Pascal's triangle: Difference between revisions

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Pascal's triangle is an arithmetic and geometric figure first imagined by Blaise Pascal.

Its first few rows look like this:

where each element of each row is either 1 or the sum of the two elements right above it.

For example, the next row would be 1 (since the first element of each row doesn't have two elements above it), 4 (1 + 3), 6 (3 + 3), 4 (3 + 1), and 1 (since the last element of each row doesn't have two elements above it).

Each row n (starting with row 0 at the top) shows the coefficients of the binomial expansion of (x + y)ⁿ.

Task

Write a function that prints out the first n rows of the triangle (with f(1) yielding the row consisting of only the element 1).

This can be done either by summing elements from the previous rows or using a binary coefficient or combination function.

Behavior for n ≤ 0 does not need to be uniform, but should be noted.

See also

Evaluate binomial coefficients

360 Assembly

Translation of: PL/I

<lang 360asm>* Pascal's triangle 25/10/2015 PASCAL CSECT

        USING  PASCAL,R15         set base register
        LA     R7,1               n=1

LOOPN C R7,=A(M) do n=1 to m

        BH     ELOOPN             if n>m then goto 
        MVC    U,=F'1'            u(1)=1
        LA     R8,PG              pgi=@pg
        LA     R6,1               i=1

LOOPI CR R6,R7 do i=1 to n

        BH     ELOOPI             if i>n then goto 
        LR     R1,R6              i
        SLA    R1,2               i*4
        L      R3,T-4(R1)         t(i)
        L      R4,T(R1)           t(i+1)
        AR     R3,R4              t(i)+t(i+1)
        ST     R3,U(R1)           u(i+1)=t(i)+t(i+1)
        LR     R1,R6              i
        SLA    R1,2               i*4
        L      R2,U-4(R1)         u(i)
        XDECO  R2,XD              edit u(i)
        MVC    0(4,R8),XD+8       output u(i):4
        LA     R8,4(R8)           pgi=pgi+4
        LA     R6,1(R6)           i=i+1
        B      LOOPI              end i

ELOOPI MVC T((M+1)*(L'T)),U t=u

        XPRNT  PG,80              print
        LA     R7,1(R7)           n=n+1
        B      LOOPN              end n

ELOOPN XR R15,R15 set return code

        BR     R14                return to caller

M EQU 11 <== input T DC (M+1)F'0' t(m+1) init 0 U DC (M+1)F'0' u(m+1) init 0 PG DC CL80' ' pg init ' ' XD DS CL12 temp

        YREGS
        END    PASCAL</lang>

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

8th

One way, using array operations: <lang forth> \ print the array

.arr \ a -- a

 ( . space ) a:each ;

pasc \ a --

 \ print the row
 .arr cr
 dup 
 \ create two rows from the first, one with a leading the other with a trailing 0
 [0] 0 a:insert swap 0 a:push
 \ add the arrays together to make the new one
 ' n:+ a:op ;

\ print the first 16 rows: [1] ' pasc 16 times </lang>

Another way, using the relation between element 'n' and element 'n-1' in a row: <lang forth>

ratio \ m n -- num denom

 tuck n:- n:1+ swap ;

\ one item in the row: n m

pascitem \ n m -- n

r@ swap ratio n:*/ n:round int dup . space ;

\ One row of Pascal's triangle

pascline \ n --

>r 1 int dup . space ' pascitem 1 r@ loop rdrop drop cr ;

\ Calculate the first 'n' rows of Pascal's triangle:

pasc \ n

' pascline 0 rot loop cr ;

15 pasc </lang>

Ada

The specification of auxiliary package "Pascal". "First_Row" outputs a row with a single "1", "Next_Row" computes the next row from a given row, and "Length" gives the number of entries in a row. The package is also used for the Catalan numbers solution [[1]]

<lang ada>package Pascal is

  type Row is array (Natural range <>) of Natural;
  
  function Length(R: Row) return Positive;
  
  function First_Row(Max_Length: Positive) return Row;
  
  function Next_Row(R: Row) return Row;

end Pascal;</lang>

The implementation of that auxiliary package "Pascal":

<lang Ada>package body Pascal is

  function First_Row(Max_Length: Positive) return Row is
     R: Row(0 .. Max_Length) := (0 | 1 => 1, others => 0);
  begin
     return R;
  end First_Row;
  
  function Next_Row(R: Row) return Row is
     S: Row(R'Range);
  begin
     S(0) := Length(R)+1;
     S(Length(S)) := 1;
     for J in reverse 2 .. Length(R) loop
        S(J) := R(J)+R(J-1);
     end loop;
     S(1) := 1;
     return S;
  end Next_Row;
  
  function Length(R: Row) return Positive is
  begin
     return R(0);
  end Length;

end Pascal;</lang>

The main program, using "Pascal". It prints the desired number of rows. The number is read from the command line.

<lang Ada>with Ada.Text_IO, Ada.Integer_Text_IO, Ada.Command_Line, Pascal; use Pascal;

procedure Triangle is

  Number_Of_Rows: Positive := Integer'Value(Ada.Command_Line.Argument(1));
  Row: Pascal.Row := First_Row(Number_Of_Rows);

begin

  loop
     -- print one row
     for J in 1 .. Length(Row) loop

Ada.Integer_Text_IO.Put(Row(J), 5);

     end loop;
     Ada.Text_IO.New_Line;
     exit when Length(Row) >= Number_Of_Rows;
     Row := Next_Row(Row);
  end loop;

end Triangle;</lang>

Output:

>./triangle 12
    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

ALGOL 68

<lang algol68>PRIO MINLWB = 8, MAXUPB = 8; OP MINLWB = ([]INT a,b)INT: (LWB a<LWB b|LWB a|LWB b),

  MAXUPB = ([]INT a,b)INT: (UPB a>UPB b|UPB a|UPB b);

OP + = ([]INT a,b)[]INT:(

 [a MINLWB b:a MAXUPB b]INT out; FOR i FROM LWB out TO UPB out DO out[i]:= 0 OD;
 out[LWB a:UPB a] := a; FOR i FROM LWB b TO UPB b DO out[i]+:= b[i] OD;
 out

);

INT width = 4, stop = 9; FORMAT centre = $n((stop-UPB row+1)*width OVER 2)(q)$;

FLEX[1]INT row := 1; # example of rowing # FOR i WHILE

 printf((centre, $g(-width)$, row, $l$));

WHILE # i < stop DO

 row := row[AT 1] + row[AT 2]

OD</lang>

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

APL

Pascal' s triangle of order ⍵

example

AutoHotkey

ahk forum: discussion <lang AutoHotkey>n := 8, p0 := "1" ; 1+n rows of Pascal's triangle Loop %n% {

  p := "p" A_Index, %p% := v := 1, q := "p" A_Index-1
  Loop Parse, %q%, %A_Space%
     If (A_Index > 1)
        %p% .= " " v+A_LoopField, v := A_LoopField
  %p% .= " 1"

}

                        ; Triangular Formatted output

VarSetCapacity(tabs,n,Asc("`t")) t .= tabs "`t1" Loop %n% {

  t .= "`n" SubStr(tabs,A_Index)
  Loop Parse, p%A_Index%, %A_Space%
     t .= A_LoopField "`t`t"

} Gui Add, Text,, %t% ; Show result in a GUI Gui Show Return

GuiClose:

 ExitApp</lang>

Alternate

Works with: AutoHotkey L

<lang AutoHotkey>Msgbox % format(pascalstriangle()) Return

format(o) ; converts object to string { For k, v in o s .= IsObject(v) ? format(v) "`n" : v " " Return s } pascalstriangle(n=7) ; n rows of Pascal's triangle { p := Object(), z:=Object() Loop, % n Loop, % row := A_Index col := A_Index , p[row, col] := row = 1 and col = 1 ? 1 : (p[row-1, col-1] = "" ; math operations on blanks return blanks; I want to assume zero ? 0 : p[row-1, col-1]) + (p[row-1, col] = "" ? 0 : p[row-1, col]) Return p }</lang> n <= 0 returns empty

AWK

<lang awk>$ awk 'BEGIN{for(i=0;i<6;i++){c=1;r=c;for(j=0;j<i;j++){c*=(i-j)/(j+1);r=r" "c};print r}}'</lang>

Output:

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1

BASIC

Summing from Previous Rows

Works with: FreeBASIC

This implementation uses an array to store one row of the triangle. DIM initializes the array values to zero. For first row, "1" is then stored in the array. To calculate values for next row, the value in cell (i-1) is added to each cell (i). This summing is done from right to left so that it can be done on-place, without using a tmp buffer. Because of symmetry, the values can be displayed from left to right.

Space for max 5 digit numbers is reserved when formatting the display. The maximum size of triangle is 100 rows, but in practice it is limited by screen space. If the user enters value less than 1, the first row is still always displayed.

<lang freebasic>DIM i AS Integer DIM row AS Integer DIM nrows AS Integer DIM values(100) AS Integer

INPUT "Number of rows: "; nrows values(1) = 1 PRINT TAB((nrows)*3);" 1" FOR row = 2 TO nrows

   PRINT TAB((nrows-row)*3+1);
   FOR i = row TO 1 STEP -1
       values(i) = values(i) + values(i-1)
       PRINT USING "##### "; values(i);
   NEXT i
   PRINT

NEXT row</lang>

Batch File

Based from the Fortran Code. <lang dos>@echo off setlocal enabledelayedexpansion

The Main Thing...

cls echo. set row=15 call :pascal echo. pause exit /b 0

/The Main Thing.

The Functions...

pascal

set /a prev=%row%-1 for /l %%I in (0,1,%prev%) do ( set c=1&set r= for /l %%K in (0,1,%row%) do ( if not !c!==0 ( call :numstr !c! set r=!r!!space!!c! ) set /a c=!c!*^(%%I-%%K^)/^(%%K+1^) ) echo !r! ) goto :EOF

numstr

::This function returns the number of whitespaces to be applied on each numbers. set cnt=0&set proc=%1&set space= :loop set currchar=!proc:~%cnt%,1! if not "!currchar!"=="" set /a cnt+=1&goto loop set /a numspaces=5-!cnt! for /l %%A in (1,1,%numspaces%) do set "space=!space! " goto :EOF

/The Functions.</lang>

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

Press any key to continue . . .

