Randomizing until 0

Question

Challenge

Sandbox post

Given a positive integer (K) Output a uniformly-random integer (Y) between [0, K).

If Y > 0 Assume K = Y and repeat the process until Y = 0.

Rules

• Input must be printed at first
• Output format as you wish
• Your program must finish.
• 0 must be the final output, Optionally an empty line instead 0

Answer 1

Pyth, 6 5 4 bytes

.uOW

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How it works

.uOW    Full program. Takes an integer from  STDIN and outputs a list to STDOUT.
.u      Cumulative fixed-point. Apply the given function with a given initial value,
        which is implicitly assigned to the input, until a result that has occurred 
        before is found. Returns the list of intermediate results.
   W    Conditional application. If the argument (the current value) is truthy, then
        apply the function below, otherwise leave it unchanged.
  O     Random integer in the range [0, N).
        IOW: at each iteration of .u, assign a variable N to the current value, starting
        with the input. If N is not 0, then choose a random integer in [0, N), else
        return N unchanged. Whenever we encounter a 0, the next iteration must also
        result in a 0, and therefore the loop stops there.

Answer 2

C (gcc), 42 bytes

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

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Uses short-circuited logical and.

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 (w/o printing initial value)

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

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Uses short-circuited logical and.

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

Answer 3

R, 66 60 56 43 41 bytes

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

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Answer 4

MATL, 6 bytes

`tYrqt

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Explanation

`        % 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)

Answer 5

Pepe, 25 bytes

Pepe is a programming language made by user Soaku.

REeErEErReEEreeEREEeEEree 

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Explanation:

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

Answer 6

Perl 6, 18 bytes

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

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Anonymous code block that returns a list of values. If you don't mind the numbers being ranges, you can do:

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

for 17 bytes. Funnily enough, another builtin random function, roll, has the same behaviour in this instance for the same amount of bytes.

Answer 7

Perl 5.10.0 -pl, 26 bytes

say;say while$_=int rand$_

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Answer 8

Jelly,  4  3 bytes

XƬ0

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

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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.

Answer 9

05AB1E, 8 7 bytes

Δ=L<Ω0M

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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

Answer 10

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>

Answer 11

Brachylog, 8 bytes

ẉ?ℕ₁-₁ṙ↰

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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.

Answer 12

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).

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Answer 13

C, 38 bytes

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

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Ungolfed

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

Answer 14

Pyth, 4 bytes

W
~O

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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

Answer 15

Python 2, 64 62 60 bytes

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

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---

Saved

• -2 bytes, thanks to Jonathan Allan

Answer 16

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.

{⌊?⎕←⍵}⍣=

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{…}⍣= 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)

Answer 17

C++ (gcc), 98 bytes

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

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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.

Answer 18

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.

Answer 19

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  

Answer 20

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

Answer 21

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

Answer 22

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...

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Answer 23

><>, 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

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><>'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

Answer 24

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.

Answer 25

PHP, 43 42 bytes

<?while($a=$a?rand(0,$a-1):$argn)echo$a?>0

To run it:

echo '<input>' | php -nF <filename>

Or Try it online!

Example output:

10019128410

-1 byte thanks to Titus's suggestion.

Answer 26

Pyth, 6 7 bytes

QWQ=OQQ

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+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 :)

Answer 27

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

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Answer 28

Java, 56 bytes

Object f(int n){return n<1?0:n+","+f(n*=Math.random());}

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Acknowledgments

• -6 bytes thanks to Kevin Cruijssen

Answer 29

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:

Answer 30

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

Answer 31

PowerShell, 35 bytes

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

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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).

Answer 32

Lua, 58 bytes

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

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For some more Lua love here :)

Answer 33

C# (Visual C# Interactive Compiler), 84 82 bytes

var k=int.Parse(ReadLine());for(Write(k);k>0;)Write($",{k=new Random().Next(k)}");

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-2 Bytes thanks to charliefox2

Explenation:

var k = int.Parse(ReadLine());   //1. Read a line from STDIN and convert it to int
for(Write(k);                    //2. Write the original value of k to STDOUT
        k>0;)                    //3. Loop while k > 0
    Write($",{                   //6. Write the separator and the new value to STDOUT
        k =                      //5. Assign it to k
        new Random().Next(k)}"); //4. Get a random int between 0 (inclusive) and k (exclusive)

