g​o​l​f​ ​a​t​a​n​2

Question

Sometimes it really is a struggle to convert Cartesian coordinates (x,y) to Polar coordinates (r,phi). While you can calculate r = sqrt(x^2+y^2) quite easily, you often need some distinction of cases when calculating the angle phi because arcsin,arccos and arctan and all other trigonometric functions have a co-domain that each only spans half the circle.

In many languages there are built-ins for converting rectangular to polar coordinates, or at least have an atan2 function, which - given (x,y) - calculate the angle phi.

Task

Your task is to write a program/function that takes two (floating point, not both zero) Cartesian coordinates (x,y), and outputs the corresponding polar angle phi, where phi has to be in degrees, radians or grades (with grades I mean gradians which are 1/400 of the full circle), whichever is more convenient for you.

The angle is measured in positive orientation, and we have the zero angle for (1,0).

Details

You may not use built-ins that calculate the angle phi given two coordinates, including atan2,rect2polar,argOfComplexNumber and similar functions. However you can use the usual trigonometric functions and their reverses, that only take one argument. Any unit symbols are optional.

The radius r must be non-negative, and phi must be in the range [-360°, 360°] (it does not matter whether you output 270° or -90°).

Examples

Input       Output
(1,1)       45°
(0,3)       90°
(-1,1)      135°
(-5,0)      180°
(-2,-2)     225°
(0,-1.5)    270°
(4,-5)      308.66°

Answer 1

MATL, 12 bytes

yYy/X;0G0<?_

The result is in radians.

Try it online! Or verify all test cases.

Explanation

MATL doesn't have an atan function (it has atan2, but it can't be used for this challenge). So I resorted to acos.

y     % Take x, y implicitly. Duplicate x onto the top of the stack
Yy    % Compute hypothenuse from x, y
/     % Divide x by hypothenuse
X;    % Arccosine (inverse of cosine function)
0G    % Push y again
0<    % Is it negative?
?_    % If so, change sign. Implicitly end conditional branch. Implicitly display
      

Answer 2

JavaScript (ES6), 50 40 bytes

(x,y)=>(Math.atan(y/x)||0)+Math.PI*(x<0)

The result is in radians. Edit: Saved 10 bytes when I noticed that it's allowed for the result to be between -90° and 270°. Previous version with -Math.PI<=result<Math.PI:

(x,y)=>(Math.atan(y/x)||0)+Math.PI*(x<0)*(y>0||-1)

Answer 3

MATLAB / Octave, 24 bytes

@(x,y)atan(y/x)+pi*(x<0)

This defines an anonymous function that produces the result in radians.

Try it on ideone.

Answer 4

Javascript ES6, 54 bytes

(x,y,m=Math)=>x<0&!y?m.PI:m.atan(y/(m.hypot(x,y)+x))*2

Uses radians.

Answer 5

Python 3, 75 67 bytes

8 bytes thanks to Dennis.

from math import*
lambda x,y:pi*(x<0==y)or atan(y/(hypot(x,y)+x))*2

Ideone it!

Answer 6

APL (Dyalog Unicode), 12 10 bytes^SBCS

-2 thanks to ngn.

Anonymous tacit infix function. Uses alephalpha's formula. Takes x as right argument and y as left argument. Result is in radians.

11○∘⍟0J1⊥,

Try it online!

, concatenate the y and x

0J1⊥ Evaluate as base i digits (i.e. yi¹+xi⁰)

⍟ natural logarithm of that

∘ then

11○ imaginary part of that

Answer 7

Jelly, 11 bytes

<0×ØP+÷@ÆṬ¥

Output is in radians. Jelly had a sign bug in its division atoms which was fixed.

Try it online! or verify all test cases (converted to degrees).

How it works

<0×ØP+÷@ÆṬ¥  Main link. Left argument x. Right argument: y

<0           Compare x with 0.
  ×ØP        Multiply the resulting Boolean by Pi.
          ¥  Combine the two links to the left into a dyadic chain.
      ÷@     Divide y by x.
        ÆṬ   Apply arctan to the result.
     +       Add the results to both sides.

Answer 8

Mathematica, 16 bytes

I am not sure whether Log is considered as a built-in that calculates the angle given two coordinates.

N@Im@Log[#+I#2]&

Example:

In[1]:= N@Im@Log[#+I#2]&[1,1]

Out[1]= 0.785398

In[2]:= N@Im@Log[#+I#2]&[4,-5]

Out[2]= -0.896055

Answer 9

x86 machine language (32 bit Linux), 25 13 bytes (noncompeting)

0:       55                      push   %ebp
1:       89 e5                   mov    %esp,%ebp
3:       dd 45 08                fldl   0x8(%ebp)
6:       dd 45 10                fldl   0x10(%ebp)
9:       d9 f3                   fpatan  
b:       c9                      leave
c:       c3                      ret

To try it online, compile the following C program (don't forget -m32 flag on x86_64)

#include<stdio.h>
#include<math.h>
const char j[]="U\x89\xe5\335E\b\335E\20\xd9\xf3\xc9\xc3";
int main(){
  for(double f=-1;f<1;f+=.1){
    for(double g=-1;g<1;g+=.1){
      printf("%.2f %.2f %f %f\n",f,g,atan2(f,g),((double(*)(double,double))j)(f,g));
    }
  }
}

Answer 10

J, 10 bytes

Anonymous tacit infix function. Uses alephalpha's formula. Takes x as left argument and y as right argument. Result is in radians.

11 o.^.@j.

Try it online!

j. calculate x + y×i

@ then

^. natural logarithm of that

11 o. imaginary part of that

Answer 11

Pyth, 26 bytes

AQ?|>G0Hy.tcH+@s^R2Q2G5.n0

theta in radians.

Test suite.

Answer 12

𝔼𝕊𝕄𝕚𝕟, 13 chars / 17 bytes

î<0)*π+МǍ í/î

Try it here (ES6 browsers only).

Uses (x<0)*pi+tan(y/x).

Answer 13

Python 3, 65 bytes

from math import*
f=lambda x,y:atan(y/x if x else y*inf)+pi*(x<0)

This outputs radians in the range [-π/2, 3π/2), equivalent to [-90, 270) degrees.

Answer 14

Axiom, 58 bytes

f(a,b)==(a=0 and b=0=>%i;sign(b)*acos(a*1./sqrt(a^2+b^2)))

test (i use only acos() it returns radiants )

(40) -> [[a,b,f(a,b)*180/%pi] for a in [1,0,-1,-5,-2,0,4] for b in [1,3,1,0,-2,-1.5,-5] ]
   (40)
   [[1.0,1.0,45.0], [0.0,3.0,90.0], [- 1.0,1.0,135.0], [- 5.0,0.0,180.0],
    [- 2.0,- 2.0,- 135.0], [0.0,- 1.5,- 90.0],
    [4.0,- 5.0,- 51.3401917459 09909396]]
                                            Type: List List Complex Float

Answer 15

Python 2, 59 bytes

lambda x,y:acos(x/hypot(x,y))/(-1,1)[y>0]
from math import*

Try it online!

Outputs in radians in range [-pi,pi)