# Random point on a sphere

## The Challenge

Write a program or function that takes no input and outputs a 3-dimensional vector of length $$\1\$$ in a theoretically uniform random direction.

This is equivalent to a random point on the sphere described by $$x^2+y^2+z^2=1$$

resulting in a distribution like such ### Output

Three floats from a theoretically uniform random distribution for which the equation $$\x^2+y^2+z^2=1\$$ holds true to precision limits.

## Challenge remarks

• The random distribution needs to be theoretically uniform. That is, if the pseudo-random number generator were to be replaced with a true RNG from the real numbers, it would result in a uniform random distribution of points on the sphere.
• Generating three random numbers from a uniform distribution and normalizing them is invalid: there will be a bias towards the corners of the three-dimensional space.
• Similarly, generating two random numbers from a uniform distribution and using them as spherical coordinates is invalid: there will be a bias towards the poles of the sphere.
• Proper uniformity can be achieved by algorithms including but not limited to:
1. Generate three random numbers $$\x\$$, $$\y\$$ and $$\z\$$ from a normal (Gaussian) distribution around $$\0\$$ and normalize them.
2. Generate three random numbers $$\x\$$, $$\y\$$ and $$\z\$$ from a uniform distribution in the range $$\(-1,1)\$$. Calculate the length of the vector by $$\l=\sqrt{x^2+y^2+z^2}\$$. Then, if $$\l>1\$$, reject the vector and generate a new set of numbers. Else, if $$\l \leq 1\$$, normalize the vector and return the result.
3. Generate two random numbers $$\i\$$ and $$\j\$$ from a uniform distribution in the range $$\(0,1)\$$ and convert them to spherical coordinates like so:\begin{align}\theta &= 2 \times \pi \times i\\\\\phi &= \cos^{-1}(2\times j -1)\end{align}so that $$\x\$$, $$\y\$$ and $$\z\$$ can be calculated by \begin{align}x &= \cos(\theta) \times \sin(\phi)\\\\y &= \sin(\theta) \times \sin(\phi)\\\\z &= \cos(\phi)\end{align}
• Provide in your answer a brief description of the algorithm that you are using.
• Read more on sphere point picking on MathWorld.

## Output examples

[ 0.72422852 -0.58643067  0.36275628]
[-0.79158628 -0.17595886  0.58517488]
[-0.16428481 -0.90804027  0.38532243]
[ 0.61238768  0.75123833 -0.24621596]
[-0.81111161 -0.46269121  0.35779156]


## General remarks

• Is it okay to pick 3 reals uniformly in [-1, 1], then reject them (and repeat) if the sum of their squares isn't 1? Sep 9 '19 at 13:06
• @Grimy I like that loophole. No, it is not allowed, because there is a theoretically zero chance of any output. Sep 9 '19 at 13:25
• Isn't @Grimy's suggestion similar to the second example implementation mentioned by you? That solution also has theoretically zero chance of producing any output Sep 9 '19 at 20:59
• @SaswatPadhi No, that has a chance pi/6 ≈ 0.5236 of producing an output. That's the area of the sphere inscribed in the unit-area cube Sep 9 '19 at 21:41
• @LuisMendo I see, right. The probability is ~0.5 in that case, as you mentioned. For Grimy's proposal, it's ~0. Sep 9 '19 at 21:47

# Wolfram Language (Mathematica), 20 bytes

RandomPoint@Sphere[]


Try it online!

Does exactly what it says on the tin.

# R, 23 bytes

x=rnorm(3)
x/(x%*%x)^.5


Try it online!

Generates 3 realizations of the $$\\mathcal N(0,1)\$$ distribution and normalizes the resulting vector.