Answer 34

CJam, 7 9 bytes

ri{_pmr}h

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Annotated

r       e# read input token
i       e# convert to int
{       e# do {
    _   e# duplicate topOfStack
    p   e# pop topOfStack and print it
    mr  e# rand(0, topOfStack)
}h      e# } while(topOfStack != 0)
        e# implicitly convert stack to string and print it

Answer 35

Bash, 65 58 56 50 bytes

f()(echo ${a=$1};for((;a;)){ echo $[a=RANDOM%a];})

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(Improved thanks to manatwork)

Recursive approach

50 28

f()(echo $1&&f $[RANDOM%$1])

Try

Answer 36

Atari 400/800 6502 Assembler – 16 bytes

K set initially to #$FF (but can set to any byte value), then calls the POKEY PRNG at $D20A, if greater than or equal, try again, else save as the new upper limit. Keep going until it reaches zero.

define K $FF
* = $600
    LDA #K
.1: STA $80
    BEQ .3
.2: LDA $D20A
    CMP $80
    BCC .1
    BCS .2
.3: BRK       ; if you assume memory is cleared, can omit for 15 bytes

Which, when assembled, is:

a9 ff 85 80 f0 09 ad 0a d2 c5 80 90 f5 b0 f7 00

Output is by running a monitor, single stepping, and spying on $80! The rules indicated “output format as you wish”!

Answer 37

C# (.NET Core), 71 bytes

Using recursion.

static int f(int k){Console.WriteLine(k);return k==0?k:f(r.Next(0,k));}

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Output:

100
9
3
0

Answer 38

><>, 147 + 2 bytes

 :0=?\:0&:0=?\:2%-2,&1+&80.
   ;n/       \r:r&:0=?\1-&2*\
          /oan:{/?(}:~/44+\ v
<         \:0=?;>{~00.\*1.^0x
                          \1/

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The idea is the same as @Sasha's—generate random bits, making sure not to exceed the original number—but just in a different layout. My solution uses the register to record the number of bits of n.

There is unfortunately a certain amount of wasted space, particularly on the last line.

How it works:

 :0=?\:0&                      n = 0?   No: duplicate n; let reg = 0
   ;n/                                  Yes: print then end

         :0=?\:2%-2,&1+&80.    n = 0?   No: n = floor(n / 2); increment reg; loop to (8,0)
             \                          Yes: enter loop to construct random number r

              r:r&:0=?\1-&2*\  reg = 0? No: r = 2 * r + (0 or 1); loop to (16, 1)
                      /44+\ v           Yes: check if r <= n
<                     \*1.^0x           
                          \1/

          /oan:{/?(}:~         r > n?   No: print r and newline; 
          \:0=?;>

                 {~00.         Go back to (0,0), using either r or n

Answer 39

Kotlin, 60 bytes

fun f(k:Int){println(k);if(k>0)f((Math.random()*k).toInt())}

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Solution with recursion.

Answer 40

J, 8 bytes

?^:*^:a:

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How it works

f^:g^:_ is a J idiom for "do while", which means to repeatedly apply f to the input while g gives true. g should always give 0 or 1. Change the last _ (infinity) to a: (boxed empty), and the resulting verb gives the list of intermediate values.

Monadic ? is "roll", i.e. given N, generates a random integer between 0 and N-1 inclusive. Monadic * is "signum", i.e. gives 1 for positive, 0 for zero, -1 for negative input.

Answer 41

Charcoal, 11 bytes

ＩθＷ⊟θＩ⊞Ｏθ‽ι

Try it online! Link is to verbose version of code. Slightly unusual input format. Explanation:

Ｉθ

Cast the input to string and print it.

Ｗ⊟θ

While the input is non-zero...

⊞Ｏθ‽ι

... replace it with a random number in the implicit range...

Ｉ

... and cast the result to string and print it.

13-byte version with more standard input format:

θＮθＷθ«≔‽θθ⸿Ｉθ

Try it online! Link is to verbose version of code.

