Pascal's triangle: Difference between revisions
| Line 1,919: | Line 1,919: | ||
pascaltriangle(4);</lang> |
pascaltriangle(4);</lang> |
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=={{header|Oforth}}== |
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No result if n <= 0 |
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<lang Oforth>func: pascal(n) { [ 1 ] #[ dup 0 + 0 rot + zipWith(#+) dup println ] times(n) drop }</lang> |
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{{out}} |
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<pre> |
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pascal(10) |
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[1, 1] |
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[1, 2, 1] |
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[1, 3, 3, 1] |
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[1, 4, 6, 4, 1] |
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[1, 5, 10, 10, 5, 1] |
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[1, 6, 15, 20, 15, 6, 1] |
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[1, 7, 21, 35, 35, 21, 7, 1] |
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[1, 8, 28, 56, 70, 56, 28, 8, 1] |
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[1, 9, 36, 84, 126, 126, 84, 36, 9, 1] |
|||
[1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1] |
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</pre> |
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=={{header|Oz}}== |
=={{header|Oz}}== |
||
Revision as of 21:14, 4 March 2015
You are encouraged to solve this task according to the task description, using any language you may know.
Pascal's triangle is an arithmetic and geometric figure first imagined by Blaise Pascal.
Its first few rows look like this:
1 1 1 1 2 1 1 3 3 1
where each element of each row is either 1 or the sum of the two elements right above it. For example, the next row would be 1 (since the first element of each row doesn't have two elements above it), 4 (1 + 3), 6 (3 + 3), 4 (3 + 1), and 1 (since the last element of each row doesn't have two elements above it). Each row n (starting with row 0 at the top) shows the coefficients of the binomial expansion of (x + y)n.
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:
Ada
The specification of auxiliary package "Pascal". "First_Row" outputs a row with a single "1", "Next_Row" computes the next row from a given row, and "Length" gives the number of entries in a row. The package is also used for the Catalan numbers solution [[1]]
<lang ada>package Pascal is
type Row is array (Natural range <>) of Natural; function Length(R: Row) return Positive; function First_Row(Max_Length: Positive) return Row; function Next_Row(R: Row) return Row;
end Pascal;</lang>
The implementation of that auxiliary package "Pascal":
<lang Ada>package body Pascal is
function First_Row(Max_Length: Positive) return Row is
R: Row(0 .. Max_Length) := (0 | 1 => 1, others => 0);
begin
return R;
end First_Row;
function Next_Row(R: Row) return Row is
S: Row(R'Range);
begin
S(0) := Length(R)+1;
S(Length(S)) := 1;
for J in reverse 2 .. Length(R) loop
S(J) := R(J)+R(J-1);
end loop;
S(1) := 1;
return S;
end Next_Row;
function Length(R: Row) return Positive is
begin
return R(0);
end Length;
end Pascal;</lang>
The main program, using "Pascal". It prints the desired number of rows. The number is read from the command line.
<lang Ada>with Ada.Text_IO, Ada.Integer_Text_IO, Ada.Command_Line, Pascal; use Pascal;
procedure Triangle is
Number_Of_Rows: Positive := Integer'Value(Ada.Command_Line.Argument(1));
Row: Pascal.Row := First_Row(Number_Of_Rows);
begin
loop
-- print one row
for J in 1 .. Length(Row) loop
Ada.Integer_Text_IO.Put(Row(J), 5);
end loop;
Ada.Text_IO.New_Line;
exit when Length(Row) >= Number_Of_Rows;
Row := Next_Row(Row);
end loop;
end Triangle;</lang>
- Output:
>./triangle 12
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
ALGOL 68
<lang algol68>PRIO MINLWB = 8, MAXUPB = 8; OP MINLWB = ([]INT a,b)INT: (LWB a<LWB b|LWB a|LWB b),
MAXUPB = ([]INT a,b)INT: (UPB a>UPB b|UPB a|UPB b);
OP + = ([]INT a,b)[]INT:(
[a MINLWB b:a MAXUPB b]INT out; FOR i FROM LWB out TO UPB out DO out[i]:= 0 OD; out[LWB a:UPB a] := a; FOR i FROM LWB b TO UPB b DO out[i]+:= b[i] OD; out
);
INT width = 4, stop = 9; FORMAT centre = $n((stop-UPB row+1)*width OVER 2)(q)$;
FLEX[1]INT row := 1; # example of rowing # FOR i WHILE
printf((centre, $g(-width)$, row, $l$));
- WHILE # i < stop DO
row := row[AT 1] + row[AT 2]
OD</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
APL
Pascal' s triangle of order ⍵
<lang apl> {A←0,⍳⍵ ⋄ ⍉A∘.!A} </lang>
example
<lang apl> {A←0,⍳⍵ ⋄ ⍉A∘.!A} 3 </lang>
1 0 0 0 1 1 0 0 1 2 1 0 1 3 3 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>
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
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++
<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#
Produces no output when n is less than or equal to zero.
