Catalan numbers/Pascal's triangle
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.
- Sequence A000108 on OEIS has a lot of information on Catalan Numbers.
- Related Tasks
11l
<lang 11l>V n = 15 V t = [0] * (n + 2) t[1] = 1 L(i) 1 .. n
L(j) (i .< 1).step(-1)
t[j] += t[j - 1]
t[i + 1] = t[i]
L(j) (i + 1 .< 1).step(-1)
t[j] += t[j - 1]
print(t[i + 1] - t[i], end' ‘ ’)</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
360 Assembly
For maximum compatibility, this program uses only the basic instruction set. <lang 360asm>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</lang>
- Output:
00000001 00000002 00000005 00000014 00000042 00000132 00000429 00001430 00004862 00016796 00058786 00208012 00742900 02674440 09694845
Ada
Uses package Pascal from the Pascal triangle solution[[1]]
<lang Ada>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;</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
ALGOL 68
<lang algol68>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</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
ALGOL W
<lang algolw>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.</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
APL
<lang apl>
⍝ Based heavily on the J solution
CATALAN←{¯1↓↑-/1 ¯1↓¨(⊂⎕IO+0 0)⍉¨0 2⌽¨⊂(⎕IO-⍨⍳N){+\⍣⍺⊢⍵}⍤0 1⊢1⍴⍨N←⍵+2}
</lang>
- Output:
CATALAN 15 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
AutoHotkey
<lang AutoHotkey>/* 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</lang>
- Produces:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
AWK
<lang AWK>
- syntax: GAWK -f CATALAN_NUMBERS_PASCALS_TRIANGLE.AWK
- converted from C
BEGIN {
printf("1")
for (n=2; n<=15; n++) {
num = den = 1
for (k=2; k<=n; k++) {
num *= (n + k)
den *= k
catalan = num / den
}
printf(" %d",catalan)
}
printf("\n")
exit(0)
} </lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Batch File
<lang dos>@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</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
C
<lang c> //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 from 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;
} </lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
C#
<lang csharp> 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]));
} </lang>
- Produces:
1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845
C++
<lang cpp>// 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;
}</lang>
- Produces:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Common Lisp
<lang Lisp>(defun catalan (n)
"Return the n-th Catalan number"
(if (<= n 1) 1
(let ((result 2))
(dotimes (k (- n 2) result)
(setq result (* result (/ (+ n k 2) (+ k 2)))) ))))
(dotimes (n 15)
(print (catalan (1+ n))) )</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
D
<lang d>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], ' ');
}
}</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Delphi
See Pascal.
EchoLisp
<lang scheme> (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)) </lang>
- Output:
<lang scheme>
- 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
</lang>
EDSAC order code
Rather than store a triangular array, this solution stores the right-hand half of the current row and updates it in place. It uses the third method in the link, e.g. once we have the half row (70, 56, 28, 8, 1), the Catalan number 42 appears as 70 - 28. <lang edsac>
[Catalan numbers from Pascal triangle, Rosetta Code website.
EDSAC program, Initial Orders 2]
..PZ [blank tape and terminator]
T54K [refer to working array with 'C']
P300F [address of working array]
T46K [to call print subroutine with 'G N']
P56F [address of print subroutine]
[Modification of library subroutine P7.
Prints non-negative integer, up to 10 digits, right-justified.
55 locations, load at even address.]
