Sierpinski triangle: Difference between revisions
(→{{header|Perl}}: ++ pascal) |
(add Tcl) |
||
| Line 573: | Line 573: | ||
(- n 1)))))) |
(- n 1)))))) |
||
</lang> |
</lang> |
||
=={{header|Tcl}}== |
|||
{{trans|Perl}} |
|||
<lang tcl>package require Tcl 8.5 |
|||
proc map {lambda list} { |
|||
foreach elem $list { |
|||
lappend result [apply $lambda $elem] |
|||
} |
|||
return $result |
|||
} |
|||
proc sierpinski_triangle n { |
|||
set down [list *] |
|||
set space " " |
|||
for {set i 1} {$i <= $n} {incr i} { |
|||
set down [concat \ |
|||
[map [subst -nocommands {x {expr {"$space[set x]$space"}}}] $down] \ |
|||
[map {x {expr {"$x $x"}}} $down] \ |
|||
] |
|||
append space $space |
|||
} |
|||
return [join $down \n] |
|||
} |
|||
puts [sierpinski_triangle 4]</lang> |
|||
=={{header|Vedit macro language}}== |
=={{header|Vedit macro language}}== |
||
Revision as of 10:48, 1 May 2009
You are encouraged to solve this task according to the task description, using any language you may know.
Produce an ASCII representation of a Sierpinski triangle of order N. For example, the Sierpinski triangle of order 4 should look like this:
*
* *
* *
* * * *
* *
* * * *
* * * *
* * * * * * * *
* *
* * * *
* * * *
* * * * * * * *
* * * *
* * * * * * * *
* * * * * * * *
* * * * * * * * * * * * * * * *
See also Sierpinski carpet
Ada
This Ada example creates a string of the binary value for each line, converting the '0' values to spaces. <lang ada>with Ada.Text_Io; use Ada.Text_Io; with Ada.Strings.Fixed; with Interfaces; use Interfaces;
procedure Sieteri_Triangles is
subtype Practical_Order is Unsigned_32 range 0..4;
function Pow(X : Unsigned_32; N : Unsigned_32) return Unsigned_32 is
begin
if N = 0 then
return 1;
else
return X * Pow(X, N - 1);
end if;
end Pow;
procedure Print(Item : Unsigned_32) is
use Ada.Strings.Fixed;
package Ord_Io is new Ada.Text_Io.Modular_Io(Unsigned_32);
use Ord_Io;
Temp : String(1..36) := (others => ' ');
First : Positive;
Last : Positive;
begin
Put(To => Temp, Item => Item, Base => 2);
First := Index(Temp, "#") + 1;
Last := Index(Temp(First..Temp'Last), "#") - 1;
for I in reverse First..Last loop
if Temp(I) = '0' then
Put(' ');
else
Put(Temp(I));
end if;
end loop;
New_Line;
end Print;
procedure Sierpinski (N : Practical_Order) is
Size : Unsigned_32 := Pow(2, N);
V : Unsigned_32 := Pow(2, Size);
begin
for I in 0..Size - 1 loop
Print(V);
V := Shift_Left(V, 1) xor Shift_Right(V,1);
end loop;
end Sierpinski;
begin
for N in Practical_Order loop
Sierpinski(N);
end loop;
end Sieteri_Triangles;</lang>
ALGOL 68
<lang algol>PROC sierpinski = (INT n)[]STRING: (
FLEX[0]STRING d := "*";
FOR i TO n DO
[UPB d * 2]STRING next;
STRING sp := " " * (2 ** (i-1));
FOR x TO UPB d DO
STRING dx = d[x];
next[x] := sp+dx+sp;
next[UPB d+x] := dx+" "+dx
OD;
d := next
OD;
d
);
printf(($gl$,sierpinski(4)))</lang>
BASIC
<lang freebasic>SUB triangle (x AS Integer, y AS Integer, length AS Integer, n AS Integer)
IF n = 0 THEN
LOCATE y,x: PRINT "*";
ELSE
triangle (x, y+length, length/2, n-1)
triangle (x+length, y, length/2, n-1)
triangle (x+length*2, y+length, length/2, n-1) END IF
END SUB
CLS triangle 1,1,16,5</lang>
Note: The total height of the triangle is 2 * parameter length. It should be power of two so that the pattern matches evenly with the character cells. Value 16 will thus create pattern of 32 lines.
C
This solution uses a cellular automaton (rule 90) with a proper initial status.
