Factorial Algorithms in different languages

Question

I want to see all the different ways you can come up with, for a factorial subroutine, or program. The hope is that anyone can come here and see if they might want to learn a new language.

Ideas:

• Procedural
• Functional
• Object Oriented
• One liners
• Obfuscated
• Oddball
• Bad Code
• Polyglot

Basically I want to see an example, of different ways of writing an algorithm, and what they would look like in different languages.

Please limit it to one example per entry. I will allow you to have more than one example per answer, if you are trying to highlight a specific style, language, or just a well thought out idea that lends itself to being in one post.

The only real requirement is it must find the factorial of a given argument, in all languages represented.

Be Creative!

Recommended Guideline:

# Language Name: Optional Style type

   - Optional bullet points

    Code Goes Here

Other informational text goes here

I will ocasionally go along and edit any answer that does not have decent formatting.

Answer 1

Perl 6: Functional

multi factorial ( Int $n where { $n <= 0 } ){
  return 1;
}
multi factorial ( Int $n ){
   return $n * factorial( $n-1 );
}

This will also work:

multi factorial(0) { 1 }
multi factorial(Int $n) { $n * factorial($n - 1) }

Check Jonathan Worthington's journal on use.perl.org, for more information about the last example.

Answer 2

Perl 6:Procedural

sub factorial ( int $n ){

  my $result = 1;

  loop ( ; $n > 0; $n-- ){

    $result *= $n;

  }

  return $result;
}

Answer 3

C:

Edit: Actually C++ I guess, because of the variable declaration in the for loop.

 int factorial(int x) {
      int product = 1;

      for (int i = x; i > 0; i--) {
           product *= i;
      }

      return product;
 }

Answer 4

Javascript:

factorial = function( n )
{
   return n > 0 ? n * factorial( n - 1 ) : 1;
}

I'm not sure what a Factorial is but that does what the other programs do in javascript.

Answer 5

Haskell:

ones = 1 : ones
integers   = head ones     : zipWith (+) integers   (tail ones)
factorials = head integers : zipWith (*) factorials (tail integers)

Answer 6

Scheme

Here is a simple recursive definition:

(define (factorial x)
  (if (= x 0) 1
      (* x (factorial (- x 1)))))

In Scheme tail-recursive functions use constant stack space. Here is a version of factorial that is tail-recursive:

(define factorial
  (letrec ((fact (lambda (x accum)
                   (if (= x 0) accum
                       (fact (- x 1) (* accum x))))))
    (lambda (x)
      (fact x 1))))

Answer 7

C++

factorial(int n)
{
    for(int i=1, f = 1; i<=n; i++)
        f *= i;
    return f;
}

Answer 8

C/C++: Procedural

unsigned long factorial(int n)
{
    unsigned long factorial = 1;
    int i;

    for (i = 2; i <= n; i++)
     factorial *= i;

    return factorial;
}

PHP: Procedural

function factorial($n)
{
    for ($factorial = 1, $i = 2; $i <= $n; $i++)
     $factorial *= $i;

    return $factorial;
}

@Niyaz: You didn't specify return type for the function

Answer 9

lolcode:

sorry I couldn't resist xD

HAI
CAN HAS STDIO?
I HAS A VAR
I HAS A INT
I HAS A CHEEZBURGER
I HAS A FACTORIALNUM
IM IN YR LOOP
    UP VAR!!1
    TIEMZD INT!![CHEEZBURGER]
    UP FACTORIALNUM!!1
    IZ VAR BIGGER THAN FACTORIALNUM? GTFO
IM OUTTA YR LOOP
U SEEZ INT
KTHXBYE

Answer 10

D Templates: Functional

template factorial(int n : 1)
{
  const factorial = 1;
}

template factorial(int n)
{
  const factorial =
     n * factorial!(n-1);
}

or

template factorial(int n)
{
  static if(n == 1)
    const factorial = 1;
  else 
    const factorial =
       n * factorial!(n-1);
}

Used like this:

factorial!(5)

Answer 11

Python:

Recursive

def fact(x): 
    return (1 if x==0 else x * fact(x-1))

Using iterator

import operator

def fact(x):
    return reduce(operator.mul, xrange(1, x+1))

Answer 12

two of many Mathematica solutions (although ! is built-in and efficient):

(* returns pure function *)
(FixedPoint[(If[#[[2]]>1,{#[[1]]*#[[2]],#[[2]]-1},#])&,{1,n}][[1]])&

(* not using built-in, returns pure function, don't use: might build 1..n list *)
(Times @@ Range[#])&

Answer 13

Java: functional

int factorial(int x) {
    return x == 0 ? 1 : x * factorial(x-1);
}

Answer 14

Mathematica : using pure recursive functions

(If[#>1,# #0[#-1],1])&

Answer 15

Ruby: functional

def factorial(n)
    return 1 if n == 1
    n * factorial(n -1)
end

Answer 16

Lua

function factorial (n)
  if (n <= 1) then return 1 end
  return n*factorial(n-1)
end

And here is a stack overflow caught in the wild:

> print (factorial(234132))
stdin:3: stack overflow
stack traceback:
    stdin:3: in function 'factorial'
    stdin:3: in function 'factorial'
    stdin:3: in function 'factorial'
    stdin:3: in function 'factorial'
    stdin:3: in function 'factorial'
    stdin:3: in function 'factorial'
    stdin:3: in function 'factorial'
    stdin:3: in function 'factorial'
    stdin:3: in function 'factorial'
    stdin:3: in function 'factorial'
    ...
    stdin:3: in function 'factorial'
    stdin:3: in function 'factorial'
    stdin:3: in function 'factorial'
    stdin:3: in function 'factorial'
    stdin:3: in function 'factorial'
    stdin:3: in function 'factorial'
    stdin:3: in function 'factorial'
    stdin:3: in function 'factorial'
    stdin:1: in main chunk
    [C]: ?

Answer 17

F#: Functional

Straight forward:

let rec fact x = 
    if   x < 0 then failwith "Invalid value."
    elif x = 0 then 1
    else x * fact (x - 1)

Getting fancy:

let fact x = [1 .. x] |> List.fold_left ( * ) 1

Answer 18

Haskell: Functional

 fact 0 = 1
 fact n = n * fact (n-1)

Answer 19

C++: Template Metaprogramming

Uses the classic enum hack.

template<unsigned int n>
struct factorial {
    enum { result = n * factorial<n - 1>::result };
};

template<>
struct factorial<0> {
    enum { result = 1 };
};

Usage.

const unsigned int x = factorial<4>::result;

Factorial is calculated completely at compile time based on the template parameter n. Therefore, factorial<4>::result is a constant once the compiler has done its work.

Answer 20

This one not only calculates n!, it is also O(n!). It may have problems if you want to calculate anything "big" though.

long f(long n)
{
    long r=1;
    for (long i=1; i<n; i++)
        r=r*i;
    return r;
}

long factorial(long n)
{
    // iterative implementation should be efficient
    long result;
    for (long i=0; i<f(n); i++)
        result=result+1;
    return result;
}

Answer 21

x86-64 Assembly: Procedural

You can call this from C (only tested with GCC on linux amd64). Assembly was assembled with nasm.

section .text
    global factorial
; factorial in x86-64 - n is passed in via RDI register
; takes a 64-bit unsigned integer
; returns a 64-bit unsigned integer in RAX register
; C declaration in GCC:
;   extern unsigned long long factorial(unsigned long long n);
factorial:
    enter 0,0
    ; n is placed in rdi by caller
    mov rax, 1 ; factorial = 1
    mov rcx, 2 ; i = 2
loopstart:
    cmp rcx, rdi
    ja loopend
    mul rcx ; factorial *= i
    inc rcx
    jmp loopstart
loopend:
    leave
    ret

Answer 22

Visual Basic: Linq

<Extension()> _
Public Function Product(ByVal xs As IEnumerable(Of Integer)) As Integer
    Return xs.Aggregate(1, Function(a, b) a * b)
End Function

Public Function Fact(ByVal n As Integer) As Integer
    Return Aggregate x In Enumerable.Range(1, n) Into Product()
End Function

This shows how to use the Aggregate keyword in VB. C# can't do this (although C# can of course call the extension method directly).

Answer 23

PowerShell

function factorial( [int] $n ) 
{ 
    $result = 1; 

    if ( $n -gt 1 ) 
    { 
        $result = $n * ( factorial ( $n - 1 ) ) 
    } 

    $result 
}

Here's a one-liner:

$n..1 | % {$result = 1}{$result *= $_}{$result}

Answer 24

Recursive Prolog

fac(0,1).
fac(N,X) :- N1 is N -1, fac(N1, T), X is N * T.

Tail Recursive Prolog

fac(0,N,N).
fac(X,N,T) :- A is N * X, X1 is X - 1, fac(X1,A,T).
fac(N,T) :- fac(N,1,T).

Answer 25

Bourne Shell: Functional

factorial() {
  if [ $1 -eq 0 ]
  then
    echo 1
    return
  fi

  a=`expr $1 - 1`
  expr $1 \* `factorial $a`
}

Also works for Korn Shell and Bourne Again Shell. :-)

Answer 26

Lisp recursive:

(defun factorial (x) 
   (if (<= x 1) 
       1 
       (* x (factorial (- x 1)))))

Answer 27

JavaScript Using anonymous functions:

var f = function(n){
  if(n>1){
    return arguments.callee(n-1)*n;
  }
  return 1;
}

Answer 28

Oddball examples? What about using the gamma function! Since, Gamma n = (n-1)!.

