Factorial
You are encouraged to solve this task according to the task description, using any language you may know.
The Factorial Function of a positive integer, n, is defined as the product of the sequence n, n-1, n-2, ...1 and the factorial of zero, 0, is defined as being 1.
Write a function to return the factorial of a number. Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion). Support for trapping negative n errors is optional.
ActionScript
Iterative
<lang actionscript> public static function factorial(n:int):int {
if (n < 0)
return 0;
var fact:int = 1;
for (var i:int = 1; i <= n; i++)
fact *= i;
return fact;
} </lang>
Recursive
<lang actionscript> public static function factorial(n:int):int {
if (n < 0)
return 0;
if (n == 0)
return 1;
return n * factorial(n - 1);
} </lang>
Ada
Iterative
<lang ada> function Factorial (N : Positive) return Positive is
Result : Positive := N; Counter : Natural := N - 1;
begin
for I in reverse 1..Counter loop
Result := Result * I;
end loop;
return Result;
end Factorial; </lang>
Recursive
<lang ada> function Factorial(N : Positive) return Positive is
Result : Positive := 1;
begin
if N > 1 then
Result := N * Factorial(N - 1);
end if;
return Result;
end Factorial; </lang>
Numerical Approximation
<lang ada> with Ada.Numerics.Generic_Complex_Types; with Ada.Numerics.Generic_Complex_Elementary_Functions; with Ada.Numerics.Generic_Elementary_Functions; with Ada.Text_IO.Complex_Io; with Ada.Text_Io; use Ada.Text_Io;
procedure Factorial_Numeric_Approximation is
type Real is digits 15;
package Complex_Pck is new Ada.Numerics.Generic_Complex_Types(Real);
use Complex_Pck;
package Complex_Io is new Ada.Text_Io.Complex_Io(Complex_Pck);
use Complex_IO;
package Cmplx_Elem_Funcs is new Ada.Numerics.Generic_Complex_Elementary_Functions(Complex_Pck);
use Cmplx_Elem_Funcs;
function Gamma(X : Complex) return Complex is
package Elem_Funcs is new Ada.Numerics.Generic_Elementary_Functions(Real);
use Elem_Funcs;
use Ada.Numerics;
-- Coefficients used by the GNU Scientific Library
G : Natural := 7;
P : constant array (Natural range 0..G + 1) of Real := (
0.99999999999980993, 676.5203681218851, -1259.1392167224028,
771.32342877765313, -176.61502916214059, 12.507343278686905,
-0.13857109526572012, 9.9843695780195716e-6, 1.5056327351493116e-7);
Z : Complex := X;
Cx : Complex;
Ct : Complex;
begin
if Re(Z) < 0.5 then
return Pi / (Sin(Pi * Z) * Gamma(1.0 - Z));
else
Z := Z - 1.0;
Set_Re(Cx, P(0));
Set_Im(Cx, 0.0);
for I in 1..P'Last loop
Cx := Cx + (P(I) / (Z + Real(I)));
end loop;
Ct := Z + Real(G) + 0.5;
return Sqrt(2.0 * Pi) * Ct**(Z + 0.5) * Exp(-Ct) * Cx;
end if;
end Gamma;
function Factorial(N : Complex) return Complex is
begin
return Gamma(N + 1.0);
end Factorial;
Arg : Complex;
begin
Put("factorial(-0.5)**2.0 = ");
Set_Re(Arg, -0.5);
Set_Im(Arg, 0.0);
Put(Item => Factorial(Arg) **2.0, Fore => 1, Aft => 8, Exp => 0);
New_Line;
for I in 0..9 loop
Set_Re(Arg, Real(I));
Set_Im(Arg, 0.0);
Put("factorial(" & Integer'Image(I) & ") = ");
Put(Item => Factorial(Arg), Fore => 6, Aft => 8, Exp => 0);
New_Line;
end loop;
end Factorial_Numeric_Approximation; </lang> Output:
factorial(-0.5)**2.0 = (3.14159265,0.00000000) factorial( 0) = ( 1.00000000, 0.00000000) factorial( 1) = ( 1.00000000, 0.00000000) factorial( 2) = ( 2.00000000, 0.00000000) factorial( 3) = ( 6.00000000, 0.00000000) factorial( 4) = ( 24.00000000, 0.00000000) factorial( 5) = ( 120.00000000, 0.00000000) factorial( 6) = ( 720.00000000, 0.00000000) factorial( 7) = ( 5040.00000000, 0.00000000) factorial( 8) = ( 40320.00000000, 0.00000000) factorial( 9) = (362880.00000000, 0.00000000)
ALGOL 68
Iterative
PROC factorial = (INT upb n)LONG LONG INT:( LONG LONG INT z := 1; FOR n TO upb n DO z *:= n OD; z );
