Evaluate binomial coefficients
From Rosetta Code
You are encouraged to solve this task according to the task description, using any language you may know.
This programming task, is to calculate ANY binomial coefficient.
However, it has to be able to output 
, which is 10.
This formula is recommended:
[edit] ACL2
(defun fac (n)
(if (zp n)
1
(* n (fac (1- n)))))
(defun binom (n k)
(/ (fac n) (* (fac (- n k)) (fac k)))
[edit] Ada
with Ada.Text_IO; use Ada.Text_IO;
procedure Test_Binomial is
function Binomial (N, K : Natural) return Natural is
Result : Natural := 1;
M : Natural;
begin
if N < K then
raise Constraint_Error;
end if;
if K > N/2 then -- Use symmetry
M := N - K;
else
M := K;
end if;
for I in 1..M loop
Result := Result * (N - M + I) / I;
end loop;
return Result;
end Binomial;
begin
for N in 0..17 loop
for K in 0..N loop
Put (Integer'Image (Binomial (N, K)));
end loop;
New_Line;
end loop;
end Test_Binomial;
Sample output:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 1 11 55 165 330 462 462 330 165 55 11 1 1 12 66 220 495 792 924 792 495 220 66 12 1 1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1 1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1 1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1 1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1 1 17 136 680 2380 6188 12376 19448 24310 24310 19448 12376 6188 2380 680 136 17 1
[edit] ALGOL 68
[edit] Iterative - unoptimised
- note: This specimen retains the original C coding style.PROC factorial = (INT n)INT:
(
INT result;
result := 1;
FOR i TO n DO
result *:= i
OD;
result
);
PROC choose = (INT n, INT k)INT:
(
INT result;
# Note: code can be optimised here as k < n #
result := factorial(n) OVER (factorial(k) * factorial(n - k));
result
);
test:(
print((choose(5, 3), new line))
)
Output:
+10
[edit] AppleScript
set n to 5
set k to 3
on calculateFactorial(val)
set partial_factorial to 1 as integer
repeat with i from 1 to val
set factorial to i * partial_factorial
set partial_factorial to factorial
end repeat
return factorial
end calculateFactorial
set n_factorial to calculateFactorial(n)
set k_factorial to calculateFactorial(k)
set n_minus_k_factorial to calculateFactorial(n - k)
return n_factorial / (n_minus_k_factorial) * 1 / (k_factorial) as integer
[edit] AutoHotkey
MsgBox, % Round(BinomialCoefficient(5, 3))
;---------------------------------------------------------------------------
BinomialCoefficient(n, k) {
;---------------------------------------------------------------------------
r := 1
Loop, % k < n - k ? k : n - k {
r *= n - A_Index + 1
r /= A_Index
}
Return, r
}
Message box shows:
10
[edit] BBC BASIC
@%=&1010
PRINT "Binomial (5,3) = "; FNbinomial(5, 3)
PRINT "Binomial (100,2) = "; FNbinomial(100, 2)
PRINT "Binomial (33,17) = "; FNbinomial(33, 17)
END
DEF FNbinomial(N%, K%)
LOCAL R%, D%
R% = 1 : D% = N% - K%
IF D% > K% THEN K% = D% : D% = N% - K%
WHILE N% > K%
R% *= N%
N% -= 1
WHILE D% > 1 AND (R% MOD D%) = 0
R% /= D%
D% -= 1
ENDWHILE
ENDWHILE
= R%
Output:
Binomial (5,3) = 10 Binomial (100,2) = 4950 Binomial (33,17) = 1166803110
[edit] Bracmat
(binomial=
n k coef
. !arg:(?n,?k)
& (!n+-1*!k:<!k:?k|)
& 1:?coef
& whl
' ( !k:>0
& !coef*!n*!k^-1:?coef
& !k+-1:?k
& !n+-1:?n
)
& !coef
);
binomial$(5,3)
10
[edit] Burlesque
blsq ) 5 3nr
10
[edit] C
#include <stdio.h>
#include <limits.h>
typedef unsigned long ULONG;
ULONG binomial(ULONG n, ULONG k)
{