OCaml: Using Gamma

let rec gamma z =
    let pi = 4.0 *. atan 1.0 in
    if z < 0.5 then
        pi /. ((sin (pi*.z)) *. (gamma (1.0 -. z)))
    else
        let consts = [| 0.99999999999980993; 676.5203681218851; -1259.1392167224028;
                        771.32342877765313; -176.61502916214059; 12.507343278686905;
                 -0.13857109526572012; 9.9843695780195716e-6; 1.5056327351493116e-7;
                     |] 
        in
        let z = z -. 1.0 in
        let results = Array.fold_right 
                          (fun x y -> x +. y)
                          (Array.mapi 
                              (fun i x -> if i = 0 then x else x /. (z+.(float i)))
                              consts
                          )
                          0.0
        in
        let x = z +. (float (Array.length consts)) -. 1.5 in
        let final = (sqrt (2.0*.pi)) *. 
                    (x ** (z+.0.5)) *.
                    (exp (-.x)) *. result
        in
        final

let factorial_gamma n = int_of_float (gamma (float (n+1)))

Answer 29

C: One liner, procedural

int f(int n) { for (int i = n - 1; i > 0; n *= i, i--); return n ? n : 1; }

I used int's for brevity; use other types to support larger numbers.

Answer 30

BASIC: old school

10 HOME
20 INPUT N
30 LET ANS = 1
40 FOR I = 1 TO N
50   ANS = ANS * I
60 NEXT I
70 PRINT ANS

Answer 31

Java 1.6: recursive, memoized (for subsequent calls)

private static Map<BigInteger, BigInteger> _results = new HashMap()

public static BigInteger factorial(BigInteger n){
    if (0 >= n.compareTo(BigInteger.ONE))
       return BigInteger.ONE.max(n);
    if (_results.containsKey(n))
       return _results.get(n);
    BigInteger result = factorial(n.subtract(BigInteger.ONE)).multiply(n);
    _results.put(n, result);
    return result;
}

Answer 32

Scheme : Functional - Tail Recursive

(define (factorial n)
  (define (fac-times n acc)
    (if (= n 0)
        acc
        (fac-times (- n 1) (* acc n))))
  (if (< n 0)
      (display "Wrong argument!")
      (fac-times n 1)))

Answer 33

Batch (NT):

@echo off

set n=%1
set result=1

for /l %%i in (%n%, -1, 1) do (
    set /a result=result * %%i
)

echo %result%

Usage: C:>factorial.bat 15

Answer 34

Haskell : Functional - Tail Recursive

factorial n = factorial' n 1

factorial' 0 a = a
factorial' n a = factorial' (n-1) (n*a)

Answer 35

Agda 2: Functional, dependently typed.

data Nat = zero | suc (m::Nat)

add (m::Nat) (n::Nat) :: Nat
 = case m of
     (zero ) -> n
     (suc p) -> suc (add p n)

mul (m::Nat) (n::Nat)::Nat
   = case m of
      (zero ) -> zero
      (suc p) -> add n (mul p n)

factorial (n::Nat)::Nat 
 = case n of
    (zero ) -> suc zero
    (suc p) -> mul n (factorial p)

Answer 36

C# Lookup:

Nothing to calculate really, just look it up. To extend it,add another 8 numbers to the table and 64 bit integers are at at their limit. Beyond that, a BigNum class is called for.

public static int Factorial(int f)
{ 
    if (f<0 || f>12)
    {
        throw new ArgumentException("Out of range for integer factorial");
    }
    int [] fact={1,1,2,6,24,120,720,5040,40320,362880,3628800,
                 39916800,479001600};
    return fact[f];
}

Answer 37

Whitespace

   	.
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  	.
   	.
			 .
 .
	 	 .
	  .
   	.
 .
  .
 			 .
		  			 .
 .
	.
.
  	 .
 .
.
	.
 	.
.
.
.

It was hard to get it to show here properly, but now I tried copying it from the preview and it works. You need to input the number and press enter.

Answer 38

Delphi

facts: array[2..12] of integer;

function TForm1.calculate(f: integer): integer;
begin
    if f = 1 then
      Result := f
    else if f > High(facts) then
      Result := High(Integer)
    else if (facts[f] > 0) then
      Result := facts[f]
    else begin
      facts[f] := f * Calculate(f-1);
      Result := facts[f];
    end;
end;

initialize

  for i := Low(facts) to High(facts) do
    facts[i] := 0;

After the first time a factorial higher or equal to the desired value has been calculated, this algorithm just returns the factorial in constant time O(1).

It takes in account that int32 only can hold up to 12!

Answer 39

Lazy K

Your pure functional programming nightmares come true!

The only Esoteric Turing-complete Programming Language that has:

• A purely functional foundation, core, and libraries---in fact, here's the complete API: S K I
• No lambdas even!
• No numbers or lists needed or allowed
• No explicit recursion but yet, allows recursion
• A simple infinite lazy stream-based I/O mechanism

Here's the Factorial code in all its parenthetical glory:

K(SII(S(K(S(S(KS)(S(K(S(KS)))(S(K(S(KK)))(S(K(S(K(S(K(S(K(S(SI(K(S(K(S(S(KS)K)I))
 (S(S(KS)K)(SII(S(S(KS)K)I))))))))K))))))(S(K(S(K(S(SI(K(S(K(S(SI(K(S(K(S(S(KS)K)I))
 (S(S(KS)K)(SII(S(S(KS)K)I))(S(S(KS)K))(S(SII)I(S(S(KS)K)I))))))))K)))))))
 (S(S(KS)K)(K(S(S(KS)K)))))))))(K(S(K(S(S(KS)K)))K))))(SII))II)

Features:

• No subtraction or conditionals
• Prints all factorials (if you wait long enough)
• Uses a second layer of Church numerals to convert the Nth factorial to N! asterisks followed by a newline
• Uses the Y combinator for recursion

In case you are interested in trying to understand it, here is the Scheme source code to run through the Lazier compiler:

(lazy-def '(fac input)
   '((Y (lambda (f n a) ((lambda (b) ((cons 10) ((b (cons 42)) (f (1+ n) b))))
       (* a n)))) 1 1))

(for suitable definitions of Y, cons, 1, 10, 42, 1+, and *).

EDIT:

Lazy K Factorial in Decimal

(10KB of gibberish or else I would paste it). For example, at the Unix prompt:

    $ echo "4" | ./lazy facdec.lazy
    24
    $ echo "5" | ./lazy facdec.lazy
    120

Rather slow for numbers above, say, 5.

The code is sort of bloated because we have to include library code for all of our own primitives (code written in Hazy, a lambda calculus interpreter and LC-to-Lazy K compiler written in Haskell).

Answer 40

C#: LINQ

    public static int factorial(int n)
    {
        return (Enumerable.Range(1, n).Aggregate(1, (previous, value) => previous * value));
    }

Answer 41

The problem with most of the above is that they will run out of precision at about 25! (12! with 32 bit ints) or just overflow. Here's a c# implementation to break through these limits!

class Number
{
  public Number ()
  {
    m_number = "0";
  }

  public Number (string value)
  {
    m_number = value;
  }

  public int this [int column]
  {
    get
    {
      return column < m_number.Length ? m_number [m_number.Length - column - 1] - '0' : 0;
    }
  }

  public static implicit operator Number (string rhs)
  {
    return new Number (rhs);
  }

  public static bool operator == (Number lhs, Number rhs)
  {
    return lhs.m_number == rhs.m_number;
  }

  public static bool operator != (Number lhs, Number rhs)
  {
    return lhs.m_number != rhs.m_number;
  }

  public override bool Equals (object obj)
  {
     return this == (Number) obj;
  }

  public override int GetHashCode ()
  {
    return m_number.GetHashCode ();
  }

  public static Number operator + (Number lhs, Number rhs)
  {
    StringBuilder
      result = new StringBuilder (new string ('0', lhs.m_number.Length + rhs.m_number.Length));

    int
      carry = 0;

    for (int i = 0 ; i < result.Length ; ++i)
    {
      int
        sum = carry + lhs [i] + rhs [i],
        units = sum % 10;

      carry = sum / 10;

      result [result.Length - i - 1] = (char) ('0' + units);
    }

    return TrimLeadingZeros (result);
  }

  public static Number operator * (Number lhs, Number rhs)
  {
    StringBuilder
      result = new StringBuilder (new string ('0', lhs.m_number.Length + rhs.m_number.Length));

    for (int multiplier_index = rhs.m_number.Length - 1 ; multiplier_index >= 0 ; --multiplier_index)
    {
      int
        multiplier = rhs.m_number [multiplier_index] - '0',
        column = result.Length - rhs.m_number.Length + multiplier_index;

      for (int i = lhs.m_number.Length - 1 ; i >= 0 ; --i, --column)
      {
        int
          product = (lhs.m_number [i] - '0') * multiplier,
          units = product % 10,
          tens = product / 10,
          hundreds = 0,
          unit_sum = result [column] - '0' + units;

        if (unit_sum > 9)
        {
          unit_sum -= 10;
          ++tens;
        }

        result [column] = (char) ('0' + unit_sum);

        int
          tens_sum = result [column - 1] - '0' + tens;

        if (tens_sum > 9)
        {
          tens_sum -= 10;
          ++hundreds;
        }

        result [column - 1] = (char) ('0' + tens_sum);

        if (hundreds > 0)
        {
          int
            hundreds_sum = result [column - 2] - '0' + hundreds;

          result [column - 2] = (char) ('0' + hundreds_sum);
        }
      }
    }

    return TrimLeadingZeros (result);
  }

  public override string ToString ()
  {
    return m_number;
  }

  static string TrimLeadingZeros (StringBuilder number)
  {
    while (number [0] == '0' && number.Length > 1)
    {
      number.Remove (0, 1);
    }

    return number.ToString ();
  }

  string
    m_number;
}

static void Main (string [] args)
{
  Number
    a = new Number ("1"),
    b = new Number (args [0]),
    one = new Number ("1");

  for (Number c = new Number ("1") ; c != b ; )
  {
    c = c + one;
    a = a * c;
  }

  Console.WriteLine (string.Format ("{0}! = {1}", new object [] { b, a }));
}

FWIW: 10000! is over 35500 character long.