<lang csharp>using System;
namespace RosettaCode {
class PascalsTriangle {
public static void CreateTriangle(int n) {
if (n > 0) {
for (int i = 0; i < n; i++) {
int c = 1;
Console.Write(" ".PadLeft(2 * (n - 1 - i)));
for (int k = 0; k <= i; k++) {
Console.Write("{0}", c.ToString().PadLeft(3));
c = c * (i - k) / (k + 1);
}
Console.WriteLine();
}
}
}
public static void Main() {
CreateTriangle(8);
}
}
}</lang>
Clojure
For n < 1, prints nothing, always returns nil. Copied from the Common Lisp implementation below, but with local functions and explicit tail-call-optimized recursion (recur). <lang lisp>(defn pascal [n]
(let [newrow (fn newrow [lst ret]
(if lst
(recur (rest lst)
(conj ret (+ (first lst) (or (second lst) 0))))
ret))
genrow (fn genrow [n lst]
(when (< 0 n)
(do (println lst)
(recur (dec n) (conj (newrow lst []) 1)))))]
(genrow n [1])))
(pascal 4)</lang> And here's another version, using the partition function to produce the sequence of pairs in a row, which are summed and placed between two ones to produce the next row: <lang lisp> (defn nextrow [row]
(vec (concat [1] (map #(apply + %) (partition 2 1 row)) [1] )))
(defn pascal [n]
(assert (and (integer? n) (pos? n)))
(let [triangle (take n (iterate nextrow [1]))]
(doseq [row triangle]
(println row))))
</lang> The assert form causes the pascal function to throw an exception unless the argument is (integral and) positive.
Here's a third version using the iterate function <lang lisp> (def pascal
(iterate
(fn [prev-row]
(->>
(concat (first prev-row) (partition 2 1 prev-row) (last prev-row))
(map (partial apply +) ,,,)))
[1]))
</lang>
Another short version which returns an infinite pascal triangle as a list, using the iterate function.
<lang lisp> (def pascal
(iterate #(concat [1]
(map + % (rest %))
[1])
[1]))
</lang>
One can then get the first n rows using the take function
<lang lisp> (take 10 pascal) ; returns a list of the first 10 pascal rows </lang>
Also, one can retrieve the nth row using the nth function
<lang lisp> (nth pascal 10) ;returns the nth row </lang>
CoffeeScript
This version assumes n is an integer and n >= 1. It efficiently computes binomial coefficients. <lang coffeescript> pascal = (n) ->
width = 6
for r in [1..n]
s = ws (width/2) * (n-r) # center row
output = (n) -> s += pad width, n
cell = 1
output cell
# Compute binomial coefficients as you go
# across the row.
for c in [1...r]
cell *= (r-c) / c
output cell
console.log s
ws = (n) ->
s = s += ' ' for i in [0...n] s
pad = (cnt, n) ->
s = n.toString() # There is probably a better way to do this. cnt -= s.length right = Math.floor(cnt / 2) left = cnt - right ws(left) + s + ws(right)
pascal(7)
</lang>
- Output:
> coffee pascal.coffee
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Common Lisp
To evaluate, call (pascal n). For n < 1, it simply returns nil.
