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

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 2

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 3

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 4

PHP, 41

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

Answer 5

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 6

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 7

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 8

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 9

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 10

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 11

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 12

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 13

Ruby: 22 characters

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

Answer 14

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 15

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 16

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 17

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 18

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 19

Clojure/ClojureScript, 26 bytes

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

Answer 20

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 21

R 27 Bytes

function(n)prod(seq_len(n))

Answer 22

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 23

PlatyPar, 8 bytes

c?1,_p\1

Try it online!

Explanation:

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

Answer 24

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 25

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 26

Lua, 47 bytes

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

Answer 27

PARI/GP, 16 bytes

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

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

Answer 28

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

Answer 29

DUP, 19 bytes

[$[$1-a;!*][%1]?]a:

Try it here!

A recursive lambda that leaves result on the stack. Usage:

6[$[$1-a;!*][%1]?]a:a;!

Explanation

[               ]a: {set a to lambda}
 $                  {check if top of stack >0}
  [       ][  ]?    {conditional}
   $1-a;!*          {if so, top of stack *a(top of stack -1)}
            %1      {otherwise, replace top of stack with 1}

Answer 30

Hoon, 29 bytes

|=(@ (reel (gulf [1 +<]) mul)

Hoon's native number is a bignum, so it works fine with 125 (or even 2000). It also correctly gives 1 for 0.

It uses +< in order to access the sample of the gate. This is axis navigation syntax: It means to access the tail of the subject, and then the head, which is where the sample is stored in the binary tree model Hoon uses.

Urbit drops you into a shell and Hoon REPL when you start it, :dojo. To test this, simply enter %. 125 on one line and then the snippet for 125! Note there are two spaces between the dot and 1.