Pascal's triangle

From Rosetta Code
Task
Pascal's triangle
You are encouraged to solve this task according to the task description, using any language you may know.

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

Its first few rows look like this:

    1
   1 1
  1 2 1
 1 3 3 1

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

For example, the next row of the triangle 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)
  1   (since the last element of each row doesn't have two elements above it)

So the triangle now looks like this:

    1
   1 1
  1 2 1
 1 3 3 1
1 4 6 4 1

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



Contents

360 Assembly[edit]

Translation of: PL/I
*        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 [email protected]
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
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[edit]

One way, using array operations:

 
\ 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
 

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

 
: 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
 

Ada[edit]

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]]

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;

The implementation of that auxiliary package "Pascal":

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;

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

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;
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[edit]

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
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[edit]

Pascal' s triangle of order ⍵

 
{A←0,⍳⍵ ⋄ ⍉A∘.!A}
 

example

 
{A←0,⍳⍵ ⋄ ⍉A∘.!A} 3
 
1 0 0 0
1 1 0 0
1 2 1 0
1 3 3 1


AppleScript[edit]

-- pascal :: Int -> [[Int]]
on pascal(intRows)
 
script addRow
on nextRow(row)
script add
on lambda(a, b)
a + b
end lambda
end script
 
zipWith(add, [0] & row, row & [0])
end nextRow
 
on lambda(xs)
xs & {nextRow(item -1 of xs)}
end lambda
end script
 
foldr(addRow, {{1}}, range(1, intRows - 1))
end pascal
 
 
-- TEST
 
on run
set lstTriangle to pascal(7)
 
script spaced
on lambda(xs)
script rightAlign
on lambda(x)
text -4 thru -1 of (" " & x)
end lambda
end script
 
intercalate("", map(rightAlign, xs))
end lambda
end script
 
script indented
on lambda(a, x)
set strIndent to leftSpace of a
 
{rows:strIndent & x & linefeed & rows of a, leftSpace:leftSpace of a & " "}
end lambda
end script
 
rows of foldr(indented, {rows:"", leftSpace:""}, map(spaced, lstTriangle))
end run
 
 
 
-- GENERIC LIBRARY FUNCTIONS
 
-- foldr :: (a -> b -> a) -> a -> [b] -> a
on foldr(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from lng to 1 by -1
set v to lambda(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldr
 
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to lambda(item i of xs, i, xs)
end repeat
return lst
end tell
end map
 
-- zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
on zipWith(f, xs, ys)
set nx to length of xs
set ny to length of ys
if nx < 1 or ny < 1 then
{}
else
set lng to cond(nx < ny, nx, ny)
set lst to {}
tell mReturn(f)
repeat with i from 1 to lng
set end of lst to lambda(item i of xs, item i of ys)
end repeat
return lst
end tell
end if
end zipWith
 
-- cond :: Bool -> (a -> b) -> (a -> b) -> (a -> b)
on cond(bool, f, g)
if bool then
f
else
g
end if
end cond
 
-- intercalate :: Text -> [Text] -> Text
on intercalate(strText, lstText)
set {dlm, my text item delimiters} to {my text item delimiters, strText}
set strJoined to lstText as text
set my text item delimiters to dlm
return strJoined
end intercalate
 
-- range :: Int -> Int -> [Int]
on range(m, n)
set lng to (n - m) + 1
set base to m - 1
set lst to {}
repeat with i from 1 to lng
set end of lst to i + base
end repeat
return lst
end range
 
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property lambda : f
end script
end if
end mReturn
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

AutoHotkey[edit]

ahk forum: discussion

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
Alternate
Works with: AutoHotkey L
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
}

n <= 0 returns empty

AWK[edit]

$ 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}}'
Output:

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

BASIC[edit]

Summing from Previous Rows[edit]

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.

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

Batch File[edit]

Based from the Fortran Code.

@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.
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[edit]

      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%
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[edit]

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

( 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
)
&
)
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[edit]

 
blsq ) {1}{1 1}{^^2CO{p^?+}m[1+]1[+}15E!#s<-spbx#S
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
 


C[edit]

Translation of: Fortran
#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;
}

Recursive[edit]

#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);
}

Adding previous row values[edit]

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);
}

C++[edit]

#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);
}
 
}

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

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.

