20190226, 04:36  #1 
"特朗普trump"
Feb 2019
朱晓丹没人草
2^{3}×17 Posts 
what does sigma mean in ECM?
I recently take great interest in ECM factorization ,
https://members.loria.fr/PZimmermann...ds/ecmnet.html what sigma can be used for ? 
20190226, 11:25  #2 
Mar 2018
3×43 Posts 
I'll keep it simple.
When factoring with ECM, you run the algorithm on the numer using multiple different curves of the same "size", each of which can yield a factor (or not). It is known how many (different) curves you need at each factor size to get a certain probability to find a factor (of that size). The sigma parameter is a number, from which the parameters (like coefficients) of the curve are derived. Testing with the same number with the same sigma, b1 and b2 would yield absolutely the same result. Running 100 curves means randomly chosing 100 sigmas. You might also notice that sigmas start with "0:" or "1:" – those are different parametizations. Basically, also just different ways to get curves. Last fiddled with by DukeBG on 20190226 at 11:26 
20190227, 01:07  #3 
"特朗普trump"
Feb 2019
朱晓丹没人草
210_{8} Posts 
https://members.loria.fr/PZimmermann...s/ecm/go.magma
I found it, but why do no use A and B instead of sigma (where y^2=x^3+A*x+B)? Is sigma much more useful than A and B? 
20190308, 02:12  #4 
"特朗普trump"
Feb 2019
朱晓丹没人草
2^{3}·17 Posts 
Code:
FindGroupOrder := function (p, sigma) K := GF(p); v := K ! (4*sigma); u := K ! (sigma^25); x := u^3; b := 4*x*v; a := (vu)^3*(3*u+v); A := a/b2; 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:=40122362455616221971122353; sigma:=2334843726764251; FindGroupOrder(p,sigma); magma calculator online output [ <2, 3>, <3, 1>, <5, 1>, <19, 1>, <71, 1>, <281, 1>, <331, 1>, <4271, 1>, <8887, 1>, <70206001, 1> ] Last fiddled with by bbb120 on 20190308 at 02:14 
20190308, 02:17  #5 
"特朗普trump"
Feb 2019
朱晓丹没人草
2^{3}×17 Posts 
https://members.loria.fr/PZimmermann...op102018.html
Code:
FindGroupOrder := function (p, sigma) K := GF(p); v := K ! (4*sigma); u := K ! (sigma^25); x := u^3; b := 4*x*v; a := (vu)^3*(3*u+v); A := a/b2; 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:=444391024295554825813920762553875384889500352609895126972409492191251; sigma:=2807183577; FindGroupOrder(p,sigma); 1>, <22098383, 1>, <136265083, 1>, <396868981, 1>, <809136473, 1>, <6258955966441, 1> ] 
20201015, 17:16  #6  
Random Account
Aug 2009
Not U. + S.A.
2^{2}·13·53 Posts 
Quote:


20210126, 13:44  #7 
Aug 2006
Monza, Italy
73 Posts 
You want the cardinality of the group generated by your elliptic curve to have as many small factors as possible. The specific form of the curves generated by the sigma parameters ensure that such cardinality is divisible by 4 (if I remember correctly, for old versions) or 12 (new versions). Sigma starting with "0:" or "1:" denote the old and new version, respectively.

20210126, 16:18  #8 
Random Account
Aug 2009
Not U. + S.A.
2^{2}×13×53 Posts 
I would take the above to mean the more sigma's applied to the same candidate, the higher the chances are of finding a factor. There still needs to be increasing values of B1 and B2 during the process, otherwise a high enough magnitude may not be reached to find a factor, depending on the candidate.

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