ULONG r = 1, d = n - k;
/* choose the smaller of k and n - k */
if (d > k) { k = d; d = n - k; }
while (n > k) {
if (r >= UINT_MAX / n) return 0; /* overflown */
r *= n--;
/* divide (n - k)! as soon as we can to delay overflows */
while (d > 1 && !(r % d)) r /= d--;
}
return r;
}
int main()
{
printf("%lu\n", binomial(5, 3));
return 0;
}
Output:
10
[edit] C++
double Factorial(double nValue)
{
double result = nValue;
double result_next;
double pc = nValue;
do
{
result_next = result*(pc-1);
result = result_next;
pc--;
}while(pc>2);
nValue = result;
return nValue;
}
double EvaluateBinomialCoefficient(double nValue, double nValue2)
{
double result;
if(nValue2 == 1)return nValue;
result = (Factorial(nValue))/(Factorial(nValue2)*Factorial((nValue - nValue2)));
nValue2 = result;
return nValue2;
}
Implementation:
int main()
{
cout<<"The Binomial Coefficient of 5, and 3, is equal to: "<< EvaluateBinomialCoefficient(5,3);
cin.get();
}
Output:
The Binomial Coefficient of 5, and 3, is equal to: 10
[edit] C#
using System;
namespace BinomialCoefficients
{
class Program
{
static void Main(string[] args)
{
ulong n = 1000000, k = 3;
ulong result = biCoefficient(n, k);
Console.WriteLine("The Binomial Coefficient of {0}, and {1}, is equal to: {2}", n, k, result);
Console.ReadLine();
}
static int fact(int n)
{
if (n == 0) return 1;
else return n * fact(n - 1);
}
static ulong biCoefficient(ulong n, ulong k)
{
if (k > n - k)
{
k = n - k;
}
ulong c = 1;
for (uint i = 0; i < k; i++)
{
c = c * (n - i);
c = c / (i + 1);
}
return c;
}
}
}
[edit] Clojure
(defn binomial-coefficient [n k]
(let [rprod (fn [a b] (reduce * (range a (inc b))))]
(/ (rprod (- n k -1) n) (rprod 1 k))))
[edit] CoffeeScript
binomial_coefficient = (n, k) ->
result = 1
for i in [0...k]
result *= (n - i) / (i + 1)
result
n = 5
for k in [0..n]
console.log "binomial_coefficient(#{n}, #{k}) = #{binomial_coefficient(n,k)}"
output
> coffee binomial.coffee
binomial_coefficient(5, 0) = 1
binomial_coefficient(5, 1) = 5
binomial_coefficient(5, 2) = 10
binomial_coefficient(5, 3) = 10
binomial_coefficient(5, 4) = 5
binomial_coefficient(5, 5) = 1
[edit] Common Lisp
(defun choose (n k)
(labels ((prod-enum (s e)
(do ((i s (1+ i)) (r 1 (* i r))) ((> i e) r)))
(fact (n) (prod-enum 1 n)))
(/ (prod-enum (- (1+ n) k) n) (fact k))))
[edit] D
import std.stdio, std.bigint;
T binomial(T)(/*in*/ T n, T k) /*pure nothrow*/ {
if (k > (n / 2))
k = n - k;
T bc = 1;
foreach (T i; cast(T)2 .. k + 1)
bc = (bc * (n - k + i)) / i;
return bc;
}
void main() {
foreach (d; [[5, 3], [100, 2], [100, 98]])
writefln("(%3d %3d) = %s", d[0], d[1], binomial(d[0], d[1]));
writeln("(100 50) = ", binomial(BigInt(100), BigInt(50)));
}
- Output:
( 5 3) = 2 (100 2) = 50 (100 98) = 50 (100 50) = 1976664223067613962806675336
[edit] dc
[sx1q]sz[d0=zd1-lfx*]sf[skdlfxrlk-lfxlklfx*/]sb
Demonstration:
5 3lbxp
10
Annotated version:
[ macro z: factorial base case when n is (z)ero ]sx
[sx [ x is our dump register; get rid of extraneous copy of n we no longer need]sx
1 [ return value is 1 ]sx
q] [ abort processing of calling macro ]sx
sz
[ macro f: factorial ]sx [
d [ duplicate the input (n) ]sx
0 =z [ if n is zero, call z, which stops here and returns 1 ]sx
d [ otherwise, duplicate n again ]sx
1 - [ subtract 1 ]sx
lfx [ take the factorial ]sx
* [ we have (n-1)!; multiply it by the copy of n to get n! ]sx
] sf
[ macro b(n,k): binomial function (n choose k).
