Pascal's triangle
You are encouraged to solve this task according to the task description, using any language you may know.
Pascal's triangle is an arithmetic and geometric figure often associated with the name of Blaise Pascal, but also studied centuries earlier in India, Persia, China and elsewhere.
Its first few rows look like this:
1 1 1 1 2 1 1 3 3 1
where each element of each row is either 1 or the sum of the two elements right above it.
For example, the next row of the triangle would be:
- 1 (since the first element of each row doesn't have two elements above it)
- 4 (1 + 3)
- 6 (3 + 3)
- 4 (3 + 1)
- 1 (since the last element of each row doesn't have two elements above it)
So the triangle now looks like this:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1
Each row n (starting with row 0 at the top) shows the coefficients of the binomial expansion of (x + y)n.
- Task
Write a function that prints out the first n rows of the triangle (with f(1) yielding the row consisting of only the element 1).
This can be done either by summing elements from the previous rows or using a binary coefficient or combination function.
Behavior for n ≤ 0 does not need to be uniform, but should be noted.
- See also
11l
<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)</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
360 Assembly
<lang 360asm>* Pascal's triangle 25/10/2015 PASCAL CSECT
USING PASCAL,R15 set base register
LA R7,1 n=1
LOOPN C R7,=A(M) do n=1 to m
BH ELOOPN if n>m then goto
MVC U,=F'1' u(1)=1
LA R8,PG pgi=@pg
LA R6,1 i=1
LOOPI CR R6,R7 do i=1 to n
BH ELOOPI if i>n then goto
LR R1,R6 i
SLA R1,2 i*4
L R3,T-4(R1) t(i)
L R4,T(R1) t(i+1)
AR R3,R4 t(i)+t(i+1)
ST R3,U(R1) u(i+1)=t(i)+t(i+1)
LR R1,R6 i
SLA R1,2 i*4
L R2,U-4(R1) u(i)
XDECO R2,XD edit u(i)
MVC 0(4,R8),XD+8 output u(i):4
LA R8,4(R8) pgi=pgi+4
LA R6,1(R6) i=i+1
B LOOPI end i
ELOOPI MVC T((M+1)*(L'T)),U t=u
XPRNT PG,80 print
LA R7,1(R7) n=n+1
B LOOPN end n
ELOOPN XR R15,R15 set return code
BR R14 return to caller
M EQU 11 <== input T DC (M+1)F'0' t(m+1) init 0 U DC (M+1)F'0' u(m+1) init 0 PG DC CL80' ' pg init ' ' XD DS CL12 temp
YREGS
END PASCAL</lang>
- Output:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1
8th
One way, using array operations: <lang forth> \ print the array
- .arr \ a -- a
( . space ) a:each ;
- pasc \ a --
\ print the row .arr cr dup \ create two rows from the first, one with a leading the other with a trailing 0 [0] 0 a:insert swap 0 a:push \ add the arrays together to make the new one ' n:+ a:op ;
\ print the first 16 rows: [1] ' pasc 16 times </lang>
Another way, using the relation between element 'n' and element 'n-1' in a row: <lang forth>
- ratio \ m n -- num denom
tuck n:- n:1+ swap ;
\ one item in the row: n m
- pascitem \ n m -- n
r@ swap ratio n:*/ n:round int dup . space ;
\ One row of Pascal's triangle
- pascline \ n --
>r 1 int dup . space ' pascitem 1 r@ loop rdrop drop cr ;
\ Calculate the first 'n' rows of Pascal's triangle:
- pasc \ n
' pascline 0 rot loop cr ;
15 pasc </lang>
Action!
<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</lang>
- Output:
Screenshot from Atari 8-bit computer
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
Ada
The specification of auxiliary package "Pascal". "First_Row" outputs a row with a single "1", "Next_Row" computes the next row from a given row, and "Length" gives the number of entries in a row. The package is also used for the Catalan numbers solution [[1]]
<lang ada>package Pascal is
type Row is array (Natural range <>) of Natural; function Length(R: Row) return Positive; function First_Row(Max_Length: Positive) return Row; function Next_Row(R: Row) return Row;
end Pascal;</lang>
The implementation of that auxiliary package "Pascal":
<lang Ada>package body Pascal is
function First_Row(Max_Length: Positive) return Row is
R: Row(0 .. Max_Length) := (0 | 1 => 1, others => 0);
begin
return R;
end First_Row;
function Next_Row(R: Row) return Row is
S: Row(R'Range);
begin
S(0) := Length(R)+1;
S(Length(S)) := 1;
for J in reverse 2 .. Length(R) loop
S(J) := R(J)+R(J-1);
end loop;
S(1) := 1;
return S;
end Next_Row;
function Length(R: Row) return Positive is
begin
return R(0);
end Length;
end Pascal;</lang>
The main program, using "Pascal". It prints the desired number of rows. The number is read from the command line.
<lang Ada>with Ada.Text_IO, Ada.Integer_Text_IO, Ada.Command_Line, Pascal; use Pascal;
procedure Triangle is
Number_Of_Rows: Positive := Integer'Value(Ada.Command_Line.Argument(1));
Row: Pascal.Row := First_Row(Number_Of_Rows);
begin
loop
-- print one row
for J in 1 .. Length(Row) loop
Ada.Integer_Text_IO.Put(Row(J), 5);
end loop;
Ada.Text_IO.New_Line;
exit when Length(Row) >= Number_Of_Rows;
Row := Next_Row(Row);
end loop;
end Triangle;</lang>
- Output:
>./triangle 12
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
ALGOL 68
<lang algol68>PRIO MINLWB = 8, MAXUPB = 8; OP MINLWB = ([]INT a,b)INT: (LWB a<LWB b|LWB a|LWB b),
MAXUPB = ([]INT a,b)INT: (UPB a>UPB b|UPB a|UPB b);
OP + = ([]INT a,b)[]INT:(
[a MINLWB b:a MAXUPB b]INT out; FOR i FROM LWB out TO UPB out DO out[i]:= 0 OD; out[LWB a:UPB a] := a; FOR i FROM LWB b TO UPB b DO out[i]+:= b[i] OD; out
);
INT width = 4, stop = 9; FORMAT centre = $n((stop-UPB row+1)*width OVER 2)(q)$;
FLEX[1]INT row := 1; # example of rowing # FOR i WHILE
printf((centre, $g(-width)$, row, $l$));
- WHILE # i < stop DO
row := row[AT 1] + row[AT 2]
OD</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
ALGOL W
<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.</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
APL
Pascal' s triangle of order ⍵
Dyalog APL
<lang apl> {⍕0~¨⍨(-⌽A)⌽↑,/0,¨⍉A∘.!A←0,⍳⍵} </lang>
example
<lang apl> {⍕0~¨⍨(-⌽A)⌽↑,/0,¨⍉A∘.!A←0,⍳⍵}5 </lang>
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
GNU APL
GNU APL doesn't allow multiple statements within lambdas so the solution is phrased differently:
<lang apl> {{⍉⍵∘.!⍵} 0,⍳⍵} </lang>
example
<lang apl> {{⍉⍵∘.!⍵} 0,⍳⍵} 3 </lang>
1 0 0 0 1 1 0 0 1 2 1 0 1 3 3 1
AppleScript
Drawing n rows from a generator: <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</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Arturo
<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
]</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
AutoHotkey
ahk forum: discussion <lang AutoHotkey>n := 8, p0 := "1" ; 1+n rows of Pascal's triangle Loop %n% {
p := "p" A_Index, %p% := v := 1, q := "p" A_Index-1
Loop Parse, %q%, %A_Space%
If (A_Index > 1)
%p% .= " " v+A_LoopField, v := A_LoopField
%p% .= " 1"
}
; Triangular Formatted output
VarSetCapacity(tabs,n,Asc("`t")) t .= tabs "`t1" Loop %n% {
t .= "`n" SubStr(tabs,A_Index)
Loop Parse, p%A_Index%, %A_Space%
t .= A_LoopField "`t`t"
} Gui Add, Text,, %t% ; Show result in a GUI Gui Show Return
GuiClose:
ExitApp</lang>
Alternate
<lang AutoHotkey>Msgbox % format(pascalstriangle()) Return
format(o) ; converts object to string { For k, v in o s .= IsObject(v) ? format(v) "`n" : v " " Return s } pascalstriangle(n=7) ; n rows of Pascal's triangle { p := Object(), z:=Object() Loop, % n Loop, % row := A_Index col := A_Index , p[row, col] := row = 1 and col = 1 ? 1 : (p[row-1, col-1] = "" ; math operations on blanks return blanks; I want to assume zero ? 0 : p[row-1, col-1]) + (p[row-1, col] = "" ? 0 : p[row-1, col]) Return p }</lang> n <= 0 returns empty
AWK
<lang awk>$ awk 'BEGIN{for(i=0;i<6;i++){c=1;r=c;for(j=0;j<i;j++){c*=(i-j)/(j+1);r=r" "c};print r}}'</lang>
- Output:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1
BASIC
Summing from Previous Rows
This implementation uses an array to store one row of the triangle. DIM initializes the array values to zero. For first row, "1" is then stored in the array. To calculate values for next row, the value in cell (i-1) is added to each cell (i). This summing is done from right to left so that it can be done on-place, without using a tmp buffer. Because of symmetry, the values can be displayed from left to right.
Space for max 5 digit numbers is reserved when formatting the display. The maximum size of triangle is 100 rows, but in practice it is limited by screen space. If the user enters value less than 1, the first row is still always displayed.
<lang freebasic>DIM i AS Integer DIM row AS Integer DIM nrows AS Integer DIM values(100) AS Integer
INPUT "Number of rows: "; nrows values(1) = 1 PRINT TAB((nrows)*3);" 1" FOR row = 2 TO nrows
PRINT TAB((nrows-row)*3+1);
FOR i = row TO 1 STEP -1
values(i) = values(i) + values(i-1)
PRINT USING "##### "; values(i);
NEXT i
PRINT
NEXT row</lang>
Batch File
Based from the Fortran Code. <lang dos>@echo off setlocal enabledelayedexpansion
- The Main Thing...
cls echo. set row=15 call :pascal echo. pause exit /b 0
- /The Main Thing.
- The Functions...
- pascal
set /a prev=%row%-1 for /l %%I in (0,1,%prev%) do ( set c=1&set r= for /l %%K in (0,1,%row%) do ( if not !c!==0 ( call :numstr !c! set r=!r!!space!!c! ) set /a c=!c!*^(%%I-%%K^)/^(%%K+1^) ) echo !r! ) goto :EOF
- numstr
::This function returns the number of whitespaces to be applied on each numbers. set cnt=0&set proc=%1&set space= :loop set currchar=!proc:~%cnt%,1! if not "!currchar!"=="" set /a cnt+=1&goto loop set /a numspaces=5-!cnt! for /l %%A in (1,1,%numspaces%) do set "space=!space! " goto :EOF
- /The Functions.</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
Press any key to continue . . .
BBC BASIC
<lang bbcbasic> nrows% = 10
colwidth% = 4
@% = colwidth% : REM Set column width
FOR row% = 1 TO nrows%
PRINT SPC(colwidth%*(nrows% - row%)/2);
acc% = 1
FOR element% = 1 TO row%
PRINT acc%;
acc% = acc% * (row% - element%) / element% + 0.5
NEXT
PRINT
NEXT row%</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
BCPL
<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')
$)
let start() be pascal(8)</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
Befunge
<lang Befunge>0" :swor fo rebmuN">:#,_&> 55+, v v01*p00-1:g00.:<1p011p00:\-1_v#:< >g:1+10p/48*,:#^_$ 55+,1+\: ^>$$@</lang>
- Output:
Number of rows: 10 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1
BQN
Displays n rows.
