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MathGroup Archive 2004

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Re: Elliptic Curves and Cryptography Questions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47661] Re: Elliptic Curves and Cryptography Questions
  • From: jlgpardo at yahoo.es (Jose Luis Gomez Pardo)
  • Date: Wed, 21 Apr 2004 05:22:59 -0400 (EDT)
  • References: <c5tepb$hv0$1@smc.vnet.net> <c603fc$ce8$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Yesterday I posted a follow-up message in this thread and after
sending it I noticed that, probably due to some cut and paste trick,
the "not equal" sign in the line of code:

S = Select[Range[23] - 1, JacobiSymbol[#^3 + # + 4, 23] != -1 &];

was not correctly displayed in my browser. If the line in my post is
different from the one above, please replace it by this one.

Thanks,

Jose Luis.



Paul Abbott <paul at physics.uwa.edu.au> wrote in message news:<c603fc$ce8$1 at smc.vnet.net>...
> In article <c5tepb$hv0$1 at smc.vnet.net>,
>  "flip" <flip_alpha at safebunch.com> wrote:
> 
> > I am working on a presentation with Elliptic Curves/Cryptography and had a
> > few questions I was hoping someone could answer.  I use Mathematica version 
> > 4.0 (wish I could afford 5.0).
> > 
> > The elliptic curve I am using is E: y^2 == x^3 + x + 4 mod 23 (defined over
> > F{23}, that is p = 23).
> > 
> > Questions:
> > 
> > 1.  There are 29 solution set points ((x, y) pairs plus the point at
> > infinity) to this elliptic curve.
> > 
> > {{0,2},
> > {0,21},{1,11},{1,12},{4,7},{4,16},{7,3},{7,20},{8,8},{8,15},{9,11},{9,12},{1
> > 0,5},{10,18},{11,9},{11,14},{13,11},{13,12},{14,5},{14,18},{15,6},{15,17},{1
> > 7,9},{17,14},{18,9},{18,14},{22,5},{22,19}}
> > 
> > Manually, one can check the quadratic residues and then determine if y^2 is
> > in that set (a cumbersome approach).
> > 
> > Is there a command in Mathematica to find this solution set?
> 
> The last point in your solution set is not correct. Here is the correct 
> answer:
> 
>   Reduce[y^2 == x^3 + x + 4, {x, y}, Modulus -> 23]
> 
>   pts = {x, y} /. {ToRules[%]}
> 
> {{0, 2}, {0, 21}, {1, 11}, {1, 12}, {4, 7}, {4, 16}, {7, 3}, {7, 20}, 
> {8, 8}, {8, 15}, {9, 11}, {9, 12}, {10, 5},  {10, 18}, {11, 9}, {11, 
> 14}, {13, 11}, {13, 12}, {14, 5}, {14, 18}, {15, 6}, {15, 17}, {17, 9}, 
> {17, 14}, {18, 9},  {18, 14}, {22, 5}, {22, 18}}
> 
> Checking the solution is straighfoward:
> 
>   f = Function[{x, y}, Mod[y^2 - (x^3 + x + 4), 23] == 0]; 
> 
>   f @@@ pts // Union
> 
> > ...  text omitted ...
> >
> > 4.  When I do,
> > 
> > << Graphics`ImplicitPlot`
> > 
> > ImplicitPlot[y^2 == x^3+x+4, {x, -2,2}]
> > 
> > I am wondering if there is a way to show the elliptic curve E and to
> > superimpose the points above and their values so the points are visible
> 
> Actually, because you want to show a family of curves, ContourPlot is 
> probably better:
> 
>   ContourPlot[y^2 - ( x^3 + x + 4), {x, 0, 12}, {y, 0, 23},     
>    Contours -> 23 Range[-60, 20], ContourShading -> None,     
>    Epilog -> {Hue[1], AbsolutePointSize[5], Point /@ pts}];
> 
> > and the plot is useable in a powerpoint slide?
> 
> Others will tell you how to do this, and it has appeared in this group 
> previously: do a google search on powerpoint at
> 
> http://groups.google.com/groups?hl=en&lr=&group=comp.soft-sys.math.mathem
> atica
> 
> Personally, I prefer to use the Mathematica SlideShow mechanism, 
> especially if you have a number of Mathematica graphics and equations. 
> You can obtain the SlideShow tools for free for Mathematica 4.2 from the 
> Wolfram website.
> 
> Cheers,
> Paul


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