Skizz

Answer 42

Python: Functional, One-liner

factorial = lambda n: reduce(lambda x,y: x*y, range(1, n+1), 1)

NOTE:

• It supports big integers. Example:

  print factorial(100) 93326215443944152681699238856266700490715968264381621468592963895217599993229915\ 608941463976156518286253697920827223758251185210916864000000000000000000000000

• It does not work for n < 0.

Answer 43

This is one of the faster algorithms, up to 170!. It fails inexplicably beyond 170!, and it's relatively slow for small factorials, but for factorials between 80 and 170 it's blazingly fast compared to many algorithms.

curl http://www.google.com/search?q=170!

There's also an online interface, try it out now!

Let me know if you find a bug, or faster implementation for large factorials.

---

EDIT:

This algorithm is slightly slower, but gives results beyond 170:

curl http://www58.wolframalpha.com/input/?i=171!

It also simplifies them into various other representations.

Answer 44

Lambda Calculus

Input and output are Church numerals (i.e. natural number k is \f n. f^k n; so 3 = \f n. f (f (f n)))

(\x. x x) (\y f. f (y y f)) (\y n. n (\x y z. z) (\x y. x) (\f n. f n) (\f. n (y (\f m. n (\g h. h (g f)) (\x. m) (\x. x)) f)))

Answer 45

Ruby: Iterative

def factorial(n)
  (1 .. n).inject{|a, b| a*b}
end

Ruby: Recursive

def factorial(n)
  n == 1 ? 1 : n * factorial(n-1)
end

Answer 46

Nemerle: Functional

def fact(n) {
    | 0 => 1
    | x => x * fact(x-1)
}

Answer 47

#Language: T-SQL
#Style: Recursive, divide and conquer

Just for fun - in T-SQL using a divide and conquer recursive method. Yes, recursive - in SQL without stack overflow.

create function factorial(@b int=1, @e int) returns float as begin
  return case when @b>=@e then @e else 
      convert(float,dbo.factorial(@b,convert(int,@b+(@e-@b)/2)))
    * convert(float,dbo.factorial(convert(int,@b+1+(@e-@b)/2),@e)) end
end

call it like this:

print dbo.factorial(1,170) -- the 1 being the starting number

Answer 48

#Language: T-SQL
#Style: Big Numbers

Here's another T-SQL solution -- supports big numbers in a most Rube Goldbergian manner. Lots of set-based ops. Tried to keep it uniquely SQL. Horrible performance (400! took 33 seconds on a Dell Latitude D830)

create function bigfact(@x varchar(max)) returns varchar(max) as begin
  declare @c int
  declare @n table(n int,e int)
  declare @f table(n int,e int)

  set @c=0
  while @c<len(@x) begin
    set @c=@c+1
    insert @n(n,e) values(convert(int,substring(@x,@c,1)),len(@x)-@c)
  end

  -- our current factorial
  insert @f(n,e) select 1,0

  while 1=1 begin
    declare @p table(n int,e int)
    delete @p
    -- product
    insert @p(n,e) select sum(f.n*n.n), f.e+n.e from @f f cross join @n n group by f.e+n.e

    -- normalize
    while 1=1 begin
      delete @f
      insert @f(n,e) select sum(n),e from (
        select (n % 10) as n,e from @p union all 
        select (n/10) % 10,e+1 from @p union all 
        select (n/100) %10,e+2 from @p union all 
        select (n/1000)%10,e+3 from @p union all 
        select (n/10000) % 10,e+4 from @p union all 
        select (n/100000)% 10,e+5 from @p union all 
        select (n/1000000)%10,e+6 from @p union all 
        select (n/10000000) % 10,e+7 from @p union all 
        select (n/100000000)% 10,e+8 from @p union all 
        select (n/1000000000)%10,e+9 from @p
      ) f group by e having sum(n)>0

      set @c=0
      select @c=count(*) from @f where n>9
      if @c=0 break
      delete @p
      insert @p(n,e) select n,e from @f
    end

    -- decrement
    update @n set n=n-1 where e=0

    -- normalize
    while 1=1 begin
      declare @e table(e int)
      delete @e
      insert @e(e) select e from @n where n<0
      if @@rowcount=0 break

      update @n set n=n+10 where e in (select e from @e)
      update @n set n=n-1 where e in (select e+1 from @e)
    end  

    set @c=0
    select @c=count(*) from @n where n>0
    if @c=0 break
  end

  select @c=max(e) from @f
  set @x=''
  declare @l varchar(max)
  while @c>=0 begin
    set @l='0'
    select @l=convert(varchar(max),n) from @f where e=@c
    set @x=@x+@l
    set @c=@c-1
  end
  return @x
end

Example:

print dbo.bigfact('69')

returns:

171122452428141311372468338881272839092270544893520369393648040923257279754140647424000000000000000

Answer 49

I find the following implementations just hilarious:

The Evolution of a Haskell Programmer

Evolution of a Python programmer

Enjoy!

Answer 50

ruby recursive

(factorial=Hash.new{|h,k|k*h[k-1]})[1]=1

usage:

factorial[5]
 => 120

Answer 51

Language Name: ChucK

Moog moog => dac;
4.0 => moog.gain;

for (0 => int i; i < 8; i++) {
    <<< factorial(i) >>>;
}

fun int factorial(int n) {
    1 => int result;
    if (n != 0) {
        n * factorial(n - 1) => result;
    }

    Std.mtof(result % 128) => moog.freq;
    0.25::second => now;

    return result;
}

And it sounds like this. Not terribly interesting, but, hey, it's just a factorial function!

Answer 52

Brainf*ck

+++++
>+<[[->>>>+<<<<]>>>>[-<<<<+>>+>>]<<<<>[->>+<<]<>>>[-<[->>+<<]>>[-<<+<+>>>]<]<[-]><<<-]

Written by Michael Reitzenstein.

Answer 53

APL (oddball/one-liner):

×/⍳X

1. ⍳X expands X into an array of the integers 1..X
2. ×/ multiplies every element in the array

Or with the built-in operator:

!X

Source: http://www.webber-labs.com/mpl/lectures/ppt-slides/01.ppt

Answer 54

Here is an interesting Ruby version. On my laptop it will find 30000! in under a second. (It takes longer for Ruby to format it for printing than to calculate it.) This is significantly faster than the naive solution of just multiplying the numbers in order.

def factorial (n)
  return multiply_range(1, n)
end

def multiply_range(n, m)
  if (m < n)
    return 1
  elsif (n == m)
    return m
  else
    i = (n + m) / 2
    return multiply_range(n, i) * multiply_range(i+1, m)
  end
end

Answer 55

Simple solutions are the best:

#include <stdexcept>;

long fact(long f)
{
    static long fact [] = { 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 1932053504, 1278945280, 2004310016, 2004189184 };
    static long max     = sizeof(fact)/sizeof(long);

    if ((f < 0) || (f >= max))
    {   throw std::range_error("Factorial Range Error");
    }

    return fact[f];
}

Answer 56

Common Lisp: Lisp as God intended it to be used (that is, with LOOP)

(defun fact (n)
  (loop for i from 1 to n
        for acc = 1 then (* acc i)
        finally (return acc)))

Now, if someone can come up with a version based on FORMAT...

Answer 57

Common Lisp: FORMAT (obfuscated)

Okay, so I'll give it a try myself.

(defun format-fact (stream arg colonp atsignp &rest args)
  (destructuring-bind (n acc) arg
    (format stream
            "~[~A~:;~*~/format-fact/~]"
            (1- n)
            acc
            (list (1- n) (* acc n)))))

(defun fact (n)
  (parse-integer (format nil "~/format-fact/" (list n 1))))

There has to be a nicer, even more obscure FORMAT-based implementation. This one is pretty straight-forward and boring, simply using FORMAT as an IF replacement. Obviously, I'm not a FORMAT expert.