<lang c>#include <stdio.h>
- include <stdlib.h>
- include <stdbool.h>
- include <string.h>
- ifndef _POSIX_C_SOURCE
char *strdup(char *s) {
int l = strlen(s); char *r = malloc(l+1); memcpy(r, s, l+1); return r;
}
- endif
- define truth(X) ((X)=='*'?true:false)
void rule_90(char *evstr) {
int i; int l = strlen(evstr); bool s[3]; char *cp = strdup(evstr);
for(i=0;i < l; i++) {
s[1] = truth(cp[i]);
s[0] = (i-1) < 0 ? false : truth(cp[i-1]);
s[2] = (i+1) < l ? truth(cp[i+1]) : false;
if ( (s[0] && !s[2]) || (!s[0] && s[2]) ) {
evstr[i] = '*';
} else {
evstr[i] = ' ';
}
}
free(cp);
}</lang>
<lang c>void sierpinski_triangle(int n) {
int i; int l = 1<<(n+1); char *b = malloc(l+1);
memset(b, ' ', l); b[l] = 0; b[l>>1] = '*';
printf("%s\n", b);
for(i=0; i < l/2-1;i++) {
rule_90(b);
printf("%s\n", b);
}
free(b);
}</lang>
<lang c>int main() {
sierpinski_triangle(4); return EXIT_SUCCESS;
}</lang>
Common Lisp
<lang lisp> (defun print-sierpinski (order)
(loop with size = (expt 2 order)
repeat size
for v = (expt 2 (1- size)) then (logxor (ash v -1) (ash v 1))
do (fresh-line)
(loop for i below (integer-length v)
do (princ (if (logbitp i v) "*" " ")))))</lang>
Printing each row could also be done by printing the integer in base 2 and replacing zeroes with spaces: (princ (substitute #\Space #\0 (format nil "~%~2,vR" (1- (* 2 size)) v)))
Replacing the iteration with for v = 1 then (logxor v (ash v 1)) produces a "right" triangle instead of an "equilateral" one.
D
Adapted from Java version (this version is slower than the Python one). <lang d>import std.stdio, std.string;
string[] sierpinski(int n) {
string[] parts = ["*"];
string space = " ";
for (int i; i < n; i++) {
string[] parts2;
foreach (x; parts)
parts2 ~= space ~ x ~ space;
foreach (x; parts)
parts2 ~= x ~ " " ~ x;
parts = parts2;
space ~= space;
}
return parts;
}
void main() {
writefln(sierpinski(4).join("\n"));
}</lang>
That sierpinski() function can run at compile time too, so with a compile-time join it can compute the whole result at compile-time:
<lang d> string[] sierpinski(int n) {
string[] parts = ["*"];
string space = " ";
for (int i; i < n; i++) {
string[] parts2;
foreach (x; parts)
parts2 ~= space ~ x ~ space;
foreach (x; parts)
parts2 ~= x ~ " " ~ x;
parts = parts2;
space ~= space;
}
return parts;
}
string joinCT(string[] parts, char sep) {
string result;
if (parts.length) {
foreach (part; parts[0 .. $-1]) {
result ~= part;
result ~= sep;
}
result ~= parts[$-1];
}
return result;
}
pragma(msg, sierpinski(4).joinCT('\n'));
void main() {} </lang>
Forth
: stars ( mask -- )
begin
dup 1 and if [char] * else bl then emit
1 rshift dup
while space repeat drop ;
: triangle ( order -- )
1 swap lshift ( 2^order )
1 over 0 do
cr over i - spaces dup stars
dup 2* xor
loop 2drop ;
5 triangle
Haskell
<lang haskell> sierpinski 0 = ["*"]
sierpinski (n+1) = map ((space ++) . (++ space)) down
++ map (unwords . replicate 2) down
where down = sierpinski n
space = replicate (2^n) ' '
printSierpinski = mapM_ putStrLn . sierpinski</lang>
IDL
The only 'special' thing here is that the math is done in a byte array, filled with the numbers 32 and 42 and then output through a "string(array)" which prints the ascii representation of each individual element in the array.
