31
\$\begingroup\$

Your task is to make a program that takes in an integer n > 1, and outputs the roll of a single n-sided die. However, this dice follows the rules for exploding dice.

When you roll the die, check what value you rolled. If you got the maximum for that kind of die (on a standard d4 that would be 4, or 6 on a d6, etc.), roll again and add the new roll to that total. Each roll continues adding to the total, until you don't roll the max number anymore. That final number is still added though.

Your program should take in a single integer n, and roll the exploding n-sided die. Here's an example distribution to show what it should look like for n=4. Note that you should never output any multiples of n, since they will always explode.

You can assume the stack size for any recursion you do is infinite, and your random function must meet our standards for randomness (built-in random generator or time/date). Your random function should also be as uniform as possible, vs. something like a geometric distribution, since these are dice we're talking about.

\$\endgroup\$
  • 1
    \$\begingroup\$ does the program have to be perfect? Like can its distribution be off by some extremely low amount? \$\endgroup\$ – Maltysen Apr 12 at 14:38
  • \$\begingroup\$ To: Riker; RE: @Maltysen's comment above; or extremely high amount? \$\endgroup\$ – Artemis Fowl Apr 12 at 15:09
  • 2
    \$\begingroup\$ @ArtemisFowl See our standards for randomness. Also, here. \$\endgroup\$ – Rɪᴋᴇʀ Apr 12 at 22:46

49 Answers 49

36
\$\begingroup\$

x86 Machine Code (for Intel Ivy Bridge and later), 17 bytes

31 C9 0F C7 F0 31 D2 F7 F6 42 01 D1 39 F2 74 F2 C3

The above bytes of code define a function that simulates an exploding die. It takes a single input, passed in the ESI register, indicating the maximum number of the die. It returns a single value in the ECX register, which is the result of the rolls.

Internally, it uses the RDRAND instruction to generate a random number. This uses a random number generator (RNG) that is built into the hardware on Intel Ivy Bridge processors and later (some AMD CPUs also support this instruction).

The logic of the function is otherwise quite straightforward. The generated random number is scaled to lie within the desired range using the standard technique ((rand % dieSize) + 1), and then it is checked to see if it should cause an explosion. The final result is kept in an accumulator register.

Here is an annotated version showing the assembly language mnemonics:

           unsigned int RollExplodingDie(unsigned int dieSize)
31 C9        xor     ecx, ecx    ; zero-out ECX, which accumulates the final result
           Roll:
0F C7 F0     rdrand  eax         ; generate a random number in EAX
31 D2        xor     edx, edx    ; zero-out EDX (in preparation for unsigned division)
F7 F6        div     esi         ; divide EDX:EAX by ESI (the die size)
                                 ;   EAX receives the quotient; EDX receives the remainder
42           inc     edx         ; increment the remainder
01 D1        add     ecx, edx    ; add this roll result to the accumulator
39 F2        cmp     edx, esi    ; see if this roll result should cause an explosion
74 F2        jz      Roll        ; if so, re-roll; otherwise, fall through
C3           ret                 ; return, with result in ECX register

I am cheating a bit. All standard x86 calling conventions return a function's result in the EAX register. But, in true machine code, there are no calling conventions. You can use any registers you want for input/output. Using ECX for the output register saved me 1 byte. If you want to use EAX, insert a 1-byte XCHG eax, ecx instruction immediately before the ret instruction. This swaps the values of the EAX and ECX registers, effectively copying the result from ECX into EAX, and trashing ECX with the old value of EAX.

Try it online!

Here's the equivalent function transcribed in C, using the __builtin_ia32_rdrand32_step intrinsic supported by GCC, Clang, and ICC to generate the RDRAND instruction:

#include <immintrin.h>

unsigned int RollExplodingDie(unsigned int dieSize)
{
    unsigned int result = 0;
Roll:
    unsigned int roll;
    __builtin_ia32_rdrand32_step(&roll);
    roll    = ((roll % dieSize) + 1);
    result += roll;
    if (roll == dieSize)   goto Roll;
    return result;
}

Interestingly, GCC with the -Os flag transforms this into almost exactly the same machine code. It takes the input in EDI instead of ESI, which is completely arbitrary and changes nothing of substance about the code. It must return the result in EAX, as I mentioned earlier, and it uses the more efficient (but larger) MOV instruction to do this immediately before the RET. Otherwise, samezies. It's always fun when the process is fully reversible: write the code in assembly, transcribe it into C, run it through a C compiler, and get your original assembly back out!

