Skip to main content

front end - Why one needs two Unevaluated to show 1+1 correctly in TreeForm?


The task sounds simple, give the tree plot of 1+1. However, after some trying, I can only make this possible by using two Unevaluated.


TreeForm@Unevaluated@Unevaluated[1+1]


TreeForm Plot



Since Head@Unevaluate[1+1] gives Plus, as expected, my first guess is that one Unevaluated is enough, however


TreeForm@Unevaluated[1+1]


gives 2.


Looks like somehow the front-end evaluates 1+1 again during the plot.


So my question is how to understand this behavior? Or are there any other conceptually correct ways to do the plot?



Answer



This issue is known and it is a defect of TreeForm like Leonid already said. Let me give an illustrative example that shows the same behavior. Let us assume you want a function that just returns its input unevaluated.


Without thinking we put down the function


f1[arg_] := HoldForm[arg]

Now, what seems like a clever idea has one flaw: when you evaluate f1[1+1] the 1+1 is already evaluated before f1 is called, because this is how the standard evaluation in Mathematica works. It evaluates all arguments of a function. Let us look at the Trace:



Trace[f1[1+1]]
(* {{1+1,2},f1[2],2} *)

You see, the 1+1 is turned into 2 and after that, f1 is called. There is no chance your HoldForm could do anything useful.


But now you say, I can prevent exactly this evaluation by using Unevaluated. This is by the way the use-case for Unevaluated: You want to stop the evaluation of an argument to make that it reaches the body of your function unharmed:


Trace[f1[Unevaluated[1+1]]]
(* {f1[1+1],1+1} *)

Perfect. But let's assume you rather want to turn your expression into a String instead of returning it with HoldForm. No problem you will say, because now you know how it works:


f2[arg_] := ToString[arg]

f2[Unevaluated[1+1]]
(* 2 *)

You don't have to think hard why this doesn't work when you have read carefully up to here. On every function call, the arguments are evaluated if this evaluation was not prevented by something. Let us look at the trace:


Trace[f2[Unevaluated[1+1]]]
(* {f2[1+1],ToString[1+1],{1+1,2},ToString[2],2} *)

You see the 1+1 makes it unharmed to ToString but gets evaluated before ToString does its action. How could we prevent this? We could wrap another layer of Unevaluated around our expression:


Trace[f2[Unevaluated@Unevaluated[1+1]]]
(* {f2[Unevaluated[1+1]],ToString[Unevaluated[1+1]],ToString[1+1],1 + 1} *)


The first Unevaluated brings our expression unharmed inside f2, the second Unevaluated makes it survive the call of ToString.


Something similar happens in TreeForm and it is the reason, why a double Unevaluated works.


The final question



Why did f1 work then? There, the one Unevaluated should be eaten by the f1 call and HoldForm should evaluate 1+1 just like ToString did.



Exactly, instead of it doesn't because it has another way to say I don't evaluate my arguments.


Attributes[HoldForm]
(* {HoldAll,Protected} *)


The function HoldForm has the attribute HoldAll which says: "I don't evaluate any of my arguments". And it is true, if you think about it with your knowledge now, the only reason why this


In[22]:= HoldForm[1+1]
Out[22]= 1+1

returns 1+1, has to be something special about HoldForm. Therefore, here comes a third function that concludes the explanation:


SetAttributes[f3,{HoldAll}];
f3[arg_]:=ToString[Unevaluated[arg]]
f3[1+1]
(* 1 + 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]],