<lang lisp>(defun pascal (n)
(genrow n '(1)))
(defun genrow (n l)
(when (< 0 n)
(print l)
(genrow (1- n) (cons 1 (newrow l)))))
(defun newrow (l)
(if (> 2 (length l))
'(1)
(cons (+ (car l) (cadr l)) (newrow (cdr l)))))</lang>
D
Less functional Version
<lang d>int[][] pascalsTriangle(in int rows) pure nothrow {
auto tri = new int[][rows];
foreach (r; 0 .. rows) {
int v = 1;
foreach (c; 0 .. r+1) {
tri[r] ~= v;
v = (v * (r - c)) / (c + 1);
}
}
return tri;
}
void main() {
immutable t = pascalsTriangle(10);
assert(t == [[1],
[1, 1],
[1, 2, 1],
[1, 3, 3, 1],
[1, 4, 6, 4, 1],
[1, 5, 10, 10, 5, 1],
[1, 6, 15, 20, 15, 6, 1],
[1, 7, 21, 35, 35, 21, 7, 1],
[1, 8, 28, 56, 70, 56, 28, 8, 1],
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]]);
}</lang>
More functional Version
<lang d>import std.stdio, std.algorithm, std.range;
auto pascal() pure nothrow {
return [1].recurrence!q{ zip(a[n - 1] ~ 0, 0 ~ a[n - 1])
.map!q{ a[0] + a[1] }
.array };
}
void main() {
pascal.take(5).writeln;
}</lang>
- Output:
[[1], [1, 1], [1, 2, 1], [1, 3, 3, 1], [1, 4, 6, 4, 1]]
Alternative Version
There is similarity between Pascal's triangle and Sierpinski triangle. Their difference are the initial line and the operation that act on the line element to produce next line. The following is a generic pascal's triangle implementation for positive number of lines output (n). <lang d>import std.stdio, std.string, std.array, std.format;
string Pascal(alias dg, T, T initValue)(int n) {
string output;
void append(in T[] l) {
output ~= " ".replicate((n - l.length + 1) * 2);
foreach (e; l)
output ~= format("%4s", format("%4s", e));
output ~= "\n";
}
if (n > 0) {
T[][] lines = initValue;
append(lines[0]);
foreach (i; 1 .. n) {
lines ~= lines[i - 1] ~ initValue; // length + 1
foreach (int j; 1 .. lines[i-1].length)
lines[i][j] = dg(lines[i-1][j], lines[i-1][j-1]);
append(lines[i]);
}
}
return output;
}
string delegate(int n) genericPascal(alias dg, T, T initValue)() {
mixin Pascal!(dg, T, initValue); return &Pascal;
}
void main() {
auto pascal = genericPascal!((int a, int b) => a + b, int, 1)();
static char xor(char a, char b) { return a == b ? '_' : '*'; }
auto sierpinski = genericPascal!(xor, char, '*')();
foreach (i; [1, 5, 9])
writef(pascal(i));
// an order 4 sierpinski triangle is a 2^4 lines generic
// Pascal triangle with xor operation
foreach (i; [16])
writef(sierpinski(i));
}</lang>
- Output:
1
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
*
* *
* _ *
* * * *
* _ _ _ *
* * _ _ * *
* _ * _ * _ *
* * * * * * * *
* _ _ _ _ _ _ _ *
* * _ _ _ _ _ _ * *
* _ * _ _ _ _ _ * _ *
* * * * _ _ _ _ * * * *
* _ _ _ * _ _ _ * _ _ _ *
* * _ _ * * _ _ * * _ _ * *
* _ * _ * _ * _ * _ * _ * _ *
* * * * * * * * * * * * * * * *
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:
11 1 1 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 11 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>
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]]
Euphoria
Summing from Previous Rows
<lang Euphoria>sequence row row = {} for m = 1 to 10 do
row = row & 1
for n = length(row)-1 to 2 by -1 do
row[n] += row[n-1]
end for
print(1,row)
puts(1,'\n')
end for</lang>
- Output:
{1}
{1,1}
{1,2,1}
{1,3,3,1}
{1,4,6,4,1}
{1,5,10,10,5,1}
{1,6,15,20,15,6,1}
{1,7,21,35,35,21,7,1}
{1,8,28,56,70,56,28,8,1}
{1,9,36,84,126,126,84,36,9,1}
F#
<lang fsharp>let rec nextrow l =
match l with | [] -> [] | h :: [] -> [1] | h :: t -> h + t.Head :: nextrow t
let pascalTri n = List.scan(fun l i -> 1 :: nextrow l) [1] [1 .. n]
for row in pascalTri(10) do
for i in row do
printf "%s" (i.ToString() + ", ")
printfn "%s" "\n"
</lang>
Factor
This implementation works by summing the previous line content. Result for n < 1 is the same as for n == 1.
<lang factor>USING: grouping kernel math sequences ;
- (pascal) ( seq -- newseq )
dup last 0 prefix 0 suffix 2 <clumps> [ sum ] map suffix ;
- pascal ( n -- seq )
1 - { { 1 } } swap [ (pascal) ] times ;</lang>
It works as:
<lang factor>5 pascal . { { 1 } { 1 1 } { 1 2 1 } { 1 3 3 1 } { 1 4 6 4 1 } }</lang>
Fantom
<lang fantom> class Main {
Int[] next_row (Int[] row)
{
new_row := [1]
(row.size-1).times |i|
{
new_row.add (row[i] + row[i+1])
}
new_row.add (1)
return new_row }
Void print_pascal (Int n) // no output for n <= 0
{
current_row := [1]
n.times
{
echo (current_row.join(" "))
current_row = next_row (current_row)
}
}
Void main ()
{
print_pascal (10)
}
} </lang>
Forth
<lang forth>: init ( n -- )
here swap cells erase 1 here ! ;
- .line ( n -- )
cr here swap 0 do dup @ . cell+ loop drop ;
- next ( n -- )
here swap 1- cells here + do i @ i cell+ +! -1 cells +loop ;
- pascal ( n -- )
dup init 1 .line 1 ?do i next i 1+ .line loop ;</lang>
This is a bit more efficient.
