Draw A Sierpinski Triangle

The Sierpinsky Triangle is a fractal created by taking a triangle, decreasing the height and width by 1/2, creating 3 copies of the resulting triangle, and place them such each triangle touches the other two on a corner. This process is repeated over and over again with the resulting triangles to produce the Sierpinski triangle, as illustrated below.

Write a program to generate a Sierpinski triangle. You can use any method you want to generate the pattern, either by drawing the actual triangles, or by using a random algorithm to generate the picture. You can draw in pixels, ascii art, or whatever you want, so long as the output looks similar to the last picture shown above. Fewest characters wins.

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See also the old Stack Overflow version: stackoverflow.com/questions/1726698/… – dmckee Jun 9 '12 at 3:12
I got the idea for this after seeing the pascal's triangle question, and remembering the example program for this in my TI-86 manual. I decided to convert it to QBasic and then code golf it. – Kibbee Jun 9 '12 at 3:22
There is no problem with running a challenge here that was already run on Stack Overflow, but many people will not want to present the same material again. So I link them for the edification of later visitors. – dmckee Jun 9 '12 at 3:23
To avoid duplication, perhaps you should change to rules to allow only graphical implementations. – primo Jun 9 '12 at 3:45
Lots of ideas from wolfram: wolframscience.com/nksonline/page-931 – luser droog Feb 6 '14 at 6:06

HTML, 150 characters (see notes for 126 characters)

<title></title><canvas></canvas><script>for(x=k=128;x--;)for(y=k;y--;)x&y||document.body.firstChild.getContext("2d").fillRect(x-~y/2,k-y,1,1)</script>

The core of it is applying the rule of coloring pixels for which x & y == 0 by the conditional x&y||, which produces a “Sierpinski right triangle”; and x-~y/2,k-y are a coordinate transformation to produce the approximately equilateral display.

A less correct (HTML-wise) version is 126 characters:

<canvas><script>for(x=k=128;x--;)for(y=k;y--;)x&y||document.body.firstChild.getContext("2d").fillRect(x-~y/2,k-y,1,1)</script>

Three characters can be saved by eliminating k in favor of the constant 64, at the cost of a smaller result; I wouldn't count the 8 option as it has insufficient detail.

Note that a size of 256 or higher requires attributes on the <canvas> to increase the canvas size from the default.

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Nobody cares if your HTML validates at codegolf :-) Some improvements: <canvas id=c> and then c.getContext. Shorten loops: for(x=k=128;x--;)for(y=k;y--;) – copy Jun 9 '12 at 15:47
ids being turned into global variables is a horrible misfeature which I refuse to acknowledge, and WebKit does not implement it in standards mode. Thank you for the loop trick. – Kevin Reid Jun 9 '12 at 16:53
Minor improvement: x&y?0: can be replaced with x&y|| Otherwise nice solution. – primo Jun 10 '12 at 11:20
Bravo, this is just wonderful. – boothby Jun 14 '12 at 7:17
May I ask, how does this work? – Derek 朕會功夫 Jul 8 '14 at 5:54

GolfScript (43 42 chars)

' /\ /__\ '4/){.+\.{[2\$.]*}%\{.+}%+\}3*;n*

Output:

/\
/__\
/\  /\
/__\/__\
/\      /\
/__\    /__\
/\  /\  /\  /\
/__\/__\/__\/__\
/\              /\
/__\            /__\
/\  /\          /\  /\
/__\/__\        /__\/__\
/\      /\      /\      /\
/__\    /__\    /__\    /__\
/\  /\  /\  /\  /\  /\  /\  /\
/__\/__\/__\/__\/__\/__\/__\/__\

Change the "3" to a larger number for a larger triangle.

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Python (234)

Maximal golfing, tiny image:

#!/usr/bin/env python3
from cairo import*
s=SVGSurface('_',97,84)
g=Context(s)
g.scale(97,84)
def f(w,x,y):
v=w/2
if w>.1:f(v,x,y);f(v,x+w/4,y-v);f(v,x+v,y)
else:g.move_to(x,y);g.line_to(x+v,y-w);g.line_to(x+w,y);g.fill()
f(1,0,1)
s.write_to_png('s.png')

Requires python3-cairo.