Answer 58

Recursively in Inform 7

(it reminds you of COBOL because it's for writing text adventures; proportional font is deliberate):

To decide what number is the factorial of (n - a number):
    if n is zero, decide on one;
    otherwise decide on the factorial of (n minus one) times n.

If you want to actually call this function ("phrase") from a game you need to define an action and grammar rule:

"The factorial game" [this must be the first line of the source]

There is a room. [there has to be at least one!]

Factorialing is an action applying to a number.

Understand "factorial [a number]" as factorialing.

Carry out factorialing:
    Let n be the factorial of the number understood;
    Say "It's [n]".

Answer 59

Scala: Recursive

• Should compile to being tail recursive. Should!

.

def factorial( value: BigInt ): BigInt = value match {
  case 0 => 1
  case _ => value * factorial( value - 1 )
}

Answer 60

Occam-pi

PROC subprocess(MOBILE CHAN INT parent.out!,parent.in?)
INT value:
  SEQ
    parent.in ? value
      IF 
        value = 1
          SEQ
            parent.out ! value
        OTHERWISE
          INITIAL MOBILE CHAN INT child.in IS MOBILE CHAN INT:
          INITIAL MOBILE CHAN INT child.out IS MOBILE CHAN INT:
          FORKING
            INT newvalue:
            SEQ
              FORK subprocess(child.in!,child.out?)
              child.out ! (value-1)
              child.in ? newvalue
              parent.out ! (newalue*value)
:

PROC main(CHAN BYTE in?,src!,kyb?)
INITIAL INT value IS 0:
INITIAL MOBILE CHAN INT child.out is MOBILE CHAN INT
INITIAL MOBILE CHAN INT child.in is MOBILE CHAN INT
SEQ 
  WHILE TRUE
    SEQ
      subprocess(child.in!,child.out?)
      child.out ! value
      child.in ? value
      src ! value:
      value := value + 1
:

Answer 61

Icon

Recursive function

procedure factorial(n)
  return (0<n) * factorial(n-1) | 1
end

I've cheated a bit allowing negatives to return 1. If you want it to fail given a negative argument it's slightly less concise:

  return (0<n) * factorial(n-1) | (n=0 & 1)

Then

write(factorial(3))
write(factorial(-1))
write(factorial(20))

prints

6
2432902008176640000

Iterative generator

procedure factorials()
  local f,n
  f := 1; n := 0
  repeat suspend f *:= (n +:= 1)
end

Then

every write(factorials() \ 5)

prints

1
2
6
24
120

To understand this: evaluation is goal-directed and backtracks on failure. There is no boolean type, and binary operators which would return a boolean in other languages, either fail or return their second argument - with the exception of |, which in a single-value context returns its first argument if it succeeds, otherwise tries its second argument. (in a multiple-value context it returns its first argument then its second argument)

suspend is like yield in other languages, except that a generator is not explicitly called multiple times to return its results. Instead, every asks its argument for all values but doesn't return anything by default; it's useful with side-effects (in this case I/O).

\ limits the number of values returned by a generator, which in the case of factorials would be infinite.

Answer 62

FoxPro:

function factorial
    parameters n
return iif( n>0, n*factorial(n-1), 1)

Answer 63

OCaml

Lest anyone believe OCaml and oddball go hand-in-hand, I thought I would provide a sane implementation of factorial.

# let rec factorial n =
    if n=0 then 1 else n * factorial(n - 1);;

I don't think I made my case very well...

Answer 64

AWK

#!/usr/bin/awk -f
{
    result=1;
    for(i=$1;i>0;i--){
        result=result*i;
    }
    print result;
}

Answer 65

Genuinely functional Java:

public final class Factorial {

  public static void main(String[] args) {
    final int n = Integer.valueOf(args[0]);
    System.out.println("Factorial of " + n + " is " + create(n).apply());
  }

  private static Function create(final int n) {
    return n == 0 ? new ZeroFactorialFunction() : new NFactorialFunction(n);
  }

  interface Function {
    int apply();
  }

  private static class NFactorialFunction implements Function {
    private final int n;
    public NFactorialFunction(final int n) {
      this.n = n;
    }
    @Override
    public int apply() {
      return n * Factorial.create(n - 1).apply();
    }
  }

  private static class ZeroFactorialFunction implements Function {
    @Override
    public int apply() {
      return 1;
    }
  }

}

Answer 66

Erlang: tail recursive

fac(0) -> 1;
fac(N) when N > 0 -> fac(N, 1).

fac(1, R) -> R;
fac(N, R) -> fac(N - 1, R * N).

Answer 67

#Language: T-SQL, C#
#Style: Custom Aggregate

Another crazy way would be to create a custom aggregate and apply it over a temporary table of the integers 1..n.

/* ProductAggregate.cs */
using System;
using System.Data.SqlTypes;
using Microsoft.SqlServer.Server;

[Serializable]
[SqlUserDefinedAggregate(Format.Native)]
public struct product {
  private SqlDouble accum;
  public void Init() { accum = 1; }
  public void Accumulate(SqlDouble value) { accum *= value; }
  public void Merge(product value) { Accumulate(value.Terminate()); }
  public SqlDouble Terminate() { return accum; }
}

add this to sql

create assembly ProductAggregate from 'ProductAggregate.dll' with permission_set=safe -- mod path to point to actual dll location on disk.

create aggregate product(@a float) returns float external name ProductAggregate.product

create the table (there should be a built-in way to do this in SQL -- hmm. a question for SO?)

select 1 as n into #n union select 2 union select 3 union select 4 union select 5

then finally

select dbo.product(n) from #n

Answer 68

The code below is tongue in cheek, however when you consider that the return value is limited to n < 34 for uint32, <65 uint64 before we run out of space for the return value with a uint, hard coding 33 values isn't that crazy :)

public static int Factorial(int n)
{
    switch (n)
    {
     case 1:
      return 1;
     case 2:
      return 2;
     case 3:
      return 6;
     case 4:
      return 24;
     default:
      throw new Exception("Sorry, I can only count to 4");
    }

}

Answer 69

C# factorial using recursion in a single line

private static int factorial(int n){ if (n == 0)return 1;else return n * factorial(n - 1); }

Answer 70

Perl6

sub factorial ($n) { [*] 1..$n }

I hardly know about Perl6. But I guess this [*] operator is same as Haskell's product.

This code runs on Pugs, and maybe Parrot (I didn't check it.)

Edit

This code also works.

sub postfix:<!> ($n) { [*] 1..$n }

# This function(?) call like below ... It looks like mathematical notation.
say 10!;

Answer 71

Haskell:

factorial n = product [1..n]

Answer 72

Eiffel

class
    APPLICATION
inherit
    ARGUMENTS

create
    make

feature -- Initialization

    make is
            -- Run application.
        local
            l_fact: NATURAL_64
        do
            l_fact := factorial(argument(1).to_natural_64)
            print("Result is: " + l_fact.out)
        end

    factorial(n: NATURAL_64): NATURAL_64 is
            --
        require
            positive_n: n >= 0
        do
            if n = 0 then
                Result := 1
            else
                Result := n * factorial(n-1)
            end
        end

end -- class APPLICATION

Answer 73

PostScript: Tail Recursive

/fact0 { dup 2 lt { pop } { 2 copy mul 3 1 roll 1 sub exch pop fact0 } ifelse } def
/fact { 1 exch fact0 } def

Answer 74

befunge-93

                                    v
>v"Please enter a number (1-16) : "0<
,:             >$*99g1-:99p#v_.25*,@
^_&:1-99p>:1-:!|10          < 
         ^     <

An esoteric language by Chris Pressey of Cat's Eye Technologies.

Answer 75

dc

Note: clobbers the e and f registers:

[2++d]se[d1-d_1<fd0>e*]sf

To use, put the value you want to take the factorial of on the top of the stack and then execute lfx (load the f register and execute it), which then pops the top of the stack and pushes that value's factorial.

Explanation: if the top of the stack is x, then the first part makes the top of the stack look like (x, x-1). If the new top-of-stack is non-negative, it calls factorial recursively, so now the stack is (x, (x-1)!) for x >= 1, or (0, -1) for x = 0. Then, if the new top-of-stack is negative, it executes 2++d, which replaces the (0, -1) with (1, 1). Finally, it multiplies the top two values on the stack.

Answer 76

R - using S4 methods (recursively)

setGeneric( 'fct', function( x ) { standardGeneric( 'fct' ) } )
setMethod( 'fct', 'numeric', function( x ) { 
    lapply( x, function(a) { 
        if( a == 0 ) 1 else a * fact( a - 1 ) 
    } )
} )

Has the advantage that you can pass arrays of numbers in, and it will work them all out...

eg:

> fct( c( 3, 5, 6 ) )
[[1]]
[1] 6

[[2]]
[1] 120

[[3]]
[1] 720

Answer 77

Perl (Y-combinator/Functional)

print sub {
  my $f = shift;
  sub {
    my $f1 = shift;
    $f->( sub { $f1->( $f1 )->( @_ ) } )
  }->( sub {
    my $f2 = shift;
    $f->( sub { $f2->( $f2 )->( @_ ) } )
  } )
}->( sub {
  my $h = shift;
  sub {
    my $n = shift;
    return 1 if $n <=1;
    return $n * $h->($n-1);
  }
})->(5);

Everything after 'print' and before the '->(5)' represents the subroutine. The factorial part is in the final "sub {...}". Everything else is to implement the Y-combinator.