<lang bqn>Pascal ← {(0⊸∾+∾⟜0)⍟(↕𝕩)⋈1}
•Show¨Pascal 6</lang> <lang>⟨ 1 ⟩ ⟨ 1 1 ⟩ ⟨ 1 2 1 ⟩ ⟨ 1 3 3 1 ⟩ ⟨ 1 4 6 4 1 ⟩ ⟨ 1 5 10 10 5 1 ⟩</lang>
Bracmat
<lang bracmat>( out$"Number of rows? " & get':?R & -1:?I & whl
' ( 1+!I:<!R:?I
& 1:?C
& -1:?K
& !R+-1*!I:?tabs
& whl'(!tabs+-1:>0:?tabs&put$\t)
& whl
' ( 1+!K:~>!I:?K
& put$(!C \t\t)
& !C*(!I+-1*!K)*(!K+1)^-1:?C
)
& put$\n
)
& )</lang>
- Output:
Number of rows?
7
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Burlesque
<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 </lang>
C
<lang c>#include <stdio.h>
void pascaltriangle(unsigned int n) {
unsigned int c, i, j, k;
for(i=0; i < n; i++) {
c = 1;
for(j=1; j <= 2*(n-1-i); j++) printf(" ");
for(k=0; k <= i; k++) {
printf("%3d ", c);
c = c * (i-k)/(k+1);
}
printf("\n");
}
}
int main() {
pascaltriangle(8); return 0;
}</lang>
Recursive
<lang c>#include <stdio.h>
- define D 32
int pascals(int *x, int *y, int d) { int i; for (i = 1; i < d; i++) printf("%d%c", y[i] = x[i - 1] + x[i], i < d - 1 ? ' ' : '\n');
return D > d ? pascals(y, x, d + 1) : 0; }
int main() { int x[D] = {0, 1, 0}, y[D] = {0}; return pascals(x, y, 0); }</lang>
Adding previous row values
<lang c>void triangleC(int nRows) {
if (nRows <= 0) return;
int *prevRow = NULL;
for (int r = 1; r <= nRows; r++) {
int *currRow = malloc(r * sizeof(int));
for (int i = 0; i < r; i++) {
int val = i==0 || i==r-1 ? 1 : prevRow[i-1] + prevRow[i];
currRow[i] = val;
printf(" %4d", val);
}
printf("\n");
free(prevRow);
prevRow = currRow;
}
free(prevRow);
}</lang>
C#
Produces no output when n is less than or equal to zero.
<lang csharp>using System;
namespace RosettaCode {
class PascalsTriangle {
public static void CreateTriangle(int n) {
if (n > 0) {
for (int i = 0; i < n; i++) {
int c = 1;
Console.Write(" ".PadLeft(2 * (n - 1 - i)));
for (int k = 0; k <= i; k++) {
Console.Write("{0}", c.ToString().PadLeft(3));
c = c * (i - k) / (k + 1);
}
Console.WriteLine();
}
}
}
public static void Main() {
CreateTriangle(8);
}
}
}</lang>
Arbitrarily large numbers (BigInteger), arbitrary row selection
<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); } } }</lang>
Example: <lang csharp>static void Main() { IEnumerable<BigInteger[]> triangle = PascalsTriangle.GetTriangle(20); string output = PascalsTriangle.FormatTriangleString(triangle) Console.WriteLine(output); }</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
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
C++
<lang cpp>#include <iostream>
- include <algorithm>
- include<cstdio>
using namespace std; void Pascal_Triangle(int size) {
int a[100][100]; int i, j;
//first row and first coloumn has the same value=1 for (i = 1; i <= size; i++) { a[i][1] = a[1][i] = 1; }
//Generate the full Triangle for (i = 2; i <= size; i++) { for (j = 2; j <= size - i; j++) { if (a[i - 1][j] == 0 || a[i][j - 1] == 0) { break; } a[i][j] = a[i - 1][j] + a[i][j - 1]; } }
/* 1 1 1 1 1 2 3 1 3 1
first print as above format-->
for (i = 1; i < size; i++) { for (j = 1; j < size; j++) { if (a[i][j] == 0) { break; } printf("%8d",a[i][j]); } cout<<"\n\n"; }*/
// standard Pascal Triangle Format
int row,space; for (i = 1; i < size; i++) { space=row=i; j=1;
while(space<=size+(size-i)+1){ cout<<" "; space++; }
while(j<=i){ if (a[row][j] == 0){ break; }
if(j==1){ printf("%d",a[row--][j++]); } else printf("%6d",a[row--][j++]); } cout<<"\n\n"; }
}
int main() { //freopen("out.txt","w",stdout);
int size; cin>>size; Pascal_Triangle(size); }
}</lang>
C++11 (with dynamic and semi-static vectors)
Constructs the whole triangle in memory before printing it. Uses vector of vectors as a 2D array with variable column size. Theoretically, semi-static version should work a little faster. <lang cpp>// Compile with -std=c++11
- include<iostream>
- include<vector>
using namespace std; void print_vector(vector<int> dummy){ for (vector<int>::iterator i = dummy.begin(); i != dummy.end(); ++i) cout<<*i<<" "; cout<<endl; } void print_vector_of_vectors(vector<vector<int>> dummy){ for (vector<vector<int>>::iterator i = dummy.begin(); i != dummy.end(); ++i) print_vector(*i); cout<<endl; } vector<vector<int>> dynamic_triangle(int dummy){ vector<vector<int>> result; if (dummy > 0){ // if the argument is 0 or negative exit immediately vector<int> row; // The first row row.push_back(1); result.push_back(row); // The second row if (dummy > 1){ row.clear(); row.push_back(1); row.push_back(1); result.push_back(row); } // The other rows if (dummy > 2){ for (int i = 2; i < dummy; i++){ row.clear(); row.push_back(1); for (int j = 1; j < i; j++) row.push_back(result.back().at(j - 1) + result.back().at(j)); row.push_back(1); result.push_back(row); } } } return result; } vector<vector<int>> static_triangle(int dummy){ vector<vector<int>> result; if (dummy > 0){ // if the argument is 0 or negative exit immediately vector<int> row; result.resize(dummy); // This should work faster than consecutive push_back()s // The first row row.resize(1); row.at(0) = 1; result.at(0) = row; // The second row if (result.size() > 1){ row.resize(2); row.at(0) = 1; row.at(1) = 1; result.at(1) = row; } // The other rows if (result.size() > 2){ for (int i = 2; i < result.size(); i++){ row.resize(i + 1); // This should work faster than consecutive push_back()s row.front() = 1; for (int j = 1; j < row.size() - 1; j++) row.at(j) = result.at(i - 1).at(j - 1) + result.at(i - 1).at(j); row.back() = 1; result.at(i) = row; } } } return result; } int main(){ vector<vector<int>> triangle; int n; cout<<endl<<"The Pascal's Triangle"<<endl<<"Enter the number of rows: "; cin>>n; // Call the dynamic function triangle = dynamic_triangle(n); cout<<endl<<"Calculated using dynamic vectors:"<<endl<<endl; print_vector_of_vectors(triangle); // Call the static function triangle = static_triangle(n); cout<<endl<<"Calculated using static vectors:"<<endl<<endl; print_vector_of_vectors(triangle); return 0; }
</lang>
C++11 (with a class)
A full fledged example with a class definition and methods to retrieve data, worthy of the title object-oriented. <lang cpp>// Compile with -std=c++11
- include<iostream>
- include<vector>
using namespace std; class pascal_triangle{ vector<vector<int>> data; // This is the actual data void print_row(vector<int> dummy){ for (vector<int>::iterator i = dummy.begin(); i != dummy.end(); ++i) cout<<*i<<" "; cout<<endl; } public: pascal_triangle(int dummy){ // Everything is done on the construction phase if (dummy > 0){ // if the argument is 0 or negative exit immediately vector<int> row; data.resize(dummy); // Theoretically this should work faster than consecutive push_back()s // The first row row.resize(1); row.at(0) = 1; data.at(0) = row; // The second row if (data.size() > 1){ row.resize(2); row.at(0) = 1; row.at(1) = 1; data.at(1) = row; } // The other rows if (data.size() > 2){ for (int i = 2; i < data.size(); i++){ row.resize(i + 1); // Theoretically this should work faster than consecutive push_back()s row.front() = 1; for (int j = 1; j < row.size() - 1; j++) row.at(j) = data.at(i - 1).at(j - 1) + data.at(i - 1).at(j); row.back() = 1; data.at(i) = row; } } } } ~pascal_triangle(){ for (vector<vector<int>>::iterator i = data.begin(); i != data.end(); ++i) i->clear(); // I'm not sure about the necessity of this loop! data.clear(); } void print_row(int dummy){ if (dummy < data.size()) for (vector<int>::iterator i = data.at(dummy).begin(); i != data.at(dummy).end(); ++i) cout<<*i<<" "; cout<<endl; } void print(){ for (int i = 0; i < data.size(); i++) print_row(i); } int get_coeff(int dummy1, int dummy2){ int result = 0; if ((dummy1 < data.size()) && (dummy2 < data.at(dummy1).size())) result = data.at(dummy1).at(dummy2); return result; } vector<int> get_row(int dummy){ vector<int> result; if (dummy < data.size()) result = data.at(dummy); return result; } }; int main(){ int n; cout<<endl<<"The Pascal's Triangle with a class!"<<endl<<endl<<"Enter the number of rows: "; cin>>n; pascal_triangle myptri(n); cout<<endl<<"The whole triangle:"<<endl; myptri.print(); cout<<endl<<"Just one row:"<<endl; myptri.print_row(n/2); cout<<endl<<"Just one coefficient:"<<endl; cout<<myptri.get_coeff(n/2, n/4)<<endl<<endl; return 0; }
</lang>
Clojure
For n < 1, prints nothing, always returns nil. Copied from the Common Lisp implementation below, but with local functions and explicit tail-call-optimized recursion (recur). <lang lisp>(defn pascal [n]
(let [newrow (fn newrow [lst ret]
(if lst
(recur (rest lst)
(conj ret (+ (first lst) (or (second lst) 0))))
ret))
genrow (fn genrow [n lst]
(when (< 0 n)
(do (println lst)
(recur (dec n) (conj (newrow lst []) 1)))))]
(genrow n [1])))
(pascal 4)</lang> And here's another version, using the partition function to produce the sequence of pairs in a row, which are summed and placed between two ones to produce the next row: <lang lisp> (defn nextrow [row]
(vec (concat [1] (map #(apply + %) (partition 2 1 row)) [1] )))
(defn pascal [n]
(assert (and (integer? n) (pos? n)))
(let [triangle (take n (iterate nextrow [1]))]
(doseq [row triangle]
(println row))))
</lang> The assert form causes the pascal function to throw an exception unless the argument is (integral and) positive.
Here's a third version using the iterate function <lang lisp> (def pascal
(iterate
(fn [prev-row]
(->>
(concat (first prev-row) (partition 2 1 prev-row) (last prev-row))
(map (partial apply +) ,,,)))
[1]))
</lang>
Another short version which returns an infinite pascal triangle as a list, using the iterate function.