To get a nice large image I needed 239 characters.

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import cairo as c whould save you a few characters – quasimodo Jun 17 '12 at 13:33
this answer needs more upvotes – ixtmixilix Jan 1 '13 at 23:47

Mathematica - 32 characters

Nest[Subsuperscript[#,#,#]&,0,5]

Mathematica - 37 characters

Grid@CellularAutomaton[90,{{1},0},31]

This will produce a 2D table of 0 and 1, where 1s are drawing Sierpinski Triangle.

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At the cost of 5 additional characters, your second solution will display better with ArrayPlot@CellularAutomaton[90, {{1}, 0}, 31] or MatrixPlot@CellularAutomaton[90, {{1}, 0}, 31]. – DavidC Jan 2 '13 at 18:25
...or with ReliefPlot@... – DavidC Jan 2 '13 at 18:29
I get this. How did you get the output without all the brackets? – Mr.Wizard Mar 27 '13 at 5:52
@Mr.Wizard hmm... where in the world brackets are from? It even works here: mathics.net Try and let me know. – Vitaliy Kaurov Mar 27 '13 at 6:23
@Mr.Wizard Also take a look at this: wolfram.com/xid/0dek1w-y16 – Vitaliy Kaurov Mar 27 '13 at 6:27

Python, 101 86

Uses the rule 90 automaton.

x=' '*31
x+='.'+x
exec"print x;x=''.join(' .'[x[i-1]!=x[i-62]]for i in range(63));"*32

This is longer, but prettier.

x=' '*31
x+=u'Δ'+x
exec u"print x;x=''.join(u' Δ'[x[i-1]!=x[i-62]]for i in range(63));"*32

Edit: playing with strings directly, got rid of obnoxiously long slicing, made output prettier.

Output:

Δ
Δ Δ
Δ   Δ
Δ Δ Δ Δ
Δ       Δ
Δ Δ     Δ Δ
Δ   Δ   Δ   Δ
Δ Δ Δ Δ Δ Δ Δ Δ
Δ               Δ
Δ Δ             Δ Δ
Δ   Δ           Δ   Δ
Δ Δ Δ Δ         Δ Δ Δ Δ
Δ       Δ       Δ       Δ
Δ Δ     Δ Δ     Δ Δ     Δ Δ
Δ   Δ   Δ   Δ   Δ   Δ   Δ   Δ
Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ
Δ                               Δ
Δ Δ                             Δ Δ
Δ   Δ                           Δ   Δ
Δ Δ Δ Δ                         Δ Δ Δ Δ
Δ       Δ                       Δ       Δ
Δ Δ     Δ Δ                     Δ Δ     Δ Δ
Δ   Δ   Δ   Δ                   Δ   Δ   Δ   Δ
Δ Δ Δ Δ Δ Δ Δ Δ                 Δ Δ Δ Δ Δ Δ Δ Δ
Δ               Δ               Δ               Δ
Δ Δ             Δ Δ             Δ Δ             Δ Δ
Δ   Δ           Δ   Δ           Δ   Δ           Δ   Δ
Δ Δ Δ Δ         Δ Δ Δ Δ         Δ Δ Δ Δ         Δ Δ Δ Δ
Δ       Δ       Δ       Δ       Δ       Δ       Δ       Δ
Δ Δ     Δ Δ     Δ Δ     Δ Δ     Δ Δ     Δ Δ     Δ Δ     Δ Δ
Δ   Δ   Δ   Δ   Δ   Δ   Δ   Δ   Δ   Δ   Δ   Δ   Δ   Δ   Δ   Δ
Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ
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That looks really cool :D – beary605 Jun 11 '12 at 5:14

J

,/.(,~,.~)^:6,'o'

Not ideal, since the triangle is lopsided and followed by a lot of whitespace - but interesting nonetheless I thought.

