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Oct 2011
7×97 Posts |
I found a TF factor that P-1 would never find (~2.6 million GHzD) and I was wondering approx how long it should take for ECM to find it. The P-1 parameters would be B1=3 and B2=~1.42e+13.
I know the 'normal' B1/B2 for a ~25 digit number is 50,000/5,000,000. Would this indicate this factor could not be found by ECM? |
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#2 | |
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"Ben"
Feb 2007
DBC16 Posts |
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#3 | |
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Nov 2003
164448 Posts |
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READ what has been written, they wouldn't have to ask such questions. My joint paper with Sam Wagstaff Jr. should be REQUIRED READING for anyone who wants to discuss ECM. I ask: why wouldn't anyone interested in this subject want to read it? Is the answer: laziness? Or is it that the claim of interest is a shallow one? |
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#4 | |
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"Ben"
Feb 2007
22×3×293 Posts |
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Now pretend that the instruction manual to their remote control exists only in scavenger hunt form, where they follow one clue to another clue to another clue, etc., until they can learn how to turn on their TV. That is close to how it is with advanced number theory software - tools that are fun to use but that have widely-dispersed, partially-hidden, and variable-content instruction manuals. This forum provides a short-cut through much of that. If someone following the short cut discovers something shiny along the way, and goes off and learns some number theory as a result, then that's a bonus, but not necessarily the primary purpose of this forum. As such your link to your paper is great, but your comments not so much. [/IMO] Last fiddled with by bsquared on 2013-04-29 at 18:18 |
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#5 | |
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Nov 2003
22×5×373 Posts |
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run the software. He was asking about the capabilities of the algorithm!!!! |
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#6 | |
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"Ben"
Feb 2007
22·3·293 Posts |
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But like I said, I'm not taking exception with you trying to get people to learn about the algorithm, I'm taking exception with your methods. |
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#7 | |
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Oct 2011
7·97 Posts |
Quote:
![]() So, to put it simply yet verbosely, while trial factoring 63186931 I found a 72.184 bit factor of length 22 digits. In the expression 2kp-1, k = 3 × 14154475893511. From scratch this TF would take a little over 30 GHzD to be found. As I stated P-1 would take over 2.6 million GHzD. While I understand ECM is largely dependant on luck time-wise when it comes to finding a factor since it could be found on Curve 1 or Curve 1001, I was curious if this factor could even be found under normal 25 digit length B1/B2 parameters. When I plugged this exponent into P95 with a B1 of 50,000 and 300 curves (which seems to be the recommended amount) it said the run would last 63 days on my laptop. I didn't feel like waiting that long (or wasting those resources) to see if it would find it, so I felt it was time to ask the experts. You may also note I did not ask this question in the MATH section, where I was sure I'd get a lecture. But, hey, lectures seem to find me. Last fiddled with by bcp19 on 2013-04-29 at 19:28 |
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#8 | ||
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Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17×251 Posts |
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The answer to "is ECM at digit level x the best general-purpose way a good way to find this factor of size y?" is yes if x is close to y (no more than 3** apart). Quote:
If you knew in advance that the factor was 22 digits, you could choose a B1 that optimized that digit level, and maybe find it in about expected 150** GHz-days. * "expected" being the same as what bsquared defined earlier: there is about a 37% chance (1 / exp(1)) of missing the factor after running enough curves to expect to find the factor. ** Numbers marked with this have been made up on the spot, and may be wildly inaccurate, but can help give you an idea of what you're dealing with. Last fiddled with by Mini-Geek on 2013-04-29 at 19:41 |
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#9 |
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Einyen
Dec 2003
Denmark
61268 Posts |
Now that you know the factor, you use the Magma Calculator to check which B1/B2 would be required for random sigma values:
http://magma.maths.usyd.edu.au/calc/ Just paste this code in and change the sigma value s at the bottom to a random (32 bit I think) number. Code:
FindGroupOrder2 := function (p, s) K := GF(p); v := K ! (4*s); u := K ! (s^2-5); x := u^3; b := 4*x*v; a := (v-u)^3*(3*u+v); A := a/b-2; x := x/v^3; b := x^3 + A*x^2 + x; E := EllipticCurve([0,b*A,0,b^2,0]); return FactoredOrder(E); end function; p := 5366267349746657428447; s := 10000; FindGroupOrder2(p, s); [ <2, 4>, <3, 1>, <5, 1>, <57223, 1>, <16530659, 1>, <23637431, 1> ] The last 2 digits is the B1 and B2 value required to find the factor with sigma = 10000, so B1 = 17e6 and B2=24e6. So you can just try random sigma values to see how low you can find B1/B2. You can also download the Magma Calculator but I haven't tried that. |
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#10 | |
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Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
426710 Posts |
Quote:
I had momentarily confused your case for the more commonly-noted case of a large P-1 factor, where P-1 is the quick way and TF is the slow way. E.g. this factor I found that, at 41 digits, would take 1.24 GHz-days of P-1 or 18 quintillion (1.8*10^19) GHz-days of TF. |
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#11 | |
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Nov 2003
22·5·373 Posts |
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discusses HOW to select parameters when the size of the factor is unknown. After running some trials one uses information from the ECM FAILURE, and the theoretical distribution of factor sizes along with BAYES THEOREM to reselect parameters in an optimal way. |
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