BBC BASIC

<lang bbcbasic> nrows% = 10

     colwidth% = 4
     @% = colwidth% : REM Set column width
     FOR row% = 1 TO nrows%
       PRINT SPC(colwidth%*(nrows% - row%)/2);
       acc% = 1
       FOR element% = 1 TO row%
         PRINT acc%;
         acc% = acc% * (row% - element%) / element% + 0.5
       NEXT
       PRINT
     NEXT row%</lang>

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

Befunge

<lang Befunge>0" :swor fo rebmuN">:#,_&> 55+, v v01*p00-1:g00.:<1p011p00:\-1_v#:< >g:1+10p/48*,:#^_$ 55+,1+\: ^>$$@</lang>

Output:

Number of rows: 10

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

Bracmat

<lang bracmat>( out$"Number of rows? " & get':?R & -1:?I & whl

 ' ( 1+!I:<!R:?I
   & 1:?C
   & -1:?K
   & !R+-1*!I:?tabs
   & whl'(!tabs+-1:>0:?tabs&put$\t)
   &   whl
     ' ( 1+!K:~>!I:?K
       & put$(!C \t\t)
       & !C*(!I+-1*!K)*(!K+1)^-1:?C
       )
   & put$\n
   )

& )</lang>

Output:

Number of rows?
7
                                                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

Burlesque

C

Translation of: Fortran

<lang c>#include <stdio.h>

void pascaltriangle(unsigned int n) {

 unsigned int c, i, j, k;

 for(i=0; i < n; i++) {
   c = 1;
   for(j=1; j <= 2*(n-1-i); j++) printf(" ");
   for(k=0; k <= i; k++) {
     printf("%3d ", c);
     c = c * (i-k)/(k+1);
   }
   printf("\n");
 }

}

int main() {

 pascaltriangle(8);
 return 0;

}</lang>

Recursive

<lang c>#include <stdio.h>

define D 32

int pascals(int *x, int *y, int d) { int i; for (i = 1; i < d; i++) printf("%d%c", y[i] = x[i - 1] + x[i], i < d - 1 ? ' ' : '\n');

return D > d ? pascals(y, x, d + 1) : 0; }

int main() { int x[D] = {0, 1, 0}, y[D] = {0}; return pascals(x, y, 0); }</lang>

Adding previous row values

<lang c>void triangleC(int nRows) {

   if (nRows <= 0) return;
   int *prevRow = NULL;
   for (int r = 1; r <= nRows; r++) {
       int *currRow = malloc(r * sizeof(int));
       for (int i = 0; i < r; i++) {
           int val = i==0 || i==r-1 ? 1 : prevRow[i-1] + prevRow[i];
           currRow[i] = val;
           printf(" %4d", val);
       }
       printf("\n");
       free(prevRow);
       prevRow = currRow;
   }
   free(prevRow);

}</lang>

C++

<lang cpp>#include <iostream>

include <algorithm>
include<cstdio>

using namespace std; void Pascal_Triangle(int size) {

int a[100][100]; int i, j;

//first row and first coloumn has the same value=1 for (i = 1; i <= size; i++) { a[i][1] = a[1][i] = 1; }

//Generate the full Triangle for (i = 2; i <= size; i++) { for (j = 2; j <= size - i; j++) { if (a[i - 1][j] == 0 || a[i][j - 1] == 0) { break; } a[i][j] = a[i - 1][j] + a[i][j - 1]; } }

/* 1 1 1 1 1 2 3 1 3 1

first print as above format-->

for (i = 1; i < size; i++) { for (j = 1; j < size; j++) { if (a[i][j] == 0) { break; } printf("%8d",a[i][j]); } cout<<"\n\n"; }*/

// standard Pascal Triangle Format

int row,space; for (i = 1; i < size; i++) { space=row=i; j=1;

while(space<=size+(size-i)+1){ cout<<" "; space++; }

while(j<=i){ if (a[row][j] == 0){ break; }

if(j==1){ printf("%d",a[row--][j++]); } else printf("%6d",a[row--][j++]); } cout<<"\n\n"; }

}

int main() { //freopen("out.txt","w",stdout);

int size; cin>>size; Pascal_Triangle(size); }

}</lang>

C++11 (with dynamic and semi-static vectors)

Constructs the whole triangle in memory before printing it. Uses vector of vectors as a 2D array with variable column size. Theoretically, semi-static version should work a little faster. <lang cpp>// Compile with -std=c++11

include<iostream>
include<vector>

using namespace std; void print_vector(vector<int> dummy){ for (vector<int>::iterator i = dummy.begin(); i != dummy.end(); ++i) cout<<*i<<" "; cout<<endl; } void print_vector_of_vectors(vector<vector<int>> dummy){ for (vector<vector<int>>::iterator i = dummy.begin(); i != dummy.end(); ++i) print_vector(*i); cout<<endl; } vector<vector<int>> dynamic_triangle(int dummy){ vector<vector<int>> result; if (dummy > 0){ // if the argument is 0 or negative exit immediately vector<int> row; // The first row row.push_back(1); result.push_back(row); // The second row if (dummy > 1){ row.clear(); row.push_back(1); row.push_back(1); result.push_back(row); } // The other rows if (dummy > 2){ for (int i = 2; i < dummy; i++){ row.clear(); row.push_back(1); for (int j = 1; j < i; j++) row.push_back(result.back().at(j - 1) + result.back().at(j)); row.push_back(1); result.push_back(row); } } } return result; } vector<vector<int>> static_triangle(int dummy){ vector<vector<int>> result; if (dummy > 0){ // if the argument is 0 or negative exit immediately vector<int> row; result.resize(dummy); // This should work faster than consecutive push_back()s // The first row row.resize(1); row.at(0) = 1; result.at(0) = row; // The second row if (result.size() > 1){ row.resize(2); row.at(0) = 1; row.at(1) = 1; result.at(1) = row; } // The other rows if (result.size() > 2){ for (int i = 2; i < result.size(); i++){ row.resize(i + 1); // This should work faster than consecutive push_back()s row.front() = 1; for (int j = 1; j < row.size() - 1; j++) row.at(j) = result.at(i - 1).at(j - 1) + result.at(i - 1).at(j); row.back() = 1; result.at(i) = row; } } } return result; } int main(){ vector<vector<int>> triangle; int n; cout<<endl<<"The Pascal's Triangle"<<endl<<"Enter the number of rows: "; cin>>n; // Call the dynamic function triangle = dynamic_triangle(n); cout<<endl<<"Calculated using dynamic vectors:"<<endl<<endl; print_vector_of_vectors(triangle); // Call the static function triangle = static_triangle(n); cout<<endl<<"Calculated using static vectors:"<<endl<<endl; print_vector_of_vectors(triangle); return 0; }

</lang>

C++11 (with a class)

A full fledged example with a class definition and methods to retrieve data, worthy of the title object-oriented. <lang cpp>// Compile with -std=c++11

include<iostream>
include<vector>

using namespace std; class pascal_triangle{ vector<vector<int>> data; // This is the actual data void print_row(vector<int> dummy){ for (vector<int>::iterator i = dummy.begin(); i != dummy.end(); ++i) cout<<*i<<" "; cout<<endl; } public: pascal_triangle(int dummy){ // Everything is done on the construction phase if (dummy > 0){ // if the argument is 0 or negative exit immediately vector<int> row; data.resize(dummy); // Theoretically this should work faster than consecutive push_back()s // The first row row.resize(1); row.at(0) = 1; data.at(0) = row; // The second row if (data.size() > 1){ row.resize(2); row.at(0) = 1; row.at(1) = 1; data.at(1) = row; } // The other rows if (data.size() > 2){ for (int i = 2; i < data.size(); i++){ row.resize(i + 1); // Theoretically this should work faster than consecutive push_back()s row.front() = 1; for (int j = 1; j < row.size() - 1; j++) row.at(j) = data.at(i - 1).at(j - 1) + data.at(i - 1).at(j); row.back() = 1; data.at(i) = row; } } } } ~pascal_triangle(){ for (vector<vector<int>>::iterator i = data.begin(); i != data.end(); ++i) i->clear(); // I'm not sure about the necessity of this loop! data.clear(); } void print_row(int dummy){ if (dummy < data.size()) for (vector<int>::iterator i = data.at(dummy).begin(); i != data.at(dummy).end(); ++i) cout<<*i<<" "; cout<<endl; } void print(){ for (int i = 0; i < data.size(); i++) print_row(i); } int get_coeff(int dummy1, int dummy2){ int result = 0; if ((dummy1 < data.size()) && (dummy2 < data.at(dummy1).size())) result = data.at(dummy1).at(dummy2); return result; } vector<int> get_row(int dummy){ vector<int> result; if (dummy < data.size()) result = data.at(dummy); return result; } }; int main(){ int n; cout<<endl<<"The Pascal's Triangle with a class!"<<endl<<endl<<"Enter the number of rows: "; cin>>n; pascal_triangle myptri(n); cout<<endl<<"The whole triangle:"<<endl; myptri.print(); cout<<endl<<"Just one row:"<<endl; myptri.print_row(n/2); cout<<endl<<"Just one coefficient:"<<endl; cout<<myptri.get_coeff(n/2, n/4)<<endl<<endl; return 0; }

</lang>

C#

Translation of: Fortran

Produces no output when n is less than or equal to zero.

<lang csharp>using System;

namespace RosettaCode {

   class PascalsTriangle {

       public static void CreateTriangle(int n) {
           if (n > 0) {
               for (int i = 0; i < n; i++) {
                   int c = 1;
                   Console.Write(" ".PadLeft(2 * (n - 1 - i)));
                   for (int k = 0; k <= i; k++) {
                       Console.Write("{0}", c.ToString().PadLeft(3));
                       c = c * (i - k) / (k + 1);
                   }
                   Console.WriteLine();
               }
           }
       }

       public static void Main() {
           CreateTriangle(8);
       }
   }

}</lang>

Clojure

For n < 1, prints nothing, always returns nil. Copied from the Common Lisp implementation below, but with local functions and explicit tail-call-optimized recursion (recur). <lang lisp>(defn pascal [n]

 (let [newrow (fn newrow [lst ret]
                  (if lst
                      (recur (rest lst)
                             (conj ret (+ (first lst) (or (second lst) 0))))
                      ret))
       genrow (fn genrow [n lst]
                  (when (< 0 n)
                    (do (println lst)
                        (recur (dec n) (conj (newrow lst []) 1)))))]
   (genrow n [1])))

(pascal 4)</lang> And here's another version, using the partition function to produce the sequence of pairs in a row, which are summed and placed between two ones to produce the next row: <lang lisp> (defn nextrow [row]

 (vec (concat [1] (map #(apply + %) (partition 2 1 row)) [1] )))

(defn pascal [n]

 (assert (and (integer? n) (pos? n)))
 (let [triangle (take n (iterate nextrow [1]))]
   (doseq [row triangle]
     (println row))))

</lang> The assert form causes the pascal function to throw an exception unless the argument is (integral and) positive.