E25KTN
GKA3FT42@A47@T31@ADE10@T31@A48@T31@SDTDH44#@NDYFLDT4DS43@
TFH17@S17@A43@G23@UFS43@T1FV4DAFG50@SFLDUFXFOFFFSFL4FT4DA49@
T31@A1FA43@G20@XFP1024FP610D@524D!FO46@O26@XFO46@SFL8FT4DE39@
[Main program]
PK T200K GK
[Constants]
[0] PD [short constant 1]
[1] P2F [to inc address by 2]
[2] T#C [used in manufacturing EDSAC orders]
[3] MF [add to T order to make A order with same address]
[4] #F [set figures]
[5] &F [line feed]
[6] @F [carriage return]
[7] P7D [maximum n = 15]
[Variable]
[8] PF [n]
[Enter with acc = 0]
[9] O4@ [set teleprinter to figures]
T4#C T2#C T#C A@ TC [initialize first 3 terms to 1, 0, 0]
T8@ E58@ [set n := 0; jump to inc n and print C_n]
[Outer loop; here with n updated]
[17] TF A8@ [acc := latest n]
L1F A2@ T22@ [make and store order 'T 2n #C']
[22] T#C [sets term := 0; also used to test for end of loop]
A2@ [load 'T#C', initial value of order 31]
[Loop to convert e.g. (20, 15, 6, 1) to (35, 21, 7, 1); works left to right]
[24] U31@ A3@ U29@ A1@ T30@ [set up orders on next line]
[29] A#C A#C T#C [replaced by manufactured orders]
A31@ A1@ S22@ E38@ [inc address in order 31, jump out if done]
A22@ E24@ [not done, loop back]
[38] A22@ T48@ [initialize order 48]
[Loop to convert e.g. (35, 21, 7, 1) to (70, 56, 28, 8, 1); works right to left]
[40] TF A48@ A3@ U46@ S1@ T47@ [set up orders on next line]
[46] A#C A#C T#C [replaced by manufactured orders]
A48@ S1@ T48@ [dec address in order 48]
A2@ S48@ G40@ [test for done, loop back if not]
A#C LD T#C [double first term, e.g. 35 -> 70 (not done in loop)]
[Increment n and print Catalan number C_n]
[58] TD [clear 0D, ensures sandwich bit = 0]
A8@ A@ U8@ TF [inc n; set 0D := n by setting 0F := n]
A63@ GN [print n]
A#C S4#C TD A68@ GN [print Catalan number C_n, e.g. C_5 = 70 - 28 = 42]
O6@ O5@ [print CR, LF]
A8@ S7@ G17@ [test for maximum n, loop back if not]
[75] O4@ ZF [flush printer buffer; stop]
E9Z PF [define entry point; enter with acc = 0]
</lang>
- 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
Elixir
<lang elixir>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)</lang>
- Output:
[1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
Erlang
<lang erlang> -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). </lang>
ERRE
<lang ERRE> 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 </lang>
- 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#
<lang fsharp> 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 ", "
</lang>
Factor
<lang factor>USING: arrays grouping io kernel math prettyprint sequences ; IN: rosetta-code.catalan-pascal
- next-row ( seq -- seq' )
2 clump [ sum ] map 1 prefix 1 suffix ;
- pascal ( n -- seq )
1 - { { 1 } } swap [ dup last next-row suffix ] times ;
15 2 * pascal [ length odd? ] filter [
dup length 1 = [ 1 ] [ dup midpoint@ dup 1 + 2array swap nths first2 - ] if pprint bl
] each drop</lang>
- Output:
1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440
FreeBASIC
<lang freebasic>' 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</lang>
- 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
Go
<lang go>package main
import "fmt"
func main() {
const n = 15
t := [n + 2]uint64{0, 1}
for i := 1; i <= n; i++ {
for j := i; j > 1; j-- {
t[j] += t[j-1]
}
t[i+1] = t[i]
for j := i + 1; j > 1; j-- {
t[j] += t[j-1]
}
fmt.Printf("%2d : %d\n", i, t[i+1]-t[i])
}
}</lang>
- 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
Groovy
<lang Groovy> 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);
}
}
} </lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Haskell
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. <lang haskell>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</lang>
- Output:
./catalan 15 [1,1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440]
Icon and Unicon
The following works in both languages. It avoids computing elements in Pascal's triangle that aren't used.