<lang lisp> (def pascal
(iterate #(concat [1]
(map + % (rest %))
[1])
[1]))
</lang>
One can then get the first n rows using the take function
<lang lisp> (take 10 pascal) ; returns a list of the first 10 pascal rows </lang>
Also, one can retrieve the nth row using the nth function
<lang lisp> (nth pascal 10) ;returns the nth row </lang>
CoffeeScript
This version assumes n is an integer and n >= 1. It efficiently computes binomial coefficients. <lang coffeescript> pascal = (n) ->
width = 6
for r in [1..n]
s = ws (width/2) * (n-r) # center row
output = (n) -> s += pad width, n
cell = 1
output cell
# Compute binomial coefficients as you go
# across the row.
for c in [1...r]
cell *= (r-c) / c
output cell
console.log s
ws = (n) ->
s = s += ' ' for i in [0...n] s
pad = (cnt, n) ->
s = n.toString() # There is probably a better way to do this. cnt -= s.length right = Math.floor(cnt / 2) left = cnt - right ws(left) + s + ws(right)
pascal(7)
</lang>
- Output:
> coffee pascal.coffee
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Commodore BASIC
<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</lang>
Output: <lang>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. </lang>
Common Lisp
To evaluate, call (pascal n). For n < 1, it simply returns nil.
<lang lisp>(defun pascal (n)
(genrow n '(1)))
(defun genrow (n l)
(when (plusp n) (print l) (genrow (1- n) (cons 1 (newrow l)))))
(defun newrow (l)
(if (null (rest l))
'(1)
(cons (+ (first l) (second l))
(newrow (rest l)))))</lang>
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)).
<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))</lang>
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.
<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)))
</lang>
For example:
<lang lisp>(print-pascal-triangle 4)</lang> <lang>
1 1 1 1 2 1
1 3 3 1 </lang> <lang lisp>(print-pascal-triangle 8)</lang> <lang>
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1 </lang>
Component Pascal
<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. </lang>
Execute: ^Q PascalTriangle.Do 0 1 2 3 4 5 6 7 8 9 10 11 12~
- Output:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 1 11 55 165 330 462 462 330 165 55 11 1 1 12 66 220 495 792 924 792 495 220 66 12 1
D
Less functional Version
<lang d>int[][] pascalsTriangle(in int rows) pure nothrow {
auto tri = new int[][rows];
foreach (r; 0 .. rows) {
int v = 1;
foreach (c; 0 .. r+1) {
tri[r] ~= v;
v = (v * (r - c)) / (c + 1);
}
}
return tri;
}
void main() {
immutable t = pascalsTriangle(10);
assert(t == [[1],
[1, 1],
[1, 2, 1],
[1, 3, 3, 1],
[1, 4, 6, 4, 1],
[1, 5, 10, 10, 5, 1],
[1, 6, 15, 20, 15, 6, 1],
[1, 7, 21, 35, 35, 21, 7, 1],
[1, 8, 28, 56, 70, 56, 28, 8, 1],
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]]);
}</lang>
More functional Version
<lang d>import std.stdio, std.algorithm, std.range;
auto pascal() pure nothrow {
return [1].recurrence!q{ zip(a[n - 1] ~ 0, 0 ~ a[n - 1])
.map!q{ a[0] + a[1] }
.array };
}
void main() {
pascal.take(5).writeln;
}</lang>
- Output:
[[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1]]
Alternative Version
There is similarity between Pascal's triangle and Sierpinski triangle. Their difference are the initial line and the operation that act on the line element to produce next line. The following is a generic pascal's triangle implementation for positive number of lines output (n). <lang d>import std.stdio, std.string, std.array, std.format;
string Pascal(alias dg, T, T initValue)(int n) {
string output;
void append(in T[] l) {
output ~= " ".replicate((n - l.length + 1) * 2);
foreach (e; l)
output ~= format("%4s", format("%4s", e));
output ~= "\n";
}
if (n > 0) {
T[][] lines = initValue;
append(lines[0]);
foreach (i; 1 .. n) {
lines ~= lines[i - 1] ~ initValue; // length + 1
foreach (int j; 1 .. lines[i-1].length)
lines[i][j] = dg(lines[i-1][j], lines[i-1][j-1]);
append(lines[i]);
}
}
return output;
}
string delegate(int n) genericPascal(alias dg, T, T initValue)() {
mixin Pascal!(dg, T, initValue); return &Pascal;
}
void main() {
auto pascal = genericPascal!((int a, int b) => a + b, int, 1)();
static char xor(char a, char b) { return a == b ? '_' : '*'; }
auto sierpinski = genericPascal!(xor, char, '*')();
foreach (i; [1, 5, 9])
writef(pascal(i));
// an order 4 sierpinski triangle is a 2^4 lines generic
// Pascal triangle with xor operation
foreach (i; [16])
writef(sierpinski(i));
}</lang>
- Output:
1
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
*
* *
* _ *
* * * *
* _ _ _ *
* * _ _ * *
* _ * _ * _ *
* * * * * * * *
* _ _ _ _ _ _ _ *
* * _ _ _ _ _ _ * *
* _ * _ _ _ _ _ * _ *
* * * * _ _ _ _ * * * *
* _ _ _ * _ _ _ * _ _ _ *
* * _ _ * * _ _ * * _ _ * *
* _ * _ * _ * _ * _ * _ * _ *
* * * * * * * * * * * * * * * *
Dart
<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);
}
</lang>
Delphi
<lang delphi>program PascalsTriangle;
procedure Pascal(r:Integer); var
i, c, k:Integer;
begin
for i := 0 to r - 1 do
begin
c := 1;
for k := 0 to i do
begin
Write(c:3);
c := c * (i - k) div (k + 1);
end;
Writeln;
end;
end;
begin
Pascal(9);
end.</lang>
DWScript
Doesn't print anything for negative or null values. <lang delphi>procedure Pascal(r : Integer); var
i, c, k : Integer;
begin
for i:=0 to r-1 do begin
c:=1;
for k:=0 to i do begin
Print(Format('%4d', [c]));
c:=(c*(i-k)) div (k+1);
end;
PrintLn();
end;
end;
Pascal(9);</lang>
- Output:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1
E
So as not to bother with text layout, this implementation generates a HTML fragment. It uses a single mutable array, appending one 1 and adding to each value the preceding value.
<lang e>def pascalsTriangle(n, out) {
def row := [].diverge(int)
out.print("
") for y in 1..n { out.print("") row.push(1) def skip := n - y if (skip > 0) { out.print(``) }
for x => v in row {
out.print(``)
}
for i in (1..!y).descending() {
row[i] += row[i - 1]
}
out.println("")
}
out.print("| $v |
")
}</lang>
<lang e>def out := <file:triangle.html>.textWriter() try {
pascalsTriangle(15, out)
} finally {
out.close()
} makeCommand("yourFavoriteWebBrowser")("triangle.html")</lang>
Eiffel
<lang eiffel> note description : "Prints pascal's triangle" output : "[
Per requirements of the RosettaCode example, execution will print the first n rows of pascal's triangle ]"
date : "19 December 2013" authors : "Sandro Meier", "Roman Brunner" revision : "1.0" libraries : "Relies on HASH_TABLE from EIFFEL_BASE library" implementation : "[ Recursive implementation to calculate the n'th row. ]" warning : "[ Will not work for large n's (INTEGER_32) ]"
class APPLICATION
inherit ARGUMENTS
create make
feature {NONE} -- Initialization
make local n:INTEGER do create {HASH_TABLE[ARRAY[INTEGER],INTEGER]}pascal_lines.make (n) --create the hash_table object io.new_line n:=25 draw(n) end feature line(n:INTEGER):ARRAY[INTEGER] --Calculates the n'th line local upper_line:ARRAY[INTEGER] i:INTEGER do if n=1 then --trivial case first line create Result.make_filled (0, 1, n+2) Result.put (0, 1) Result.put (1, 2) Result.put (0, 3) elseif pascal_lines.has (n) then --checks if the result was already calculated Result := pascal_lines.at (n) else --calculates the n'th line recursively create Result.make_filled(0,1,n+2) --for caluclation purposes add a 0 at the beginning of each line Result.put (0, 1) upper_line:=line(n-1) from i:=1 until i>upper_line.count-1 loop Result.put(upper_line[i]+upper_line[i+1],i+1) i:=i+1 end Result.put (0, n+2) --for caluclation purposes add a 0 at the end of each line pascal_lines.put (Result, n) end end
draw(n:INTEGER) --draw n lines of pascal's triangle local space_string:STRING width, i:INTEGER
do space_string:=" " --question of design: add space_string at the beginning of each line width:=line(n).count space_string.multiply (width) from i:=1 until i>n loop space_string.remove_tail (1) io.put_string (space_string) across line(i) as c loop if c.item/=0 then io.put_string (c.item.out+" ") end end io.new_line i:=i+1 end end
feature --Access pascal_lines:HASH_TABLE[ARRAY[INTEGER],INTEGER] --Contains all already calculated lines end </lang>
Elixir
<lang elixir>defmodule Pascal do
def triangle(n), do: triangle(n,[1])
def triangle(0,list), do: list
def triangle(n,list) do
IO.inspect list
new_list = Enum.zip([0]++list, list++[0]) |> Enum.map(fn {a,b} -> a+b end)
triangle(n-1,new_list)
end
end
Pascal.triangle(8)</lang>
- Output:
[1] [1, 1] [1, 2, 1] [1, 3, 3, 1] [1, 4, 6, 4, 1] [1, 5, 10, 10, 5, 1] [1, 6, 15, 20, 15, 6, 1] [1, 7, 21, 35, 35, 21, 7, 1]
Emacs Lisp
Using mapcar and append, returing a list of rows
<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)))))</lang>
- Output:
Call the function from the REPL, IELM:
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))
Translation from Pascal
<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))))</lang>
- Output:
From the REPL:
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
Returning a string
Same as the translation from Pascal, but now returning a string. <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))</lang>
- Output:
Now, since this one returns a string, it is possible to insert the result in the current buffer:
(insert (pascal 6)) 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1
Erlang
<lang erlang> -import(lists). -export([pascal/1]).
pascal(1)-> 1; pascal(N) ->
L = pascal(N-1), [H|_] = L, [lists:zipwith(fun(X,Y)->X+Y end,[0]++H,H++[0])|L].