Here's a third version using the iterate function <lang lisp> (def pascal

 (iterate
   (fn [prev-row]
     (->>
       (concat (first prev-row) (partition 2 1 prev-row) (last prev-row))
       (map (partial apply +) ,,,)))
    [1]))

</lang>

Another short version which returns an infinite pascal triangle as a list, using the iterate function.

<lang lisp> (def pascal

 (iterate #(concat [1] 
                   (map + % (rest %)) 
                   [1]) 
          [1]))

</lang>

One can then get the first n rows using the take function

<lang lisp> (take 10 pascal) ; returns a list of the first 10 pascal rows </lang>

Also, one can retrieve the nth row using the nth function

<lang lisp> (nth pascal 10) ;returns the nth row </lang>

CoffeeScript

This version assumes n is an integer and n >= 1. It efficiently computes binomial coefficients. <lang coffeescript> pascal = (n) ->

 width = 6
 for r in [1..n]
   s = ws (width/2) * (n-r) # center row
   output = (n) -> s += pad width, n
   cell = 1
   output cell
   # Compute binomial coefficients as you go
   # across the row.
   for c in [1...r]
     cell *= (r-c) / c
     output cell
   console.log s

ws = (n) ->

 s = 
 s += ' ' for i in [0...n]
 s

pad = (cnt, n) ->

 s = n.toString()
 # There is probably a better way to do this.
 cnt -= s.length
 right = Math.floor(cnt / 2)
 left = cnt - right
 ws(left) + s + ws(right)

pascal(7)

</lang>

Output:

> coffee pascal.coffee 
                     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

Common Lisp

To evaluate, call (pascal n). For n < 1, it simply returns nil.

<lang lisp>(defun pascal (n)

  (genrow n '(1)))

(defun genrow (n l)

  (when (< 0 n)
      (print l)
      (genrow (1- n) (cons 1 (newrow l)))))

(defun newrow (l)

  (if (> 2 (length l))
     '(1)
     (cons (+ (car l) (cadr l)) (newrow (cdr l)))))</lang>

D

Less functional Version

<lang d>int[][] pascalsTriangle(in int rows) pure nothrow {

   auto tri = new int[][rows];
   foreach (r; 0 .. rows) {
       int v = 1;
       foreach (c; 0 .. r+1) {
           tri[r] ~= v;
           v = (v * (r - c)) / (c + 1);
       }
   }
   return tri;

}

void main() {

   immutable t = pascalsTriangle(10);
   assert(t == [[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]]);

}</lang>

More functional Version

<lang d>import std.stdio, std.algorithm, std.range;

auto pascal() pure nothrow {

   return [1].recurrence!q{ zip(a[n - 1] ~ 0, 0 ~ a[n - 1])
                            .map!q{ a[0] + a[1] }
                            .array };

}

void main() {

   pascal.take(5).writeln;

}</lang>

Output:

[[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1]]

Alternative Version

There is similarity between Pascal's triangle and Sierpinski triangle. Their difference are the initial line and the operation that act on the line element to produce next line. The following is a generic pascal's triangle implementation for positive number of lines output (n). <lang d>import std.stdio, std.string, std.array, std.format;

string Pascal(alias dg, T, T initValue)(int n) {

   string output;

   void append(in T[] l) {
       output ~= " ".replicate((n - l.length + 1) * 2);
       foreach (e; l)
           output ~= format("%4s", format("%4s", e));
       output ~= "\n";
   }

   if (n > 0) {
       T[][] lines = initValue;
       append(lines[0]);
       foreach (i; 1 .. n) {
           lines ~= lines[i - 1] ~ initValue; // length + 1
           foreach (int j; 1 .. lines[i-1].length)
               lines[i][j] = dg(lines[i-1][j], lines[i-1][j-1]);
           append(lines[i]);
       }
   }
   return output;

}

string delegate(int n) genericPascal(alias dg, T, T initValue)() {

   mixin Pascal!(dg, T, initValue);
   return &Pascal;

}

void main() {

   auto pascal = genericPascal!((int a, int b) => a + b, int, 1)();
   static char xor(char a, char b) { return a == b ? '_' : '*'; }
   auto sierpinski = genericPascal!(xor, char, '*')();

   foreach (i; [1, 5, 9])
       writef(pascal(i));
   // an order 4 sierpinski triangle is a 2^4 lines generic
   // Pascal triangle with xor operation
   foreach (i; [16])
       writef(sierpinski(i));

}</lang>

Output:

     1
             1
           1   1
         1   2   1
       1   3   3   1
     1   4   6   4   1
                     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
                                   *
                                 *   *
                               *   _   *
                             *   *   *   *
                           *   _   _   _   *
                         *   *   _   _   *   *
                       *   _   *   _   *   _   *
                     *   *   *   *   *   *   *   *
                   *   _   _   _   _   _   _   _   *
                 *   *   _   _   _   _   _   _   *   *
               *   _   *   _   _   _   _   _   *   _   *
             *   *   *   *   _   _   _   _   *   *   *   *
           *   _   _   _   *   _   _   _   *   _   _   _   *
         *   *   _   _   *   *   _   _   *   *   _   _   *   *
       *   _   *   _   *   _   *   _   *   _   *   _   *   _   *
     *   *   *   *   *   *   *   *   *   *   *   *   *   *   *   *

Delphi

<lang delphi>program PascalsTriangle;

procedure Pascal(r:Integer); var

 i, c, k:Integer;

begin

 for i := 0 to r - 1 do
 begin
   c := 1;
   for k := 0 to i do
   begin
     Write(c:3);
     c := c * (i - k) div (k + 1);
   end;
   Writeln;
 end;

end;

begin

 Pascal(9);

end.</lang>

DWScript

Doesn't print anything for negative or null values. <lang delphi>procedure Pascal(r : Integer); var

  i, c, k : Integer;

begin

  for i:=0 to r-1 do begin
     c:=1;
     for k:=0 to i do begin
        Print(Format('%4d', [c]));
        c:=(c*(i-k)) div (k+1);
     end;
     PrintLn();
  end;

end;

Pascal(9);</lang>

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

E

So as not to bother with text layout, this implementation generates a HTML fragment. It uses a single mutable array, appending one 1 and adding to each value the preceding value.

<lang e>def pascalsTriangle(n, out) {

   def row := [].diverge(int)

out.print("

") for y in 1..n { out.print("") row.push(1) def skip := n - y if (skip > 0) { out.print(``)

       }
       for x => v in row {

out.print(``)

       }
       for i in (1..!y).descending() {
           row[i] += row[i - 1]
       }

out.println("") } out.print("

$v

")

}</lang>

<lang e>def out := <file:triangle.html>.textWriter() try {

   pascalsTriangle(15, out)

} finally {

   out.close()

} makeCommand("yourFavoriteWebBrowser")("triangle.html")</lang>

Eiffel

<lang eiffel> note description : "Prints pascal's triangle" output : "[

   			   Per requirements of the RosettaCode example, execution will print the first n rows of pascal's triangle
   			  ]"

date : "19 December 2013" authors : "Sandro Meier", "Roman Brunner" revision : "1.0" libraries : "Relies on HASH_TABLE from EIFFEL_BASE library" implementation : "[ Recursive implementation to calculate the n'th row. ]" warning : "[ Will not work for large n's (INTEGER_32) ]"

class APPLICATION

inherit ARGUMENTS

create make

feature {NONE} -- Initialization

make local n:INTEGER do create {HASH_TABLE[ARRAY[INTEGER],INTEGER]}pascal_lines.make (n) --create the hash_table object io.new_line n:=25 draw(n) end feature line(n:INTEGER):ARRAY[INTEGER] --Calculates the n'th line local upper_line:ARRAY[INTEGER] i:INTEGER do if n=1 then --trivial case first line create Result.make_filled (0, 1, n+2) Result.put (0, 1) Result.put (1, 2) Result.put (0, 3) elseif pascal_lines.has (n) then --checks if the result was already calculated Result := pascal_lines.at (n) else --calculates the n'th line recursively create Result.make_filled(0,1,n+2) --for caluclation purposes add a 0 at the beginning of each line Result.put (0, 1) upper_line:=line(n-1) from i:=1 until i>upper_line.count-1 loop Result.put(upper_line[i]+upper_line[i+1],i+1) i:=i+1 end Result.put (0, n+2) --for caluclation purposes add a 0 at the end of each line pascal_lines.put (Result, n) end end

draw(n:INTEGER) --draw n lines of pascal's triangle local space_string:STRING width, i:INTEGER

do space_string:=" " --question of design: add space_string at the beginning of each line width:=line(n).count space_string.multiply (width) from i:=1 until i>n loop space_string.remove_tail (1) io.put_string (space_string) across line(i) as c loop if c.item/=0 then io.put_string (c.item.out+" ") end end io.new_line i:=i+1 end end

feature --Access pascal_lines:HASH_TABLE[ARRAY[INTEGER],INTEGER] --Contains all already calculated lines end </lang>

Elixir

<lang elixir>defmodule Pascal do

 def triangle(n), do: triangle(n,[1])
 
 def triangle(0,list), do: list
 def triangle(n,list) do
   IO.inspect list
   new_list = Enum.zip([0]++list, list++[0]) |> Enum.map(fn {a,b} -> a+b end)
   triangle(n-1,new_list)
 end

end

Pascal.triangle(8)</lang>

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]

Erlang

<lang erlang> -import(lists). -export([pascal/1]).

pascal(1)-> 1; pascal(N) ->

   L = pascal(N-1),
   [H|_] = L,
   [lists:zipwith(fun(X,Y)->X+Y end,[0]++H,H++[0])|L].

</lang>

Output:

  Eshell V5.5.5  (abort with ^G)
  1> pascal:pascal(5).
  [[1,4,6,4,1],[1,3,3,1],[1,2,1],[1,1],[1]]

ERRE

<lang ERRE> PROGRAM PASCAL_TRIANGLE

PROCEDURE PASCAL(R%)

 LOCAL I%,C%,K%
   FOR I%=0 TO R%-1 DO
     C%=1
     FOR K%=0 TO I% DO
       WRITE("###";C%;)
       C%=(C%*(I%-K%)) DIV (K%+1)
     END FOR
     PRINT
  END FOR

END PROCEDURE

BEGIN

 PASCAL(9)

END PROGRAM </lang> 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

Euphoria

Summing from Previous Rows

<lang Euphoria>sequence row row = {} for m = 1 to 10 do

   row = row & 1
   for n = length(row)-1 to 2 by -1 do
       row[n] += row[n-1]
   end for
   print(1,row)
   puts(1,'\n')

end for</lang>

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}

F#

<lang fsharp>let rec nextrow l =

   match l with
   | []      -> []
   | h :: [] -> [1]
   | h :: t  -> h + t.Head :: nextrow t

let pascalTri n = List.scan(fun l i -> 1 :: nextrow l) [1] [1 .. n]

for row in pascalTri(10) do

   for i in row do
       printf "%s" (i.ToString() + ", ")
   printfn "%s" "\n"

</lang>

Factor

This implementation works by summing the previous line content. Result for n < 1 is the same as for n == 1.