straightforward RPN version of formula.]sx [
sk [ remember k. stack: n ]sx
d [ duplicate: n n ]sx
lfx [ call factorial: n n! ]sx
r [ swap: n! n ]sx
lk [ load k: n! n k ]sx
- [ subtract: n! n-k ]sx
lfx [ call factorial: n! (n-k)! ]sx
lk [ load k: n! (n-k)! k ]sx
lfx [ call factorial; n! (n-k)! k! ]sx
* [ multiply: n! (n-k)!k! ]sx
/ [ divide: n!/(n-k)!k! ]sx
] sb
5 3 lb x p [print(5 choose 3)]sx
[edit] Delphi
program Binomial;
{$APPTYPE CONSOLE}
function BinomialCoff(N, K: Cardinal): Cardinal;
var
L: Cardinal;
begin
if N < K then
Result:= 0 // Error
else begin
if K > N - K then
K:= N - K; // Optimization
Result:= 1;
L:= 0;
while L < K do begin
Result:= Result * (N - L);
Inc(L);
Result:= Result div L;
end;
end;
end;
begin
Writeln('C(5,3) is ', BinomialCoff(5, 3));
ReadLn;
end.
[edit] Erlang
choose(N, K) when is_integer(N), is_integer(K), (N >= 0), (K >= 0), (N >= K) ->
choose(N, K, 1, 1).
choose(N, K, K, Acc) ->
(Acc * (N-K+1)) div K;
choose(N, K, I, Acc) ->
choose(N, K, I+1, (Acc * (N-I+1)) div I).
[edit] F#
let choose n k = List.fold (fun s i -> s * (n-i+1)/i ) 1 [1..k]
[edit] Forth
: choose ( n k -- nCk ) 1 swap 0 ?do over i - i 1+ */ loop nip ;
5 3 choose . \ 10
33 17 choose . \ 1166803110
[edit] Fortran
program test_choose
implicit none
write (*, '(i0)') choose (5, 3)
contains
function factorial (n) result (res)
implicit none
integer, intent (in) :: n
integer :: res
integer :: i
res = product ((/(i, i = 1, n)/))
end function factorial
function choose (n, k) result (res)
implicit none
integer, intent (in) :: n
integer, intent (in) :: k
integer :: res
res = factorial (n) / (factorial (k) * factorial (n - k))
end function choose
end program test_choose
Output:
10
[edit] Frink
Frink has a built-in efficient function to find binomial coefficients. It produces arbitrarily-large integers.
println[binomial[5,3]]
[edit] GAP
# Built-in
Binomial(5, 3);
# 10
[edit] Go
package main
import "fmt"
import "math/big"
func main() {
fmt.Println(new(big.Int).Binomial(5, 3))
}
[edit] Golfscript
Actually evaluating n!/(k! (n-k)!):
;5 3 # Set up demo input
{),(;{*}*}:f; # Define a factorial function
.f@.f@/\@-f/
But Golfscript is meant for golfing, and it's shorter to calculate 
:
;5 3 # Set up demo input
1\,@{1$-@\*\)/}+/
[edit] Groovy
Solution:
def factorial = { x ->
assert x > -1
x == 0 ? 1 : (1..x).inject(1G) { BigInteger product, BigInteger factor -> product *= factor }
}
def combinations = { n, k ->
assert k >= 0
assert n >= k
factorial(n).intdiv(factorial(k)*factorial(n-k))
}
Test:
assert combinations(20, 0) == combinations(20, 20)
assert combinations(20, 10) == (combinations(19, 9) + combinations(19, 10))
assert combinations(5, 3) == 10
println combinations(5, 3)
Output:
10
[edit] Haskell
The only trick here is realizing that everything's going to divide nicely, so we can use div instead of (/).