// 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;
}
 
 

C++11 (with a class)[edit]

A full fledged example with a class definition and methods to retrieve data, worthy of the title object-oriented.

// 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;
}
 
 

C#[edit]

Translation of: Fortran

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

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);
}
}
}

Clojure[edit]

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).

(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)

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:

 
(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))))
 

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

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

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

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

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

 
(take 10 pascal) ; returns a list of the first 10 pascal rows
 

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

 
(nth pascal 10) ;returns the nth row
 

CoffeeScript[edit]

This version assumes n is an integer and n >= 1. It efficiently computes binomial coefficients.

 
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)
 
 
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[edit]

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

(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)))))

An iterative solution with loop, using nconc instead of collect to keep track of the last cons. Otherwise, it would be necessary to traverse the list to do a (rplacd (last a) (list 1)).

(defun pascal-next-row (a)
(loop :for q :in a
:and p = 0 :then q
:as s = (list (+ p q))
:nconc s :into a
:finally (rplacd s (list 1))
(return a)))
 
(defun pascal (n)
(loop :for a = (list 1) :then (pascal-next-row a)
:repeat n
:collect a))

D[edit]

Less functional Version[edit]

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]]);
}

More functional Version[edit]

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;
}
Output:
[[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1]]

Alternative Version[edit]

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).

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));
}
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[edit]

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.

DWScript[edit]

Doesn't print anything for negative or null values.

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);
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[edit]

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.

def pascalsTriangle(n, out) {
def row := [].diverge(int)
out.print("<table style='text-align: center; border: 0; border-collapse: collapse;'>")
for y in 1..n {
out.print("<tr>")
row.push(1)
def skip := n - y
if (skip > 0) {
out.print(`<td colspan="$skip"></td>`)
}
for x => v in row {
out.print(`<td>$v</td><td></td>`)
}
for i in (1..!y).descending() {
row[i] += row[i - 1]
}
out.println("</tr>")
}
out.print("</table>")
}
def out := <file:triangle.html>.textWriter()
try {
pascalsTriangle(15, out)
} finally {
out.close()
}
makeCommand("yourFavoriteWebBrowser")("triangle.html")

Eiffel[edit]

 
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
 

Elixir[edit]

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)
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[edit]

 
-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].
 
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[edit]

 
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
 

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[edit]

Summing from Previous Rows[edit]

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
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#[edit]

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"
 

Factor[edit]

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

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 ;

It works as:

5 pascal .
{ { 1 } { 1 1 } { 1 2 1 } { 1 3 3 1 } { 1 4 6 4 1 } }

Fantom[edit]

 
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)
}
}
 

Forth[edit]

: 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 ;

This is a bit more efficient.

Translation of: C
: 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

Fortran[edit]

Works with: Fortran version 90 and later

Prints nothing for n<=0. Output formatting breaks down for n>20

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

FreeBASIC[edit]

' FB 1.05.0 Win64
 
Sub pascalTriangle(n As UInteger)
If n = 0 Then Return
Dim prevRow(1 To n) As UInteger
Dim currRow(1 To n) As UInteger
Dim start(1 To n) As UInteger ''stores starting column for each row
start(n) = 1
For i As Integer = n - 1 To 1 Step -1
start(i) = start(i + 1) + 3
Next
prevRow(1) = 1
Print Tab(start(1));
Print 1U
For i As UInteger = 2 To n
For j As UInteger = 1 To i
If j = 1 Then
Print Tab(start(i)); "1";
currRow(1) = 1
ElseIf j = i Then
Print " 1"
currRow(i) = 1
Else
currRow(j) = prevRow(j - 1) + prevRow(j)
Print Using "######"; currRow(j); " ";
End If
Next j
For j As UInteger = 1 To i
prevRow(j) = currRow(j)
Next j
Next i
End Sub
 
pascalTriangle(14)
Print
Print "Press any key to quit"
Sleep
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

