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{{task|Arithmetic operations}}
[[wp:Pascal's triangle|Pascal's triangle]] is an arithmetic and geometric figure often associated with the name of [[wp:Blaise Pascal|Blaise Pascal]], but also studied centuries earlier in India, Persia, China and elsewhere.
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:
- [[Evaluate binomial coefficients]]
360 Assembly
{{trans|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 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
{{out}}
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:
\ 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
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 [[http://rosettacode.org/wiki/Catalan_numbers/Pascal%27s_triangle]]
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;
{{out}}
>./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
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
{{Out}}
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
ALGOL W
begin
% prints the first n lines of Pascal's triangle lines %
% if n is <= 0, no output is produced %
procedure printPascalTriangle( integer value n ) ;
if n > 0 then begin
integer array pascalLine ( 1 :: n );
pascalLine( 1 ) := 1;
for line := 1 until n do begin
for i := line - 1 step - 1 until 2 do pascalLine( i ) := pascalLine( i - 1 ) + pascalLine( i );
pascalLine( line ) := 1;
write( s_w := 0, " " );
for i := line until n do writeon( s_w := 0, " " );
for i := 1 until line do writeon( i_w := 6, s_w := 0, pascalLine( i ) )
end for_line ;
end printPascalTriangle ;
printPascalTriangle( 8 )
end.
{{out}}
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
APL
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
Drawing n rows from a generator:
-- pascal :: Generator [[Int]]
on pascal()
script nextRow
on |λ|(row)
zipWith(my plus, {0} & row, row & {0})
end |λ|
end script
iterate(nextRow, {1})
end pascal
on run
showPascal(take(7, pascal()))
end run
-- showPascal :: [[Int]] -> String
on showPascal(xs)
set w to length of intercalate(" ", item -1 of xs)
script align
on |λ|(x)
|center|(w, space, intercalate(" ", x))
end |λ|
end script
unlines(map(align, xs))
end showPascal
-- GENERIC ABSTRACTIONS -------------------------------------------------------
-- center :: Int -> Char -> String -> String
on |center|(n, cFiller, strText)
set lngFill to n - (length of strText)
if lngFill > 0 then
set strPad to replicate(lngFill div 2, cFiller) as text
set strCenter to strPad & strText & strPad
if lngFill mod 2 > 0 then
cFiller & strCenter
else
strCenter
end if
else
strText
end if
end |center|
-- intercalate :: String -> [String] -> String
on intercalate(sep, xs)
set {dlm, my text item delimiters} to {my text item delimiters, sep}
set s to xs as text
set my text item delimiters to dlm
return s
end intercalate
-- iterate :: (a -> a) -> a -> Generator [a]
on iterate(f, x)
script
property v : missing value
property g : mReturn(f)'s |λ|
on |λ|()
if missing value is v then
set v to x
else
set v to g(v)
end if
return v
end |λ|
end script
end iterate
-- length :: [a] -> Int
on |length|(xs)
set c to class of xs
if list is c or string is c then
length of xs
else
2 ^ 30 -- (simple proxy for non-finite)
end if
end |length|
-- 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 |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- min :: Ord a => a -> a -> a
on min(x, y)
if y < x then
y
else
x
end if
end min
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- plus :: Num -> Num -> Num
on plus(a, b)
a + b
end plus
-- Egyptian multiplication - progressively doubling a list, appending
-- stages of doubling to an accumulator where needed for binary
-- assembly of a target length
-- replicate :: Int -> a -> [a]
on replicate(n, a)
set out to {}
if n < 1 then return out
set dbl to {a}
repeat while (n > 1)
if (n mod 2) > 0 then set out to out & dbl
set n to (n div 2)
set dbl to (dbl & dbl)
end repeat
return out & dbl
end replicate
-- take :: Int -> [a] -> [a]
-- take :: Int -> String -> String
on take(n, xs)
set c to class of xs
if list is c then
if 0 < n then
items 1 thru min(n, length of xs) of xs
else
{}
end if
else if string is c then
if 0 < n then
text 1 thru min(n, length of xs) of xs
else
""
end if
else if script is c then
set ys to {}
repeat with i from 1 to n
set end of ys to xs's |λ|()
end repeat
return ys
else
missing value
end if
end take
-- unlines :: [String] -> String
on unlines(xs)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, linefeed}
set str to xs as text
set my text item delimiters to dlm
str
end unlines
-- unwords :: [String] -> String
on unwords(xs)
set {dlm, my text item delimiters} to {my text item delimiters, space}
set s to xs as text
set my text item delimiters to dlm
return s
end unwords
-- zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
on zipWith(f, xs, ys)
set lng to min(|length|(xs), |length|(ys))
if 1 > lng then return {}
set xs_ to take(lng, xs) -- Allow for non-finite
set ys_ to take(lng, ys) -- generators like cycle etc
set lst to {}
tell mReturn(f)
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs_, item i of ys_)
end repeat
return lst
end tell
end zipWith
{{Out}}
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
ahk forum: [http://www.autohotkey.com/forum/viewtopic.php?p=276617#276617 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
$ 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}}'
{{Out}}
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.