Plot of 1000 realizations: • Can you justify 3 axis distributed normally resulting in uniform distribution over the sphere ? (I don't see it) Sep 9 '19 at 20:17
• @Jeffrey It's pretty well-known in probability/statistics; but the proof for 2D (which extends neatly to 3 dimensions) is approximately: let $X,Y\sim N(0,1)$ and independent. Then $f_X(x)=Ke^{-\frac{1}{2}x^2}$ and $f_Y(y)=Ke^{-\frac{1}{2}y^2}$, so by independence $f_{XY}(x,y)=K^2e^{-\frac{1}{2}(x^2+y^2)}=f_Z(z)=K^2e^{-\frac{1}{2}\lVert z\rVert^2}$ where $z=(x,y)$, so it's clear that the distribution of $z$ depends solely on the magnitude of $z$, and thus the direction is uniformly distributed. Sep 9 '19 at 21:07
• So, the normal distribution gives us uniformly distributed points around the circle, and dividing by the magnitude ensures the points lie on the circle Sep 9 '19 at 21:18

# x86-64 Machine Code - 63 62 55 49 bytes

6A 4F                push        4Fh
68 00 00 80 3F       push        3F800000h
C4 E2 79 18 4C 24 05 vbroadcastss xmm1,dword ptr [rsp+5]
rand:
0F C7 F0             rdrand      eax
73 FB                jnc         rand
66 0F 6E C0          movd        xmm0,eax
greaterThanOne:
66 0F 38 DC C0       aesenc      xmm0,xmm0
0F 5B C0             cvtdq2ps    xmm0,xmm0
0F 5E C1             divps       xmm0,xmm1
C4 E3 79 40 D0 7F    vdpps       xmm2,xmm0,xmm0,7Fh
0F 2F 14 24          comiss      xmm2,dword ptr [rsp]
75 E9                jne         greaterThanOne
58                   pop         rax
58                   pop         rax
C3                   ret



Uses the second algorithm, modified. Returns vector of [x, y, z, 0] in xmm0.

Explanation:

push 4Fh
push 3f800000h


Pushes the value for 1 and 2^31 as a float to the stack. The data overlap due to the sign extension, saving a few bytes.

vbroadcastss xmm1,dword ptr [rsp+5] Loads the value for 2^31 into 4 positions of xmm1.

rdrand      eax
jnc         rand
movd        xmm0,eax


Generates random 32-bit integer and loads it to bottom of xmm0.

aesenc      xmm0,xmm0
cvtdq2ps    xmm0,xmm0
divps       xmm0,xmm1


Generates a random 32 bit integer, convert it to float (signed) and divide by 2^31 to get numbers between -1 and 1.

vdpps xmm2,xmm0,xmm0,7Fh adds the squares of the lower 3 floats using a dot product by itself, masking out the top float. This gives the length

comiss      xmm2,dword ptr [rsp]
jne          rand+9h (07FF7A1DE1C9Eh)



Compares the length squared with 1 and rejects the values if it is not equal to 1. If the length squared is one, then the length is also one. That means the vector is already normalised and saves a square root and divide.

pop         rax
pop         rax


Restore the stack.

ret returns value in xmm0

• +1 Using aesenc to produce 128 "random" bits is just beautiful. Sep 10 '19 at 5:39

# Python 2, 86 bytes

from random import*;R=random
z=R()*2-1
a=(1-z*z)**.5*1j**(4*R())
print a.real,a.imag,z


Try it online!

Generates the z-coordinate uniformly from -1 to 1. Then the x and y coordinates are sampled uniformly on a circle of radius (1-z*z)**.5.

It might not be obvious that the spherical distribution is in factor uniform over the z coordinate (and so over every coordinate). This is something special for dimension 3. See this proof that the surface area of a horizontal slice of a sphere is proportional to its height. Although slices near the equator have a bigger radius, slices near the pole are titled inward more, and it turns out these two effects exactly cancel.

To generate a random angle on this circle, we raise the imaginary unit 1j to a uniformly random power between 0 and 4, which saves us from needing trig functions, pi, or e, any of which would need an import. We then extract the real imaginary part. If we can output a complex number for two of the coordinates, the last line could just be print a,z.