<lang factor>USING: grouping kernel math sequences ;

(pascal) ( seq -- newseq )

   dup last 0 prefix 0 suffix 2 <clumps> [ sum ] map suffix ;

pascal ( n -- seq )

   1 - { { 1 } } swap [ (pascal) ] times ;</lang>

It works as:

<lang factor>5 pascal . { { 1 } { 1 1 } { 1 2 1 } { 1 3 3 1 } { 1 4 6 4 1 } }</lang>

Fantom

<lang fantom> class Main {

 Int[] next_row (Int[] row)
 {
   new_row := [1]
   (row.size-1).times |i|
   {
     new_row.add (row[i] + row[i+1])
   }
   new_row.add (1)

   return new_row
 }

 Void print_pascal (Int n)  // no output for n <= 0
 {
   current_row := [1]
   n.times 
   {
     echo (current_row.join(" "))
     current_row = next_row (current_row)
   }
 }

 Void main ()
 {
   print_pascal (10)
 }

} </lang>

Forth

<lang forth>: init ( n -- )

 here swap cells erase  1 here ! ;

.line ( n -- )

 cr here swap 0 do dup @ . cell+ loop drop ;

next ( n -- )

 here swap 1- cells here + do
   i @ i cell+ +!
 -1 cells +loop ;

pascal ( n -- )

     dup init   1  .line
 1 ?do i next i 1+ .line loop ;</lang>

This is a bit more efficient.

Translation of: C

<lang forth>: PascTriangle

 cr dup 0
 ?do
    1 over 1- i - 2* spaces i 1+ 0 ?do dup 4 .r j i - * i 1+ / loop cr drop
 loop drop

13 PascTriangle</lang>

Fortran

Works with: Fortran version 90 and later

Prints nothing for n<=0. Output formatting breaks down for n>20 <lang fortran>PROGRAM Pascals_Triangle

 CALL Print_Triangle(8)

END PROGRAM Pascals_Triangle

SUBROUTINE Print_Triangle(n)

 IMPLICIT NONE
 INTEGER, INTENT(IN) :: n
 INTEGER :: c, i, j, k, spaces

 DO i = 0, n-1
    c = 1
    spaces = 3 * (n - 1 - i)
    DO j = 1, spaces
       WRITE(*,"(A)", ADVANCE="NO") " "
    END DO
    DO k = 0, i
       WRITE(*,"(I6)", ADVANCE="NO") c
       c = c * (i - k) / (k + 1)
    END DO
    WRITE(*,*)
 END DO

END SUBROUTINE Print_Triangle</lang>

FunL

Summing from Previous Rows

Translation of: Scala

<lang funl>import lists.zip

def

 pascal( 1 ) = [1]
 pascal( n ) = [1] + map( (a, b) -> a + b, zip(pascal(n-1), pascal(n-1).tail()) ) + [1]</lang>

Combinations

Translation of: Haskell

<lang funl>import integers.choose

def pascal( n ) = [choose( n - 1, k ) | k <- 0..n-1]</lang>

Pascal's Triangle

<lang funl>def triangle( height ) =

 width = max( map(a -> a.toString().length(), pascal(height)) )

 if 2|width
   width++

 for n <- 1..height
   print( ' '*((width + 1)\2)*(height - n) )
   println( map(a -> format('%' + width + 'd ', a), pascal(n)).mkString() )

triangle( 10 )</lang>

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

GAP

<lang gap>Pascal := function(n) local i, v; v := [1]; for i in [1 .. n] do Display(v); v := Concatenation([0], v) + Concatenation(v, [0]); od; end;

Pascal(9);

[ 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 ]</lang>

Go

No output for n < 1. Otherwise, output formatted left justified. <lang go> package main

import "fmt"

func printTriangle(n int) {

   // degenerate cases
   if n <= 0 {
       return
   }
   fmt.Println(1)
   if n == 1 {
       return
   }
   // iterate over rows, zero based
   a := make([]int, (n+1)/2)
   a[0] = 1
   for row, middle := 1, 0; row < n; row++ {
       // generate new row
       even := row&1 == 0
       if even {
           a[middle+1] = a[middle] * 2
       }
       for i := middle; i > 0; i-- {
           a[i] += a[i-1]
       }
       // print row
       for i := 0; i <= middle; i++ {
           fmt.Print(a[i], " ")
       }
       if even {
           middle++
       }
       for i := middle; i >= 0; i-- {
           fmt.Print(a[i], " ")
       }
       fmt.Println("")
   }

}

func main() {

   printTriangle(4)

} </lang> Output:

Groovy

Recursive

In the spirit of the Haskell "think in whole lists" solution here is a list-driven, minimalist solution: <lang groovy>def pascal = { n -> (n <= 1) ? [1] : GroovyCollections.transpose([[0] + pascal(n - 1), pascal(n - 1) + [0]]).collect { it.sum() } }</lang> However, this solution is horribly inefficient (O(n**2)). It slowly grinds to a halt on a reasonably powerful PC after about line 25 of the triangle.

Test program: <lang groovy>def count = 15 (1..count).each { n ->

   printf ("%2d:", n); (0..(count-n)).each { print "    " }; pascal(n).each{ printf("%6d  ", it) }; println ""

}</lang>

Output:

 1:                                                                 1  
 2:                                                             1       1  
 3:                                                         1       2       1  
 4:                                                     1       3       3       1  
 5:                                                 1       4       6       4       1  
 6:                                             1       5      10      10       5       1  
 7:                                         1       6      15      20      15       6       1  
 8:                                     1       7      21      35      35      21       7       1  
 9:                                 1       8      28      56      70      56      28       8       1  
10:                             1       9      36      84     126     126      84      36       9       1  
11:                         1      10      45     120     210     252     210     120      45      10       1  
12:                     1      11      55     165     330     462     462     330     165      55      11       1  
13:                 1      12      66     220     495     792     924     792     495     220      66      12       1  
14:             1      13      78     286     715    1287    1716    1716    1287     715     286      78      13       1  
15:         1      14      91     364    1001    2002    3003    3432    3003    2002    1001     364      91      14       1

GW-BASIC

<lang qbasic>10 INPUT "Number of rows? ",R 20 FOR I=0 TO R-1 30 C=1 40 FOR K=0 TO I 50 PRINT USING "####";C; 60 C=C*(I-K)/(K+1) 70 NEXT 80 PRINT 90 NEXT</lang>

Output:

Number of rows? 7
   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

Haskell

An approach using the "think in whole lists" principle: Each row in the triangle can be calculated from the previous row by adding a shifted version of itself to it, keeping the ones at the ends. The Prelude function zipWith can be used to add two lists, but it won't keep the old values when one list is shorter. So we need a similar function

<lang haskell>zapWith :: (a -> a -> a) -> [a] -> [a] -> [a] zapWith f xs [] = xs zapWith f [] ys = ys zapWith f (x:xs) (y:ys) = f x y : zapWith f xs ys</lang>

Now we can shift a list and add it to itself, extending it by keeping the ends:

<lang haskell>extendWith f [] = [] extendWith f xs@(x:ys) = x : zapWith f xs ys</lang>

And for the whole (infinite) triangle, we just iterate this operation, starting with the first row:

<lang haskell>pascal = iterate (extendWith (+)) [1]</lang>

For the first n rows, we just take the first n elements from this list, as in

<lang haskell>*Main> take 6 pascal [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1],[1,5,10,10,5,1]]</lang>

A shorter approach, plagiarized from [2] <lang haskell>-- generate next row from current row nextRow row = zipWith (+) ([0] ++ row) (row ++ [0])

-- returns the first n rows pascal = iterate nextRow [1]</lang>

With binomial coefficients: <lang haskell>fac = product . enumFromTo 1

binCoef n k = (fac n) `div` ((fac k) * (fac $ n - k))

pascal n = map (binCoef $ n - 1) [0..n-1]</lang>

Example: <lang haskell>*Main> putStr $ unlines $ map unwords $ map (map show) $ pascal 10 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 </lang>

HicEst

<lang HicEst> CALL Pascal(30)

SUBROUTINE Pascal(rows)

  CHARACTER fmt*6
  WRITE(Text=fmt, Format='"i", i5.5') 1+rows/4

  DO row = 0, rows-1
    n = 1
    DO k = 0, row
      col = rows*(rows-row+2*k)/4
      WRITE(Row=row+1, Column=col, F=fmt) n
      n = n * (row - k) / (k + 1)
    ENDDO
  ENDDO

END</lang>

Icon and Unicon

The code below is slightly modified from the library version of pascal which prints 0's to the full width of the carpet. It also presents the data as an isoceles triangle. <lang Icon>link math

procedure main(A) every n := !A do { # for each command line argument

  n := integer(\n) | &null
  pascal(n)
  }

end

procedure pascal(n) #: Pascal triangle

  /n := 16
  write("width=", n, " height=", n)	# carpet header
  fw := *(2 ^ n)+1
  every i := 0 to n - 1 do {
     writes(repl(" ",fw*(n-i)/2))
     every j := 0 to n - 1 do
        writes(center(binocoef(i, j),fw) | break)
     write()
     }

end</lang>

Library: Icon Programming Library

math provides binocoef math provides the original version of pascal

Sample output:

->pascal 1 4 8
width=1 height=1
 1 
width=4 height=4
       1 
     1  1 
    1  2  1 
  1  3  3  1 
width=8 height=8
                 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  
->

IDL

<lang IDL>Pro Pascal, n

n is the number of lines of the triangle to be displayed

r=[1]
print, r
 for i=0, (n-2) do begin
   pascalrow,r
 endfor

End

Pro PascalRow, r

 for i=0,(n_elements(r)-2) do begin 
   r[i]=r[i]+r[i+1]
 endfor

r= [1, r] print, r

End</lang>

J

<lang j> (-@|. |."_1 [: ":@-.&0"1 !~/~)@i. 5

    1    
   1 1   
  1 2 1  
 1 3 3 1 
1 4 6 4 1</lang>

See the talk page for explanation of earlier version

Java

Summing from Previous Rows

Works with: Java version 1.5+

<lang java>import java.util.ArrayList; ...//class definition, etc. public static void genPyrN(int rows){ if(rows < 0) return; //save the last row here ArrayList<Integer> last = new ArrayList<Integer>(); last.add(1); System.out.println(last); for(int i= 1;i <= rows;++i){ //work on the next row ArrayList<Integer> thisRow= new ArrayList<Integer>(); thisRow.add(last.get(0)); //beginning for(int j= 1;j < i;++j){//loop the number of elements in this row //sum from the last row thisRow.add(last.get(j - 1) + last.get(j)); } thisRow.add(last.get(0)); //end last= thisRow;//save this row System.out.println(thisRow); } }</lang>

