Msieve / lattice siever with degree 7/8 poly

[frmky]

Originally Posted by Batalov

Also note that all 'currently doable' reciprocal octics are better re-dressed as quartics

I was just doing a little housekeeping on my Cunningham spreadsheet and noticed 2,2370L C177. As a quartic it has SNFS difficulty 237.8, but as an octic it has difficulty 190.3. Is there a reason you haven't jumped on it?

[frmky]

Here are a few (including 2,2370L) that seem doable as an octic.

Number      SNFS Difficulty
2,2370L     190.3
10,750L     200.0
10,530L     212.0
10,530M     212.0
10,550M     220.0
10,590M     236.0

[Batalov]

Nope, just left them for others. Why have all the fun to ourselves?
(The 2nd ever octic was now also finished.)
I am not sure about 10s, but 2,2370L is definitely doable as octic.

[henryzz]

Originally Posted by Batalov

Nope, just left them for others. Why have all the fun to ourselves?
(The 2nd ever octic was now also finished.)
I am not sure about 10s, but 2,2370L is definitely doable as octic.

vaguely what equivilent gnfs would 2,2370L be?
it normally would be .7*snfs difficulty = 133.21
but this is an octic
i might chose to make this my first(and probably only for some time:)) cunningham factorization

[Batalov]

It is doable for NFS@Home, not for individual users.

Take the poly from 2,2190L logs; adapt it for this number (1 FBLIM bit should be added, I think, and "sim"med, but later). Run factMsieve.pl with the line siever option turned on. This will give you the Murphy E value most easily, within ~ a minute of typing/editing. (Or prepare the .fb file and all others and run msieve with proper parameters to obtain the E.) The E may be transformed into a "feels-like" sextic number of digits. Off the top of my head, I expect it to be around 245 digits which still easily beats the quartic-237 speed and yield (which will be abysmal; note: the same E estimate can be done for it). E value is a proxy to simulations. I expect them to lead to the same conclusions.

P.S. 0.7 (or rather 0.68) ratio works the opposite way.

P.P.S. oh. I see - you are used to gnfs jobs and want to know what it will feel like? About gnfs-170 :-) ...the number is 177-digits, so gnfs on it will be harder, too.

[Andi47]

Originally Posted by Batalov

P.P.S. oh. I see - you are used to gnfs jobs and want to know what it will feel like? About gnfs-170 :-) ...the number is 177-digits, so gnfs on it will be harder, too.

I almost agree:
The "octic" OPN-factorization which I did was just below the 13e/14e-crossover (faster with 13e, but high duplication rate and large Q-range had to be sieved), near the crossover of lpb27/28 and needed ~26M relations. This also applies to ~GNFS-138 ... GNFS-140, so the ratio would be somewhere between 0.9 and 0.92. So I'd expect an "octic" SNFS-190.3 to feel like GNFS-172 or even GNFS-175.

(but... we need some more datapoints to get a better estimation of the ratio between "octic" SNFS and GNFS.)

[henryzz]

i somewhat underjudged the difference between optimal snfs polynomials and octics
i suppose there probably arent any cunningham numbers of difficuly equivalent to <140 gnfs digits
i am currently considering attempting to set a huge personal nfs record
i suppose that gnfs is preferable for me anyway as i think in terms of gnfs difficulty having only done about 3 snfses
i will do one in the aliquot project instead unless an obvious candidate comes up
my problem with doing my personal record for aliquot sequences is i will do one factorization and then leave the sequence very shortly after as i dont want any other large factorizations
what i really could do with to do larger jobs would be a properly 64-bit siever for windows
i will have to go on linux to set my record
most of what i do is on windows unfortunately
if i had any news of ggnfs v2 then i would wait for it to come but there has been way too long with no word
my currently personal record gnfs is ~127 digits i think but i know i can do 135-140 based on how much effort is needed in the aliquot team sieves
it will be a one off though doing a number that big

[Andi47]

@Henry: If you are looking for a c135...140 number for GNFS, then you can e.g. take one of the Homogeneous Cunningham Numbers. (Currently there are plenty of them available in the range of GNFS-118 to GNFS-170+ and SNFS-157 to SNFS-252. Note: In this table, difficulty printed in green font indicates whether GNFS or SNFS would be easier.)

[jrk]

Originally Posted by henryzz

i will do one in the aliquot project instead unless an obvious candidate comes up
my problem with doing my personal record for aliquot sequences is i will do one factorization and then leave the sequence very shortly after as i dont want any other large factorizations

You can do the current c137 from the aliquot 4788 thread. I just posted a polynomial for it there.

[henryzz]

Originally Posted by jrk

You can do the current c137 from the aliquot 4788 thread. I just posted a polynomial for it there.

i would do one for 4788 except that i dont want to hold that sequence up
i probably will take a couple of weeks to a month to finish the facorization

i have decided to take up Andi47's suggestion of doing a Homogenous Cunningham number
i have reserved 5^319+4^319 on the page he mentioned
this should be a nice challenge
BTW Andi47 read posts 17,18 and 20 of this thread

edit: when i came to calculate the number there were lots of addtional factors that werent mentioned here including a p20 that i had to find myself

[10metreh]

Originally Posted by henryzz

edit: when i came to calculate the number there were lots of addtional factors that werent mentioned here including a p20 that i had to find myself

They are algebraic factors: the p20 is from 5^29+4^29, the 23 * 256147 are from 5^11+4^11 and the 3^2 is from 5^1+4^1.