# Python 2, 85 bytes

from random import*
a,b,c=eval('gauss(0,1),'*3)
R=(a*a+b*b+c*c)**.5
print a/R,b/R,c/R


Try it online!

Generates three normals and scales the result. Thanks to ovs for saving 1 byte with eval in the second line.

Python 2 with numpy, 57 bytes

from numpy import*
a=random.randn(3)
print a/sum(a*a)**.5


Try it online!

sum(a*a)**.5 is shorter than linalg.norm(a). We could also do dot(a,a) for the same length as sum(a*a). In Python 3, this can be shortened to a@a using the new operator @.

• I like your first approach. I'm having trouble understanding how a bias towards the equator is avoided if z, from a uniform distribution, is left unmodified. Sep 9 '19 at 8:12
• @Jitse The spherical distribution is in factor uniform over each coordinate. This is something special for dimension 3. See for instance this proof that the surface area of a slice of a sphere is proportional to its height. Regarding the intuition that this is biased to the equator, note that while slices near the equator have a bigger radius, ones near the pole are titled inward more which gives more area, and it turns out these two effects exactly cancel.
– xnor
Sep 9 '19 at 8:14
• Very nice! Thanks for the clarification and the reference. Sep 9 '19 at 8:23
• @Jitse Thanks, I added it to the body. I realized though that I was only sampling positive z though, and fixed that for a few bytes.
– xnor
Sep 9 '19 at 8:26
• @Jitse Indeed, the surface area of a sphere equals the lateral surface area of the enclosing cylinder!
– Neil
Sep 9 '19 at 8:50

# Octave, 40 33 22 bytes

We sample form a 3d standard normal distribution and normalize the vector:

(x=randn(1,3))/norm(x)


Try it online!

• For Octave only (i.e. not MATLAB), you can save a byte with this Sep 10 '19 at 8:40
• @TomCarpenter Thanks! In this case as it is just one expression we can even omit the disp:) Sep 10 '19 at 8:42

# Unity C#, 34 bytes

f=>UnityEngine.Random.onUnitSphere


Unity has a builtin for unit sphere random values, so I thought I'd post it.

• Good use of a built in +1, You could just submit a function to be a bit shorter f=>Random.onUnitSphere Sep 10 '19 at 10:51
• @LiefdeWen I knew about lambdas, I just wasn't sure if that was enough (in terms of validity on Code Golf) because it isn't declaring f's Type; using var only works inside a method and System.Func<Vector3> was longer. Sep 10 '19 at 13:21
• In codegolf returning a function is perfectly fine, and you don't have to count the declaration either meaning you can do sneaky things with dynamic parameters. You also don't count the last semi-colon. You do however count all using statements you add. so your byte count needs to include the using. But f=>Random.onUnitSphere is a perfectly valid submission Sep 10 '19 at 14:37
• @LiefdeWen Yeah, I just wasn't sure how the declaration was handled and wasn't really feeling up to "searching meta." Sep 10 '19 at 15:35
• f=>UnityEngine.Random.onUnitSphere saves you the using  Sep 11 '19 at 10:17

# MATL, 10 bytes

1&3Xrt2&|/


Try it online!

### Explanation

This uses the first approach described in the challenge.

1&3Xr  % Generate a 1×3 vector of i.i.d standard Gaussian variables
t      % Duplicate
2&|    % Compute the 2-norm
/      % Divide, element-wise. Implicitly display


# Ruby, 34 50 49 bytes

->{[z=rand*2-1]+((1-z*z)**0.5*1i**(rand*4)).rect}


Try it online!

Returns an array of 3 numbers [z,y,x].

x and y are generated by raising i (square root of -1) to a random power between 0 and 4. This complex number needs to be scaled appropriately according to the z value in accordance with Pythagoras theorem: (x**2 + y**2) + z**2 = 1.

The z coordinate (which is generated first) is simply a uniformly distributed number between -1 and 1. Though not immediately obvious, dA/dz for a slice through a sphere is constant (and equal to the perimeter of a circle of the same radius as the whole sphere.) .