<lang unicon>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</lang>
Sample run:
->cn
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
->
J
<lang j> Catalan=. }:@:(}.@:((<0 1)&|:) - }:@:((<0 1)&|:@:(2&|.)))@:(i. +/\@]^:[ #&1)@:(2&+)</lang>
- Example use:
<lang j> Catalan 15 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
</lang>
A structured derivation of Catalan follows:
<lang j> 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&+)</lang>
Java
<lang java>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]);
}
}
}</lang>
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
JavaScript
ES5
Iteration
<lang javascript>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]);
}</lang>
- Output:
1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845
ES6
Functional composition
<lang JavaScript>(() => {
'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));
})();</lang>
- Output:
<lang JavaScript>[1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845]</lang>
jq
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. <lang jq>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; $n-1);</lang>
Example:
(range(0;16), 30, 31, 510, 511) | [., catalan_by_pascal]
- Output:
<lang sh>$ 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]</lang>
Julia
<lang julia># v0.6
function pascal(n::Int)
r = ones(Int, n, n)
for i in 2:n, j in 2:n
r[i, j] = r[i-1, j] + r[i, j-1]
end
return r
end
function catalan_num(n::Int)
p = pascal(n + 2) p[n+4:n+3:end-1] - diag(p, 2)
end
@show catalan_num(15)</lang>
- Output:
catalan_num(15) = [1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
Kotlin
<lang scala>// version 1.1.2
import java.math.BigInteger
val ONE = BigInteger.ONE
fun pascal(n: Int, k: Int): BigInteger {
if (n == 0 || k == 0) return ONE
val num = (k + 1..n).fold(ONE) { acc, i -> acc * BigInteger.valueOf(i.toLong()) }
val den = (2..n - k).fold(ONE) { acc, i -> acc * BigInteger.valueOf(i.toLong()) }
return num / den
}
fun catalanFromPascal(n: Int) {
for (i in 1 until n step 2) {
val mi = i / 2 + 1
val catalan = pascal(i, mi) - pascal(i, mi - 2)
println("${"%2d".format(mi)} : $catalan")
}
}
fun main(args: Array<String>) {
val n = 15 catalanFromPascal(n * 2)
}</lang>
- Output:
1 : 1 2 : 2 3 : 5 4 : 14 5 : 42 6 : 132 7 : 429 8 : 1430 9 : 4862 10 : 16796 11 : 58786 12 : 208012 13 : 742900 14 : 2674440 15 : 9694845
Lua
For each line of odd-numbered length from Pascal's triangle, print the middle number minus the one immediately to its right. This solution is heavily based on the Lua code to generate Pascal's triangle from the page for that task. <lang Lua>function nextrow (t)
local ret = {}
t[0], t[#t + 1] = 0, 0
for i = 1, #t do ret[i] = t[i - 1] + t[i] end
return ret
end
function catalans (n)
local t, middle = {1}
for i = 1, n do
middle = math.ceil(#t / 2)
io.write(t[middle] - (t[middle + 1] or 0) .. " ")
t = nextrow(nextrow(t))
end
end
catalans(15)</lang>
- Output:
1 1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440
M2000 Interpreter
We have to add -1 in For x=2 to rows, because in FreeBasic when x=rows then inner loop never happen because end value for y is 1, so lower than start value 2. In M2000 this should run from 2 to 1, so we have to exclude this situation from outer loop, adding -1, and before loop we have to include en exit from sub if rows are less than 2.
We can define integer variables (16 bit), and we can use integer literals numbers using % as last char.
Inside triangle array we use decimal numbers, using @ for first literals, so all additions next produce decimals too.
We use & to pass by reference, here anarray, to sub, but because a sub can see anything in module we can change array name inside sub to same as triangle and we can remove arguments (including size).
