Skip to main content

bugs - Sometimes Region can't build right region


During my work with NDSolveFEM I found strange behavior of function Region.


a = 600;

reg1 = Rectangle[{0, 0}, {6 a, a}];
hole1 = Disk[{3 a, 0.75 a}, a/8];
hole2 = Disk[{3 a, 0.25 a}, a/8];
hole3 = Disk[{2.5 a, 0.5 a}, a/8];
resreg = RegionDifference[reg1, RegionUnion[hole1, hole2, hole3]]

So, it is really simple region with 3 holes near the centre of rectangle 6a * a.


resreg is:


BooleanRegion[#1 && ! (#2 || #3 || #4) &, {Rectangle[{0, 0}, 
{3600, 600}], Disk[{1800, 450.}, 75], Disk[{1800, 150.}, 75], Disk[{1500., 300.}, 75]}]


If I try to plot it by Region on my first computer, I will receive the simple Quad region 1 * 1: enter image description here


So, this region is unmeshable and I can't understand the reason of this behavior.


If I try it on my notebook. I will receive the right plot and meshable region. enter image description here


Mesh of region: enter image description here


Both computers have Windows 10 with Mathematica v 11.3. How I can make Region plot right image?



Answer



This is a bug and you should report it to support@wolfram.com. Here is why - simply because DiscretizeRegion has the same problem and does not give a message about it:


a = 600;
reg1 = Rectangle[{0, 0}, {6 a, a}];

hole1 = Disk[{3 a, 0.75 a}, a/8];
hole2 = Disk[{3 a, 0.25 a}, a/8];
hole3 = Disk[{2.5 a, 0.5 a}, a/8];
resreg = RegionDifference[reg1, RegionUnion[hole1, hole2, hole3]];

enter image description here


The finite element functions have the same problem, but they give a message about what is going on. Start with a fresh kernel:


Quit

And then:



a = 600;
reg1 = Rectangle[{0, 0}, {6 a, a}];
hole1 = Disk[{3 a, 0.75 a}, a/8];
hole2 = Disk[{3 a, 0.25 a}, a/8];
hole3 = Disk[{2.5 a, 0.5 a}, a/8];
resreg = RegionDifference[reg1, RegionUnion[hole1, hole2, hole3]];
NDSolve`FEM`ToElementMesh[resreg]

This gives a message:


enter image description here



So now we know that the RegionBounds computation either timed out or was not able to compute the bounds. However, when you call


RegionBounds[resreg]

You will get the bounds. Once the bounds are computed they are cached. That's why we needed the Quit above to make sure that we start from the same state.


As a workaround you can compute the RegionBounds before calling Region, or, if you happen to already have a FEM mesh just use:


 Region[MeshRegion[myFEMMesh]]

or


MeshRegion[myFEMMesh]

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....