This was apparently discovered by Archimedes who described it in a very non-calculus-like way, and it is known as Archimedes Hat-Box theorem. See https://brilliant.org/wiki/surface-area-sphere/

Another reference from comments on xnor's answer. A surprisingly short URL, describing a surprisingly simple formula: http://mathworld.wolfram.com/Zone.html

• @Jitse I forgot to scale back the x and y at high values of z. Effectively the points defined a cylinder. It's fixed now but it cost a lot of bytes! I could save a few if the the output can be expressed with a complex number [z, x+yi] I'll leave it as it is unless you say that's OK. Sep 9 '19 at 9:58
• Looks good! I really like this approach. For consistency, the required output is three floats, so I suggest leaving it like this. Sep 9 '19 at 10:09
• Why not use z*z instead of z**2? Sep 9 '19 at 21:06
• @ValueInk yeah thanks I realised I'd missed that z*z. I've edited it in now. The other thing I could do is replace rand*4 with something like z*99 or x*9E9 (effectively limiting the possible values to a very fine spiral on the sphere) but I think that reduces the quality of the random. Sep 9 '19 at 22:25

# TI-BASIC, 15 bytes *

:randNorm(0,1,3
:Ans/√(sum(Ans²


Using the algorithm "generate 3 normally distributed values and normalize that vector".

Ending a program with an expression automatically prints the result on the Homescreen after the program terminates, so the result is actually shown, not just generated and blackholed.

*: randNorm( is a two-byte token, the rest are one-byte tokens. I've counted the initial (unavoidable) :, without that it would be 14 bytes. Saved as a program with a one-letter name, it takes 24 bytes of memory, which includes the 9 bytes of file-system overhead.

# 05AB1E, 23 22 bytes

[тε5°x<Ýs/<Ω}DnOtDî#}/


Implements the 2nd algorithm.

Explanation:

NOTE: 05AB1E doesn't have a builtin to get a random decimal value in the range $$\[0,1)\$$. Instead, I create a list in increments of $$\0.00001\$$, and pick random values from that list. This increment could be changed to $$\0.000000001\$$ by changing the 5 to 9 in the code (although it would become rather slow..).

[            # Start an infinite loop:
тε          #  Push 100, and map (basically, create a list with 3 values):
5°        #   Push 100,000 (10**5)
x       #   Double it to 200,000 (without popping)
<      #   Decrease it by 1 to 199,999
Ý     #   Create a list in the range [0, 199,999]
s/   #   Swap to get 100,000 again, and divide each value in the list by this
<  #   And then decrease by 1 to change the range [0,2) to [-1,1)
Ω #   And pop and push a random value from this list
}          #  After the map, we have our three random values
D         #   Duplicate this list
n        #   Square each inner value
O       #   Take the sum of these squares
t      #   Take the square-root of that
D     #   Duplicate that as well
î    #   Ceil it, and if it's now exactly 1:
#   #    Stop the infinite loop
}/           # After the infinite loop: normalize by dividing
# (after which the result is output implicitly)

• Using $l<1$ is equally as valid as $l\leq1$. The only criterion for $l\leq x$ is that $0< x\leq 1$. You may just as well accept vectors with $l<0.5$ if it would save any bytes. Any value smaller than or equal to $1$ removes the bias. Sep 9 '19 at 9:28
• @Jitse Ok, implemented the normalization in both my Java and 05AB1E answers. I hope everything is correct now. Sep 9 '19 at 9:53
• @Jitse Actually saved a byte by checking $v\leq1$ as $\lceil v\rceil==1$, instead of $v\lt1$. But thanks for the clarification that only $0\lt x\leq1$ is a requirement, and there isn't a strict $\leq$ requirement on $l$, as long as it's $\leq1$. Sep 9 '19 at 10:11

# JavaScript (ES7),  77 76  75 bytes

Implements the 3rd algorithm, using $$\\sin(\phi)=\sin(\cos^{-1}(z))=\sqrt{1-z^2}\$$.

with(Math)f=_=>[z=2*(r=random)()-1,cos(t=2*PI*r(q=(1-z*z)**.5))*q,sin(t)*q]


Try it online!

