Skip to main content

computational geometry - Diagonals of a regular octagon


We had a little activity after school today and one of the questions was:


All diagonals are drawn in a regular octagon. At how many distinct points in the interior of the octagon (not on the boundary) do two or more diagonals intersect?


So I came home and I wanted to draw the image, which I managed to do, but not very sophisticated code. :-)


pts = Table[{Cos[t], Sin[t]}, {t, 0, 7 Pi/4, Pi/4}];
diags1 = Table[Line[{pts[[1]], pts[[j]]}], {j, {3, 4, 5, 6, 7}}];

diags2 = Table[Line[{pts[[2]], pts[[j]]}], {j, {4, 5, 6, 7, 8}}];
diags3 = Table[Line[{pts[[3]], pts[[j]]}], {j, {5, 6, 7, 8}}];
diags4 = Table[Line[{pts[[4]], pts[[j]]}], {j, {6, 7, 8}}];
diags5 = Table[Line[{pts[[5]], pts[[j]]}], {j, {7, 8}}];
diags6 = Table[Line[{pts[[6]], pts[[j]]}], {j, {8}}];
Show[
Graphics[{
EdgeForm[Black], White, Polygon[pts]
}],
Graphics[{

diags1, diags2, diags3, diags4, diags5, diags6
}]
]

enter image description here


I am looking for a more sophisticated manner to draw the image. Also, is there something in Mathematica that would easily include the points of intersection and count the number of points of intersection?



Answer



Here's one way to use Mathematica to calculate the positions of the internal intersections for you:


(* CirclePoints is an alternative to manually generating the vertex list *)
pts = CirclePoints[{1, 90 Degree}, 8];


(* Generate all edges and internal diagonals as Line objects *)
lines = Line /@ Subsets[pts, {2}];

(* List all possible pairs of lines from the set above *)
linepairs = Subsets[lines, {2}];

(* Find unique intersection points between lines in each pair *)
intersectionpts =
DeleteDuplicatesBy[N]@

Simplify@
DeleteCases[RegionIntersection /@ linepairs, _EmptyRegion];

(* Remove intersection points that are also vertices *)
internalpts = Complement[intersectionpts, Point[{#}] & /@ pts];

RegionIntersection (docs) returns a list of Point objects at which the pairs of diagonals and edges intersect. This list contains duplicates and EmptyRegions corresponding to those line pairs that do not intersect (docs). The EmptyRegions are removed by DeleteCases. The results of the calculation are expressed analytically, and the expressions are not always returned in their simplest form; Simplify then brings them all to a comparable format. Rather than attempting to compare the analytical forms directly, it is more reliable to remove the duplicate points by comparison of the numerical values of their coordinates, which is taken care of by DeleteDuplicatesBy (docs).




internalpts contains a list of Point objects representing those unique intersections. Therefore the number of internal intersection points is:


Length@internalpts

(* Out: 49 *)



Finally we can show the results graphically:


(* Draw results *)
Graphics[{
lines,
Red, PointSize[0.02], internalpts
}]


Mathematica graphics


Comments

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....