Herausforderung

Sandkastenpfosten

Bei einer positiven Ganzzahl (K)Geben Sie eine gleichmäßig zufällige Ganzzahl (Y)zwischen aus [0, K).

Wenn Y > 0Angenommen , K = Yund wiederholen Sie den Vorgang , bis Y = 0.

Regeln

• Die Eingabe muss zuerst gedruckt werden
• Ausgabeformat wie Sie möchten
• Ihr Programm muss beendet sein.
• 0 muss die endgültige Ausgabe sein, optional eine Leerzeile 0

Pyth , 6 5 4 Bytes

.uOW

Probieren Sie es hier aus!

Wie es funktioniert

.uOW Volles Programm. Nimmt eine Ganzzahl aus STDIN und gibt eine Liste an STDOUT aus.
.u Kumulativer Fixpunkt. Wenden Sie die gegebene Funktion mit einem gegebenen Anfangswert an,
        die implizit dem Eingang zugeordnet ist, bis ein Ergebnis vorliegt 
        bevor gefunden wird. Gibt die Liste der Zwischenergebnisse zurück.
   W Bedingte Anwendung. Wenn das Argument (der aktuelle Wert) wahr ist, dann
        Wenden Sie die unten stehende Funktion an, andernfalls lassen Sie sie unverändert.
  O Zufällige ganze Zahl im Bereich [0, N).
        IOW: Ordnen Sie bei jeder Iteration von .u dem aktuellen Wert beginnend eine Variable N zu
        mit der Eingabe. Wenn N nicht 0 ist, wählen Sie eine zufällige ganze Zahl in [0, N), sonst
        N unverändert zurückgeben. Immer wenn wir auf eine 0 stoßen, muss auch die nächste Iteration erfolgen
        führen zu einer 0, und daher stoppt die Schleife dort.

C (gcc) , 42 Bytes

f(_){printf("%d\n",_);(_=rand()%_)&&f(_);}

Probieren Sie es online!

Verwendet kurzgeschlossene logische und.

f(_){                 // f(int _) {
    printf("%d\n",_); // print argument and a newline
    (_=rand()%_)      // set _ to rand()%_
    &&f(_);}          // short-circuit AND to recursively call f if _ not zero

C (gcc) , 40 Bytes (ohne Druckanfangswert)

f(_){printf("%d\n",_=rand()%_);_&&f(_);}

Probieren Sie es online!

Verwendet kurzgeschlossene logische und.

f(_){              // f(int _) {
    printf("%d\n", // print an integer and a newline 
    _=             // The integer is _ which we set to...
    rand()%_);     // a random value modulo the input _
    _&&f(_);}      // short-circuit AND to recursively call f if _ not zero

R , 66 60 56 43 41 Bytes

function(n)while(print(n))n=sample(n,1)-1

Probieren Sie es online!

MATL , 6 Bytes

`tYrqt

Probieren Sie es online!

Erläuterung

`        % Do...while
  t      %   Duplicate. Takes input (implicit) the first time
  Yr     %   Uniform random integer from 1 to n, included
  q      %   Subtract 1
  t      %   Duplicate. This will be used as loop condition
         % End (implicit). Proceeds with next iteration if non-zero
         % Display stack (implicit)

Pepe , 25 Bytes

Pepe ist eine Programmiersprache, die vom Benutzer Soaku erstellt wurde .

REeErEErReEEreeEREEeEEree 

Probieren Sie es online!

Erläuterung:

REeErEErReEEreeEREEeEEree # full program

REeE                      # input as num, in stack 1
    rEE                   # create loop in stack 2 with name 0
       rReEE              # - output and preserve the number in stack 1
            reeE          # - output a newline "\n"
                REEeEE    # - random number by 0 to input
                      ree # goto loop with name 0 if stack 1 is not equal
                            to stack 2

Perl 6 , 18 Bytes

{$_,(^*).pick...0}

Probieren Sie es online!

