Find the Factorial!

Question

Create the shortest program or function that finds the factorial of a non-negative integer.

The factorial, represented with ! is defined as such

$$n!:=\begin{cases}1 & n=0\\n\cdot(n-1)!&n>0\end{cases}$$

In plain English the factorial of 0 is 1 and the factorial of n, where n is larger than 0 is n times the factorial of one less than n.

Your code should perform input and output using a standard methods.

Requirements:

• Does not use any built-in libraries that can calculate the factorial (this includes any form of eval)
• Can calculate factorials for numbers up to 125
• Can calculate the factorial for the number 0 (equal to 1)
• Completes in under a minute for numbers up to 125

The shortest submission wins, in the case of a tie the answer with the most votes at the time wins.

Answer 1

Scala, 39

def f(x:BigInt)=(BigInt(1)to x).product

Answer 2

In Q (18 characters)

f:{(*/)9h$1+til x}

Computes in less than one millisecond.

q)\t f 125
0

Answer 3

F# based on cfern's 63 36 characters

His didn't work on 125 for me. Adapted to use BigInteger

let f n:BigInteger=Seq.fold(*)BigInteger.One{BigInteger.One..n}

Edit: I just realized that double works too.

let f n:double=Seq.fold(*)1.{1.0..n}

Answer 4

C++11 (35 chars)

Here's the function version:

int f(int x){return x?x*f(x-1):1;}

C++11 template version (103 chars)

And here's the template version:

template<int I>struct f{static const int v=I*f<I-1>::v;};template<>struct f<0>{static const int v=1;};

Answer 5

Golfscript — 16

{.!+,{(}%{*}*}:f

The way I handle 0! is to do this trick: .!+:

• 0 + 0! = 0 + 1 = 1
• a + a! = a + 0 = a (for every a != 0)

or:

{),{)}%);{*}*}:f

Here, I start of by increasing the argument by 1. But before I factor the array, I drop the last element.

Answer 6

PHP, 41

function f($i){return $i==1?:$i*f($i-1);}

Answer 7

C 20 characters

x(){while(n)f*=n--;}

Assuming f and n are global variables. Here is the entire program :

double n=5,f=1;

x(){while(n)f*=n--;}

main(){
x();
printf("%f",f);
}

Answer 8

C 37 characters

double f(int n){return n?n*f(n-1):1;}

This returns the value but is slightly longer than my
previous answer which used global variables.

Answer 9

Julia - 17

!n=n>1?n*!(n-1):1

This defines !n as !(n-1)*n if n>1, 1 otherwise. To make it work with big numbers you just need to make "n" a BigInt type (build in Julia).

And if its permitted (13 chars.):

!n=gamma(n+1)

with gamma equals to:

In the particular case that z its an integer the gamma function would be equal to:

Like its not a build in factorial it must not break the rules, but Im not posting it as solution just in case it does.

Answer 10

JavaScript (ES6) - 17 Characters

f=x=>x?x*f(x-1):1

Or:

f=x=>!x||x*f(x-1)

JavaScript - 17 Characters (not a function)

for(a=1;n;)a*=n--

Assumes that the variable n contains the number you want the factorial for and outputs the answer to the console and stores it in the variable a.

Answer 11

the minimum solutions is already given using C# lamada. But just try to this another way.

        var seq = Enumerable.Range(1, 5).ToList();
        int O=1;    //O will contain Factorial or ouput
        seq.ForEach(x=>O*=x); 

Answer 12

C#, just the relevant code, 59

(assuming the argument variable is called a)

Enumerable.Range(1,int.Parse(a[0])).Aggregate(1,(x,y)=>x*y)

With boilerplate, 122

using System.Linq;class A{static int Main(string[] a){return Enumerable.Range(1,int.Parse(a[0])).Aggregate(1,(x,y)=>x*y);}

(note that this solution returns the result)

Answer 13

C#, 30

double a=1;while(p>0){a*=p--;}

With spaces for ease of reading:

double a = 1;
        while(p > 0)
        {
            a *= p--;
        }

a is the factorial result, while p is the number for which factorial is computed.

Answer 14

Python 2.7.5 - 29 characters

f=lambda n:n and n*f(n-1)or 1

29 characters. It's still mathematically sound for a negative input value, since (-n)! = ∞ and therefore the program gives a Runtime Error maximum recursion depth exceeded.

