You are encouraged to solve this task according to the task description, using any language you may know.
- Task
Print out the first 15 Catalan numbers by extracting them from Pascal's triangle.
- See
- Catalan Numbers and the Pascal Triangle. This method enables calculation of Catalan Numbers using only addition and subtraction.
- Catalan's Triangle for a Number Triangle that generates Catalan Numbers using only addition.
- Related Tasks
Contents
- 1 360 Assembly
- 2 Ada
- 3 ALGOL 68
- 4 ALGOL W
- 5 APL
- 6 AutoHotkey
- 7 Batch File
- 8 C
- 9 C++
- 10 C#
- 11 D
- 12 EchoLisp
- 13 Elixir
- 14 Erlang
- 15 ERRE
- 16 F#
- 17 FreeBASIC
- 18 Groovy
- 19 Haskell
- 20 Icon and Unicon
- 21 J
- 22 Java
- 23 JavaScript
- 24 jq
- 25 Lua
- 26 Mathematica / Wolfram Language
- 27 MATLAB / Octave
- 28 Nim
- 29 OCaml
- 30 Oforth
- 31 PARI/GP
- 32 Pascal
- 33 Perl
- 34 Perl 6
- 35 Phix
- 36 PicoLisp
- 37 Python
- 38 Racket
- 39 REXX
- 40 Ring
- 41 Ruby
- 42 Run BASIC
- 43 Rust
- 44 Scilab
- 45 Seed7
- 46 Sidef
- 47 smart BASIC
- 48 Tcl
- 49 TI-83 BASIC
- 50 VBScript
- 51 Visual Basic
- 52 zkl
- 53 ZX Spectrum Basic
360 Assembly[edit]
For maximum compatibility, this program uses only the basic instruction set.
CATALAN CSECT
USING CATALAN,R13,R12
SAVEAREA B STM-SAVEAREA(R15)
DC 17F'0'
DC CL8'CATALAN'
STM STM R14,R12,12(R13)
ST R13,4(R15)
ST R15,8(R13)
LR R13,R15
LA R12,4095(R13)
LA R12,1(R12)
* ---- CODE
LA R0,1
ST R0,T t(1)=1
LA R4,0 ix:i=1
LA R6,1 by 1
LH R7,N to n
LOOPI BXH R4,R6,ENDLOOPI loop i
LR R5,R4 ix:j=i+1
LA R5,2(R5) i+2
LA R8,0
BCTR R8,0 by -1
LA R9,1 to 2
LOOP1J BXLE R5,R8,ENLOOP1J loop j
LR R10,R5 j
BCTR R10,0
SLA R10,2
L R2,T(R10) r2=t(j)
LR R1,R10 j
SH R1,=H'4'
L R3,T(R1) r3=t(j-1)
AR R2,R3 r2=r2+r3
ST R2,T(R10) t(j)=t(j)+t(j-1)
B LOOP1J
ENLOOP1J EQU *
LR R1,R4 i
BCTR R1,0
SLA R1,2
L R3,T(R1) t(i)
LA R1,4(R1)
ST R3,T(R1) t(i+1)
LR R5,R4 ix:j=i+2
LA R5,3(R5) i+3
LA R8,0
BCTR R8,0 by -1
LA R9,1 to 2
LOOP2J BXLE R5,R8,ENLOOP2J loop j
LR R10,R5 j
BCTR R10,0
SLA R10,2
L R2,T(R10) r2=t(j)
LR R1,R10 j
SH R1,=H'4'
L R3,T(R1) r3=t(j-1)
AR R2,R3 r2=r2+r3
ST R2,T(R10) t(j)=t(j)+t(j-1)
B LOOP2J
ENLOOP2J EQU *
LR R1,R4 i
BCTR R1,0
SLA R1,2
L R2,T(R1) t(i)
LA R1,4(R1)
L R3,T(R1) t(i+1)
SR R3,R2
CVD R3,P
UNPK Z,P
MVC C,Z
OI C+L'C-1,X'F0'
MVC WTOBUF(8),C+8
WTO MF=(E,WTOMSG)
B LOOPI
ENDLOOPI EQU *
* ---- END CODE
CNOP 0,4
L R13,4(0,R13)
LM R14,R12,12(R13)
XR R15,R15
BR R14
* ---- DATA
N DC H'15'
T DC 17F'0'
P DS PL8
Z DS ZL16
C DS CL16
WTOMSG DS 0F
DC H'80'
DC H'0'
WTOBUF DC CL80' '
YREGS
END
- Output:
00000001 00000002 00000005 00000014 00000042 00000132 00000429 00001430 00004862 00016796 00058786 00208012 00742900 02674440 09694845
Ada[edit]
Uses package Pascal from the Pascal triangle solution[[1]]
with Ada.Text_IO, Pascal;
procedure Catalan is
Last: Positive := 15;
Row: Pascal.Row := Pascal.First_Row(2*Last+1);
begin
for I in 1 .. Last loop
Row := Pascal.Next_Row(Row);
Row := Pascal.Next_Row(Row);
Ada.Text_IO.Put(Integer'Image(Row(I+1)-Row(I+2)));
end loop;
end Catalan;
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
ALGOL 68[edit]
INT n = 15;
[ 0 : n + 1 ]INT t;
t[0] := 0;
t[1] := 1;
FOR i TO n DO
FOR j FROM i BY -1 TO 2 DO t[j] := t[j] + t[j-1] OD;
t[i+1] := t[i];
FOR j FROM i+1 BY -1 TO 2 DO t[j] := t[j] + t[j-1] OD;
print( ( whole( t[i+1] - t[i], 0 ), " " ) )
OD
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
ALGOL W[edit]
begin
% print the first 15 Catalan numbers from Pascal's triangle %
integer n;
n := 15;
begin
integer array pascalLine ( 1 :: n + 1 );
% the Catalan numbers are the differences between the middle and middle - 1 numbers of the odd %
% lines of Pascal's triangle (lines with 3 or more numbers) %
% note - we only need to calculate the left side of the triangle %
pascalLine( 1 ) := 1;
for c := 2 until n + 1 do begin
% even line %
for i := c - 1 step -1 until 2 do pascalLine( i ) := pascalLine( i - 1 ) + pascalLine( i );
pascalLine( c ) := pascalLine( c - 1 );
% odd line %
for i := c step -1 until 2 do pascalLine( i ) := pascalLine( i - 1 ) + pascalLine( i );
writeon( i_w := 1, s_w := 0, " ", pascalLine( c ) - pascalLine( c - 1 ) )
end for_c
end
end.