Anonymer Codeblock, der eine Liste von Werten zurückgibt. Wenn es Ihnen nichts ausmacht, dass die Zahlen Bereiche sind, können Sie Folgendes tun:

{^$_,^*.pick...0}

für 17 Bytes. Komischerweise hat eine andere eingebaute Zufallsfunktion rollin dieser Instanz das gleiche Verhalten für die gleiche Anzahl von Bytes.

Perl 5.10.0 -pl, 26 bytes

say;say while$_=int rand$_

Try it online!

Jelly,  4  3 bytes

XƬ0

This is a monadic link (function) that prints an array and returns 0.

Try it online!

How it works

XƬ0  Monadic link. Argument: n

XƬ   Pseudo-randomly pick (X) an integer k in [1, ..., n], set n = k, and repeat.
     Do this 'til (Ƭ) the results are no longer unique and return the array of
     unique results, including the initial value of n.
     This stops once X returns k with argument k. The second k will be omitted
     from the return value.
  0  Print the resulting array and set the return value to 0.

Brachylog, 8 bytes

ẉ?ℕ₁-₁ṙ↰

Try it online!

Explanation

ẉ          Write the input followed by a linebreak
 ?ℕ₁       The input must be in [1, …, +∞)
    -₁ṙ    Generate an integer in [0, …, input - 1] uniformly at random
       ↰   Recursive call with that random integer as the new input

The recursion will stop once ?ℕ₁ fails, that is, when the input is 0.

05AB1E, 8 7 bytes

Δ=L<Ω0M

Try it online!

Explanation

Δ         # loop until value doesn't change
 =        # print current value
  L<Ω     # push a random number in ([1 ... X] - 1)
          # will return -1 when X=0
     0M   # push max of that and 0

J, 13 bytes

[:}:? ::]^:a:

On the subway, so apologies for lack of TIO (hopefully there isn’t a lack of correctness).

Outputs a list of values.

Presumably the APL approach will be shorter, but this is what I thought of.

How it works

^:a: apply repeatedly until convergence, storing intermediate results in an array.

? random integer in range [0, K) for K greater than 0. For 0, it gives a random integer in range (0,1). For a floating point number, it errors.

::] catch an error for an input to ? and instead of erroring, output the input that caused the error.

}: get rid of the last value in the array (this is so that a floating point number isn’t output).

Try it online!

JavaScript (ES6), 38 37 bytes

-1 byte thanks to @Arnauld

f=n=>[n,...n?f(Math.random()*n|0):[]]

f=n=>[n,...n?f(Math.random()*n|0):[]]

<input type=number id=a><button onclick="console.log(f(+a.value))">Test</button>

C, 38 bytes

f(k){printf("%d ",k);k?f(rand()%k):0;}

Try it online

Ungolfed

void f(int k){
    printf("%d ",k);
    if(k)
        f(rand()%k);
}

Pyth, 4 bytes

W
~O

Try it online!

This basically implements the algorithm:

$\begin{align} % Unknown environment algorithm :( &Q \gets \text{input} \\ &\mathbf{Repeat} \\ &\begin{array}{cl} 1. & temp \gets Q \\ 2. & Q \gets \text{unif}\{ 0, Q - 1 \} \\ 3. & \mathtt{Print}(temp) \end{array} \\ &\mathbf{Until} \quad temp = 0 \end{align}$

To translate the Pyth into the algorithm, we can mostly just examine what each character means. Since Pyth is written in prefix notation (i.e. * + 1 2 3 is (1 + 2) * 3) we can start from the left and fill in the arguments as we go.

W begins a traditional while loop. The first statement after it is the loop condition and the second statement after it is the loop body. If the second statement is empty it becomes a no-op. This while works exactly like the Python while, so it will evaluate non-zero integers as True and zero as false.

The first statement after the while begins with the newline character. This corresponds to Pyth's "print and return with a newline" function. This takes one argument, which is then printed and also returned unmodified. This allows us to print the intermediate steps while also performing the needed operations.

The argument passed to this print function begins with ~ which is a bit special. If the character immediately after ~ is a variable it takes two arguments, otherwise it takes one. Since O is not a variable ~ will consume only one argument. ~ functions a bit like += does in many conventional languages, though the closest operator would be the post-increment operator ++ from C. You may know that x++ will be like using x as the current value, but thereafter x will be x+1. ~ is the same idea, but generalised to whatever the result of the first argument is. How it picks what variable to assign to will be addressed later.