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
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.
{{Out}}
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
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%
{{Out}}
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
0" :swor fo rebmuN">:#,_&> 55+, v
v01*p00-1:g00.:<1p011p00:\-1_v#:<
>g:1+10p/48*,:#^_$ 55+,1+\: ^>$$@
{{Out}}
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
( 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
)
&
)
{{Out}}
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
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
{{trans|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
#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
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++
#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)=== 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) === 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#
{{trans|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);
}
}
}
===Arbitrarily large numbers (BigInteger), arbitrary row selection===
using System;
using System.Linq;
using System.Numerics;
using System.Collections.Generic;
namespace RosettaCode
{
public static class PascalsTriangle
{
public static IEnumerable<BigInteger[]> GetTriangle(int quantityOfRows)
{
IEnumerable<BigInteger> range = Enumerable.Range(0, quantityOfRows).Select(num => new BigInteger(num));
return range.Select(num => GetRow(num).ToArray());
}
public static IEnumerable<BigInteger> GetRow(BigInteger rowNumber)
{
BigInteger denominator = 1;
BigInteger numerator = rowNumber;
BigInteger currentValue = 1;
for (BigInteger counter = 0; counter <= rowNumber; counter++)
{
yield return currentValue;
currentValue = BigInteger.Multiply(currentValue, numerator--);
currentValue = BigInteger.Divide(currentValue, denominator++);
}
yield break;
}
public static string FormatTriangleString(IEnumerable<BigInteger[]> triangle)
{
int maxDigitWidth = triangle.Last().Max().ToString().Length;
IEnumerable<string> rows = triangle.Select(arr =>
string.Join(" ", arr.Select(array => CenterString(array.ToString(), maxDigitWidth)) )
);
int maxRowWidth = rows.Last().Length;
return string.Join(Environment.NewLine, rows.Select(row => CenterString(row, maxRowWidth)));
}
private static string CenterString(string text, int width)
{
int spaces = width - text.Length;
int padLeft = (spaces / 2) + text.Length;
return text.PadLeft(padLeft).PadRight(width);
}
}
}
Example:
static void Main()
{
IEnumerable<BigInteger[]> triangle = PascalsTriangle.GetTriangle(20);
string output = PascalsTriangle.FormatTriangleString(triangle)
Console.WriteLine(output);
}
{{out}}
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
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).
(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
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)
{{Out}}
> 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
Commodore BASIC
10 INPUT "HOW MANY";N
20 IF N<1 THEN END
30 DIM C(N)
40 DIM D(N)
50 LET C(1)=1
60 LET D(1)=1
70 FOR J=1 TO N
80 FOR I=1 TO N-J+1
90 PRINT " ";
100 NEXT I
110 FOR I=1 TO J
120 PRINT C(I)" ";
130 NEXT I
140 PRINT
150 IF J=N THEN END
160 C(J+1)=1
170 D(J+1)=1
180 FOR I=1 TO J-1
190 D(I+1)=C(I)+C(I+1)
200 NEXT I
210 FOR I=1 TO J
220 C(I)=D(I)
230 NEXT I
240 NEXT J
Output:
## Common Lisp
To evaluate, call (pascal n). For n < 1, it simply returns nil.
```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)))))
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))
Another iterative solution, this time using pretty-printing to automatically print the triangle in the shape of a triangle in the terminal. The pascal-print function determines the length of the final row and uses it to decide how wide the triangle should be.
(defun next-pascal (l)
`(1 ,@(loop for i from 0 to (- (length l) 2)
collect (+ (nth i l) (nth (1+ i) l)))
1))
(defun pascal-print (r)
(let* ((pasc (loop with p = (list (list 1))
repeat r do (nconc p (list (apply #'next-pascal (last p))))
finally (return p)))
(len (length (format nil "~A" (car (last pasc))))))
(format t (format nil "~~{~~~D:@<~~{~~A ~~}~~>~~%~~}" len) pasc)))
For example:
(pascal-print 4)
```lisp
(pascal-print 8)
## Component Pascal
{{Works with|BlackBox Component Builder}}
```oberon2
MODULE PascalTriangle;
IMPORT StdLog, DevCommanders, TextMappers;
TYPE
Expansion* = POINTER TO ARRAY OF LONGINT;
PROCEDURE Show*(e: Expansion);
VAR
i: INTEGER;
BEGIN
i := 0;
WHILE (i < LEN(e)) & (e[i] # 0) DO
StdLog.Int(e[i]);
INC(i)
END;
StdLog.Ln
END Show;
PROCEDURE GenFor*(p: LONGINT): Expansion;
VAR
expA,expB: Expansion;
i,j: LONGINT;
PROCEDURE Swap(VAR x,y: Expansion);
VAR
swap: Expansion;
BEGIN
swap := x; x := y; y := swap
END Swap;
BEGIN
ASSERT(p >= 0);
NEW(expA,p + 2);NEW(expB,p + 2);
FOR i := 0 TO p DO
IF i = 0 THEN expA[0] := 1
ELSE
FOR j := 0 TO i DO
IF j = 0 THEN
expB[j] := expA[j]
ELSE
expB[j] := expA[j - 1] + expA[j]
END
END;
Swap(expA,expB)
END;
END;
expB := NIL; (* for the GC *)
RETURN expA
END GenFor;
PROCEDURE Do*;
VAR
s: TextMappers.Scanner;
exp: Expansion;
BEGIN
s.ConnectTo(DevCommanders.par.text);
s.SetPos(DevCommanders.par.beg);
s.Scan;
WHILE (~s.rider.eot) DO
IF (s.type = TextMappers.char) & (s.char = '~') THEN
RETURN
ELSIF (s.type = TextMappers.int) THEN
exp := GenFor(s.int);
Show(exp)
END;
s.Scan
END
END Do;
END PascalTriangle.