Answer 78

Forth (recursive):

: factorial ( n -- n )
    dup 1 > if
        dup 1 - recurse *
    else
        drop 1
     then
;

Answer 79

XSLT 1.0

The input file, factorial.xml:

<?xml version="1.0"?>
<?xml-stylesheet href="factorial.xsl" type="text/xsl" ?>
<n>
  20
</n>

The XSLT file, factorial.xsl:

<?xml version="1.0"?>
<xsl:stylesheet version="1.0"                     
                xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
                xmlns:msxsl="urn:schemas-microsoft-com:xslt" >
  <xsl:output method="text"/>
  <!-- 0! = 1 -->
  <xsl:template match="text()[. = 0]">
    1
  </xsl:template>
  <!-- n! = (n-1)! * n-->
  <xsl:template match="text()[. > 0]">
    <xsl:variable name="x">
      <xsl:apply-templates select="msxsl:node-set( . - 1 )/text()"/>
    </xsl:variable>
    <xsl:value-of select="$x * ."/>
  </xsl:template>
  <!-- Calculate n! -->
  <xsl:template match="/n">
    <xsl:apply-templates select="text()"/>
  </xsl:template>
</xsl:stylesheet>

Save both files in the same directory and open factorial.xml in IE.

Answer 80

J

   fact=. verb define
*/ >:@i. y
)

Answer 81

Iswim/Lucid:

factorial = 1 fby factorial * (time+1);

Answer 82

Python, one liner:

A bit more clean than the other python answer. This, and the previous answer, will fail if the input is less than 1.

def fact(n): return reduce(int.mul,xrange(2,n))

Answer 83

Clojure

Tail-recursive

(defn fact 
  ([n] (fact n 1))
  ([n acc] (if (= n 0) 
               acc 
               (recur (- n 1) (* acc n)))))

Short and simple

 (defn fact [n] (apply * (range 1 (+ n 1))))

Answer 84

Factor

USE: math.ranges

: factorial ( n -- n! ) 1 [a,b] product ;

Answer 85

Scala

The factorial can be defined functionally as:

def fact(n: Int): BigInt = 1 to n reduceLeft(_*_)

or more traditionally as

def fact(n: Int): BigInt = if (n == 0) 1 else fact(n-1) * n

and we can make ! a valid method on Ints:

object extendBuiltins extends Application {

  class Factorizer(n: Int) {
    def ! = 1 to n reduceLeft(_*_)
  }

  implicit def int2fact(n: Int) = new Factorizer(n)

  println("10! = " + (10!))
}

Answer 86

Compile time in C++

template<unsigned i>
struct factorial
{ static const unsigned value = i * factorial<i-1>::value; };

template<>
struct factorial<0>
{ static const unsigned value = 1; };

Use in code as:

Factorial<5>::value

Answer 87

Haskell

factorial n = product [1..n]

Answer 88

Freshman Haskell programmer

fac n = if n == 0 
           then 1
           else n * fac (n-1)

Sophomore Haskell programmer, at MIT (studied Scheme as a freshman)

fac = (\(n) ->
        (if ((==) n 0)
            then 1
            else ((*) n (fac ((-) n 1)))))

Junior Haskell programmer (beginning Peano player)

fac  0    =  1
fac (n+1) = (n+1) * fac n

Another junior Haskell programmer (read that n+k patterns are “a disgusting part of Haskell” [1] and joined the “Ban n+k patterns”-movement [2])

fac 0 = 1
fac n = n * fac (n-1)

Senior Haskell programmer (voted for Nixon Buchanan Bush — “leans right”)

fac n = foldr (*) 1 [1..n]

Another senior Haskell programmer (voted for McGovern Biafra Nader — “leans left”)

fac n = foldl (*) 1 [1..n]

Yet another senior Haskell programmer (leaned so far right he came back left again!)

-- using foldr to simulate foldl

fac n = foldr (\x g n -> g (x*n)) id [1..n] 1

Memoizing Haskell programmer (takes Ginkgo Biloba daily)

facs = scanl (*) 1 [1..]

fac n = facs !! n

Pointless (ahem) “Points-free” Haskell programmer (studied at Oxford)

fac = foldr (*) 1 . enumFromTo 1

Iterative Haskell programmer (former Pascal programmer)

fac n = result (for init next done)
        where init = (0,1)
              next   (i,m) = (i+1, m * (i+1))
              done   (i,_) = i==n
              result (_,m) = m

for i n d = until d n i

Iterative one-liner Haskell programmer (former APL and C programmer)

fac n = snd (until ((>n) . fst) (\(i,m) -> (i+1, i*m)) (1,1))

Accumulating Haskell programmer (building up to a quick climax)

facAcc a 0 = a
facAcc a n = facAcc (n*a) (n-1)

fac = facAcc 1

Continuation-passing Haskell programmer (raised RABBITS in early years, then moved to New Jersey)

facCps k 0 = k 1
facCps k n = facCps (k . (n *)) (n-1)

fac = facCps id

Boy Scout Haskell programmer (likes tying knots; always “reverent,” he belongs to the Church of the Least Fixed-Point [8])

y f = f (y f)

fac = y (\f n -> if (n==0) then 1 else n * f (n-1))

Combinatory Haskell programmer (eschews variables, if not obfuscation; all this currying’s just a phase, though it seldom hinders)

s f g x = f x (g x)

k x y   = x

b f g x = f (g x)

c f g x = f x g

y f     = f (y f)

cond p f g x = if p x then f x else g x

fac  = y (b (cond ((==) 0) (k 1)) (b (s (*)) (c b pred)))

List-encoding Haskell programmer (prefers to count in unary)

arb = ()    -- "undefined" is also a good RHS, as is "arb" :)

listenc n = replicate n arb
listprj f = length . f . listenc

listprod xs ys = [ i (x,y) | x<-xs, y<-ys ]
                 where i _ = arb

facl []         = listenc  1
facl n@(_:pred) = listprod n (facl pred)

fac = listprj facl

Interpretive Haskell programmer (never “met a language” he didn't like)

-- a dynamically-typed term language

data Term = Occ Var
          | Use Prim
          | Lit Integer
          | App Term Term
          | Abs Var  Term
          | Rec Var  Term

type Var  = String
type Prim = String


-- a domain of values, including functions

data Value = Num  Integer
           | Bool Bool
           | Fun (Value -> Value)

instance Show Value where
  show (Num  n) = show n
  show (Bool b) = show b
  show (Fun  _) = ""

prjFun (Fun f) = f
prjFun  _      = error "bad function value"

prjNum (Num n) = n
prjNum  _      = error "bad numeric value"

prjBool (Bool b) = b
prjBool  _       = error "bad boolean value"

binOp inj f = Fun (\i -> (Fun (\j -> inj (f (prjNum i) (prjNum j)))))


-- environments mapping variables to values

type Env = [(Var, Value)]

getval x env =  case lookup x env of
                  Just v  -> v
                  Nothing -> error ("no value for " ++ x)


-- an environment-based evaluation function

eval env (Occ x) = getval x env
eval env (Use c) = getval c prims
eval env (Lit k) = Num k
eval env (App m n) = prjFun (eval env m) (eval env n)
eval env (Abs x m) = Fun  (\v -> eval ((x,v) : env) m)
eval env (Rec x m) = f where f = eval ((x,f) : env) m


-- a (fixed) "environment" of language primitives

times = binOp Num  (*)

minus = binOp Num  (-)
equal = binOp Bool (==)
cond  = Fun (\b -> Fun (\x -> Fun (\y -> if (prjBool b) then x else y)))

prims = [ ("*", times), ("-", minus), ("==", equal), ("if", cond) ]


-- a term representing factorial and a "wrapper" for evaluation

facTerm = Rec "f" (Abs "n" 
              (App (App (App (Use "if")
                   (App (App (Use "==") (Occ "n")) (Lit 0))) (Lit 1))
                   (App (App (Use "*")  (Occ "n"))
                        (App (Occ "f")  
                             (App (App (Use "-") (Occ "n")) (Lit 1))))))

fac n = prjNum (eval [] (App facTerm (Lit n)))

Static Haskell programmer (he does it with class, he’s got that fundep Jones! After Thomas Hallgren’s “Fun with Functional Dependencies” [7])

-- static Peano constructors and numerals

data Zero
data Succ n

type One   = Succ Zero
type Two   = Succ One
type Three = Succ Two
type Four  = Succ Three


-- dynamic representatives for static Peanos

zero  = undefined :: Zero
one   = undefined :: One
two   = undefined :: Two
three = undefined :: Three
four  = undefined :: Four


-- addition, a la Prolog

class Add a b c | a b -> c where
  add :: a -> b -> c

instance              Add  Zero    b  b
instance Add a b c => Add (Succ a) b (Succ c)


-- multiplication, a la Prolog

class Mul a b c | a b -> c where
  mul :: a -> b -> c

instance                           Mul  Zero    b Zero
instance (Mul a b c, Add b c d) => Mul (Succ a) b d


-- factorial, a la Prolog

class Fac a b | a -> b where
  fac :: a -> b

instance                                Fac  Zero    One
instance (Fac n k, Mul (Succ n) k m) => Fac (Succ n) m