</lang>
- Output:
Eshell V5.5.5 (abort with ^G) 1> pascal:pascal(5). [[1,4,6,4,1],[1,3,3,1],[1,2,1],[1,1],[1]]
ERRE
<lang ERRE> PROGRAM PASCAL_TRIANGLE
PROCEDURE PASCAL(R%)
LOCAL I%,C%,K%
FOR I%=0 TO R%-1 DO
C%=1
FOR K%=0 TO I% DO
WRITE("###";C%;)
C%=(C%*(I%-K%)) DIV (K%+1)
END FOR
PRINT
END FOR
END PROCEDURE
BEGIN
PASCAL(9)
END PROGRAM </lang> Output:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1
Euphoria
Summing from Previous Rows
<lang Euphoria>sequence row row = {} for m = 1 to 10 do
row = row & 1
for n = length(row)-1 to 2 by -1 do
row[n] += row[n-1]
end for
print(1,row)
puts(1,'\n')
end for</lang>
- Output:
{1}
{1,1}
{1,2,1}
{1,3,3,1}
{1,4,6,4,1}
{1,5,10,10,5,1}
{1,6,15,20,15,6,1}
{1,7,21,35,35,21,7,1}
{1,8,28,56,70,56,28,8,1}
{1,9,36,84,126,126,84,36,9,1}
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 LAMBDA: The ultimate Excel worksheet function)
<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))
)
)</lang>
- Output:
| fx | =PASCAL(A2) | |||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| A | B | C | D | E | F | G | H | I | J | K | ||
| 1 | Row number | PASCAL's TRIANGLE | ||||||||||
| 2 | 1 | 1 | ||||||||||
| 3 | 2 | 1 | 1 | |||||||||
| 4 | 3 | 1 | 2 | 1 | ||||||||
| 5 | 4 | 1 | 3 | 3 | 1 | |||||||
| 6 | 5 | 1 | 4 | 6 | 4 | 1 | ||||||
| 7 | 6 | 1 | 5 | 10 | 10 | 5 | 1 | |||||
| 8 | 7 | 1 | 6 | 15 | 20 | 15 | 6 | 1 | ||||
| 9 | 8 | 1 | 7 | 21 | 35 | 35 | 21 | 7 | 1 | |||
| 10 | 9 | 1 | 8 | 28 | 56 | 70 | 56 | 28 | 8 | 1 | ||
| 11 | 10 | 1 | 9 | 36 | 84 | 126 | 126 | 84 | 36 | 9 | 1 | |
Or defining the whole triangle as a single grid, by binding the name TRIANGLE to an additional lambda:
<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),
""
)
)
)</lang>
- Output:
| fx | =TRIANGLE(10) | |||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| A | B | C | D | E | F | G | H | I | J | K | ||
| 1 | PASCAL's TRIANGLE | |||||||||||
| 2 | 1 | |||||||||||
| 3 | 1 | 1 | ||||||||||
| 4 | 1 | 2 | 1 | |||||||||
| 5 | 1 | 3 | 3 | 1 | ||||||||
| 6 | 1 | 4 | 6 | 4 | 1 | |||||||
| 7 | 1 | 5 | 10 | 10 | 5 | 1 | ||||||
| 8 | 1 | 6 | 15 | 20 | 15 | 6 | 1 | |||||
| 9 | 1 | 7 | 21 | 35 | 35 | 21 | 7 | 1 | ||||
| 10 | 1 | 8 | 28 | 56 | 70 | 56 | 28 | 8 | 1 | |||
| 11 | 1 | 9 | 36 | 84 | 126 | 126 | 84 | 36 | 9 | 1 | ||
F#
<lang fsharp>let rec nextrow l =
match l with | [] -> [] | h :: [] -> [1] | h :: t -> h + t.Head :: nextrow t
let pascalTri n = List.scan(fun l i -> 1 :: nextrow l) [1] [1 .. n]
for row in pascalTri(10) do
for i in row do
printf "%s" (i.ToString() + ", ")
printfn "%s" "\n"
</lang>
Factor
This implementation works by summing the previous line content. Result for n < 1 is the same as for n == 1.
<lang factor>USING: grouping kernel math sequences ;
- (pascal) ( seq -- newseq )
dup last 0 prefix 0 suffix 2 <clumps> [ sum ] map suffix ;
- pascal ( n -- seq )
1 - { { 1 } } swap [ (pascal) ] times ;</lang>
It works as:
<lang factor>5 pascal . { { 1 } { 1 1 } { 1 2 1 } { 1 3 3 1 } { 1 4 6 4 1 } }</lang>
Fantom
<lang fantom> class Main {
Int[] next_row (Int[] row)
{
new_row := [1]
(row.size-1).times |i|
{
new_row.add (row[i] + row[i+1])
}
new_row.add (1)
return new_row }
Void print_pascal (Int n) // no output for n <= 0
{
current_row := [1]
n.times
{
echo (current_row.join(" "))
current_row = next_row (current_row)
}
}
Void main ()
{
print_pascal (10)
}
} </lang>
FOCAL
<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)</lang>
- Output:
= 1 = 1= 1 = 1= 2= 1 = 1= 3= 3= 1 = 1= 4= 6= 4= 1 = 1= 5= 10= 10= 5= 1 = 1= 6= 15= 20= 15= 6= 1 = 1= 7= 21= 35= 35= 21= 7= 1 = 1= 8= 28= 56= 70= 56= 28= 8= 1 = 1= 9= 36= 84= 126= 126= 84= 36= 9= 1 = 1= 10= 45= 120= 210= 252= 210= 120= 45= 10= 1
Forth
<lang forth>: init ( n -- )
here swap cells erase 1 here ! ;
- .line ( n -- )
cr here swap 0 do dup @ . cell+ loop drop ;
- next ( n -- )
here swap 1- cells here + do i @ i cell+ +! -1 cells +loop ;
- pascal ( n -- )
dup init 1 .line 1 ?do i next i 1+ .line loop ;</lang>
This is a bit more efficient.
<lang forth>: PascTriangle
cr dup 0
?do
1 over 1- i - 2* spaces i 1+ 0 ?do dup 4 .r j i - * i 1+ / loop cr drop
loop drop
13 PascTriangle</lang>
Fortran
Prints nothing for n<=0. Output formatting breaks down for n>20 <lang fortran>PROGRAM Pascals_Triangle
CALL Print_Triangle(8)
END PROGRAM Pascals_Triangle
SUBROUTINE Print_Triangle(n)
IMPLICIT NONE INTEGER, INTENT(IN) :: n INTEGER :: c, i, j, k, spaces
DO i = 0, n-1
c = 1
spaces = 3 * (n - 1 - i)
DO j = 1, spaces
WRITE(*,"(A)", ADVANCE="NO") " "
END DO
DO k = 0, i
WRITE(*,"(I6)", ADVANCE="NO") c
c = c * (i - k) / (k + 1)
END DO
WRITE(*,*)
END DO
END SUBROUTINE Print_Triangle</lang>
FreeBASIC
<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</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
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. <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] </lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
FunL
Summing from Previous Rows
<lang funl>import lists.zip
def
pascal( 1 ) = [1] pascal( n ) = [1] + map( (a, b) -> a + b, zip(pascal(n-1), pascal(n-1).tail()) ) + [1]</lang>
Combinations
<lang funl>import integers.choose
def pascal( n ) = [choose( n - 1, k ) | k <- 0..n-1]</lang>
Pascal's Triangle
<lang funl>def triangle( height ) =
width = max( map(a -> a.toString().length(), pascal(height)) )
if 2|width width++
for n <- 1..height
print( ' '*((width + 1)\2)*(height - n) )
println( map(a -> format('%' + width + 'd ', a), pascal(n)).mkString() )
triangle( 10 )</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
Fōrmulæ
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.
Programs in Fōrmulæ are created/edited online in its website, However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used.
In this page you can see the program(s) related to this task and their results.
GAP
<lang gap>Pascal := function(n) local i, v; v := [1]; for i in [1 .. n] do Display(v); v := Concatenation([0], v) + Concatenation(v, [0]); od; end;
Pascal(9);
- [ 1 ]
- [ 1, 1 ]
- [ 1, 2, 1 ]
- [ 1, 3, 3, 1 ]
- [ 1, 4, 6, 4, 1 ]
- [ 1, 5, 10, 10, 5, 1 ]
- [ 1, 6, 15, 20, 15, 6, 1 ]
- [ 1, 7, 21, 35, 35, 21, 7, 1 ]
- [ 1, 8, 28, 56, 70, 56, 28, 8, 1 ]</lang>
Go
No output for n < 1. Otherwise, output formatted left justified. <lang go> package main
import "fmt"
func printTriangle(n int) {
// degenerate cases
if n <= 0 {
return
}
fmt.Println(1)
if n == 1 {
return
}
// iterate over rows, zero based
a := make([]int, (n+1)/2)
a[0] = 1
for row, middle := 1, 0; row < n; row++ {
// generate new row
even := row&1 == 0
if even {
a[middle+1] = a[middle] * 2
}
for i := middle; i > 0; i-- {
a[i] += a[i-1]
}
// print row
for i := 0; i <= middle; i++ {
fmt.Print(a[i], " ")
}
if even {
middle++
}
for i := middle; i >= 0; i-- {
fmt.Print(a[i], " ")
}
fmt.Println("")
}
}
func main() {
printTriangle(4)
} </lang> Output:
1 1 1 1 2 1 1 3 3 1
Groovy
Recursive
In the spirit of the Haskell "think in whole lists" solution here is a list-driven, minimalist solution: <lang groovy>def pascal pascal = { n -> (n <= 1) ? [1] : [[0] + pascal(n - 1), pascal(n - 1) + [0]].transpose().collect { it.sum() } }</lang> However, this solution is horribly inefficient (O(n**2)). It slowly grinds to a halt on a reasonably powerful PC after about line 25 of the triangle.
Test program: <lang groovy>def count = 15 (1..count).each { n ->
printf ("%2d:", n); (0..(count-n)).each { print " " }; pascal(n).each{ printf("%6d ", it) }; println ""
}</lang>
- Output:
1: 1 2: 1 1 3: 1 2 1 4: 1 3 3 1 5: 1 4 6 4 1 6: 1 5 10 10 5 1 7: 1 6 15 20 15 6 1 8: 1 7 21 35 35 21 7 1 9: 1 8 28 56 70 56 28 8 1 10: 1 9 36 84 126 126 84 36 9 1 11: 1 10 45 120 210 252 210 120 45 10 1 12: 1 11 55 165 330 462 462 330 165 55 11 1 13: 1 12 66 220 495 792 924 792 495 220 66 12 1 14: 1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1 15: 1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
GW-BASIC
<lang qbasic>10 INPUT "Number of rows? ",R 20 FOR I=0 TO R-1 30 C=1 40 FOR K=0 TO I 50 PRINT USING "####";C; 60 C=C*(I-K)/(K+1) 70 NEXT 80 PRINT 90 NEXT</lang>
Output:
Number of rows? 7 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1
Haskell
An approach using the "think in whole lists" principle: Each row in the triangle can be calculated from the previous row by adding a shifted version of itself to it, keeping the ones at the ends. The Prelude function zipWith can be used to add two lists, but it won't keep the old values when one list is shorter. So we need a similar function
<lang haskell>zapWith :: (a -> a -> a) -> [a] -> [a] -> [a] zapWith f xs [] = xs zapWith f [] ys = ys zapWith f (x:xs) (y:ys) = f x y : zapWith f xs ys</lang>
Now we can shift a list and add it to itself, extending it by keeping the ends:
<lang haskell>extendWith f [] = [] extendWith f xs@(x:ys) = x : zapWith f xs ys</lang>
And for the whole (infinite) triangle, we just iterate this operation, starting with the first row:
<lang haskell>pascal = iterate (extendWith (+)) [1]</lang>
For the first n rows, we just take the first n elements from this list, as in
<lang haskell>*Main> take 6 pascal [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1],[1,5,10,10,5,1]]</lang>
A shorter approach, plagiarized from [2] <lang haskell>-- generate next row from current row nextRow row = zipWith (+) ([0] ++ row) (row ++ [0])
-- returns the first n rows pascal = iterate nextRow [1]</lang>
Alternatively, using list comprehensions:
<lang haskell> pascal :: Integer pascal =
(1 : [ 0 | _ <- head pascal]) : [zipWith (+) (0:row) row | row <- pascal]
</lang>
<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]] </lang>
With binomial coefficients: <lang haskell>fac = product . enumFromTo 1
binCoef n k = fac n `div` (fac k * fac (n - k))
pascal = ((fmap . binCoef) <*> enumFromTo 0) . pred</lang>
Example: <lang haskell>*Main> putStr $ unlines $ map unwords $ map (map show) $ pascal 10 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 </lang>
HicEst
<lang HicEst> CALL Pascal(30)
SUBROUTINE Pascal(rows)
CHARACTER fmt*6 WRITE(Text=fmt, Format='"i", i5.5') 1+rows/4
DO row = 0, rows-1
n = 1
DO k = 0, row
col = rows*(rows-row+2*k)/4
WRITE(Row=row+1, Column=col, F=fmt) n
n = n * (row - k) / (k + 1)
ENDDO
ENDDO
END</lang>
Icon and Unicon
The code below is slightly modified from the library version of pascal which prints 0's to the full width of the carpet. It also presents the data as an isoceles triangle. <lang Icon>link math
procedure main(A) every n := !A do { # for each command line argument
n := integer(\n) | &null pascal(n) }
end
procedure pascal(n) #: Pascal triangle
/n := 16
write("width=", n, " height=", n) # carpet header
fw := *(2 ^ n)+1
every i := 0 to n - 1 do {
writes(repl(" ",fw*(n-i)/2))
every j := 0 to n - 1 do
writes(center(binocoef(i, j),fw) | break)
write()
}
end</lang>
math provides binocoef math provides the original version of pascal
Sample output:
->pascal 1 4 8
width=1 height=1
1
width=4 height=4
1
1 1
1 2 1
1 3 3 1
width=8 height=8
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
->
IDL
<lang IDL>Pro Pascal, n
- n is the number of lines of the triangle to be displayed
r=[1] print, r for i=0, (n-2) do begin pascalrow,r endfor
End
Pro PascalRow, r
for i=0,(n_elements(r)-2) do begin r[i]=r[i]+r[i+1] endfor
r= [1, r] print, r
End</lang>
IS-BASIC
<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</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
ivy
<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
</lang>
J
<lang j> !~/~ i.5 1 0 0 0 0 1 1 0 0 0 1 2 1 0 0 1 3 3 1 0 1 4 6 4 1</lang>
<lang j> ([: ":@-.&0"1 !~/~)@i. 5 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1</lang>
<lang j> (-@|. |."_1 [: ":@-.&0"1 !~/~)@i. 5
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1</lang>
See the talk page for explanation of earlier version
See also Pascal matrix generation and Sierpinski triangle.