<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>
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
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 = { n -> (n <= 1) ? [1] : GroovyCollections.transpose([[0] + pascal(n - 1), pascal(n - 1) + [0]]).collect { it.sum() } }</lang> However, this solution is horribly inefficient (O(n**2)). It slowly grinds to a halt on a reasonably powerful PC after about line 25 of the triangle.
Test program: <lang groovy>def count = 15 (1..count).each { n ->
printf ("%2d:", n); (0..(count-n)).each { print " " }; pascal(n).each{ printf("%6d ", it) }; println ""
}</lang>
Output:
1: 1 2: 1 1 3: 1 2 1 4: 1 3 3 1 5: 1 4 6 4 1 6: 1 5 10 10 5 1 7: 1 6 15 20 15 6 1 8: 1 7 21 35 35 21 7 1 9: 1 8 28 56 70 56 28 8 1 10: 1 9 36 84 126 126 84 36 9 1 11: 1 10 45 120 210 252 210 120 45 10 1 12: 1 11 55 165 330 462 462 330 165 55 11 1 13: 1 12 66 220 495 792 924 792 495 220 66 12 1 14: 1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1 15: 1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
GW-BASIC
<lang qbasic>10 INPUT "Number of rows? ",R 20 FOR I=0 TO R-1 30 C=1 40 FOR K=0 TO I 50 PRINT USING "####";C; 60 C=C*(I-K)/(K+1) 70 NEXT 80 PRINT 90 NEXT</lang>
Output:
Number of rows? 7 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1
Haskell
An approach using the "think in whole lists" principle: Each row in the triangle can be calculated from the previous row by adding a shifted version of itself to it, keeping the ones at the ends. The Prelude function zipWith can be used to add two lists, but it won't keep the old values when one list is shorter. So we need a similar function
<lang haskell>zapWith :: (a -> a -> a) -> [a] -> [a] -> [a] zapWith f xs [] = xs zapWith f [] ys = ys zapWith f (x:xs) (y:ys) = f x y : zapWith f xs ys</lang>
Now we can shift a list and add it to itself, extending it by keeping the ends:
<lang haskell>extendWith f [] = [] extendWith f xs@(x:ys) = x : zapWith f xs ys</lang>
And for the whole (infinite) triangle, we just iterate this operation, starting with the first row:
<lang haskell>pascal = iterate (extendWith (+)) [1]</lang>
For the first n rows, we just take the first n elements from this list, as in
<lang haskell>*Main> take 6 pascal [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1],[1,5,10,10,5,1]]</lang>
A shorter approach, plagiarized from [2] <lang haskell>-- generate next row from current row nextRow row = zipWith (+) ([0] ++ row) (row ++ [0])
-- returns the first n rows pascal = iterate nextRow [1]</lang>
With binomial coefficients:
<lang haskell>fac = product . enumFromTo 1
binCoef n k = (fac n) `div` ((fac k) * (fac $ n - k))
pascal n = map (binCoef $ n - 1) [0..n-1]</lang>
Example: <lang haskell>*Main> putStr $ unlines $ map unwords $ map (map show) $ pascal 10 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 </lang>
HicEst
<lang HicEst> CALL Pascal(30)
SUBROUTINE Pascal(rows)
CHARACTER fmt*6 WRITE(Text=fmt, Format='"i", i5.5') 1+rows/4
DO row = 0, rows-1
n = 1
DO k = 0, row
col = rows*(rows-row+2*k)/4
WRITE(Row=row+1, Column=col, F=fmt) n
n = n * (row - k) / (k + 1)
ENDDO
ENDDO
END</lang>
Icon and Unicon
The code below is slightly modified from the library version of pascal which prints 0's to the full width of the carpet. It also presents the data as an isoceles triangle. <lang Icon>link math
procedure main(A) every n := !A do { # for each command line argument
n := integer(\n) | &null pascal(n) }
end
procedure pascal(n) #: Pascal triangle