### Commented

with(Math)                       // use Math
f = _ =>                         //
[ z = 2 * (r = random)() - 1,  // z = 2 * j - 1
cos(                         //
t =                        // θ =
2 * PI *                 //   2 * π * i
r(q = (1 - z * z) ** .5) // q = sin(ɸ) = sin(arccos(z)) = √(1 - z²)
// NB: it is safe to compute q here because
//     Math.random ignores its parameter(s)
) * q,                       // x = cos(θ) * sin(ɸ)
sin(t) * q                   // y = sin(θ) * sin(ɸ)
]                              //


# JavaScript (ES6), 79 bytes

Implements the 2nd algorithm.

f=_=>(n=Math.hypot(...v=[0,0,0].map(_=>Math.random()*2-1)))>1?f():v.map(x=>x/n)


Try it online!

### Commented

f = _ =>                         // f is a recursive function taking no parameter
( n = Math.hypot(...           // n is the Euclidean norm of
v =                        // the vector v consisting of:
[0, 0, 0].map(_ =>       //
Math.random() * 2 - 1  //   3 uniform random values in [-1, 1]
)                        //
)) > 1 ?                       // if n is greater than 1:
f()                          //   try again until it's not
:                              // else:
v.map(x => x / n)            //   return the normalized vector


# Processing 26 bytes

Full program

print(PVector.random3D());


This is the implementation https://github.com/processing/processing/blob/master/core/src/processing/core/PVector.java

  static public PVector random3D(PVector target, PApplet parent) {
float angle;
float vz;
if (parent == null) {
angle = (float) (Math.random()*Math.PI*2);
vz    = (float) (Math.random()*2-1);
} else {
angle = parent.random(PConstants.TWO_PI);
vz    = parent.random(-1,1);
}
float vx = (float) (Math.sqrt(1-vz*vz)*Math.cos(angle));
float vy = (float) (Math.sqrt(1-vz*vz)*Math.sin(angle));
if (target == null) {
target = new PVector(vx, vy, vz);
//target.normalize(); // Should be unnecessary
} else {
target.set(vx,vy,vz);
}
return target;
}

• You might want to make it clearer that the implementation is not part of your byte count. I missed it on first reading, then did a double-take. Sep 9 '19 at 18:35
• I like that the implementation uses essentially the same approach as me, though Sep 9 '19 at 18:38

# Python 2, 86 bytes

from random import*
x,y,z=map(gauss,*3,*3);l=(x*x+y*y+z*z)**.5
print x/l,y/l,z/l


Try it online!

Implements the first algorithm.

# Python 2, 107 103 bytes

from random import*
l=2
while l>1:x,y,z=map(uniform,[-1]*3,*3);l=(x*x+y*y+z*z)**.5
print x/l,y/l,z/l


Try it online!

Implements the second algorithm.

• @RobinRyder This implementation rejects vectors with an initial length >1, which is valid as specified in the challenge. Sep 9 '19 at 7:50
• @Jitse Right, sorry. I misread the code. Sep 9 '19 at 7:51

import System.Random
f=mapM(\_->randomRIO(-1,1))"lol">>= \a->last$f:[pure$(/n)<$>a|n<-[sqrt.sum$map(^2)a::Double],n<1]


Try it online!

Does three uniforms randoms and rejection sampling.

• It looks like your randoms are from the distribution (0,1) instead of (-1,1), so that only 1/8 of the sphere is covered. Sep 9 '19 at 8:21
• @Jitse gotcha, thanks for noticing.
– Angs
Sep 9 '19 at 8:24

# JavaScript, 95 bytes

f=(a=[x,y,z]=[0,0,0].map(e=>Math.random()*2-1))=>(s=Math.sqrt(x*x+y*y+z*z))>1?f():a.map(e=>e/s)


You don't need not to input a.

