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I am trying to understand the following code in this answer regarding Möbius transformation:

(* Projection from the sphere to the plane *)
stereo = Compile[{{xyz, _Real, 1}, {XYZ, _Real, 1}}, Module[{
     r = Sqrt[(xyz[[1]] - XYZ[[1]])^2 + (xyz[[2]] - XYZ[[2]])^2],
     theta = ArcTan[(xyz[[1]] - XYZ[[1]]), (xyz[[2]] - XYZ[[2]])]},
    {(r (1 + xyz[[3]]))/(1 - XYZ[[3]] + xyz[[3]]) Cos[theta + Pi] + xyz[[1]],
     (r (1 + xyz[[3]]))/(1 - XYZ[[3]] + xyz[[3]]) Sin[theta + Pi] + xyz[[2]], 0}]];

I have read the documents for the functions Compile and Module, but still cannot figure out how the code works.

Could anyone elaborate the following?

  • What does {xyz,_Real,1} do and what does xyz[[1]] mean?

    [Added:] In the document for Compile, it is said that

    Compile[{{x1,t1,n1},…},expr] assumes that xi is a rank ni array of objects, each of a type that matches ti.

    But what is 1 in {xyz,_Real,1}? If it means rank 1, then why later it is written that xyz[[1]], xyz[[2]], xyz[[3]], which suggests that xyz is an array?

  • How do Compile and Module work together to give the definition of a function?

  • What is the formula for the function that this code is really defining? What is the input and what it is the output?
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  • $\begingroup$ Part and Compile will pretty quickly help to explain your first bulleted question. $\endgroup$
    – ktm
    Feb 13, 2017 at 20:07
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    $\begingroup$ Be sure to always read under "Details and Options" on documentation pages. You can select [[ and press F1 (or Command-Shift-F) to see what it is (it's array indexing). The formula is exactly as written there, just substitute in r and theta from the first part of the Module. xyz in this case is a three-element real array representing a 3D vector with x, y, z components. $\endgroup$
    – Szabolcs
    Feb 13, 2017 at 20:07
  • $\begingroup$ @Szabolcs: Thank you for your comment. I'm slowly learning things and I might come back later with my own answer. I would like to leave the question undeleted (if it is not a very stupid trivial one :-)). $\endgroup$
    – user664
    Feb 13, 2017 at 21:13
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    $\begingroup$ Rank 1 is an array, you're thinking about rank 0 which is a point. Rank 2 is a matrix, etc. Module does the same thing inside Compile as it does outside Compile. $\endgroup$
    – C. E.
    Feb 14, 2017 at 2:19
  • $\begingroup$ @C.E.: the rank is indeed confusing at first. Thanks for pointing that out. $\endgroup$
    – user664
    Feb 14, 2017 at 2:29

1 Answer 1

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  • As the document says,

    Compile[{{x1,t1,n1},…},expr] assumes that xi is a rank ni array of objects, each of a type that matches ti.

    Hence, {xyz,_Real,1} means xyz is a rank 1 array of "real numbers". Since xyz is a vector, xyz[[1]] means the first component of xyz. Therefore, if xyz={2,4,3}, then xyz[[1]]=2.

  • The expr in Compile[{{x1,t1,n1},…},expr] is given by the block of Module, a simple example of which for seeing how it works could be

    f=Module[{r=x,theta=y},{r,theta,r+theta}];
    
  • Here is how the function really works. Given $x=(x_1,x_2,x_3)$ and $y=(y_1,y_2,y_3)$ in $\mathbb{R}^3$, the output of the function is $$ \bigg(\frac{r(1+x_3)}{1+x_3-y_3}\cos(\theta+\pi)+x_1, \frac{r(1+x_3)}{1+x_3-y_3}\sin(\theta+\pi)+x_2,0\bigg) $$ where $$ r:=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2},\quad \theta:=\arctan\frac{x_2-y_2}{x_1-y_1}. $$

    The coordinate of the center of the sphere $S$ is given by $x=(x_1,x_2,x_3)$ and the radius of the sphere is $1$. This makes sense since the output of the function is of the form $(R\cos\psi+x_1,R\sin\psi+x_2,0)$, which implies that $x_1,x_2$ are the first two coordinates of the center of the sphere.

    On the other hand, $y=(y_1,y_2,y_3)$ is the coordinate of a point on the sphere. The output of the function gives the coordinate of the "sterographic projection" of $y$ (the intersection of the line (which goes through $y$ and the north pole of the sphere) with the complex plane). When $x=(0,0,0),$ the function coincides with the one defined in Wikipedia.


One needs to be careful with the definition $\theta$ since in Mathematica, ArcTan[x,y] is different from the function $\arctan\frac{y}{x}$. For instance one has in Mathematica ArcTan[1,1]=$\frac{\pi}{4}$ and ArcTan[-1,-1]=$\frac{-3\pi}{4}$. This is why $\theta+\pi$ instead of $\theta$ is used in the output of the function. For the sake of a function, one could define $\theta$ as the angle of the vector $(y_1-x_1,y_2-x_2,0)$ with the $X_1$-axis and replace $\theta+\pi$ with $\theta$ in the output of the function.

In the code, one could replace theta = ArcTan[(xyz[[1]] - XYZ[[1]]), (xyz[[2]] - XYZ[[2]])] with theta = ArcTan[(XYZ[[1]] - xyz[[1]]), (XYZ[[2]] - xyz[[2]])] and theta + Pi in the Cos and Sin functions with theta.

Having not checked it though, I think one can simply replace Cos[theta+Pi] with XYZ[[1]]-xyz[[1]] and Sin[theta+Pi] with XYZ[[2]]-xyz[[2]] so that one does not need to introduce the variable theta at all.

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