Answer 42

Jelly, 6 bytes

X’$¹Ð¿

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Answer 43

JavaScript, 34 bytes

Outputs a space-delimited string of integers.

f=n=>n&&n+" "+f(Math.random()*n|0)

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Answer 44

Ruby, 31 29 bytes

->n{n=rand p n until 1>n;p 0}

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Answer 45

Red, 41 bytes

func[n][print n if 0 < n[f random n - 1]]

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Answer 46

Batch, 54 bytes

@echo %1&if %1 gtr 0 set/an=%random%%%%1&call %0 %%n%%

Four %s in a row... just a typical day for Batch. Only works up to 32768 (and not terribly uniform at that) due to limitations of Batch's random number generator. See my Batch answer to Pick a random number between 0 and n using a constant source of randomness for large uniform randomness.

Answer 47

Japt, 10 bytes

Outputs an array of integers.

_Ì}f@NpU=ö

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---

Explanation

               :Implicit input of integer U
  }f           :Loop until left function returns falsey and return final value of right function
    @          :Right function
         ö     :  Random integer in the range [0,U)
       U=      :  Reassign to U
      p        :  Push
     N         :   To the array of inputs
_              :Left function
 Ì             :Get the last element of N. If it's 0, which is falsey, then the loop will be broken

---

Alternative, 8 bytes

A direct port of my JavaScript solution.

©U+S+ßUö

Try it

Answer 48

Excel VBA, 42 bytes

An anonymous function that takes input from range [A1] and outputs to the console.

k=[A1]:?k:While k:y=Int(k*Rnd):?y:k=y:Wend

Answer 49

Python 2.7, 112 bytes

from random import randint
k=int(raw_input())
def r(k):
    y=randint(0, k-1)
    print y
    if y!=0:
        r(y)
r(k)

Answer 50

F#, 70 bytes

let r=System.Random()
let rec x K=printfn"%i"K;if K>0 then r.Next K|>x

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Answer 51

Julia 0.6, 33 bytes

~n=(println(n);n>0&&~rand(0:n-1))

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Defines the function as an operator ~ (to save bytes). Print the given number, and if we haven't reached 0, call itself recursively with a new uniformly random number from range 0:n-1.

(Could've used @show instead of println since OP says "output format as you wish" (@show prints the variable name before the value every time, eg. n = 8 instead of 8), but the 2 byte saving doesn't seem worth it relative to total bytecount, and I like this neater output.)

Answer 52

AWK, 56 bytes

{print($0);k=$0;for(srand();k>0;)print(k=int(k*rand()))}

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Answer 53

Pascal (FPC), 89 bytes

var K:word;begin Randomize;read(K);repeat writeln(K);K:=Random(K)until K<1;writeln(K)end.

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Not only Pascal code is as long as usual, you also need to Randomize; beforehand...

Instead of writing of the initial value before the loop, 0 is written after the loop. This approach saves a byte because 1end. isn't valid; instead od separating this into 2 tokens, it tries to find a meaningful scientific notation.

Answer 54

Ruby, 23 bytes

f=->k{k<1||f[rand p k]}

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Non-recursive, 25 bytes

->k{k=rand p k while k>0}

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->k{(k=rand p k)>0&&redo}

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Answer 55

MATLAB, 36 bytes

K
Y=1;while Y>0,Y=randi(K)-1;K=Y,end

Sample output: For K = 25

K =

    25


K =

    15


K =

    11


K =

     6


K =

     4


K =

     1


K =

     0

Answer 56

Racket, 52 bytes

(define(f n)(println n)(unless(= n 0)(f(random n))))

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Answer 57

Julia 0.6, 49 bytes

~x=(println(x);x-1)
!x=(y=rand(0:~x))>0 ? !y : ~0

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If one were willing to accept outputting all output except the last to STDOUT, and the last 0 as returning false, then it can be

~x=(println(x);x-1)
!x=(y=rand(0:~x))>0&&!y

Which is a bit short. But I think mixing outputs like that is not really in spirit.

Answer 58

K4, 13 12 11 bytes

Solution:

(*1?)\[0<;]

Examples:

q)k)(*1?)\[0<;] 100
100 77 30 17 12 2 0
q)k)(*1?)\[0<;] 100
100 37 28 20 2 0
q)k)(*1?)\[0<;] 100
100 77 61 55 53 6 2 0
q)k)(*1?)\[0<;] 100
100 12 1 0

Explanation:

Iterate over expression while x is greater than zero.