-- try, for "instance" (sorry):
-- 
--     :t fac four

Beginning graduate Haskell programmer (graduate education tends to liberate one from petty concerns about, e.g., the efficiency of hardware-based integers)

-- the natural numbers, a la Peano

data Nat = Zero | Succ Nat


-- iteration and some applications

iter z s  Zero    = z
iter z s (Succ n) = s (iter z s n)

plus n = iter n     Succ
mult n = iter Zero (plus n)


-- primitive recursion

primrec z s  Zero    = z
primrec z s (Succ n) = s n (primrec z s n)


-- two versions of factorial

fac  = snd . iter (one, one) (\(a,b) -> (Succ a, mult a b))
fac' = primrec one (mult . Succ)


-- for convenience and testing (try e.g. "fac five")

int = iter 0 (1+)

instance Show Nat where
  show = show . int

(zero : one : two : three : four : five : _) = iterate Succ Zero

Origamist Haskell programmer (always starts out with the “basic Bird fold”)

-- (curried, list) fold and an application

fold c n []     = n
fold c n (x:xs) = c x (fold c n xs)

prod = fold (*) 1


-- (curried, boolean-based, list) unfold and an application

unfold p f g x = 
  if p x 
     then [] 
     else f x : unfold p f g (g x)

downfrom = unfold (==0) id pred


-- hylomorphisms, as-is or "unfolded" (ouch! sorry ...)

refold  c n p f g   = fold c n . unfold p f g

refold' c n p f g x = 
  if p x 
     then n 
     else c (f x) (refold' c n p f g (g x))


-- several versions of factorial, all (extensionally) equivalent

fac   = prod . downfrom
fac'  = refold  (*) 1 (==0) id pred
fac'' = refold' (*) 1 (==0) id pred

Cartesianally-inclined Haskell programmer (prefers Greek food, avoids the spicy Indian stuff; inspired by Lex Augusteijn’s “Sorting Morphisms” [3])

-- (product-based, list) catamorphisms and an application

cata (n,c) []     = n
cata (n,c) (x:xs) = c (x, cata (n,c) xs)

mult = uncurry (*)
prod = cata (1, mult)


-- (co-product-based, list) anamorphisms and an application

ana f = either (const []) (cons . pair (id, ana f)) . f

cons = uncurry (:)

downfrom = ana uncount

uncount 0 = Left  ()
uncount n = Right (n, n-1)


-- two variations on list hylomorphisms

hylo  f  g    = cata g . ana f

hylo' f (n,c) = either (const n) (c . pair (id, hylo' f (c,n))) . f

pair (f,g) (x,y) = (f x, g y)


-- several versions of factorial, all (extensionally) equivalent

fac   = prod . downfrom
fac'  = hylo  uncount (1, mult)
fac'' = hylo' uncount (1, mult)

Ph.D. Haskell programmer (ate so many bananas that his eyes bugged out, now he needs new lenses!)

-- explicit type recursion based on functors

newtype Mu f = Mu (f (Mu f))  deriving Show

in      x  = Mu x
out (Mu x) = x


-- cata- and ana-morphisms, now for *arbitrary* (regular) base functors

cata phi = phi . fmap (cata phi) . out
ana  psi = in  . fmap (ana  psi) . psi


-- base functor and data type for natural numbers,
-- using a curried elimination operator

data N b = Zero | Succ b  deriving Show

instance Functor N where
  fmap f = nelim Zero (Succ . f)

nelim z s  Zero    = z
nelim z s (Succ n) = s n

type Nat = Mu N


-- conversion to internal numbers, conveniences and applications

int = cata (nelim 0 (1+))

instance Show Nat where
  show = show . int

zero = in   Zero
suck = in . Succ       -- pardon my "French" (Prelude conflict)

plus n = cata (nelim n     suck   )
mult n = cata (nelim zero (plus n))


-- base functor and data type for lists

data L a b = Nil | Cons a b  deriving Show

instance Functor (L a) where
  fmap f = lelim Nil (\a b -> Cons a (f b))

lelim n c  Nil       = n
lelim n c (Cons a b) = c a b

type List a = Mu (L a)


-- conversion to internal lists, conveniences and applications

list = cata (lelim [] (:))

instance Show a => Show (List a) where
  show = show . list

prod = cata (lelim (suck zero) mult)

upto = ana (nelim Nil (diag (Cons . suck)) . out)

diag f x = f x x

fac = prod . upto

Post-doc Haskell programmer (from Uustalu, Vene and Pardo’s “Recursion Schemes from Comonads” [4])

-- explicit type recursion with functors and catamorphisms

newtype Mu f = In (f (Mu f))

unIn (In x) = x

cata phi = phi . fmap (cata phi) . unIn


-- base functor and data type for natural numbers,
-- using locally-defined "eliminators"

data N c = Z | S c

instance Functor N where
  fmap g  Z    = Z
  fmap g (S x) = S (g x)

type Nat = Mu N

zero   = In  Z
suck n = In (S n)

add m = cata phi where
  phi  Z    = m
  phi (S f) = suck f

mult m = cata phi where
  phi  Z    = zero
  phi (S f) = add m f


-- explicit products and their functorial action

data Prod e c = Pair c e

outl (Pair x y) = x
outr (Pair x y) = y

fork f g x = Pair (f x) (g x)

instance Functor (Prod e) where
  fmap g = fork (g . outl) outr


-- comonads, the categorical "opposite" of monads

class Functor n => Comonad n where
  extr :: n a -> a
  dupl :: n a -> n (n a)

instance Comonad (Prod e) where
  extr = outl
  dupl = fork id outr


-- generalized catamorphisms, zygomorphisms and paramorphisms

gcata :: (Functor f, Comonad n) =>
           (forall a. f (n a) -> n (f a))
             -> (f (n c) -> c) -> Mu f -> c

gcata dist phi = extr . cata (fmap phi . dist . fmap dupl)

zygo chi = gcata (fork (fmap outl) (chi . fmap outr))

para :: Functor f => (f (Prod (Mu f) c) -> c) -> Mu f -> c
para = zygo In


-- factorial, the *hard* way!

fac = para phi where
  phi  Z             = suck zero
  phi (S (Pair f n)) = mult f (suck n)


-- for convenience and testing

int = cata phi where
  phi  Z    = 0
  phi (S f) = 1 + f

instance Show (Mu N) where
  show = show . int

Tenured professor (teaching Haskell to freshmen)

fac n = product [1..n]

Answer 89

Smalltalk, using a closure

    fac := [ :x | x = 0 ifTrue: [ 1 ] ifFalse: [ x * (fac value: x -1) ]].

    Transcript show: (fac value: 24) "-> 620448401733239439360000"

NB does not work in Squeak, requires full closures.

Answer 90

Smalltalk, memoized

Define a method on Dictionary

Dictionary >> fac: x
    ^self at: x ifAbsentPut: [ x * (self fac: x - 1) ]

usage

 d := Dictionary new.
 d at: 0 put: 1.
 d fac: 24

Answer 91

Smalltalk, 1-Liner

(1 to: 24) inject: 1 into: [ :a :b | a * b ]

Answer 92

Java Script: Creative method using "interview question" counting bits fnc.

function nu(x)
{
  var r=0
  while( x ) {
    x &= x-1
    r++
  }
  return r
}

function fac(n)
{
  var r= Math.pow(2,n-nu(n))

  for ( var i=3 ; i <= n ; i+= 2 )
    r *= Math.pow(i,Math.floor(Math.log(n/i)/Math.LN2)+1)
  return r
}

Works up to 21! then Chrome switches to scientific notation. Inspiration thanks lack of sleep and Knuth, et al's "concrete mathematics".

Answer 93

Nothing is as fast as bash & bc:

function fac { seq $1 | paste -sd* | bc; }  
$ fac 42
1405006117752879898543142606244511569936384000000000
$

Answer 94

In MUMPS:

fact(N)
  N F,I S F=1 F  I=2:1:N S F=F*I
  QUIT F

Or, if you're a fan of indirection:

fact(N)
  N F,I S F=1 F I=2:1:N S F=F_"*"_I
  QUIT @F

Answer 95

Mathematica: non-recursive

fact[n_] := Times @@ Range[n]

Which is syntactic sugar for Apply[Times, Range[n]]. I think that's the best way to do it, not counting the built-in n!, of course. Note that that automatically uses bignums.

Answer 96

Common Lisp version:

(defun ! (n) (reduce #'* (loop for i from 2 below (+ n 1) collect i)))

Seems to be quite fast.

* (! 42)

1405006117752879898543142606244511569936384000000000

Answer 97

Common Lisp, since noone has commited that yet:

(defun factorial (n)
  (if (<= n 1)
      1 
      (* n (factorial (1- n)))))

Answer 98

Brainfuck: with bignum support!

Accepts as input a non-negative integer followed by newline, and outputs the corresponding factorial followed by newline.

