Pascal's triangle: Difference between revisions
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→{{header|Wren}}: Minor tidy and now uses binomial method in Math module.
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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: <b>
1
1 1
1 2 1
1 3 3 1 </b>
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: <b>
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1 </b>
Each row <tt> n </tt> (starting with row 0 at the top) shows the coefficients of the binomial expansion of <big><big> (x + y)<sup>n</sup>. </big></big>
;Task:
Write a function that prints out the first <tt> n </tt> rows of the triangle (with <tt> f(1) </tt> 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 <big><tt> n ≤ 0 </tt></big> does not need to be uniform, but should be noted.
;See also:
* [[Evaluate binomial coefficients]]
<br><br>
=={{header|11l}}==
{{trans|Python}}
<syntaxhighlight lang="11l">F pascal(n)
V row = [1]
V k = [0]
L 0 .< max(n, 0)
print(row.join(‘ ’).center(16))
row = zip(row [+] k, k [+] row).map((l, r) -> l + r)
pascal(7)</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|360 Assembly}}==
{{trans|PL/I}}
<syntaxhighlight lang="360asm">* Pascal's triangle 25/10/2015
PASCAL CSECT
USING PASCAL,R15 set base register
LA R7,1 n=1
LOOPN C R7,=A(M) do n=1 to m
BH ELOOPN if n>m then goto
MVC U,=F'1' u(1)=1
LA R8,PG pgi=@pg
LA R6,1 i=1
LOOPI CR R6,R7 do i=1 to n
BH ELOOPI if i>n then goto
LR R1,R6 i
SLA R1,2 i*4
L R3,T-4(R1) t(i)
L R4,T(R1) t(i+1)
AR R3,R4 t(i)+t(i+1)
ST R3,U(R1) u(i+1)=t(i)+t(i+1)
LR R1,R6 i
SLA R1,2 i*4
L R2,U-4(R1) u(i)
XDECO R2,XD edit u(i)
MVC 0(4,R8),XD+8 output u(i):4
LA R8,4(R8) pgi=pgi+4
LA R6,1(R6) i=i+1
B LOOPI end i
ELOOPI MVC T((M+1)*(L'T)),U t=u
XPRNT PG,80 print
LA R7,1(R7) n=n+1
B LOOPN end n
ELOOPN XR R15,R15 set return code
BR R14 return to caller
M EQU 11 <== input
T DC (M+1)F'0' t(m+1) init 0
U DC (M+1)F'0' u(m+1) init 0
PG DC CL80' ' pg init ' '
XD DS CL12 temp
YREGS
END PASCAL</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|8th}}==
One way, using array operations:
<syntaxhighlight lang="forth">
\ print the array
: .arr \ a -- a
( . space ) a:each ;
: pasc \ a --
\ print the row
.arr cr
dup
\ create two rows from the first, one with a leading the other with a trailing 0
[0] 0 a:insert swap 0 a:push
\ add the arrays together to make the new one
' n:+ a:op ;
\ print the first 16 rows:
[1] ' pasc 16 times
</syntaxhighlight>
Another way, using the relation between element 'n' and element 'n-1' in a row:
<syntaxhighlight lang="forth">
: ratio \ m n -- num denom
tuck n:- n:1+ swap ;
\ one item in the row: n m
: pascitem \ n m -- n
r@ swap
ratio
n:*/ n:round int
dup . space ;
\ One row of Pascal's triangle
: pascline \ n --
>r 1 int dup . space
' pascitem
1 r@ loop rdrop drop cr ;
\ Calculate the first 'n' rows of Pascal's triangle:
: pasc \ n
' pascline 0 rot loop cr ;
15 pasc
</syntaxhighlight>
=={{header|Action!}}==
<syntaxhighlight lang="action!">PROC Main()
BYTE count=[10],row,item
CHAR ARRAY s(5)
INT v
FOR row=0 TO count-1
DO
v=1
FOR item=0 TO row
DO
StrI(v,s)
Position(2*(count-row)+4*item-s(0),row+1)
Print(s)
v=v*(row-item)/(item+1)
OD
PutE()
OD
RETURN</syntaxhighlight>
{{out}}
[https://gitlab.com/amarok8bit/action-rosetta-code/-/raw/master/images/Pascal's_triangle.png Screenshot from Atari 8-bit computer]
<pre>
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
</pre>
=={{header|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]]
<syntaxhighlight lang="ada">package Pascal is
type Row is array (Natural range <>) of Natural;
function Length(R: Row) return Positive;
function First_Row(Max_Length: Positive) return Row;
function Next_Row(R: Row) return Row;
end Pascal;</syntaxhighlight>
The implementation of that auxiliary package "Pascal":
<syntaxhighlight lang="ada">package body Pascal is
function First_Row(Max_Length: Positive) return Row is
R: Row(0 .. Max_Length) := (0 | 1 => 1, others => 0);
begin
return R;
end First_Row;
function Next_Row(R: Row) return Row is
S:
begin
S(Length(S))
for J in reverse 2 ..
end loop;
return S;
end Next_Row;
function Length(R: Row) return Positive is
begin
return R(0);
end Length;
end Pascal;</syntaxhighlight>
The main program, using "Pascal". It prints the desired number of rows. The number is read from the command line.
<syntaxhighlight lang="ada">with Ada.Text_IO, Ada.Integer_Text_IO, Ada.Command_Line, Pascal; use Pascal;
procedure Triangle is
Number_Of_Rows: Positive := Integer'Value(Ada.Command_Line.Argument(1));
Row: Pascal.Row := First_Row(Number_Of_Rows);
begin
loop
-- print one row
for J in 1 .. Length(Row) loop
Ada.Integer_Text_IO.Put(Row(J), 5);
end loop;
Ada.Text_IO.New_Line;
exit when Length(Row) >= Number_Of_Rows;
Row := Next_Row(Row);
end loop;
end Triangle;</syntaxhighlight>
{{out}}
<pre>>./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</pre>
=={{header|ALGOL 68}}==
<
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);
Line 62 ⟶ 297:
# WHILE # i < stop DO
row := row[AT 1] + row[AT 2]
OD</
{{Out}}
<pre>
1
1 1
Line 73 ⟶ 309:
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
</pre>
=={{header|ALGOL W}}==
<syntaxhighlight lang="algolw">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.</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Amazing Hopper}}==
<syntaxhighlight lang="Amazing Hopper">
#include <jambo.h>
#define Mulbyandmoveto(_X_) Mul by '_X_', Move to '_X_'
Main
filas=0, Get arg numeric '2', Move to 'filas'
i=0, r=""
Loop if( var 'i' Is less than 'filas' )
c=1, j=0
Set 'c' To str, Move to 'r'
Loop if ( var 'j' Is less than 'i' )
Set 'i' Minus 'j', Plus one 'j', Div it; Mul by and move to 'c'
Multi cat ' r, "\t", Str(c) '; Move to 'r'
++j
Back
Printnl 'r'
++i
Back
End
</syntaxhighlight>
{{out}}
<pre>
$ hopper jm/pascal.jambo 14
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
</pre>
=={{header|APL}}==
Pascal' s triangle of order ⍵
=== Dyalog APL ===
<syntaxhighlight lang="apl">
{⍕0~¨⍨(-⌽A)⌽↑,/0,¨⍉A∘.!A←0,⍳⍵}
</syntaxhighlight>
example
<
{⍕0~¨⍨(-⌽A)⌽↑,/0,¨⍉A∘.!A←0,⍳⍵}5
</syntaxhighlight>
<pre>
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
</pre>
=== GNU APL ===
GNU APL doesn't allow multiple statements within lambdas so the solution is phrased differently:
<syntaxhighlight lang="apl">
{{⍉⍵∘.!⍵} 0,⍳⍵}
</syntaxhighlight>
example
<
{
</syntaxhighlight>
<pre>
Line 93 ⟶ 424:
1 3 3 1
</pre>
=={{header|AppleScript}}==
Drawing n rows from a generator:
<syntaxhighlight lang="applescript">-------------------- PASCAL'S TRIANGLE -------------------
-- 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
--------------------------- TEST -------------------------
on run
showPascal(take(7, pascal()))
end run
------------------------ FORMATTING ----------------------
-- 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 ------------------------
-- 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
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
-- 2nd class handler function lifted into 1st class script wrapper.
if script is class of f 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</syntaxhighlight>
{{Out}}
<pre> 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</pre>
=={{header|Arturo}}==
{{trans|Nim}}
<syntaxhighlight lang="rebol">pascalTriangle: function [n][
triangle: new [[1]]
loop 1..dec n 'x [
'triangle ++ @[map couple (last triangle)++[0] [0]++(last triangle) 'x -> x\[0] + x\[1]]
]
return triangle
]
loop pascalTriangle 10 'row [
print pad.center join.with: " " map to [:string] row 'x -> pad.center x 5 60
]</syntaxhighlight>
{{out}}
<pre> 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</pre>
=={{header|AutoHotkey}}==
ahk forum: [http://www.autohotkey.com/forum/viewtopic.php?p=276617#276617 discussion]
<
Loop %n% {
p := "p" A_Index, %p% := v := 1, q := "p" A_Index-1
Line 117 ⟶ 701:
GuiClose:
ExitApp</
Alternate {{works with|AutoHotkey L}}
<syntaxhighlight lang="autohotkey">Msgbox % format(pascalstriangle())
Return
format(o) ; converts object to string
{
For k, v in o
s .= IsObject(v) ? format(v) "`n" : v " "
Return s
}
pascalstriangle(n=7) ; n rows of Pascal's triangle
{
p := Object(), z:=Object()
Loop, % n
Loop, % row := A_Index
col := A_Index
, p[row, col] := row = 1 and col = 1
? 1
: (p[row-1, col-1] = "" ; math operations on blanks return blanks; I want to assume zero
? 0
: p[row-1, col-1])
+ (p[row-1, col] = ""
? 0
: p[row-1, col])
Return p
}</syntaxhighlight>
n <= 0 returns empty
=={{header|AWK}}==
<
{{Out}}<pre>
1
1 1
Line 127 ⟶ 739:
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
</pre>
=={{header|Bait}}==
<syntaxhighlight lang="bait">
// Create a Pascal's triangle with a given number of rows.
// Returns an empty array for row_nr <= 0.
fun pascals_triangle(row_nr i32) [][]i32 {
mut rows := [][]i32
// Iterate over all rows
for r := 0; r < row_nr; r += 1 {
// Store the row above the current one
mut above := rows[r - 1]
// Fill the current row. It contains r + 1 numbers
for i := 0; i <= r; i += 1 {
// First number is always 1
if i == 0 {
rows.push([1]) // Push new row
}
// Last number is always 1
else if i == r {
rows[r].push(1)
}
// Other numbers are the sum of the two numbers above them
else {
rows[r].push(above[i - 1] + above[i])
}
}
}
return rows
}
// Helper function to pretty print triangles.
// It still get's ugly once numbers have >= 2 digits.
fun print_triangle(triangle [][]i32) {
for i, row in triangle {
// Create string with leading spaces
mut s := ' '.repeat(triangle.length - i - 1)
// Add each number to the string
for n in row {
s += n.str() + ' '
}
// Print and trim the extra trailing space
println(s.trim_right(' '))
}
}
fun main() {
print_triangle(pascals_triangle(7))
}
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|BASIC}}==
Line 142 ⟶ 823:
If the user enters value less than 1, the first row is still always displayed.
<
DIM row AS Integer
DIM nrows AS Integer
Line 157 ⟶ 838:
NEXT i
PRINT
NEXT row</
=={{header|Batch File}}==
Based from the Fortran Code.
<syntaxhighlight lang="dos">@echo off
setlocal enabledelayedexpansion
::The Main Thing...
cls
echo.
set row=15
call :pascal
echo.
pause
exit /b 0
::/The Main Thing.
::The Functions...
:pascal
set /a prev=%row%-1
for /l %%I in (0,1,%prev%) do (
set c=1&set r=
for /l %%K in (0,1,%row%) do (
if not !c!==0 (
call :numstr !c!
set r=!r!!space!!c!
)
set /a c=!c!*^(%%I-%%K^)/^(%%K+1^)
)
echo !r!