Java
Summing from Previous Rows
<lang java>import java.util.ArrayList; ...//class definition, etc. public static void genPyrN(int rows){ if(rows < 0) return; //save the last row here ArrayList<Integer> last = new ArrayList<Integer>(); last.add(1); System.out.println(last); for(int i= 1;i <= rows;++i){ //work on the next row ArrayList<Integer> thisRow= new ArrayList<Integer>(); thisRow.add(last.get(0)); //beginning for(int j= 1;j < i;++j){//loop the number of elements in this row //sum from the last row thisRow.add(last.get(j - 1) + last.get(j)); } thisRow.add(last.get(0)); //end last= thisRow;//save this row System.out.println(thisRow); } }</lang>
Combinations
This method is limited to 21 rows because of the limits of long. Calling pas with an argument of 22 or above will cause intermediate math to wrap around and give false answers. <lang java>public class Pas{ public static void main(String[] args){ //usage pas(20); }
public static void pas(int rows){ for(int i = 0; i < rows; i++){ for(int j = 0; j <= i; j++){ System.out.print(ncr(i, j) + " "); } System.out.println(); } }
public static long ncr(int n, int r){ return fact(n) / (fact(r) * fact(n - r)); }
public static long fact(int n){ long ans = 1; for(int i = 2; i <= n; i++){ ans *= i; } return ans; } }</lang>
Using arithmetic calculation of each row element
This method is limited to 30 rows because of the limits of integer calculations (probably when calculating the multiplication). If m is declared as long then 62 rows can be printed. <lang java> public class Pascal { private static void printPascalLine (int n) { if (n < 1) return; int m = 1; System.out.print("1 "); for (int j=1; j<n; j++) { m = m * (n-j)/j; System.out.print(m); System.out.print(" "); } System.out.println(); }
public static void printPascal (int nRows) { for(int i=1; i<=nRows; i++) printPascalLine(i); } } </lang>
JavaScript
ES5
Imperative
<lang javascript>// Pascal's triangle object function pascalTriangle (rows) {
// Number of rows the triangle contains this.rows = rows;
// The 2D array holding the rows of the triangle this.triangle = new Array(); for (var r = 0; r < rows; r++) { this.triangle[r] = new Array(); for (var i = 0; i <= r; i++) { if (i == 0 || i == r) this.triangle[r][i] = 1; else this.triangle[r][i] = this.triangle[r-1][i-1]+this.triangle[r-1][i]; } }
// Method to print the triangle this.print = function(base) { if (!base) base = 10;
// Private method to calculate digits in number var digits = function(n,b) { var d = 0; while (n >= 1) { d++; n /= b; } return d; }
// Calculate max spaces needed var spacing = digits(this.triangle[this.rows-1][Math.round(this.rows/2)],base);
// Private method to add spacing between numbers var insertSpaces = function(s) { var buf = ""; while (s > 0) { s--; buf += " "; } return buf; }
// Print the triangle line by line for (var r = 0; r < this.triangle.length; r++) { var l = ""; for (var s = 0; s < Math.round(this.rows-1-r); s++) { l += insertSpaces(spacing); } for (var i = 0; i < this.triangle[r].length; i++) { if (i != 0) l += insertSpaces(spacing-Math.ceil(digits(this.triangle[r][i],base)/2)); l += this.triangle[r][i].toString(base); if (i < this.triangle[r].length-1) l += insertSpaces(spacing-Math.floor(digits(this.triangle[r][i],base)/2)); } print(l); } }
}
// Display 4 row triangle in base 10 var tri = new pascalTriangle(4); tri.print(); // Display 8 row triangle in base 16 tri = new pascalTriangle(8); tri.print(16);</lang> Output:
$ d8 pascal.js
1
1 1
1 2 1
1 3 3 1
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 a a 5 1
1 6 f 14 f 6 1
1 7 15 23 23 15 7 1
Functional
<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);</lang>
- Output:
| 1 | ||||||||||||
| 1 | 1 | |||||||||||
| 1 | 2 | 1 | ||||||||||
| 1 | 3 | 3 | 1 | |||||||||
| 1 | 4 | 6 | 4 | 1 | ||||||||
| 1 | 5 | 10 | 10 | 5 | 1 | |||||||
| 1 | 6 | 15 | 20 | 15 | 6 | 1 |
<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]]</lang>
ES6
<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();
})();</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
Recursive
<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) </lang>
- Output:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1
Recursive - memoized
<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) </lang>
jq
pascal(n) as defined here produces a stream of n arrays, each corresponding to a row of the Pascal triangle. The implementation avoids any arithmetic except addition. <lang jq># pascal(n) for n>=0; pascal(0) emits an empty stream. def pascal(n):
def _pascal: # input: the previous row
. as $in
| .,
if length >= n then empty
else
reduce range(0;length-1) as $i
([1]; . + [ $in[$i] + $in[$i + 1] ]) + [1] | _pascal
end;
if n <= 0 then empty else [1] | _pascal end ;</lang>
Example:
pascal(5)
- Output:
<lang sh>$ jq -c -n -f pascal_triangle.jq [1] [1,1] [1,2,1] [1,3,3,1] [1,4,6,4,1]</lang>
Using recurse/1
Here is an equivalent implementation that uses the built-in filter, recurse/1, instead of the inner function. <lang jq>def pascal(n):
if n <= 0 then empty
else [1]
| recurse( if length >= n then empty
else . as $in
| reduce range(0;length-1) as $i
([1]; . + [ $in[$i] + $in[$i + 1] ]) + [1]
end)
end;</lang>
Julia
function pascal(n)
if n<=0
print("n has to have a positive value")
end
x=0
while x<=n
for a=0:x
print(binomial(x,a)," ")
end
println("")
x+=1
end
end
pascal(5) 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1
Another solution using matrix exponentiation.
<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
</lang>
- Output:
pascal(5) 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1
Yet another solution using a static vector
<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 </lang>
- Output:
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
K
<lang K> pascal:{(x-1){+':x,0}\1} pascal 6 (1
1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1)</lang>
Kotlin
<lang kotlin>fun pas(rows: Int) {
for (i in 0..rows - 1) {
for (j in 0..i)
print(ncr(i, j).toString() + " ")
println()
}
}
fun ncr(n: Int, r: Int) = fact(n) / (fact(r) * fact(n - r))
fun fact(n: Int) : Long {
var ans = 1.toLong()
for (i in 2..n)
ans *= i
return ans
}
fun main(args: Array<String>) = pas(args[0].toInt())</lang>
Lambdatalk
<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 </lang>
Liberty BASIC
<lang lb>input "How much rows would you like? "; n dim a$(n)
for i= 0 to n
c = 1
o$ =""
for k =0 to i
o$ =o$ ; c; " "
c =c *(i-k)/(k+1)
next k
a$(i)=o$
next i
maxLen = len(a$(n)) for i= 0 to n
print space$((maxLen-len(a$(i)))/2);a$(i)
next i
end</lang>
Locomotive Basic
<lang locobasic>10 CLS 20 INPUT "Number of rows? ", rows:GOSUB 40 30 END 40 FOR i=0 TO rows-1 50 c=1 60 FOR k=0 TO i 70 PRINT USING "####";c; 80 c=c*(i-k)/(k+1) 90 NEXT 100 PRINT 110 NEXT 120 RETURN</lang>
Output:
Number of rows? 7 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1
Logo
<lang logo>to pascal :n
if :n = 1 [output [1]] localmake "a pascal :n-1 output (sentence first :a (map "sum butfirst :a butlast :a) last :a)
end
for [i 1 10] [print pascal :i]</lang>
Lua
<lang lua> function nextrow(t)
local ret = {}
t[0], t[#t+1] = 0, 0
for i = 1, #t do ret[i] = t[i-1] + t[i] end
return ret
end
function triangle(n)
t = {1}
for i = 1, n do
print(unpack(t))
t = nextrow(t)
end
end </lang>
Maple
<lang maple>f:=n->seq(print(seq(binomial(i,k),k=0..i)),i=0..n-1);
f(3);</lang>
1 1 1 1 2 1
Mathematica/Wolfram Language
<lang Mathematica>n=7;
Column[StringReplace[ToString /@ Replace[MatrixExp[SparseArray[
{Band[{2,1}] -> Range[n-1]},{n,n}]],{x__,0..}->{x},2] ,{"{"|"}"|","->" "}], Center]</lang>
A more graphical output with arrows would involve the plotting functionality with Graph[]: <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] </lang>
MATLAB / Octave
A matrix containing the pascal triangle can be obtained this way: <lang MATLAB>pascal(n);</lang>
>> pascal(6)
ans =
1 1 1 1 1 1
1 2 3 4 5 6
1 3 6 10 15 21
1 4 10 20 35 56
1 5 15 35 70 126
1 6 21 56 126 252
The binomial coefficients can be extracted from the Pascal triangle in this way: <lang MATLAB> binomCoeff = diag(rot90(pascal(n)))', </lang>
>> for k=1:6,diag(rot90(pascal(k)))', end
ans = 1
ans =
1 1
ans =
1 2 1
ans =
1 3 3 1
ans =
1 4 6 4 1
ans =
1 5 10 10 5 1
Another way to get a formated pascals triangle is to use the convolution method:
>>
x = [1 1] ;
y = 1;
for k=8:-1:1
fprintf(['%', num2str(k), 'c'], zeros(1,3)),
fprintf('%6d', y), fprintf('\n')
y = conv(y,x);
end
The result is:
>>
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
Maxima
<lang maxima>sjoin(v, j) := apply(sconcat, rest(join(makelist(j, length(v)), v)))$
display_pascal_triangle(n) := for i from 0 thru 6 do disp(sjoin(makelist(binomial(i, j), j, 0, i), " "));
display_pascal_triangle(6); /* "1"
"1 1" "1 2 1" "1 3 3 1" "1 4 6 4 1" "1 5 10 10 5 1" "1 6 15 20 15 6 1" */</lang>
Metafont
(The formatting starts to be less clear when numbers start to have more than two digits)
<lang metafont>vardef bincoeff(expr n, k) = save ?; ? := (1 for i=(max(k,n-k)+1) upto n: * i endfor )
/ (1 for i=2 upto min(k, n-k): * i endfor); ?