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
APL[edit]
⍝ Based heavily on the J solution
CATALAN←{¯1↓↑-/1 ¯1↓¨(⊂⎕IO+0 0)⍉¨0 2⌽¨⊂(⎕IO-⍨⍳N){+\⍣⍺⊢⍵}⍤0 1⊢1⍴⍨N←⍵+2}
- Output:
CATALAN 15
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
AutoHotkey[edit]
/* Generate Catalan Numbers
//
// smgs: 20th Feb, 2014
*/
Array := [], Array[2,1] := Array[2,2] := 1 ; Array inititated and 2nd row of pascal's triangle assigned
INI := 3 ; starts with calculating the 3rd row and as such the value
Loop, 31 ; every odd row is taken for calculating catalan number as such to obtain 15 we need 2n+1
{
if ( A_index > 2 )
{
Loop, % A_INDEX
{
old := ini-1, index := A_index, index_1 := A_index + 1
Array[ini, index_1] := Array[old, index] + Array[old, index_1]
Array[ini, 1] := Array[ini, ini] := 1
line .= Array[ini, A_index] " "
}
;~ MsgBox % line ; gives rows of pascal's triangle
; calculating every odd row starting from 1st so as to obtain catalan's numbers
if ( mod(ini,2) != 0)
{
StringSplit, res, line, %A_Space%
ans := res0//2, ans_1 := ans++
result := result . res%ans_1% - res%ans% " "
}
line :=
ini++
}
}
MsgBox % result
- Produces:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Batch File[edit]
@echo off
setlocal ENABLEDELAYEDEXPANSION
set n=15
set /A nn=n+1
for /L %%i in (0,1,%nn%) do set t.%%i=0
set t.1=1
for /L %%i in (1,1,%n%) do (
set /A ip=%%i+1
for /L %%j in (%%i,-1,1) do (
set /A jm=%%j-1
set /A t.%%j=t.%%j+t.!jm!
)
set /A t.!ip!=t.%%i
for /L %%j in (!ip!,-1,1) do (
set /A jm=%%j-1
set /A t.%%j=t.%%j+t.!jm!
)
set /A ci=t.!ip!-t.%%i
echo !ci!
)
)
pause
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
C[edit]
//This code implements the print of 15 first Catalan's Numbers
//Formula used:
// __n__
// | | (n + k) / k n>0
// k=2
#include <stdio.h>
#include <stdlib.h>
//the number of Catalan's Numbers to be printed
const int N = 15;
int main()
{
//loop variables (in registers)
register int k, n;
//necessarily ull for reach big values
unsigned long long int num, den;
//the nmmber
int catalan;
//the first is not calculated for the formula
printf("1 ");
//iterating fro 2 to 15
for (n=2; n<=N; ++n) {
//initializaing for products
num = den = 1;
//applying the formula
for (k=2; k<=n; ++k) {
num *= (n+k);
den *= k;
catalan = num /den;
}
//output
printf("%d ", catalan);
}
//the end
printf("\n");
return 0;
}
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
C++[edit]
// Generate Catalan Numbers
//
// Nigel Galloway: June 9th., 2012
//
#include <iostream>
int main() {
const int N = 15;
int t[N+2] = {0,1};
for(int i = 1; i<=N; i++){
for(int j = i; j>1; j--) t[j] = t[j] + t[j-1];
t[i+1] = t[i];
for(int j = i+1; j>1; j--) t[j] = t[j] + t[j-1];
std::cout << t[i+1] - t[i] << " ";
}
return 0;
}
- Produces:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
C#[edit]
int n = 15;
List<int> t = new List<int>() { 0, 1 };
for (int i = 1; i <= n; i++)
{
for (var j = i; j > 1; j--) t[j] += t[j - 1];
t.Add(t[i]);
for (var j = i + 1; j > 1; j--) t[j] += t[j - 1];
Console.Write(((i == 1) ? "" : ", ") + (t[i + 1] - t[i]));
}
- Produces:
1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845
D[edit]
void main() {
import std.stdio;
enum uint N = 15;
uint[N + 2] t;
t[1] = 1;
foreach (immutable i; 1 .. N + 1) {
foreach_reverse (immutable j; 2 .. i + 1)
