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.
Ada
Iterative
<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; </Ada>
Recursive
<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; </Ada>
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 ;
Fortran
A simple one-liner is sufficient
nfactorial = PRODUCT((/(i, i=1,n)/))
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 = foldl1 (*) [1..n]
Java
Iterative
<java>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;
}</java>
Recursive
<java>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);
}</java>
OCaml
Recursive
<ocaml>let rec factorial n =
if n <= 0 then 1 else n * factorial (n-1)</ocaml>
The following is tail-recursive, so it is effectively iterative: <ocaml>let factorial n =
let rec loop i accum = if i > n then accum else loop (i + 1) (accum * i) in loop 1 1</ocaml>
Iterative
It can be done using explicit state, but this is usually discouraged in a functional language: <ocaml>let factorial n =
let result = ref 1 in for i = 1 to n do result := !result * i done; !result</ocaml>
Python
Iterative
<python>def factorial(n):
if n == 0: return 1 z=n while n>1: n=n-1 z=z*n return z
</python>
Functional
<python>from operator import mul
def factorial(n):
return reduce(mul, xrange(1,n+1), 1)
</python>
Sample output:
<python>
>>> for i in range(6): print i, factorial(i) 0 1 1 1 2 2 3 6 4 24 5 120 >>>
</python>
Numerical Approximation
The following sample uses Lanczos approximation from http://en.wikipedia.org/wiki/Lanczos_approximation <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))
</python> 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
<python>def factorial(n):
z=1 if n>1: z=n*factorial(n-1) return z
</python>
Scheme
Recursive
<scheme>(define (factorial n)
(if (<= n 0) 1 (* n factorial (- n 1))))</scheme>
The following is tail-recursive, so it is effectively iterative: <scheme>(define (factorial n)
(let loop ((i 1) (accum 1)) (if (> i n) accum (loop (+ i 1) (* accum i)))))</scheme>
Iterative
<scheme>(define (factorial n)
(do ((i 1 (+ i 1)) (accum 1 (* accum i))) ((> i n) accum)))</scheme>