>>>>,----------[>>>>,----------]>>>>++<<<<<<<<[>++++++[<----
-->-]<-<<<<]>>>>[[>>+<<-]>>[<<+>+>-]<->+<[>>>>+<<<-<[-]]>[-]
>>]>[-<<<<<[<<<<]>>>>[[>>+<<-]>>[<<+>+>-]>>]>>>>[-[>+<-]+>>>
>]<<<<[<<<<]<<<<[<<<<]>>>>>[>>>[>>>>]>>>>[>>>>]<<<<[[>>>>+<<
<<-]<<<<]>>>>+<<<<<<<[<<<<]>>>>-[>>>[>>>>]>>>>[>>>>]<<<<[>>>
+<<<-]>>>[<<<+>>+>-]<-[>>+<<[-]]<<[<<<<]>>>>[>[>+<-]>[<<+>+>
-]<<[>>>+<<<-]>>>[<<<+>>+>-]<->+++++++++[-<[-[>>>>+<<<<-]]>>
>>[<<<<+>>>>-]<<<]<[>>+<<<<[-]>>[<<+>>-]]>>]<<<<[<<<<]<<<[<<
<<]>>>>-]>>>>]>>>[>[-]>>>]<<<<[>>+<<-]>>[<<+>+>-]<->+<[>-<[-
]]>[-<<-<<<<[>>+<<-]>>[<<+>+>-]<->+<[>-<[-]]>]<<[<<<<]<<<<-[
>>+<<-]>>[<<+>+>-]+<[>-<[-]]>[-<<++++++++++<<<<-[>>+<<-]>>[<
<+>+>-]+<[>-<[-]]>]<<[<<<<]>>>>[[>>+<<-]>>[<<+>+>-]<->+<[>>>
>+<<<-<[-]]>[-]>>]>]>>>[>>>>]<<<<[>+++++++[<+++++++>-]<--.<<
<<]++++++++++.

Unlike the brainf*ck answer posted earlier, this does not overflow any memory locations. (That implementation put n! in a single memory location, effectively limiting it to n less than 6 under standard bf rules.) This program will output n! for any value of n, limited only by time and memory (or bf implementation). For example, using Urban Muller's compiler on my machine, it takes 12 seconds to compute 1000! I think that's pretty good, considering the program can only move left/right and increment/decrement by one.

Believe it or not, this is the first bf program I've written; it took about 10 hours, which were mostly spent debugging. Unfortunately, I later found out that Daniel B Cristofani has written a factorial generator, which just outputs ever-larger factorials, never terminating:

>++++++++++>>>+>+[>>>+[-[<<<<<[+<<<<<]>>[[-]>[<<+>+>-]<[>+<-
]<[>+<-[>+<-[>+<-[>+<-[>+<-[>+<-[>+<-[>+<-[>+<-[>[-]>>>>+>+<
<<<<<-[>+<-]]]]]]]]]]]>[<+>-]+>>>>>]<<<<<[<<<<<]>>>>>>>[>>>>
>]++[-<<<<<]>>>>>>-]+>>>>>]<[>++<-]<<<<[<[>+<-]<<<<]>>[->[-]
++++++[<++++++++>-]>>>>]<<<<<[<[>+>+<<-]>.<<<<<]>.>>>>]

His program is much shorter, but he's practically a professional bf golfer.

Answer 99

Polyglot: 5 languages, all using bignums

So, I wrote a polyglot which works in the three languages I often write in, as well as one from my other answer to this question and one I just learned today. It's a standalone program, which reads a single line containing a nonnegative integer and prints a single line containing its factorial. Bignums are used in all languages, so the maximum computable factorial depends only on your computer's resources.

• Perl: uses built-in bignum package. Run with perl FILENAME.
• Haskell: uses built-in bignums. Run with runhugs FILENAME or your favorite compiler's equivalent.
• C++: requires GMP for bignum support. To compile with g++, use g++ -lgmpxx -lgmp -x c++ FILENAME to link against the right libraries. After compiling, run ./a.out. Or use your favorite compiler's equivalent.
• brainf*ck: I wrote some bignum support in this post. Using Muller's classic distribution, compile with bf < FILENAME > EXECUTABLE. Make the output executable and run it. Or use your favorite distribution.
• Whitespace: uses built-in bignum support. Run with wspace FILENAME.

Edit: added Whitespace as a fifth language. Incidentally, do not wrap the code with <code> tags; it breaks the Whitespace. Also, the code looks much nicer in fixed-width.

char //# b=0+0{- |0*/; #>>>>,----------[>>>>,--------
#define	a/*#--]>>>>++<<<<<<<<[>++++++[<------>-]<-<<<
#Perl	><><><>	 <> <> <<]>>>>[[>>+<<-]>>[<<+>+>-]<->
#C++	--><><>	<><><><	> < > <	+<[>>>>+<<<-<[-]]>[-]
#Haskell >>]>[-<<<<<[<<<<]>>>>[[>>+<<-]>>[<<+>+>-]>>]
#Whitespace	>>>>[-[>+<-]+>>>>]<<<<[<<<<]<<<<[<<<<
#brainf*ck > < ]>>>>>[>>>[>>>>]>>>>[>>>>]<<<<[[>>>>*/
exp; ;//;#+<<<<-]<<<<]>>>>+<<<<<<<[<<<<][.POLYGLOT^5.
#include <gmpxx.h>//]>>>>-[>>>[>>>>]>>>>[>>>>]<<<<[>>
#define	eval int	main()//>+<<<-]>>>[<<<+>>+>->
#include <iostream>//<]<-[>>+<<[-]]<<[<<<<]>>>>[>[>>>
#define	print std::cout	<< // >	<+<-]>[<<+>+>-]<<[>>>
#define	z std::cin>>//<< +<<<-]>>>[<<<+>>+>-]<->+++++
#define c/*++++[-<[-[>>>>+<<<<-]]>>>>[<<<<+>>>>-]<<*/
#define	abs int $n //><	<]<[>>+<<<<[-]>>[<<+>>-]]>>]<
#define	uc mpz_class fact(int	$n){/*<<<[<<<<]<<<[<<
use bignum;sub#<<]>>>>-]>>>>]>>>[>[-]>>>]<<<<[>>+<<-]
z{$_[0+0]=readline(*STDIN);}sub fact{my($n)=shift;#>>
#[<<+>+>-]<->+<[>-<[-]]>[-<<-<<<<[>>+<<-]>>[<<+>+>+*/
uc;if($n==0){return 1;}return $n*fact($n-1);	}//;#
eval{abs;z($n);print fact($n);print("\n")/*2;};#-]<->
'+<[>-<[-]]>]<<[<<<<]<<<<-[>>+<<-]>>[<<+>+>-]+<[>-+++
-}--	<[-]]>[-<<++++++++++<<<<-[>>+<<-]>>[<<+>+>-++
fact 0	= 1 -- ><><><><	> <><><	]+<[>-<[-]]>]<<[<<+ +
fact	n=n*fact(n-1){-<<]>>>>[[>>+<<-]>>[<<+>+++>+-}
main=do{n<-readLn;print(fact n)}-- +>-]<->+<[>>>>+<<+
{-x<-<[-]]>[-]>>]>]>>>[>>>>]<<<<[>+++++++[<+++++++>-]
<--.<<<<]+written+by+++A+Rex+++2009+.';#+++x-}--x*/;}

Answer 100

ActionScript: Procedural/OOP

function f(n) {
    var result = n>1 ? arguments.callee(n-1)*n : 1;
    return result;
}
// function call
f(3);

Answer 101

Common Lisp

I'm fairly sure this could be more effieicnet. It is my first lisp function other than "hello, world" and typing in the example code in the third chapter. Practical Common Lisp is a great text. This function does seem to handle large factorials well.

(defun factorial (x)
  (if (< x 2) (return-from factorial (print 1)))
  (let ((tempx 1) (ans 1))
  (loop until (equalp x tempx) do
       (incf tempx)
       (setf ans (* tempx ans)))
  (list ans)))

Answer 102

Delphi iterative

While recursion can be the only decent solution to a problem, for factorials it is not. To describe it, yes. To program it, no. Iteration is cheapest.

This function calculates factorials for somewhat larger arguments.

function Factorial(aNumber: Int64): String;
var
  F: Double;
begin
  F := 0;
  while aNumber > 1 do begin
    F := F + log10(aNumber);
    dec(aNumber);
  end;
  Result := FloatToStr(Power(10, Frac(F))) + ' * 10^' + IntToStr(Trunc(F));
end;

1000000! = 8.2639327850046 * 10^5565708

Answer 103

Hmm... no TCL

proc factorial {n} {
  if { $n == 0 } { return 1 }
  return [expr {$n*[factorial [expr {$n-1}]]}]
}
puts [factorial 6]

But of course that doesn't work for a damn for large values of n.... we can do better with tcllib!

package require math::bignum
proc factorial {n} {
  if { $n == 0 } { return 1 }
  return [ ::math::bignum::tostr [ ::math::bignum::mul [
    ::math::bignum::fromstr $n] [ ::math::bignum::fromstr [
      factorial [expr {$n-1} ]
    ]]]]
}
puts [factorial 60]

Look at all those ]'s at the end. This is practically LISP!

I'll leave the version for values of n>2^32 as an excersize for the reader

Answer 104

Logo

? to factorial :n
> ifelse :n = 0 [output 1] [output :n * factorial :n - 1]
> end

And to invoke:

? print factorial 5
120

This is using the UCBLogo dialect of logo.

