[QUOTE=fivemack;187953]General rule for siever choice: if you get less than 2000 relations from a range of 1000 Q, use a bigger siever; if you get more than 6000 use a smaller siever.

General rule for relation counting: 0.1 * 2 ^ large prime bound[/QUOTE]

Thanks.

Sieving from Q=33M to 33M+500 on the -r side...

13e: total yield: 816, q=33000547 (0.22109 sec/rel)
14e: total yield: 1628, q=33000547 (0.26377 sec/rel)

...so it would be better to take 14e, although 13e is sieving faster?

Repeating the questions:

1.) Did I understand this right, SNFS-193 would take approx. as long as GNFS-133? (193*0.69=133 point something)
2.) Did I get the parameters right?

Yes, SNFS-193 should be comparable to GNFS-133; I'd expect the sieving to take four to six weeks on your machine.

I think 2^25-1 is a bit large as a small-prime bound, 2^24-1 might be quicker; also see whether 29-bit large primes get you enough more relations per second to make up for needing twice as many.

That's eight experiments (13e vs 14e, 16777215 vs 33554431, lpb=28 vs 29); sieve the same range, presumably 2^25 .. 2^25+1000, for each of them and see what works better.

If the yield with sieving -a is much less than -r (which means the numbers on the R side are larger and are benefitting more from the 'free' factor that lattice sieving provides), consider setting alim lower than rlim, or even lpba lower than lpbr.

[quote=fivemack;187953]General rule for siever choice: if you get less than 2000 relations from a range of 1000 Q, use a bigger siever; if you get more than 6000 use a smaller siever.

General rule for relation counting: 0.1 * 2 ^ large prime bound[/quote]
Could you collect together these rules in one place? You keep on posting them all over the place and i keep on losing them.

[QUOTE=henryzz;187966]Could you collect together these rules in one place? You keep on posting them all over the place and i keep on losing them.[/QUOTE]Why not just bookmark the earlier post? Why not just copy and paste them to a file on your machine? Why not just write them down on a piece of paper?

In short, why not exert a miniscule amount of effort [i]yourself[/i] to satisfy one of [i]your[/i] desires instead of asking others to do it for you?

Tom has already given you more than he needs to.


Paul

[QUOTE=fivemack;187925]If someone is committing to a calculation that takes at least 24 hours, I don't think having to look twice to see what colour some number is counts as an onerous burden. If they can't see it on the screen it's clearly there in the HTML.[/QUOTE]


Note that one number has been reserved for over 500 days!

Shouldn't it be released so someone can do it?????

It's not as if there's a shortage of candidate numbers.

I'm a bit wary of releasing long-reserved numbers when I've personally had two out for nearly a year, but I've emailed Justin to ask if I should unreserve his number; he submitted two factors this April.

I just spotted this one:

[code]10^248+9^248 110.4 248.0 Reserved by E van Dijk -0.6 hours ago: unreserve[/code]

[B]minus[/B] 0.6 hours ago? so he will reserve in 36 minutes from now? :grin:

[QUOTE=Andi47;187977]I just spotted this one:

[code]10^248+9^248 110.4 248.0 Reserved by E van Dijk -0.6 hours ago: unreserve[/code]

[B]minus[/B] 0.6 hours ago? so he will reserve in 36 minutes from now? :grin:[/QUOTE]

One bit of the code was using localtime and another bit was using gmtime. Fixed now.

[QUOTE=fivemack;187959]That's eight experiments (13e vs 14e, 16777215 vs 33554431, lpb=28 vs 29); sieve the same range, presumably 2^25 .. 2^25+1000, for each of them and see what works better.[/QUOTE]

Results (on my C2D laptop, not on my very busy P4 as yesterday) (sieving from ~2^24 to ~2^24+1000 (ooops), mfb = 2*lp)

I calculated a "28-score" to compare the "lp29" results with the "lp28" ones; just yield/2 and sec/rel*2.

