Catalan numbers/Pascal's triangle

Revision as of 01:51, 26 June 2014 by rosettacode>Craigd (Added zkl)

The task is to print out the first 15 Catalan numbers by extracting them from Pascal's triangle, see Catalan Numbers and the Pascal Triangle.

Task
Catalan numbers/Pascal's triangle
You are encouraged to solve this task according to the task description, using any language you may know.

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

AutoHotkey

Works with: AutoHotkey_L

<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

C++

<lang cpp>// Generate Catalan Numbers // // Nigel Galloway: June 9th., 2012 //

  1. 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

D

Translation of: C++

<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 

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>

Mathematica

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>p = pascal(17); diag(p(2:end-1,2:end-1))-diag(p,2)</lang> Output:

ans =
         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]

Perl

Translation of: C++

<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

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

Python

Translation of: C++

<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>

Works with: Python version 2.7

<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>

Racket

<lang Racket>

  1. 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>

REXX

<lang rexx>/*REXX program obtains Catalan numbers from Pascal's triangle. */ numeric digits 200 /*might have large Catalan nums. */ parse arg N .; if N== then N=15 /*Any args? No, then use default*/ @.=0; @.1=1 /*stem array default, 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
 end   /*i*
                                      /*stick a fork in it, we're done.*/</lang>

output when using the default input:

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 

Ruby

<lang tcl>def catalan(num)

 t = [0, 1] #grows as needed
 1.upto(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

puts catalan(15).join(", ")</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 

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

zkl

Translation of: PARI/GP

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)