/n := 16
write("width=", n, " height=", n) # carpet header
fw := *(2 ^ n)+1
every i := 0 to n - 1 do {
writes(repl(" ",fw*(n-i)/2))
every j := 0 to n - 1 do
writes(center(binocoef(i, j),fw) | break)
write()
}
end</lang>
math provides binocoef math provides the original version of pascal
Sample output:
->pascal 1 4 8
width=1 height=1
1
width=4 height=4
1
1 1
1 2 1
1 3 3 1
width=8 height=8
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
->
IDL
<lang IDL>Pro Pascal, n
- n is the number of lines of the triangle to be displayed
r=[1] print, r for i=0, (n-2) do begin pascalrow,r endfor
End
Pro PascalRow, r
for i=0,(n_elements(r)-2) do begin r[i]=r[i]+r[i+1] endfor
r= [1, r] print, r
End</lang>
J
<lang j> !~/~ 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
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
<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
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>
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>
Liberty BASIC
<lang lb>input "How much rows would you like? "; n dim a$(n)
for i= 0 to n
c = 1
o$ =""
for k =0 to i
o$ =o$ ; c; " "
c =c *(i-k)/(k+1)
next k
a$(i)=o$
next i
maxLen = len(a$(n)) for i= 0 to n
print space$((maxLen-len(a$(i)))/2);a$(i)
next i
end</lang>
Locomotive Basic
<lang locobasic>10 CLS 20 INPUT "Number of rows? ", rows:GOSUB 40 30 END 40 FOR i=0 TO rows-1 50 c=1 60 FOR k=0 TO i 70 PRINT USING "####";c; 80 c=c*(i-k)/(k+1) 90 NEXT 100 PRINT 110 NEXT 120 RETURN</lang>
Output:
Number of rows? 7 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1
Logo
<lang logo>to pascal :n
if :n = 1 [output [1]] localmake "a pascal :n-1 output (sentence first :a (map "sum butfirst :a butlast :a) last :a)
end
for [i 1 10] [print pascal :i]</lang>
Lua
<lang lua> function nextrow(t)
local ret = {}
t[0], t[#t+1] = 0, 0
for i = 1, #t do ret[i] = t[i-1] + t[i] end
return ret
end
function triangle(n)
t = {1}
for i = 1, n do
print(unpack(t))
t = nextrow(t)
end
end </lang>
Mathematica
<lang Mathematica>Column[StringReplace[ToString /@ Replace[MatrixExp[SparseArray[
{Band[{2,1}] -> Range[n-1]},{n,n}]],{x__,0..}->{x},2] ,{"{"|"}"|","->" "}], Center]</lang>
MATLAB / Octave
A matrix containing the pascal triangle can be obtained this way: <lang MATLAB>pascal(n);</lang>
>> pascal(6)
ans =
1 1 1 1 1 1
1 2 3 4 5 6
1 3 6 10 15 21
1 4 10 20 35 56
1 5 15 35 70 126
1 6 21 56 126 252
The binomial coefficients can be extracted from the Pascal triangle in this way: <lang MATLAB> binomCoeff = diag(rot90(pascal(n)))', </lang>
>> for k=1:6,diag(rot90(pascal(k)))', end
ans = 1
ans =
1 1
ans =
1 2 1
ans =
1 3 3 1
ans =
1 4 6 4 1
ans =
1 5 10 10 5 1
Another way to get a formated pascals triangle is to use the convolution method:
>>
x = [1 1] ;
y = 1;
for k=8:-1:1
fprintf(['%', num2str(k), 'c'], zeros(1,3)),
fprintf('%6d', y), fprintf('\n')
y = conv(y,x);
end
The result is:
>>
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
Maxima
<lang maxima>sjoin(v, j) := apply(sconcat, rest(join(makelist(j, length(v)), v)))$
display_pascal_triangle(n) := for i from 0 thru 6 do disp(sjoin(makelist(binomial(i, j), j, 0, i), " "));
display_pascal_triangle(6); /* "1"
"1 1" "1 2 1" "1 3 3 1" "1 4 6 4 1" "1 5 10 10 5 1" "1 6 15 20 15 6 1" */</lang>
Metafont
(The formatting starts to be less clear when numbers start to have more than two digits)
<lang metafont>vardef bincoeff(expr n, k) = save ?; ? := (1 for i=(max(k,n-k)+1) upto n: * i endfor )
/ (1 for i=2 upto min(k, n-k): * i endfor); ?