choose :: (Integral a) => a -> a -> a
choose n k = product [k+1..n] `div` product [1..n-k]
> 5 `choose` 3
10
Or, generate the binomial coefficients iteratively to avoid computing with big numbers:
choose :: (Integral a) => a -> a -> a
choose n k = foldl (\z i -> (z * (n-i+1)) `div` i) 1 [1..k]
[edit] HicEst
WRITE(Messagebox) BinomCoeff( 5, 3) ! displays 10
FUNCTION factorial( n )
factorial = 1
DO i = 1, n
factorial = factorial * i
ENDDO
END
FUNCTION BinomCoeff( n, k )
BinomCoeff = factorial(n)/factorial(n-k)/factorial(k)
END
[edit] Icon and Unicon
link math, factors
procedure main()
write("choose(5,3)=",binocoef(5,3))
end
Output:
choose(5,3)=10
math provides binocoef and factors provides factorial.
procedure binocoef(n, k) #: binomial coefficient
k := integer(k) | fail
n := integer(n) | fail
if (k = 0) | (n = k) then return 1
if 0 <= k <= n then
return factorial(n) / (factorial(k) * factorial(n - k))
else fail
end
procedure factorial(n) #: return n! (n factorial)
local i
n := integer(n) | runerr(101, n)
if n < 0 then fail
i := 1
every i *:= 1 to n
return i
end
[edit] J
Solution:
The dyadic form of the primitive ! ([Out of]) evaluates binomial coefficients directly.
Example usage:
3 ! 5
10
[edit] Java
public class Binom {
static long combinations(int n, int k) {
long coeff = 1;
for (int i = n - k + 1; i <= n; i++) {
coeff *= i;
}
for (int i = 1; i <= k; i++) {
coeff /= i;
}
return coeff;
}
public static void main(String[] args){
System.out.println(combinations(5, 3));
}
}
Output:
10.0
public class Binom {
public static double binomCoeff(double n, double k) {
double result = 1;
for (int i = 1; i < k + 1; i++) {
result *= (n - i + 1) / i;
}
return result;
}
public static void main(String[] args) {
System.out.println(binomCoeff(5, 3));
}
}
Output:
10.0
[edit] JavaScript
function binom(n, k) {
var coeff = 1;
for (var i = n-k+1; i <= n; i++) coeff *= i;
for (var i = 1; i <= k; i++) coeff /= i;
return coeff;
}
print(binom(5,3));
10
[edit] K
{[n;k]_(*/(k-1)_1+!n)%(*/1+!k)} . 5 3
10
Alternative version:
{[n;k]i:!(k-1);_*/((n-i)%(i+1))} . 5 3
10
Using Pascal's triangle:
pascal:{x{+':0,x,0}\1}
pascal 5
(1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1)
{[n;k](pascal n)[n;k]} . 5 3
10
[edit] Logo
to choose :n :k
if :k = 0 [output 1]
output (choose :n :k-1) * (:n - :k + 1) / :k
end
show choose 5 3 ; 10
show choose 60 30 ; 1.18264581564861e+17
[edit] Lua
function Binomial( n, k )
if k > n then return nil end
if k > n/2 then k = n - k end -- (n k) = (n n-k)
numer, denom = 1, 1
for i = 1, k do
numer = numer * ( n - i + 1 )
denom = denom * i
end
return numer / denom
end
[edit] Liberty BASIC
' [RC] Binomial Coefficients
print "Binomial Coefficient of "; 5; " and "; 3; " is ",BinomialCoefficient( 5, 3)
n =1 +int( 10 *rnd( 1))
k =1 +int( n *rnd( 1))
print "Binomial Coefficient of "; n; " and "; k; " is ",BinomialCoefficient( n, k)
end
function BinomialCoefficient( n, k)
BinomialCoefficient =factorial( n) /factorial( n -k) /factorial( k)
end function
function factorial( n)
if n <2 then
f =1
else
f =n *factorial( n -1)
end if
factorial =f
end function
[edit] Mathematica
(Local) In[1]:= Binomial[5,3]
(Local) Out[1]= 10
[edit] MATLAB / Octave
This is a built-in function in MATLAB called "nchoosek(n,k)". But, this will only work for scalar inputs. If "n" is a vector then "nchoosek(v,k)" finds all combinations of choosing "k" elements out of the "v" vector (see Combinations#MATLAB).