)
goto :EOF
:numstr
::This function returns the number of whitespaces to be applied on each numbers.
set cnt=0&set proc=%1&set space=
:loop
set currchar=!proc:~%cnt%,1!
if not "!currchar!"=="" set /a cnt+=1&goto loop
set /a numspaces=5-!cnt!
for /l %%A in (1,1,%numspaces%) do set "space=!space! "
goto :EOF
::/The Functions.</syntaxhighlight>
{{Out}}
<pre> 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 . . .</pre>
=={{header|BBC BASIC}}==
<
colwidth% = 4
Line 172 ⟶ 913:
NEXT
PRINT
NEXT row%</
{{Out}}
<pre> 1
1 1
Line 184 ⟶ 925:
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1</pre>
=={{header|BCPL}}==
<syntaxhighlight lang="bcpl">get "libhdr"
let pascal(n) be
for i=0 to n-1
$( let c = 1
for j=1 to 2*(n-1-i) do wrch(' ')
for k=0 to i
$( writef("%I3 ",c)
c := c*(i-k)/(k+1)
$)
wrch('*N')
$)
let start() be pascal(8)</syntaxhighlight>
{{out}}
<pre> 1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1</pre>
=={{header|Befunge}}==
<syntaxhighlight lang="befunge">0" :swor fo rebmuN">:#,_&> 55+, v
v01*p00-1:g00.:<1p011p00:\-1_v#:<
>g:1+10p/48*,:#^_$ 55+,1+\: ^>$$@</syntaxhighlight>
{{Out}}
<pre>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 </pre>
=={{header|BQN}}==
Displays n rows.
<syntaxhighlight lang="bqn">Pascal ← {(0⊸∾+∾⟜0)⍟(↕𝕩)⋈1}
•Show¨Pascal 6</syntaxhighlight>
<syntaxhighlight lang="text">⟨ 1 ⟩
⟨ 1 1 ⟩
⟨ 1 2 1 ⟩
⟨ 1 3 3 1 ⟩
⟨ 1 4 6 4 1 ⟩
⟨ 1 5 10 10 5 1 ⟩</syntaxhighlight>
=={{header|Bracmat}}==
<
& get':?R
& -1:?I
Line 203 ⟶ 1,001:
)
&
)</
{{Out}}
<pre>Number of rows?
7
Line 214 ⟶ 1,012:
1 5 10 10 5 1
1 6 15 20 15 6 1</pre>
=={{header|Burlesque}}==
<syntaxhighlight lang="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
</syntaxhighlight>
=={{header|C}}==
Line 219 ⟶ 1,040:
{{trans|Fortran}}
<
void pascaltriangle(unsigned int n)
Line 240 ⟶ 1,061:
pascaltriangle(8);
return 0;
}</
===Recursive===
<
#define D 32
Line 260 ⟶ 1,081:
int x[D] = {0, 1, 0}, y[D] = {0};
return pascals(x, y, 0);
}</
===Adding previous row values===
<syntaxhighlight lang="c">void
if (
for (int r = 1; r <= nRows; r++) {
int *currRow = malloc(r * sizeof(int));
for (int i = 0; i < r; i++) {
printf(" %4d", val);
}
printf("\n");
free(prevRow);
prevRow = currRow;
}
free(prevRow);
}</syntaxhighlight>
=={{header|C sharp|C#}}==
Line 294 ⟶ 1,106:
Produces no output when n is less than or equal to zero.
<
namespace RosettaCode {
class PascalsTriangle {
public static void CreateTriangle(int
if (n >
int
Console.Write(" ".PadLeft(2
for
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===
<syntaxhighlight lang="csharp">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);
}
}
}</syntaxhighlight>
Example:
<syntaxhighlight lang="csharp">static void Main()
{
IEnumerable<BigInteger[]> triangle = PascalsTriangle.GetTriangle(20);
string output = PascalsTriangle.FormatTriangleString(triangle)
Console.WriteLine(output);
}</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|C++}}==
<syntaxhighlight lang="cpp">#include <iostream>
#include <algorithm>
#include<cstdio>
using namespace std;
void Pascal_Triangle(int size) {
int a[100][100];
int i, j;
//first row and first coloumn has the same value=1
for (i = 1; i <= size; i++) {
a[i][1] = a[1][i] = 1;
}
//Generate the full Triangle
for (i = 2; i <= size; i++) {
for (j = 2; j <= size - i; j++) {
if (a[i - 1][j] == 0 || a[i][j - 1] == 0) {
break;
}
a[i][j] = a[i - 1][j] + a[i][j - 1];
}
}
/*
1 1 1 1
1 2 3
1 3
1
first print as above format-->
for (i = 1; i < size; i++) {
for (j = 1; j < size; j++) {
if (a[i][j] == 0) {
break;
}
printf("%8d",a[i][j]);
}
cout<<"\n\n";
}*/
// standard Pascal Triangle Format
int row,space;
for (i = 1; i < size; i++) {
space=row=i;
j=1;
while(space<=size+(size-i)+1){
cout<<" ";
space++;
}
while(j<=i){
if (a[row][j] == 0){
break;
}
if(j==1){
printf("%d",a[row--][j++]);
}
else
printf("%6d",a[row--][j++]);
}
cout<<"\n\n";
}
}
int main()
{
//freopen("out.txt","w",stdout);
int size;
cin>>size;
Pascal_Triangle(size);
}
}</syntaxhighlight>
===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.
<syntaxhighlight lang="cpp">// Compile with -std=c++11
#include<iostream>
#include<vector>
using namespace std;
void print_vector(vector<int> dummy){
for (vector<int>::iterator i = dummy.begin(); i != dummy.end(); ++i)
cout<<*i<<" ";
cout<<endl;
}
void print_vector_of_vectors(vector<vector<int>> dummy){
for (vector<vector<int>>::iterator i = dummy.begin(); i != dummy.end(); ++i)
print_vector(*i);
cout<<endl;
}
vector<vector<int>> dynamic_triangle(int dummy){
vector<vector<int>> result;
if (dummy > 0){ // if the argument is 0 or negative exit immediately
vector<int> row;
// The first row
row.push_back(1);
result.push_back(row);
// The second row
if (dummy > 1){
row.clear();
row.push_back(1); row.push_back(1);
result.push_back(row);
}
// The other rows
if (dummy > 2){
for (int i = 2; i < dummy; i++){
row.clear();
row.push_back(1);
for (int j = 1; j < i; j++)
row.push_back(result.back().at(j - 1) + result.back().at(j));
row.push_back(1);
result.push_back(row);
}
}
}
return result;
}
vector<vector<int>> static_triangle(int dummy){
vector<vector<int>> result;
if (dummy > 0){ // if the argument is 0 or negative exit immediately
vector<int> row;
result.resize(dummy); // This should work faster than consecutive push_back()s
// The first row
row.resize(1);
row.at(0) = 1;
result.at(0) = row;
// The second row
if (result.size() > 1){
row.resize(2);
row.at(0) = 1; row.at(1) = 1;
result.at(1) = row;
}
// The other rows
if (result.size() > 2){
for (int i = 2; i < result.size(); i++){
row.resize(i + 1); // This should work faster than consecutive push_back()s
row.front() = 1;
for (int j = 1; j < row.size() - 1; j++)
row.at(j) = result.at(i - 1).at(j - 1) + result.at(i - 1).at(j);
row.back() = 1;
result.at(i) = row;
}
}
}
return result;
}
int main(){
vector<vector<int>> triangle;
int n;
cout<<endl<<"The Pascal's Triangle"<<endl<<"Enter the number of rows: ";
cin>>n;
// Call the dynamic function
triangle = dynamic_triangle(n);
cout<<endl<<"Calculated using dynamic vectors:"<<endl<<endl;
print_vector_of_vectors(triangle);
// Call the static function
triangle = static_triangle(n);
cout<<endl<<"Calculated using static vectors:"<<endl<<endl;
print_vector_of_vectors(triangle);
return 0;
}
</syntaxhighlight>
===C++11 (with a class) ===
A full fledged example with a class definition and methods to retrieve data, worthy of the title object-oriented.
<syntaxhighlight lang="cpp">// Compile with -std=c++11
#include<iostream>
#include<vector>
using namespace std;
class pascal_triangle{
vector<vector<int>> data; // This is the actual data
void print_row(vector<int> dummy){
for (vector<int>::iterator i = dummy.begin(); i != dummy.end(); ++i)
cout<<*i<<" ";
cout<<endl;
}
public:
pascal_triangle(int dummy){ // Everything is done on the construction phase
if (dummy > 0){ // if the argument is 0 or negative exit immediately
vector<int> row;
data.resize(dummy); // Theoretically this should work faster than consecutive push_back()s
// The first row
row.resize(1);
row.at(0) = 1;
data.at(0) = row;
// The second row
if (data.size() > 1){
row.resize(2);
row.at(0) = 1; row.at(1) = 1;
data.at(1) = row;
}
// The other rows
if (data.size() > 2){
for (int i = 2; i < data.size(); i++){
row.resize(i + 1); // Theoretically this should work faster than consecutive push_back()s
row.front() = 1;
for (int j = 1; j < row.size() - 1; j++)
row.at(j) = data.at(i - 1).at(j - 1) + data.at(i - 1).at(j);
row.back() = 1;
data.at(i) = row;
}
}
}
}
~pascal_triangle(){
for (vector<vector<int>>::iterator i = data.begin(); i != data.end(); ++i)
i->clear(); // I'm not sure about the necessity of this loop!
data.clear();
}
void print_row(int dummy){
if (dummy < data.size())
for (vector<int>::iterator i = data.at(dummy).begin(); i != data.at(dummy).end(); ++i)
cout<<*i<<" ";
cout<<endl;
}
void print(){
for (int i = 0; i < data.size(); i++)
print_row(i);
}
int get_coeff(int dummy1, int dummy2){
int result = 0;
if ((dummy1 < data.size()) && (dummy2 < data.at(dummy1).size()))
result = data.at(dummy1).at(dummy2);
return result;
}
vector<int> get_row(int dummy){
vector<int> result;
if (dummy < data.size())
result = data.at(dummy);
return result;
}
};
int main(){
int n;
cout<<endl<<"The Pascal's Triangle with a class!"<<endl<<endl<<"Enter the number of rows: ";
cin>>n;
pascal_triangle myptri(n);
cout<<endl<<"The whole triangle:"<<endl;
myptri.print();
cout<<endl<<"Just one row:"<<endl;
myptri.print_row(n/2);
cout<<endl<<"Just one coefficient:"<<endl;
cout<<myptri.get_coeff(n/2, n/4)<<endl<<endl;
return 0;
}
</syntaxhighlight>
=={{header|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).
<
(let [newrow (fn newrow [lst ret]
(if lst
Line 332 ⟶ 1,483:
(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] )))
Line 343 ⟶ 1,494:
(doseq [row triangle]
(println row))))
</syntaxhighlight>
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
Line 355 ⟶ 1,506:
(map (partial apply +) ,,,)))
[1]))
</syntaxhighlight>
Another short version which returns an infinite pascal triangle as a list, using the iterate function.
<syntaxhighlight lang="lisp">
(def pascal
(iterate #(concat [1]
(map + % (rest %))
[1])
[1]))
</syntaxhighlight>
One can then get the first n rows using the take function
<syntaxhighlight lang="lisp">
(take 10 pascal) ; returns a list of the first 10 pascal rows
</syntaxhighlight>
Also, one can retrieve the nth row using the nth function
<syntaxhighlight lang="lisp">
(nth pascal 10) ;returns the nth row
</syntaxhighlight>
=={{header|CoffeeScript}}==
This version assumes n is an integer and n >= 1. It efficiently computes binomial coefficients.
<
pascal = (n) ->
width = 6
Line 389 ⟶ 1,562:
pascal(7)
</syntaxhighlight>
{{Out}}
<pre>
> coffee pascal.coffee
Line 402 ⟶ 1,575:
1 6 15 20 15 6 1
</pre>
=={{header|Commodore BASIC}}==
<syntaxhighlight lang="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</syntaxhighlight>
Output:
<syntaxhighlight lang="text">RUN
HOW MANY? 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
1 8 28 56 70 56 28 8 1
READY.
</syntaxhighlight>
=={{header|Common Lisp}}==
To evaluate, call (pascal n). For n < 1, it simply returns nil.
<
(defun genrow (n l)
(defun newrow (l)
'(1)
(cons (+ (
(newrow (rest l)))))</syntaxhighlight>
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))''.