enddef;
def pascaltr expr c =
string s_;
for i := 0 upto (c-1):
s_ := "" for k=0 upto (c-i): & " " endfor;
s_ := s_ for k=0 upto i: & decimal(bincoeff(i,k))
& " " if bincoeff(i,k)<9: & " " fi endfor;
message s_;
endfor
enddef;
pascaltr(4); end</lang>
Microsoft Small Basic
<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 </lang>
Output:
Number of rows? 7 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1
Modula-2
<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.</lang>
NetRexx
<lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols nobinary
numeric digits 1000 -- allow very large numbers parse arg rows . if rows = then rows = 11 -- default to 11 rows printPascalTriangle(rows) return
-- ----------------------------------------------------------------------------- method printPascalTriangle(rows = 11) public static
lines = mx = (factorial(rows - 1) / factorial(rows % 2) / factorial(rows - 1 - rows % 2)).length() -- width of widest number
loop row = 1 to rows
n1 = 1.center(mx)
line = n1
loop col = 2 to row
n2 = col - 1
n1 = n1 * (row - n2) / n2
line = line n1.center(mx)
end col
lines[row] = line.strip()
end row
-- display triangle ml = lines[rows].length() -- length of longest line loop row = 1 to rows say lines[row].centre(ml) end row
return
-- ----------------------------------------------------------------------------- method factorial(n) public static
fac = 1 loop n_ = 2 to n fac = fac * n_ end n_ return fac /*calc. factorial*/
</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
Nial
Like J
(pascal.nial) <lang nial>factorial is recur [ 0 =, 1 first, pass, product, -1 +] combination is fork [ > [first, second], 0 first,
/ [factorial second, * [factorial - [second, first], factorial first] ]
] pascal is transpose each combination cart [pass, pass] tell</lang> Using it <lang nial>|loaddefs 'pascal.nial' |pascal 5</lang>
Nim
<lang nim>import sequtils, 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)</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
A more optimized solution that doesn't require importing, but produces, naturally, uglier output, would look like this: <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)</lang>
- Output:
@[1] @[1, 1] @[1, 2, 1] @[1, 3, 3, 1] @[1, 4, 6, 4, 1] @[1, 5, 10, 10, 5, 1] @[1, 6, 15, 20, 15, 6, 1] @[1, 7, 21, 35, 35, 21, 7, 1] @[1, 8, 28, 56, 70, 56, 28, 8, 1] @[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]
OCaml
<lang ocaml>(* generate next row from current row *) let next_row row =
List.map2 (+) ([0] @ row) (row @ [0])
(* returns the first n rows *) let pascal n =
let rec loop i row = if i = n then [] else row :: loop (i+1) (next_row row) in loop 0 [1]</lang>
Octave
<lang octave>function pascaltriangle(h)
for i = 0:h-1
for k = 0:h-i
printf(" ");
endfor
for j = 0:i
printf("%3d ", bincoeff(i, j));
endfor
printf("\n");
endfor
endfunction
pascaltriangle(4);</lang>
Oforth
No result if n <= 0
<lang Oforth>: pascal(n) [ 1 ] #[ dup println dup 0 + 0 rot + zipWith(#+) ] times(n) drop ;</lang>
- Output:
10 pascal [1] [1, 1] [1, 2, 1] [1, 3, 3, 1] [1, 4, 6, 4, 1] [1, 5, 10, 10, 5, 1] [1, 6, 15, 20, 15, 6, 1] [1, 7, 21, 35, 35, 21, 7, 1] [1, 8, 28, 56, 70, 56, 28, 8, 1] [1, 9, 36, 84, 126, 126, 84, 36, 9, 1]
Oz
<lang oz>declare
fun {NextLine Xs}
{List.zip 0|Xs {Append Xs [0]}
fun {$ Left Right}
Left + Right
end}
end
fun {Triangle N}
{List.take {Iterate [1] NextLine} N}
end
fun lazy {Iterate I F}
I|{Iterate {F I} F}
end
%% Only works nicely for N =< 5.
proc {PrintTriangle T}
N = {Length T}
in
for
Line in T
Indent in N-1..0;~1
do
for _ in 1..Indent do {System.printInfo " "} end
for L in Line do {System.printInfo L#" "} end
{System.printInfo "\n"}
end
end
in
{PrintTriangle {Triangle 5}}</lang>
For n = 0, prints nothing. For negative n, throws an exception.
PARI/GP
<lang parigp>pascals_triangle(N)= { my(row=[],prevrow=[]); for(x=1,N,
if(x>5,break(1));
row=eval(Vec(Str(11^(x-1))));
print(row));
prevrow=row; for(y=6,N,
for(p=2,#prevrow,
row[p]=prevrow[p-1]+prevrow[p]);
row=concat(row,1);
prevrow=row;
print(row);
);
}</lang>
Pascal
<lang pascal>Program PascalsTriangle(output);
procedure Pascal(r : Integer);
var
i, c, k : Integer;
begin
for i := 0 to r-1 do
begin
c := 1;
for k := 0 to i do
begin
write(c:3);
c := (c * (i-k)) div (k+1);
end;
writeln;
end;
end;
begin
Pascal(9)
end.</lang> Output:
% ./PascalsTriangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1
Perl
These functions perform as requested in the task: they print out the first n lines. If n <= 0, they print nothing. The output is simple (no fancy formatting). <lang perl>sub pascal {
my $rows = shift;
my @next = (1);
for my $n (1 .. $rows) {
print "@next\n";
@next = (1, (map $next[$_]+$next[$_+1], 0 .. $n-2), 1);
}
}</lang>
If you want more than 68 rows, then use either "use bigint" or "use Math::GMP qw/:constant/" inside the function to enable bigints. We can also use a binomial function which will expand to bigints if many rows are requested:
<lang perl>use ntheory qw/binomial/; sub pascal {
my $rows = shift;
for my $n (0 .. $rows-1) {
print join(" ", map { binomial($n,$_) } 0 .. $n), "\n";
}
}</lang>
Here is a non-obvious version using bignum, which is limited to the first 23 rows because of the algorithm used: <lang perl>use bignum; sub pascal_line { $_[0] ? unpack "A(A6)*", 1000001**$_[0] : 1 } sub pascal { print "@{[map -+-$_, pascal_line $_]}\n" for 0..$_[0]-1 }</lang>
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 Tartaglia's triangle
<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($_),"\n"}
print "\n\n";
print tartaglia(3,3),"\n";
my @third = tartaglia_row(5);
print "@third\n";
</lang>
which output
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 20 1 5 10 10 5 1
Phix
sequence row = {} for m = 1 to 13 do row = row & 1 for n=length(row)-1 to 2 by -1 do row[n] += row[n-1] end for printf(1,repeat(' ',(13-m)*2)) for i=1 to length(row) do printf(1," %3d",row[i]) end for puts(1,'\n') end for
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
"Reflected" Pascal's triangle, it uses symmetry property to "mirror" second part. It determines even and odd strings. automatically.
PHP
<lang php> <?php
//Author Ivan Gavryshin @dcc0
function tre($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($n);
echo "
";
}
?>
</lang> ==PHP==
<lang php>function pascalsTriangle($num){
$c = 1;
$triangle = Array();
for($i=0;$i<=$num;$i++){
$triangle[$i] = Array();
if(!isset($triangle[$i-1])){
$triangle[$i][] = $c;
}else{
for($j=0;$j<count($triangle[$i-1])+1;$j++){
$triangle[$i][] = (isset($triangle[$i-1][$j-1]) && isset($triangle[$i-1][$j])) ? $triangle[$i-1][$j-1] + $triangle[$i-1][$j] : $c;
}
}
}
return $triangle;
}
$tria = pascalsTriangle(8);
foreach($tria as $val){
foreach($val as $value){
echo $value . ' ';
}
echo '
';
}</lang>
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
Picat
<lang Picat> 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. [1] [1,1] [1,2,1] [1,3,3,1] [1,4,6,4,1] [1,5,10,10,5,1] [1,6,15,20,15,6,1] [1,7,21,35,35,21,7,1] [1,8,28,56,70,56,28,8,1] [1,9,36,84,126,126,84,36,9,1] [1,10,45,120,210,252,210,120,45,10,1] </lang>
PicoLisp
<lang PicoLisp>(de pascalTriangle (N)
(for I N
(space (* 2 (- N I)))
(let C 1
(for K I
(prin (align 3 C) " ")
(setq C (*/ C (- I K) K)) ) )
(prinl) ) )</lang>
PL/I
<lang PL/I> declare (t, u)(40) fixed binary; declare (i, n) fixed binary;
t,u = 0; get (n); if n <= 0 then return;
do n = 1 to n;
u(1) = 1;
do i = 1 to n;
u(i+1) = t(i) + t(i+1);
end;
put skip edit ((u(i) do i = 1 to n)) (col(40-2*n), (n+1) f(4));
t = u;
end; </lang>
<lang>
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
</lang>
Potion
<lang potion>printpascal = (n) :
if (n < 1) :
1 print
(1)
. else :
prev = printpascal(n - 1)
prev append(0)
curr = (1)
n times (i):
curr append(prev(i) + prev(i + 1))
.
"\n" print
curr join(", ") print
curr
.
.
printpascal(read number integer)</lang>
PowerShell
<lang powershell> $Infinity = 1 $NewNumbers = $null $Numbers = $null $Result = $null $Number = $null $Power = $args[0]
Write-Host $Power
For(
$i=0;
$i -lt $Infinity;
$i++
)
{
$Numbers = New-Object Object[] 1
$Numbers[0] = $Power
For(
$k=0;
$k -lt $NewNumbers.Length;
$k++
)
{
$Numbers = $Numbers + $NewNumbers[$k]
}
If(
$i -eq 0
)
{
$Numbers = $Numbers + $Power
}
$NewNumbers = New-Object Object[] 0
Try
{
For(
$j=0;
$j -lt $Numbers.Length;
$j++
)
{
$Result = $Numbers[$j] + $Numbers[$j+1]
$NewNumbers = $NewNumbers + $Result
}
}
Catch [System.Management.Automation.RuntimeException]
{
Write-Warning "Value was too large for a Decimal. Script aborted."
Break;
}
Foreach(
$Number in $Numbers
)
{
If(
$Number.ToString() -eq "+unendlich"
)
{
Write-Warning "Value was too large for a Decimal. Script aborted."
Exit
}
}
Write-Host $Numbers
$Infinity++
}
</lang>
Save the above code to a .ps1 script file and start it by calling its name and providing N.
PS C:\> & '.\Pascals Triangle.ps1' 1 ---- 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 1 11 55 165 330 462 462 330 165 55 11 1 1 12 66 220 495 792 924 792 495 220 66 12 1 1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1 1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1 1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
Prolog
Difference-lists are used to make quick append. <lang Prolog>pascal(N) :- pascal(1, N, [1], [[1]|X]-X, L), maplist(my_format, L).
pascal(Max, Max, L, LC, LF) :- !, make_new_line(L, NL), append_dl(LC, [NL|X]-X, LF-[]).
pascal(N, Max, L, NC, LF) :- build_new_line(L, NL), append_dl(NC, [NL|X]-X, NC1), N1 is N+1, pascal(N1, Max, NL, NC1, LF).
build_new_line(L, R) :- build(L, 0, X-X, R).
build([], V, RC, RF) :- append_dl(RC, [V|Y]-Y, RF-[]).
build([H|T], V, RC, R) :- V1 is V+H, append_dl(RC, [V1|Y]-Y, RC1), build(T, H, RC1, R).
append_dl(X1-X2, X2-X3, X1-X3).