FunL[edit]

Summing from Previous Rows[edit]

Translation of: Scala
import lists.zip
 
def
pascal( 1 ) = [1]
pascal( n ) = [1] + map( (a, b) -> a + b, zip(pascal(n-1), pascal(n-1).tail()) ) + [1]

Combinations[edit]

Translation of: Haskell
import integers.choose
 
def pascal( n ) = [choose( n - 1, k ) | k <- 0..n-1]

Pascal's Triangle[edit]

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 )
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[edit]

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 ]

Go[edit]

No output for n < 1. Otherwise, output formatted left justified.

 
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)
}
 

Output:

1
1 1
1 2 1
1 3 3 1

Groovy[edit]

Recursive[edit]

In the spirit of the Haskell "think in whole lists" solution here is a list-driven, minimalist solution:

def pascal
pascal = { n -> (n <= 1) ? [1] : [[0] + pascal(n - 1), pascal(n - 1) + [0]].transpose().collect { it.sum() } }

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:

def count = 15
(1..count).each { n ->
printf ("%2d:", n); (0..(count-n)).each { print " " }; pascal(n).each{ printf("%6d ", it) }; println ""
}
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[edit]

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

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[edit]

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

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

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

extendWith f [] = []
extendWith f xs@(x:ys) = x : zapWith f xs ys

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

pascal = iterate (extendWith (+)) [1]

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

*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]]

A shorter approach, plagiarized from [2]

-- generate next row from current row
nextRow row = zipWith (+) ([0] ++ row) (row ++ [0])
 
-- returns the first n rows
pascal = iterate nextRow [1]


With binomial coefficients:

fac = product . enumFromTo 1
 
binCoef n k = (fac n) `div` ((fac k) * (fac $ n - k))
 
pascal n = map (binCoef $ n - 1) [0..n-1]

Example:

*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
 

HicEst[edit]

   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

Icon and Unicon[edit]

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.

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

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[edit]

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

J[edit]

   !~/~ i.5
1 0 0 0 0
1 1 0 0 0
1 2 1 0 0
1 3 3 1 0
1 4 6 4 1
   ([: ":@-.&0"1 !~/~)@i. 5
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
   (-@|. |."_1 [: ":@-.&0"1 !~/~)@i. 5
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

See the talk page for explanation of earlier version

See also Pascal matrix generation and Sierpinski triangle.

Java[edit]

Summing from Previous Rows[edit]

Works with: Java version 1.5+
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);
}
}

Combinations[edit]

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.

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;
}
}

Using arithmetic calculation of each row element[edit]

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.

 
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);
}
}
 

JavaScript[edit]

ES5[edit]

Imperative[edit]

Works with: SpiderMonkey
Works with: V8
// 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);

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[edit]

Translation of: Haskell
(function (n) {
'use strict';
 
// A Pascal triangle of n rows
 
// pascal :: Int -> [[Int]]
function pascal(n) {
return range(1, n - 1)
.reduce(function (a) {
var lstPreviousRow = a.slice(-1)[0];
 
return a
.concat(
[zipWith(
function (a, b) {
return a + b
},
[0].concat(lstPreviousRow),
lstPreviousRow.concat(0)
)]
);
}, [[1]]);
}
 
 
 
// GENERIC FUNCTIONS
 
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
function zipWith(f, xs, ys) {
return xs.length === ys.length ? (
xs.map(function (x, i) {
return f(x, ys[i]);
})
) : undefined;
}
 
// range :: Int -> Int -> [Int]
function range(m, n) {
return Array.apply(null, Array(n - m + 1))
.map(function (x, i) {
return m + i;
});
}
 
// TEST
var lstTriangle = pascal(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);

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],[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]]


ES6[edit]

(() => {
'use strict';
 
// pascal :: Int -> [[Int]]
let pascal = n =>
range(1, n - 1)
.reduce(a => {
let lstPreviousRow = a.slice(-1)[0];
 
return a
.concat([zipWith((a, b) => a + b,
[0].concat(lstPreviousRow),
lstPreviousRow.concat(0)
)]);
}, [
[1]
]);
 