<syntaxhighlight lang="lisp">(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))</syntaxhighlight>
Another iterative solution, this time using pretty-printing to automatically print the triangle in the shape of a triangle in the terminal. The print-pascal-triangle function computes and uses the length of the printed last row to decide how wide the triangle should be.
<syntaxhighlight lang="lisp">
(defun next-pascal-triangle-row (list)
`(1
,.(mapcar #'+ list (rest list))
1))
(defun pascal-triangle (number-of-rows)
(loop repeat number-of-rows
for row = '(1) then (next-pascal-triangle-row row)
collect row))
(defun print-pascal-triangle (number-of-rows)
(let* ((triangle (pascal-triangle number-of-rows))
(max-row-length (length (write-to-string (first (last triangle))))))
(format t
(format nil "~~{~~~D:@<~~{~~A ~~}~~>~~%~~}" max-row-length)
triangle)))
</syntaxhighlight>
For example:
<syntaxhighlight lang="lisp">(print-pascal-triangle 4)</syntaxhighlight>
<syntaxhighlight lang="text">
1
1 1
1 2 1
1 3 3 1
</syntaxhighlight>
<syntaxhighlight lang="lisp">(print-pascal-triangle 8)</syntaxhighlight>
<syntaxhighlight lang="text">
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
</syntaxhighlight>
=={{header|Component Pascal}}==
{{Works with|BlackBox Component Builder}}
<syntaxhighlight lang="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.
</syntaxhighlight>
<pre>Execute: ^Q PascalTriangle.Do 0 1 2 3 4 5 6 7 8 9 10 11 12~</pre>
{{out}}
<pre>
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
</pre>
=={{header|D}}==
===Less functional Version===
<
auto tri = new int[][rows];
foreach (r; 0 .. rows) {
Line 445 ⟶ 1,810:
[1, 8, 28, 56, 70, 56, 28, 8, 1],
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]]);
}</
===More functional Version===
<
auto pascal(
return [1].recurrence!q{ zip(a[n - 1] ~ 0, 0 ~ a[n - 1])
.map!q{ a[0] + a[1] }
}
void main() {
}</
{{out}}
<pre>[[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1]]</pre>
===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).
<syntaxhighlight lang="d">import std.stdio, std.string, std.array, std.format;
string Pascal(alias dg, T, T initValue)(int n) {
Line 504 ⟶ 1,871:
foreach (i; [16])
writef(sierpinski(i));
}</
{{out}}
<pre> 1
Line 537 ⟶ 1,904:
* _ * _ * _ * _ * _ * _ * _ *
* * * * * * * * * * * * * * * *</pre>
=={{header|Dart}}==
<syntaxhighlight lang="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);
}
</syntaxhighlight>
=={{header|Delphi}}==
<syntaxhighlight lang="delphi">program PascalsTriangle;
procedure Pascal(r:Integer);
var
i, c, k:Integer;
begin
for i := 0 to r - 1 do
begin
c := 1;
for k := 0 to i do
begin
Write(c:3);
c := c * (i - k) div (k + 1);
end;
Writeln;
end;
end;
begin
Pascal(9);
end.</syntaxhighlight>
=={{header|DWScript}}==
Doesn't print anything for negative or null values.
<
var
i, c, k : Integer;
Line 554 ⟶ 1,990:
end;
Pascal(9);</
{{Out}}
<pre> 1
1 1
1 2 1
Line 569 ⟶ 2,006:
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 row := [].diverge(int)
out.print("<table style='text-align: center; border: 0; border-collapse: collapse;'>")
Line 588 ⟶ 2,025:
}
out.print("</table>")
}</
<
try {
pascalsTriangle(15, out)
Line 596 ⟶ 2,033:
out.close()
}
makeCommand("yourFavoriteWebBrowser")("triangle.html")</
=={{header|EasyLang}}==
<syntaxhighlight>
numfmt 0 4
proc pascal n . .
r[] = [ 1 ]
for i to n
rn[] = [ ]
l = 0
for j to n - len r[]
write " "
.
for r in r[]
write r
rn[] &= l + r
l = r
.
print ""
rn[] &= l
swap r[] rn[]
.
.
pascal 13
</syntaxhighlight>
=={{header|Eiffel}}==
<syntaxhighlight lang="eiffel">
note
description : "Prints pascal's triangle"
output : "[
Per requirements of the RosettaCode example, execution will print the first n rows of pascal's triangle
]"
date : "19 December 2013"
authors : "Sandro Meier", "Roman Brunner"
revision : "1.0"
libraries : "Relies on HASH_TABLE from EIFFEL_BASE library"
implementation : "[
Recursive implementation to calculate the n'th row.
]"
warning : "[
Will not work for large n's (INTEGER_32)
]"
class
APPLICATION
inherit
ARGUMENTS
create
make
feature {NONE} -- Initialization
make
local
n:INTEGER
do
create {HASH_TABLE[ARRAY[INTEGER],INTEGER]}pascal_lines.make (n) --create the hash_table object
io.new_line
n:=25
draw(n)
end
feature
line(n:INTEGER):ARRAY[INTEGER]
--Calculates the n'th line
local
upper_line:ARRAY[INTEGER]
i:INTEGER
do
if n=1 then --trivial case first line
create Result.make_filled (0, 1, n+2)
Result.put (0, 1)
Result.put (1, 2)
Result.put (0, 3)
elseif pascal_lines.has (n) then --checks if the result was already calculated
Result := pascal_lines.at (n)
else --calculates the n'th line recursively
create Result.make_filled(0,1,n+2) --for caluclation purposes add a 0 at the beginning of each line
Result.put (0, 1)
upper_line:=line(n-1)
from
i:=1
until
i>upper_line.count-1
loop
Result.put(upper_line[i]+upper_line[i+1],i+1)
i:=i+1
end
Result.put (0, n+2) --for caluclation purposes add a 0 at the end of each line
pascal_lines.put (Result, n)
end
end
draw(n:INTEGER)
--draw n lines of pascal's triangle
local
space_string:STRING
width, i:INTEGER
do
space_string:=" " --question of design: add space_string at the beginning of each line
width:=line(n).count
space_string.multiply (width)
from
i:=1
until
i>n
loop
space_string.remove_tail (1)
io.put_string (space_string)
across line(i) as c
loop
if
c.item/=0
then
io.put_string (c.item.out+" ")
end
end
io.new_line
i:=i+1
end
end
feature --Access
pascal_lines:HASH_TABLE[ARRAY[INTEGER],INTEGER]
--Contains all already calculated lines
end
</syntaxhighlight>
=={{header|Elixir}}==
<syntaxhighlight lang="elixir">defmodule Pascal do
def triangle(n), do: triangle(n,[1])
def triangle(0,list), do: list
def triangle(n,list) do
IO.inspect list
new_list = Enum.zip([0]++list, list++[0]) |> Enum.map(fn {a,b} -> a+b end)
triangle(n-1,new_list)
end
end
Pascal.triangle(8)</syntaxhighlight>
{{out}}
<pre>
[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]
</pre>
=={{header|Emacs Lisp}}==
===Using mapcar and append, returing a list of rows===
<syntaxhighlight lang="lisp">(require 'cl-lib)
(defun next-row (row)
(cl-mapcar #'+ (cons 0 row)
(append row '(0))))
(defun triangle (row rows)
(if (= rows 0)
'()
(cons row (triangle (next-row row) (- rows 1)))))</syntaxhighlight>
{{Out}}
Call the function from the REPL, IELM:
<pre>
ELISP> (triangle (list 1) 6)
((1)
(1 1)
(1 2 1)
(1 3 3 1)
(1 4 6 4 1)
(1 5 10 10 5 1))
</pre>
===Translation from Pascal===
<syntaxhighlight lang="lisp">(defun pascal (r)
(dotimes (i r)
(let ((c 1))
(dotimes (k (+ i 1))
(princ (format "%d " c))
(setq c (/ (* c (- i k))
(+ k 1))))
(terpri))))</syntaxhighlight>
{{Out}}
From the REPL:
<pre>
ELISP> (princ (pascal 6))
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
</pre>
===Returning a string===
Same as the translation from Pascal, but now returning a string.
<syntaxhighlight lang="lisp">(defun pascal (r)
(let ((out ""))
(dotimes (i r)
(let ((c 1))
(dotimes (k (+ i 1))
(setq out (concat out (format "%d " c)))
(setq c (/ (* c (- i k))
(+ k 1))))
(setq out (concat out "\n"))))
out))</syntaxhighlight>
{{Out}}
Now, since this one returns a string, it is possible to insert the result in the current buffer:
<pre>
(insert (pascal 6))
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
</pre>
=={{header|Erlang}}==
<
-import(lists).
-export([pascal/1]).
Line 609 ⟶ 2,271:
[H|_] = L,
[lists:zipwith(fun(X,Y)->X+Y end,[0]++H,H++[0])|L].