Numerical Approximation
# Coefficients used by the GNU Scientific Library #
INT g = 7;
[]REAL p = []REAL(0.99999999999980993, 676.5203681218851, -1259.1392167224028,
771.32342877765313, -176.61502916214059, 12.507343278686905,
-0.13857109526572012, 9.9843695780195716e-6, 1.5056327351493116e-7)[@0];
PROC complex gamma = (COMPL in z)COMPL: (
# Reflection formula #
COMPL z := in z;
IF re OF z < 0.5 THEN
pi / (complex sin(pi*z)*complex gamma(1-z))
ELSE
z -:= 1;
COMPL x := p[0];
FOR i TO g+1 DO x +:= p[i]/(z+i) OD;
COMPL t := z + g + 0.5;
complex sqrt(2*pi) * t**(z+0.5) * complex exp(-t) * x
FI
);
OP ** = (COMPL z, p)COMPL: ( z=0|0|complex exp(complex ln(z)*p) );
PROC factorial = (COMPL n)COMPL: complex gamma(n+1);
FORMAT compl fmt = $g(-16, 8)"⊥"g(-10, 8)$;
printf(($q"factorial(-0.5)**2="f(compl fmt)l$, factorial(-0.5)**2));
FOR i TO 9 DO
printf(($q"factorial("d")="f(compl fmt)l$, i, factorial(i)))
OD
Output:
factorial(-0.5)**2= 3.14159265⊥0.00000000 factorial(1)= 1.00000000⊥0.00000000 factorial(2)= 2.00000000⊥0.00000000 factorial(3)= 6.00000000⊥0.00000000 factorial(4)= 24.00000000⊥0.00000000 factorial(5)= 120.00000000⊥0.00000000 factorial(6)= 720.00000000⊥0.00000000 factorial(7)= 5040.00000000⊥0.00000000 factorial(8)= 40320.00000000⊥0.00000000 factorial(9)= 362880.00000000⊥0.00000000
Recursive
PROC factorial = (INT n)LONG LONG INT:
CASE n+1 IN
1,1,2,6,24,120,720 # a brief lookup #
OUT
n*factorial(n-1)
ESAC
;
BASIC
Iterative
<lang freebasic> FUNCTION factorial (n AS Integer) AS Integer
DIM f AS Integer, i AS Integer
f = 1
FOR i = 2 TO n
f = f*i
NEXT i
factorial = f
END FUNCTION </lang>
Recursive
<lang freebasic> FUNCTION factorial (n AS Integer) AS Integer
IF n < 2 THEN
factorial = 1
ELSE
factorial = n * factorial(n-1)
END IF
END FUNCTION </lang>
Befunge
&1\> :v v *<
^-1:_$>\:|
@.$<
C
Iterative
<lang c> int factorial(int n) {
int result = 1; for (int i = 1; i <= n; ++i) result *= i; return result;
} </lang>
Recursive
<lang c> int factorial(int n) {
if (n == 0) return 1; else return n*factorial(n-1);
} </lang>
C++
The C versions work unchanged with C++, however, here is another possibility using the STL and boost: <lang cpp>
- include <boost/iterator/counting_iterator.hpp>
- include <algorithm>
int factorial(int n) {
// last is one-past-end return std::accumulate(boost::counting_iterator<int>(1), boost::counting_iterator<int>(n+1), 1, std::multiplies<int>());
} </lang>
Common Lisp
Warning: Written without an interpreter available. May have mismatched parentheses. <lang lisp> (defun fact(n) (cond (or (= n 1) (= n 0)) (1) (t (* n (fact(- n 1)))) ) ) </lang>
Forth
: fac ( n -- n! ) 1 swap 1+ 1 ?do i * loop ;
Fortran
A simple one-liner is sufficient
nfactorial = PRODUCT((/(i, i=1,n)/))
<lang fortran>
INTEGER FUNCTION FACT(N)
INTEGER N, I
FACT = 1
DO 10, I = 1, N
FACT = FACT * I
10 CONTINUE
RETURN
END
</lang>
Haskell
The simplest description: factorial is the product of the numbers from 1 to n:
factorial n = product [1..n]
Or, written explicitly as a fold:
factorial n = foldl (*) 1 [1..n]
See also: The Evolution of a Haskell Programmer
J
Operator
! 8 NB. Built in factorial operator 40320
Iterative / Functional
*/1+i.8 40320
Recursive
(*$:@:<:)^:(1&<) 8 40320
Generalization
Factorial, like all J's primitives, is generalized:
! 8 0.8 _0.8 NB. Generalizes as the gamma function 40320 0.931384 4.59084 ! 800x NB. Also arbitrarily large 7710530113353860041446393977750283605955564018160102391634109940339708518270930693670907697955390330926478612242306774446597851526397454014801846531749097625044706382742591201733097017026108750929188168469858421505936237186038616420630788341172340985137252...