Execute: ^Q PascalTriangle.Do 0 1 2 3 4 5 6 7 8 9 10 11 12~
{{out}}
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
D
Less functional Version
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
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;
}
{{out}}
[[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).
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));
}
{{out}}
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
*
* *
* _ *
* * * *
* _ _ _ *
* * _ _ * *
* _ * _ * _ *
* * * * * * * *
* _ _ _ _ _ _ _ *
* * _ _ _ _ _ _ * *
* _ * _ _ _ _ _ * _ *
* * * * _ _ _ _ * * * *
* _ _ _ * _ _ _ * _ _ _ *
* * _ _ * * _ _ * * _ _ * *
* _ * _ * _ * _ * _ * _ * _ *
* * * * * * * * * * * * * * * *
Dart
import 'dart:io';
pascal(n) {
if(n<=0) print("Not defined");
else if(n==1) print(1);
else {
List<List<int>> matrix = new List<List<int>>();
matrix.add(new List<int>());
matrix.add(new List<int>());
matrix[0].add(1);
matrix[1].add(1);
matrix[1].add(1);
for (var i = 2; i < n; i++) {
List<int> list = new List<int>();
list.add(1);
for (var j = 1; j<i; j++) {
list.add(matrix[i-1][j-1]+matrix[i-1][j]);
}
list.add(1);
matrix.add(list);
}
for(var i=0; i<n; i++) {
for(var j=0; j<=i; j++) {
stdout.write(matrix[i][j]);
stdout.write(' ');
}
stdout.write('\n');
}
}
}
void main() {
pascal(0);
pascal(1);
pascal(3);
pascal(6);
}
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.
DWScript
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);
{{Out}}
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.
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>")
}
.textWriter()
try {
pascalsTriangle(15, out)
} finally {
out.close()
}
makeCommand("yourFavoriteWebBrowser")("triangle.html")
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
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)
{{out}}
[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
-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].
{{Out}}
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
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
Summing from Previous Rows
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
{{Out}}
{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}
=={{header|F Sharp|F#}}==
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
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
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)
}
}
=={{header|Fōrmulæ}}==
In [http://wiki.formulae.org/Pascal%27s_triangle this] page you can see the solution of this task.
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text ([http://wiki.formulae.org/Editing_F%C5%8Drmul%C3%A6_expressions more info]). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.
The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.
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 ;
This is a bit more efficient. {{trans|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
{{works with|Fortran|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
' 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
{{out}}
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
Summing from Previous Rows
{{trans|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
{{trans|Haskell}}
import integers.choose
def pascal( n ) = [choose( n - 1, k ) | k <- 0..n-1]
=== Pascal's Triangle ===
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 )
{{out}}
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
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
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
Recursive
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 ""
}
{{out}}
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
=={{header|GW-BASIC}}==
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
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 [http://www.haskell.org/haskellwiki/Blow_your_mind]
-- generate next row from current row
nextRow row = zipWith (+) ([0] ++ row) (row ++ [0])
-- returns the first n rows
pascal = iterate nextRow [1]
Alternatively, using list comprehensions:
pascal :: [[Integer]]
pascal =
(1 : [ 0 | _ <- head pascal])
: [zipWith (+) (0:row) row | row <- pascal]
*Pascal> take 5 <$> (take 5 $ triangle)
[[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]]
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
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
=={{header|Icon}} and {{header|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.
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
{{libheader|Icon Programming Library}} [http://www.cs.arizona.edu/icon/library/src/procs/math.icn math provides binocoef] [http://www.cs.arizona.edu/icon/library/src/procs/pascal.icn 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
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
=={{header|IS-BASIC}}==
{{out}}
```txt
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
J
!~/~ 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:Pascal's_triangle#J_Explanation|talk page]] for explanation of earlier version
See also [[Pascal_matrix_generation#J|Pascal matrix generation]] and [[Sierpinski_triangle#J|Sierpinski triangle]].