</syntaxhighlight>
{{Out}}
<pre>
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]]
</pre>
=={{header|ERRE}}==
<syntaxhighlight lang="erre">
PROGRAM PASCAL_TRIANGLE
PROCEDURE PASCAL(R%)
LOCAL I%,C%,K%
FOR I%=0 TO R%-1 DO
C%=1
FOR K%=0 TO I% DO
WRITE("###";C%;)
C%=(C%*(I%-K%)) DIV (K%+1)
END FOR
PRINT
END FOR
END PROCEDURE
BEGIN
PASCAL(9)
END PROGRAM
</syntaxhighlight>
Output:
<pre>
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
</pre>
=={{header|Euphoria}}==
===Summing from Previous Rows===
<
row = {}
for m = 1 to 10 do
Line 628 ⟶ 2,324:
print(1,row)
puts(1,'\n')
end for</
{{Out}}
<pre>
{1}
{1,1}
Line 641 ⟶ 2,338:
{1,8,28,56,70,56,28,8,1}
{1,9,36,84,126,126,84,36,9,1}
</pre>
=={{header|Excel}}==
===LAMBDA===
Binding the names PASCAL and BINCOEFF to the following lambda expressions in the Name Manager of the Excel WorkBook, to define Pascal's triangle in terms of binomial coefficients:
(See [https://www.microsoft.com/en-us/research/blog/lambda-the-ultimatae-excel-worksheet-function/ LAMBDA: The ultimate Excel worksheet function])
{{Works with|Office 365 betas 2021}}
<syntaxhighlight lang="lisp">PASCAL
=LAMBDA(n,
BINCOEFF(n - 1)(
SEQUENCE(1, n, 0, 1)
)
)
BINCOEFF
=LAMBDA(n,
LAMBDA(k,
QUOTIENT(FACT(n), FACT(k) * FACT(n - k))
)
)</syntaxhighlight>
{{Out}}
{| class="wikitable"
|-
|||style="text-align:right; font-family:serif; font-style:italic; font-size:120%;"|fx
! colspan="11" style="text-align:left; vertical-align: bottom; font-family:Arial, Helvetica, sans-serif !important;"|=PASCAL(A2)
|- style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff;"
|
| A
| B
| C
| D
| E
| F
| G
| H
| I
| J
| K
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 1
| style="font-style:italic" | Row number
| colspan="10" style="font-weight:bold" | PASCAL's TRIANGLE
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 2
| style="text-align:right; font-style:italic" | 1
| style="text-align:center; background-color:#cbcefb" | 1
|
|
|
|
|
|
|
|
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 3
| style="text-align:right; font-style:italic" | 2
| style="text-align:center" | 1
| style="text-align:center" | 1
|
|
|
|
|
|
|
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 4
| style="text-align:right; font-style:italic" | 3
| style="text-align:center" | 1
| style="text-align:center" | 2
| style="text-align:center" | 1
|
|
|
|
|
|
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 5
| style="text-align:right; font-style:italic" | 4
| style="text-align:center" | 1
| style="text-align:center" | 3
| style="text-align:center" | 3
| style="text-align:center" | 1
|
|
|
|
|
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 6
| style="text-align:right; font-style:italic" | 5
| style="text-align:center" | 1
| style="text-align:center" | 4
| style="text-align:center" | 6
| style="text-align:center" | 4
| style="text-align:center" | 1
|
|
|
|
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 7
| style="text-align:right; font-style:italic" | 6
| style="text-align:center" | 1
| style="text-align:center" | 5
| style="text-align:center" | 10
| style="text-align:center" | 10
| style="text-align:center" | 5
| style="text-align:center" | 1
|
|
|
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 8
| style="text-align:right; font-style:italic" | 7
| style="text-align:center" | 1
| style="text-align:center" | 6
| style="text-align:center" | 15
| style="text-align:center" | 20
| style="text-align:center" | 15
| style="text-align:center" | 6
| style="text-align:center" | 1
|
|
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 9
| style="text-align:right; font-style:italic" | 8
| style="text-align:center" | 1
| style="text-align:center" | 7
| style="text-align:center" | 21
| style="text-align:center" | 35
| style="text-align:center" | 35
| style="text-align:center" | 21
| style="text-align:center" | 7
| style="text-align:center" | 1
|
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 10
| style="text-align:right; font-style:italic" | 9
| style="text-align:center" | 1
| style="text-align:center" | 8
| style="text-align:center" | 28
| style="text-align:center" | 56
| style="text-align:center" | 70
| style="text-align:center" | 56
| style="text-align:center" | 28
| style="text-align:center" | 8
| style="text-align:center" | 1
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 11
| style="text-align:right; font-style:italic" | 10
| style="text-align:center" | 1
| style="text-align:center" | 9
| style="text-align:center" | 36
| style="text-align:center" | 84
| style="text-align:center" | 126
| style="text-align:center" | 126
| style="text-align:center" | 84
| style="text-align:center" | 36
| style="text-align:center" | 9
| style="text-align:center" | 1
|}
Or defining the whole triangle as a single grid, by binding the name TRIANGLE to an additional lambda:
<syntaxhighlight lang="lisp">TRIANGLE
=LAMBDA(n,
LET(
ixs, SEQUENCE(n, n, 0, 1),
x, MOD(ixs, n),
y, QUOTIENT(ixs, n),
IF(x <= y,
BINCOEFF(y)(x),
""
)
)
)</syntaxhighlight>
{{Out}}
{| class="wikitable"
|-
|||style="text-align:right; font-family:serif; font-style:italic; font-size:120%;"|fx
! colspan="11" style="text-align:left; vertical-align: bottom; font-family:Arial, Helvetica, sans-serif !important;"|=TRIANGLE(10)
|- style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff;"
|
| A
| B
| C
| D
| E
| F
| G
| H
| I
| J
| K
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 1
| style="font-style:italic" |
| colspan="10" style="font-weight:bold" | PASCAL's TRIANGLE
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 2
| style="font-style:italic" |
| style="text-align:center; background-color:#cbcefb" | 1
|
|
|
|
|
|
|
|
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 3
| style="font-style:italic" |
| style="text-align:center" | 1
| style="text-align:center" | 1
|
|
|
|
|
|
|
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 4
| style="font-style:italic" |
| style="text-align:center" | 1
| style="text-align:center" | 2
| style="text-align:center" | 1
|
|
|
|
|
|
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 5
| style="font-style:italic" |
| style="text-align:center" | 1
| style="text-align:center" | 3
| style="text-align:center" | 3
| style="text-align:center" | 1
|
|
|
|
|
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 6
| style="font-style:italic" |
| style="text-align:center" | 1
| style="text-align:center" | 4
| style="text-align:center" | 6
| style="text-align:center" | 4
| style="text-align:center" | 1
|
|
|
|
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 7
| style="font-style:italic" |
| style="text-align:center" | 1
| style="text-align:center" | 5
| style="text-align:center" | 10
| style="text-align:center" | 10
| style="text-align:center" | 5
| style="text-align:center" | 1
|
|
|
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 8
| style="font-style:italic" |
| style="text-align:center" | 1
| style="text-align:center" | 6
| style="text-align:center" | 15
| style="text-align:center" | 20
| style="text-align:center" | 15
| style="text-align:center" | 6
| style="text-align:center" | 1
|
|
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 9
| style="font-style:italic" |
| style="text-align:center" | 1
| style="text-align:center" | 7
| style="text-align:center" | 21
| style="text-align:center" | 35
| style="text-align:center" | 35
| style="text-align:center" | 21
| style="text-align:center" | 7
| style="text-align:center" | 1
|
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 10
| style="font-style:italic" |
| style="text-align:center" | 1
| style="text-align:center" | 8
| style="text-align:center" | 28
| style="text-align:center" | 56
| style="text-align:center" | 70
| style="text-align:center" | 56
| style="text-align:center" | 28
| style="text-align:center" | 8
| style="text-align:center" | 1
|
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 11
| style="font-style:italic" |
| style="text-align:center" | 1
| style="text-align:center" | 9
| style="text-align:center" | 36
| style="text-align:center" | 84
| style="text-align:center" | 126
| style="text-align:center" | 126
| style="text-align:center" | 84
| style="text-align:center" | 36
| style="text-align:center" | 9
| style="text-align:center" | 1
|}
=={{header|F Sharp|F#}}==
<
match l with
| [] -> []
Line 655 ⟶ 2,699:
printf "%s" (i.ToString() + ", ")
printfn "%s" "\n"
</syntaxhighlight>
=={{header|Factor}}==
Line 661 ⟶ 2,705:
This implementation works by summing the previous line content. Result for n < 1 is the same as for n == 1.
<
: (pascal) ( seq -- newseq )
Line 667 ⟶ 2,711:
: pascal ( n -- seq )
1 - { { 1 } } swap [ (pascal) ] times ;</
It works as:
<
{ { 1 } { 1 1 } { 1 2 1 } { 1 3 3 1 } { 1 4 6 4 1 } }</
=={{header|Fantom}}==
<
class Main
{
Line 706 ⟶ 2,750:
}
}
</syntaxhighlight>
=={{header|FOCAL}}==
<syntaxhighlight lang="focal">1.1 S OLD(1)=1; T %4.0, 1, !
1.2 F N=1,10; D 2
1.3 Q
2.1 S NEW(1)=1
2.2 F X=1,N; S NEW(X+1)=OLD(X)+OLD(X+1)
2.3 F X=1,N+1; D 3
2.4 T !
3.1 S OLD(X)=NEW(X)
3.2 T %4.0, OLD(X)</syntaxhighlight>
{{output}}
<pre>
= 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
</pre>
=={{header|Forth}}==
<
here swap cells erase 1 here ! ;
: .line ( n -- )
Line 719 ⟶ 2,790:
: pascal ( n -- )
dup init 1 .line
1 ?do i next i 1+ .line loop ;</
This is a bit more efficient.
{{trans|C}}
<
cr dup 0
?do
Line 729 ⟶ 2,800:
;
13 PascTriangle</
=={{header|Fortran}}==
{{works with|Fortran|90 and later}}
Prints nothing for n<=0. Output formatting breaks down for n>20
<
CALL Print_Triangle(8)
Line 759 ⟶ 2,830:
END DO
END SUBROUTINE Print_Triangle</
=={{header|FreeBASIC}}==
<syntaxhighlight lang="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</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Frink}}==
This version takes a little effort to automatically format the tree based upon the width of the largest numbers in the bottom row. It automatically calculates this easily using Frink's builtin function for efficiently calculating (even large) binomial coefficients with cached factorials and binary splitting.
<syntaxhighlight lang="frink">
pascal[rows] :=
{
widest = length[toString[binomial[rows-1, (rows-1) div 2]]]
for row = 0 to rows-1
{
line = repeat[" ", round[(rows-row)* (widest+1)/2]]
for col = 0 to row
line = line + padRight[binomial[row, col], widest+1, " "]
println[line]
}
}
pascal[10]
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|FunL}}==
=== Summing from Previous Rows ===
{{trans|Scala}}
<syntaxhighlight lang="funl">import lists.zip
def
pascal( 1 ) = [1]
pascal( n ) = [1] + map( (a, b) -> a + b, zip(pascal(n-1), pascal(n-1).tail()) ) + [1]</syntaxhighlight>
=== Combinations ===
{{trans|Haskell}}
<syntaxhighlight lang="funl">import integers.choose
def pascal( n ) = [choose( n - 1, k ) | k <- 0..n-1]</syntaxhighlight>
=== Pascal's Triangle ===
<syntaxhighlight lang="funl">def triangle( height ) =
width = max( map(a -> a.toString().length(), pascal(height)) )
if 2|width
width++
for n <- 1..height
print( ' '*((width + 1)\2)*(height - n) )
println( map(a -> format('%' + width + 'd ', a), pascal(n)).mkString() )
triangle( 10 )</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Fōrmulæ}}==
{{FormulaeEntry|page=https://formulae.org/?script=examples/Pascal%27s_triangle}}
'''Solution'''
[[File:Fōrmulæ - Pascal's triangle 01.png]]
'''Test case'''
[[File:Fōrmulæ - Pascal's triangle 02.png]]
[[File:Fōrmulæ - Pascal's triangle 03.png]]
=={{header|GAP}}==
<
local i, v;
v := [1];
Line 780 ⟶ 2,998:
# [ 1, 6, 15, 20, 15, 6, 1 ]
# [ 1, 7, 21, 35, 35, 21, 7, 1 ]
# [ 1, 8, 28, 56, 70, 56, 28, 8, 1 ]</
=={{header|Go}}==
No output for n < 1. Otherwise, output formatted left justified.
<syntaxhighlight lang="go">
package main
Line 827 ⟶ 3,045:
printTriangle(4)
}
</syntaxhighlight>
Output:
<pre>
Line 839 ⟶ 3,057:
=== Recursive ===
In the spirit of the Haskell "think in whole lists" solution here is a list-driven, minimalist solution:
<syntaxhighlight lang="groovy">def pascal
pascal = { n -> (n <= 1) ? [1] : [[0] + pascal(n - 1), pascal(n - 1) + [0]].transpose().collect { it.sum() } }</syntaxhighlight>
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:
<
(1..count).each { n ->
printf ("%2d:", n); (0..(count-n)).each { print " " }; pascal(n).each{ printf("%6d ", it) }; println ""
}</
{{out}}
<pre> 1: 1
2: 1 1
Line 866 ⟶ 3,085:
=={{header|GW-BASIC}}==
<
20 FOR I=0 TO R-1
30 C=1
Line 874 ⟶ 3,093:
70 NEXT
80 PRINT
90 NEXT</
Output:
Line 896 ⟶ 3,115:
similar function
<
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 xs@(x:ys) = x : zapWith f xs ys</
And for the whole (infinite) triangle, we just iterate this operation,
starting with the first row:
<
For the first ''n'' rows, we just take the first ''n'' elements from this
list, as in
<
[[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]
<
nextRow row = zipWith (+) ([0] ++ row) (row ++ [0])
-- returns the first n rows
pascal = iterate nextRow [1]</
Alternatively, using list comprehensions:
<syntaxhighlight lang="haskell">
pascal :: [[Integer]]
pascal =
(1 : [ 0 | _ <- head pascal])
: [zipWith (+) (0:row) row | row <- pascal]
</syntaxhighlight>
<syntaxhighlight lang="haskell">
*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]]
</syntaxhighlight>
With binomial coefficients:
<syntaxhighlight lang="haskell">fac = product . enumFromTo 1
binCoef n k = fac n `div` (fac k * fac (n - k))
pascal = ((fmap . binCoef) <*> enumFromTo 0) . pred</syntaxhighlight>
Example:
<
1
1 1
Line 948 ⟶ 3,177:
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
</syntaxhighlight>
=={{header|HicEst}}==
<
SUBROUTINE Pascal(rows)
Line 965 ⟶ 3,194:
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.