The argument of ~ is O which is very simple. When its one argument is an integer O returns a value from 0 to one less than that integer uniformly at random.

Now you may have noticed O does not have an argument. Here the Pyth interpreter kindly fills in a guess, which here is the variable Q. Q has a special meaning in Pyth: whenever it is present in a program the Pyth program begins with assigning Q to the input of the program. Since this is the first variable occurring in ~'s argument Q is also now the variable that ~ will assign a value to.

Summed up our "readable" program might look like:

while print_and_return( assign_variable( Q, unif(0, Q-1) ) ):
    pass

And one sample "run-through" might look like:

1. Q = 5
2. O returns 3, ~ returns 5, \n returns and prints 5 which is true
3. Q = 3
4. O returns 0, ~ returns 3, \n returns and prints 3 which is true
5. Q=0
6. O returns something irrelevant, ~ returns 0, \n returns and prints 0 which is false
7. Q = something irrelevant
8. Terminate

APL (Dyalog Unicode), 12 9 bytes

Anonymous tacit prefix function. Assumes ⎕IO (Index Origin) to be 0, which is default on many systems. Returns the final value (0) in addition to printing while run.

{⌊?⎕←⍵}⍣=

Try it online!

{…}⍣= apply the following function until stable:

 ⎕←⍵ output the argument

 ? return a uniformly distributed random number in the range 0 through that–1

 ⌊ round down (because ?0 gives a (0,1) float)

C (gcc), 40 42 bytes

Some idiot™ forgot to print the initial value first.

f(K){while(K)printf("%d\n",K,K=rand()%K);}

Don't panic.

x86 + rdrand, 19 bytes

Straightforward implementation. Takes input K in ecx and outputs to a buffer in ebx.

0000000a <start>:
   a:   0f c7 f0                rdrand %eax
   d:   31 d2                   xor    %edx,%edx
   f:   f7 f1                   div    %ecx
  11:   89 13                   mov    %edx,(%ebx)
  13:   83 c3 04                add    $0x4,%ebx
  16:   89 d1                   mov    %edx,%ecx
  18:   85 c9                   test   %ecx,%ecx
  1a:   75 ee                   jne    a <start>
  1c:   c3                      ret  

Python 3, 39 bytes

f=lambda k:print(k)or f(hash('.'*k)%k)

Probably not the most cryptographically secure random number generator but to the human eye it looks random enough...

Try it online!

TI-Basic (TI-84 Plus CE), 17 13 bytes

While Ans
Disp Ans
int(randAns
End
Ans

-4 bytes from Misha Lavrov

Takes input in Ans as 50:prgmNAME.

TI-Basic is a tokenized language. All tokens used here are one byte.

Explanation:

While Ans    # 3 bytes, While the number we hold is not zero:
Disp Ans     # 3 bytes,   Display it on its own line
int(randAns  # 4 bytes,   and replace it with a number randomly
                        # chosen from 0 to one less than it (inclusive)
End          # 2 bytes, end While loop
Ans          # 1 byte,  Display (and return) zero

An 11-byte solution suggested by Misha Lavrov that requires pressing enter after each line following the first.

Ans
While Ans
Pause int(randAns
End

Python 2, 64 62 60 bytes

from random import*
k=input()
while k:print k;k=randrange(k)

Try it online!

Saved

• -2 bytes, thanks to Jonathan Allan

C++ (gcc), 98 bytes

#import<cstdio>
#import<cstdlib>
#define d printf("%i ",x 
int p(int x){d);while(x>0)d=rand()%x);}

Try it here!

Usage

int main() {
    p(100);
}

This is my first code golf attempt. Any feedback or remarks are welcome.

Edit: Removed the main function as suggested to make it valid.

><>, 92+2 Bytes

:nao:0=?;0_1>:{:}(?\\
}(?!\$2*1>x~\$+1$*2/\~00:{{:@}
8+1.\~:{:}\+>$1+f
~~@~~47*0.\(a2*1@@?!.