t[j] += t[j - 1];
t[i + 1] = t[i];
foreach_reverse (immutable j; 2 .. i + 2)
t[j] += t[j - 1];
write(t[i + 1] - t[i], ' ');
}
}
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
EchoLisp[edit]
(define dim 100)
(define-syntax-rule (Tidx i j) (+ i (* dim j)))
;; generates Catalan's triangle
;; T (i , j) = T(i-1,j) + T (i, j-1)
(define (T n)
(define i (modulo n dim))
(define j (quotient n dim))
(cond
((zero? i) 1) ;; left column = 1
((= i j) (T (Tidx (1- i) j))) ;; diagonal value = left value
(else (+ (T (Tidx (1- i) j)) (T (Tidx i (1- j)))))))
(remember 'T #(1))
- Output:
;; take elements on diagonal = Catalan numbers
(for ((i (in-range 0 16))) (write (T (Tidx i i))))
→ 1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Elixir[edit]
defmodule Catalan do
def numbers(num) do
{result,_} = Enum.reduce(1..num, {[],{0,1}}, fn i,{list,t0} ->
t1 = numbers(i, t0)
t2 = numbers(i+1, Tuple.insert_at(t1, i+1, elem(t1, i)))
{[elem(t2, i+1) - elem(t2, i) | list], t2}
end)
Enum.reverse(result)
end
defp numbers(0, t), do: t
defp numbers(n, t), do: numbers(n-1, put_elem(t, n, elem(t, n-1) + elem(t, n)))
end
IO.inspect Catalan.numbers(15)
- Output:
[1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
Erlang[edit]
-module(catalin).
-compile(export_all).
mul(N,D,S,S)->
N2=N*(S+S),
D2=D*S,
K = N2 div D2 ;
mul(N,D,S,L)->
N2=N*(S+L),
D2=D*L,
K = mul(N2,D2,S,L+1).
catl(Ans,16) -> Ans;
catl(D,S)->
C=mul(1,1,S,2),
catl([D|C],S+1).
main()->
Ans=catl(1,2).
ERRE[edit]
PROGRAM CATALAN
!$DOUBLE
DIM CATALAN[50]
FUNCTION ODD(X)
ODD=FRC(X/2)<>0
END FUNCTION
PROCEDURE GETCATALAN(L)
LOCAL J,K,W
LOCAL DIM PASTRI[100]
L=L*2
PASTRI[0]=1
J=0
WHILE J<L DO
J+=1
K=INT((J+1)/2)
PASTRI[K]=PASTRI[K-1]
FOR W=K TO 1 STEP -1 DO
PASTRI[W]+=PASTRI[W-1]
END FOR
IF NOT(ODD(J)) THEN
K=INT(J/2)
CATALAN[K]=PASTRI[K]-PASTRI[K-1]
END IF
END WHILE
END PROCEDURE
BEGIN
LL=15
GETCATALAN(LL)
FOR I=1 TO LL DO
WRITE("### ####################";I;CATALAN[I])
END FOR
END PROGRAM
- Output:
1 1 2 2 3 5 4 14 5 42 6 132 7 429 8 1430 9 4862 10 16796 11 58786 12 208012 13 742900 14 2674440 15 9694845
F#[edit]
let mutable nm=uint64(1)
let mutable dm=uint64(1)
let mutable a=uint64(1)
printf "1, "
for i = 2 to 15 do
nm<-uint64(1)
dm<-uint64(1)
for k = 2 to i do
nm <-uint64( uint64(nm) * (uint64(i)+uint64(k)))
dm <-uint64( uint64(dm) * uint64(k))
let a = uint64(uint64(nm)/uint64(dm))
printf "%u"a
if(i<>15) then
printf ", "
FreeBASIC[edit]
' version 15-09-2015
' compile with: fbc -s console
#Define size 31 ' (N * 2 + 1)
Sub pascal_triangle(rows As Integer, Pas_tri() As ULongInt)
Dim As Integer x, y
For x = 1 To rows
Pas_tri(1,x) = 1
Pas_tri(x,1) = 1
Next
For x = 2 To rows
For y = 2 To rows + 1 - x
Pas_tri(x, y) = pas_tri(x - 1 , y) + pas_tri(x, y - 1)
Next
Next
End Sub
' ------=< MAIN >=------
Dim As Integer count, row
Dim As ULongInt triangle(1 To size, 1 To size)
pascal_triangle(size, triangle())
' 1 1 1 1 1 1
' 1 2 3 4 5 6
' 1 3 6 10 15 21
' 1 4 10 20 35 56
' 1 5 15 35 70 126
' 1 6 21 56 126 252
' The Pascal triangle is rotated 45 deg.
' to find the Catalan number we need to follow the diagonal
' for top left to bottom right
' take the number on diagonal and subtract the number in de cell
' one up and one to right
' 1 (2 - 1), 2 (6 - 4), 5 (20 - 15) ...