Java
Summing from Previous Rows
{{works with|Java|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
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
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
ES5
=Imperative=
{{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=
{{Trans|Haskell}}
(function (n) {
'use strict';
// PASCAL TRIANGLE --------------------------------------------------------
// pascal :: Int -> [[Int]]
function pascal(n) {
return foldl(function (a) {
var xs = a.slice(-1)[0]; // Previous row
return append(a, [zipWith(
function (a, b) {
return a + b;
},
append([0], xs),
append(xs, [0])
)]);
}, [
[1] // Initial seed row
], enumFromTo(1, n - 1));
};
// GENERIC FUNCTIONS ------------------------------------------------------
// (++) :: [a] -> [a] -> [a]
function append(xs, ys) {
return xs.concat(ys);
};
// enumFromTo :: Int -> Int -> [Int]
function enumFromTo(m, n) {
return Array.from({
length: Math.floor(n - m) + 1
}, function (_, i) {
return m + i;
});
};
// foldl :: (b -> a -> b) -> b -> [a] -> b
function foldl(f, a, xs) {
return xs.reduce(f, a);
};
// foldr (a -> b -> b) -> b -> [a] -> b
function foldr(f, a, xs) {
return xs.reduceRight(f, a);
};
// map :: (a -> b) -> [a] -> [b]
function map(f, xs) {
return xs.map(f);
};
// min :: Ord a => a -> a -> a
function min(a, b) {
return b < a ? b : a;
};
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
function zipWith(f, xs, ys) {
return Array.from({
length: min(xs.length, ys.length)
}, function (_, i) {
return f(xs[i], ys[i]);
});
};
// TEST and FORMAT --------------------------------------------------------
var lstTriangle = pascal(n);
// [[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);
{{Out}} {| class="wikitable" style="text-align:center;width:26em;height:26em;table-layout:fixed;" |- | || || || || || || 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
(() => {
'use strict';
const main = () =>
showPascal(take(7, pascal()));
// pascal :: Generator [[Int]]
const pascal = () =>
iterate(
xs => zipWith(
plus,
append([0], xs), append(xs, [0])
),
[1]
);
// showPascal :: [[Int]] -> String
const showPascal = xs => {
const
w = length(intercalate(' ', last(xs))),
align = xs => center(w, ' ', intercalate(' ', xs));
return unlines(map(align, xs));
};
// GENERIC FUNCTIONS ----------------------------------
// Tuple (,) :: a -> b -> (a, b)
const Tuple = (a, b) => ({
type: 'Tuple',
'0': a,
'1': b,
length: 2
});
// append (++) :: [a] -> [a] -> [a]
// append (++) :: String -> String -> String
const append = (xs, ys) => xs.concat(ys);
// Size of space -> filler Char -> String -> Centered String
// center :: Int -> Char -> String -> String
const center = (n, c, s) => {
const
qr = quotRem(n - s.length, 2),
q = qr[0];
return replicateString(q, c) +
s + replicateString(q + qr[1], c);
};
// intercalate :: String -> [String] -> String
const intercalate = (s, xs) =>
xs.join(s);
// iterate :: (a -> a) -> a -> Generator [a]
function* iterate(f, x) {
let v = x;
while (true) {
yield(v);
v = f(v);
}
}
// last :: [a] -> a
const last = xs =>
0 < xs.length ? xs.slice(-1)[0] : undefined;
// Returns Infinity over objects without finite length
// this enables zip and zipWith to choose the shorter
// argument when one non-finite like cycle, repeat etc
// length :: [a] -> Int
const length = xs => xs.length || Infinity;
// map :: (a -> b) -> [a] -> [b]
const map = (f, xs) => xs.map(f);
// plus :: Num -> Num -> Num
const plus = (a, b) => a + b;
// quotRem :: Int -> Int -> (Int, Int)
const quotRem = (m, n) =>
Tuple(Math.floor(m / n), m % n);
// replicateString :: Int -> String -> String
const replicateString = (n, s) => s.repeat(n);
// take :: Int -> [a] -> [a]
// take :: Int -> String -> String
const take = (n, xs) =>
xs.constructor.constructor.name !== 'GeneratorFunction' ? (
xs.slice(0, n)
) : [].concat.apply([], Array.from({
length: n
}, () => {
const x = xs.next();
return x.done ? [] : [x.value];
}));
// unlines :: [String] -> String
const unlines = xs => xs.join('\n');
// Use of `take` and `length` here allows zipping with non-finite lists
// i.e. generators like cycle, repeat, iterate.
