At the WWDC 2003 keynote, Theodore Gray took the stage, along with Phill Schiller, and demoed Mathematica 5, comparing performance on G5 and Xeon processors.

enter image description here

In the presentation, a 40-step fractal battle is conducted and I'd like to recreate this on my new Macbook. So I was wondering what function is being used to make the DensityPlots (if that's what they are)?

enter image description here

  • 4
    $\begingroup$ You don't, by chance, have any more pixels, do you? $\endgroup$
    – Jason B.
    Commented Nov 21, 2019 at 20:44
  • $\begingroup$ I looked but the best YouTube video is only 480p and the text is unreadable $\endgroup$
    – M.R.
    Commented Nov 22, 2019 at 0:41
  • $\begingroup$ Looks like it computing some high order polynomial roots $\endgroup$
    – M.R.
    Commented Nov 22, 2019 at 2:43
  • $\begingroup$ @jasonb ok I added a gif of the screen capture... $\endgroup$
    – M.R.
    Commented Dec 3, 2019 at 7:23
  • 3
    $\begingroup$ From Michael Trott: "I darkly remember making something for Theo. But I don't think I have it anymore." $\endgroup$
    – Jason B.
    Commented Dec 3, 2019 at 15:34

1 Answer 1


Yes, that would be ListDensityPlot.

Some example code:

SqrtFractalSetup[{xmin_, xmax_}, {ymin_, ymax_}, n_] := 
     With[{xminn = N[xmin], yminn = N[ymin], 
             xstep = N[(xmax - xmin)/n], ystep = N[(ymax - ymin)/n]}, 
       re0 = Table[r, {r, xminn, xmax, xstep}, {i, yminn, ymax, ystep}]; 
       im0 = Table[i, {r, xminn, xmax, xstep}, {i, yminn, ymax, ystep}]];

SqrtFractalDraw[c_, {xmin_, xmax_}, {ymin_, ymax_}, steps_] := 
     Module[{quadrant}, {re, im} = {re0, im0}; 
        Do[{re, im} = {(c - Sqrt[Abs[re]] + Sqrt[Abs[im]])^2, 
                (1 - Sqrt[Abs[im]] - Sqrt[Abs[re]])^2}; , {steps}]; 
        quadrant = Abs[re + I*im]; 
        ListDensityPlot[quadrant, Mesh -> False, ColorFunction -> Hue, PlotRange -> All]]; 

SqrtFractalSetup[{-1, 1}, {-1, 1}, 400]; 

  N[0.6 + 0.6*Cos[c] + (0.25 + 0.25*Sin[c])*I], {-1, 1}, {-1, 1}, 7], 
    {c, (2*Pi)/9, 2*Pi - (2*Pi)/9, (2*Pi)/36}]

enter image description here

and here is a related video from WWDC 2003:

enter image description here

  • 1
    $\begingroup$ Any words on the provenance of this awesome answer? $\endgroup$
    – Chris K
    Commented Dec 18, 2019 at 8:27

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