<lang M2000 Interpreter> Module CatalanNumbers {
Def Integer count, t_row, size=31
Dim triangle(1 to size, 1 to size)
\\ call sub
pascal_triangle(size, &triangle())
Print "The first 15 Catalan numbers are"
count = 1% : t_row = 2%
Do {
Print Format$("{0:0:-3}:{1:0:-15}", count, triangle(t_row, t_row) - triangle(t_row +1, t_row -1))
t_row++
count++
} Until count > 15
End
Sub pascal_triangle(rows As Integer, &Pas_tri())
Local x=0%, y=0%
For x = 1 To rows
Pas_tri( 1%, x ) = 1@
Pas_tri( x, 1% ) = 1@
Next x
if rows<2 then exit sub
For x = 2 To rows-1
For y = 2 To rows + 1 - x
Pas_tri(x, y) = pas_tri(x - 1 , y) + pas_tri(x, y - 1)
Next y
Next x
End Sub
} CatalanNumbers </lang>
- 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
Maple
<lang maple>catalan:=proc(n)
local i,a:=[1],C:=[1]; for i to n do a:=[0,op(a)]+[op(a),0]; a:=[0,op(a)]+[op(a),0]; C:=[op(C),a[i+1]-a[i]]; od; C
end:
catalan(10);
- [1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796]</lang>
Mathematica / Wolfram Language
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. <lang Mathematica>nextrow[lastrow_] := Module[{output},
output = ConstantArray[1, Length[lastrow] + 1]; Do[ outputi + 1 = lastrowi + lastrowi + 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]</lang>
- Output:
{1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845}
MATLAB / Octave
<lang MATLAB>n = 15; p = pascal(n + 2); p(n + 4 : n + 3 : end - 1)' - diag(p, 2)</lang>
- Output:
ans =
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
Nim
<lang nim>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], " "
stdout.write '\n'</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
OCaml
<lang ocaml> 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
<lang Oforth>import: mapping
- pascal( n -- [] )
[ 1 ] n #[ dup [ 0 ] + [ 0 ] rot + zipWith( #+ ) ] times ;
- catalan( n -- m )
n 2 * pascal at( n 1+ ) n 1+ / ;</lang>
- Output:
>#catalan 15 seq map . [1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
PARI/GP
<lang parigp>vector(15,n,binomial(2*n,n)-binomial(2*n,n+1))</lang>
- Output:
%1 = [1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
Pascal
<lang pascal> Program CatalanNumbers 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.</lang>
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
<lang Perl>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], " ";
}</lang>
- 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:
<lang Perl>use ntheory qw/binomial/; print join(" ", map { binomial( 2*$_, $_) / ($_+1) } 1 .. 1000), "\n";</lang>
The Math::Pari module also has a binomial, but isn't as fast and overflows its stack after 3400.
Phix
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 (gmp 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 <lang Phix>-- 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)</lang>
gmp version
with javascript_semantics include builtins\mpfr.e function catalanB(integer n) -- very very fast! sequence catalan = mpz_inits(n), p = mpz_inits(n+1,1) mpz p1 = mpz_init(1) if n=0 then return p1 end if for i=1 to n do mpz_mul_si(p1,p[1],2) mpz_sub(catalan[i],p1,p[2]) for j=1 to n-i+1 do mpz_add(p1,p1,p[j+1]) mpz_set(p[j],p1) end for end for return catalan[n] end function printf(1,"%d: %s\n",{100,shorten(mpz_get_str(catalanB(100)))}) printf(1,"%d: %s\n",{250,shorten(mpz_get_str(catalanB(250)))})
- Output:
100: 89651994709013149668...20837521538745909320 (57 digits) 250: 46511679596923379649...69029457769094808256 (147 digits)
The above is significantly faster than the equivalent(s) on Catalan_numbers, a quick comparison showing the latter getting exponentially worse (then again I memoised the slowest recursive version):
800 2000 4000 8000 catalanB: 0.6s 3.5s 14.5s 64s catalan2m: 0.7s 7.0s 64.9s 644s
PicoLisp
<lang PicoLisp>(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)</lang>
PureBasic
<lang PureBasic>#MAXNUM = 15 Declare catalan()
If OpenConsole("Catalan numbers")
catalan() Input() End 0
Else
End -1
EndIf
Procedure catalan()
Define k.i, n.i, num.d, den.d, cat.d
Print("1 ")
For n=2 To #MAXNUM
num=1 : den =1
For k=2 To n
num * (n+k)
den * k
cat = num / den
Next
Print(Str(cat)+" ")
Next
EndProcedure</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Python
Procedural
<lang python>>>> 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
>>> </lang>
<lang python>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]</lang>
Composition of pure functions
Note that sequence A000108 on OEIS (referenced in the task description) confirms that the first four Catalan numbers are indeed 1, 1, 2, 5 ...