(*1?)\[0<;] / the solution
(   )\[0<;] / iterate over brackets whilst 0< evaluates true
  1?        / 1 choose (returns a 1-item list)
 *          / take the first

Bonus:

A 14 byte solution that also works in K (oK); this one is an if/else version of the same theme:

{$[x;*1?x;x]}\

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Answer 59

Elixir, 157 148 bytes

-9 bytes from Okx

defmodule N do def f(0,k)do IO.puts 0 end;def f(n,k)do IO.puts n;f Enum.random(0..k),k end end;k=String.to_integer IO.gets"";N.f Enum.random(0..k),k

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Formatted:

defmodule N do 
    def f(0,k) do
      IO.puts n 
    end

    def f(n,k) do
      IO.puts n
      f Enum.random(0..k), k
    end
end
{k,_} = Integer.parse IO.gets""
N.f Enum.random(0..k),k

We define a module with 2 functions - one is recursive, the other is the base case with a guard so it only executes when our random is 0 (Elixir doesn't necessarily assign anything, it does pattern matching for e.g. arguments- and thus f(0,k) only matches when n=0, otherwise f(n,k) matches). After defining that module (since functions can't be defined outside a module), we parse an integer from input and start our recursive looping.

Notably, k=String.to_integer IO.gets"" is the same length as the other method I've found to parse integers from input, {k,_}=Integer.parse IO.gets"", which is kinda neat.

Answer 60

Common Lisp - 52 Bytes

(loop while (/= a 0) do (print (setf a (random a))))

Assumes a is already defined.

Test case:

(setf a 5000000)
(loop while (/= a 0) do (print (setf a (random a))))

Answer 61

Elixir, 87 bytes

fn n->[n]++Enum.take_while(Stream.repeatedly(fn->Enum.random(0..n)end),&(&1>1))++[0]end

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Answer 62

PHP, 44 bytes

<?=$a=$argn;while($a--)echo _,$a=rand(0,$a);

or

<?for(;$a=&$argn;$a=rand(0,$a))echo$a--,_?>0

print integers delimited by underscore. Save to file, run as pipe with -nF.

Answer 63

Noether, 9 bytes

I~n0(nRP)

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Output is given in the form

735920

Explanation:

I~n        - Store input in the variable n
   0(   ) - Loop until the top of the stack equals zero
     nR    - Push a random integer between 0 and n
       P   - Print the top of the stack

Answer 64

Swift 4, 30 bytes

while n>0{n=rand()%n;print(n)}

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n is the value of input

Answer 65

Forth (gforth), 57 bytes

include random.fs
: f begin dup . random dup 0= until . ;

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Explanation

Get Random integer in range, loop until result is 0.

Code Explanation

include random.fs     \ import the file that implements the "random" word
:f                    \ start a word defintion
  begin               \ start an indefinite loop
    dup .             \ duplicate the top of the stack and print it
    random            \ get random int between 0 and n - 1
    dup 0=            \ duplicate the top of the stack and then check if it's 0
  until               \ if it's zero end the loop
  .                   \ output the top of the stack (always 0)
;                     \ end the word definition

Answer 66

Julia 1.0, 53 bytes

function q(N)println(N);if N>0;q(rand(0:N-1));end;end

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Answer 67

Pyt, 5 bytes

`0⇹ɾł

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Explanation:

         Implicit input
`   ł    Do ... while top of stack is not 0:
 0          Push 0
  ⇹         Swap top two elements on stack
   ɾ        Get random number in [0,top of stack)
         Implicit print

Answer 68

Fortran 90, 56 bytes

read(*,*)i
do while(i>0)
i=rand(i)*i
print*,i
enddo
end

Trying to golf Fortran is always fun.

Also, because seeding the RNG in Fortran is a pain and costs a lot of bytes (and RAND() uses a particularly poor algorithm but has a nice short name), the output is pretty much deterministic - but I'm hoping to get away with it.

Example

Input:

10000000

Output:

 2636923
 1681125
  264113
   17707
    2453
      47
       0

Answer 69

PHP 5.6

$k = $y = 5; do { echo $k. "\n"; $k = rand(0, $y-1); } while($k > 0)