pro sierp,n
s = (t = bytarr(3+2^(n+1))+32b)
t[2^n+1] = 42b
for lines = 1,2^n do begin
print,string( (s = t) )
for i=1,n_elements(t)-2 do if s[i-1] eq s[i+1] then t[i]=32b else t[i]=42b
end
end
J
There are any number of succinct ways to produce this in J. Here's one that exploits self-similarity:
|._31]\,(,.~,])^:4,:'* '
Here's one that leverages the relationship between Sierpinski's and Pascal's triangles:
' *'{~'1'=(-|."_1[:":2|!/~)i.-16
Java
<lang java>public static void triangle(int n){
n= 1 << n;
StringBuilder line= new StringBuilder(); //use a "mutable String"
char t= 0;
char u= 0; // avoid warnings
for(int i= 0;i <= 2 * n;++i)
line.append(" "); //start empty
line.setCharAt(n, '*'); //with the top point of the triangle
for(int i= 0;i < n;++i){
System.out.println(line);
u= '*';
for(int j= n - i;j < n + i + 1;++j){
t= (line.charAt(j - 1) == line.charAt(j + 1) ? ' ' : '*');
line.setCharAt(j - 1, u);
u= t;
}
line.setCharAt(n + i, t);
line.setCharAt(n + i + 1, '*');
}
}</lang>
<lang java> import java.util.*;
public class Sierpinski {
public static List<String> sierpinski(int n)
{
List<String> down = Arrays.asList("*");
String space = " ";
for (int i = 0; i < n; i++) {
List<String> newDown = new ArrayList<String>();
for (String x : down)
newDown.add(space + x + space);
for (String x : down)
newDown.add(x + " " + x);
down = newDown;
space += space;
}
return down;
}
public static void main(String[] args)
{
for (String x : sierpinski(4))
System.out.println(x);
}
} </lang>
JavaScript
<lang javascript>
function triangle(o) {
var n = 1<<o, line = new Array(2*n), i,j,t,u;
for (i=0; i<line.length; ++i) line[i] = ' ';
line[n] = '*';
for (i=0; i<n; ++i) {
document.write(line.join()+"\n");
u ='*';
for(j=n-i; j<n+i+1; ++j) {
t = (line[j-1] == line[j+1] ? ' ' : '*');
line[j-1] = u;
u = t;
}
line[n+i] = t;
line[n+i+1] = '*';
}
}
document.write("
\n");
triangle(6);
document.write("");
</lang>
Logo
This will draw a graphical Sierpinski gasket using turtle graphics.
to sierpinski :n :length if :n = 0 [stop] repeat 3 [sierpinski :n-1 :length/2 fd :length rt 120] end seth 30 sierpinski 5 200
OCaml
<lang ocaml> let sierpinski n =
let rec loop down space n =
if n = 0 then
down
else
loop (List.map (fun x -> space ^ x ^ space) down @
List.map (fun x -> x ^ " " ^ x) down)
(space ^ space)
(n - 1)
in loop ["*"] " " n
let () =
List.iter print_endline (sierpinski 4)
</lang>
Pascal
<lang pascal>program Sierpinski;
function ipow(b, n : Integer) : Integer; var
i : Integer;
begin
ipow := 1;
for i := 1 to n do
ipow := ipow * b
end;
function truth(a : Char) : Boolean; begin
if a = '*' then
truth := true
else
truth := false
end;</lang>
<lang pascal>function rule_90(ev : String) : String; var
l, i : Integer; cp : String; s : Array[0..1] of Boolean;
begin
l := length(ev);
cp := copy(ev, 1, l);
for i := 1 to l do begin
if (i-1) < 1 then
s[0] := false
else
s[0] := truth(ev[i-1]);
if (i+1) > l then
s[1] := false
else
s[1] := truth(ev[i+1]);
if ( (s[0] and not s[1]) or (s[1] and not s[0]) ) then
cp[i] := '*'
else
cp[i] := ' ';
end; rule_90 := cp
end;
procedure triangle(n : Integer); var
i, l : Integer; b : String;
begin
l := ipow(2, n+1);
b := ' ';
for i := 1 to l do
b := concat(b, ' ');
b[round(l/2)] := '*';
writeln(b);
for i := 1 to (round(l/2)-1) do begin
b := rule_90(b);
writeln(b)
end
end;</lang>
<lang pascal>begin
triangle(4)
end.</lang>
Perl
<lang perl> sub sierpinski {
my ($n) = @_;
my @down = '*';
my $space = ' ';
foreach (1..$n) {
@down = (map("$space$_$space", @down), map("$_ $_", @down));
$space = "$space$space";
}
return @down;
}
print "$_\n" foreach sierpinski 4; </lang>
Pop11
Solution using line buffer in an integer array oline, 0 represents ' ' (space), 1 represents '*' (star).