enddef;
def pascaltr expr c =
string s_;
for i := 0 upto (c-1):
s_ := "" for k=0 upto (c-i): & " " endfor;
s_ := s_ for k=0 upto i: & decimal(bincoeff(i,k))
& " " if bincoeff(i,k)<9: & " " fi endfor;
message s_;
endfor
enddef;
pascaltr(4); end</lang>
NetRexx
<lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols nobinary
numeric digits 1000 -- allow very large numbers parse arg rows . if rows = then rows = 11 -- default to 11 rows printPascalTriangle(rows) return
-- ----------------------------------------------------------------------------- method printPascalTriangle(rows = 11) public static
lines = mx = (factorial(rows - 1) / factorial(rows % 2) / factorial(rows - 1 - rows % 2)).length() -- width of widest number
loop row = 1 to rows
n1 = 1.center(mx)
line = n1
loop col = 2 to row
n2 = col - 1
n1 = n1 * (row - n2) / n2
line = line n1.center(mx)
end col
lines[row] = line.strip()
end row
-- display triangle ml = lines[rows].length() -- length of longest line loop row = 1 to rows say lines[row].centre(ml) end row
return
-- ----------------------------------------------------------------------------- method factorial(n) public static
fac = 1 loop n_ = 2 to n fac = fac * n_ end n_ return fac /*calc. factorial*/
</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
Nial
Like J
(pascal.nial) <lang nial>factorial is recur [ 0 =, 1 first, pass, product, -1 +] combination is fork [ > [first, second], 0 first,
/ [factorial second, * [factorial - [second, first], factorial first] ]
] pascal is transpose each combination cart [pass, pass] tell</lang> Using it <lang nial>|loaddefs 'pascal.nial' |pascal 5</lang>
Nim
<lang nim>import sequtils
proc pascal(n: int) =
var row = @[1] for r in 1..n: echo row row = zip(row & @[0], @[0] & row).mapIt(int, it[0] + it[1])
pascal(10)</lang>
OCaml
<lang ocaml>(* generate next row from current row *) let next_row row =
List.map2 (+) ([0] @ row) (row @ [0])
(* returns the first n rows *) let pascal n =
let rec loop i row = if i = n then [] else row :: loop (i+1) (next_row row) in loop 0 [1]</lang>
Octave
<lang octave>function pascaltriangle(h)
for i = 0:h-1
for k = 0:h-i
printf(" ");
endfor
for j = 0:i
printf("%3d ", bincoeff(i, j));
endfor
printf("\n");
endfor
endfunction
pascaltriangle(4);</lang>
Oforth
No result if n <= 0
<lang Oforth>func: pascal(n) { [ 1 ] #[ dup 0 + 0 rot + zipWith(#+) dup println ] times(n) drop }</lang>
- Output:
pascal(10) [1, 1] [1, 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]
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>
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, @^p Z+ @^p, 0] } ... * }
.say for pascal[^10];</lang>
(See this blog post for a partial explanation of how this one-liner works.)
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, which gives the compiler more latitude in whether to cache or recalculate intermediate values:
<lang perl6>constant Pascal = [1], { [0, @^p Z+ @^p, 0] } ... *;
.say for Pascal[^10];</lang>
(A desperate garbage collector may still trim an overly long lazy constant, since it can always be recreated later, assuming the programmer didn't lie about its constancy.)
Non-positive inputs throw an index-out-of-bounds error.
recursive
<lang perl6>multi pascal (1) { [1] } multi 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 .. $row - 2), 1;
say @last;
}
}
pascal 10;</lang>
Non-positive inputs throw a type check error.
- Output:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1
PHP
<lang php>function pascalsTriangle($num){ $c = 1; $triangle = Array(); for($i=0;$i<=$num;$i++){ $triangle[$i] = Array(); if(!isset($triangle[$i-1])){ $triangle[$i][] = $c; }else{ for($j=0;$j<count($triangle[$i-1])+1;$j++){ $triangle[$i][] = (isset($triangle[$i-1][$j-1]) && isset($triangle[$i-1][$j])) ? $triangle[$i-1][$j-1] + $triangle[$i-1][$j] : $c; } } } return $triangle; }
$tria = pascalsTriangle(8);
foreach($tria as $val){
foreach($val as $value){
echo $value . ' ';
}
echo '
';
}</lang>
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
PL/I
<lang PL/I> declare (t, u)(40) fixed binary; declare (i, n) fixed binary;
t,u = 0; get (n); if n <= 0 then return;
do n = 1 to n;
u(1) = 1;
do i = 1 to n;
u(i+1) = t(i) + t(i+1);
end;
put skip edit ((u(i) do i = 1 to n)) (col(40-2*n), (n+1) f(4));
t = u;
end; </lang>
<lang>
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
</lang>
PicoLisp
<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>
PowerShell
<lang powershell> $Infinity = 1 $NewNumbers = $null $Numbers = $null $Result = $null $Number = $null $Power = $args[0]
Write-Host $Power
For(
$i=0;
$i -lt $Infinity;
$i++
)
{
$Numbers = New-Object Object[] 1
$Numbers[0] = $Power
For(
$k=0;
$k -lt $NewNumbers.Length;
$k++
)
{
$Numbers = $Numbers + $NewNumbers[$k]
}
If(
$i -eq 0
)
{
$Numbers = $Numbers + $Power
}
$NewNumbers = New-Object Object[] 0
Try
{
For(
$j=0;
$j -lt $Numbers.Length;
$j++
)
{
$Result = $Numbers[$j] + $Numbers[$j+1]
$NewNumbers = $NewNumbers + $Result
}
}
Catch [System.Management.Automation.RuntimeException]
{
Write-Warning "Value was too large for a Decimal. Script aborted."