# Julia 1.0, 24 bytes

x=randn(3)
x/hypot(x...)


Try it online!

Draws a vector of 3 values, drawn from a normal distribution around 0 with standard deviation 1. Then just normalizes them.

• randn(), from a couple of quick tests, doesn't seem to be bound to the required range. Also, this doesn't include a check for hypot() returning a value >1, which should be rejected. Sep 10 '19 at 21:02
• @Shaggy it would appear randn simulates from a standard normal distribution rather than a uniform(0,1) one, so this approach is identical to the R one. Sep 10 '19 at 21:04
• @Giuseppe Yes, exactly! Sep 10 '19 at 21:05
• @Giuseppe, I think that I may not have a proper grasp on the maths behind this challenge but, if I'm understanding you correctly, you're saying that if any of the floats are outside the bounds of [-1,1) then dividing by them by the hypotenuse, which will be >1, offsets that? That leads me to wonder if the ternary in my solution is necessary ... Sep 10 '19 at 23:15
• @Shaggy no, the normal/Gaussian distribution has some properties (specifically, rotational invariance) that the uniform doesn't have, see this comment, for example Sep 10 '19 at 23:19

# MathGolf, 2119 18 bytes

{╘3Éƒ∞(ß_²Σ√_1>}▲/


Implementation of the 2nd algorithm.

Explanation:

{              }▲   # Do-while true by popping the value:
╘                  #  Discard everything on the stack to clean up previous iterations
3É                #  Loop 3 times, executing the following three operations:
ƒ               #   Push a random value in the range [0,1]
∞              #   Double it to make the range [0,2]
(             #   Decrease it by 1 to make the range [-1,1]
ß            #  Wrap these three values into a list
_           #  Duplicate the list of random values
²          #  Square each value in the list
Σ         #  Sum them
√        #  And take the square-root of that
_       #  Duplicate it as well
1>     #  And check if it's larger than 1
/  # After the do-while, divide to normalize
# (after which the entire stack joined together is output implicitly,
#  which is why we need the ╘ to cleanup after every iteration)


# Java 8 (@Arnauld's modified 3rd algorithm), 131126119111 109 bytes

v->{double k=2*M.random()-1,t=M.sqrt(1-k*k),r[]={k,M.cos(k=2*M.PI*M.random())*t,M.sin(k)*t};return r;}


Port of @Arnauld's JavaScript answer, so make sure to upvote him!
-2 bytes thanks to @OlivierGrégoire.

This is implemented as:

$$\k = N\cap[-1,1)\$$
$$\t=\sqrt{1-k^2}\$$
$$\u=2\pi×(N\cap[0,1))\$$
$$\x,y,z = \{k, \cos(u)×t, \sin(u)×t\}\$$

Try it online.

Previous 3rd algorithm implementation (131 126 119 bytes):

Math M;v->{double k=2*M.random()-1,t=2*M.PI*M.random();return k+","+M.cos(t)*M.sin(k=M.acos(k))+","+M.sin(t)*M.sin(k);}


Implemented as:

$$\k = N\cap[-1,1)\$$
$$\t=2\pi×(N\cap[0,1))\$$
$$\x,y,z = \{k, \cos(t)×\sin(\arccos(k)), \sin(t)×\sin(\arccos(k))\}\$$

Try it online.

### Explanation:

Math M;                         // Math on class-level to use for static calls to save bytes
v->{                            // Method with empty unused parameter & double-array return
double k=2*M.random()-1,      //  Get a random value in the range [-1,1)
t=M.sqrt(1-k*k),       //  Calculate the square-root of 1-k^2
r[]={                       //  Create the result-array, containing:
k,                     //   X: the random value k
M.cos(k=2*M.PI         //   Y: first change k to TAU (2*PI)
*M.random()//       multiplied by a random [0,1) value
)               //      Take the cosine of that
*t,            //      and multiply it by t
M.sin(k)               //   Z: Also take the sine of the new k (TAU * random)
*t};          //      And multiply it by t as well
return r;}                    //  Return this array as result


# Java 8 (2nd algorithm), 153 143 bytes

v->{double x=2,y=2,z=2,l;for(;(l=Math.sqrt(x*x+y*y+z*z))>1;y=m(),z=m())x=m();return x/l+","+y/l+","+z/l;};double m(){return Math.random()*2-1;}


Try it online.