<
procedure main(A)
Line 988 ⟶ 3,218:
write()
}
end</
{{libheader|Icon Programming Library}}
Line 1,014 ⟶ 3,244:
=={{header|IDL}}==
<
;n is the number of lines of the triangle to be displayed
r=[1]
Line 1,030 ⟶ 3,260:
print, r
End</
=={{header|IS-BASIC}}==
<syntaxhighlight lang="is-basic">100 PROGRAM "PascalTr.bas"
110 TEXT 80
120 LET ROW=12
130 FOR I=0 TO ROW
140 LET C=1
150 PRINT TAB(37-I*3);
160 FOR K=0 TO I
170 PRINT USING " #### ":C;
180 LET C=C*(I-K)/(K+1)
190 NEXT
200 PRINT
210 NEXT</syntaxhighlight>
{{out}}
<pre> 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</pre>
=={{header|ivy}}==
<syntaxhighlight lang="ivy">
op pascal N = transp (0 , iota N) o.! -1 , iota N
pascal 5
1 0 0 0 0 0
1 1 0 0 0 0
1 2 1 0 0 0
1 3 3 1 0 0
1 4 6 4 1 0
1 5 10 10 5 1
</syntaxhighlight>
=={{header|J}}==
<
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</
<
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</
However, multi-digit numbers take up additional space, which looks slightly odd. But we can work around that by adding additional padding and shifting the lines a bit more:
<syntaxhighlight lang=J> (|."_1~ 0-3*i.@-@#) ;@((<'%6d') sprintf each -.&0)"1 !~/~i.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
</syntaxhighlight>
Also... when we mix positive and negative numbers it stops being a triangle:
<syntaxhighlight lang=J> i:5
_5 _4 _3 _2 _1 0 1 2 3 4 5
!~/~i:5
1 0 0 0 0 1 _5 15 _35 70 _126
_4 1 0 0 0 1 _4 10 _20 35 _56
6 _3 1 0 0 1 _3 6 _10 15 _21
_4 3 _2 1 0 1 _2 3 _4 5 _6
1 _1 1 _1 1 1 _1 1 _1 1 _1
0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 1 0 0 0 0
0 0 0 0 0 1 2 1 0 0 0
0 0 0 0 0 1 3 3 1 0 0
0 0 0 0 0 1 4 6 4 1 0
0 0 0 0 0 1 5 10 10 5 1
!/~i:5
1 _4 6 _4 1 0 0 0 0 0 0
0 1 _3 3 _1 0 0 0 0 0 0
0 0 1 _2 1 0 0 0 0 0 0
0 0 0 1 _1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 1
_5 _4 _3 _2 _1 0 1 2 3 4 5
15 10 6 3 1 0 0 1 3 6 10
_35 _20 _10 _4 _1 0 0 0 1 4 10
70 35 15 5 1 0 0 0 0 1 5
_126 _56 _21 _6 _1 0 0 0 0 0 1</syntaxhighlight>
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]].
=={{header|Java}}==
===Summing from Previous Rows===
{{works with|Java|1.5+}}
<
...//class definition, etc.
public static void genPyrN(int rows){
Line 1,075 ⟶ 3,394:
thisRow.add(last.get(j - 1) + last.get(j));
}
thisRow.add(last.get(
last= thisRow;//save this row
System.out.println(thisRow);
}
}</
===Combinations===
This method is limited to 21 rows because of the limits of <tt>long</tt>. Calling <tt>pas</tt> with an argument of 22 or above will cause intermediate math to wrap around and give false answers.
<
public static void main(String[] args){
//usage
Line 1,108 ⟶ 3,428:
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) {
Line 1,132 ⟶ 3,452:
}
}
</syntaxhighlight>
=={{header|JavaScript}}==
===ES5===
====Imperative====
{{works with|SpiderMonkey}}
{{works with|V8}}
<
function pascalTriangle (rows) {
Line 1,207 ⟶ 3,529:
// Display 8 row triangle in base 16
tri = new pascalTriangle(8);
tri.print(16);</
Output:
<pre>$ d8 pascal.js
Line 1,221 ⟶ 3,543:
1 5 a a 5 1
1 6 f 14 f 6 1
1 7 15 23 23 15 7 1</pre>
====Functional====
{{Trans|Haskell}}
<syntaxhighlight lang="javascript">(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);</syntaxhighlight>
{{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
|}
<syntaxhighlight lang="javascript">[[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]]</syntaxhighlight>
===ES6===
<syntaxhighlight lang="javascript">(() => {
"use strict";
// ---------------- PASCAL'S TRIANGLE ----------------
// pascal :: Generator [[Int]]
const pascal = () =>
iterate(
xs => zipWith(
a => b => a + b
)(
[0, ...xs]
)(
[...xs, 0]
)
)([1]);
// ---------------------- TEST -----------------------
// main :: IO ()
const main = () =>
showPascal(
take(10)(
pascal()
)
);
// showPascal :: [[Int]] -> String
const showPascal = xs => {
const w = last(xs).join(" ").length;
return xs.map(
ys => center(w)(" ")(ys.join(" "))
)
.join("\n");
};
// ---------------- GENERIC FUNCTIONS ----------------
// center :: Int -> Char -> String -> String
const center = n =>
// Size of space -> filler Char ->
// String -> Centered String
c => s => {
const gap = n - s.length;
return 0 < gap ? (() => {
const
margin = c.repeat(Math.floor(gap / 2)),
dust = c.repeat(gap % 2);
return `${margin}${s}${margin}${dust}`;
})() : s;
};
// iterate :: (a -> a) -> a -> Gen [a]
const iterate = f =>
// An infinite list of repeated
// applications of f to x.
function* (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;
// take :: Int -> [a] -> [a]
// take :: Int -> String -> String
const take = n =>
// The first n elements of a list,
// string of characters, or stream.
xs => "GeneratorFunction" !== xs
.constructor.constructor.name ? (
xs.slice(0, n)
) : Array.from({
length: n
}, () => {
const x = xs.next();
return x.done ? [] : [x.value];
}).flat();
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
const zipWith = f =>
// A list constructed by zipping with a
// custom function, rather than with the
// default tuple constructor.
xs => ys => xs.map(
(x, i) => f(x)(ys[i])
).slice(
0, Math.min(xs.length, ys.length)
);
// MAIN ---
return main();
})();</syntaxhighlight>
{{Out}}
<pre> 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</pre>
====Recursive====
<syntaxhighlight lang="javascript">
const aux = n => {
if(n <= 1) return [1]
const prevLayer = aux(n - 1)
const shifted = [0, ...prevLayer]
return shifted.map((x, i) => (prevLayer[i] || 0) + x)
}
const pascal = n => {
for(let i = 1; i <= n; i++) {
console.log(aux(i).join(' '))
}
}
pascal(8)
</syntaxhighlight>
{{Out}}
<pre>
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
</pre>
====Recursive - memoized====
<syntaxhighlight lang="javascript">
const aux = (() => {
const layers = [[1], [1]]
return n => {
if(layers[n]) return layers[n]
const prevLayer = aux(n - 1)
const shifted = [0, ...prevLayer]
layers[n] = shifted.map((x, i) => (prevLayer[i] || 0) + x)
return layers[n]
}
})()
const pascal = n => {
for(let i = 1; i <= n; i++) {
console.log(aux(i).join(' '))
}
}
pascal(8)
</syntaxhighlight>
=={{header|
{{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.
<syntaxhighlight lang="jq"># pascal(n) for n>=0; pascal(0) emits an empty stream.
def pascal(n):
def _pascal: # input: the previous row
. as $in
| .,
if length >= n then empty
else
reduce range(0;length-1) as $i
([1]; . + [ $in[$i] + $in[$i + 1] ]) + [1] | _pascal
end;
if n <= 0 then empty else [1] | _pascal end ;</syntaxhighlight>
'''Example''':
pascal(5)
{{ Out }}
<syntaxhighlight lang="sh">$ jq -c -n -f pascal_triangle.jq
[1]
[1,1]
[1,2,1]
[1,3,3,1]
[1,4,6,4,1]</syntaxhighlight>
'''Using recurse/1'''
Here is an equivalent implementation that uses the built-in filter, recurse/1, instead of the inner function.
<syntaxhighlight lang="jq">def pascal(n):
if n <= 0 then empty
else [1]
| recurse( if length >= n then empty
else . as $in
| reduce range(0;length-1) as $i
([1]; . + [ $in[$i] + $in[$i + 1] ]) + [1]
end)
end;</syntaxhighlight>
=={{header|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
<pre>
pascal(5)
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
</pre>
Another solution using matrix exponentiation.
<syntaxhighlight lang="julia">
iround(x) = round(Int64, x)
triangle(n) = iround.(exp(diagm(-1=> 1:n)))
function pascal(n)
t=triangle(n)
println.(join.([filter(!iszero, t[i,:]) for i in 1:(n+1)], " "))
end
</syntaxhighlight>
{{Out}}
<pre>
pascal(5)
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
</pre>
Yet another solution using a static vector
<syntaxhighlight lang="julia">
function pascal(n)
(n<=0) && error("Pascal trinalge can not have zero or negative rows")
r=Vector{Int}(undef,n)
pr=Vector{Int}(undef,n)
pr[1]=r[1]=1
println(@view pr[1])
for i=2:n
r[1]=r[i]=1
for j=2:i-1
r[j]=pr[j-1]+pr[j]
end
println(join(view(r,1:i), " "))
r,pr=pr,r
end
end
</syntaxhighlight>
{{Out}}
<pre>
pascal(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
</pre>
=={{header|K}}==
<syntaxhighlight lang="k">
pascal:{(x-1){+':x,0}\1}
pascal 6
(1
1 1
Line 1,233 ⟶ 3,970:
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1)</
=={{header|Kotlin}}==
<syntaxhighlight lang="kotlin">fun pas(rows: Int) {
for (i in 0..rows - 1) {
for (j in 0..i)
print(ncr(i, j).toString() + " ")
println()
}
}
fun ncr(n: Int, r: Int) = fact(n) / (fact(r) * fact(n - r))
fun fact(n: Int) : Long {
var ans = 1.toLong()
for (i in 2..n)
ans *= i
return ans
}
fun main(args: Array<String>) = pas(args[0].toInt())</syntaxhighlight>
=={{header|Lambdatalk}}==
<syntaxhighlight lang="scheme">
1) Based on this expression of pascalian binomial:
Cnp = [n*(n-1)...(n-p+1)]/[p*(p-1)...2*1]
2) we define the following function:
{def C
{lambda {:n :p}
{/ {* {S.serie :n {- :n :p -1} -1}}
{* {S.serie :p 1 -1}}}}}
{C 16 8}
-> 12870
3) Writing
1{S.map {lambda {:n} {br}1
{S.map {C :n} {S.serie 1 {- :n 1}}} 1}
{S.serie 2 16}}
displays:
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
</syntaxhighlight>
=={{header|Liberty BASIC}}==
<
dim a$(n)
Line 1,254 ⟶ 4,052:
next i
end</
=={{header|Locomotive Basic}}==
<
20 INPUT "Number of rows? ", rows:GOSUB 40
30 END
Line 1,269 ⟶ 4,067:
100 PRINT
110 NEXT
120 RETURN</
Output:
Line 1,285 ⟶ 4,083:
=={{header|Logo}}==
<
if :n = 1 [output [1]]
localmake "a pascal :n-1
Line 1,291 ⟶ 4,089:
end
for [i 1 10] [print pascal :i]</
=={{header|Logtalk}}==
Our implementation will have an object <code>pascals</code> with work done in the method <code>triangle/2</code>. We will be caching results for time efficiency at the cost of space efficiency,and the <code>reset/0</code> method will flush that cache should it grow to be a problem. The resulting object looks like this:
<syntaxhighlight lang="logtalk">
:- object(pascals).
:- uses(integer, [plus/3, succ/2]).
:- public(reset/0).
reset :-
retractall(triangle_(_,_,_)).
:- private(triangle_/3).
:- dynamic(triangle_/3).
:- public(triangle/2).
triangle(N, Lines) :-
triangle(N, _, DiffLines),
difflist::as_list(DiffLines, Lines).
% Shortcut with cached value if it exists.
triangle(N, Line, DiffLines) :- triangle_(N, Line, DiffLines), !.
triangle(N, Line, DiffLines) :-
succ(N0, N),
triangle(N0, Line0, DiffLines0),
ZL = [0|Line0],
list::append(Line0, [0], ZR),
meta::map(plus, ZL, ZR, Line),
difflist::add(Line, DiffLines0, DiffLines),
asserta(triangle_(N, Line, DiffLines)).
triangle(1, [1], [[1]|X]-X).
:- end_object.
</syntaxhighlight>
{{Out}}
Using the SWI-Prolog back-end:
<pre>
?- logtalk_load([meta(loader), types(loader), pascals], [optimize(on)]).
% messages elided
true.
?- pascals::triangle(17, Ls), logtalk::print_message(information, user, Ls).
% - [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]
Ls = [[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4|...], [1, 5, 10|...], [1, 6|...], [1|...], [...|...]|...].