Break;
}
Foreach(
$Number in $Numbers
)
{
If(
$Number.ToString() -eq "+unendlich"
)
{
Write-Warning "Value was too large for a Decimal. Script aborted."
Exit
}
}
Write-Host $Numbers
$Infinity++
}
</lang>
Save the above code to a .ps1 script file and start it by calling its name and providing N.
PS C:\> & '.\Pascals Triangle.ps1' 1 ---- 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 1 11 55 165 330 462 462 330 165 55 11 1 1 12 66 220 495 792 924 792 495 220 66 12 1 1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1 1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1 1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
Prolog
Difference-lists are used to make quick append. <lang Prolog>pascal(N) :- pascal(1, N, [1], [[1]|X]-X, L), maplist(my_format, L).
pascal(Max, Max, L, LC, LF) :- !, make_new_line(L, NL), append_dl(LC, [NL|X]-X, LF-[]).
pascal(N, Max, L, NC, LF) :- build_new_line(L, NL), append_dl(NC, [NL|X]-X, NC1), N1 is N+1, pascal(N1, Max, NL, NC1, LF).
build_new_line(L, R) :- build(L, 0, X-X, R).
build([], V, RC, RF) :- append_dl(RC, [V|Y]-Y, RF-[]).
build([H|T], V, RC, R) :- V1 is V+H, append_dl(RC, [V1|Y]-Y, RC1), build(T, H, RC1, R).
append_dl(X1-X2, X2-X3, X1-X3).
% to have a correct output ! my_format([H|T]) :- write(H), maplist(my_writef, T), nl.
my_writef(X) :- writef(' %5r', [X]). </lang>
Output : <lang Prolog> ?- pascal(15). 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 1 11 55 165 330 462 462 330 165 55 11 1 1 12 66 220 495 792 924 792 495 220 66 12 1 1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1 1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1 1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1 true. </lang>
PureBasic
<lang PureBasic>Procedure pascaltriangle( n.i)
For i= 0 To n
c = 1
For k=0 To i
Print(Str( c)+" ")
c = c * (i-k)/(k+1);
Next ;k
PrintN(" "); nächste zeile
Next ;i
EndProcedure
OpenConsole() Parameter.i = Val(ProgramParameter(0)) pascaltriangle(Parameter); Input()</lang>
Python
<lang python>def pascal(n):
"""Prints out n rows of Pascal's triangle.
It returns False for failure and True for success."""
row = [1]
k = [0]
for x in range(max(n,0)):
print row
row=[l+r for l,r in zip(row+k,k+row)]
return n>=1</lang>
Or, by creating a scan function: <lang python>def scan(op, seq, it):
a = [] result = it a.append(it) for x in seq: result = op(result, x) a.append(result) return a
def pascal(n):
def nextrow(row, x):
return [l+r for l,r in zip(row+[0,],[0,]+row)]
return scan(nextrow, range(n-1), [1,])
for row in pascal(4):
print(row)</lang>
q
<lang q> pascal:{(x-1){0+':x,0}\1} pascal 5 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 </lang>
Qi
<lang Qi> (define iterate
_ _ 0 -> [] F V N -> [V|(iterate F (F V) (1- N))])
(define next-row
R -> (MAPCAR + [0|R] (append R [0])))
(define pascal
N -> (iterate next-row [1] N))
</lang>
R
<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))))
(let-values
([(previous-rows final-row)
(for/fold ([triangle null]
[row '(1)])
([row-number (in-range 1 n)])
(values (cons row triangle)
(next-row row)))])
(reverse (cons final-row previous-rows))))
</lang>
RapidQ
Summing from Previous Rows
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>
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 pratical limit for this version, triangles up to 46 rows have been generated in a
window 620 pixels wide.
If the number specified is negative, the output is written to a file instead. Triangles up to
1000 rows have been created. The file created is named: PASCALS.n
where n is the absolute value of the number entered.
