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I am trying to draw a picture using Mathematica of three loops linking together in 3D as follows.

3D visualization

Module[{r = 0.03, col1, col2, col3}, {col1, col2, col3} = 
ColorData["HTML"] /@ {"Firebrick", "ForestGreen", "RoyalBlue"};
    Rotate[#, π/12, {0, 0, 1}, {0, 0, 0}]}, {col3, 
    Rotate[#, -π/12, {0, 0, 1}, {0, 0, 0}]}} &@
    0.5 {Cos[θ], 0, Sin[θ]}, {θ, 0, 
     2 π, π/24}], r], {1, 0, 0}]}, {col1, 
Tube[Table[{Cos[θ], Sin[θ], 0}, {θ, 0, 
   2 π, π/24}], r]}},
ViewPoint -> {5, 0, 2}, Boxed -> False, Lighting -> "Neutral"]]

I used Graphics3D and Tube to draw the above picture. But I found that it is not easy to see which line is in front of which.

So then I use Graphics and Circle to draw the following 2D picture, in which the overlapping relation is represented by a small gap of the underlying line at the intersection.

2D visualization

Module[{col1, col2, col3},
{col1, col2, col3} = 
ColorData["HTML"] /@ {"Firebrick", "ForestGreen", "RoyalBlue"}; 
Graphics[{{col1, Circle[{0, 0}, {2, 1}]}, 
Translate[{{White, Disk[{-0.47, -0.08}, 0.15]}, col2, 
 Circle[{0, 0}, {0.5, 1}, {0.07 π, 1.97 π}]}, {0.6, -0.9}],
Translate[{{White, Disk[{-0.47, 0.12}, 0.15]}, col3, 
 Circle[{0, 0}, {0.5, 1}, {0.02 π, 
   1.92 π}]}, {-0.8, -0.9}]}]]

I think the 2D picture is nicer and can be saved as the vectorized image with a much smaller size compared to the 3D version. However I need to explicitly tell Mathematica where and how to break the lines.

My question is: is there a method to have Mathematica automatically draw links or knots in the 2D style with the underlying lines broken at the intersections.

share|improve this question
Please share your code for the images, it might help people get started. –  Pickett May 20 at 7:01
If you need 3D figure, you can try making torus using ParametricPlot3D[] reference.wolfram.com/mathematica/ref/ParametricPlot3D.html use the third example and reduce the tube radius. –  Sumit May 20 at 8:46
I must agree with Pickett. It would be much easier to provide a useful answer if we knew your input. –  Mark McClure May 20 at 15:31

2 Answers 2

up vote 17 down vote accepted

For 3D curves, you can use an old trick sometimes used for toon-style rendering. Render each curve twice: once normally to show the curve itself; once thicker and in pure white, with only the backward-facing polygons drawn, creating an outline around the curve that occludes other curves passing behind it.

(P.S. The trick is called the two-pass method in Gooch et al.'s survey of silhouette algorithms.)

torusKnot[p_, q_, t_] := With[{r = Cos[q t] + 2}, {r Cos[p t], r Sin[p t], Sin[q t]}]
points = Table[torusKnot[2, 3, t], {t, 0, 2 π, 2 π/200}];
  Lighter@Orange, Tube[points, 0.05], 
  FaceForm[None, Glow[White]], Tube[points, 0.15]}]

enter image description here

You can even rotate this interactively and the gaps still work.

(P.P.S. If you want a 2D flat-colour look, replace Lighter@Orange with Glow[Orange], Black.)

share|improve this answer

There must be a better way to do this, but...