?-
</pre>
=={{header|Lua}}==
<
function nextrow(t)
local ret = {}
Line 1,308 ⟶ 4,177:
end
end
</syntaxhighlight>
=={{header|Maple}}==
<syntaxhighlight lang="maple">f:=n->seq(print(seq(binomial(i,k),k=0..i)),i=0..n-1);
f(3);</syntaxhighlight>
1
1 1
1 2 1
=={{header|Mathematica}}/{{header|Wolfram Language}}==
<syntaxhighlight lang="mathematica">n=7;
Column[StringReplace[ToString /@ Replace[MatrixExp[SparseArray[
{Band[{2,1}] -> Range[n-1]},{n,n}]],{x__,0..}->{x},2] ,{"{"|"}"|","->" "}], Center]</syntaxhighlight>
[[File:MmaPascal.png]]
A more graphical output with arrows would involve the plotting functionality with Graph[]:
<syntaxhighlight lang="mathematica">nmax := 10;
pascal[nmax_] := Module[
{vals = Table[Binomial[n, k], {n, 0, nmax}, {k, 0, n}],
ids = Table[{n, k}, {n, 0, nmax}, {k, 0, n}],
labels, left, right, leftright, edgeLabels
},
labels = Flatten[Thread /@ (Thread[ids -> vals]), 1];
left = DeleteCases[Flatten[Flatten[ids, 1] /. {n_, k_} /; (n >= k + 1) :> {{n, k + 1} -> {n + 1, k + 1}}, 1], _?NumberQ];
right = DeleteCases[Flatten[Flatten[ids, 1] /. {n_, k_} /; (n > k) :> {{n, k} -> {n + 1, k + 1}}, 1], _?NumberQ];
leftright = DeleteCases[left \[Union] right, _ -> {b_, _} /; b > nmax];
edgeLabels = (# -> Style["+", Medium] & /@ leftright);
Graph[Flatten[ids, 1], leftright
, VertexLabels -> MapAt[Placed[#, Center] &, labels, {All, 2}]
, GraphLayout -> "SpringEmbedding"
, VertexSize -> 0.8, EdgeLabels -> edgeLabels
, PlotLabel -> "Pascal's Triangle"
]
];
pascal[nmax]
</syntaxhighlight>
=={{header|MATLAB}} / {{header|Octave}}==
A matrix containing the pascal triangle can be obtained this way:
<syntaxhighlight lang
<pre>>> pascal(6)
Line 1,333 ⟶ 4,233:
The binomial coefficients can be extracted from the Pascal triangle in this way:
<
<pre>>> for k=1:6,diag(rot90(pascal(k)))', end
Line 1,357 ⟶ 4,257:
1 5 10 10 5 1
</pre>
Another way to get a formated pascals triangle is to use the convolution method:
<pre>>>
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
</pre>
The result is:
<pre>>>
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
</pre>
=={{header|Maxima}}==
<syntaxhighlight lang="maxima">
sjoin(v, j) := apply(sconcat, rest(join(makelist(j, length(v)), v)))$
display_pascal_triangle(n) := for i from 0 thru 6 do disp(sjoin(makelist(binomial(i, j), j, 0, i), " "));
Line 1,371 ⟶ 4,295:
"1 4 6 4 1"
"1 5 10 10 5 1"
"1 6 15 20 15 6 1" */
</syntaxhighlight>
=={{header|Metafont}}==
Line 1,378 ⟶ 4,302:
(The formatting starts to be less clear when numbers start to have more than two digits)
<
save ?;
? := (1 for i=(max(k,n-k)+1) upto n: * i endfor )
Line 1,395 ⟶ 4,319:
pascaltr(4);
end</
=={{header|Microsoft Small Basic}}==
{{trans|GW-BASIC}}
<syntaxhighlight lang="microsoftsmallbasic">
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
</syntaxhighlight>
Output:
<pre>
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
</pre>
=={{header|Modula-2}}==
<syntaxhighlight lang="modula2">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.</syntaxhighlight>
=={{header|NetRexx}}==
<syntaxhighlight lang="netrexx">/* NetRexx */
options replace format comments java crossref symbols nobinary
numeric digits 1000 -- allow very large numbers
parse arg rows .
if rows = '' then rows = 11 -- default to 11 rows
printPascalTriangle(rows)
return
-- -----------------------------------------------------------------------------
method printPascalTriangle(rows = 11) public static
lines = ''
mx = (factorial(rows - 1) / factorial(rows % 2) / factorial(rows - 1 - rows % 2)).length() -- width of widest number
loop row = 1 to rows
n1 = 1.center(mx)
line = n1
loop col = 2 to row
n2 = col - 1
n1 = n1 * (row - n2) / n2
line = line n1.center(mx)
end col
lines[row] = line.strip()
end row
-- display triangle
ml = lines[rows].length() -- length of longest line
loop row = 1 to rows
say lines[row].centre(ml)
end row
return
-- -----------------------------------------------------------------------------
method factorial(n) public static
fac = 1
loop n_ = 2 to n
fac = fac * n_
end n_
return fac /*calc. factorial*/
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Nial}}==
Line 1,401 ⟶ 4,445:
(pascal.nial)
<
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
<
|pascal 5</
=={{header|Nim}}==
<syntaxhighlight lang="nim">import sequtils, strutils
proc printPascalTriangle(n: int) =
## Print a Pascal triangle.
# Build the triangle.
var triangle: seq[seq[int]]
triangle.add @[1]
for _ in 1..<n:
triangle.add zip(triangle[^1] & @[0], @[0] & triangle[^1]).mapIt(it[0] + it[1])
# Build the lines to display.
let length = len($max(triangle[^1])) # Maximum length of number.
var lines: seq[string]
for row in triangle:
lines.add row.mapIt(($it).center(length)).join(" ")
# Display the lines.
let lineLength = lines[^1].len # Length of largest line (the last one).
for line in lines:
echo line.center(lineLength)
printPascalTriangle(10)</syntaxhighlight>
{{out}}
<pre> 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 </pre>
A more optimized solution that doesn't require importing, but produces, naturally, uglier output, would look like this:
<syntaxhighlight lang="nim">const ROWS = 10
const TRILEN = toInt(ROWS * (ROWS + 1) / 2) # Sum of arth progression
var triangle = newSeqOfCap[Natural](TRILEN) # Avoid reallocations
proc printPascalTri(row: Natural, result: var seq[Natural]) =
add(result, 1)
for i in 2..row-1: add(result, result[^row] + result[^(row-1)])
add(result, 1)
echo result[^row..^1]
if row + 1 <= ROWS: printPascalTri(row + 1, result)
printPascalTri(1, triangle)</syntaxhighlight>
{{out}}
<pre>@[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]</pre>
=={{header|OCaml}}==
<
let next_row row =
List.map2 (+) ([0] @ row) (row @ [0])
Line 1,421 ⟶ 4,529:
if i = n then []
else row :: loop (i+1) (next_row row)
in loop 0 [1]</
=={{header|Octave}}==
<
for i = 0:h-1
for k = 0:h-i
Line 1,436 ⟶ 4,544:
endfunction
pascaltriangle(4);</
=={{header|Oforth}}==
No result if n <= 0
<syntaxhighlight lang="oforth">: pascal(n) [ 1 ] #[ dup println dup 0 + 0 rot + zipWith(#+) ] times(n) drop ;</syntaxhighlight>
{{out}}
<pre>
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]
</pre>
=={{header|Oz}}==
<
fun {NextLine Xs}
{List.zip 0|Xs {Append Xs [0]}
Line 1,469 ⟶ 4,598:
end
in
{PrintTriangle {Triangle 5}}</
For n = 0, prints nothing. For negative n, throws an exception.
=={{header|PARI/GP}}==
<
my(row=[],prevrow=[]);
for(x=1,N,
Line 1,487 ⟶ 4,617:
print(row);
);
}</
=={{header|Pascal}}==
<
procedure Pascal(r : Integer);
Line 1,510 ⟶ 4,640:
begin
Pascal(9)
end.</
Output:
<pre>% ./PascalsTriangle
Line 1,525 ⟶ 4,655:
=={{header|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).
<syntaxhighlight lang="perl">sub pascal {
my $rows = shift;
for my $n
print "
@next = (1, (map $next[$_]+$next[$_+1], 0 .. $n-2), 1);
}
}</syntaxhighlight>
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}}
<syntaxhighlight lang="perl">use ntheory qw/binomial/;
sub pascal {
my $rows = shift;
for my $n (0 .. $rows-1) {
print join(" ", map { binomial($n,$_) } 0 .. $n), "\n";
}
}</syntaxhighlight>
Here is a non-obvious version using bignum, which is limited to the first 23 rows because of the algorithm used:
<syntaxhighlight lang="perl">use bignum;
sub pascal_line { $_[0] ? unpack "A(A6)*", 1000001**$_[0] : 1 }
sub pascal { print "@{[map -+-$_, pascal_line $_]}\n" for 0..$_[0]-1 }</syntaxhighlight>
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]
<syntaxhighlight lang="perl">
#!/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";
</syntaxhighlight>
which output
<pre>
1
1 1
1 2 1
Line 1,594 ⟶ 4,727:
1 4 6 4 1
1 5 10 10 5 1
20
1
</pre>
=={{header|Phix}}==
<!--<syntaxhighlight lang="phix">(phixonline)-->
<span style="color: #004080;">sequence</span> <span style="color: #000000;">row</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{}</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">m</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">1</span> <span style="color: #008080;">to</span> <span style="color: #000000;">13</span> <span style="color: #008080;">do</span>
<span style="color: #000000;">row</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">row</span> <span style="color: #0000FF;">&</span> <span style="color: #000000;">1</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">=</span><span style="color: #7060A8;">length</span><span style="color: #0000FF;">(</span><span style="color: #000000;">row</span><span style="color: #0000FF;">)-</span><span style="color: #000000;">1</span> <span style="color: #008080;">to</span> <span style="color: #000000;">2</span> <span style="color: #008080;">by</span> <span style="color: #0000FF;">-</span><span style="color: #000000;">1</span> <span style="color: #008080;">do</span>
<span style="color: #000000;">row</span><span style="color: #0000FF;">[</span><span style="color: #000000;">n</span><span style="color: #0000FF;">]</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">row</span><span style="color: #0000FF;">[</span><span style="color: #000000;">n</span><span style="color: #0000FF;">-</span><span style="color: #000000;">1</span><span style="color: #0000FF;">]</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #7060A8;">repeat</span><span style="color: #0000FF;">(</span><span style="color: #008000;">' '</span><span style="color: #0000FF;">,(</span><span style="color: #000000;">13</span><span style="color: #0000FF;">-</span><span style="color: #000000;">m</span><span style="color: #0000FF;">)*</span><span style="color: #000000;">2</span><span style="color: #0000FF;">))</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">i</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1</span> <span style="color: #008080;">to</span> <span style="color: #7060A8;">length</span><span style="color: #0000FF;">(</span><span style="color: #000000;">row</span><span style="color: #0000FF;">)</span> <span style="color: #008080;">do</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">" %3d"</span><span style="color: #0000FF;">,</span><span style="color: #000000;">row</span><span style="color: #0000FF;">[</span><span style="color: #000000;">i</span><span style="color: #0000FF;">])</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #7060A8;">puts</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">'\n'</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<!--</syntaxhighlight>-->
{{out}}
<pre style="font-size: 8px">
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
</pre>
"Reflected" Pascal's triangle, it uses symmetry property to "mirror" second part. It determines even and odd strings. automatically.
=={{header|PHP}}==
<syntaxhighlight lang="php">
<?php
//Author Ivan Gavryshin @dcc0
function tre($n) {
$ck=1;
$kn=$n+1;
if($kn%2==0) {
$kn=$kn/2;
$i=0;
}
else
{
$kn+=1;
$kn=$kn/2;
$i= 1;
}
for ($k = 1; $k <= $kn-1; $k++) {
$ck = $ck/$k*($n-$k+1);
$arr[] = $ck;
echo "+" . $ck ;
}
if ($kn>1) {
echo $arr[i];
$arr=array_reverse($arr);
for ($i; $i<= $kn-1; $i++) {
echo "+" . $arr[$i] ;
}
}
}
//set amount of strings here
while ($n<=20) {
++$n;
echo tre($n);
echo "<br/>";
}
?>
</syntaxhighlight> =={{header|PHP}}==
<syntaxhighlight lang="php">function pascalsTriangle($num){
$c = 1;
$triangle = Array();
Line 1,622 ⟶ 4,837:
}
echo '<br>';
}</
1
1 1
Line 1,632 ⟶ 4,847:
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
=={{header|Picat}}==
<syntaxhighlight lang="picat">
%Author: Petar Kabashki
spatr([]) = [].
spatr([_|T]) = A, T = [] => A = [].
spatr([H|T]) = A, T = [TH|_] => A = [H+TH] ++ spatr(T).
table
patr(0) = [1].
patr(1) = [1, 1].
patr(N) = A, N > 1 => Apre = patr(N-1), A = [1] ++ spatr(Apre) ++ [1].
foreach(I in 0 .. 10) println(patr(I)) end.