% to have a correct output ! my_format([H|T]) :- write(H), maplist(my_writef, T), nl.
my_writef(X) :- writef(' %5r', [X]). </lang>
Output : <lang Prolog> ?- pascal(15). 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 1 11 55 165 330 462 462 330 165 55 11 1 1 12 66 220 495 792 924 792 495 220 66 12 1 1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1 1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1 1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1 true. </lang>
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. <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(_).</lang>
- Output*:
<lang prolog>?- pascal(5). [1] [1,1] [1,2,1] [1,3,3,1] [1,4,6,4,1]</lang>
PureBasic
<lang PureBasic>Procedure pascaltriangle( n.i)
For i= 0 To n
c = 1
For k=0 To i
Print(Str( c)+" ")
c = c * (i-k)/(k+1);
Next ;k
PrintN(" "); nächste zeile
Next ;i
EndProcedure
OpenConsole() Parameter.i = Val(ProgramParameter(0)) pascaltriangle(Parameter); Input()</lang>
Python
Procedural
<lang python>def pascal(n):
"""Prints out n rows of Pascal's triangle.
It returns False for failure and True for success."""
row = [1]
k = [0]
for x in range(max(n,0)):
print row
row=[l+r for l,r in zip(row+k,k+row)]
return n>=1</lang>
or by creating a scan function: <lang Python>def scan(op, seq, it):
a = [] result = it a.append(it) for x in seq: result = op(result, x) a.append(result) return a
def pascal(n):
def nextrow(row, x):
return [l+r for l,r in zip(row+[0,],[0,]+row)]
return scan(nextrow, range(n-1), [1,])
for row in pascal(4):
print(row)</lang>
Functional
Deriving finite and non-finite lists of pascal rows from a simple nextPascal step function:
<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()</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
q
<lang q> pascal:{(x-1){0+':x,0}\1} pascal 5 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 </lang>
Qi
<lang Qi> (define iterate
_ _ 0 -> [] F V N -> [V|(iterate F (F V) (1- N))])
(define next-row
R -> (MAPCAR + [0|R] (append R [0])))
(define pascal
N -> (iterate next-row [1] N))
</lang>
Quackery
The behaviour of pascal for values less than 1 is the same as its behaviour for 1.
<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</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
R
<lang R>pascalTriangle <- function(h) {
for(i in 0:(h-1)) {
s <- ""
for(k in 0:(h-i)) s <- paste(s, " ", sep="")
for(j in 0:i) {
s <- paste(s, sprintf("%3d ", choose(i, j)), sep="")
}
print(s)
}
}</lang>
Here's an R version:
<lang R>pascalTriangle <- function(h) {
lapply(0:h, function(i) choose(i, 0:i))
}</lang>
Racket
Iterative version by summing rows up to .
<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))))
</lang>
Raku
(formerly Perl 6)
using a lazy sequence generator
The following routine returns a lazy list of lines using the sequence operator (...). With a lazy result you need not tell the routine how many you want; you can just use a slice subscript to get the first N lines: <lang perl6>sub pascal {
[1], { [0, |$_ Z+ |$_, 0] } ... *
}
.say for pascal[^10];</lang>
One problem with the routine above is that it might recalculate the sequence each time you call it. Slightly more idiomatic would be to define the sequence as a lazy constant. Here we use the @ sigil to indicate that the sequence should cache its values for reuse, and use an explicit parameter $prev for variety:
<lang perl6>constant @pascal = [1], -> $prev { [0, |$prev Z+ |$prev, 0] } ... *;
.say for @pascal[^10];</lang>
Since we use ordinary subscripting, non-positive inputs throw an index-out-of-bounds error.
recursive
<lang perl6>multi sub pascal (1) { $[1] } multi sub pascal (Int $n where 2..*) {
my @rows = pascal $n - 1; |@rows, [0, |@rows[*-1] Z+ |@rows[*-1], 0 ];
}
.say for pascal 10;</lang>
Non-positive inputs throw a multiple-dispatch error.
iterative
<lang perl6>sub pascal ($n where $n >= 1) {
say my @last = 1;
for 1 .. $n - 1 -> $row {
@last = 1, |map({ @last[$_] + @last[$_ + 1] }, 0 .. $row - 2), 1;
say @last;
}
}
pascal 10;</lang>
Non-positive inputs throw a type check error.
- Output:
[1] [1 1] [1 2 1] [1 3 3 1] [1 4 6 4 1] [1 5 10 10 5 1] [1 6 15 20 15 6 1] [1 7 21 35 35 21 7 1] [1 8 28 56 70 56 28 8 1] [1 9 36 84 126 126 84 36 9 1]
RapidQ
Summing from Previous Rows
The main difference to BASIC implementation is the output formatting. RapidQ does not support PRINT USING. Instead, function FORMAT$() is used. TAB() is not supported, so SPACE$() was used instead.
Another difference is that in RapidQ, DIM does not clear array values to zero. But if dimensioning is done with DEF..., you can give the initial values in curly braces. If less values are given than there are elements in the array, the remaining positions are initialized to zero. RapidQ does not require simple variables to be declared before use.
<lang rapidq>DEFINT values(100) = {0,1}
INPUT "Number of rows: "; nrows PRINT SPACE$((nrows)*3);" 1" FOR row = 2 TO nrows
PRINT SPACE$((nrows-row)*3+1);
FOR i = row TO 1 STEP -1
values(i) = values(i) + values(i-1)
PRINT FORMAT$("%5d ", values(i));
NEXT i
PRINT
NEXT row</lang>
Using binary coefficients
<lang rapidq>INPUT "Number of rows: "; nrows FOR row = 0 TO nrows-1
c = 1
PRINT SPACE$((nrows-row)*3);
FOR i = 0 TO row
PRINT FORMAT$("%5d ", c);
c = c * (row - i) / (i+1)
NEXT i
PRINT
NEXT row</lang>
Red
<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
]
]</lang> Output:
pascal-triangle 7 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1
Retro
<lang Retro>2 elements i j
- pascalTriangle
cr dup [ dup !j 1 swap 1+ [ !i dup putn space @j @i - * @i 1+ / ] iter cr drop ] iter drop
13 pascalTriangle</lang>
REXX
There is no practical limit for this REXX version, triangles up to 46 rows have been generated (without wrapping) in a screen window with a width of 620 characters.
If the number (of rows) specified is negative, the output is written to a (disk) file
instead. Triangles with over a 1,000 rows have been easily created.
The output file created (that is written to disk) is named PASCALS.n
where n is the absolute value of the number entered.
Note: Pascal's triangle is also known as:
- Khayyam's triangle
- Khayyam─Pascal's triangle
- Tartaglia's triangle
- Yang Hui's triangle
<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*/</lang>
- output when using the input of: 11
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
- output when using the input of: 22
(Output shown at 4/5 size.)
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1
1 17 136 680 2380 6188 12376 19448 24310 24310 19448 12376 6188 2380 680 136 17 1
1 18 153 816 3060 8568 18564 31824 43758 48620 43758 31824 18564 8568 3060 816 153 18 1
1 19 171 969 3876 11628 27132 50388 75582 92378 92378 75582 50388 27132 11628 3876 969 171 19 1
1 20 190 1140 4845 15504 38760 77520 125970 167960 184756 167960 125970 77520 38760 15504 4845 1140 190 20 1
1 21 210 1330 5985 20349 54264 116280 203490 293930 352716 352716 293930 203490 116280 54264 20349 5985 1330 210 21 1
Ring
<lang ring> row = 5 for i = 0 to row - 1
col = 1
see left(" ",row-i)
for k = 0 to i
see "" + col + " "
col = col*(i-k)/(k+1)
next
see nl
next </lang> Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
Ruby
<lang ruby>def pascal(n)
raise ArgumentError, "must be positive." if n < 1 yield ar = [1] (n-1).times do ar.unshift(0).push(0) # tack a zero on both ends yield ar = ar.each_cons(2).map(&:sum) end
end
pascal(8){|row| puts row.join(" ").center(20)}</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
Or for more or less a translation of the two line Haskell version (with inject being abused a bit I know):
<lang ruby>def next_row(row) ([0] + row).zip(row + [0]).collect {|l,r| l + r } end
def pascal(n) n.times.inject([1]) {|x,_| next_row x } end
8.times{|i| p pascal(i)}</lang>
- Output:
[1] [1, 1] [1, 2, 1] [1, 3, 3, 1] [1, 4, 6, 4, 1] [1, 5, 10, 10, 5, 1] [1, 6, 15, 20, 15, 6, 1] [1, 7, 21, 35, 35, 21, 7, 1]
Run BASIC
<lang runbasic>input "number of rows? ";r for i = 0 to r - 1
c = 1
print left$(" ",(r*2)-(i*2));
for k = 0 to i
print using("####",c);
c = c*(i-k)/(k+1)
next
print
next</lang>Output:
Number of rows? ?5
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
Rust
<lang rust> fn pascal_triangle(n: u64) {
for i in 0..n {
let mut c = 1;
for _j in 1..2*(n-1-i)+1 {
print!(" ");
}
for k in 0..i+1 {
print!("{:2} ", c);
c = c * (i-k)/(k+1);
}
println!();
}
} </lang>
Scala
Functional solutions
Summing: Recursive row definition
<lang scala>
def tri(row: Int): List[Int] =
row match {
case 1 => List(1)
case n: Int => 1 +: ((tri(n - 1) zip tri(n - 1).tail) map { case (a, b) => a + b }) :+ 1
}</lang>
Function to pretty print n rows: <lang scala>def prettyTri(n:Int) = (1 to n) foreach {i => print(" "*(n-i)); tri(i) map (c => print(c + " ")); println}
prettyTri(5)</lang>
- Output:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1
Summing: Scala Stream (Recursive & Memoization)
<lang Scala>object Blaise extends App {
def pascalTriangle(): Stream[Vector[Int]] = Vector(1) #:: Stream.iterate(Vector(1, 1))(1 +: _.sliding(2).map(_.sum).toVector :+ 1)
val output = pascalTriangle().take(15).map(_.mkString(" "))
val longest = output.last.length
println("Pascal's Triangle")
output.foreach(line => println(s"${" " * ((longest - line.length) / 2)}$line"))
}</lang>
- Output:
See it in running in your browser by ScalaFiddle (JavaScript) or by Scastie (JVM).
Scheme
<lang scheme>(define (next-row row)
(map + (cons 0 row) (append row '(0))))
(define (triangle row rows)
(if (= rows 0)
'()
(cons row (triangle (next-row row) (- rows 1)))))
(triangle (list 1) 5) </lang> Output: <lang>((1) (1 1) (1 2 1) (1 3 3 1) (1 4 6 4 1))</lang>
Seed7
<lang seed7>$ include "seed7_05.s7i";
const proc: main is func
local
var integer: numRows is 0;
var array integer: values is [] (0, 1);
var integer: row is 0;
var integer: index is 0;
begin
write("Number of rows: ");
readln(numRows);
writeln("1" lpad succ(numRows) * 3);
for row range 2 to numRows do
write("" lpad (numRows - row) * 3);
values &:= [] 0;
for index range succ(row) downto 2 do
values[index] +:= values[pred(index)];
write(" " <& values[index] lpad 5);
end for;
writeln;
end for;
end func;</lang>
Sidef
<lang ruby>func pascal(rows) {
var row = [1]
{ | n|
say row.join(' ')
row = [1, {|i| row[i] + row[i+1] }.map(0 .. n-2)..., 1]
} << 1..rows
} pascal(10)</lang>
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).