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12
\$\begingroup\$

R, 39 bytes

n=scan();n*rgeom(1,1-1/n)+sample(n-1,1)

Try it online!

Explanation: this solution avoids recursion/while loops by directly calculating the distribution of the number of explosions that will occur. Let \$n\$ be the number of sides on the die. If you denote success as rolling an \$n\$ and failure as rolling anything else, then you have probability \$\frac1n\$ of success. The total number of explosions is the number of successes before the first failure. This corresponds to a \$\mathrm{Geometric}\left(1-\frac{1}{n}\right)\$ distribution (see the wikipedia page, which defines success and failure the other way round). Each explosion brings \$n\$ to the total. The final roll follows a \$\mathrm{Uniform}(1,2,\ldots,n-1)\$ distribution which we add to the total.

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  • \$\begingroup\$ very nice! Gotta love the builtin distributions for random challenges! \$\endgroup\$ – Giuseppe Apr 12 at 15:35
  • \$\begingroup\$ Does sample meet the standards for randomness, given its bias? \$\endgroup\$ – Xi'an Apr 25 at 8:37
  • \$\begingroup\$ @Xi'an Pretty sure it does: it is the built-in random generator for discrete random variables. \$\endgroup\$ – Robin Ryder Apr 25 at 22:37
  • \$\begingroup\$ I know, I know, but check the link I put: the discretisation inherent to sample leads to a lack of uniformity that gives a ratio of max to min probability as high as 1.03... Shocking, isn't it?! \$\endgroup\$ – Xi'an Apr 26 at 7:28
  • \$\begingroup\$ Yes, it is shocking. But then again, how often do you use sample with \$m\approx 2^{31}\$? ;-) \$\endgroup\$ – Robin Ryder Apr 26 at 22:28
11
\$\begingroup\$

Python 2, 66 64 61 bytes

-3 bytes thanks to xnor

f=lambda n,c=0:c%n or c+f(n,randint(1,n))
from random import*

Try it online!

The previous roll is stored in c, allowing us to access it multiple times without having to store it to a variable, which can't be done in a Python lambda. Each recursion, we check if we rolled exploding dice.

c is initialised to zero, so c%n is falsey there. In the next iterations, it will only be falsey if exploding dice were rolled.

Python 2, 55 bytes

f=lambda n:randint(1,n)%n or n+f(n)
from random import*

Try it online!

My other answer seems to be a bit overengineered, since this appears to work as well... I'll leave it anyway.

\$\endgroup\$
  • 2
    \$\begingroup\$ Recursive functions where the breaking condition is based on randomness will always have a non-zero chance of stack overflow. Statistically insignificant chance, but still... \$\endgroup\$ – mypetlion Apr 12 at 19:23
  • 3
    \$\begingroup\$ Typically, stack size is assumed to be infinite in code golfing challenges in my experience. As the stack size increases to infinity, the likelihood of stack overflow quickly converges to zero. \$\endgroup\$ – ArBo Apr 12 at 20:07
  • \$\begingroup\$ ArBo do @mypetlion before you type in your comment so you can ping the user \$\endgroup\$ – MilkyWay90 Apr 13 at 2:04
  • 1
    \$\begingroup\$ I think c*(c<n) can be c%n. \$\endgroup\$ – xnor Apr 13 at 2:28
  • \$\begingroup\$ @xnor Of course, I'm an idiot... \$\endgroup\$ – ArBo Apr 13 at 3:40
9
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Perl 6, 26 bytes

{sum {roll 1..$_:}...*-$_}

Try it online!

Explanation

{                        } # Anonymous block
                  ...      # Sequence constructor
     {roll 1..$_:}         #   Next elem is random int between 1 and n
                           #   (Called as 0-ary function with the original
                           #   $_ for the 1st elem, then as 1-ary function
                           #   with $_ set to the previous elem which
                           #   equals n.)
                     *-$_  #   Until elem not equal to n (non-zero difference)
 sum                       # Sum all elements
\$\endgroup\$
9
\$\begingroup\$

J, 16 11 bytes

(+$:)^:=1+?

Try it online!