Answer 105

Mathematica, Memoized

f[n_ /; n < 2] := 1
f[n_] := (f[n] = n*f[n - 1])

Mathematica supports n! natively, but this shows how to make definitions on the fly. When you execute f[2], this code will make a definition f[2]=2 which will subsequently be executed no differently than if you'd hard-coded it; no need for an internal data structure; you just use the language's own function definition machinery.

Answer 106

Lisp : tail-recursive

(defun factorial(x)
  (labels((f (x acc)
             (if (> x 1)
                 (f (1- x)(* x acc))
                 acc)))
         (f x 1)))

Answer 107

Agda2

It is Agda2, using the very nice Agda2 syntax.

module fac where

data Nat : Set where        -- Peano numbers
  zero : Nat
  suc : Nat -> Nat
{-# BUILTIN NATURAL Nat #-}
{-# BUILTIN SUC suc #-}
{-# BUILTIN ZERO zero #-}

infixl 10 _+_               -- Addition over Peano numbers
_+_ : Nat -> Nat -> Nat
zero + n    = n
(suc n) + m = suc (n + m)

infixl 20 _*_               -- Multiplication over Peano numbers
_*_ : Nat -> Nat -> Nat
zero * n = zero
n * zero = zero
(suc n) * (suc m) = suc n + (suc n * m)

_! : Nat -> Nat             -- Factorial function, syntax: "x !"
zero ! = suc zero
(suc n) ! = (suc n) * (n !)

Answer 108

Another ruby one.

class Integer
  def fact
    return 1 if self.zero?
    (1..self).to_a.inject(:*)
  end
end

This works if to_proc is supported on symbols.

Answer 109

Perl, pessimal:

# Because there are just so many other ways to get programs wrong...
use strict;
use warnings;

sub factorial {
    my ($x)=@_;

    for(my $f=1;;$f++) {
        my $tmp=$f;
        foreach my $g (1..$x) {
           $tmp/=$g;
        }
        return $f if $tmp == 1;
    }
}

I trust I get extra points for not using the '*' operator...

Answer 110

Here's my proposal. Runs in Mathematica, works fine:

gen[f_, n_] := Module[{id = -1, val = Table[Null, {n}], visit},
  visit[k_] := Module[{t},
    id++; If[k != 0, val[[k]] = id];
    If[id == n, f[val]];
    Do[If[val[[t]] == Null, visit[t]], {t, 1, n}];
    id--; val[[k]] = Null;];
  visit[0];
  ]

Factorial[n_] := Module[{res=0}, gen[res++&, n]; res]

Update Ok, here's how it works: the visit function is from Sedgewick's Algorithm book, it "visits" all permutations of length n. Upon the visit, it calls function f with the permutation as an argument.

So, Factorial enumerates all permutations of length n, and for each permutation the counter res is increased, thus computing n! in O(n+1)! time.

Answer 111

Python:

def factorial(n):
    return reduce(lambda x, y: x * y,range(1, n + 1))

Answer 112

PHP - 59 chars

function f($n){return array_reduce(range(1,$n),'bcmul',1);}

Answer 113

*NIX Shell

Linux version:

seq -s'*' 42 | bc

BSD version:

jot -s'*' 42 | bc

Answer 114

SETL

...where Haskell and Python borrowed their list comprehensions from.

proc factorial(n);
    return 1 */ {1..n};
end factorial;

And the built-in INTEGER type is arbitrary-precision, so this will work for any positive n.

Answer 115

Befunge:

0&>:1-:v v *_$.@ 
  ^    _$>\:^

Answer 116

CLOS

I see Common Lisp solutions abusing recursion, LOOP, and even FORMAT. I guess it's time for somebody to write a solution that abuses CLOS!

(defgeneric factorial (n))
(defmethod factorial ((n (eql 0))) 1)
(defmethod factorial ((n integer)) (* n (factorial (1- n))))

(Can your favorite language's object system dispatcher do that?)

Answer 117

Golfscript: designed for golfing, of course

~),1>{*}*

• ~ evaluates the input string (to an integer)
• ) increments the number
• , is range (4, becomes [0 1 2 3])
• 1> selects values whose index is 1 or bigger
• {*}* folds multiplication over the list
• Stack contents are printed when the program terminates.

To run:

echo 5 | ruby gs.rb fact.gs

Answer 118

FORTH, iterative 1 liner

: FACT 1 SWAP 1 + 1 DO I * LOOP ;

Answer 119

Oh fork() its another example in Perl

This will make use of your multiple core CPUs... although perhaps not in the most effective manner. The open statement clones the process with fork and opens a pipe from the child process to the parent. The work of multiplying numbers 2 at a time is split among a tree of very short lived processes. Of course, this example is a bit silly. The point is that if you actually had more difficult calculations to do then this example illustrates one way to divide up the work in parallel.

#!/usr/bin/perl -w

use strict;
use bigint;

print STDOUT &main::rangeProduct(1,$ARGV[0])."\n";

sub main::rangeProduct {
    my($l, $h) = @_;
    return $l    if ($l==$h);
    return $l*$h if ($l==($h-1));
    # arghhh - multiplying more than 2 numbers at a time is too much work
    # find the midpoint and split the work up :-)
    my $m = int(($h+$l)/2);
    my $pid = open(my $KID, "-|");
      if ($pid){ # parent
        my $X = &main::rangeProduct($l,$m);
        my $Y = <$KID>;
        chomp($Y);
        close($KID);
        die "kid failed" unless defined $Y;
        return $X*$Y;
      } else {
        # kid
        print STDOUT &main::rangeProduct($m+1,$h)."\n";
        exit(0);
    }
}

Answer 120

In Io:

factorial := method(n,
    if (list(0, 1) contains(n),
       1,
       n * factorial(n - 1)
    )
)

Answer 121

Scheme evolution

Regular Scheme program:

(define factorial
   (lambda (n)
     (if (= n 0)
         1
         (* n (factorial (- n 1))))))

Should work, but notice that calling this function on large numbers will extend the stack on every recursion, which is bad in languages like C and Java.

Continuation-passing style

(define factorial
  (lambda (n)
    (factorial_cps n (lambda (k) k))))

(define factorial_cps
  (lambda (n k)
    (if (zero? n)
        (k 1)
        (factorial (- n 1) (lambda (v)
                             (k (* n v)))))))

Ah, this way, we don't grow our stack every recursion because we can extend a continuation instead. However, C doesn't have continuations.

Representation-independent CPS

(define factorial
  (lambda (n)
    (factorial_cps n (k_))))

(define factorial_cps
  (lambda (n k)
    (if (zero? n)
        (apply_k 1)
        (factorial (- n 1) (k_extend n k))))

(define apply_k
  (lambda (ko v)
    (ko v)))
(define kt_empty
  (lambda ()
    (lambda (v) v)))
(define kt_extend 
  (lambda ()
    (lambda (v)
      (apply_k k (* n v)))))

Notice that responsibility for representation of the continuations used in the original CPS program has been shifted to the kt_ helper procedures.

Representation-independent CPS using ParentheC unions

Since representation of the continuations is in the helper procedures, we can switch to using ParentheC instead, with kt_ being a type designator.

(define factorial
  (lambda (n)
    (factorial_cps n (kt_empty))))

(define factorial_cps
  (lambda (n k)
    (if (zero? n)
        (apply_k 1)
        (factorial (- n 1) (kt_extend n k))))

(define-union kt
  (empty)
  (extend n k))
(define apply_k
  (lambda ()
    (union-case kh kt
      [(empty) v]
      [(extend n k) (begin
                      (set! kh k)
                      (set! v (* n v))
                      (apply_k))])))

Trampolined, registerized ParentheC program

That's not enough. We now replace all function calls by instead setting global variables and a program counter. Procedures are now labels suitable for GOTO statements.

(define-registers n k kh v)
(define-program-counter pc)

(define-label main
  (begin
    (set! n 5) ; what is the factorial of 5??
    (set! pc factorial_cps)
    (mount-trampoline kt_empty k pc)
    (printf "Factorial of 5: ~d\n" v)))

(define-label factorial_cps
  (if (zero? n)
      (begin
        (set! kh k)
        (set! v 1)
        (set! pc apply_k))
      (begin
        (set! k (kt_extend n k))
        (set! n (- n 1))
        (set! pc factorial_cps))))

(define-union kt
  (empty dismount) ; get off the trampoline!
  (extend n k))

(define-label apply_k
  (union-case kh kt
    [(empty dismount) (dismount-trampoline dismount)]
    [(extend n k) (begin
                    (set! kh k)
                    (set! v (* n v))
                    (set! pc apply_k))]))

Oh look, we have a main procedure now too. Now all that's left to do is save this file as fact5.pc and run it through ParentheC's pc2c:

> (load "pc2c.ss")
> (pc2c "fact5.pc" "fact5.c" "fact5.h")

Could it be? We got fact5.c and fact5.h. Let's see...

$ gcc fact5.c -o fact5
$ ./fact5
Factorial of 5: 120

Success! We have converted a recursive Scheme program into a non-recursive C program! And it only took several hours and many forehead-shaped impressions in the wall to do it! For convenience, fact5.c and and fact5.h.

Answer 122

Python: functional, recursive one-liner using short circuit boolean evaluation.

    factorial = lambda n: ((n <= 1) and 1) or (factorial(n-1) * n)