2nd algorithm:

v->{                              // Method with empty unused parameter & String return-type
double x=2,y=2,z=2,l;           //  Start results a,b,c all at 2
for(;(l=Math.sqrt(x*x+y*y+z*z)) //  Loop as long as the hypotenuse of x,y,z
>1;                        //  is larger than 1
y=m(),z=m())x=m();            //   Calculate a new x, y, and z
return x/l+","+y/l+","+z/l;}    //  And return the normalized x,y,z as result
double m(){                       // Separated method to reduce bytes, which will:
return Math.random()*2-1;}      //  Return a random value in the range [-1,1)

• Using sqrt(1-k*k) actually saves more bytes in Java than it does in JS. :) Sep 9 '19 at 11:11
• @Arnauld Yep. Instead of 3x M.sin, 1x M.cos and 1x M.acos, your approach uses 2x M.sin and 1x M.sqrt, which is where the additional saved bytes mostly come from. :) Sep 9 '19 at 11:18
• 108 bytes Uses a modified 2nd algorithm where I only allow values where s == 1 (instead of s<=1 and then normalizing). It sometimes gives an answer but mostly doesn't because of the timeout. Edit: Oops, I forgot to Math.sqrt the result Sep 11 '19 at 12:41
• Actually, no, no need to sqrt because sqrt(1)==1. So I stand with my golf suggestion. Sep 11 '19 at 12:48
• 109 bytes (you can use your string output instead of double[] as that doesn't change the byte-count.) Sep 11 '19 at 13:35

# Nim, 105101100 99 bytes

import math,random
randomize()
let
t=TAU.rand
u=2.0.rand-1
v=sqrt 1-u*u
echo [t.cos*v,t.sin*v,u]


Try it online!

Uses the third method with a bit of substitution:

$$u = \cos(\phi) = 2j - 1$$ $$v = \sin(\phi) = \sqrt{1-u^2}$$

# Vyxal, 20 19 bytes

{≥|3ʁƛ∆Ṙd⌐;:²∑√:1}/


Try it Online!

Uses the second method.

## Explanation

{≥|3ʁƛ∆Ṙd⌐;:²∑√:1}/
{                }  While
≥                    Greater or equal
|                 do:
3ʁ                 [0,1,2]
ƛ    ;           Map:
∆Ṙ                Random number from 0 to 1
d               Double
⌐              Subtract from 1
:          Duplicate
²         Square
∑        Sum
√       Square root
:      Duplicate
1     Push 1
/ Divide [the vector by the length]


# Japt, 20 bytes

Port of Arnauld's implementation of the 2nd algorithm.

MhV=3ÆMrJ1
>1?ß:V®/U


Test it

MhV=3ÆMrJ1
Mh             :Get the hypotenuse of
V=           :  Assign to V
3Æ         :  Map the range [0,3)
Mr       :    Random float
J1     :    In range [-1,1)
>1?ß:V®/U      :Assign result to U
>1?            :If U is greater than 1
ß           :  Run the programme again
:V®/U      :Else map V, dividing all elements by U


# Pyth, 24 bytes

W<1Ks^R2JmtO2.0 3;cR@K2J


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Uses algorithm #2

W                         # while
<1                       #   1 <
Ks                     #       K := sum(
^R2                  #               map(lambda x:x**2,
Jm      3         #                    J := map(                            , range(3))
tO2.0           #                             lambda x: random(0, 2.0) - 1           )):
;        #   pass
R   J  # [return] map(lambda x:            , J)
c @K2   #                        x / sqrt(K)


# OCaml, 11099 95 bytes

(fun f a c s->let t,p=f 4.*.a 0.,a(f 2.-.1.)in[c t*.s p;s t*.s p;c p])Random.float acos cos sin


EDIT: Shaved off some bytes by inlining $$\ i \$$ and $$\ j \$$, replacing the first let ... in with a fun, and taking advantage of operator associativity to avoid some parens ().