Print "The first 15 Catalan numbers are" : print
count = 1 : row = 2
Do
Print Using "###: #########"; count; triangle(row, row) - triangle(row +1, row -1)
row = row + 1
count = count + 1
Loop Until count > 15
' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
- Output:
The first 15 Catalan numbers are 1: 1 2: 2 3: 5 4: 14 5: 42 6: 132 7: 429 8: 1430 9: 4862 10: 16796 11: 58786 12: 208012 13: 742900 14: 2674440 15: 9694845
Groovy[edit]
class Catalan
{
public static void main(String[] args)
{
BigInteger N = 15;
BigInteger k,n,num,den;
BigInteger catalan;
print(1);
for(n=2;n<=N;n++)
{
num = 1;
den = 1;
for(k=2;k<=n;k++)
{
num = num*(n+k);
den = den*k;
catalan = num/den;
}
print(" " + catalan);
}
}
}
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Haskell[edit]
As required by the task this implementation extracts the Catalan numbers from Pascal's triangle, rather than calculating them directly. Also, note that it (correctly) produces [1, 1] as the first two numbers.
import System.Environment (getArgs)
-- Pascal's triangle.
pascal :: [[Integer]]
pascal = [1] : map (\row -> 1 : zipWith (+) row (tail row) ++ [1]) pascal
-- The Catalan numbers from Pascal's triangle. This uses a method from
-- http://www.cut-the-knot.org/arithmetic/algebra/CatalanInPascal.shtml
-- (see "Grimaldi").
catalan :: [Integer]
catalan = map (diff . uncurry drop) $ zip [0..] (alt pascal)
where alt (x:_:zs) = x : alt zs -- every other element of an infinite list
diff (x:y:_) = x - y
diff (x:_) = x
main :: IO ()
main = do
ns <- fmap (map read) getArgs :: IO [Int]
mapM_ (print . flip take catalan) ns
- Output:
./catalan 15 [1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440]
Icon and Unicon[edit]
The following works in both languages. It avoids computing elements in Pascal's triangle that aren't used.
link math
procedure main(A)
limit := (integer(A[1])|15)+1
every write(right(binocoef(i := 2*seq(0)\limit,i/2)-binocoef(i,i/2+1),30))
end
Sample run:
->cn
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
->
J[edit]
Catalan=. }:@:(}.@:((<0 1)&|:) - }:@:((<0 1)&|:@:(2&|.)))@:(i. +/\@]^:[ #&1)@:(2&+)
- Example use:
Catalan 15
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
A structured derivation of Catalan follows:
o=. @: NB. Composition of verbs (functions)
( PascalTriangle=. i. ((+/\@]^:[)) #&1 ) 5
1 1 1 1 1
1 2 3 4 5
1 3 6 10 15
1 4 10 20 35
1 5 15 35 70
( MiddleDiagonal=. (<0 1)&|: ) o PascalTriangle 5
1 2 6 20 70
( AdjacentLeft=. MiddleDiagonal o (2&|.) ) o PascalTriangle 5
1 4 15 1 5
( Catalan=. }: o (}. o MiddleDiagonal - }: o AdjacentLeft) o PascalTriangle o (2&+) f. ) 5
1 2 5 14 42
Catalan
}:@:(}.@:((<0 1)&|:) - }:@:((<0 1)&|:@:(2&|.)))@:(i. +/\@]^:[ #&1)@:(2&+)
Java[edit]
public class Test {
public static void main(String[] args) {
int N = 15;
int[] t = new int[N + 2];
t[1] = 1;
for (int i = 1; i <= N; i++) {
for (int j = i; j > 1; j--)
t[j] = t[j] + t[j - 1];
t[i + 1] = t[i];
for (int j = i + 1; j > 1; j--)
t[j] = t[j] + t[j - 1];
System.out.printf("%d ", t[i + 1] - t[i]);
}
}
}
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
JavaScript[edit]
ES5[edit]
Iteration
var n = 15;
for (var t = [0, 1], i = 1; i <= n; i++) {
for (var j = i; j > 1; j--) t[j] += t[j - 1];
t[i + 1] = t[i];
for (var j = i + 1; j > 1; j--) t[j] += t[j - 1];
document.write(i == 1 ? '' : ', ', t[i + 1] - t[i]);
}
- Output:
1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845
ES6[edit]
Functional composition
(() => {
'use strict';
// CATALAN
// catalanSeries :: Int -> [Int]
let catalanSeries = n => {
let alternate = xs => xs.reduce(
(a, x, i) => i % 2 === 0 ? a.concat([x]) : a, []
),
diff = xs => xs.length > 1 ? xs[0] - xs[1] : xs[0];
return alternate(pascal(n * 2))
.map((xs, i) => diff(drop(i, xs)));
}
// PASCAL
// pascal :: Int -> [[Int]]
let pascal = n => until(
m => m.level <= 1,
m => {
let nxt = zipWith(
(a, b) => a + b, [0].concat(m.row), m.row.concat(0)
);
return {
row: nxt,
triangle: m.triangle.concat([nxt]),
level: m.level - 1
}
}, {
level: n,
row: [1],
triangle: [
[1]
]
}
)
.triangle;
// GENERIC FUNCTIONS
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
let zipWith = (f, xs, ys) =>
xs.length === ys.length ? (
xs.map((x, i) => f(x, ys[i]))
) : undefined;
// until :: (a -> Bool) -> (a -> a) -> a -> a
let until = (p, f, x) => {
let v = x;
while (!p(v)) v = f(v);
return v;
}
// drop :: Int -> [a] -> [a]
let drop = (n, xs) => xs.slice(n);
// tail :: [a] -> [a]
let tail = xs => xs.length ? xs.slice(1) : undefined;
return tail(catalanSeries(16));
})();
- Output:
[1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845]
jq[edit]
The first identity (C(2n,n) - C(2n, n-1)) given in the reference is used in accordance with the task description, but it would of course be more efficient to factor out C(2n,n) and use the expression C(2n,n)/(n+1). See also Catalan_numbers#jq for other alternatives.