Combinations

This method is limited to 21 rows because of the limits of long. Calling pas with an argument of 22 or above will cause intermediate math to wrap around and give false answers. <lang java>public class Pas{ public static void main(String[] args){ //usage pas(20); }

public static void pas(int rows){ for(int i = 0; i < rows; i++){ for(int j = 0; j <= i; j++){ System.out.print(ncr(i, j) + " "); } System.out.println(); } }

public static long ncr(int n, int r){ return fact(n) / (fact(r) * fact(n - r)); }

public static long fact(int n){ long ans = 1; for(int i = 2; i <= n; i++){ ans *= i; } return ans; } }</lang>

Using arithmetic calculation of each row element

This method is limited to 30 rows because of the limits of integer calculations (probably when calculating the multiplication). If m is declared as long then 62 rows can be printed. <lang java> public class Pascal { private static void printPascalLine (int n) { if (n < 1) return; int m = 1; System.out.print("1 "); for (int j=1; j<n; j++) { m = m * (n-j)/j; System.out.print(m); System.out.print(" "); } System.out.println(); }

public static void printPascal (int nRows) { for(int i=1; i<=nRows; i++) printPascalLine(i); } } </lang>

JavaScript

Imperative

Works with: SpiderMonkey

Works with: V8

<lang javascript>// Pascal's triangle object function pascalTriangle (rows) {

// Number of rows the triangle contains this.rows = rows;

// The 2D array holding the rows of the triangle this.triangle = new Array(); for (var r = 0; r < rows; r++) { this.triangle[r] = new Array(); for (var i = 0; i <= r; i++) { if (i == 0 || i == r) this.triangle[r][i] = 1; else this.triangle[r][i] = this.triangle[r-1][i-1]+this.triangle[r-1][i]; } }

// Method to print the triangle this.print = function(base) { if (!base) base = 10;

// Private method to calculate digits in number var digits = function(n,b) { var d = 0; while (n >= 1) { d++; n /= b; } return d; }

// Calculate max spaces needed var spacing = digits(this.triangle[this.rows-1][Math.round(this.rows/2)],base);

// Private method to add spacing between numbers var insertSpaces = function(s) { var buf = ""; while (s > 0) { s--; buf += " "; } return buf; }

// Print the triangle line by line for (var r = 0; r < this.triangle.length; r++) { var l = ""; for (var s = 0; s < Math.round(this.rows-1-r); s++) { l += insertSpaces(spacing); } for (var i = 0; i < this.triangle[r].length; i++) { if (i != 0) l += insertSpaces(spacing-Math.ceil(digits(this.triangle[r][i],base)/2)); l += this.triangle[r][i].toString(base); if (i < this.triangle[r].length-1) l += insertSpaces(spacing-Math.floor(digits(this.triangle[r][i],base)/2)); } print(l); } }

}

// Display 4 row triangle in base 10 var tri = new pascalTriangle(4); tri.print(); // Display 8 row triangle in base 16 tri = new pascalTriangle(8); tri.print(16);</lang> Output:

$ d8 pascal.js 
   1
  1 1
 1 2 1
1 3 3 1
              1
            1   1
          1   2   1
        1   3   3   1
      1   4   6   4   1
    1   5   a   a   5   1
  1   6   f   14  f   6   1
1   7   15  23  23  15  7   1

Functional (ES5)

<lang JavaScript>(function (n) {

 // A Pascal triangle of n rows
 // n --> n
 function pascalTriangle(n) {

   // Sums of each consecutive pair of numbers
   // [n] --> [n]
   function pairSums(lst) {
     return lst.reduce(function (acc, n, i, l) {
       var iPrev = i ? i - 1 : 0;
       return i ? acc.concat(l[iPrev] + l[i]) : acc
     }, []);
   }

   // Next line in a Pascal triangle series
   // [n] --> [n]
   function nextPascal(lst) {
     return lst.length ? [1].concat(
       pairSums(lst)
     ).concat(1) : [1];
   }

   // Each row is a function of the preceding row
   return n ? Array.apply(null, Array(n - 1)).reduce(
     function (a, _, i) {
       return a.concat([nextPascal(a[i])]);
     }, 1) : [];
 }

 // TEST
 var lstTriangle = pascalTriangle(n);

 // FORMAT OUTPUT AS WIKI TABLE

 // a -> bool -> s -> s
 function wikiTable(lstRows, blnHeaderRow, strStyle) {
   return '{| class="wikitable" ' + (
     strStyle ? 'style="' + strStyle + '"' : 
   ) + lstRows.map(function (lstRow, iRow) {
     var strDelim = ((blnHeaderRow && !iRow) ? '!' : '|');

     return '\n|-\n' + strDelim + ' ' + lstRow.map(function (v) {
       return typeof v === 'undefined' ? ' ' : v;
     }).join(' ' + strDelim + strDelim + ' ');
   }).join() + '\n|}';
 }

 var lstLastLine = lstTriangle.slice(-1)[0],
   lngBase = (lstLastLine.length * 2) - 1,
   nWidth = lstLastLine.reduce(function (a, x) {
     var d = x.toString().length;
     return d > a ? d : a;
   }, 1) * lngBase;

 return [
   wikiTable(
     lstTriangle.map(function (lst) {
       return lst.join(';;').split(';');
     }).map(function (line, i) {
       var lstPad = Array((lngBase - line.length) / 2);
       return lstPad.concat(line).concat(lstPad);
     }),
     false,
     'text-align:center;width:' + nWidth + 'em;height:' + nWidth +
     'em;table-layout:fixed;'
   ),

   JSON.stringify(lstTriangle)
 ].join('\n\n');

})(7);</lang>

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

jq

Works with: jq version 1.4

pascal(n) as defined here produces a stream of n arrays, each corresponding to a row of the Pascal triangle. The implementation avoids any arithmetic except addition. <lang jq># pascal(n) for n>=0; pascal(0) emits an empty stream. def pascal(n):

 def _pascal:  # input: the previous row
   . as $in
   | .,
     if length >= n then empty
     else
       reduce range(0;length-1) as $i
         ([1]; . + [ $in[$i] + $in[$i + 1] ]) + [1] | _pascal
     end;
 if n <= 0 then empty else [1] | _pascal end ;</lang>

Example:

pascal(5)

Output:

<lang sh>$ jq -c -n -f pascal_triangle.jq [1] [1,1] [1,2,1] [1,3,3,1] [1,4,6,4,1]</lang>

Using recurse/1

Here is an equivalent implementation that uses the built-in filter, recurse/1, instead of the inner function. <lang jq>def pascal(n):

 if n <= 0 then empty
 else [1]
 | recurse( if length >= n then empty
            else . as $in 
            | reduce range(0;length-1) as $i
                ([1]; . + [ $in[$i] + $in[$i + 1] ]) + [1]
            end)
 end;</lang>

K

<lang K> pascal:{(x-1){+':x,0}\1} pascal 6 (1

1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1)</lang>

Kotlin

<lang kotlin>fun pas(rows: Int) {

   for (i in 0..rows - 1) {
       for (j in 0..i)
           print(ncr(i, j).toString() + " ")
       println()
   }

}

fun ncr(n: Int, r: Int) = fact(n) / (fact(r) * fact(n - r))

fun fact(n: Int) : Long {

   var ans = 1.toLong()
   for (i in 2..n)
       ans *= i
   return ans

}

fun main(args: Array<String>) = pas(args[0].toInt())</lang>

Liberty BASIC

<lang lb>input "How much rows would you like? "; n dim a$(n)

for i= 0 to n

      c = 1
      o$ =""
      for k =0 to i
            o$ =o$ ; c; " "
            c =c *(i-k)/(k+1)
      next k
      a$(i)=o$

next i

maxLen = len(a$(n)) for i= 0 to n

   print space$((maxLen-len(a$(i)))/2);a$(i)

next i

end</lang>

Locomotive Basic

<lang locobasic>10 CLS 20 INPUT "Number of rows? ", rows:GOSUB 40 30 END 40 FOR i=0 TO rows-1 50 c=1 60 FOR k=0 TO i 70 PRINT USING "####";c; 80 c=c*(i-k)/(k+1) 90 NEXT 100 PRINT 110 NEXT 120 RETURN</lang>

Output:

Number of rows? 7
   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

Logo

<lang logo>to pascal :n

 if :n = 1 [output [1]]
 localmake "a pascal :n-1
 output (sentence first :a (map "sum butfirst :a butlast :a) last :a)

end

for [i 1 10] [print pascal :i]</lang>

Lua

<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 triangle(n)

 t = {1}
 for i = 1, n do
   print(unpack(t))
   t = nextrow(t)
 end

end </lang>

Mathematica

<lang Mathematica>Column[StringReplace[ToString /@ Replace[MatrixExp[SparseArray[ {Band[{2,1}] -> Range[n-1]},{n,n}]],{x__,0..}->{x},2] ,{"{"|"}"|","->" "}], Center]</lang>

MATLAB / Octave

A matrix containing the pascal triangle can be obtained this way: <lang MATLAB>pascal(n);</lang>

>> pascal(6)
ans =

     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 binomial coefficients can be extracted from the Pascal triangle in this way: <lang MATLAB> binomCoeff = diag(rot90(pascal(n)))', </lang>

>> for k=1:6,diag(rot90(pascal(k)))', end
ans =  1
ans =

   1   1

ans =

   1   2   1

ans =

   1   3   3   1

ans =

   1   4   6   4   1

ans =

    1    5   10   10    5    1

Another way to get a formated pascals triangle is to use the convolution method:

>>
x = [1  1] ;      
y = 1;                   
for k=8:-1:1
    fprintf(['%', num2str(k), 'c'], zeros(1,3)), 
    fprintf('%6d', y), fprintf('\n')             
   y = conv(y,x);                        
                                         
end

The result is:

>>

                          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

Maxima

<lang maxima>sjoin(v, j) := apply(sconcat, rest(join(makelist(j, length(v)), v)))$

display_pascal_triangle(n) := for i from 0 thru 6 do disp(sjoin(makelist(binomial(i, j), j, 0, i), " "));

display_pascal_triangle(6); /* "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" */</lang>