Explanation

TL;DR 1+? performs the die roll, (+$:)^:= reiterates only when it equals the input.


The function is a train of 4 verbs:

             ┌─ + 
         ┌───┴─ $:
  ┌─ ^: ─┴─ =     
  │               
──┤      ┌─ 1     
  └──────┼─ +     
         └─ ?     

A train is when 2 or more verbs are concatenated. Here, the answer is of the form f g h j:

(+$:)^:=  1  +  ?
    f     g  h  j

A so-called "4-train" is parsed as a hook and a fork:

f g h j   ⇔   f (g h j)

Thus, the answer is equivalent to:

(+$:)^:= (1 + ?)

Hooks: (f g) x and x (f g) y

A monadic (one-argument) hook of two verbs, given an argument x, the following equivalence holds:

(f g) x   ⇔   x f (g x)

For example, (* -) 5 evaluates to 5 * (- 5), which evaluates to _25.

This means that our 4-train, a hook of f and (g h j), is equivalent to:

(f (g h j)) x   ⇔   x f ((g h j) x)

But what does f do here? (+$:)^:= is a conjunction of two verbs using the Power conjunction ^:: another hook ((+$:)) and a verb (=). Note here that f is dyadic—it has two arguments (x and (g h j) x). So we have to look at how ^: behaves. The power conjunction f^:o takes a verb f and either a verb or a noun o (a noun is just a piece of data) and applies f o times. For example, take o = 3. The following equivalences holds:

(f^:3) x     ⇔   f (f (f x))
x (f^:3) y   ⇔   x f (x f (x f y))

If o is a verb, the power conjunction will simply evaluate o over the arguments and use the noun result as the repeat count.

For our verb, o is =, the equality verb. It evaluates to 0 for differing arguments and to 1 for equal arguments. We repeat the hook (+$:) once for equal arguments and no times for differing ones. For ease of notation for the explanation, let y ⇔ ((g h j) x). Remember that our initial hook is equivalent to this:

x   (+$:)^:=   ((g h j) x)
x   (+$:)^:=   y

Expanding the conjunction, this becomes:

x ((+$:)^:(x = y)) y

If x and y are the same, this becomes:

x (+$:)^:1 y   ⇔   x (+$:) y

Otherwise, this becomes:

x (+$:)^:0 y   ⇔   y

Now, we've seen monadic forks. Here, we have a dyadic fork:

x (f g) y   ⇔   x f (g y)

So, when x and y are the same, we get:

x (+$:) y   ⇔   x + ($: y)

What is $:? It refers to the entire verb itself and allows for recursion. This means that, when x and yare the same, we apply the verb toyand addx` to it.

Forks: (g h j) x

Now, what does the inner fork do? This was y in our last example. For a monadic fork of three verbs, given an argument x, the following equivalence hold:

(g h j) x   ⇔   (g x) h (j x)

For this next example, suppose we have verbs named SUM, DIVIDE, and LENGTH, which do what you suppose they might. If we concatenate the three into a fork, we get:

(SUM DIVIDE LENGTH) x   ⇔   (SUM x) DIVIDE (LENGTH x)

This fork evaluates to the average of x (assuming x is a list of numbers). In J, we'd actually write this as example as +/ % #.

One last thing about forks. When the leftmost "tine" (in our symbolic case above, g) is a noun, it is treated as a constant function returning that value.

With all this in place, we can now understand the above fork:

(1 + ?) x   ⇔   (1 x) + (? x)
            ⇔   1 + (? x)

? here gives a random integer in the range \$[0,x)\$, so we need to transform the range to represent dice; incrementing yields the range \$[1, x]\$.

Putting it all together

Given all these things, our verb is equivalent to:

((+$:)^:=1+?) x   ⇔   ((+$:)^:= 1 + ?) x
                  ⇔   ((+$:)^:= (1 + ?)) x
                  ⇔   x ((+$:)^:=) (1 + ?) x
                  ⇔   x ((+$:)^:=) (1 + (? x))
                  ⇔   x (+$:)^:(x = (1 + (? x))
(let y = 1 + (? x))
if x = y          ⇒   x + $: y
otherwise         ⇒   y

This expresses the desired functionality.