Warning: jq uses IEEE 754 64-bit arithmetic, so the algorithm used here for Catalan numbers loses precision for n > 30 and fails completely for n > 510.
def binomial(n; k):
if k > n / 2 then binomial(n; n-k)
else reduce range(1; k+1) as $i (1; . * (n - $i + 1) / $i)
end;
# Direct (naive) computation using two numbers in Pascal's triangle:
def catalan_by_pascal: . as $n | binomial(2*$n; $n) - binomial(2*$n; $n-1);
Example:
(range(0;16), 30, 31, 510, 511) | [., catalan_by_pascal]
- Output:
$ jq -n -c -f Catalan_numbers_Pascal.jq
[0,0]
[1,1]
[2,2]
[3,5]
[4,14]
[5,42]
[6,132]
[7,429]
[8,1430]
[9,4862]
[10,16796]
[11,58786]
[12,208012]
[13,742900]
[14,2674440]
[15,9694845]
[30,3814986502092304]
[31,14544636039226880]
[510,5.491717746183512e+302]
[511,null]
Lua[edit]
For each line of odd-numbered length from Pascal's triangle, print the middle number minus the one immediately to its right. This solution is heavily based on the Lua code to generate Pascal's triangle from the page for that task.
function nextrow (t)
local ret = {}
t[0], t[#t + 1] = 0, 0
for i = 1, #t do ret[i] = t[i - 1] + t[i] end
return ret
end
function catalans (n)
local t, middle = {1}
for i = 1, n do
middle = math.ceil(#t / 2)
io.write(t[middle] - (t[middle + 1] or 0) .. " ")
t = nextrow(nextrow(t))
end
end
catalans(15)
- Output:
1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440
Mathematica / Wolfram Language[edit]
This builds the entire Pascal triangle that's needed and holds it in memory. Very inefficienct, but seems to be what is asked in the problem.
nextrow[lastrow_] := Module[{output},
output = ConstantArray[1, Length[lastrow] + 1];
Do[
output[[i + 1]] = lastrow[[i]] + lastrow[[i + 1]];
, {i, 1, Length[lastrow] - 1}];
output
]
pascaltriangle[size_] := NestList[nextrow, {1}, size]
catalannumbers[length_] := Module[{output, basetriangle},
basetriangle = pascaltriangle[2 length];
list1 = basetriangle[[# *2 + 1, # + 1]] & /@ Range[length];
list2 = basetriangle[[# *2 + 1, # + 2]] & /@ Range[length];
list1 - list2
]
(* testing *)
catalannumbers[15]
- Output:
{1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845}
MATLAB / Octave[edit]
n = 15;
p = pascal(n + 2);
p(n + 4 : n + 3 : end - 1)' - diag(p, 2)
- Output:
ans =
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
Nim[edit]
const n = 15
var t = newSeq[int](n + 2)
t[1] = 1
for i in 1..n:
for j in countdown(i, 1): t[j] += t[j-1]
t[i+1] = t[i]
for j in countdown(i+1, 1): t[j] += t[j-1]
stdout.write t[i+1] - t[i], " "
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
OCaml[edit]
let catalan : int ref = ref 0 in Printf.printf "%d ," 1 ; for i = 2 to 9 do let nm : int ref = ref 1 in let den : int ref = ref 1 in for k = 2 to i do nm := (!nm)*(i+k); den := (!den)*k; catalan := (!nm)/(!den) ; done; print_int (!catalan); print_string "," ; done;; </lang>
- Output:
OUTPUT: 1 ,2,5,14,42,132,429,1430,4862
Oforth[edit]
: pascal(n) [ 1 ] #[ dup 0 + 0 rot + zipWith(#+) ] times(n) ;
: catalan(n) pascal(n 2 * ) at(n 1+) n 1+ / ;
- Output:
>15 seq map(#catalan) . [1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
PARI/GP[edit]
vector(15,n,binomial(2*n,n)-binomial(2*n,n+1))
- Output:
%1 = [1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
Pascal[edit]
type
tElement = Uint64;
var
Catalan : array[0..50] of tElement;
procedure GetCatalan(L:longint);
var
PasTri : array[0..100] of tElement;
j,k: longInt;
begin
l := l*2;
PasTri[0] := 1;
j := 0;
while (j<L) do
begin
inc(j);
k := (j+1) div 2;
PasTri[k] :=PasTri[k-1];
For k := k downto 1 do
inc(PasTri[k],PasTri[k-1]);
IF NOT(Odd(j)) then
begin
k := j div 2;
Catalan[k] :=PasTri[k]-PasTri[k-1];
end;
end;
end;
var
i,l: longint;
Begin
l := 15;
GetCatalan(L);
For i := 1 to L do
Writeln(i:3,Catalan[i]:20);
end.
