20190108, 19:37  #1 
Dec 2018
2 Posts 
Noob question about YAFU usage
Hello, as a math and prime number lover i've been following and using yafu for about 3 years but because of my insufficent computer skills sometimes im not able to use it.
I better ask the question directly, Lets say we have a '214 digit' number and we know that there are only 2 factors of it and the factors are both 108 digits. For example, '108 digits x 108 digits = 214 digits'. In this situation, how can i tell yafu to look out for only 108 digits numbers to factorize the 214 digits number? I mean i know that there are 2 factors only and they are 108 digits. With this knowledge, there is no need to lose time and look for other numbers. Is there any usage option for this? ps, sorry for my bad english but i hope you understand what i want to say :) 
20190108, 19:40  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3·11·281 Posts 
Technically, first thing: a product of two 108digit numbers will have at least 215 digits.

20190108, 19:47  #3  
"Ben"
Feb 2007
3×19×59 Posts 
Quote:
Short answer: If you know the size of the factors and they are both large, then you can probably skip directly to the number field sieve. In yafu you would do this with the command "nfs(your_number)". But you shouldn't do this (see below). Longer answer: Assuming this isn't a number with a special form (e.g., equal to something like a^b  c, with a,b,c "smal"), which it probably isn't because of the equal size of the factors, then it would rank among the most difficult factorizations ever attempted by humans. Yafu will not be successful in factoring it. To attempt it, you'll need hundreds of computer cores sustained over a period of months to years, along with a fairly manual process using different tools (CADO or combination of msieve, ggnfs) in a way that yafu isn't designed to optimize. Others on the forum have more experience and could maybe provide more specific guidance if you are determined enough to proceed. 

20190108, 19:54  #4 
Dec 2018
2 Posts 
wow i wasn't really expecting answers that fast :)
@Batalov thanks for correcting i was trying to give examples from nowhere but you are right my example was wrong. @bsquared, ok i think understand it. thank you for your kindness and time to give explanation. 
20190108, 20:05  #5 
"Ben"
Feb 2007
D23_{16} Posts 
Here is a reference point: the factorization of RSA220
It took ~370 CPUyears of sieving effort, in 2016 computeryears. And that's just the sieving. The arguably more difficult matrixsolve stage took weeks on an Infinibandinterconnected cluster. 
20190108, 20:22  #6 
Jul 2018
11111_{2} Posts 
Do you know anything more about these factors, e.g. that their difference is fairly small...?

20190108, 20:40  #7 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
10010000000111_{2} Posts 
Sounds like encryption keys or some such. Why might one want to factor them????

20190109, 20:16  #8  
Aug 2006
174A_{16} Posts 
Quote:


20190109, 21:00  #9  
"Ben"
Feb 2007
3·19·59 Posts 
Quote:
+5 years and +5 digits and maybe one could do the OP's number for about the same? 

20190109, 23:59  #10 
Apr 2012
2·181 Posts 
I'll second B^2's opinion of your assessment. You covered all bases. I wasn't aware of Passmark so your post provided me with a bit of unexpected help also. Ty.
Last fiddled with by jwaltos on 20190110 at 00:01 
20190110, 03:43  #11  
Aug 2006
174A_{16} Posts 
Quote:


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