\$\endgroup\$
  • 1
    \$\begingroup\$ (+$:)^:=1+?­­ \$\endgroup\$ – ngn Apr 13 at 1:26
  • \$\begingroup\$ @ngn Thanks! Incorporated. \$\endgroup\$ – Conor O'Brien Apr 14 at 18:00
7
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Jelly, 7 bytes

X+ß}¥=¡

Try it online!

Uses recursion. Runs the program again (ß) and adds (+) if (¡) the random number (X) is equal (=) to the program input. } makes ß act on the program input and ¥ combines +ß} into a single link for ¡ to consume.

Here a distribution of 1000 outputs for n=6 which I collected using this program. Plotted with python/matplotlib. histogram

Here is a 5000 data points from n=3 on a semilog plot which shows the (approximately?) exponential distribution. enter image description here

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  • \$\begingroup\$ Nice plots! The distribution you get is a geometric distribution (see my R answer), which is closely related to the exponential distribution. \$\endgroup\$ – Robin Ryder Apr 14 at 21:42
6
\$\begingroup\$

Pyth - 12 11 bytes

Uses functional while. I feel like there should be a smarter answer that just simulates the distribution.

-.W!%HQ+hOQ

-         (Q)         Subtract Q. This is because we start Z at Q to save a char
 .W                   While, functionally
  !                   Logical not. In this case, it checks for 0
   %HQ                Current val mod input
  +     (Z)           Add to current val
   h                  Plus 1
    OQ                Random val in [0, input)

Try it online.

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4
\$\begingroup\$

Python 3, 80 bytes

import random as r,math
lambda n:int(-math.log(r.random(),n))*n+r.randint(1,n-1)

Try it online!

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  • 1
    \$\begingroup\$ There's a slight chance for failure if r.random() happens to return 0. 1-r.random() should work, though. \$\endgroup\$ – nwellnhof Apr 12 at 16:38
  • \$\begingroup\$ Though technically that chance is 0 \$\endgroup\$ – Quintec Apr 12 at 20:45
  • 1
    \$\begingroup\$ A rare case where import ... as _ is shortest! \$\endgroup\$ – xnor Apr 13 at 2:24
  • \$\begingroup\$ @xnor indeed! The only other time I remember that winning out in an answer of mine is here \$\endgroup\$ – Lynn Apr 13 at 15:48
4
\$\begingroup\$

05AB1E, 10 bytes

[ILΩDIÊ#}O

Try it online or verify the lists.

10 bytes alternative:

[LΩDˆÊ#}¯O

Try it online or verify the lists.

Although I like the top one more because it got the 'word' DIÊ in it, which suits the challenge.

Explanation:

[         # Start an infinite loop:
 IL       #  Create a list in the range [1, input]
   Ω      #  Pop and push a random value from this list
    D     #  Duplicate it
     IÊ   #  If it's NOT equal to the input:
       #  #   Stop the infinite loop
}O        # After the loop: sum all values on the stack
          # (which is output implicitly as result)

[         # Start an infinite loop
 L        #  Create a list in the range [1, (implicit) input]
  Ω       #  Pop and push a random value from this list
   Dˆ     #  Add a copy to the global_array
     Ê    #  If it's NOT equal to the (implicit) input:
      #   #   Stop the infinite loop
}¯        # After the loop: push the global_array
  O       # Pop and push its sum
          # (which is output implicitly as result)  
\$\endgroup\$
  • \$\begingroup\$ Was trying to think of a way to use or something. \$\endgroup\$ – Magic Octopus Urn Apr 22 at 17:24
3
\$\begingroup\$

Wolfram Language (Mathematica), 50 bytes

R@#//.x_/;x~Mod~#==0:>x+R@#&
R=RandomChoice@*Range

Try it online!

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3
\$\begingroup\$

R, 47 42 bytes

function(n){while(!F%%n)F=F+sample(n,1)
F}

Try it online!

Credit to ArBo's approach.

Still a byte longer than Robin Ryder's, go upvote his!

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  • \$\begingroup\$ Interesting, I reworked this to a recursive if for 46 bytes, but ended up getting a 52 on one roll which shouldn't be possible with n=4, so I don't know if there's a weird low recursion limit thing happening, but I think it may be buggy. Try it online! \$\endgroup\$ – CriminallyVulgar Apr 12 at 14:45
  • \$\begingroup\$ I tried a recursive and got a 54 byte solution. Then tried something similar to yours for 44 Try it online! \$\endgroup\$ – Aaron Hayman Apr 12 at 16:48
3
\$\begingroup\$

Ruby, 35 bytes

->n,s=0{s+=x=1+rand(n);x<n||redo;s}

Try it online!