Solution:
>> nchoosek(5,3)
ans =
10
Alternative implementations are:
function r = binomcoeff1(n,k)
r = diag(rot90(pascal(n+1))); % vector of all binomial coefficients for order n
r = r(k);
end;
function r = binomcoeff2(n,k)
prod((n-k+1:n)./(1:k))
end;
function r = binomcoeff3(n,k)
m = pascal(max(n-k,k)+1);
r = m(n-k+1,k+1);
end;
If you want a vectorized function that returns multiple binomial coefficients given vector inputs, you must define that function yourself. A sample implementation is given below. This function takes either scalar or vector inputs for "n" and "v" and returns either a: scalar, vector, or matrix. Where the columns are indexed by the "k" vector and the rows indexed by the "n" vector. binomialCoeff.m:
function coefficients = binomialCoeff(n,k)
coefficients = zeros(numel(n),numel(k)); %Preallocate memory
columns = (1:numel(k)); %Preallocate row and column counters
rows = (1:numel(n));
%Iterate over every row and column. The rows represent the n number,
%and the columns represent the k number. If n is ever greater than k,
%the nchoosek function will throw an error. So, we test to make sure
%it isn't, if it is then we leave that entry in the coefficients matrix
%zero. Which makes sense combinatorically.
for row = rows
for col = columns
if k(col) <= n(row)
coefficients(row,col) = nchoosek(n(row),k(col));
end
end
end
end %binomialCoeff
Sample Usage:
>> binomialCoeff((0:5),(0:5))
ans =
1 0 0 0 0 0
1 1 0 0 0 0
1 2 1 0 0 0
1 3 3 1 0 0
1 4 6 4 1 0
1 5 10 10 5 1
>> binomialCoeff([1 0 3 2],(0:3))
ans =
1 1 0 0
1 0 0 0
1 3 3 1
1 2 1 0
>> binomialCoeff(3,(0:3))
ans =
1 3 3 1
>> binomialCoeff((0:3),2)
ans =
0
0
1
3
>> binomialCoeff(5,3)
ans =
10
[edit] Maxima
binomial( 5, 3); /* 10 */
binomial(-5, 3); /* -35 */
binomial( 5, -3); /* 0 */
binomial(-5, -3); /* 0 */
binomial( 3, 5); /* 0 */
binomial(x, 3); /* ((x - 2)*(x - 1)*x)/6 */
binomial(3, 1/2); /* binomial(3, 1/2) */
makegamma(%); /* 32/(5*%pi) */
binomial(a, b); /* binomial(a, b) */
makegamma(%); /* gamma(a + 1)/(gamma(-b + a + 1)*gamma(b + 1)) */
[edit] OCaml
let binomialCoeff n p =
let p = if p < n -. p then p else n -. p in
let rec cm res num denum =
(* this method partially prevents overflow.
* float type is choosen to have increased domain on 32-bits computer,
* however algorithm ensures an integral result as long as it is possible
*)
if denum <= p then cm ((res *. num) /. denum) (num -. 1.) (denum +. 1.)
else res in
cm 1. n 1.