// GENERIC FUNCTIONS
 
// Int -> Int -> Maybe Int -> [Int]
let range = (m, n, step) => {
let d = (step || 1) * (n >= m ? 1 : -1);
return Array.from({
length: Math.floor((n - m) / d) + 1
}, (_, i) => m + (i * d));
},
 
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
zipWith = (f, xs, ys) =>
xs.length === ys.length ? (
xs.map((x, i) => f(x, ys[i]))
) : undefined;
 
// TEST
return pascal(7)
.reduceRight((a, x) => {
let strIndent = a.indent;
 
return {
rows: strIndent + x
.map(n => (' ' + n).slice(-4))
.join('') + '\n' + a.rows,
indent: strIndent + ' '
};
}, {
rows: '',
indent: ''
}).rows;
})();
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[edit]

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.

# 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 ;

Example:

pascal(5)
Output:
$ jq -c -n -f pascal_triangle.jq
[1]
[1,1]
[1,2,1]
[1,3,3,1]
[1,4,6,4,1]

Using recurse/1

Here is an equivalent implementation that uses the built-in filter, recurse/1, instead of the inner function.

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;

K[edit]

 
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)

Kotlin[edit]

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())

Liberty BASIC[edit]

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

Locomotive Basic[edit]

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

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

[edit]

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]

Lua[edit]

 
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
 

Maple[edit]

f:=n->seq(print(seq(binomial(i,k),k=0..i)),i=0..n-1);
 
f(3);
   1
  1 1
 1 2 1

Mathematica[edit]

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

MmaPascal.png

MATLAB / Octave[edit]

A matrix containing the pascal triangle can be obtained this way:

pascal(n);
>> 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:

  binomCoeff = diag(rot90(pascal(n)))', 
>> 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[edit]

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" */


Metafont[edit]

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

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

NetRexx[edit]

/* 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*/
 
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[edit]

Like J

(pascal.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

Using it

|loaddefs 'pascal.nial'
|pascal 5

Nim[edit]

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)

OCaml[edit]

(* 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]

Octave[edit]

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);

Oforth[edit]

No result if n <= 0

: pascal(n)  [ 1 ] #[ dup println dup 0 + 0 rot + zipWith(#+) ] times(n) drop ;
Output:
10 pascal
[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[edit]

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}}

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

PARI/GP[edit]

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);
);
}

Pascal[edit]

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.

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[edit]

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).

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);
}
}

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

Here is a non-obvious version using bignum, which is limited to the first 23 rows because of the algorithm used:

use bignum;
sub pascal_line { $_[0] ? unpack "A(A6)*", 1000001**$_[0] : 1 }
sub pascal { print "@{[map -+-$_, pascal_line $_]}\n" for 0..$_[0]-1 }

Perl 6[edit]

Works with: rakudo version 2015-10-03

using a lazy sequence generator[edit]

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:

sub pascal {
[1], { [0, |$_ Z+ |$_, 0] } ... *
}
 
.say for pascal[^10];

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, and use an explicit parameter $prev for variety:

constant @pascal = [1], -> $prev { [0, |$prev Z+ |$prev, 0] } ... *;
 
.say for @pascal[^10];

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

recursive[edit]

Translation of: Haskell
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;

Non-positive inputs throw a multiple-dispatch error.

iterative[edit]

Translation of: Perl
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;

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[edit]

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
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

"Reflected" Pascal's triangle, it uses symmetry property to "mirror" second part. It determines even and odd strings. automatically.