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
const zipWith = (f, xs, ys) => {
const
lng = Math.min(length(xs), length(ys)),
as = take(lng, xs),
bs = take(lng, ys);
return Array.from({
length: lng
}, (_, i) => f(as[i], bs[i], i));
};
// MAIN ---
return main();
})();
{{Out}}
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|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) {{ Out }}
$ 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;
Julia
function pascal(n) if n<=0 print("n has to have a positive value") end x=0 while x<=n for a=0:x print(binomial(x,a)) end println("") x+=1 end end
pascal(4)
1
11
121
1331
14641
Another solution using matrix exponentiation.
iround(x) = round(Int64, x)
triangle(n) = iround.(expm(diagm(1:n, -1)))
function pascal(n)
println.(join.([filter(!iszero, triangle(n)[i,:]) for i in 1:(n+1)], " "))
return
end
{{Out}}
pascal(5)
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
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)
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())
Liberty BASIC
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
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
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]
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
Maple
f:=n->seq(print(seq(binomial(i,k),k=0..i)),i=0..n-1);
f(3);
1
1 1 1 2 1
Mathematica
Column[StringReplace[ToString /@ Replace[MatrixExp[SparseArray[
{Band[{2,1}] -> Range[n-1]},{n,n}]],{x__,0..}->{x},2] ,{"{"|"}"|","->" "}], Center]
[[File:MmaPascal.png]]
=={{header|MATLAB}} / {{header|Octave}}==
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
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
(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
Microsoft Small Basic
{{trans|GW-BASIC}}
TextWindow.Write("Number of rows? ")
r = TextWindow.ReadNumber()
For i = 0 To r - 1
c = 1
For k = 0 To i
TextWindow.CursorLeft = (k + 1) * 4 - Text.GetLength(c)
TextWindow.Write(c)
c = c * (i - k) / (k + 1)
EndFor
TextWindow.WriteLine("")
EndFor
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
=={{header|Modula-2}}==
MODULE Pascal;
FROM FormatString IMPORT FormatString;
FROM Terminal IMPORT WriteString,WriteLn,ReadChar;
PROCEDURE PrintLine(n : INTEGER);
VAR
buf : ARRAY[0..63] OF CHAR;
m,j : INTEGER;
BEGIN
IF n<1 THEN RETURN END;
m := 1;
WriteString("1 ");
FOR j:=1 TO n-1 DO
m := m * (n - j) DIV j;
FormatString("%i ", buf, m);
WriteString(buf)
END;
WriteLn
END PrintLine;
PROCEDURE Print(n : INTEGER);
VAR i : INTEGER;
BEGIN
FOR i:=1 TO n DO
PrintLine(i)
END
END Print;
BEGIN
Print(10);
ReadChar
END Pascal.
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*/
{{out}}
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)
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
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
(* 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
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
No result if n <= 0
: pascal(n) [ 1 ] #[ dup println dup 0 + 0 rot + zipWith(#+) ] times(n) drop ;
{{out}}
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
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
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
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
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: {{libheader|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 }
This triangle is build using the 'sock' or 'hockey stick' pattern property. Here I use the word tartaglia and not pascal because in my country it's called after the Niccolò Fontana, known also as Tartaglia. A full graphical implementation of 16 properties that can be found in the triangle can be found at mine [https://github.com/LorenzoTa/Tartaglia-s-triangle Tartaglia's triangle]
#!/usr/bin/perl
use strict;
use warnings;
{
my @tartaglia ;
sub tartaglia {
my ($x,$y) = @_;
if ($x == 0 or $y == 0) { $tartaglia[$x][$y]=1 ; return 1};
my $ret ;
foreach my $yps (0..$y){
$ret += ( $tartaglia[$x-1][$yps] || tartaglia($x-1,$yps) );
}
$tartaglia[$x][$y] = $ret;
return $ret;
}
}
sub tartaglia_row {
my $y = shift;
my $x = 0;
my @row;
$row[0] = &tartaglia($x,$y+1);
foreach my $pos (0..$y-1) {push @row, tartaglia(++$x,--$y)}
return @row;
}
for (0..5) {print join ' ', tartaglia_row($_),"\n"}
print "\n\n";
print tartaglia(3,3),"\n";
my @third = tartaglia_row(5);
print "@third\n";
which output
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
20
1 5 10 10 5 1
Perl 6
{{works with|rakudo|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:
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
{{trans|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
{{trans|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
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
{{out}}
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
```php
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 "
";
}
?>
```
## PHP
```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
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
{{trans|C}}
```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) ) )
```
## Potion
```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)
```
## PowerShell
```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++
}
```
Save the above code to a .ps1 script file and start it by calling its name and providing N.
```txt
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.
```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]).
```
Output :
```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.
```
### An alternative
The above use of difference lists is a really innovative example of late binding. Here's an alternative source which, while possibly not as efficient (or as short) as the previous example, may be a little easier to read and understand.
```prolog
%~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
% Produce a pascal's triangle of depth N
%~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
% Prolog is declarative. The predicate pascal/3 below says that to produce
% a row of depth N, we can do so by first producing the row at depth(N-1),
% and then adding the paired values in that row. The triangle is produced
% by prepending the row at N-1 to the preceding rows as recursion unwinds.
% The triangle produced by pascal/3 is upside down and lacks the last row,
% so pascal/2 prepends the last row to the triangle and reverses it.
% Finally, pascal/1 produces the triangle, iterates each row and prints it.