[edit] Alternate version using big integers
#load "nums.cma";;
open Num;;
let binomial n p =
let m = min p (n - p) in
if m < 0 then Int 0 else
let rec a j v =
if j = m then v
else a (succ j) ((v */ (Int (n - j))) // (Int (succ j)))
in a 0 (Int 1)
;;
[edit] Simple recursive version
open Num;;
let rec binomial n k = if n = k then Int 1 else ((binomial (n-1) k) */ Int n) // Int (n-k)
[edit] Oz
declare
fun {BinomialCoeff N K}
{List.foldL {List.number 1 K 1}
fun {$ Z I}
Z * (N-I+1) div I
end
1}
end
in
{Show {BinomialCoeff 5 3}}
[edit] PARI/GP
binomial(5,3)
[edit] Pascal
See Delphi
[edit] Perl
sub binomial {
use bigint;
my ($r, $n, $k) = (1, @_);
for (1 .. $k) { $r *= $n + 1 - $_, $r /= $_ }
$r;
}
print binomial(30, 13);
[edit] Perl 6
multi sub postfix:<!>(Int $a) {
[*] 1..$a;
}
sub binomialcoefficient($n, $k) {
$n! / (($n - $k)! * $k!);
}
say binomialcoefficient(5, 3);
Output:
10
This particular piece of code implements the factorial function as a custom operator, which allows you to use it as you would in mathematical equations. The multi keyword on line 1 is to allow operator overloading.
[edit] PL/I
binomial_coefficients:
procedure options (main);
declare (n, k) fixed;
get (n, k);
put (coefficient(n, k));
coefficient: procedure (n, k) returns (fixed decimal (15));
declare (n, k) fixed;
return (fact(n)/ (fact(n-k) * fact(k)) );
end coefficient;
fact: procedure (n) returns (fixed decimal (15));
declare n fixed;
declare i fixed, f fixed decimal (15);
f = 1;
do i = 1 to n;
f = f * i;
end;
return (f);
end fact;
end binomial_coefficients;
Output:
10
[edit] PHP
<?php
$n=5;
$k=3;
function factorial($val){
for($f=2;$val-1>1;$f*=$val--);
return $f;
}
$binomial_coefficient=factorial($n)/(factorial($k)*factorial($n-$k));
echo $binomial_coefficient;
?>
Alternative version, not based on factorial
function binomial_coefficient($n, $k) {
if ($k == 0) return 1;
$result = 1;
foreach (range(0, $k - 1) as $i) {
$result *= ($n - $i) / ($i + 1);
}
return $result;
}
[edit] PicoLisp
(de binomial (N K)
(let f '((N) (apply * (range 1 N)))
(/ (f N) (* (f (- N K)) (f K))) ) )
Output:
: (binomial 5 3) -> 10
[edit] PureBasic
Procedure Factor(n)
Protected Result=1
While n>0
Result*n
n-1
Wend
ProcedureReturn Result
EndProcedure
Macro C(n,k)
(Factor(n)/(Factor(k)*factor(n-k)))
EndMacro
If OpenConsole()
Print("Enter value n: "): n=Val(Input())
Print("Enter value k: "): k=Val(Input())
PrintN("C(n,k)= "+str(C(n,k)))
Print("Press ENTER to quit"): Input()
CloseConsole()
EndIf
Example
Enter value n: 5 Enter value k: 3 C(n,k)= 10
[edit] Python
Straight-forward implementation:
def binomialCoeff(n, k):
result = 1
for i in range(1, k+1):
result = result * (n-i+1) / i
return result
if __name__ == "__main__":
print(binomialCoeff(5, 3))
Output:
10
Alternate implementation
from operator import mul
def comb(n,r):
''' calculate nCr - the binomial coefficient
>>> comb(3,2)
3
>>> comb(9,4)
126
>>> comb(9,6)
84
>>> comb(20,14)
38760
'''
if r > n-r: # for smaller intermediate values
r = n-r
return int( reduce( mul, range((n-r+1), n+1), 1) /
reduce( mul, range(1,r+1), 1) )
[edit] R
R's built-in choose() function evaluates binomial coefficients:
choose(5,3)
Output:
[1] 10
[edit] Racket
#lang racket
(require math)
(binomial 10 5)
[edit] REXX
The task is to compute ANY binomial coefficient(s), but this example is limited to 100k digits.