<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) }</lang>
Now print the Pascal triangle.
<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</lang>
Swift
<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) </lang>
Tcl
Summing from Previous Rows
<lang tcl>proc pascal_iterative n {
if {$n < 1} {error "undefined behaviour for n < 1"}
set row [list 1]
lappend rows $row
set i 1
while {[incr i] <= $n} {
set prev $row
set row [list 1]
for {set j 1} {$j < [llength $prev]} {incr j} {
lappend row [expr {[lindex $prev [expr {$j - 1}]] + [lindex $prev $j]}]
}
lappend row 1
lappend rows $row
}
return $rows
}
puts [join [pascal_iterative 6] \n]</lang>
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1
Using binary coefficients
<lang tcl>proc pascal_coefficients n {
if {$n < 1} {error "undefined behaviour for n < 1"}
for {set i 0} {$i < $n} {incr i} {
set c 1
set row [list $c]
for {set j 0} {$j < $i} {incr j} {
set c [expr {$c * ($i - $j) / ($j + 1)}]
lappend row $c
}
lappend rows $row
}
return $rows
}
puts [join [pascal_coefficients 6] \n]</lang>
Combinations
Thanks to Tcl 8.5's arbitrary precision integer arithmetic, this solution is not limited to a couple of dozen rows. Uses a caching factorial calculator to improve performance. <lang tcl>package require Tcl 8.5
proc pascal_combinations n {
if {$n < 1} {error "undefined behaviour for n < 1"}
for {set i 0} {$i < $n} {incr i} {
set row [list]
for {set j 0} {$j <= $i} {incr j} {
lappend row [C $i $j]
}
lappend rows $row
}
return $rows
}
proc C {n k} {
expr {[ifact $n] / ([ifact $k] * [ifact [expr {$n - $k}]])}
}
set fact_cache {1 1} proc ifact n {
global fact_cache
if {$n < [llength $fact_cache]} {
return [lindex $fact_cache $n]
}
set i [expr {[llength $fact_cache] - 1}]
set sum [lindex $fact_cache $i]
while {$i < $n} {
incr i
set sum [expr {$sum * $i}]
lappend fact_cache $sum
}
return $sum
}
puts [join [pascal_combinations 6] \n]</lang>
Comparing Performance
<lang tcl>set n 100 puts "calculate $n rows:" foreach proc {pascal_iterative pascal_coefficients pascal_combinations} {
puts "$proc: [time [list $proc $n] 100]"
}</lang>
- Output:
calculate 100 rows: pascal_iterative: 2800.14 microseconds per iteration pascal_coefficients: 8760.98 microseconds per iteration pascal_combinations: 38176.66 microseconds per iteration
TI-83 BASIC
Using Addition of Previous Rows
<lang ti83b>PROGRAM:PASCALTR
- Lbl IN
- ClrHome
- Disp "NUMBER OF ROWS"
- Input N
- If N < 1:Goto IN
- {N,N}→dim([A])
- "CHEATING TO MAKE IT FASTER"
- For(I,1,N)
- 1→[A](1,1)
- End
- For(I,2,N)
- For(J,2,I)
- [A](I-1,J-1)+[A](I-1,J)→[A](I,J)
- End
- End
- [A]</lang>
Using nCr Function
<lang ti83b>PROGRAM:PASCALTR
- Lbl IN
- ClrHome
- Disp "NUMBER OF ROWS"
- Input N
- If N < 1:Goto IN
- {N,N}→dim([A])
- For(I,2,N)
- For(J,2,I)
- (I-1) nCr (J-1)→[A](I,J)
- End
- End
- [A]</lang>
Turing
<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)</lang>
Output:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1
TypeScript
<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); </lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
uBasic/4tH
<lang>Input "Number Of Rows: "; N @(1) = 1 Print Tab((N+1)*3);"1"
For R = 2 To N
Print Tab((N-R)*3+1);
For I = R To 1 Step -1
@(I) = @(I) + @(I-1)
Print Using "______";@(i);
Next
Next
Print End</lang> Output:
Number Of Rows: 10
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
0 OK, 0:380
UNIX Shell
Any n <= 1 will print the "1" row. <lang bash>#! /bin/bash pascal() {
local -i n=${1:-1}
if (( n <= 1 )); then
echo 1
else
local output=$( $FUNCNAME $((n - 1)) )
set -- $( tail -n 1 <<<"$output" ) # previous row
echo "$output"
printf "1 "
while -n $1 ; do
printf "%d " $(( $1 + ${2:-0} ))
shift
done
echo
fi
} pascal "$1"</lang>
Ursala
Zero maps to the empty list. Negatives are inexpressible. This solution uses a library function for binomial coefficients. <lang Ursala>#import std
- import nat
pascal = choose**ziDS+ iota*t+ iota+ successor</lang> This solution uses direct summation. The algorithm is to insert zero at the head of a list (initially the unit list <1>), zip it with its reversal, map the sum over the list of pairs, iterate n times, and return the trace. <lang Ursala>#import std
- import nat
pascal "n" = (next"n" sum*NiCixp) <1></lang> test program: <lang Ursala>#cast %nLL
example = pascal 10</lang>
- Output:
< <1>, <1,1>, <1,2,1>, <1,3,3,1>, <1,4,6,4,1>, <1,5,10,10,5,1>, <1,6,15,20,15,6,1>, <1,7,21,35,35,21,7,1>, <1,8,28,56,70,56,28,8,1>, <1,9,36,84,126,126,84,36,9,1>>
VBA
<lang vb>Option Base 1 Private Sub pascal_triangle(n As Integer)
Dim odd() As String
Dim eve() As String
ReDim odd(1)
ReDim eve(2)
odd(1) = " 1"
For i = 1 To n
If i Mod 2 = 1 Then
Debug.Print String$(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</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
VBScript
Derived from the BASIC version. <lang vb>Pascal_Triangle(WScript.Arguments(0)) Function Pascal_Triangle(n) Dim values(100) values(1) = 1 WScript.StdOut.Write values(1) WScript.StdOut.WriteLine For row = 2 To n For i = row To 1 Step -1 values(i) = values(i) + values(i-1) WScript.StdOut.Write values(i) & " " Next WScript.StdOut.WriteLine Next End Function</lang>
- Output:
Invoke from a command line.
F:\VBScript>cscript /nologo rosettacode-pascals_triangle.vbs 6 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1
Vedit macro language
Summing from Previous Rows
Vedit macro language does not have actual arrays (edit buffers are normally used for storing larger amounts of data). However, a numeric register can be used as index to access another numeric register. For example, if #99 contains value 2, then #@99 accesses contents of numeric register #2.
<lang vedit>#100 = Get_Num("Number of rows: ", STATLINE)
- 0=0; #1=1
Ins_Char(' ', COUNT, #100*3-2) Num_Ins(1) for (#99 = 2; #99 <= #100; #99++) {
Ins_Char(' ', COUNT, (#100-#99)*3)
#@99 = 0
for (#98 = #99; #98 > 0; #98--) {
#97 = #98-1 #@98 += #@97 Num_Ins(#@98, COUNT, 6)
} Ins_Newline
}</lang>
Using binary coefficients
<lang vedit>#1 = Get_Num("Number of rows: ", STATLINE) for (#2 = 0; #2 < #1; #2++) {
#3 = 1
Ins_Char(' ', COUNT, (#1-#2-1)*3)
for (#4 = 0; #4 <= #2; #4++) {
Num_Ins(#3, COUNT, 6) #3 = #3 * (#2-#4) / (#4+1)
} Ins_Newline
}</lang>
Visual Basic
<lang vb>Sub pascaltriangle()
'Pascal's triangle
Const m = 11
Dim t(40) As Integer, u(40) As Integer
Dim i As Integer, n As Integer, s As String, ss As String
ss = ""
For n = 1 To m
u(1) = 1
s = ""
For i = 1 To n
u(i + 1) = t(i) + t(i + 1)
s = s & u(i) & " "
t(i) = u(i)
Next i
ss = ss & s & vbCrLf
Next n
MsgBox ss, , "Pascal's triangle"
End Sub 'pascaltriangle</lang>
- Output:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1
Visual Basic .NET
<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</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
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
Wren
<lang ecmascript>import "/fmt" for Fmt import "/math" for Int
var binomial = Fn.new { |n, k|
if (n == k) return 1
var prod = 1
var i = n - k + 1
while (i <= n) {
prod = prod * i
i = i + 1
}
return prod / Int.factorial(k)
}
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("$3d ", binomial.call(i, j))
}
System.print()
}
}
pascalTriangle.call(13)</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
X86 Assembly
uses: io.inc - Macro library from SASM <lang asm> %include "io.inc"
section .text global CMAIN CMAIN: mov ebx, 7 ;size call mloop ret
mloop: mov edx, 0 ;edx stands for the nth line looping: push ebx push edx call line pop edx pop ebx inc edx cmp edx, ebx jl looping xor eax, eax ret
line: mov ecx, 0 ;ecx stands for the nth character in each line mlp: push ecx
push edx
call nCk
pop edx
pop ecx PRINT_DEC 4, eax ;print returned number
PRINT_STRING " "
inc ecx cmp ecx, edx ;if jle mlp NEWLINE ret
nCk: ;ecx : j ;edx : i mov esi, edx call fac ;i! push eax ;save i! mov esi, ecx call fac ;j! push eax ;save j! mov ebx, edx sub ebx, ecx ;(i-j) mov esi, ebx call fac ;(i-j)! pop ebx ;(i-j)! is in eax mul ebx ;(i-j)! * j! mov ecx, eax pop eax ; get i! div ecx ; ; last step : i! divided by (i-j)! * j! ret
fac: push ecx
push edx
mov eax, 1 mov ecx, esi cmp ecx, 0 ; 0! returns 1 je facz lp: mul ecx ;multiplies eax by ecx and then decrements ecx until ecx is 0 dec ecx cmp ecx, 0 jg lp jmp end facz: mov eax, 1 end:
pop edx
pop ecx ret </lang>
- Output:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1
XBasic
<lang xbasic> PROGRAM "pascal" VERSION "0.0001"
DECLARE FUNCTION Entry()
FUNCTION Entry()
r@@ = UBYTE(INLINE$("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 </lang>
- Output:
Number of rows? 7 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1
XPL0
<lang XPL0>include c:\cxpl\codes;
proc Pascal(N); \Display the first N rows of Pascal's triangle int N; \if N<=0 then nothing is displayed int Row, I, Old(40), New(40); [for Row:= 0 to N-1 do
[New(0):= 1;
for I:= 1 to Row do New(I):= Old(I-1) + Old(I);
for I:= 1 to (N-Row-1)*2 do ChOut(0, ^ );
for I:= 0 to Row do
[if New(I)<100 then ChOut(0, ^ );
IntOut(0, New(I));
if New(I)<10 then ChOut(0, ^ );
ChOut(0, ^ );
];
New(Row+1):= 0;
I:= Old; Old:= New; New:= I;
CrLf(0);
];
];
Pascal(13)</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
zkl
<lang zkl>fcn pascalTriangle(n){ // n<=0-->""
foreach i in (n){
c := 1;
print(" "*(2*(n-1-i)));
foreach k in (i+1){
print("%3d ".fmt(c));
c = c * (i-k)/(k+1);
}
println();
}
}
pascalTriangle(8);</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
ZX Spectrum Basic
In edit mode insert: <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</lang>
Then in command mode (basically don't put a number in front): <lang BASIC>RUN</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
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