</syntaxhighlight>
<syntaxhighlight lang="picat">
[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]
</syntaxhighlight>
=={{header|PicoLisp}}==
{{trans|C}}
<syntaxhighlight lang="picolisp">(de pascalTriangle (N)
(for I N
(space (* 2 (- N I)))
(let C 1
(for K I
(prin (align 3 C) " ")
(setq C (*/ C (- I K) K)) ) )
(prinl) ) )</syntaxhighlight>
=={{header|PL/I}}==
<syntaxhighlight lang="pl/i">
declare (t, u)(40) fixed binary;
declare (i, n) fixed binary;
Line 1,650 ⟶ 4,904:
t = u;
end;
</syntaxhighlight>
<syntaxhighlight lang="text">
1
1 1
Line 1,664 ⟶ 4,918:
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
</syntaxhighlight>
=={{header|
<syntaxhighlight lang="potion">printpascal = (n) :
if (n < 1) :
(
. else
prev =
prev
curr
curr append(prev(i) + prev(i + 1))
.
"\n" print
curr join(", ") print
curr
.
.
printpascal(read number integer)</syntaxhighlight>
=={{header|PowerShell}}==
<
$Infinity = 1
$NewNumbers = $null
Line 1,743 ⟶ 5,006:
$Infinity++
}
</syntaxhighlight>
Save the above code to a .ps1 script file and start it by calling its name and providing N.
Line 1,772 ⟶ 5,035:
=={{header|Prolog}}==
Difference-lists are used to make quick append.
<
pascal(1, N, [1], [[1]|X]-X, L),
maplist(my_format, L).
Line 1,808 ⟶ 5,071:
my_writef(X) :-
writef(' %5r', [X]).
</syntaxhighlight>
Output :
<
1
1 1
Line 1,829 ⟶ 5,092:
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
true.
</syntaxhighlight>
===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.
<syntaxhighlight lang="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(_).</syntaxhighlight>
*Output*:
<syntaxhighlight lang="prolog">?- pascal(5).
[1]
[1,1]
[1,2,1]
[1,3,3,1]
[1,4,6,4,1]</syntaxhighlight>
=={{header|PureBasic}}==
<
For i= 0 To n
Line 1,848 ⟶ 5,150:
Parameter.i = Val(ProgramParameter(0))
pascaltriangle(Parameter);
Input()</
=={{header|Python}}==
===Procedural===
<syntaxhighlight lang="python">def pascal(n):
"""Prints out n rows of Pascal's triangle.
It returns False for failure and True for success."""
Line 1,859 ⟶ 5,162:
print row
row=[l+r for l,r in zip(row+k,k+row)]
return n>=1</
<
a = []
result = it
Line 1,878 ⟶ 5,181:
for row in pascal(4):
print(row)</
===Functional===
Deriving finite and non-finite lists of pascal rows from a simple '''nextPascal''' step function:
{{Works with|Python|3.7}}
<syntaxhighlight lang="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 list(
map(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.'''
return accumulate(
chain(
[[1]], range(1, n)
),
lambda a, _: nextPascal(a)
)
# ------------------------ 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('\n'.join(map(
showPascal,
[
islice(
pascalTriangle(), # Non finite,
7
),
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 go
# MAIN ---
if __name__ == '__main__':
main()</syntaxhighlight>
{{Out}}
<pre> 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</pre>
=={{header|q}}==
<syntaxhighlight lang="q">
pascal:{(x-1){0+':x,0}\1}
pascal 5
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
</syntaxhighlight>
=={{header|Qi}}==
{{trans|Haskell}}
<syntaxhighlight lang="qi">
(define iterate
_ _ 0 -> []
Line 1,892 ⟶ 5,324:
(define pascal
N -> (iterate next-row [1] N))
</syntaxhighlight>
=={{header|Quackery}}==
The behaviour of <code>pascal</code> for values less than 1 is the same as its behaviour for 1.
<syntaxhighlight lang="quackery"> [ over size -
space swap of
swap join ] is justify ( $ n --> )
[ witheach
[ number$
5 justify echo$ ]
cr ] is echoline ( [ --> )
[ [] 0 rot 0 join
witheach
[ tuck +
rot join swap ]
drop ] is nextline ( [ --> [ )
[ ' [ 1 ] swap
1 - times
[ dup echoline
nextline ]
echoline ] is pascal ( n --> )
16 pascal</syntaxhighlight>
{{out}}
<pre> 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
</pre>
=={{header|R}}==
{{trans|Octave}}
<
for(i in 0:(h-1)) {
s <- ""
Line 1,905 ⟶ 5,384:
print(s)
}
}</
Here's an R version:
<
lapply(0:h, function(i) choose(i, 0:i))
}</
=={{header|Racket}}==
Iterative version by summing rows up to <math>n</math>.
<syntaxhighlight lang="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))))
</syntaxhighlight>
=={{header|Raku}}==
(formerly 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 (<tt>...</tt>). 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:
<syntaxhighlight lang="raku" line>sub pascal {
[1], { [0, |$_ Z+ |$_, 0] } ... *
}
.say for pascal[^10];</syntaxhighlight>
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 <tt>@</tt> sigil to indicate that the sequence should cache its values for reuse, and use an explicit parameter <tt>$prev</tt> for variety:
<syntaxhighlight lang="raku" line>constant @pascal = [1], -> $prev { [0, |$prev Z+ |$prev, 0] } ... *;
.say for @pascal[^10];</syntaxhighlight>
Since we use ordinary subscripting, non-positive inputs throw an index-out-of-bounds error.
=== recursive ===
{{trans|Haskell}}
<syntaxhighlight lang="raku" line>multi sub pascal (1) { $[1] }
multi sub pascal (Int $n where 2..*) {
my @rows = pascal $n - 1;
|@rows, [0, |@rows[*-1] Z+ |@rows[*-1], 0 ];
}
.say for pascal 10;</syntaxhighlight>
Non-positive inputs throw a multiple-dispatch error.
=== iterative ===
{{trans|Perl}}
<syntaxhighlight lang="raku" line>sub pascal ($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;</syntaxhighlight>
Non-positive inputs throw a type check error.
{{Output}}
<pre>[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]</pre>
=={{header|RapidQ}}==
Line 1,926 ⟶ 5,484:
RapidQ does not require simple variables to be declared before use.
<
INPUT "Number of rows: "; nrows
Line 1,937 ⟶ 5,495:
NEXT i
PRINT
NEXT row</
===Using binary coefficients===
{{trans|BASIC}}
<
FOR row = 0 TO nrows-1
c = 1
Line 1,950 ⟶ 5,508:
NEXT i
PRINT
NEXT row</
=={{header|Red}}==
<syntaxhighlight lang="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
]
]</syntaxhighlight>
Output:
<pre>
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
</pre>
=={{header|Retro}}==
<
: pascalTriangle
cr dup
[ dup !j 1 swap 1+ [ !i dup putn space @j @i - * @i 1+ / ] iter cr drop ] iter drop
;
13 pascalTriangle</
=={{header|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 been easily created.
<br>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
<syntaxhighlight lang="rexx">/*REXX program displays (or writes to a file) Pascal's triangle (centered/formatted).*/
numeric digits 3000 /*be able to handle gihugeic triangles.*/
parse arg nn . /*obtain the optional argument from CL.*/
if nn=='' | nn=="," then nn= 10 /*Not specified? Then use the default.*/
n= abs(nn) /*N is the number of rows in triangle.*/
w= length( !(n-1) % !(n%2) % !(n - 1 - n%2)) /*W: the width of biggest integer. */
ww= (n-1) * (W + 1) + 1 /*WW: " " " triangle's last row.*/
@.= 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.*/
do r=1 for n; $$= center($.r, ww) /*center this particular Pascals' row. */
if nn>0 then say $$ /*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*/</syntaxhighlight>
{{out|output|text= when using the input of: <tt> 11 </tt>}}
<pre>
1
1 1
Line 2,003 ⟶ 5,598:
1 10 45 120 210 252 210 120 45 10 1
</pre>
(Output shown at <big>'''<sup>4</sup>/<sub>5</sub>'''</big> size.)
<pre style="font-size:80%">
1
1 1
Line 2,029 ⟶ 5,626:
</pre>
=={{header|
<syntaxhighlight lang="ring">
row = 5
for i
col = 1
see left(" ",row-i)
for k = 0 to i
see "" + col + " "
col = col*(i-k)/(k+1)
next
see nl
next
</syntaxhighlight>
Output:
<pre>
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
</pre>
=={{header|RPL}}==
« 0 SWAP '''FOR''' n
"" 0 n '''FOR''' p
'''NEXT'''
n 1 + DISP
'''NEXT'''
7 FREEZE
» '<span style="color:blue">PASCAL</span>' STO
8 <span style="color:blue">PASCAL</span>
{{out}}
<pre>
1
1 1
1 2 1
1 3 3 1
1 4 6 4
1 5 10 10 5 1
1 6 15 20 15 6
1 7 21 35
1 8 28 56 70 56 28 8 …
</pre>
RPL screens are limited to 22 characters.
=={{header|Ruby}}==
<syntaxhighlight lang="ruby">def pascal(n)
raise ArgumentError, "must be positive." if n < 1
yield ar = [1]
(n-1).times do
ar.unshift(0).push(0) # tack a zero on both ends
yield ar = ar.each_cons(2).map(&:sum)
end
end
pascal(8){|row| puts row.join(" ").center(20)}</syntaxhighlight>
{{out}}
<pre>
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
</pre>
Or for more or less a translation of the two line Haskell version (with inject being abused a bit I know):
<
def pascal(n) n.times.inject([1]) {|x,_| next_row x } end
8.times{|i| p pascal(i)}</syntaxhighlight>
{{out}}
<pre>
[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]
</pre>
=={{header|Run BASIC}}==
<
for i = 0 to r - 1
c = 1
Line 2,076 ⟶ 5,725:
next
print
next</
<pre>Number of rows? ?5
1
Line 2,084 ⟶ 5,733:
1 4 6 4 1</pre>
=={{header|
{{trans|C}}
<syntaxhighlight lang="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!();
}
}
</syntaxhighlight>
=={{header|Scala}}==
===Functional solutions===
====Summing: Recursive row definition====
<syntaxhighlight lang="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
}</syntaxhighlight>
Function to pretty print n rows:
<syntaxhighlight lang="scala">def prettyTri(n:Int) = (1 to n) foreach {i => print(" "*(n-i)); tri(i) map (c => print(c + " ")); println}
prettyTri(5)</syntaxhighlight>
{{Out}}
<pre> 1
1 1
1 2 1
1 3 3 1
1 4 6 4 1</pre>
====Summing: Scala Stream (Recursive & Memoization)====
<syntaxhighlight lang="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"))
}</syntaxhighlight>
{{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)].
=={{header|Scheme}}==
{{Works with|Scheme|R<math>^5</math>RS}}
<
(map + (cons 0 row) (append row '(0))))
Line 2,118 ⟶ 5,796:
(triangle (list 1) 5)
</syntaxhighlight>
Output:
<syntaxhighlight lang="text">((1) (1 1) (1 2 1) (1 3 3 1) (1 4 6 4 1))</
=={{header|Seed7}}==
<
const proc: main is func
Line 2,144 ⟶ 5,822:
writeln;
end for;
end func;</
=={{header|Sidef}}==
<syntaxhighlight lang="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)</syntaxhighlight>
=={{header|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).
<syntaxhighlight lang="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<n; i++) {
a[i+1,i] = i
}
s = p = I(n)
k = 1
for (i=0; i<n; i++) {
p = p*a/k++
s = s+p
}
return(s)
}</syntaxhighlight>
Now print the Pascal triangle.