Answer 15

Ruby: 22 characters

n.downto(1).reduce(:*)

Answer 16

Simplefunge, 87 chars including whitespace

v

     v  *&<
     >   &V
     `    &
 v     <  o
     ^H^  @
v>>!1-^
>iV    
  >1o@

I don't actually have time to test this right now, but it should work. If it doesn't work, I'll fix it tomorrow.

Answer 17

Scala, 28 (39 w/ recursion)

Solution:

def f(n:Int)=(1 to n)product

Recursive solution:

def f(n:Int):Int=if(n<2)1 else n*f(n-1)

Answer 18

R - 33

function(x) ifelse(x,prod(1:x),1)

> (function(x) ifelse(x,prod(1:x),1))(0)
[1] 1
> (function(x) ifelse(x,prod(1:x),1))(5)
[1] 120
> (function(x) ifelse(x,prod(1:x),1))(120)
[1] 6.689503e+198

Answer 19

Powershell, 31

$a=1;$args[0]..1|%{$a=$_*$a};$a

usage

powershell -nologo .\fact125.ps1 0
0
powershell -nologo .\fact125.ps1 1
1
owershell -nologo .\fact125.ps1 5
120
powershell -nologo .\fact125.ps1 125
1.88267717688893E+209

Answer 20

Python, 43 38

import math
f=lambda n:math.gamma(n+1)

Explanation: The gamma function is a very quickly-growing complex function which, at integer values, is equal to the factorial of one less than the number. So we add one to n and take the gamma function of it.

I hope this isn't considered cheating, since the gamma function is not technically able to directly calculate the factorial.

Answer 21

Clojure/ClojureScript, 26 bytes

#(apply *(range 1(inc %)))

Answer 22

Bash/coreutils/dc, 25

dc<<<"1 `seq -f%g* $1`p"

This forms a dc script and evaluates it. So ,with input of 5, we evaluate

1 1*
2*
3*
4*
5*p

It took my machine 2.05 seconds to compute 10000! here (that's factorial ten-thousand, with 36693 digits), so seems to scale reasonably well. For the zero case, seq produces no output, so the dc script is just 1 p which produces the correct output 1.

Answer 23

R 27 Bytes

function(n)prod(seq_len(n))

Answer 24

APL (13)

∇R←F X
R←×/ιX
∇

May need a ⎕IO←1 line to be sure ι starts at 1 - it's been awhile since I last used APL.

Answer 25

PlatyPar, 8 bytes

c?1,_p\1

Try it online!

Explanation:

c?        ## if (n != 0)
  1,_p     ## product [1..n]
       \  ## else
        1  ## 1

Answer 26

JavaScript, 34 bytes

function f(n){return n?n*f(n-1):1}

_(or)

function f(n){return n?n*f(--n):1}

Explanation

Function takes in a value, returns itself multiplied by
if n != 0: the same function on the number decreased by one
if n == 0: 1

The final f(0) returns first with 1, times 1, times 2, etc.

Terminator removed, may upset use strict.

Answer 27

𝔼𝕊𝕄𝕚𝕟, 9 chars / 19 bytes (noncompetitive)

+!ï⋎⨴⩤⁽1ï

Try it here (Firefox only).

Ay, 19th byte!

Great thing about this is that it also calculates factorials up to 171 instantly without returning Infinity.

Bonus solution!

+!ï⋎⨴МĂ⩤⁽1ï

Try it here (Firefox only).

This one allows you to calculate past 171 without getting Infinity. Still superbly fast!

Answer 28

Lua, 47 bytes

function f(n)return(n<1 and 1 or n*f(n-1))end

Answer 29

PARI/GP, 16 bytes

n->prod(i=2,n,i)

The shortest answer would be the native ! which is disallowed.

Answer 30

PHP

Short, 58

<?$r=$i=$argv[1];while($i>1){$i--;$r=$r*$i;}echo($r==0?1:$r);

Tests

0 -> 1

1 -> 1

5 -> 120

125 -> 1.8826771768889E+209

170 -> 7.257415615308E+306

171 -> INF

Executes in microseconds.

Ungolfed

<?php
$r = $i = $argv[1]; // Set $r and $i to Arg.
while($i > 1) // Calculate while $i bigger than 1
{
    $i--; // Decrement $i (so it's not infinite)
    $r = $r * $i; // Calculation the Factorial
 }
 echo ($r==0 ? 1: $r); // Output and make 0! = 1
 ?>

Slighty Longer, 86

<?$r=$i=(isset($argv[1])?$argv[1]:0);while($i>1){$i--;$r=$r*$i;}echo($r==0?1:$r)."\n";

Improvements

• Output with \n
• Doesn't throw error if no arg defined