\$\endgroup\$
  • \$\begingroup\$ Save a byte by dropping the x variable: Try it online! \$\endgroup\$ – benj2240 Apr 12 at 22:55
3
\$\begingroup\$

APL (Dyalog Unicode), 15 14 bytes

{×r←?⍵:r⋄⍵+∇⍵}

Try it online!

\$\endgroup\$
3
\$\begingroup\$

Haskell, 77 76 bytes

import System.Random
f x=randomRIO(1,x)>>=(x!)
x!y|y<x=pure y|0<1=(y+)<$>f x

Try it online!

Thanks to killmous for one byte.

If <|> were in the prelude, we could do better with MonadComprehensions:

Haskell, non-competing, 66 bytes

import System.Random
f x=do y<-randomRIO(1,x);[y|y<x]<|>(y+)<$>f x

Try it online!

\$\endgroup\$
  • 1
    \$\begingroup\$ You can save a byte if you define g as an infix function. \$\endgroup\$ – killmous Apr 12 at 19:38
  • \$\begingroup\$ @killmous, thanks. At first glance I figured that would be the same or worse, but it's better. \$\endgroup\$ – dfeuer Apr 12 at 21:02
3
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Python 2, 53 bytes

f=lambda n:random()*n//1or n+f(n)
from random import*

Try it online!

Uses the or short-circuiting idea from ArBo's answer. The expression random()*n//1 generates a number from 0 to n-1, with 0 taking the place of a roll of n. The or takes the that number, except if it's zero (Falsey) it continues on to n+f(n).

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  • \$\begingroup\$ It seems your answer was already up when I edited in my shorter one... I didn't see this, but if you want me to delete it because it's quite alike I will. \$\endgroup\$ – ArBo Apr 13 at 18:05
  • \$\begingroup\$ @ArBo Not at all, independently duplicated answers are fine. \$\endgroup\$ – xnor Apr 13 at 18:07
3
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Japt, 13 bytes

ö)g@¶°X?X+ß:X

Try it

Port of Arnauld's answer. Figured out how to make a recursive call ;)

Transpiled JS:

// U: Implicit input
// ö: generate a random number [0,U)
(U.ö())
  // g: run the result through a function
  .g(function(X, Y, Z) {
    // increment the result and compare to input
    return U === (++X)
      // if they are the same, roll again and add to current roll
      ? (X + rp())
      // if they are different, use current roll
      : X
   })
\$\endgroup\$
  • 1
    \$\begingroup\$ Very nice use of N.g(f) :) \$\endgroup\$ – Shaggy Apr 13 at 11:56
  • \$\begingroup\$ Took a stab at this meself and ended up with 12 bytes but I don't want to post it 'cause I like your solution too much! \$\endgroup\$ – Shaggy Apr 13 at 19:15
  • \$\begingroup\$ Post it as a different answer :) \$\endgroup\$ – dana Apr 13 at 19:19
  • \$\begingroup\$ It may be shorter, but it's a hell of a lot uglier than yours: petershaggynoble.github.io/Japt-Interpreter/… \$\endgroup\$ – Shaggy Apr 13 at 19:21
  • \$\begingroup\$ I see - yeah I was trying to come up with a way not to pollute U. Skipping a line seems to work as well. That's a good trick :) \$\endgroup\$ – dana Apr 13 at 19:37
3
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Japt, 12 bytes

It may be shorter than dana's solution, but it's a hell of a lot uglier. I'm only posting it 'cause it seems like forever since we had a Japt solution that started with an empty line.


ö
>°V©VªV+ß

Try it

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2
\$\begingroup\$

PowerShell, 49 bytes

for($a=$l="$args";$a-eq$l){$o+=$l=1..$a|Random}$o

Try it online!

Iterative method. Sets the input $args to $a and the $last roll (done so we enter the loop at least once). Then, so long as the last roll is -equal to the input, we keep rolling. Inside the loop we accumulate into $o the last roll, which is updated by creating a range from 1 to input $a and picking a Random element thereof. (Honestly, I'm a little surprised that $o+=$l= works.) Once we're out of the loop, we leave $o on the pipeline and output is implicit.