Metafont

(The formatting starts to be less clear when numbers start to have more than two digits)

<lang metafont>vardef bincoeff(expr n, k) = save ?; ? := (1 for i=(max(k,n-k)+1) upto n: * i endfor )

    / (1 for i=2 upto min(k, n-k): * i endfor); ?

enddef;

def pascaltr expr c =

 string s_;
 for i := 0 upto (c-1):
   s_ := "" for k=0 upto (c-i): & "  " endfor;
   s_ := s_ for k=0 upto i: & decimal(bincoeff(i,k))
            & "  " if bincoeff(i,k)<9: & " " fi endfor;
   message s_;
 endfor

enddef;

pascaltr(4); end</lang>

NetRexx

<lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols nobinary

numeric digits 1000 -- allow very large numbers parse arg rows . if rows = then rows = 11 -- default to 11 rows printPascalTriangle(rows) return

-- ----------------------------------------------------------------------------- method printPascalTriangle(rows = 11) public static

 lines = 
 mx = (factorial(rows - 1) / factorial(rows % 2) / factorial(rows - 1 - rows % 2)).length() -- width of widest number

 loop row = 1 to rows
   n1 = 1.center(mx)
   line = n1
   loop col = 2 to row
     n2 = col - 1
     n1 = n1 * (row - n2) / n2
     line = line n1.center(mx)
     end col
   lines[row] = line.strip()
   end row

 -- display triangle
 ml = lines[rows].length() -- length of longest line
 loop row = 1 to rows
   say lines[row].centre(ml)
   end row

 return

-- ----------------------------------------------------------------------------- method factorial(n) public static

 fac = 1
 loop n_ = 2 to n
   fac = fac * n_
   end n_
 return fac /*calc. factorial*/

</lang>

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

Nial

Like J

(pascal.nial) <lang nial>factorial is recur [ 0 =, 1 first, pass, product, -1 +] combination is fork [ > [first, second], 0 first,

  / [factorial second, * [factorial - [second, first], factorial first] ]

] pascal is transpose each combination cart [pass, pass] tell</lang> Using it <lang nial>|loaddefs 'pascal.nial' |pascal 5</lang>

Nim

<lang nim>import sequtils

proc pascal(n: int) =

 var row = @[1]
 for r in 1..n:
   echo row
   row = zip(row & @[0], @[0] & row).mapIt(int, it[0] + it[1])

pascal(10)</lang>

OCaml

<lang ocaml>(* generate next row from current row *) let next_row row =

 List.map2 (+) ([0] @ row) (row @ [0])

(* returns the first n rows *) let pascal n =

 let rec loop i row =
   if i = n then []
   else row :: loop (i+1) (next_row row)
 in loop 0 [1]</lang>

Octave

<lang octave>function pascaltriangle(h)

 for i = 0:h-1
   for k = 0:h-i
     printf("  ");
   endfor
   for j = 0:i
     printf("%3d ", bincoeff(i, j));
   endfor
   printf("\n");
 endfor

endfunction

pascaltriangle(4);</lang>

Oforth

No result if n <= 0

<lang Oforth>func: pascal(n) { [ 1 ] #[ dup println dup 0 + 0 rot + zipWith(#+) ] times(n) drop }</lang>

Output:

pascal(10)
[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]

Oz

<lang oz>declare

 fun {NextLine Xs}
    {List.zip 0|Xs {Append Xs [0]}
     fun {$ Left Right}
        Left + Right
     end}
 end

 fun {Triangle N}
    {List.take {Iterate [1] NextLine} N}
 end

 fun lazy {Iterate I F}
    I|{Iterate {F I} F}
 end

 %% Only works nicely for N =< 5.
 proc {PrintTriangle T}
    N = {Length T}
 in
    for
       Line in T
       Indent in N-1..0;~1
    do
       for _ in 1..Indent do {System.printInfo " "} end
       for L in Line do {System.printInfo L#" "} end
       {System.printInfo "\n"}
    end
 end

in

 {PrintTriangle {Triangle 5}}</lang>

For n = 0, prints nothing. For negative n, throws an exception.

PARI/GP

<lang parigp>pascals_triangle(N)= { my(row=[],prevrow=[]); for(x=1,N,

   if(x>5,break(1));
        row=eval(Vec(Str(11^(x-1))));
        print(row));

prevrow=row; for(y=6,N,

  for(p=2,#prevrow,
        row[p]=prevrow[p-1]+prevrow[p]);
        row=concat(row,1);
        prevrow=row;
        print(row);
    );

}</lang>

Pascal

<lang pascal>Program PascalsTriangle(output);

procedure Pascal(r : Integer);

 var
   i, c, k : Integer;
 begin
   for i := 0 to r-1 do
   begin
     c := 1;
     for k := 0 to i do
     begin
       write(c:3);
       c := (c * (i-k)) div (k+1);
     end;
     writeln;
  end;

end;

begin

 Pascal(9)

end.</lang> Output:

% ./PascalsTriangle 
  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

Perl

These functions perform as requested in the task: they print out the first n lines. If n <= 0, they print nothing. The output is simple (no fancy formatting). <lang perl>sub pascal {

 my $rows = shift;
 my @next = (1);
 for my $n (1 .. $rows) {
   print "@next\n";
   @next = (1, (map $next[$_]+$next[$_+1], 0 .. $n-2), 1);
 }

}</lang>

If you want more than 68 rows, then use either "use bigint" or "use Math::GMP qw/:constant/" inside the function to enable bigints. We can also use a binomial function which will expand to bigints if many rows are requested:

Library: ntheory

<lang perl>use ntheory qw/binomial/; sub pascal {

 my $rows = shift;
 for my $n (0 .. $rows-1) {
   print join(" ", map { binomial($n,$_) } 0 .. $n), "\n";
 }

}</lang>

Here is a non-obvious version using bignum, which is limited to the first 23 rows because of the algorithm used: <lang perl>use bignum; sub pascal_line { $_[0] ? unpack "A(A6)*", 1000001**$_[0] : 1 } sub pascal { print "@{[map -+-$_, pascal_line $_]}\n" for 0..$_[0]-1 }</lang>

Perl 6

Works with: rakudo version 2015-10-03

using a lazy sequence generator

The following routine returns a lazy list of lines using the sequence operator (...). With a lazy result you need not tell the routine how many you want; you can just use a slice subscript to get the first N lines: <lang perl6>sub pascal { [1], -> $prev { [0, |$prev Z+ |$prev, 0] } ... * }

.say for pascal[^10];</lang>

One problem with the routine above is that it might recalculate the sequence each time you call it. Slightly more idiomatic would be to define the sequence as a lazy constant. Here we use the @ sigil to indicate that the sequence should cache its values for reuse:

<lang perl6>constant @pascal = [1], -> $prev { [0, |$prev Z+ |$prev, 0] } ... *;

.say for @pascal[^10];</lang>

Since we use ordinary subscripting, non-positive inputs throw an index-out-of-bounds error.

recursive

Translation of: Haskell

<lang perl6>multi sub pascal (1) { $[1] } multi sub pascal (Int $n where 2..*) {

   my @rows = pascal $n - 1;
   |@rows, [0, |@rows[*-1] Z+ |@rows[*-1], 0 ];

}

.say for pascal 10;</lang>

Non-positive inputs throw a multiple-dispatch error.

iterative

Translation of: Perl

<lang perl6>sub pascal ($n where $n >= 1) {

  say my @last = 1;
  for 1 .. $n - 1 -> $row {
      @last = 1, |map({ @last[$_] + @last[$_ + 1] }, 0 .. $row - 2), 1;
      say @last;
  }

}

pascal 10;</lang>

Non-positive inputs throw a type check error.

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]

Phix

<lang Phix>sequence row = {} for m = 1 to 13 do

   row = row & 1
   for n=length(row)-1 to 2 by -1 do
       row[n] += row[n-1]
   end for
   printf(1,repeat(' ',(13-m)*2))
   for i=1 to length(row) do
       printf(1," %3d",row[i])
   end for
   puts(1,'\n')

end for</lang>

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

PHP

<lang php>function pascalsTriangle($num){ $c = 1; $triangle = Array(); for($i=0;$i<=$num;$i++){ $triangle[$i] = Array(); if(!isset($triangle[$i-1])){ $triangle[$i][] = $c; }else{ for($j=0;$j<count($triangle[$i-1])+1;$j++){ $triangle[$i][] = (isset($triangle[$i-1][$j-1]) && isset($triangle[$i-1][$j])) ? $triangle[$i-1][$j-1] + $triangle[$i-1][$j] : $c; } } } return $triangle; }

$tria = pascalsTriangle(8); foreach($tria as $val){ foreach($val as $value){ echo $value . ' '; } echo '
'; }</lang>

                                       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

PL/I

<lang PL/I> declare (t, u)(40) fixed binary; declare (i, n) fixed binary;

t,u = 0; get (n); if n <= 0 then return;

do n = 1 to n;

  u(1) = 1;
  do i = 1 to n;
     u(i+1) = t(i) + t(i+1);
  end;
  put skip edit ((u(i) do i = 1 to n)) (col(40-2*n), (n+1) f(4));
  t = u;

end; </lang>

<lang>

                                       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

</lang>

PicoLisp

Translation of: C

<lang PicoLisp>(de pascalTriangle (N)

  (for I N
     (space (* 2 (- N I)))
     (let C 1
        (for K I
           (prin (align 3 C) " ")
           (setq C (*/ C (- I K) K)) ) )
     (prinl) ) )</lang>

Potion

<lang potion>printpascal = (n) :

  if (n < 1) :
     1 print
     (1)
  . else :
     prev = printpascal(n - 1)
     prev append(0)
     curr = (1)
     n times (i):
        curr append(prev(i) + prev(i + 1))
     .
     "\n" print
     curr join(", ") print
     curr
  .