1 1 2 2 3 5 4 14 5 42 6 132 7 429 8 1430 9 4862 10 16796 11 58786 12 208012 13 742900 14 2674440 15 9694845
Perl[edit]
use constant N => 15;
my @t = (0, 1);
for(my $i = 1; $i <= N; $i++) {
for(my $j = $i; $j > 1; $j--) { $t[$j] += $t[$j-1] }
$t[$i+1] = $t[$i];
for(my $j = $i+1; $j>1; $j--) { $t[$j] += $t[$j-1] }
print $t[$i+1] - $t[$i], " ";
}
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
After the 28th Catalan number, this overflows 64-bit integers. We could add use bigint; use Math::GMP ":constant"; to make it work, albeit not at a fast pace. However we can use a module to do it much faster:
use ntheory qw/binomial/;
print join(" ", map { binomial( 2*$_, $_) / ($_+1) } 1 .. 1000), "\n";
The Math::Pari module also has a binomial, but isn't as fast and overflows its stack after 3400.
Perl 6[edit]
constant @pascal = [1], -> @p { [0, |@p Z+ |@p, 0] } ... *;
constant @catalan = gather for 2, 4 ... * -> $ix {
my @row := @pascal[$ix];
my $mid = +@row div 2;
take [-] @row[$mid, $mid+1]
}
.say for @catalan[^20];
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845 35357670 129644790 477638700 1767263190 6564120420
Phix[edit]
Calculates the minimum pascal triangle in minimum memory. Inspired by the comments in, but not the code of the FreeBasic example
constant N = 15 -- accurate to 30, nan/inf for anything over 514 (bigatom version is below).
sequence catalan = {}, -- (>=1 only)
p = repeat(1,N+1)
atom p1
for i=1 to N do
p1 = p[1]*2
catalan = append(catalan,p1-p[2])
for j=1 to N-i+1 do
p1 += p[j+1]
p[j] = p1
end for
-- ?p[1..N-i+1]
end for
?catalan
- Output:
{1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845}
Explanatory comments to accompany the above
-- FreeBASIC said:
--' 1 1 1 1 1 1
--' 1 2 3 4 5 6
--' 1 3 6 10 15 21
--' 1 4 10 20 35 56
--' 1 5 15 35 70 126
--' 1 6 21 56 126 252
--' The Pascal triangle is rotated 45 deg.
--' to find the Catalan number we need to follow the diagonal
--' for top left to bottom right
--' take the number on diagonal and subtract the number in de cell
--' one up and one to right
--' 1 (2 - 1), 2 (6 - 4), 5 (20 - 15) ...
--
-- The first thing that struck me was it is twice as big as it needs to be,
-- something like this would do...
-- 1 1 1 1 1 1
-- 2 3 4 5 6
-- 6 10 15 21
-- 20 35 56
-- 70 126
-- 252
-- It is more obvious from the upper square that the diagonal on that, which is
-- that same as column 1 on this, is twice the previous, which on the second
-- diagram is in column 2. Further, once we have calculated the value for column
-- one above, we can use it immediately to calculate the next catalan number and
-- do not need to store it. Lastly we can overwrite row 1 with row 2 etc in situ,
-- and the following shows what we need for subsequent rounds:
-- 1 1 1 1 1
-- 3 4 5 6
-- 10 15 21
-- 35 56
-- 126 (unused)
The following bigatom version is over ten times faster than the equivalent on Catalan_numbers
include builtins\bigatom.e
function catalanB(integer n) -- very very fast!
sequence catalan = {},
p = repeat(1,n+1)
bigatom p1
if n=0 then return 1 end if
for i=1 to n do
p1 = ba_multiply(p[1],2)
catalan = append(catalan,ba_sub(p1,p[2]))
for j=1 to n-i+1 do
p1 = ba_add(p1,p[j+1])
p[j] = p1
end for
end for
return catalan[n]
end function
atom t0 = time()
string sc100 = ba_sprint(catalanB(100))
printf(1,"%d: %s (%3.2fs)\n",{100,sc100,time()-t0})
atom t0 = time()
string sc250 = ba_sprint(catalanB(250))
printf(1,"%d: %s (%3.2fs)\n",{250,sc250,time()-t0})
- Output:
100: 896519947090131496687170070074100632420837521538745909320 (0.08s) 250: 465116795969233796497747947259667807407291160080922096111953326525143875193659257831340309862635877995262413955019878805418475969029457769094808256 (1.01s)
PicoLisp[edit]
(de bino (N K)
(let f
'((N)
(if (=0 N) 1 (apply * (range 1 N))) )
(/
(f N)
(* (f (- N K)) (f K)) ) ) )
(for N 15
(println
(-
(bino (* 2 N) N)
(bino (* 2 N) (inc N)) ) ) )
(bye)
Python[edit]
>>> n = 15
>>> t = [0] * (n + 2)
>>> t[1] = 1
>>> for i in range(1, n + 1):
for j in range(i, 1, -1): t[j] += t[j - 1]
t[i + 1] = t[i]
for j in range(i + 1, 1, -1): t[j] += t[j - 1]
print(t[i+1] - t[i], end=' ')
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
>>>
def catalan_number(n):
nm = dm = 1
for k in range(2, n+1):
nm, dm = ( nm*(n+k), dm*k )
return nm/dm
print [catalan_number(n) for n in range(1, 16)]
[1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
Racket[edit]
#lang racket
(define (next-half-row r)
(define r1 (for/list ([x r] [y (cdr r)]) (+ x y)))
`(,(* 2 (car r1)) ,@(for/list ([x r1] [y (cdr r1)]) (+ x y)) 1 0))
(let loop ([n 15] [r '(1 0)])
(cons (- (car r) (cadr r))
(if (zero? n) '() (loop (sub1 n) (next-half-row r)))))
;; -> '(1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900
;; 2674440 9694845)
REXX[edit]
explicit subscripts[edit]
All of the REXX program examples can handle arbitrary large numbers.