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Original solution:

Random.(let a,c,s,i,j=acos,cos,sin,float 4.,float 2. in let t,p=i*.(a 0.),a (j-.1.) in[c t*.s p;s t*.s p;c p])


First I define:

$$a = \arccos,\ \ c = \cos,\ \ s = \sin \\ i \sim \textsf{unif}(0,4),\ \ j \sim \textsf{unif}(0,2)$$

OCaml's Random.float function includes the bounds. Then,

$$t = i \cdot a(0) = \frac{i\pi}{2},\ \ p = a (j-1)$$

This is very similar to the 3rd example implementation (with $$\ \phi = p \$$ and $$\ \theta = t \$$) $$\ - \$$ except that I pick $$\ i \$$ and $$\ j \$$ within larger intervals to avoid multiplication (with 2) later on.

• I'm not quite familiar with this language, but it looks like you use the random floats between 0 and 1 directly as spherical coordinates. This is incorrect, as shown in the challenge remarks 3 and 4, since you end up with a bias towards the poles of the sphere. You can correct this by applying the method shown in remark 4. Sep 10 '19 at 7:27
• Thank you! Totally missed that. Fixed the bug and updated my answer Sep 10 '19 at 8:25
• Looks good! Very nice first answer! Sep 10 '19 at 8:27
• Thank you :) I was able to reduce it to under-100 bytes! Sep 10 '19 at 8:38

# Raku, 65 bytes

{(&cos,&sin,{1})».(τ.rand)Z*(&sin,&sin,&cos)».(acos 2.rand-1)}


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This function uses the "convert two random numbers to spherical coordinates" method. τ.rand generates the first number in the range of zero to τ (tau, that is, twice pi) and acos 2.rand - 1 generates the second with the acos argument in the range of -1 to 1.

(&cos, &sin, {1})».(...) applies the cosine function, the sine function, and an anonymous function that always returns 1 to the first random parameter, producing a list of the results. Similarly, (&sin, &sin, &cos)».(...) applies the sine, sine, and cosine functions to the second random parameter, producing a second list. Finally, Z* zips those two lists together using multiplication.

# Pxem, 58 bytes (filename).

Because Pxem does not handle decimal values nor negative values (negative values can be supplied via input but cannot be generated in other ways), I am representing them as integers.

Unprintables are represented as a backslash followed by three-digit octal value of its code point.

x\003\144.w\002.r\002.!+.+.o.v21.y.c\013.r.c.t.c.!.a.m.n.m.m.!.-.v\001.-.c.a


## Output format

It outputs 10x, 10y, and 10z rather than x, y, and z.

"%c%d%c%d%c%d\n", sign, value, sign value, sign, value


Sign is as either + or -; always output. Value is multiplied by 10. Thus the precision of the program is up to 0.1.

XX.z
# memory layout: counter, l
# until counter is zero; do
.ax\003\144.wXX.z
# output sign
.a\002.r\002.!+.+.oXX.z
# while :; do
.a.v21.yXX.z
# generate a random number 0<=x<=10
.a.c\013.rXX.z
# continue if x^2>l; otherwise break
.a.c.t.c.!XX.z
# done
.a.aXX.z
# output x
.a.m.nXX.z
# update l to l minus x^2
.a.m.m.!.-XX.z
# decrement counter; done
.a.v\001.-.c.a


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# APL (Dyalog Classic), 22 bytes

⎕←{⍵÷((+/⍵*2)*0.5)}?⍳3


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• Hi, and welcome to Code Golf! Can you please provide a link so others can test it, for example Try It Online!? Jun 18 at 12:07
• Hej. Are you sure this solves the problem? As far as I can tell, it can only give 3 unique answers.