bugs - Why does one of my Manipulate integer variables get turned into a real?

Bug introduced in 9.0 or earlier and persisting through 11.0.1 or later

I'm developing code for manipulating graphs. Spot the difference in the 2 Manipulates below:

Manipulate[

 {n, m},
 {{n, 10, "no of vertices"}, 1, 18, 1},
 {{m, n, "number of edges"}, n, n (n - 1)/2 , 1}
 (* using the +/-/slider for the number of edges does not generate Reals 
    but only Integers as expected *)]

first image of Manipulate

this first example works as expected but

Manipulate[
 {n, m},

 {{n, 10, "no of vertices"}, 1, 18, 1},
 {{m, n, "number of edges"}, n - 1, n (n - 1) /2, 1}
 (* using the +/-/slider for the number of edges now generates Reals 
    not Integers: what gives? *)]

second image of Manipulate

turns my nice integers into reals as soon as I change the number of edges. I can turn this back into an integer when that's what I need, but I'd prefer not to. Naturally I have a more complex task to complete but these examples are simple enough to illustrate the issue.

Version: Mathematica 9.0.1.0

Platform: Mac

Answer

This is an example of unintended behavior in Mathematica that the developers are aware of. There are a number of workarounds. The simplest is to wrap a function around the appropriate bounds; instead of

Manipulate[m, {n, {10}}, {m, n - 1, 2 n, 1}]

which illustrates the problem, try

Manipulate[Floor@m, {n, {10}}, {m, n - 1, 2 n, 1}]

A solution can also be pushed into the controller itself:

Manipulate[m, {n, {10}}, {{m, 9}, Dynamic[Slider[#1, {n - 1, 2 n, 1}]] &}]

Thanks for this feedback and suggestions are due to Karl Isensee.

Comments

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....

Blog

Search This Blog