/*REXX program calculates binomial coefficients (aka, combinations). */
numeric digits 100000
parse arg n k .
say 'combinations('n","k') =' comb(n,k)
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────COMB subroutine─────────────────────*/
comb: procedure; parse arg x,y; return fact(x) / (fact(x-y) * fact(y))
/*──────────────────────────────────FACT subroutine─────────────────────*/
fact: procedure; parse arg z; !=1; do j=2 to z; !=!*j; end; return !
output when using the input of: 5 3
combinations(5,3) = 10
output when using the input of: 1200 120
combinations(1200,120) = 1004576581793084916475353119318331966507299414258370667602185866686463289093457468590558508056798211449853806741873396451735444387513582540860551330127062642417424083600
[edit] Ruby
class Integer
# binomial coefficient: n C k
def choose(k)
# n!/(n-k)!
pTop = (self-k+1 .. self).inject(1, &:*)
# k!
pBottom = (2 .. k).inject(1, &:*)
pTop / pBottom
end
end
p 5.choose(3)
p 60.choose(30)
result
10 118264581564861424
another implementation:
def c n, r
(0...r).inject(1) do |m,i| (m * (n - i)) / (i + 1) end
end
[edit] Run BASIC
print "binomial (5,1) = "; binomial(5, 1)Output:
print "binomial (5,2) = "; binomial(5, 2)
print "binomial (5,3) = "; binomial(5, 3)
print "binomial (5,4) = "; binomial(5,4)
print "binomial (5,5) = "; binomial(5,5)
end
function binomial(n,k)
coeff = 1
for i = n - k + 1 to n
coeff = coeff * i
next i
for i = 1 to k
coeff = coeff / i
next i
binomial = coeff
end function
binomial (5,1) = 5 binomial (5,2) = 10 binomial (5,3) = 10 binomial (5,4) = 5 binomial (5,5) = 1
[edit] Scala
object Binomial {
def main(args: Array[String]): Unit = {
val n=5
val k=3
val result=binomialCoefficient(n,k)
println("The Binomial Coefficient of %d and %d equals %d.".format(n, k, result))
}
def binomialCoefficient(n:Int, k:Int)=fact(n) / (fact(k) * fact(n-k))
def fact(n:Int):Int=if (n==0) 1 else n*fact(n-1)
}
Output:
The Binomial Coefficient of 5 and 3 equals 10.
Another (more flexible and efficient) implementation. n and k are taken from command line. The use of BigInts allows to compute coefficients of arbitrary size:
object Binomial extends App {
def binomialCoefficient(n: Int, k: Int) =
(BigInt(n - k + 1) to n).product /
(BigInt(1) to k).product
val Array(n, k) = args.map(_.toInt)
println("The Binomial Coefficient of %d and %d equals %,3d.".format(n, k, binomialCoefficient(n, k)))
}
Output:
java Binomial 100 30
The Binomial Coefficient of 100 and 30 equals 29,372,339,821,610,944,823,963,760.
Using recursive formula C(n,k) = C(n-1,k-1) + C(n-1,k):
def bico(n: Long, k: Long): Long = (n, k) match {
case (n, 0) => 1
case (0, k) => 0
case (n, k) => bico(n - 1, k - 1) + bico(n - 1, k)
}
println("bico(5,3) = " + bico(5, 3))
Output:
bico(5,3) = 10
[edit] Scheme
(define (factorial n)
(define (*factorial n acc)
(if (zero? n)
acc
(*factorial (- n 1) (* acc n))))
(*factorial n 1))
(define (choose n k)
(/ (factorial n) (* (factorial k) (factorial (- n k)))))
(display (choose 5 3))
(newline)
Output:
10
[edit] Seed7
$ include "seed7_05.s7i";
const func integer: binomial (in integer: n, in var integer: k) is func
result
var integer: binomial is 0;
local
var integer: l is 0;
begin
if n >= k then
if k > n - k then
k := n - k; # Optimization
end if;
binomial := 1;
l := 0;
while l < k do
binomial *:= n - l;
incr(l);
binomial := binomial div l;
end while;
end if;
end func;
const proc: main is func
begin
writeln("binomial coefficient of (5, 3) is " <& binomial(5, 3));
end func;
Output:
binomial coefficient of (5, 3) is 10
[edit] Tcl
This uses exact arbitrary precision integer arithmetic.