%~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
pascal_row([V], [V]). % No more value pairs to add
pascal_row([V0, V1|T], [V|Rest]) :- % Add values from preceding row
V is V0 + V1, !, pascal_row([V1|T], Rest). % Drops initial value (1).
pascal(1, [1], []). % at depth 1, this row is [1] and no preceding rows.
pascal(N, [1|ThisRow], [Last|Preceding]) :- % Produce a row of depth N
succ(N0, N), % N is the successor to N0
pascal(N0, Last, Preceding), % Get the previous row
!, pascal_row(Last, ThisRow). % Calculate this row from the previous
pascal(N, Triangle) :-
pascal(N, Last, Rows), % Retrieve row at depth N and preceding rows
!, reverse([Last|Rows], Triangle). % Add last row to triangle and reverse order
pascal(N) :-
pascal(N, Triangle), member(Row, Triangle), % Iterate and write each row
write(Row), nl, fail.
pascal(_).
```
*Output*:
```prolog
?- pascal(5).
[1]
[1,1]
[1,2,1]
[1,3,3,1]
[1,4,6,4,1]
```
## PureBasic
```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()
```
## Python
### Procedural
```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
```
or by creating a scan function:
```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)
```
### Functional
The itertools module yields a simple functional definition of '''scanl''' in terms of '''accumulate''', and '''zipWith''' can be defined in terms either of '''itertools.starmap''', or the base '''map'''.
With a scanl and a zipWith to hand, we can derive both finite and non-finite lists of pascal rows from a simple '''nextPascal''' step function:
{{Works with|Python|3.7}}
```python
'''Pascal's triangle'''
from itertools import (accumulate, chain, islice)
from operator import (add)
# nextPascal :: [Int] -> [Int]
def nextPascal(xs):
'''A row of Pascal's triangle
derived from a preceding row.'''
return zipWith(add)([0] + xs)(xs + [0])
# pascalTriangle :: Generator [[Int]]
def pascalTriangle():
'''A non-finite stream of
Pascal's triangle rows.'''
return iterate(nextPascal)([1])
# finitePascalRows :: Int -> [[Int]]
def finitePascalRows(n):
'''The first n rows of Pascal's triangle.'''
def go(a, _):
return nextPascal(a)
return scanl(go)([1])(
range(1, n)
)
# TESTS ---------------------------------------------------
# main :: IO ()
def main():
'''Test of two different approaches:
- taking from a non-finite stream of rows,
- or constructing a finite list of rows.'''
print(unlines(map(
showPascal,
[
take(7)(
pascalTriangle() # Non finite,
),
finitePascalRows(7) # finite.
]
)))
# showPascal :: [[Int]] -> String
def showPascal(xs):
'''Stringification of a list of
Pascal triangle rows.'''
ys = list(xs)
def align(w):
return lambda ns: center(w)(
' '
)(' '.join(map(str, ns)))
w = len(' '.join((map(str, ys[-1]))))
return '\n'.join(map(align(w), ys))
# GENERIC -------------------------------------------------
# center :: Int -> Char -> String -> String
def center(n):
'''String s padded with c to approximate centre,
fitting in but not truncated to width n.'''
def go(c, s):
qr = divmod(n - len(s), 2)
q = qr[0]
return (q * c) + s + ((q + qr[1]) * c)
return lambda c: lambda s: go(c, s)
# iterate :: (a -> a) -> a -> Gen [a]
def iterate(f):
'''An infinite list of repeated applications of f to x.'''
def go(x):
v = x
while True:
yield v
v = f(v)
return lambda x: go(x)
# scanl :: (b -> a -> b) -> b -> [a] -> [b]
def scanl(f):
'''scanl is like reduce, but returns a succession of
intermediate values, building from the left.'''
return lambda a: lambda xs: (
accumulate(chain([a], xs), f)
)
# take :: Int -> [a] -> [a]
# take :: Int -> String -> String
def take(n):
'''The prefix of xs of length n,
or xs itself if n > length xs.'''
return lambda xs: (
xs[0:n]
if isinstance(xs, list)
else list(islice(xs, n))
)
# unlines :: [String] -> String
def unlines(xs):
'''A single string derived by the intercalation
of a list of strings with the newline character.'''
return '\n'.join(xs)
# zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
def zipWith(f):
'''A list constructed by zipping with a
custom function, rather than with the
default tuple constructor.'''
return lambda xs: lambda ys: (
list(map(f, xs, ys))
)
# MAIN ---
if __name__ == '__main__':
main()
```
{{Out}}
```txt
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
```
## q
```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
```
## Qi
{{trans|Haskell}}
```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))
```
## R
{{trans|Octave}}
```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)
}
}
```
Here's an R version:
```R
pascalTriangle <- function(h) {
lapply(0:h, function(i) choose(i, 0:i))
}
```
## Racket
Iterative version by summing rows up to .
```Racket
#lang racket
(define (pascal n)
(define (next-row current-row)
(map + (cons 0 current-row)
(append current-row '(0))))
(reverse
(for/fold ([triangle '((1))])
([row (in-range 1 n)])
(cons (next-row (first triangle)) triangle))))
```
## RapidQ
### Summing from Previous Rows
{{trans|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.