define triangle(n);
lvars k = 2**n, j, l, oline, nline;
initv(2*k+3) -> oline;
initv(2*k+3) -> nline;
for l from 1 to 2*k+3 do 0 -> oline(l) ; endfor;
1 -> oline(k+2);
0 -> nline(1);
0 -> nline(2*k+3);
for j from 1 to k do
for l from 1 to 2*k+3 do
printf(if oline(l) = 0 then ' ' else '*' endif);
endfor;
printf('\n');
for l from 2 to 2*k+2 do
(oline(l-1) + oline(l+1)) rem 2 -> nline(l);
endfor;
(oline, nline) -> (nline, oline);
endfor;
enddefine;
triangle(4);
Alternative solution, keeping all triangle as list of strings
define triangle2(n);
lvars acc = ['*'], spaces = ' ', j;
for j from 1 to n do
maplist(acc, procedure(x); spaces >< x >< spaces ; endprocedure)
<> maplist(acc, procedure(x); x >< ' ' >< x ; endprocedure) -> acc;
spaces >< spaces -> spaces;
endfor;
applist(acc, procedure(x); printf(x, '%p\n'); endprocedure);
enddefine;
triangle2(4);
Python
<lang python> def sierpinski(n):
d = ["*"]
for i in xrange(n):
sp = " " * (2 ** i)
d = [sp+x+sp for x in d] + [x+" "+x for x in d]
return d
print "\n".join(sierpinski(4)) </lang>
Ruby
From the command line: <lang bash> ruby -le'16.times{|y|print" "*(15-y),(0..y).map{|x|~y&x>0?" ":" *"}}' </lang>
Scheme
<lang scheme> (define (sierpinski n)
(for-each
(lambda (x) (display (list->string x)) (newline))
(let loop ((acc (list (list #\*))) (spaces (list #\ )) (n n))
(if (zero? n)
acc
(loop
(append
(map (lambda (x) (append spaces x spaces)) acc)
(map (lambda (x) (append x (list #\ ) x)) acc))
(append spaces spaces)
(- n 1))))))
</lang>
Tcl
<lang tcl>package require Tcl 8.5
proc map {lambda list} {
foreach elem $list {
lappend result [apply $lambda $elem]
}
return $result
}
proc sierpinski_triangle n {
set down [list *]
set space " "
for {set i 1} {$i <= $n} {incr i} {
set down [concat \
[map [subst -nocommands {x {expr {"$space[set x]$space"}}}] $down] \
[map {x {expr {"$x $x"}}} $down] \
]
append space $space
}
return [join $down \n]
}
puts [sierpinski_triangle 4]</lang>
Vedit macro language
Iterative
The macro writes the fractal into an edit buffer where it can be viewed and saved to file if required. This allows creating images larger than screen, the size is only limited by free disk space.
#3 = 16 // size (height) of the triangle
Buf_Switch(Buf_Free) // Open a new buffer for output
Ins_Char(' ', COUNT, #3*2+2) // fill first line with spaces
Ins_Newline
Line(-1) Goto_Col(#3)
Ins_Char('*', OVERWRITE) // the top of triangle
for (#10=0; #10 < #3-1; #10++) {
BOL Reg_Copy(9,1) Reg_Ins(9) // duplicate the line
#20 = '*'
for (#11 = #3-#10; #11 < #3+#10+1; #11++) {
Goto_Col(#11-1)
if (Cur_Char==Cur_Char(2)) { #21=' ' } else { #21='*' }
Ins_Char(#20, OVERWRITE)
#20 = #21
}
Ins_Char(#21, OVERWRITE)
Ins_Char('*', OVERWRITE)
}
Recursive
Vedit macro language does not have recursive functions, so some pushing and popping is needed to implement recursion.
#1 = 1 // x
#2 = 1 // y
#3 = 16 // length (height of the triangle / 2)
#4 = 5 // depth of recursion
Buf_Switch(Buf_Free) // Open a new buffer for output
Ins_Newline(#3*2) // Create as many empty lines as needed
Call("Triangle") // Draw the triangle
BOF
Return
:Triangle:
if (#4 == 0) {
Goto_Line(#2)
EOL Ins_Char(' ', COUNT, #1-Cur_Col+1) // add spaces if needed
Goto_Col(#1)
Ins_Char('*', OVERWRITE)
} else {
Num_Push(1,4)
#2 += #3; #3 /= 2; #4--; Call("Triangle")
Num_Pop(1,4)
Num_Push(1,4)
#1 += #3; #3 /= 2; #4--; Call("Triangle")
Num_Pop(1,4)
Num_Push(1,4)
#1 += 2*#3; #2 += #3; #3 /= 2; #4--; Call("Triangle")
Num_Pop(1,4)
}
Return