Skip to main content

regions - Possible bug in RegionUnion or related functions?


Bug introduced in 11.2.0




Functions for derived geometric regions (RegionUnion and related) have been significantly updated in MMA 11.2 and now this simple example doesn't work any more as expected. Could this be a bug or I am using the functions in the wrong way?


d[radius_Real] := Disk[{0, 0}, radius];
r[size_Real] := Rectangle[-{size, size}, {size, size}]


$Version
Region[
RegionUnion[
RegionDifference[r[4.], r[3.]],
RegionDifference[d[2.], d[1.]]
],
ImageSize -> 200,
PlotRange -> All
]


In version 11.1.1 I get the expected region shape. version_11.1.1


And this is result of version 11.2. Also ToElementMesh fails to create a mesh from this region. (However, it does work if I replace the arguments of r and d with exact numbers.)


version_11.2



Answer



I think this is a bug.


Notice that applying BoundaryDiscretizeRegion to your region shows an error:



BoundaryDiscretizeRegion::defbnds: Unable to compute bounds for the region. Using default bounds of {-1, 1} in all dimensions.




Trying RegionBounds on it returns no result.


This explains the strange output.


However, changing RegionDifference[d[2.], d[1.]] to Region@RegionDifference[d[2.], d[1.]] "fixes" the problem. Now everything works fine.


Region does not seem to change the structure of RegionDifference[d[2.], d[1.]].


My conclusion is that this might be a bug. I would not expect different results when Region is or isn't used, only perhaps different performance (as Region is able to do certain checks a single time, and then guarantee that whatever it contains is a correctly defined region).


I suggest you report this to Wolfram Support.


Comments

Popular posts from this blog

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 - 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 - 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.