.

printpascal(read number integer)</lang>

PowerShell

<lang powershell> $Infinity = 1 $NewNumbers = $null $Numbers = $null $Result = $null $Number = $null $Power = $args[0]

Write-Host $Power

For(

  $i=0;
  $i -lt $Infinity;
  $i++
  )
  {
   $Numbers = New-Object Object[] 1
   $Numbers[0] = $Power
  For(
     $k=0;
     $k -lt $NewNumbers.Length;
     $k++
     )
     {
      $Numbers = $Numbers + $NewNumbers[$k]
     }
  If(
    $i -eq 0
    )
    {
     $Numbers = $Numbers + $Power
    }
   $NewNumbers = New-Object Object[] 0
  Try
  {
  For(
     $j=0;
     $j -lt $Numbers.Length;
     $j++
     )
     {
      $Result = $Numbers[$j] + $Numbers[$j+1]
      $NewNumbers = $NewNumbers + $Result
     }
  }
  Catch [System.Management.Automation.RuntimeException]
  {
   Write-Warning "Value was too large for a Decimal. Script aborted."
   Break;
  }
  Foreach(
         $Number in $Numbers
         )
         {
         If(
           $Number.ToString() -eq "+unendlich"
           )
           {
            Write-Warning "Value was too large for a Decimal. Script aborted."
            Exit
           }
         }
   Write-Host $Numbers
   $Infinity++
  }

</lang>

Save the above code to a .ps1 script file and start it by calling its name and providing N.

PS C:\> & '.\Pascals Triangle.ps1' 1

----

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

Prolog

Difference-lists are used to make quick append. <lang Prolog>pascal(N) :- pascal(1, N, [1], [[1]|X]-X, L), maplist(my_format, L).

pascal(Max, Max, L, LC, LF) :- !, make_new_line(L, NL), append_dl(LC, [NL|X]-X, LF-[]).

pascal(N, Max, L, NC, LF) :- build_new_line(L, NL), append_dl(NC, [NL|X]-X, NC1), N1 is N+1, pascal(N1, Max, NL, NC1, LF).

build_new_line(L, R) :- build(L, 0, X-X, R).

build([], V, RC, RF) :- append_dl(RC, [V|Y]-Y, RF-[]).

build([H|T], V, RC, R) :- V1 is V+H, append_dl(RC, [V1|Y]-Y, RC1), build(T, H, RC1, R).

append_dl(X1-X2, X2-X3, X1-X3).

% to have a correct output ! my_format([H|T]) :- write(H), maplist(my_writef, T), nl.

my_writef(X) :- writef(' %5r', [X]). </lang>

Output : <lang Prolog> ?- pascal(15). 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 true. </lang>

PureBasic

<lang PureBasic>Procedure pascaltriangle( n.i)

 For i=  0 To  n
      c = 1
      For k=0 To i
            Print(Str( c)+" ")
        c = c * (i-k)/(k+1);
       Next ;k
   PrintN(" "); nächste zeile
 Next ;i

EndProcedure

OpenConsole() Parameter.i = Val(ProgramParameter(0)) pascaltriangle(Parameter); Input()</lang>

Python

<lang python>def pascal(n):

  """Prints out n rows of Pascal's triangle.
  It returns False for failure and True for success."""
  row = [1]
  k = [0]
  for x in range(max(n,0)):
     print row
     row=[l+r for l,r in zip(row+k,k+row)]
  return n>=1</lang>

Or, by creating a scan function: <lang python>def scan(op, seq, it):

 a = []
 result = it
 a.append(it)
 for x in seq:
   result = op(result, x)
   a.append(result)
 return a

def pascal(n):

   def nextrow(row, x):
       return [l+r for l,r in zip(row+[0,],[0,]+row)]

   return scan(nextrow, range(n-1), [1,])

for row in pascal(4):

   print(row)</lang>

q

<lang q> pascal:{(x-1){0+':x,0}\1} pascal 5 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 </lang>

Qi

Translation of: Haskell

<lang Qi> (define iterate

 _ _ 0 -> []
 F V N -> [V|(iterate F (F V) (1- N))])

(define next-row

 R -> (MAPCAR + [0|R] (append R [0])))

(define pascal

 N -> (iterate next-row [1] N))

</lang>

R

Translation of: Octave

<lang R>pascalTriangle <- function(h) {

 for(i in 0:(h-1)) {
   s <- ""
   for(k in 0:(h-i)) s <- paste(s, "  ", sep="")
   for(j in 0:i) {
     s <- paste(s, sprintf("%3d ", choose(i, j)), sep="")
   }
   print(s)
 }

}</lang>

Here's an R version:

<lang R>pascalTriangle <- function(h) {

 lapply(0:h, function(i) choose(i, 0:i))

}</lang>

Racket

Iterative version by summing rows up to $n$ .

<lang Racket>#lang racket

(define (pascal n)

 (define (next-row current-row)
   (map + (cons 0 current-row)
          (append current-row '(0))))
 (let-values 
     ([(previous-rows final-row)
      (for/fold ([triangle null]
                 [row '(1)]) 
        ([row-number (in-range 1 n)])
        (values (cons row triangle)
                (next-row row)))])
   (reverse (cons final-row previous-rows))))

</lang>

RapidQ

Summing from Previous Rows

Translation of: BASIC

The main difference to BASIC implementation is the output formatting. RapidQ does not support PRINT USING. Instead, function FORMAT$() is used. TAB() is not supported, so SPACE$() was used instead.

Another difference is that in RapidQ, DIM does not clear array values to zero. But if dimensioning is done with DEF..., you can give the initial values in curly braces. If less values are given than there are elements in the array, the remaining positions are initialized to zero. RapidQ does not require simple variables to be declared before use.

<lang rapidq>DEFINT values(100) = {0,1}

INPUT "Number of rows: "; nrows PRINT SPACE$((nrows)*3);" 1" FOR row = 2 TO nrows

   PRINT SPACE$((nrows-row)*3+1);
   FOR i = row TO 1 STEP -1
       values(i) = values(i) + values(i-1)
       PRINT FORMAT$("%5d ", values(i));
   NEXT i
   PRINT

NEXT row</lang>

Using binary coefficients

Translation of: BASIC

<lang rapidq>INPUT "Number of rows: "; nrows FOR row = 0 TO nrows-1

   c = 1
   PRINT SPACE$((nrows-row)*3);
   FOR i = 0 TO row
       PRINT FORMAT$("%5d ", c);
       c = c * (row - i) / (i+1)
   NEXT i
   PRINT

NEXT row</lang>

Retro

<lang Retro>2 elements i j

pascalTriangle

 cr dup
 [ dup !j 1 swap 1+ [ !i dup putn space @j @i - * @i 1+ / ] iter cr drop ] iter drop

13 pascalTriangle</lang>

REXX

There is no practical limit for this REXX version, triangles up to 46 rows have been generated (without wrapping) in a screen window with a width of 620 characters.

If the number (of rows) specified is negative, the output is written to a (disk) file instead. Triangles with over a 1,000 rows have easily been created. The output file created (that is written to disk) is named PASCALS.n where n is the absolute value of the number entered.

Note: Pascal's triangle is also known as:

Yang Hui's triangle
Khayyam's triangle
Khayyam─Pascal's triangle
Tartaglia's triangle

<lang rexx>/*REXX program displays (or writes to a file) Pascal's triangle (centered/formatted).*/ numeric digits 3000 /*be able to handle gihugeic triangles.*/ parse arg nn . /*obtain the optional argument from CL.*/ if nn== | nn=="," then nn=10 /*Not specified? Then use the default.*/ N=abs(nn) /*N is the number of rows in triangle.*/ w=length( !(N-1) / !(N%2) / !(N-1-N%2) ) /*W: the width of the biggest integer.*/ @.=1; $.=@.; unity=right(1, w) /*defaults rows & lines; aligned unity.*/

                                                /* [↓]  build rows of Pascals' triangle*/
 do   r=1  for N;           rm=r-1              /*Note:  the first column is always  1.*/
   do c=2  to rm;           cm=c-1              /*build the rest of the columns in row.*/
   @.r.c= @.rm.cm + @.rm.c                      /*assign value to a specific row & col.*/
   $.r  = $.r     right(@.r.c, w)               /*and construct a line for output (row)*/
   end   /*c*/                                  /* [↑]    C  is the column being built.*/
 if r\==1  then $.r=$.r  unity                  /*for  rows≥2,  append a trailing  "1".*/
 end     /*r*/                                  /* [↑]    R  is the  row   being built.*/
                                                /* [↑]  WIDTH: for nicely looking line.*/

width=length($.N) /*width of the last (output) line (row)*/

                                                /*if NN<0, output is written to a file.*/
     do r=1  for N;     $$=center($.r, width)   /*center this particular Pascals' row. */
     if nn>0  then say                       $$ /*SAY    if   NN    is positive,  else */
              else call lineout 'PASCALS.'n, $$ /*write this Pascal's row ───►  a file.*/
     end   /*r*/

exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ !: procedure; !=1; do j=2 to arg(1); !=!*j; end /*j*/; return ! /*compute factorial*/</lang> output when using the input of: 11

                    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

output when using the input of: 22

                                                                         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
          1     18    153    816   3060   8568  18564  31824  43758  48620  43758  31824  18564   8568   3060    816    153     18      1
       1     19    171    969   3876  11628  27132  50388  75582  92378  92378  75582  50388  27132  11628   3876    969    171     19      1
   1     20    190   1140   4845  15504  38760  77520 125970 167960 184756 167960 125970  77520  38760  15504   4845   1140    190     20      1
1     21    210   1330   5985  20349  54264 116280 203490 293930 352716 352716 293930 203490 116280  54264  20349   5985   1330    210     21      1

Ring

<lang ring> row = 5 for i = 0 to row - 1

   col = 1
   see left("     ",row-i)
   for k = 0 to i
       see "" + col + " "
       col = col*(i-k)/(k+1)
   next
   see nl

next </lang> Output:

Ruby

<lang ruby>def pascal(n)

 raise ArgumentError, "must be positive." if n < 1
 yield ar = [1]
 (n-1).times do
   ar.unshift(0).push(0) # tack a zero on both ends
   yield ar = ar.each_cons(2).map{|a, b| a + b } 
 end

end

pascal(8){|row| puts row.join(" ").center(20)}</lang>

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

Or for more or less a translation of the two line Haskell version (with inject being abused a bit I know):

<lang ruby>def next_row(row) ([0] + row).zip(row + [0]).collect {|l,r| l + r } end

def pascal(n) n.times.inject([1]) {|x,_| next_row x } end

8.times{|i| p pascal(i)}</lang>

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]

Run BASIC

<lang runbasic>input "number of rows? ";r for i = 0 to r - 1

 c = 1
 print left$("                          ",(r*2)-(i*2));
 for k = 0 to i
   print using("####",c);
   c = c*(i-k)/(k+1)
 next
 print

next</lang>Output:

Number of rows? ?5
             1
           1   1
         1   2   1
       1   3   3   1
     1   4   6   4   1

Scala

Simple recursive row definition: <lang scala> def tri(row:Int):List[Int] = { row match {

 case 1 => List(1)
 case n:Int => List(1) ::: ((tri(n-1) zip tri(n-1).tail) map {case (a,b) => a+b}) ::: List(1)
 }