/*REXX program obtains and displays Catalan numbers from a Pascal's triangle. */
parse arg N . /*Obtain the optional argument from CL.*/
if N=='' | N=="," then N=15 /*Not specified? Then use the default.*/
numeric digits max(9, N%2 + N%8) /*so we can handle huge Catalan numbers*/
@.=0; @.1=1 /*stem array default; define 1st value.*/
do i=1 for N; ip=i+1
do j=i by -1 for N; jm=j-1; @.j=@.j+@.jm; end /*j*/
@.ip=@.i; do k=ip by -1 for N; km=k-1; @.k=@.k+@.km; end /*k*/
say @.ip - @.i /*display the Ith Catalan number. */
end /*i*/ /*stick a fork in it, we're all done. */
output when using the default input:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
implicit subscripts[edit]
/*REXX program obtains and displays Catalan numbers from a Pascal's triangle. */
parse arg N . /*Obtain the optional argument from CL.*/
if N=='' | N=="," then N=15 /*Not specified? Then use the default.*/
numeric digits max(9, N%2 + N%8) /*so we can handle huge Catalan numbers*/
@.=0; @.1=1 /*stem array default; define 1st value.*/
do i=1 for N; ip=i+1
do j=i by -1 for N; @.j=@.j+@(j-1); end /*j*/
@.ip=@.i; do k=ip by -1 for N; @.k=@.k+@(k-1); end /*k*/
say @.ip - @.i /*display the Ith Catalan number. */
end /*i*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
@: parse arg !; return @.! /*return the value of @.[arg(1)] */
output is the same as the 1st version.
using binomial coefficients[edit]
/*REXX program obtains and displays Catalan numbers from a Pascal's triangle. */
parse arg N . /*Obtain the optional argument from CL.*/
if N=='' | N=="," then N=15 /*Not specified? Then use the default.*/
numeric digits max(9, N%2 + N%8) /*so we can handle huge Catalan numbers*/
do j=1 for N /* [↓] display N Catalan numbers. */
say comb(j+j, j) % (j+1) /*display the Jth Catalan number. */
end /*j*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
!: procedure; parse arg z; _=1; do j=1 for arg(1); _=_*j; end; return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
comb: procedure; parse arg x,y; if x=y then return 1; if y>x then return 0
if x-y<y then y=x-y; _=1; do j=x-y+1 to x; _=_*j; end; return _/!(y)
output is the same as the 1st version.
binomial coefficients, memoized[edit]
This REXX version uses memoization for the calculation of factorials.
/*REXX program obtains and displays Catalan numbers from a Pascal's triangle. */
parse arg N . /*Obtain the optional argument from CL.*/
if N=='' | N=="," then N=15 /*Not specified? Then use the default.*/
numeric digits max(9, N%2 + N%8) /*so we can handle huge Catalan numbers*/
!.=.
do j=1 for N /* [↓] display N Catalan numbers. */
say comb(j+j, j) % (j+1) /*display the Jth Catalan number. */
end /*j*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
!: procedure expose !.; parse arg z; if !.z\==. then return !.z; _=1
do j=1 for arg(1); _=_*j; end; !.z=_; return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
comb: procedure expose !.; parse arg x,y; if x=y then return 1; if y>x then return 0
if x-y<y then y=x-y; _=1; do j=x-y+1 to x; _=_*j; end; return _/!(y)
output is the same as the 1st version.