```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
```
### Using binary coefficients
{{trans|BASIC}}
```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
```
## Red
```Red
Red[]
pascal-triangle: function [
n [ integer! ] "number of rows"
][
row: make vector! [ 1 ]
loop n [
print row
left: copy row
right: copy row
insert left 0
append right 0
row: left + right
]
]
```
Output:
```txt
pascal-triangle 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
```
## Retro
```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
```
## 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:
:::* Khayyam's triangle
:::* Khayyam─Pascal's triangle
:::* Tartaglia's triangle
:::* Yang Hui's triangle
```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*/
```
'''output''' when using the input of: 11
```txt
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
(Output shown at '''4/5''' size.)
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
```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
```
Output:
```txt
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
```
## Ruby
```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(&:sum)
end
end
pascal(8){|row| puts row.join(" ").center(20)}
```
{{out}}
```txt
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):
```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)}
```
{{out}}
```txt
[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
```runbasic
input "number of rows? ";r
for i = 0 to r - 1
c = 1
print left$(" ",(r*2)-(i*2));
for k = 0 to i
print using("####",c);
c = c*(i-k)/(k+1)
next
print
next
```
Output:
```txt
Number of rows? ?5
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
```
## Rust
{{trans|C}}
```rust
fn pascal_triangle(n: u64)
{
for i in 0..n {
let mut c = 1;
for _j in 1..2*(n-1-i)+1 {
print!(" ");
}
for k in 0..i+1 {
print!("{:2} ", c);
c = c * (i-k)/(k+1);
}
println!();
}
}
```
## Scala
### Functional solutions
### =Summing: Recursive row definition=
```scala
def tri(row: Int): List[Int] =
row match {
case 1 => List(1)
case n: Int => 1 +: ((tri(n - 1) zip tri(n - 1).tail) map { case (a, b) => a + b }) :+ 1
}
```
Function to pretty print n rows:
```scala
def prettyTri(n:Int) = (1 to n) foreach {i => print(" "*(n-i)); tri(i) map (c => print(c + " ")); println}
prettyTri(5)
```
{{Out}}
```txt
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
```
====Summing: Scala Stream (Recursive & Memoization)====
```Scala
object Blaise extends App {
def pascalTriangle(): Stream[Vector[Int]] =
Vector(1) #:: Stream.iterate(Vector(1, 1))(1 +: _.sliding(2).map(_.sum).toVector :+ 1)
val output = pascalTriangle().take(15).map(_.mkString(" "))
val longest = output.last.length
println("Pascal's Triangle")
output.foreach(line => println(s"${" " * ((longest - line.length) / 2)}$line"))
}
```
{{Out}}See it in running in your browser by [https://scalafiddle.io/sf/8VqiX0P/1 ScalaFiddle (JavaScript)] or by [https://scastie.scala-lang.org/c3dDWMCcT3eoydy6QJcWCw Scastie (JVM)].
## Scheme
{{Works with|Scheme|RRS}}
```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)
```
Output:
((1) (1 1) (1 2 1) (1 3 3 1) (1 4 6 4 1))
```
## Seed7
```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;
```
## Sidef
```ruby
func pascal(rows) {
var row = [1]
{ | n|
say row.join(' ')
row = [1, {|i| row[i] + row[i+1] }.map(0 .. n-2)..., 1]
} << 1..rows
}
pascal(10)
```
## Stata
First, a few ways to compute a "Pascal matrix". With the first, the upper triangle is made of missing values (zeros with the other two).
```stata
function pascal1(n) {
return(comb(J(1,n,0::n-1),J(n,1,0..n-1)))
}
function pascal2(n) {
a = I(n)
a[.,1] = J(n,1,1)
for (i=3; i<=n; i++) {
a[i,2..i-1] = a[i-1,2..i-1]+a[i-1,1..i-2]
}
return(a)
}
function pascal3(n) {
a = J(n,n,0)
for (i=1; i[Int]{
if n==1{
let a=[1]
print(a)
return a
}
else{
var a=pascal(n:n-1)
var temp=a
for i in 0..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:
```txt
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.
```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 "$1"
```
## Ursala
Zero maps to the empty list. Negatives are inexpressible.
This solution uses a library function for binomial coefficients.
```Ursala
#import std
#import nat
pascal = choose**ziDS+ iota*t+ iota+ successor
```
This solution uses direct summation. The algorithm is to
insert zero at the head of a list (initially the unit list <1>), zip it with its reversal,
map the sum over the list of pairs, iterate n times, and return the trace.