(Several scripts on this page appear to lose the first 1).
<lang python>Catalan numbers from Pascal's triangle
from itertools import (islice) from operator import (add)
- nCatalans :: Int -> [Int]
def nCatalans(n):
The first n Catalan numbers,
derived from Pascal's triangle.
# diff :: [Int] -> Int
def diff(xs):
Difference between the first two items in the list,
if its length is more than one.
Otherwise, the first (only) item in the list.
return (
xs[0] - (xs[1] if 1 < len(xs) else 0)
) if xs else None
return list(map(
compose(diff)(uncurry(drop)),
enumerate(map(fst, take(n)(
everyOther(
pascalTriangle()
)
)))
))
- pascalTriangle :: Gen Int
def pascalTriangle():
A non-finite stream of
Pascal's triangle rows.
return iterate(nextPascal)([1])
- nextPascal :: [Int] -> [Int]
def nextPascal(xs):
A row of Pascal's triangle
derived from a preceding row.
return zipWith(add)([0] + xs)(xs + [0])
- TEST ----------------------------------------------------
- main :: IO ()
def main():
First 16 Catalan numbers.
print(
nCatalans(16)
)
- GENERIC -------------------------------------------------
- compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
def compose(g):
Right to left function composition. return lambda f: lambda x: g(f(x))
- drop :: Int -> [a] -> [a]
- drop :: Int -> String -> String
def drop(n):
The sublist of xs beginning at
(zero-based) index n.
def go(xs):
if isinstance(xs, list):
return xs[n:]
else:
take(n)(xs)
return xs
return lambda xs: go(xs)
- everyOther :: Gen [a] -> Gen [a]
def everyOther(g):
Every other item of a generator stream.
while True:
yield take(1)(g)
take(1)(g) # Consumed, not yielded.
- fst :: (a, b) -> a
def fst(tpl):
First component of a pair. return tpl[0]
- iterate :: (a -> a) -> a -> Gen [a]
def iterate(f):
An infinite list of repeated applications of f to x.
def go(x):
v = x
while True:
yield v
v = f(v)
return lambda x: go(x)
- take :: Int -> [a] -> [a]
- take :: Int -> String -> String
def take(n):
The prefix of xs of length n,
or xs itself if n > length xs.
return lambda xs: (
xs[0:n]
if isinstance(xs, list)
else list(islice(xs, n))
)
- uncurry :: (a -> b -> c) -> ((a, b) -> c)
def uncurry(f):
A function over a tuple
derived from a curried function.
return lambda xy: f(xy[0])(
xy[1]
)
- zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
def zipWith(f):
A list constructed by zipping with a
custom function, rather than with the
default tuple constructor.
return lambda xs: lambda ys: (
list(map(f, xs, ys))
)
- MAIN ---
if __name__ == '__main__':
main()</lang>
- Output:
[1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
Racket
<lang Racket>
- 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)
</lang>
Raku
(formerly Perl 6)
<lang perl6>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];</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845 35357670 129644790 477638700 1767263190 6564120420
REXX
explicit subscripts
All of the REXX program examples can handle arbitrary large numbers. <lang rexx>/*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. */</lang>
output when using the default input:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
implicit subscripts
<lang rexx>/*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)] */</lang> output is the same as the 1st version.
using binomial coefficients
<lang rexx>/*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)</lang>
output is the same as the 1st version.
binomial coefficients, memoized
This REXX version uses memoization for the calculation of factorials. <lang rexx>/*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)</lang>
output is the same as the 1st version.