Ring[edit]
n=15
cat = list(n+2)
cat[1]=1
for i=1 to n
for j=i+1 to 2 step -1
cat[j]=cat[j]+cat[j-1]
next
cat[i+1]=cat[i]
for j=i+2 to 2 step -1
cat[j]=cat[j]+cat[j-1]
next
see "" + (cat[i+1]-cat[i]) + " "
next
Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Ruby[edit]
def catalan(num)
t = [0, 1] #grows as needed
(1..num).map do |i|
i.downto(1){|j| t[j] += t[j-1]}
t[i+1] = t[i]
(i+1).downto(1) {|j| t[j] += t[j-1]}
t[i+1] - t[i]
end
end
p catalan(15)
- Output:
[1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
Run BASIC[edit]
n = 15
dim t(n+2)
t(1) = 1
for i = 1 to n
for j = i to 1 step -1 : t(j) = t(j) + t(j-1): next j
t(i+1) = t(i)
for j = i+1 to 1 step -1: t(j) = t(j) + t(j-1 : next j
print t(i+1) - t(i);" ";
next i
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Rust[edit]
fn main()
{let n=15usize;
let mut t= [0; 17];
t[1]=1;
let mut j:usize;
for i in 1..n+1
{
j=i;
loop{
if j==1{
break;
}
t[j]=t[j] + t[j-1];
j=j-1;
}
t[i+1]= t[i];
j=i+1;
loop{
if j==1{
break;
}
t[j]=t[j] + t[j-1];
j=j-1;
}
print!("{} ", t[i+1]-t[i]);
}
}
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Scilab[edit]
n=15
t=zeros(1,n+2)
t(1)=1
for i=1:n
for j=i+1:-1:2
t(j)=t(j)+t(j-1)
end
t(i+1)=t(i)
for j=i+2:-1:2
t(j)=t(j)+t(j-1)
end
disp(t(i+1)-t(i))
end
- Output:
1.
2.
5.
14.
42.
132.
429.
1430.
4862.
16796.
58786.
208012.
742900.
2674440.
9694845.
Seed7[edit]
$ include "seed7_05.s7i";
const proc: main is func
local
const integer: N is 15;
var array integer: t is [] (1) & N times 0;
var integer: i is 0;
var integer: j is 0;
begin
for i range 1 to N do
for j range i downto 2 do
t[j] +:= t[j - 1];
end for;
t[i + 1] := t[i];
for j range i + 1 downto 2 do
t[j] +:= t[j - 1];
end for;
write(t[i + 1] - t[i] <& " ");
end for;
writeln;
end func;
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Sidef[edit]
func catalan(num) {
var t = [0, 1]
(1..num).map { |i|
flip(^i ).each {|j| t[j+1] += t[j] }
t[i+1] = t[i]
flip(^i.inc).each {|j| t[j+1] += t[j] }
t[i+1] - t[i]
}
}
say catalan(15).join(' ')
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
smart BASIC[edit]
PRINT "Catalan Numbers from Pascal's Triangle"!PRINT
x = 15
DIM t(x+2)
t(1) = 1
FOR n = 1 TO x
FOR m = n TO 1 STEP -1
t(m) = t(m) + t(m-1)
NEXT m
t(n+1) = t(n)
FOR m = n+1 TO 1 STEP -1
t(m) = t(m) + t(m-1)
NEXT m
PRINT n,"#######":t(n+1) - t(n)
NEXT n
Tcl[edit]
proc catalan n {
set result {}
array set t {0 0 1 1}
for {set i 1} {[set k $i] <= $n} {incr i} {
for {set j $i} {$j > 1} {} {incr t($j) $t([incr j -1])}
set t([incr k]) $t($i)
for {set j $k} {$j > 1} {} {incr t($j) $t([incr j -1])}
lappend result [expr {$t($k) - $t($i)}]
}
return $result
}
puts [catalan 15]
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
TI-83 BASIC[edit]
"CATALAN
15→N
seq(0,I,1,N+2)→L1
1→L1(1)
For(I,1,N)
For(J,I+1,2,-1)
L1(J)+L1(J-1)→L1(J)
End
L1(I)→L1(I+1)
For(J,I+2,2,-1)
L1(J)+L1(J-1)→L1(J)
End
Disp L1(I+1)-L1(I)
End
- Output:
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
Done
VBScript[edit]
To run in console mode with cscript.
dim t()
if Wscript.arguments.count=1 then
n=Wscript.arguments.item(0)
else
n=15
end if
redim t(n+1)
't(*)=0
t(1)=1
for i=1 to n
ip=i+1
for j = i to 1 step -1
t(j)=t(j)+t(j-1)
next 'j
t(i+1)=t(i)
for j = i+1 to 1 step -1
t(j)=t(j)+t(j-1)
next 'j
Wscript.echo t(i+1)-t(i)
next 'i
Visual Basic[edit]
Sub catalan()
Const n = 15
Dim t(n + 2) As Long
Dim i As Integer, j As Integer
t(1) = 1
For i = 1 To n
For j = i + 1 To 2 Step -1
t(j) = t(j) + t(j - 1)
Next j
t(i + 1) = t(i)
For j = i + 2 To 2 Step -1
t(j) = t(j) + t(j - 1)
Next j
Debug.Print i, t(i + 1) - t(i)
Next i
End Sub 'catalan
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
zkl[edit]
using binomial coefficients.fcn binomial(n,k){ (1).reduce(k,fcn(p,i,n){ p*(n-i+1)/i },1,n) }
(1).pump(15,List,fcn(n){ binomial(2*n,n)-binomial(2*n,n+1) })
- Output:
L(1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845)
ZX Spectrum Basic[edit]
10 LET N=15
20 DIM t(N+2)
30 LET t(2)=1
40 FOR i=2 TO N+1
50 FOR j=i TO 2 STEP -1: LET t(j)=t(j)+t(j-1): NEXT j
60 LET t(i+1)=t(i)
70 FOR j=i+1 TO 2 STEP -1: LET t(j)=t(j)+t(j-1): NEXT j
80 PRINT t(i+1)-t(i);" ";
90 NEXT i
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