<syntaxhighlight lang="stata">function print_pascal_triangle(n) {
a = pascal1(n)
for (i=1; i<=n; i++) {
for (j=1; j<=i; j++) {
printf("%10.0f",a[i,j])
}
printf("\n")
}
}
print_pascal_triangle(5)
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1</syntaxhighlight>
=={{header|Swift}}==
<syntaxhighlight lang="swift">func pascal(n:Int)->[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..<a.count{
if i+1==a.count{
temp.append(1)
break
}
temp[i+1] = a[i]+a[i+1]
}
a=temp
print(a)
return a
}
}
let waste = pascal(n:10)
</syntaxhighlight>
=={{header|Tcl}}==
===Summing from Previous Rows===
<
if {$n < 1} {error "undefined behaviour for n < 1"}
set row [list 1]
Line 2,165 ⟶ 5,928:
}
puts [join [pascal_iterative 6] \n]</
<pre>1
1 1
Line 2,174 ⟶ 5,937:
===Using binary coefficients===
{{trans|BASIC}}
<
if {$n < 1} {error "undefined behaviour for n < 1"}
for {set i 0} {$i < $n} {incr i} {
Line 2,188 ⟶ 5,951:
}
puts [join [pascal_coefficients 6] \n]</
===Combinations===
{{trans|Java}}
Thanks to Tcl 8.5's arbitrary precision integer arithmetic, this solution is not limited to a couple of dozen rows. Uses a caching factorial calculator to improve performance.
<
proc pascal_combinations n {
Line 2,226 ⟶ 5,989:
}
puts [join [pascal_combinations 6] \n]</
===Comparing Performance===
<
puts "calculate $n rows:"
foreach proc {pascal_iterative pascal_coefficients pascal_combinations} {
puts "$proc: [time [list $proc $n] 100]"
}</
{{Out}}
<pre>calculate 100 rows:
pascal_iterative: 2800.14 microseconds per iteration
pascal_coefficients: 8760.98 microseconds per iteration
pascal_combinations: 38176.66 microseconds per iteration</pre>
=={{header|TI-83 BASIC}}==
===Using Addition of Previous Rows===
<syntaxhighlight lang="ti83b">PROGRAM:PASCALTR
:Lbl IN
:ClrHome
:Disp "NUMBER OF ROWS"
:Input N
:If N < 1:Goto IN
:{N,N}→dim([A])
:"CHEATING TO MAKE IT FASTER"
:For(I,1,N)
:1→[A](1,1)
:End
:For(I,2,N)
:For(J,2,I)
:[A](I-1,J-1)+[A](I-1,J)→[A](I,J)
:End
:End
:[A]</syntaxhighlight>
===Using nCr Function===
<syntaxhighlight lang="ti83b">PROGRAM:PASCALTR
:Lbl IN
:ClrHome
:Disp "NUMBER OF ROWS"
:Input N
:If N < 1:Goto IN
:{N,N}→dim([A])
:For(I,2,N)
:For(J,2,I)
:(I-1) nCr (J-1)→[A](I,J)
:End
:End
:[A]</syntaxhighlight>
=={{header|Turing}}==
<syntaxhighlight lang="turing">proc pascal (n : int)
for i : 0 .. n
var c := 1
for k : 0 .. i
put c : 4 ..
c := c * (i - k) div (k + 1)
end for
put ""
end for
end pascal
pascal(5)</syntaxhighlight>
Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
== {{header|TypeScript}} ==
{{trans|XPL0}}
<syntaxhighlight lang="javascript">// Pascal's triangle
function pascal(n: number): void {
// Display the first n rows of Pascal's triangle
// if n<=0 then nothing is displayed
var ld: number[] = new Array(40); // Old
var nw: number[] = new Array(40); // New
for (var row = 0; row < n; row++) {
nw[0] = 1;
for (var i = 1; i <= row; i++)
nw[i] = ld[i - 1] + ld[i];
process.stdout.write(" ".repeat((n - row - 1) * 2));
for (var i = 0; i <= row; i++) {
if (nw[i] < 100)
process.stdout.write(" ");
process.stdout.write(nw[i].toString());
if (nw[i] < 10)
process.stdout.write(" ");
process.stdout.write(" ");
}
nw[row + 1] = 0;
// We do not copy data from nw to ld
// but we work with references.
var tmp = ld;
ld = nw;
nw = tmp;
console.log();
}
}
pascal(13);
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|uBasic/4tH}}==
<syntaxhighlight lang="text">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</syntaxhighlight>
Output:
<pre>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
</pre>
=={{header|UNIX Shell}}==
{{works with|Bourne Again SHell}}
Any n <= 1 will print the "1" row.
<syntaxhighlight lang="bash">#! /bin/bash
pascal() {
local -i n=${1:-1}
if (( n <= 1 )); then
echo 1
else
local output=$( $FUNCNAME $((n - 1)) )
set -- $( tail -n 1 <<<"$output" ) # previous row
echo "$output"
printf "1 "
while [[ -n $1 ]]; do
printf "%d " $(( $1 + ${2:-0} ))
shift
done
echo
fi
}
pascal "$1"</syntaxhighlight>
=={{header|Ursala}}==
Zero maps to the empty list. Negatives are inexpressible.
This solution uses a library function for binomial coefficients.
<
#import nat
pascal = choose**ziDS+ iota*t+ iota+ successor</
This solution uses direct summation. The algorithm is to
insert zero at the head of a list (initially the unit list <1>), zip it with its reversal,
map the sum over the list of pairs, iterate n times, and return the trace.
<
#import nat
pascal "n" = (next"n" sum*NiCixp) <1></
test program:
<
example = pascal 10</
{{Out}}
<pre><
<1>,
Line 2,270 ⟶ 6,194:
<1,8,28,56,70,56,28,8,1>,
<1,9,36,84,126,126,84,36,9,1>></pre>
=={{header|VBA}}==
<syntaxhighlight lang="vb">Option Base 1
Private Sub pascal_triangle(n As Integer)
Dim odd() As String
Dim eve() As String
ReDim odd(1)
ReDim eve(2)
odd(1) = " 1"
For i = 1 To n
If i Mod 2 = 1 Then
Debug.Print String$(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</syntaxhighlight>{{out}}
<pre> 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</pre>
=={{header|VBScript}}==
Derived from the BASIC version.
<syntaxhighlight lang="vb">Pascal_Triangle(WScript.Arguments(0))
Function Pascal_Triangle(n)
Dim values(100)
values(1) = 1
WScript.StdOut.Write values(1)
WScript.StdOut.WriteLine
For row = 2 To n
For i = row To 1 Step -1
values(i) = values(i) + values(i-1)
WScript.StdOut.Write values(i) & " "
Next
WScript.StdOut.WriteLine
Next
End Function</syntaxhighlight>
{{out}}
Invoke from a command line.
<pre>
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
</pre>
=={{header|Vedit macro language}}==
Line 2,279 ⟶ 6,276:
For example, if #99 contains value 2, then #@99 accesses contents of numeric register #2.
<
#0=0; #1=1
Ins_Char(' ', COUNT, #100*3-2) Num_Ins(1)
Line 2,291 ⟶ 6,288:
}
Ins_Newline
}</
===Using binary coefficients===
{{trans|BASIC}}
<
for (#2 = 0; #2 < #1; #2++) {
#3 = 1
Line 2,304 ⟶ 6,301:
}
Ins_Newline
}</
=={{header|Visual Basic}}==
{{works with|Visual Basic|VB6 Standard}}
<syntaxhighlight lang="vb">Sub pascaltriangle()
'Pascal's triangle
Const m = 11
Dim t(40) As Integer, u(40) As Integer
Dim i As Integer, n As Integer, s As String, ss As String
ss = ""
For n = 1 To m
u(1) = 1
s = ""
For i = 1 To n
u(i + 1) = t(i) + t(i + 1)
s = s & u(i) & " "
t(i) = u(i)
Next i
ss = ss & s & vbCrLf
Next n
MsgBox ss, , "Pascal's triangle"
End Sub 'pascaltriangle</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Visual Basic .NET}}==
{{trans|C#}}
<syntaxhighlight lang="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</syntaxhighlight>
{{out}}
<pre> 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</pre>
=={{header|Wren}}==
{{libheader|Wren-fmt}}
{{libheader|Wren-math}}
<syntaxhighlight lang="wren">import "./fmt" for Fmt
import "./math" for Int
var pascalTriangle = Fn.new { |n|
if (n <= 0) return
for (i in 0...n) {
System.write(" " * (n-i-1))
for (j in 0..i) {
Fmt.write("$3d ", Int.binomial(i, j))
}
System.print()
}
}
pascalTriangle.call(13)</syntaxhighlight>
{{out}}
<pre>
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
</pre>
=={{header|X86 Assembly}}==
{{works with|NASM}}
{{works with|Windows}}
<b>uses:</b> io.inc - Macro library from SASM
<syntaxhighlight lang="asm">
%include "io.inc"
section .text
global CMAIN
CMAIN:
mov ebx, 7 ;size
call mloop
ret
mloop:
mov edx, 0 ;edx stands for the nth line
looping:
push ebx
push edx
call line
pop edx
pop ebx
inc edx
cmp edx, ebx
jl looping
xor eax, eax
ret
line:
mov ecx, 0 ;ecx stands for the nth character in each line
mlp:
push ecx
push edx
call nCk
pop edx
pop ecx
PRINT_DEC 4, eax ;print returned number
PRINT_STRING " "
inc ecx
cmp ecx, edx ;if
jle mlp
NEWLINE
ret
nCk:
;ecx : j
;edx : i
mov esi, edx
call fac ;i!
push eax ;save i!
mov esi, ecx
call fac ;j!
push eax ;save j!
mov ebx, edx
sub ebx, ecx ;(i-j)
mov esi, ebx
call fac ;(i-j)!
pop ebx ;(i-j)! is in eax
mul ebx ;(i-j)! * j!
mov ecx, eax
pop eax ; get i!
div ecx ; ; last step : i! divided by (i-j)! * j!
ret
fac:
push ecx
push edx
mov eax, 1
mov ecx, esi
cmp ecx, 0 ; 0! returns 1
je facz
lp:
mul ecx ;multiplies eax by ecx and then decrements ecx until ecx is 0
dec ecx
cmp ecx, 0
jg lp
jmp end
facz:
mov eax, 1
end:
pop edx
pop ecx
ret
</syntaxhighlight>
{{out}}
<pre>
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
</pre>
=={{header|XBasic}}==
{{trans|GW-BASIC}}
{{works with|Windows XBasic}}
<syntaxhighlight lang="xbasic">
PROGRAM "pascal"
VERSION "0.0001"
DECLARE FUNCTION Entry()
FUNCTION Entry()
r@@ = UBYTE(INLINE$("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
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|XPL0}}==
<syntaxhighlight lang="xpl0">include c:\cxpl\codes;
proc Pascal(N); \Display the first N rows of Pascal's triangle
int N; \if N<=0 then nothing is displayed
int Row, I, Old(40), New(40);
[for Row:= 0 to N-1 do
[New(0):= 1;
for I:= 1 to Row do New(I):= Old(I-1) + Old(I);
for I:= 1 to (N-Row-1)*2 do ChOut(0, ^ );
for I:= 0 to Row do
[if New(I)<100 then ChOut(0, ^ );
IntOut(0, New(I));
if New(I)<10 then ChOut(0, ^ );
ChOut(0, ^ );
];
New(Row+1):= 0;
I:= Old; Old:= New; New:= I;
CrLf(0);
];
];
Pascal(13)</syntaxhighlight>
{{Out}}
<pre>
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
</pre>
=={{header|zkl}}==
{{trans|C}}
<syntaxhighlight lang="zkl">fcn pascalTriangle(n){ // n<=0-->""
foreach i in (n){
c := 1;
print(" "*(2*(n-1-i)));
foreach k in (i+1){
print("%3d ".fmt(c));
c = c * (i-k)/(k+1);
}
println();
}
}
pascalTriangle(8);</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|ZX Spectrum Basic}}==
In edit mode insert:
<syntaxhighlight lang="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</syntaxhighlight>
Then in command mode (basically don't put a number in front):
<syntaxhighlight lang="basic">RUN</syntaxhighlight>
{{out}}
<pre>
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
</pre>
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