} </lang> Function to pretty print n rows: <lang scala> def prettytri(n:Int) = (1 to n) foreach {i=>print(" "*(n-i)); tri(i) map (c=>print(c+" ")); println} prettytri(5) </lang> Outputs:

Scheme

Works with: Scheme version R

^{5}

RS

<lang scheme>(define (next-row row)

 (map + (cons 0 row) (append row '(0))))

(define (triangle row rows)

 (if (= rows 0)
     '()
     (cons row (triangle (next-row row) (- rows 1)))))

(triangle (list 1) 5) </lang> Output: <lang>((1) (1 1) (1 2 1) (1 3 3 1) (1 4 6 4 1))</lang>

Seed7

<lang seed7>$ include "seed7_05.s7i";

const proc: main is func

 local
   var integer: numRows is 0;
   var array integer: values is [] (0, 1);
   var integer: row is 0;
   var integer: index is 0;
 begin
   write("Number of rows: ");
   readln(numRows);
   writeln("1" lpad succ(numRows) * 3);
   for row range 2 to numRows do
     write("" lpad (numRows - row) * 3);
     values &:= [] 0;
     for index range succ(row) downto 2 do
       values[index] +:= values[pred(index)];
       write(" " <& values[index] lpad 5);
     end for;
     writeln;
   end for;
 end func;</lang>

Sidef

<lang ruby>func pascal(rows) {

   var row = [1];
   { | n|
       say row.join(' ');
       row = [1, 0..(n-2).map {|i| row[i] + row[i+1] }..., 1];
   } * rows;

}

pascal(10);</lang>

Tcl

Summing from Previous Rows

<lang tcl>proc pascal_iterative n {

   if {$n < 1} {error "undefined behaviour for n < 1"}
   set row [list 1]
   lappend rows $row    
   set i 1
   while {[incr i] <= $n} {
       set prev $row
       set row [list 1]
       for {set j 1} {$j < [llength $prev]} {incr j} {
           lappend row [expr {[lindex $prev [expr {$j - 1}]] + [lindex $prev $j]}]
       }
       lappend row 1
       lappend rows $row
   }
   return $rows

}

puts [join [pascal_iterative 6] \n]</lang>

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

Using binary coefficients

Translation of: BASIC

<lang tcl>proc pascal_coefficients n {

   if {$n < 1} {error "undefined behaviour for n < 1"}
   for {set i 0} {$i < $n} {incr i} {
       set c 1
       set row [list $c]
       for {set j 0} {$j < $i} {incr j} {
           set c [expr {$c * ($i - $j) / ($j + 1)}]
           lappend row $c
       }
       lappend rows $row
   }
   return $rows

}

puts [join [pascal_coefficients 6] \n]</lang>

Combinations

Translation of: Java

Thanks to Tcl 8.5's arbitrary precision integer arithmetic, this solution is not limited to a couple of dozen rows. Uses a caching factorial calculator to improve performance. <lang tcl>package require Tcl 8.5

proc pascal_combinations n {

   if {$n < 1} {error "undefined behaviour for n < 1"}
   for {set i 0} {$i < $n} {incr i} {
       set row [list]
       for {set j 0} {$j <= $i} {incr j} {
           lappend row [C $i $j]
       }
       lappend rows $row
   }
   return $rows

}

proc C {n k} {

   expr {[ifact $n] / ([ifact $k] * [ifact [expr {$n - $k}]])}

}

set fact_cache {1 1} proc ifact n {

   global fact_cache
   if {$n < [llength $fact_cache]} {
       return [lindex $fact_cache $n]
   }
   set i [expr {[llength $fact_cache] - 1}]
   set sum [lindex $fact_cache $i]
   while {$i < $n} {
       incr i
       set sum [expr {$sum * $i}]
       lappend fact_cache $sum
   }
   return $sum

}

puts [join [pascal_combinations 6] \n]</lang>

Comparing Performance

<lang tcl>set n 100 puts "calculate $n rows:" foreach proc {pascal_iterative pascal_coefficients pascal_combinations} {

   puts "$proc: [time [list $proc $n] 100]"

}</lang>

Output:

calculate 100 rows:
pascal_iterative: 2800.14 microseconds per iteration
pascal_coefficients: 8760.98 microseconds per iteration
pascal_combinations: 38176.66 microseconds per iteration

TI-83 BASIC

Using Addition of Previous Rows

<lang ti83b>PROGRAM:PASCALTR

Lbl IN

ClrHome

Disp "NUMBER OF ROWS"

Input N

If N < 1:Goto IN

{N,N}→dim([A])

"CHEATING TO MAKE IT FASTER"

For(I,1,N)

1→[A](1,1)

End

For(I,2,N)

For(J,2,I)

[A](I-1,J-1)+[A](I-1,J)→[A](I,J)

End

[A]</lang>

Using nCr Function

<lang ti83b>PROGRAM:PASCALTR

Lbl IN

ClrHome

Disp "NUMBER OF ROWS"

Input N

If N < 1:Goto IN

{N,N}→dim([A])

For(I,2,N)

For(J,2,I)

(I-1) nCr (J-1)→[A](I,J)

End

[A]</lang>

Turing

<lang turing> procedure pascal (n : int)

   for i : 0 .. n
       var c : int
       c := 1
       for k : 0 .. i
           put intstr(c) + " " ..
           c := c * (i - k) div (k + 1)
       end for
       put ""
   end for

end pascal

pascal(5)</lang>

uBasic/4tH

<lang>Input "Number Of Rows: "; N @(1) = 1 Print Tab((N+1)*3);"1"

For R = 2 To N

   Print Tab((N-R)*3+1);
   For I = R To 1 Step -1
       @(I) = @(I) + @(I-1)
       Print Using "______";@(i);
   Next

Number Of Rows: 10
                                 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

0 OK, 0:380

UNIX Shell

Works with: Bourne Again SHell

Any n <= 1 will print the "1" row. <lang bash>#! /bin/bash pascal() {

   local -i n=${1:-1}
   if (( n <= 1 )); then
       echo 1
   else
       local output=$( $FUNCNAME $((n - 1)) )
       set -- $( tail -n 1 <<<"$output" )   # previous row
       echo "$output"
       printf "1 "
       while -n $1 ; do
           printf "%d " $(( $1 + ${2:-0} ))
           shift
       done
       echo
   fi

} pascal "$1"</lang>

Ursala

Zero maps to the empty list. Negatives are inexpressible. This solution uses a library function for binomial coefficients. <lang Ursala>#import std

import nat

pascal = choose**ziDS+ iota*t+ iota+ successor</lang> This solution uses direct summation. The algorithm is to insert zero at the head of a list (initially the unit list <1>), zip it with its reversal, map the sum over the list of pairs, iterate n times, and return the trace. <lang Ursala>#import std

import nat

pascal "n" = (next"n" sum*NiCixp) <1></lang> test program: <lang Ursala>#cast %nLL

example = pascal 10</lang>

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>>

VBScript

Derived from the BASIC version. <lang vb>Pascal_Triangle(WScript.Arguments(0)) Function Pascal_Triangle(n) Dim values(100) values(1) = 1 WScript.StdOut.Write values(1) WScript.StdOut.WriteLine For row = 2 To n For i = row To 1 Step -1 values(i) = values(i) + values(i-1) WScript.StdOut.Write values(i) & " " Next WScript.StdOut.WriteLine Next End Function</lang>

Output:

Invoke from a command line.

F:\VBScript>cscript /nologo rosettacode-pascals_triangle.vbs 6
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

Vedit macro language

Summing from Previous Rows

Translation of: BASIC

Vedit macro language does not have actual arrays (edit buffers are normally used for storing larger amounts of data). However, a numeric register can be used as index to access another numeric register. For example, if #99 contains value 2, then #@99 accesses contents of numeric register #2.

<lang vedit>#100 = Get_Num("Number of rows: ", STATLINE)

0=0; #1=1

Ins_Char(' ', COUNT, #100*3-2) Num_Ins(1) for (#99 = 2; #99 <= #100; #99++) {

   Ins_Char(' ', COUNT, (#100-#99)*3)
   #@99 = 0
   for (#98 = #99; #98 > 0; #98--) {

#97 = #98-1 #@98 += #@97 Num_Ins(#@98, COUNT, 6)

   }
   Ins_Newline

}</lang>

Using binary coefficients

Translation of: BASIC

<lang vedit>#1 = Get_Num("Number of rows: ", STATLINE) for (#2 = 0; #2 < #1; #2++) {

   #3 = 1
   Ins_Char(' ', COUNT, (#1-#2-1)*3)
   for (#4 = 0; #4 <= #2; #4++) {

Num_Ins(#3, COUNT, 6) #3 = #3 * (#2-#4) / (#4+1)

   }
   Ins_Newline

}</lang>

Visual Basic

Works with: Visual Basic version VB6 Standard

<lang vb>Sub pascaltriangle()

   'Pascal's triangle
   Const m = 11
   Dim t(40) As Integer, u(40) As Integer
   Dim i As Integer, n As Integer, s As String, ss As String
   ss = ""
   For n = 1 To m
       u(1) = 1
       s = ""
       For i = 1 To n
           u(i + 1) = t(i) + t(i + 1)
           s = s & u(i) & " "
           t(i) = u(i)
       Next i
       ss = ss & s & vbCrLf
   Next n
   MsgBox ss, , "Pascal's triangle"

End Sub 'pascaltriangle</lang>

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

XPL0

<lang XPL0>include c:\cxpl\codes;

proc Pascal(N); \Display the first N rows of Pascal's triangle int N; \if N<=0 then nothing is displayed int Row, I, Old(40), New(40); [for Row:= 0 to N-1 do

       [New(0):= 1;
       for I:= 1 to Row do New(I):= Old(I-1) + Old(I);
       for I:= 1 to (N-Row-1)*2 do ChOut(0, ^ );
       for I:= 0 to Row do
               [if New(I)<100 then ChOut(0, ^ );
               IntOut(0, New(I));
               if New(I)<10 then ChOut(0, ^ );
               ChOut(0, ^ );
               ];
       New(Row+1):= 0;
       I:= Old;  Old:= New;  New:= I;
       CrLf(0);
       ];

];

Pascal(13)</lang>

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

zkl

Translation of: C

<lang zkl>fcn pascalTriangle(n){ // n<=0-->""

  foreach i in (n){
     c := 1;
     print(" "*(2*(n-1-i)));
     foreach k in (i+1){
        print("%3d ".fmt(c));
        c = c * (i-k)/(k+1);
     }
     println();
  }

}

pascalTriangle(8);</lang>

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