a3: rlim = alim = 2^25-1, lp 28, 13e
total yield: 1509, q=16779001 (0.09158 sec/rel)

b3: rlim = alim = 2^25-1, lp 29, 13e
total yield: 3128, q=16779001 (0.04460 sec/rel)
"28-score": 1564 / 0.08920

c3: rlim = alim = 2^24-1, lp 28, 13e
[B]total yield: 1306, q=16779001 (0.08176 sec/rel)[/B]

d3: rlim = alim = 2^24-1, lp 29, 13e
total yield: 2603, q=16779001 (0.04205 sec/rel)
"28-score": 1301.5 / 0.08410

a4: rlim = alim = 2^25-1, lp 28, 14e
total yield: 3082, q=16779001 (0.09659 sec/rel)

b4: rlim = alim = 2^25-1, lp 29, 14e
total yield: 6240, q=16779001 (0.04877 sec/rel)
"28-score": 3120 / 0.09754

c4: rlim = alim = 2^24-1, lp 28, 14e
total yield: 2584, q=16779001 (0.09689 sec/rel)

d4: rlim = alim = 2^24-1, lp 29, 14e
total yield: 5061, q=16779001 (0.04965 sec/rel)
"28-score": 2530 / 0.0933

So I think I will take rlim = alim = 2^24-1, lp 28, with the 13e siever - or should I repeat the experiment with Q0=2^25?
Edit: Just tweaked my script accordingly and started the experiment.

[QUOTE=Andi47;188027]
So I think I will take rlim = alim = 2^24-1, lp 28, with the 13e siever - or should I repeat the experiment with Q0=2^25?
Edit: Just tweaked my script accordingly and started the experiment.[/QUOTE]

Q0=2^25 gives the same result:

a3: rlim = alim = 2^25-1, lp 28, 13e
total yield: 1517, q=33555439 (0.13332 sec/rel)

b3: rlim = alim = 2^25-1, lp 29, 13e
total yield: 3053, q=33555439 (0.07026 sec/rel)
28-score: 1526.5 / 0.14052

c3: rlim = alim = 2^24-1, lp 28, 13e
[B]total yield: 1252, q=33555439 (0.10164 sec/rel)[/B]

d3: rlim = alim = 2^24-1, lp 29, 13e
total yield: 2403, q=33555439 (0.05264 sec/rel)
28-score: 1201.5 / 0.10521

a4: rlim = alim = 2^25-1, lp 28, 14e
total yield: 2977, q=33555439 (0.12830 sec/rel)

b4: rlim = alim = 2^25-1, lp 29, 14e
total yield: 6098, q=33555439 (0.06318 sec/rel)
28-score: 3049 / 0.12636

c4: rlim = alim = 2^24-1, lp 28, 14e
total yield: 2380, q=33555439 (0.12204 sec/rel)

d4: rlim = alim = 2^24-1, lp 29, 14e
total yield: 4674, q=33555439 (0.06316 sec/rel)
28-score: 2337 / 0.13632

Extrapolating from that says that c3 would be the fastest to get ~28M relations (I guess I would need ~27M ... 35M rels) with ~2.567M seconds (~30 CPU-days), so I will go with rlim = alim = 2^24-1, lp 28, 13e.

[QUOTE=fivemack;187541]Touché. Running them now (4 threads on i7 each, since 5pm on 23rd August) and should have factors before the first birthday of the reservation.[/QUOTE]

And done:
[code]
Wed Sep 2 02:21:31 2009 prp54 factor: 156856020082959523616587909662783965994208064240475509
Wed Sep 2 02:21:31 2009 prp110 factor: 17426077921418681977480489616045159043445721454789036743338720418172464114055358444947859361977046685927541261

Wed Sep 2 02:00:03 2009 prp81 factor: 189336367039730109801009638518058760056609393829914619289697816020709477269865093
Wed Sep 2 02:00:03 2009 prp94 factor: 8283442039618108275937758760461138397871290459364421016175942327901266305862698032989481720917
[/code]

For this size of number, sieving with 29-bit large primes takes exactly the same time, namely 770 i7/2800 thread-hours (eight days), whether you use small prime bound 3*2^23 and work backwards (to about q=12 million) or 2^24 and work forwards (to about q=30 million).