papers point to sieving the two sides separately. What is the advantage of this? Won't this need extra work, like matching (a,b)-pair of both sides, since both sides need to be smooth?
 2015-11-16, 15:21 #2 fivemack (loop (#_fork))     Feb 2006 Cambridge, England 24·3·7·19 Posts Are the papers you're looking at specifically referring to the line siever? With lattice sieving you are definitely working on one side or the other. Also, multiplying the two sides together is going to make large-prime variants more difficult - you'd be factoring larger numbers and looking for rarer events, and you're not able to take so much advantage of the nice filter 'if the residue is < p_max^2, it must be prime and therefore it's likely to be too large a prime.
 2015-11-16, 15:24 #3 paul0   Sep 2011 5710 Posts So there is an implicit "match the smooth pairs from both sides" step?
2015-11-16, 16:06   #4
R.D. Silverman

Nov 2003

1D2416 Posts

Quote:
 Originally Posted by paul0 So there is an implicit "match the smooth pairs from both sides" step?
I don't know what you mean by "match the smooth pairs".

Initialize the sieve array. You sieve one side and identify which lattice points might be smooth.
Re-initialize by setting the values of those locations that can't be smooth to -oo, then sieve
the other side. You then identify which points might be smooth on the second side. Finally,
one attempts to factor those candidates that might be smooth on both sides. (i.e. the intersection
of the sub-lattices with just smooth points)

Note also that doing one side at a time reduces memory and makes bucket sieving a lot easier.

Note also that sieving both sides at the same time will increase paging, because the cache can't
hold even one factor base, let alone two, so you will spend a lot of extra time pulling in both
factors bases during sieving (as opposed to one at a time).

Give us some data. How long do you take to process a single special-q? Specify the size of your
factor base and the size of the sieve region. We can then let you know whether your siever
competes with others in terms of performance.

The time it spends should be directly proportional to the size of the sieve array times loglog(pmax)
where pmax is the largest prime in the factor base. The constant will change according to the speed
of your processor, size of cache, size of memory and whether you have one, two, or three LP on each side,
as well as the cleverness of your code.

Last fiddled with by R.D. Silverman on 2015-11-16 at 16:13

2015-11-16, 16:11   #5
paul0

Sep 2011

3·19 Posts

Quote:
 Originally Posted by R.D. Silverman I don't know what you mean by "match the smooth pairs".
Oh, I was assuming that the two sides are sieved separately, as in two different memory allocations, producing two different sets of smooth pairs. It's clearer to me now. Thanks.

2015-11-18, 13:58   #6
paul0

Sep 2011

1110012 Posts

Quote:
 Originally Posted by R.D. Silverman Give us some data. How long do you take to process a single special-q? Specify the size of your factor base and the size of the sieve region. We can then let you know whether your siever competes with others in terms of performance.
I like to point out that for now, my NFS implementation is a learning (toy) experiment. It won't be made to compete.

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