Here's a 3D image of a trefoil knot

trefoil[t_] = {Sin[3 t], Sin[t] + 2 Sin[2 t], Cos[t] - 2 Cos[2 t]};
Show[ParametricPlot3D[trefoil[t],{t, 0, 2 Pi}, 
  ViewPoint -> {25, 0, 0}, Boxed -> False, Axes -> False] /. 
  Line[pts_] :>Tube[pts, 0.4], PlotRange -> All, ImageSize -> 500]

enter image description here

We get a plain plane image by projecting along the $x$-axis.

trefoil2d[t_] = {Sin[t] + 2 Sin[2 t], Cos[t] - 2 Cos[2 t]};
{plot2d, {ts}} = Reap[
  ParametricPlot[{Sin[t] + 2 Sin[2 t], Cos[t] - 2 Cos[2 t]},
   {t, 0, 2 Pi}, Axes -> False, EvaluationMonitor :> Sow[t], 
    ImageSize -> 500]];
ts = Sort[ts];

enter image description here

Note that ts now contains the $t$ values used by ParametricPlot to generate the image.

ts // Short

(* Out: {0., 0.00192704, 0.00385407, <<1226>>, 6.27887, 6.28103, 6.28319} *)

We'll use these to manually construct a sequence of striped strips that create the effect you want. Here is such a striped strip expressed in terms of a time pair.

{xprime[t_], yprime[t_]} = trefoil2d'[t];
normal[t_] = {yprime[t], -xprime[t]};
normal[t_] = Simplify[normal[t]/Norm[normal[t]], Element[t, Reals]];
thickness = 0.1;
strip[{t1_, t2_}] := {
   {White, EdgeForm[White], Polygon[{
      trefoil2d[t1] + thickness*normal[t1],
      trefoil2d[t2] + thickness*normal[t2],
      trefoil2d[t2] - thickness*normal[t2],
      trefoil2d[t1] - thickness*normal[t1]}]},
   {Thickness[0.004], Line[{trefoil2d[t1], trefoil2d[t2]}]}};

Now, we lay those down sorted by the $x$ component of the 3D position

pointTimePairs = Table[{trefoil[ts[[i]]], {ts[[i]], ts[[i + 1]]}}, 
  {i, 1, Length[ts] - 1}];
xSortedTimes = Last /@ SortBy[pointTimePairs, #[[1, 1]] &];
Graphics[strip /@ xSortedTimes, ImageSize -> 500]

enter image description here

Perhaps, we can see what's going on a bit better by generating a shaded background, rather than a white background and creating an animation showing the order in which the pieces are laid down.

shadedStrip[{t1_, t2_}, {x_}] := {
   {GrayLevel[1 - x/(1.2 Length[xSortedTimes])], 
    EdgeForm[GrayLevel[1 - x/(1.2 Length[xSortedTimes])]], Polygon[{
      trefoil2d[t1] + thickness*normal[t1],
      trefoil2d[t2] + thickness*normal[t2],
      trefoil2d[t2] - thickness*normal[t2],
      trefoil2d[t1] - thickness*normal[t1]}]},
   {Thickness[0.004], Line[{trefoil2d[t1], trefoil2d[t2]}]}};
pic[n_] := Graphics[MapIndexed[shadedStrip, xSortedTimes[[1 ;; n]]],
  PlotRange -> {{-3, 3}, {-3.5, 2.5}}, ImageSize -> 500];
pics = Table[pic[n], {n, 1, Length[xSortedTimes], 30}];
pics = Join[pics, Table[Last[pics], {10}]];
Export["anim.gif", pics]

enter image description here

share|improve this answer
A nice general approach for 2D curves! But if I'm not mistaken, it doesn't take into account the under/over-ness of the original knot's crossings? –  Rahul May 20 at 12:24
@RahulNarain Better? –  Mark McClure May 20 at 12:55
Looks good to me. Unfortunately I already gave you a +1. :) –  Rahul May 20 at 13:03
You might do with fewer strips if you could compute the $t$ values of the crossings themselves and draw curve segments around them. I tried doing Solve[Rest@Thread[trefoil[t1] == trefoil[t2]], {t1, t2}] but it only gives the $t_1=t_2$ solution. –  Rahul May 20 at 23:22
@RahulNarain We can find all the crossings as follows: N[{ToRules[Reduce[{trefoil2d[t1] == trefoil2d[t2], 0 < t1 < 2 Pi, 0 < t2 < 2 Pi, t1 < t2}, {t1, t2}]]}]. –  Mark McClure May 21 at 9:04

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