Die Herausforderung

Schreiben Sie ein Programm oder eine Funktion, die keine Eingabe akzeptiert und einen Vektor der Länge $1$ in einer theoretisch einheitlichen Zufallsrichtung ausgibt .

Dies entspricht einem zufälligen Punkt auf der Kugel, der durch

was zu einer Verteilung wie z

Ausgabe

Drei floats aus einer theoretisch gleichmäßigen Zufallsverteilung, für die die Gleichung $x^2+y^2+z^2=1$ für Genauigkeitsgrenzen gilt.

Herausforderungsbemerkungen

• Die Zufallsverteilung muss theoretisch gleichmäßig sein . Das heißt, wenn der Pseudozufallszahlengenerator durch ein echtes RNG aus den reellen Zahlen ersetzt würde, würde dies zu einer gleichmäßigen Zufallsverteilung der Punkte auf der Kugel führen.
• Das Erzeugen von drei Zufallszahlen aus einer Gleichverteilung und deren Normalisierung ist ungültig: In Richtung der Ecken des dreidimensionalen Raums besteht eine Tendenz.
• In ähnlicher Weise ist es ungültig, zwei Zufallszahlen aus einer gleichmäßigen Verteilung zu erzeugen und als Kugelkoordinaten zu verwenden: Es besteht eine Tendenz zu den Polen der Kugel.
• Die richtige Einheitlichkeit kann durch Algorithmen erreicht werden, einschließlich, aber nicht beschränkt auf:
  • Generieren Sie drei Zufallszahlen $x$ , $y$ und $z$ aus einer normalen (Gaußschen) Verteilung um $0$ und normalisieren Sie sie.
    • Implementierungsbeispiel
  • Generieren Sie drei Zufallszahlen $x$ , $y$ und $z$ aus einer Gleichverteilung im Bereich $(-1,1)$ . Berechnen Sie die Länge des Vektors mit $l=\sqrt{x^2+y^2+z^2}$ . Wenn dann$l>1$, lehnen Sie den Vektor ab und erzeugen Sie eine neue Menge von Zahlen. Sonst, wenn$l \leq 1$ , normalisiere den Vektor und gib das Ergebnis zurück.
    • Implementierungsbeispiel
  • Erzeugen Sie aus einer Gleichverteilung im Bereich ( 0 , 1 ) zwei Zufallszahlen $i$ und $j$ und rechnen Sie sie wie folgt in Kugelkoordinaten um:$(0,1)$ $$\begin{align}\theta &= 2 \times \pi \times i\\\\\phi &= \cos^{-1}(2\times j -1)\end{align}$$ so dass$x$,$y$und$z$durch x berechnet werden können $$\begin{align}x &= \cos(\theta) \times \sin(\phi)\\\\y &= \sin(\theta) \times \sin(\phi)\\\\z &= \cos(\phi)\end{align}$$
    • Implementierungsbeispiel
• Geben Sie in Ihrer Antwort eine kurze Beschreibung des von Ihnen verwendeten Algorithmus an.
• Lesen Sie mehr über die Auswahl von Kugelpunkten in MathWorld .

Ausgabebeispiele

[ 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]

Allgemeine Bemerkungen

• Das ist Code-Golf , also gewinnt die Antwort mit den wenigsten Bytes in jeder Sprache.
• Es gelten Standardregeln , E / A-Regeln und Regelungslücken .
• Bitte fügen Sie einen Try it Online- Link oder einen vergleichbaren Link hinzu, um die Funktionsweise Ihres Codes zu demonstrieren.
• Bitte begründen Sie Ihre Antwort mit einer Erklärung Ihres Codes.

Wolfram Language (Mathematica), 20 bytes

RandomPoint@Sphere[]

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Does exactly what it says on the tin.

R, 23 bytes

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

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

Plot of 1000 realizations:

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

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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

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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.

86 bytes

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

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Generates three normals and scales the result.

Python 2 with numpy, 57 bytes

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

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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 @.

Octave, 40 33 22 bytes

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

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

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Unity C#, 34 bytes

f=>UnityEngine.Random.onUnitSphere

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

MATL, 10 bytes

1&3Xrt2&|/

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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}

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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

05AB1E, 23 22 bytes

[тε5°x<Ýs/<Ω}DnOtDî#}/

Implements the 2nd algorithm.

Try it online or get a few more random outputs.

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)

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.

JavaScript (ES7),  77 76  75 bytes

Implements the 3^rd 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]

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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 2^nd algorithm.

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

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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;
  }

Python 2, 86 bytes

from random import*
x,y,z=map(gauss,[0]*3,[1]*3);l=(x*x+y*y+z*z)**.5
print x/l,y/l,z/l

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Implements the first algorithm.

Python 2, 107 103 bytes

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

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Implements the second algorithm.

Haskell, 125 123 119 118 bytes

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

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Does three uniforms randoms and rejection sampling.

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...)

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

MathGolf, 21 19 18 bytes

{╘3Ƀ∞(ß_²Σ√_1>}▲/

Implementation of the 2nd algorithm.

Try it online or see a few more outputs at the same time.

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), 131 126 119 111 109 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:

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

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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:

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

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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)

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, 110 99 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 ().

Try it online

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:

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

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.