\$\endgroup\$
2
\$\begingroup\$

Forth (gforth), 72 bytes

include random.fs
: f >r 0 begin i random 1+ >r i + r> i < until rdrop ;

Try it online!

Code Explanation

include random.fs      \ include library file for random
: f                    \ start a new word definition
  >r                   \ stick the input on the return stack (for easy access)
  0                    \ add a counter to hold the sum
  begin                \ start an indefinite loop
    i random 1+        \ generate a random number from 1 to n
    >r i + r>          \ add the result to the counter, use the return stack to save a few bytes
    i <                \ check if result was less than n
  until                \ end the loop if it was, otherwise go back to begin
  rdrop                \ remove n from the return stack
;                      \ end the word definition
\$\endgroup\$
2
\$\begingroup\$

Batch, 70 bytes

@set t=0
:g
@set/at+=d=%random%%%%1+1
@if %d%==%1 goto g
@echo %t%

Takes input n as a command-line parameter %1. d is the current roll, t the cumulative total. Simply keeps rolling until d is not equal to n.

\$\endgroup\$
2
\$\begingroup\$

Jelly, 9 bytes

x⁹X€Ä%ƇµḢ

Try it online!

A monadic link that takes n as its argument and returns a number generated by an exploding n-sided die. This generates 256 numbers from 1 to n and returns the first cumulative sum that is not a multiple of n. In theory this could return 256n, but even for a 2-sided die this would happen only one every \$2^{256}\$ times.

An alternative that doesn’t have this limitation is:

Jelly, 10 bytes

X³X¤+¥³ḍ¥¿

Try it online!

Note both TIO links generate 400 numbers to show the distribution.

\$\endgroup\$
2
\$\begingroup\$

Python 3, 81 72 bytes

from random import*
def f(x,a=0):
 while a%x<1:a+=randint(1,x)
 return a

Try it online!

-9 bytes thanks to ArBo

Explanation

import random             #load the random module              
def explodeDice(num):     #main function
    ans = 0                     #set answer to 0
    while a % num != 0:         #while a isn't a multiple of the input
        ans += random.randint(1, num) #add the next dice roll to answer
    return ans                  #return the answer
\$\endgroup\$
  • \$\begingroup\$ You can save 1 byte by using from random import* instead. \$\endgroup\$ – orthoplex Apr 12 at 15:48
  • 1
    \$\begingroup\$ You can get this down to 74 bytes using this recursive solution \$\endgroup\$ – squid Apr 12 at 15:51
  • 1
    \$\begingroup\$ @squid You can save 1 byte like this. \$\endgroup\$ – orthoplex Apr 12 at 16:01
  • 1
    \$\begingroup\$ @orthoplex and then you can shorten the if/else, and make it a one-liner. Starts to look like my solution then ;) \$\endgroup\$ – ArBo Apr 12 at 16:16
  • 1
    \$\begingroup\$ @ArBo Yea that's why I didn't change to recursive, didn't want to just copy you. \$\endgroup\$ – Artemis Fowl Apr 12 at 17:18
2
\$\begingroup\$

TI-BASIC, 28 23 bytes

-5 bytes thanks to this meta post!

Ans→N:0:Repeat fPart(Ans/N:Ans+randInt(1,N:End:Ans

Input is in Ans.
Output is in Ans and is implicitly printed.

Examples:

4
              4
prgmCDGF11
              5
6
              6
prgmCDGF11
              3

Explanation:

Ans→N:0:Repeat fPart(Ans/N:Ans+randInt(1,N:End:Ans   ;full logic

Ans→N                                                ;store the input in "N"
      0                                              ;leave 0 in "Ans"
        Repeat fPart(Ans/N                 End       ;loop until the sum
                                                     ; is not a multiple of
                                                     ; the input
                               randInt(1,N           ;generate a random
                                                     ; integer in [1,N]
                           Ans+                      ;then add it to "Ans"
                                               Ans   ;leave the sum in "Ans"
                                                     ;implicitly print "Ans"

Notes:

  • TI-BASIC is a tokenized language. Character count does not equal byte count.
\$\endgroup\$
  • \$\begingroup\$ Since startTmr is no longer necessary, this submission will now work for versions of TI-BASIC earlier than the TI-84+ \$\endgroup\$ – Tau Apr 13 at 0:17
2
\$\begingroup\$

PowerShell, 43 bytes (Iterative method)

param($n)do{$x+=1..$n|random}until($x%$n)$x

Try it online!


