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I am studying the maxima of function $\overline{r}(x,y)$ of closed shapes.

Suppose the curve is star-shaped with respect to this center point $\mathbf p=(u,v)$, so that any ray emanating from $\mathbf p$ meets the curve exactly once, at say point $\mathbf q$. Then $r = \|\mathbf q - \mathbf p\|$, $\theta$ is the angle between $\mathbf q-\mathbf p$ and the $x$-axis, $\overline{r}(x,y)$ is the average radius $$\overline{r}(x,y)=\frac1{2\pi}\oint_{\mathbf q\in\mathcal C}\|\mathbf q-\mathbf p\|\,\mathrm d\theta.$$

and $\mathbf{p}$ is the point maximizing $\overline{r}$.

(Conveniently, this integral can also be computed for non-star-shaped curves; for a ray that meets the curve multiple times, it amounts to taking the total length of all segments that lie in the interior of the curve.)

In a previous answer, using Mesh Coordinates, Discretization, and Euler's Distance, I could calculate the maxima $\mathbf{p}$ of $\overline{r}(x,y)$.

curve = DiscretizeRegion[
      ImplicitRegion[
       S1[x, y] == 1, {{x, -3, 3}, {y, -4, 4}}], {{-3, 3}, {-4, 4}}, 
      AccuracyGoal -> 8]
    q = MeshCoordinates[curve];
    edges = First /@ MeshCells[curve, 1];
    signedAngle[a_, b_] := Arg[(Complex @@ a)/(Complex @@ b)]
    avgRadius[p_] := 
     1/(2 \[Pi]) Abs[Sum[Module[{q1, q2, r, d\[Theta]}, q1 = q[[First@e]];
         q2 = q[[Last@e]];
         r = EuclideanDistance[p, (q1 + q2)/2];(*midpoint approximation*)
         d\[Theta] = signedAngle[q1 - p, q2 - p];
         r d\[Theta]], {e, edges}]]
    s = FindMaximum[avgRadius[{x, y}], {{x, 0}, {y, 0}}]

However, I wanted to apply a similar definition to closed shapes that contain straight lines.


Shapes With Straight Lines: Polygons

For example, for a triangle with vertices $(-1,0)$, $(0,1)$ and $(1,0)$, if I apply DescritizeRegion

curve = DiscretizeRegion[
  Line[{{-1, 0}, {0, 1}, {1, 0}, {-1, 0}}]], {{-1, 1}, {-1, 
    1}}, AccuracyGoal -> 8]

instead of showing the discretized region it gives the following.

DiscretizeRegion::regp: A correctly specified region expected at position 1 of DiscretizeRegion[,{{-1,1},{-1,1}},AccuracyGoal->8].

Assuming the positions should be labelled, I used BoundaryMeshRegion

curve = DiscretizeRegion[
  BoundaryMeshRegion[{{-1, 0}, {0, 1}, {1, 0}}, 
   Line[{1, 2, 3, 1}]], {{-1, 1}, {0, 1}}, AccuracyGoal -> 8]

but instead of descritizing the boundary, its descritizes the area.

enter image description here

How do I apply $\overline{r}(x,y)$ to the triangle? Similarily how would apply this to a dumbell shaped curve?

enter image description here

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  • 2
    $\begingroup$ Note that Graphics is not a region. You can discretize the 1-dimensional line by removing the Graphics[...] and just using DiscretizeRegion[ Line[{{-1, 0}, {0, 1}, {1, 0}, {-1, 0}}], AccuracyGoal -> 8]. $\endgroup$ – eyorble Jan 2 '18 at 5:16
  • $\begingroup$ @eyorble I tried that but I get DiscretizeRegion::regp: A correctly specified region expected at position 1 of DiscretizeRegion[Line[{-1,0},{0,1},{1,0},{-1,0}],{{-2,2},{-2,2}},AccuracyGoal->4]. $\endgroup$ – Arbuja Jan 2 '18 at 14:03
  • $\begingroup$ As far a I know, there is no dedicated method in Mathematica to discretize a non-fulldimensional MeshRegion. However, even being not the most efficient way and having requirement that the curve has to be closed, you can use curve = BoundaryMesh@ DiscretizeRegion[ BoundaryMeshRegion[{{-1, 0}, {0, 1}, {1, 0}}, Line[{1, 2, 3, 1}]], {{-1, 1}, {0, 1}}] $\endgroup$ – Henrik Schumacher Jan 2 '18 at 16:10
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    $\begingroup$ That's not an issue of the discretization but of division by zero. That happens because the initial guess for FindMinimum is {0,0} which happens to lie on the curve. Try s = FindMaximum[avgRadius[{x, y}], {{x, 0.1}, {y, 0.1}}] instead. $\endgroup$ – Henrik Schumacher Jan 2 '18 at 17:31
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    $\begingroup$ @Arbuja I am not sure that increaing AccuracyGoal in DiscretizeRegion is the right way to achieve higher accuracy here; the error of your method is proportional to the maximal edge length of curve. You can control the maximal edge length with MaxCellMeasure->{1->len} (1 stands for 1-dimensional cells, len for the maximal length you allow). Anyway, using DiscretizeRegion and applying BoundaryMesh on the result is a very expensive way to get a finely smapled curve. See my aanswer below for a more performant approach. $\endgroup$ – Henrik Schumacher Jan 2 '18 at 20:44
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Here is another way to specify a one-dimensional region:

γ = t \[Function] {Cos[t], Sin[t]} (1. + 0.25 Sin[5 t]);
n = 1000;
pts = Table[γ[t], {t, 0, 2 Pi, 2 Pi/n}];
curve = MeshRegion[pts, Line[Transpose[{Range[1, n], Range[2, n + 1]}]]];

It circumvents the need for DiscretizeRegion. You can adapt that also for triangles: Simply parameterize and sample the boundary pieces and use Join to create a single list of points. Your procedure for finding the point of maximal avarage ray length will work quite likely on regions defined this way.

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  • $\begingroup$ @HendrickSchumacer You can’t use any parametrization. For example, consider a ellipse with parametric curve $(2\cos{(t)},3\sin({t}))$ gives a different average distance then $(\frac{6\cos{(t)}}{\sqrt{4\cos^{2}(t)+9\sin^{2}(t)}},\frac{6\sin{(t)}}{\sqrt{4\cos^{2}(t)+9\sin^{2}(t)}})$ $\endgroup$ – Arbuja Jan 3 '18 at 13:01
  • $\begingroup$ @Arbuja Why on earth should the average distance (or average ray length) depend on the parameterization parameter t? Really, could you please think about the problem a bit more thoroughly before asking the next time? $\endgroup$ – Henrik Schumacher Jan 3 '18 at 17:07
  • $\begingroup$ @HenrikSchmacer I never said the average distance must depend on t. What I'm saying is if you parameterize the curve, the distance formula of some parametrizations give the incorrect average radius. $\endgroup$ – Arbuja Jan 3 '18 at 18:45
  • $\begingroup$ @HenrikSchumacer How come you graphed $r=1+0.25\sin{(5\theta)}$. Can your code work for non-polar equations. $\endgroup$ – Arbuja Jan 3 '18 at 18:48
  • $\begingroup$ @Arbuja This may work for any other sufficiently smooth and regular curve. I chose to use a curve in polar coordinates just for convenience. $\endgroup$ – Henrik Schumacher Jan 3 '18 at 20:44

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