Ring
<lang ring> 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 </lang> Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Ruby
<lang tcl>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)</lang>
- Output:
[1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]
Run BASIC
<lang runbasic>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</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Rust
<lang rust>
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]);
}
} </lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Scala
<lang Scala>def catalan(n: Int): Int =
if (n <= 1) 1 else (0 until n).map(i => catalan(i) * catalan(n - i - 1)).sum
(1 to 15).map(catalan(_))</lang>
- Output:
See it in running in your browser by Scastie (JVM).
Scilab
<lang>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</lang>
- Output:
1.
2.
5.
14.
42.
132.
429.
1430.
4862.
16796.
58786.
208012.
742900.
2674440.
9694845.
Seed7
<lang seed7>$ 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;</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Sidef
<lang ruby>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(' ')</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
smart BASIC
<lang qbasic>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</lang>
Tcl
<lang tcl>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 $result
}
puts [catalan 15]</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
TI-83 BASIC
<lang ti83b>"CATALAN 15→N seq(0,I,1,N+2)→L1 1→L1(1) For(I,1,N) For(J,I+1,2,-1) L1(J)+L1(J-1)→L1(J) End L1(I)→L1(I+1) For(J,I+2,2,-1) L1(J)+L1(J-1)→L1(J) End Disp L1(I+1)-L1(I) End</lang>
- Output:
1
2
5
14
42
132
429
1430
4862
16796
58786
208012
742900
2674440
9694845
Done
Vala
<lang vala>void main() {
const int N = 15;
uint64[] t = {0, 1};
for (int i = 1; i <= N; i++) {
for (int j = i; j > 1; j--) t[j] = t[j] + t[j - 1];
t[i + 1] = t[i];
for (int j = i + 1; j > 1; j--) t[j] = t[j] + t[j - 1];
print(@"$(t[i + 1] - t[i]) ");
}
print("\n");
}</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
VBScript
To run in console mode with cscript. <lang vbscript>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</lang>
Visual Basic
<lang vb> 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 </lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Wren
<lang ecmascript>var n = 15 var t = List.filled(n+2, 0) t[1] = 1 for (i in 1..n) {
if (i > 1) for (j in i..2) t[j] = t[j] + t[j-1]
t[i+1] = t[i]
if (i > 0) for (j in i+1..2) t[j] = t[j] + t[j-1]
System.write("%(t[i+1]-t[i]) ")
} System.print()</lang>
- Output:
1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845
Zig
Uses comptime functionality to compile Pascal's triangle into the object code, then at runtime, it's simply a table lookup. (uses code from AKS primality task.) <lang Zig> const std = @import("std"); const stdout = std.io.getStdOut().outStream();
pub fn main() !void {
var n: u32 = 1;
while (n <= 15) : (n += 1) {
const row = binomial(n * 2).?;
try stdout.print("{d:2} {d:8}\n", .{ n, row[n] - row[n + 1] });
}
}
pub fn binomial(n: u32) ?[]const u64 {
if (n >= rmax)
return null
else {
const k = n * (n + 1) / 2;
return pascal[k .. k + n + 1];
}
}
const rmax = 68;
const pascal = build: {
@setEvalBranchQuota(100_000);
var coefficients: [(rmax * (rmax + 1)) / 2]u64 = undefined;
coefficients[0] = 1;
var j: u32 = 0;
var k: u32 = 1;
var n: u32 = 1;
while (n < rmax) : (n += 1) {
var prev = coefficients[j .. j + n];
var next = coefficients[k .. k + n + 1];
next[0] = 1;
var i: u32 = 1;
while (i < n) : (i += 1)
next[i] = prev[i] + prev[i - 1];
next[i] = 1;
j = k;
k += n + 1;
}
break :build coefficients;
}; </lang>
- 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
zkl
using binomial coefficients.
<lang zkl>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) })</lang>
- Output:
L(1,2,5,14,42,132,429,1430,4862,16796,58786,208012,742900,2674440,9694845)
ZX Spectrum Basic
<lang zxbasic>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</lang>
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