PowerShell, 48 bytes (recursive method)

filter f{if($_-eq($x=1..$_|random)){$x+=$_|f}$x}

Try it online!

\$\endgroup\$
2
\$\begingroup\$

Jelly, 7 bytes

X=п⁸S_

A monadic Link accepting an integer, n, which yields an integer.

Try it online! Or see the counts of \$10^5\$ runs

How?

X=п⁸S_ - Link: integer, n
  п    - Collect up while...
 =  ⁸   - ...condition: equal to chain's left argument, n
X       - ...next value: random number in [1..n]
     S  - sum
      _ - subtract n (since the collection starts with [n])
\$\endgroup\$
2
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SmileBASIC, 41 bytes

INPUT N@L
S=S+RND(N)+1ON S MOD N GOTO@L?S

After reading:

Note that you should never output any multiples of n, since they will always explode.

I realized that rather than checking if a dice roll was n, you can just repeat while the sum is a multiple of n.

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2
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AnyDice, 36 bytes

Almost a built-in in the language:

function:f I:n{result: [explode dI]}

For this to be correct I have to abuse the infinite recursion depth assumption. AnyDice limits the recursion depth with a global property maximum function depth. the explode builtin however uses it's own; explode depth - which defaults to 2.

set "explode depth" to 99

Would add another 25 bytes; and would not really match the requirements since it's theoretically possible for a dice to explode more than 99 times.

The output of the function is a die, ie. an AnyDice built-in type that is a paring of results and probabilities for the result.

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  • 1
    \$\begingroup\$ I think I'm OK with it not exploding much, the 36 byte one is fine by me. I didn't say no builtins and I'm ok with having them here, since it's not like your 1 or 0 byte answer is winning. But welcome to the site! \$\endgroup\$ – Rɪᴋᴇʀ Apr 15 at 13:57
2
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CJam, 19 bytes

qi{__mr)_T+:T;=}g;T

Explanation:

T is pre-set to 0

qi{__mr)_T+:T;=}g;T - whole code
qi                  - read input as integer (n) | Stack: n
  {            }    - block
   __               - Duplicate twice | Stack: n n n
     mr)            - Choose a random number from 1 to n (r). Since 'mr' picks a
                      number from 0 to n-1, the number has to be incremented with ')' 
                      Stack: n n r
        _           - Duplicate |  Stack: n n r r
         T          - push T | Stack: n n r r T
          +         - add the random number to T (t) | Stack: n n r t
           :T;      - pop the value and store in T | Stack: n n r
              =     - are the top two stack values the same (c) | Stack: n c
               }
                g   - do while loop that pops the condition from the stack after each
                      iteration | Stack: n
                 ;  - pop the top stack element and discard | Stack: T
                  T - push T | Stack: T
                    - implicit output

Or in pseudocode:

input n
var sum = 0
do {
    var random_number = pick random number from 1 to n
    sum = sum + random_number
} while (random_number == n)
output n

As a flowchart:

Flowchart of code

Try it online!

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2
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Excel VBA, 46 bytes

Thanks to @TaylorScott

Do:v=-Int(-[A1]*Rnd):t=t+v:Loop While[A1]=v:?t

Executed in the command window.

As a user-defined function.

Excel VBA, 108 67 bytes

Function z(i)
Do
v=Int((i*Rnd)+1)
z=z+v
Loop While v=i
End Function
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  • \$\begingroup\$ You can get this down quite a bit by using a do..loop while loop, and converting the function into an immediate window function. - Do:v=-Int(-[A1]*Rnd):t=t+v:Loop While[A1]=v:?t - 46 Bytes \$\endgroup\$ – Taylor Scott Apr 15 at 19:03
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    \$\begingroup\$ @TaylorScott Thanks, I forgot that Do x While y existed in Excel VBA. \$\endgroup\$ – william porter Apr 15 at 20:23
2
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JavaScript (ES6),  39  30 bytes

Saved 9 bytes thanks to @tsh!

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

Try it online! or See the distribution for n=4

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  • 1
    \$\begingroup\$ 30 bytes, f=n=>Math.random()*n|0||n+f(n) \$\endgroup\$ – tsh Apr 15 at 9:18

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