Challenge
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
0 must be the final output, Optionally an empty line instead 0
.uOW
.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.
f(_){printf("%d\n",_);(_=rand()%_)&&f(_);}
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
f(_){printf("%d\n",_=rand()%_);_&&f(_);}
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
>0 and cat(n,"") (empty string) will also work.
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k=scan();while(x<-sample(1:k-1,1))k=c(x,k);cat(rev(k),0)
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n=scan();while(print(n))n=sample(n,1)-1
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`tYrqt
` % 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 is a programming language made by user Soaku.
REeErEErReEEreeEREEeEEree
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
{$_,(^*).pick...0}
Anonymous code block that returns a list of values. If you don't mind the numbers being ranges, you can do:
{^$_,^*.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.
XƬ0
This is a monadic link (function) that prints an array and returns 0.
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.
ẉ?ℕ₁-₁ṙ↰
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.
Δ=L<Ω0M
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
Δ=ݨΩ0M is equivalent.
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Commented
Jul 9, 2018 at 18:01
[:}:? ::]^: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.
^: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).
?., but I don’t think I’m using that.
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-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>
new Date%n doesn't really work here, since it doesn't change fast enough to be useful for generating multiple random numbers
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f(k){printf("%d ",k);k?f(rand()%k):0;}
Ungolfed
void f(int k){
printf("%d ",k);
if(k)
f(rand()%k);
}
&&; also, you may want to consider seeding the RNG in your main function: Try it online!
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W
~O
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:
O returns 3, ~ returns 5, \n returns and prints 5 which is trueO returns 0, ~ returns 3, \n returns and prints 3 which is trueO returns something irrelevant, ~ returns 0, \n returns and prints 0 which is false
[ [ random ] follow ]
Basically, collects the values after each iteration (including zeroth iteration) stopping when the value is f, where each iteration converts the value n to a random integer between 0 and n-1 inclusive. Calling random on zero yields f, conveniently terminating the loop.
Interestingly, the doc page for follow has an example code that solves almost the same task, but it is actually longer since it tries to exclude zero at the end.
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.
{⌊?⎕←⍵}⍣=
{…}⍣= 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)
Some idiot™ forgot to print the initial value first.
f(K){while(K)printf("%d\n",K,K=rand()%K);}
Don't panic.
f(K){while(K)printf("%d\n",K),K=rand()%K;}. Still you got my +1 for an equal solution!
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Commented
Jul 9, 2018 at 17:26
f(K){while(K)printf("%d\n",K,K=rand()%K);}
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Commented
Jul 9, 2018 at 17:47
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
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...
k=0 is acceptable.
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Commented
Jul 10, 2018 at 0:42
hash() tries to maintain but doesn't guarantee this property. For that task you should use the random module.
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Commented
Jul 10, 2018 at 12:41
random.randrange(): from random import*;f=lambda k:print(k)or f(randrange(k))
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Commented
Jul 10, 2018 at 13:29
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
int(Ansrand is shorter. Also, using Pause instead of Disp, you can make the only statement in the loop be Pause int(Ansrand, which also updates Ans.
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Commented
Jul 10, 2018 at 3:37
from random import*
k=input()
while k:print k;k=randrange(k)
Saved
while 1:print k;k=randint(0,~-k) should work (with an error at the end)
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Commented
Jul 9, 2018 at 16:28
while 1:print k;k=randrange(k) saves two.
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Commented
Jul 9, 2018 at 16:30
#import<cstdio>
#import<cstdlib>
#define d printf("%i ",x
int p(int x){d);while(x>0)d=rand()%x);}
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.
-Dd='printf("%i,",x' instead of the #define would save you some bytes (-4), and is allowed as long as you count the bytes towards your result (because it is a non-standard preprocessor directive. You can also leave out the imports (at least with -std=c++98 and -w, which don't count for bytes), and the variable types. So, you'd have p(x){d);while(x>0)d=rand()%x;} and -Dd='printf("%i,",x'.
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:nao:0=?;0_1>:{:}(?\\
}(?!\$2*1>x~\$+1$*2/\~00:{{:@}
8+1.\~:{:}\+>$1+f
~~@~~47*0.\(a2*1@@?!.
+2B for -v flag
><>'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.
: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
$1+ increment i
8+1. f Jump back to the start of the inner loop
After we've generated our number, check that it's actually below n
~ Delete i
:{:} ( ? Test that the number is less than n
a2*1 . If it's not, jump back to the start
of the number generation section
@~~ Otherwise delete the old value of n, and numberOfBits
47*0. Then jump back to the start of the program
@(k)eval('while k;disp(k);k=randi(k)-1;end;0')
Sample output:
>> @(k)eval('while k;disp(k);k=randi(k)-1;end;0')
ans(5)
ans =
@(k)eval('while k;disp(k);k=randi(k)-1;end;0')
5
3
2
1
ans =
0
k=randi(k)-1 for a few bytes less.
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Commented
Jul 11, 2018 at 13:16
.+
*
L$`.
$.`
+¶<)G?`
Try it online! Explanation:
+
Repeat until the value stops changing (i.e. 0).
¶<)
Print the value before each pass through the loop.
.+
*
Convert to unary.
L$`.
$.`
Create the range and convert to decimal.
G?`
Pick a random element.
QWQ=OQQ
+1 to print the initial input value.
While Q is truthy, set Q to be a random integer between 0 and Q and print Q.
Not the shortest Pyth answer but I'm just learning and only posting because of the recent discussion about no-one using Pyth any more :)
= and ~ use the first variable of an expression as the variable that will be assigned if one isn't specified. For example ~hT will set T to 11 while returning 10. The only other fancy trick is that the newline character prints its input and then returns that value unmodified, so we can have an empty loop body. Let me know if something else is confusing :)
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Commented
Jul 9, 2018 at 21:54
-3 bytes by actually doing what the specs say it should do.
import System.Random
f 0=pure[0]
f x=randomRIO(0::Int,x-1)>>=fmap(x:).f
o:=i;@While(i>0;i:=@Integer(i*@Random);o:=o:i);o
Field formula that takes input from another field i.
There's no TIO for formula so here's a screenshot of a sample output:
-4 bytes thanks AdmBorkBork
filter f{$_;if($_){Random $_|f}}
Testscript:
filter f{$_;if($_){Random $_|f}}
100 |f
Output:
100
61
20
8
6
3
0
Get- is implied, so you can remove it for -4 bytes Try it online!
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Commented
Jul 9, 2018 at 18:08
for($a="$args";$a;$a=Random $a){$a}
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).
"$args" - nice. I was stuck on $args[0] in this case
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