package require Tcl 8.5
proc binom {n k} {
# Compute the top half of the division; this is n!/(n-k)!
set pTop 1
for {set i $n} {$i > $n - $k} {incr i -1} {
set pTop [expr {$pTop * $i}]
}
# Compute the bottom half of the division; this is k!
set pBottom 1
for {set i $k} {$i > 1} {incr i -1} {
set pBottom [expr {$pBottom * $i}]
}
# Integer arithmetic divide is correct here; the factors always cancel out
return [expr {$pTop / $pBottom}]
}
Demonstrating:
puts "5_C_3 = [binom 5 3]"
puts "60_C_30 = [binom 60 30]"
Output:
5_C_3 = 10 60_C_30 = 118264581564861424
[edit] TI-89 BASIC
Builtin function.
nCr(n,k)
[edit] UNIX Shell
#!/bin/sh
n=5;
k=3;
calculate_factorial(){
partial_factorial=1;
for (( i=1; i<="$1"; i++ ))
do
factorial=$(expr $i \* $partial_factorial)
partial_factorial=$factorial
done
echo $factorial
}
n_factorial=$(calculate_factorial $n)
k_factorial=$(calculate_factorial $k)
n_minus_k_factorial=$(calculate_factorial `expr $n - $k`)
binomial_coefficient=$(expr $n_factorial \/ $k_factorial \* 1 \/ $n_minus_k_factorial )
echo "Binomial Coefficient ($n,$k) = $binomial_coefficient"
[edit] Ursala
A function for computing binomial coefficients (choose) is included in a standard library, but if it weren't, it could be defined in one of the following ways, starting from the most idiomatic.
#import nat
choose = ~&ar^?\1! quotient^\~&ar product^/~&al ^|R/~& predecessor~~
The standard library functions quotient, product and predecessor pertain to natural numbers in the obvious way.
-
chooseis defined using the recursive conditional combinator (^?) as a function taking a pair of numbers, with the predicate~&artesting whether the number on the right side of the pair is non-zero. - If the predicate does not hold (implying the right side is zero), then a constant value of 1 is returned.
- If the predicate holds, the function given by the rest of the expression executes as follows.
- First the
predecessorof both sides (~~) of the argument is taken. - Then a recursive call (
^|R) is made to the whole function (~&) with the pair of predecessors passed to it as an argument. - The result returned by the recursive call is multiplied (
product) by the left side of the original argument (~&al). - The product of these values is then divided (
quotient) by the right side (~&ar) of the original argument and returned as the result.
Here is a less efficient implementation more closely following the formula above.
choose = quotient^/factorial@l product+ factorial^~/difference ~&r
-
chooseis defined as thequotientof the results of a pair (^) of functions. - The left function contributing to the quotient is the
factorialof the left side (@l) of the argument, which is assumed to be a pair of natural numbers. Thefactorialfunction is provided in a standard library. - The right function contributing to the quotient is the function given by the rest of the expression, which applies to the whole pair as follows.
- It begins by forming a pair of numbers from the argument, the left being their
differenceobtained by subtraction, and the right being the a copy of the right (~&r) side of the argument. - The
factorialfunction is applied separately to both results (^~). - A composition (
+) of this function with theproductfunction effects the multiplication of the two factorials, to complete the other input to the quotient.
Here is an equivalent implementation using pattern matching, dummy variables, and only the apply-to-both (~~) operator.
choose("n","k") = quotient(factorial "n",product factorial~~ (difference("n","k"),"k"))
test program:
#cast %nL
main = choose* <(5,3),(60,30)>
output:
<10,118264581564861424>
[edit] XPL0
code ChOut=8, CrLf=9, IntOut=11;
func Binomial(N, K);
int N, K;
int M, B, I;
[M:= K;
if K>N/2 the M:= N-K;
B:=1;
for I:= 1 to M do
B:= B*(N-M+I)/I;
return B;
];
int N, K;
[for N:= 0 to 9 do
[for K:= 0 to 9 do
[if N>=K then IntOut(0, Binomial(N,K));
ChOut(0, 9\tab\);
];
CrLf(0);
];
] \Mr. Pascal's triangle!
Output:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1
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