+2B for -v flag

Try it online!

><>'s only source of randomness comes from the 'x' instruction, which sets the instruction pointer's direction to a random value. As such, generating a random number from 0 to n isn't trivial.

I first calculate how many bits are required to represent a number in the range [0,n), then generate random bits to generate a random number. This leaves the possibility that it'll generate a number slightly larger than n, in which case we just discard it and try again.

Explanation:

:nao                              Print the current number followed by a newline
    :0=?;                         If the current n is 0, terminate

Calculate how many random bits we need to be generating:
         0 1                      Initialise the variables for this loop: 
                                      numberOfBits = 0, maxValue = 1
             :{:}(?\              If maxValue >= n, break out of the loop
                 *2               maxValue *= 2
             $+1$                 numberOfBits += 1

Generate the random number:
                     ~            Delete maxValue
                      00          Initialise randomNumber = 0, i = 0
}(?!\                   :{{:@}    If i >= numberOfBits, break out of the (inner) loop
     $2*                          randomNumber *= 2
          _
        1>x~\                     The random bit: If the IP goes up or 
          \+>                     left, it'll be redirected back onto the 'x', 
                                  if it goes down, it adds one to randomNumber
                                  If it goes right, it does nothing to randomNumber

             $1+                  increment i
8+1.            f                 Jump back to the start of the inner loop

After we've generated our number, check that it's actually below n
     ~                            Delete i
      :{:} (      ?               Test that the number is less than n
            a2*1    .             If it's not, jump back to the start 
                                  of the number generation section
  @~~                             Otherwise delete the old value of n, and numberOfBits
     47*0.                        Then jump back to the start of the program

MATLAB (49 46 bytes)

@(k)eval('while k;disp(k);k=randi(k)-1;end;0')

Sample output:

>> @(k)eval('while k;disp(k);k=randi(k)-1;end;0')
ans(5)

ans = 

    @(k)eval('while k;disp(k);k=randi(k)-1;end;0')

     5    
     3    
     2    
     1   

ans =    
     0

Retina, 21 bytes

.+
*
L$`.
$.`
+¶<)G?`

Try it online! Explanation:

+

Repeat until the value stops changing (i.e. 0).

¶<)

Print the value before each pass through the loop.

.+
*

Convert to unary.

L$`.
$.`

Create the range and convert to decimal.

G?`

Pick a random element.

Pyth, 6 7 bytes

QWQ=OQQ

Try it online!

+1 to print the initial input value.

While Q is truthy, set Q to be a random integer between 0 and Q and print Q.

Not the shortest Pyth answer but I'm just learning and only posting because of the recent discussion about no-one using Pyth any more :)

Haskell, 74 71 bytes

^{-3 bytes by actually doing what the specs say it should do.}

import System.Random
f 0=pure[0]
f x=randomRIO(0::Int,x-1)>>=fmap(x:).f

Try it online!

IBM/Lotus Notes Formula, 48 bytes

o:=i;@While(i>0;i:=@Integer(i*@Random);o:=o:i);o

Field formula that takes input from another field i.

There's no TIO for formula so here's a screenshot of a sample output:

Powershell, 36 32 bytes

-4 bytes thanks AdmBorkBork

filter f{$_;if($_){Random $_|f}}

Testscript:

filter f{$_;if($_){Random $_|f}}

100 |f

Output:

100
61
20
8
6
3
0

PowerShell, 35 bytes

for($a="$args";$a;$a=Random $a){$a}

Try it online!

Full program. Takes input $args, stores it into $a, and enters a for loop. Each iteration we're checking whether $a is still positive (as 0 is falsey in PowerShell). Then we leave $a on the pipeline and move to the next iteration, where we set $a to be the result of Get-Random $a, which returns an integer in the range 0..($a-1).

(Ab)uses the fact that PowerShell outputs an additional trailing newline in lieu of outputting the final zero (allowed by the rules as currently written).

Lua, 58 bytes

p,r=print,...+0 p(r)while r>0 do r=math.random(0,r)p(r)end

Try it online!

For some more Lua love here :)