```Ursala
#import std
#import nat
pascal "n" = (next"n" sum*NiCixp) <1>
```
test program:
```Ursala
#cast %nLL
example = pascal 10
```
{{Out}}
```txt
<
<1>,
<1,1>,
<1,2,1>,
<1,3,3,1>,
<1,4,6,4,1>,
<1,5,10,10,5,1>,
<1,6,15,20,15,6,1>,
<1,7,21,35,35,21,7,1>,
<1,8,28,56,70,56,28,8,1>,
<1,9,36,84,126,126,84,36,9,1>>
```
## VBA
```vb
Option Base 1
Private Sub pascal_triangle(n As Integer)
Dim odd() As String
Dim eve() As String
ReDim odd(1)
ReDim eve(2)
odd(1) = " 1"
For i = 1 To n
If i Mod 2 = 1 Then
Debug.Print String$(2 * n - 2 * i, " ") & Join(odd, " ")
eve(1) = " 1"
ReDim Preserve eve(i + 1)
For j = 2 To i
eve(j) = Format(CStr(Val(odd(j - 1)) + Val(odd(j))), "@@@")
Next j
eve(i + 1) = " 1"
Else
Debug.Print String$(2 * n - 2 * i, " ") & Join(eve, " ")
odd(1) = " 1"
ReDim Preserve odd(i + 1)
For j = 2 To i
odd(j) = Format(CStr(Val(eve(j - 1)) + Val(eve(j))), "@@@")
Next j
odd(i + 1) = " 1"
End If
Next i
End Sub
Public Sub main()
pascal_triangle 13
End Sub
```
{{out}}
```txt
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
```
## VBScript
Derived from the BASIC version.
```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
```
{{out}}
Invoke from a command line.
```txt
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
{{trans|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.
```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
}
```
### Using binary coefficients
{{trans|BASIC}}
```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
}
```
## Visual Basic
{{works with|Visual Basic|VB6 Standard}}
```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
```
{{out}}
```txt
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
```
## Visual Basic .NET
{{trans|C#}}
```vbnet
Imports System.Numerics
Module Module1
Iterator Function GetRow(rowNumber As BigInteger) As IEnumerable(Of BigInteger)
Dim denominator As BigInteger = 1
Dim numerator = rowNumber
Dim currentValue As BigInteger = 1
For counter = 0 To rowNumber
Yield currentValue
currentValue = currentValue * numerator
numerator = numerator - 1
currentValue = currentValue / denominator
denominator = denominator + 1
Next
End Function
Function GetTriangle(quantityOfRows As Integer) As IEnumerable(Of BigInteger())
Dim range = Enumerable.Range(0, quantityOfRows).Select(Function(num) New BigInteger(num))
Return range.Select(Function(num) GetRow(num).ToArray())
End Function
Function CenterString(text As String, width As Integer)
Dim spaces = width - text.Length
Dim padLeft = (spaces / 2) + text.Length
Return text.PadLeft(padLeft).PadRight(width)
End Function
Function FormatTriangleString(triangle As IEnumerable(Of BigInteger())) As String
Dim maxDigitWidth = triangle.Last().Max().ToString().Length
Dim rows = triangle.Select(Function(arr) String.Join(" ", arr.Select(Function(array) CenterString(array.ToString(), maxDigitWidth))))
Dim maxRowWidth = rows.Last().Length
Return String.Join(Environment.NewLine, rows.Select(Function(row) CenterString(row, maxRowWidth)))
End Function
Sub Main()
Dim triangle = GetTriangle(20)
Dim output = FormatTriangleString(triangle)
Console.WriteLine(output)
End Sub
End Module
```
{{out}}
```txt
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
```
## XBasic
{{trans|GW-BASIC}}
{{works with|Windows XBasic}}
```xbasic
PROGRAM "pascal"
VERSION "0.0001"
DECLARE FUNCTION Entry()
FUNCTION Entry()
r@@ = UBYTE(INLINE$("Number of rows? "))
FOR i@@ = 0 TO r@@ - 1
c%% = 1
FOR k@@ = 0 TO i@@
PRINT FORMAT$("####", c%%);
c%% = c%% * (i@@ - k@@) / (k@@ + 1)
NEXT k@@
PRINT
NEXT i@@
END FUNCTION
END PROGRAM
```
{{out}}
```txt
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
```
## XPL0
```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)
```
{{Out}}
```txt
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
{{trans|C}}
```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);
```
{{out}}
```txt
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
```
## ZX Spectrum Basic
In edit mode insert:
```BASIC
10 INPUT "How many rows? ";n
15 IF n<1 THEN GO TO 210
20 DIM c(n)
25 DIM d(n)
30 LET c(1)=1
35 LET d(1)=1
40 FOR r=1 TO n
50 FOR i=1 TO (n-r)
60 PRINT " ";
70 NEXT i
80 FOR i=1 TO r
90 PRINT c(i);" ";
100 NEXT i
110 PRINT
120 IF r>=n THEN GO TO 140
130 LET d(r+1)=1
140 FOR i=2 TO r
150 LET d(i)=c(i-1)+c(i)
160 NEXT i
165 IF r>=n THEN GO TO 200
170 FOR i=1 TO r+1
180 LET c(i)=d(i)
190 NEXT i
200 NEXT r
```
Then in command mode (basically don't put a number in front):
```BASIC>RUN