Skip to main content

formatting - How does MakeBoxes handle an n-ary operator?


I want to format results from Reduce using the $\cup$ symbol instead of the $\lor$ symbol. For example, Reduce[x^2 > 4, x, Reals] produces $x<-2\lor x>2$ while I would like it to produce $x<-2\cup x>2$.


My current solution is to add the head xO to use in place of the Or head and associate the appropriate formatting with the xO head using MakeBoxes.


xO /: MakeBoxes[xO[x_, y_], form_] :=                                                             
RowBox[{MakeBoxes[x, form], "\[Union]", MakeBoxes[y, form]}];


When I want to invoke the special formatting, I simply substitute xO for Or.


Reduce[x^2 > 4, x, Reals] /. Or -> xO

This produces the desired result of $(x<-2)\cup (x>2)$. I am feeling like a master bit-twiddler at this point.


Of course, my next test case didn't work.


Reduce[x^3 + 2 x^2 - x - 2 == 0] /. Or -> xO
(* xO(x==-2,x==-1,x==1) *)

My MakeBoxes[xO[x_, y_]] definition above does not match the case when xO has three arguments. I added the following to limp along a little further.



xO /: MakeBoxes[xO[x_, y_, z_], form_] :=                                                         
RowBox[{MakeBoxes[x, form], "\[Union]", MakeBoxes[y, form], "\[Union]", MakeBoxes[z, form]}];

This works for three arguments, but is clearly not a robust solution. What is the general method for an n-ary function?


I tried several variations of the following:


xO /: MakeBoxes[xO[x_, y__], form_] :=                                                            
RowBox[{MakeBoxes[x, form], "\[Union]", MakeBoxes[y, form]}];

which produced the error:


MakeBoxes::argt: "MakeBoxes called with 3 arguments; 1 or 2 arguments are expected"


when applied to the Reduce[x^3 + 2 x^2 - x - 2 == 0] /. Or -> xO test case.


I am probably missing something obvious. Any ideas?



Answer



You were almost there. You just need to use the multiple-argument pattern, and generalize your code accordingly to create the internals of RowBox programmatically:


xO /: MakeBoxes[xO[x___], form_] :=
RowBox[
Riffle[
Map[MakeBoxes[#, form] &, {x}],
"\[Union]"

]
]

Note however that the above implementation leaks evaluation. It may or may not be a problem, but for example here:


x = 1;
xO[x == 1, x > 1, x < 1]

one may argue that the desired result should not be sensitive to the possible global values that x may have, so the result:


(* True \[Union] False \[Union] False *)


may be unsatisfacory. Thus, here is a more careful version:


ClearAll[xO];
SetAttributes[xO, HoldAllComplete];
xO /: MakeBoxes[xO[x___], form_] :=
RowBox@Riffle[
List @@ Replace[
HoldComplete[x],
elem_ :> With[{eval = MakeBoxes[elem, form]}, eval /; True],
{1}
],

"\[Union]"
]

which now gives


xO[x == 1, x > 1, x < 1]

(* x == 1 \[Union] x > 1 \[Union] x < 1 *)

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 - Adding a thick curve to a regionplot

Suppose we have the following simple RegionPlot: f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}] Now I'm trying to change the curve defined by $y=g[x]$ into a thick black curve, while leaving all other boundaries in the plot unchanged. I've tried adding the region $y=g[x]$ and playing with the plotstyle, which didn't work, and I've tried BoundaryStyle, which changed all the boundaries in the plot. Now I'm kinda out of ideas... Any help would be appreciated! Answer With f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 You can use Epilog to add the thick line: RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}, PlotPoints -> 50, Epilog -> (Plot[g[x], {x, 0, 2}, PlotStyle -> {Black, Thick}][[1]]), PlotStyle -> {Directive[Yellow, Opacity[0.4]], Directive[Pink, Opacity[0.4]],