PHP[edit]

 
<?php
//Author Ivan Gavryshin @dcc0
function tre($n) {
$ck=1;
$kn=$n+1;
 
if($kn%2==0) {
$kn=$kn/2;
$i=0;
}
else
{
 
$kn+=1;
$kn=$kn/2;
$i= 1;
}
 
for ($k = 1; $k <= $kn-1; $k++) {
$ck = $ck/$k*($n-$k+1);
$arr[] = $ck;
echo "+" . $ck ;
 
}
 
 
if ($kn>1) {
echo $arr[i];
$arr=array_reverse($arr);
for ($i; $i<= $kn-1; $i++) {
echo "+" . $arr[$i] ;
}
 
}
 
}
//set amount of strings here
while ($n<=20) {
++$n;
echo tre($n);
echo "<br/>";
}
 
 
?>
 

PHP[edit]

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 '<br>';
}
                                       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[edit]

 
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;
 
 
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
 

PicoLisp[edit]

Translation of: C
(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) ) )

Potion[edit]

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)

PowerShell[edit]

 
$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++
}
 

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[edit]

Difference-lists are used to make quick append.

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]).
 

Output :

 ?- 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.
 

PureBasic[edit]

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()

Python[edit]

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

Or, by creating a scan function:

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)

q[edit]

 
pascal:{(x-1){0+':x,0}\1}
pascal 5
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
 

Qi[edit]

Translation of: Haskell
 
(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))
 

R[edit]

Translation of: Octave
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)
}
}

Here's an R version:

pascalTriangle <- function(h) {
lapply(0:h, function(i) choose(i, 0:i))
}

Racket[edit]

Iterative version by summing rows up to .

#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))))
 
 

RapidQ[edit]

Summing from Previous Rows[edit]

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.

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

Using binary coefficients[edit]

Translation of: BASIC
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

Retro[edit]

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

REXX[edit]

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:

  •   Khayyam's triangle
  •   Khayyam─Pascal's triangle
  •   Tartaglia's triangle
  •   Yang Hui's triangle
/*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*/

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[edit]

 
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
 

Output:

     1
    1 1
   1 2 1
  1 3 3 1
 1 4 6 4 1

Ruby[edit]

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)}
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):

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)}
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[edit]

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

Scala[edit]

Simple recursive row definition:

 
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)
}
}
 

Function to pretty print n rows:

 
def prettytri(n:Int) = (1 to n) foreach {i=>print(" "*(n-i)); tri(i) map (c=>print(c+" ")); println}
prettytri(5)
 

Outputs:

    1 
   1 1 
  1 2 1 
 1 3 3 1 
1 4 6 4 1 

Scheme[edit]

Works with: Scheme version RRS
(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)
 

Output:

((1) (1 1) (1 2 1) (1 3 3 1) (1 4 6 4 1))

Seed7[edit]

$ 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;

Sidef[edit]

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);

Tcl[edit]

Summing from Previous Rows[edit]

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

Using binary coefficients[edit]

Translation of: BASIC
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]

Combinations[edit]

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.

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]

Comparing Performance[edit]

set n 100
puts "calculate $n rows:"
foreach proc {pascal_iterative pascal_coefficients pascal_combinations} {
puts "$proc: [time [list $proc $n] 100]"
}
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[edit]

Using Addition of Previous Rows[edit]

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
:End
:[A]

Using nCr Function[edit]

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
:End
:[A]

Turing[edit]

 
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)

uBasic/4tH[edit]

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
Next
 
Print
End

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

0 OK, 0:380

UNIX Shell[edit]

Works with: Bourne Again SHell

Any n <= 1 will print the "1" row.

#! /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"

Ursala[edit]

Zero maps to the empty list. Negatives are inexpressible. This solution uses a library function for binomial coefficients.

#import std
#import nat
 
pascal = choose**ziDS+ iota*t+ iota+ successor

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.

#import std
#import nat
 
pascal "n" = (next"n" sum*NiCixp) <1>

test program:

#cast %nLL
 
example = pascal 10
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[edit]

Derived from the BASIC version.

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
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[edit]

Summing from Previous Rows[edit]

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 [email protected] accesses contents of numeric register #2.

#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)
[email protected] = 0
for (#98 = #99; #98 > 0; #98--) {
#97 = #98-1
[email protected] += [email protected]
Num_Ins([email protected], COUNT, 6)
}
Ins_Newline
}

Using binary coefficients[edit]

Translation of: BASIC
#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
}


Visual Basic[edit]

Works with: Visual Basic version VB6 Standard
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
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[edit]

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)
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[edit]

Translation of: C
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);
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