<lang rexx>/*REXX program to display Pascal's triangle, neatly centered/formatted.*/
/*AKA: Yang Hui's ▲, Khayyam-Pascal ▲, Kyayyam ▲, Tartaglia's ▲ */
numeric digits 1000 /*let's be able to handle big ▲. */
arg nn .; if nn== then nn=10; n=abs(nn)
a. = 1 /*if NN < 0, output is to a file.*/
mx = !(n-1) / !(n%2) / !(n-1-n%2) /*MX =biggest number in triangle.*/
w = length(mx) /* W =width of biggest number. */
line. = 1
do row=1 for n; prev=row-1
a.row.1 = 1
do j=2 to row-1; jm=j-1
a.row.j = a.prev.jm + a.prev.j
line.row = line.row right(a.row.j,w)
end /*j*/
if row\==1 then line.row=line.row right(1,w) /*append the last "1".*/ end /*row*/
width=length(line.n) /*width of last line in triangle.*/
do L=1 for n /*show lines in Pascal's triangle*/
if nn>0 then say center(line.L,width) /*either SAY or write.*/
else call lineout 'PASCALS.'n, center(line.L,width)
end /*L*/
exit /*stick a fork in it, we're done.*/ /*─────────────────────────────────────! (factorial) subroutine─────────*/ !: procedure; arg x;!=1;do j=2 to x;!=!*j;end;return ! /*calc. factorial*/</lang> output when the input was given as: 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 the input was given as: 22
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1
1 17 136 680 2380 6188 12376 19448 24310 24310 19448 12376 6188 2380 680 136 17 1
1 18 153 816 3060 8568 18564 31824 43758 48620 43758 31824 18564 8568 3060 816 153 18 1
1 19 171 969 3876 11628 27132 50388 75582 92378 92378 75582 50388 27132 11628 3876 969 171 19 1
1 20 190 1140 4845 15504 38760 77520 125970 167960 184756 167960 125970 77520 38760 15504 4845 1140 190 20 1
1 21 210 1330 5985 20349 54264 116280 203490 293930 352716 352716 293930 203490 116280 54264 20349 5985 1330 210 21 1
Ruby
<lang ruby>def pascal(n)
raise ArgumentError, "must be positive." if n < 1
yield ar = [1]
(n-1).times do
ar.unshift(0).push(0) # tack a zero on both ends
yield ar = ar.each_cons(2).map{|a, b| a + b }
end
end
pascal(8){|row| puts row.join(" ").center(20)}</lang>
- Output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
Or for more or less a translation of the two line Haskell version (with inject being abused a bit I know):
<lang ruby>def next_row(row) ([0] + row).zip(row + [0]).collect {|l,r| l + r } end
def pascal(n) n.times.inject([1]) {|x,_| next_row x } end
8.times{|i| p pascal(i)}</lang>
- Output:
[1] [1, 1] [1, 2, 1] [1, 3, 3, 1] [1, 4, 6, 4, 1] [1, 5, 10, 10, 5, 1] [1, 6, 15, 20, 15, 6, 1] [1, 7, 21, 35, 35, 21, 7, 1]
Run BASIC
<lang runbasic>input "number of rows? ";r for i = 0 to r - 1
c = 1
print left$(" ",(r*2)-(i*2));
for k = 0 to i
print using("####",c);
c = c*(i-k)/(k+1)
next
print
next</lang>Output:
Number of rows? ?5
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
Scala
Simple recursive row definition: <lang scala> def tri(row:Int):List[Int] = { row match {
case 1 => List(1)
case n:Int => List(1) ::: ((tri(n-1) zip tri(n-1).tail) map {case (a,b) => a+b}) ::: List(1)
}
} </lang> Function to pretty print n rows: <lang scala> def prettytri(n:Int) = (1 to n) foreach {i=>print(" "*(n-i)); tri(i) map (c=>print(c+" ")); println} prettytri(5) </lang> Outputs:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1
Scheme
<lang scheme>(define (next-row row)
(map + (cons 0 row) (append row '(0))))
(define (triangle row rows)
(if (= rows 0)
'()
(cons row (triangle (next-row row) (- rows 1)))))
(triangle (list 1) 5) </lang> Output: <lang>((1) (1 1) (1 2 1) (1 3 3 1) (1 4 6 4 1))</lang>
Seed7
<lang seed7>$ 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>
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> procedure pascal (n : int)
for i : 0 .. n
var c : int
c := 1
for k : 0 .. i
put intstr(c) + " " ..
c := c * (i - k) div (k + 1)
end for
put ""
end for
end pascal
pascal(5)</lang>
uBasic/4tH
<lang>Input "Number Of Rows: "; N @(1) = 1 Print Tab((N+1)*3);"1"
For R = 2 To N
Print Tab((N-R)*3+1);
For I = R To 1 Step -1
@(I) = @(I) + @(I-1)
Print Using "______";@(i);
Next
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
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>>
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>
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
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