Java
Iterative
<lang java5>public static long fact(int n){
if(n < 0){
System.err.println("No negative numbers");
return 0;
}
long ans = 1;
for(int i = 1;i <= n;i++){
ans *= i;
}
return ans;
}</lang>
Recursive
<lang java5>public static long fact(int n){
if(n < 0){
System.err.println("No negative numbers");
return 0;
}
if(n == 0) return 1;
return n * fact(n-1);
}</lang>
JavaScript
<lang javascript>function factorial(n) {
return n<2 ? 1 : n * factorial(n-1);
}</lang>
Logo
to factorial :n if :n < 2 [output 1] output :n * factorial :n-1 end
Mathematica
Note that Mathematica already comes with a factorial function, which can be used as e.g. 5! (gives 120). So the following implementations are only of pedagogical value.
Recursive
<lang mathematica> factorial[n_Integer] := n*factorial[n-1] factorial[0] = 1 </lang>
Iterative (direct loop)
<lang mathematica> factorial[n_Integer] :=
Block[{i, result = 1}, For[i = 1, i <= n, ++i, result *= i]; result]
</lang>
Iterative (list)
factorial[n_Integer] := Block[{i}, Times @@ Table[i, {i, n}]]
Nial
(from Nial help file)
fact is recur [ 0 =, 1 first, pass, product, -1 +]
Using it
|fact 4 =24
OCaml
Recursive
<lang ocaml>let rec factorial n =
if n <= 0 then 1 else n * factorial (n-1)</lang>
The following is tail-recursive, so it is effectively iterative: <lang ocaml>let factorial n =
let rec loop i accum = if i > n then accum else loop (i + 1) (accum * i) in loop 1 1</lang>
Iterative
It can be done using explicit state, but this is usually discouraged in a functional language: <lang ocaml>let factorial n =
let result = ref 1 in for i = 1 to n do result := !result * i done; !result</lang>
Pascal
Iterative
<lang pascal> function factorial(n: integer): integer;
var i, result: integer; begin result := 1; for i := 1 to n do result := result * i; factorial := result end;
</lang>
Recursive
<lang pascal> function factorial(n: integer): integer;
begin if n = 0 then factorial := 1 else factorial := n*factorial(n-1) end;
</lang>
Perl
Iterative
<lang perl>sub factorial {
my $n = shift;
my $result = 1;
for (my $i = 1; $i <= $n; ++$i)
{
$result *= $i;
};
$result;
}</lang>
Recursive
<lang perl>sub factorial {
my $n = shift; ($n == 0)? 1 : $n*factorial($n-1);
}</lang>
Functional
<lang perl>use List::Util qw(reduce); sub factorial {
my $n = shift;
reduce { $a * $b } 1, 1 .. $n
}</lang>
Python
Iterative
<lang python>def factorial(n):
if n == 0:
return 1
z=n
while n>1:
n=n-1
z=z*n
return z
</lang>
Functional
<lang python>from operator import mul
def factorial(n):
return reduce(mul, xrange(1,n+1), 1)
</lang>
Sample output:
<lang python>
>>> for i in range(6):
print i, factorial(i)
0 1
1 1
2 2
3 6
4 24
5 120
>>>
</lang>
Numerical Approximation
The following sample uses Lanczos approximation from wp:Lanczos_approximation <lang python> from cmath import *
- Coefficients used by the GNU Scientific Library
g = 7 p = [0.99999999999980993, 676.5203681218851, -1259.1392167224028,
771.32342877765313, -176.61502916214059, 12.507343278686905,
-0.13857109526572012, 9.9843695780195716e-6, 1.5056327351493116e-7]
def gamma(z):
z = complex(z)
# Reflection formula
if z.real < 0.5:
return pi / (sin(pi*z)*gamma(1-z))
else:
z -= 1
x = p[0]
for i in range(1, g+2):
x += p[i]/(z+i)
t = z + g + 0.5
return sqrt(2*pi) * t**(z+0.5) * exp(-t) * x
def factorial(n):
return gamma(n+1)
print "factorial(-0.5)**2=",factorial(-0.5)**2 for i in range(10):
print "factorial(%d)=%s"%(i,factorial(i))
</lang> Output:
factorial(-0.5)**2= (3.14159265359+0j) factorial(0)=(1+0j) factorial(1)=(1+0j) factorial(2)=(2+0j) factorial(3)=(6+0j) factorial(4)=(24+0j) factorial(5)=(120+0j) factorial(6)=(720+0j) factorial(7)=(5040+0j) factorial(8)=(40320+0j) factorial(9)=(362880+0j)
Recursive
<lang python>def factorial(n):
z=1
if n>1:
z=n*factorial(n-1)
return z
</lang>
Scheme
Recursive
<slang cheme>(define (factorial n)
(if (<= n 0)
1
(* n factorial (- n 1))))</lang>
The following is tail-recursive, so it is effectively iterative: <lang scheme>(define (factorial n)
(let loop ((i 1)
(accum 1))
(if (> i n)
accum
(loop (+ i 1) (* accum i)))))</lang>
Iterative
<lang scheme>(define (factorial n)
(do ((i 1 (+ i 1))
(accum 1 (* accum i)))
((> i n) accum)))</lang>
UNIX Shell
fact ()
{
if [ $1 -eq 0 ];
then echo 1;
else echo $(($1 * $(fact $(($1-1)) ) ));
fi;
}