Output:

o
oo
o o
oooo
o   o
oo  oo
o o o o
oooooooo
o       o
oo      oo
o o     o o
oooo    oooo
o   o   o   o
oo  oo  oo  oo
o o o o o o o o
oooooooooooooooo
o               o
oo              oo
o o             o o
oooo            oooo
o   o           o   o
oo  oo          oo  oo
o o o o         o o o o
oooooooo        oooooooo
o       o       o       o
oo      oo      oo      oo
o o     o o     o o     o o
oooo    oooo    oooo    oooo
o   o   o   o   o   o   o   o
oo  oo  oo  oo  oo  oo  oo  oo
o o o o o o o o o o o o o o o o
oooooooooooooooooooooooooooooooo
o                               o
oo                              oo
o o                             o o
oooo                            oooo
o   o                           o   o
oo  oo                          oo  oo
o o o o                         o o o o
oooooooo                        oooooooo
o       o                       o       o
oo      oo                      oo      oo
o o     o o                     o o     o o
oooo    oooo                    oooo    oooo
o   o   o   o                   o   o   o   o
oo  oo  oo  oo                  oo  oo  oo  oo
o o o o o o o o                 o o o o o o o o
oooooooooooooooo                oooooooooooooooo
o               o               o               o
oo              oo              oo              oo
o o             o o             o o             o o
oooo            oooo            oooo            oooo
o   o           o   o           o   o           o   o
oo  oo          oo  oo          oo  oo          oo  oo
o o o o         o o o o         o o o o         o o o o
oooooooo        oooooooo        oooooooo        oooooooo
o       o       o       o       o       o       o       o
oo      oo      oo      oo      oo      oo      oo      oo
o o     o o     o o     o o     o o     o o     o o     o o
oooo    oooo    oooo    oooo    oooo    oooo    oooo    oooo
o   o   o   o   o   o   o   o   o   o   o   o   o   o   o   o
oo  oo  oo  oo  oo  oo  oo  oo  oo  oo  oo  oo  oo  oo  oo  oo
o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o
oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

A quick explanation:

The verb (,~,.~) is what's doing the work here. It's a hook which first stitches ,. the argument to itself (o -> oo) and then appends the original argument to the output:

oo

becomes

oo
o

This verb is repeated 6 times ^:6 with the output of each iteration becoming the input of the next iteration. So

oo
o

becomes

oooo
o o
oo
o

which in turn becomes

oooooooo
o o o o
oo  oo
o   o
oooo
o o
oo
o

etc. I've then used the oblique adverb on append ,/. to read the rows diagonally to straighten(ish) the triangle. I didn't need to do this, as randomra points out. I could have just reversed |. the lot to get the same result. Even better, I could have just used (,,.~)^:6,'o' to save the reverse step completely.

Ah well, you live and learn. :-)

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Could you briefly explain how it works? I'm not familiar with J – aditsu Feb 27 '13 at 10:18
|.(,~,.~)^:6,'o' is shorter and without extra spaces. And (,~,.~)^:6,1 also gives decent input in just 12 characters! – randomra Feb 27 '13 at 10:19
@aditsu I've added an explanation. – Gareth Feb 27 '13 at 12:20
So if I get it, that operator concatenates two 2d arrays? – Dokkat Mar 23 '13 at 10:56

Wolfram Rule 126 - 1 character

1

Of course the form of the input depends heavily on the implementation of rule 126, but the pattern generated by an input that looks like a binary 1 will resemble the Sierpinsky triangle: http://www.wolframalpha.com/input/?i=rule+126

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I'm not very good at golfing haskell codes.

solve n = tri (putStrLn "") [2^n] n
tri m xs 1 =
do putStrLn (l1 1 xs "/\\" 0)
putStrLn (l1 1 xs "/__\\" 1)
m
tri m xs n=tri m' xs (n-1)
where m'=tri m (concat[[x-o,x+o]|x<-xs]) (n-1)
o=2^(n-1)
l1 o [] s t=""
l1 o (x:xs) s t=replicate (x-o-t) ' '++s++l1 (x+2+t) xs s t

Output of solve 4 is:

/\
/__\
/\  /\
/__\/__\
/\      /\
/__\    /__\
/\  /\  /\  /\
/__\/__\/__\/__\
/\              /\
/__\            /__\
/\  /\          /\  /\
/__\/__\        /__\/__\
/\      /\      /\      /\
/__\    /__\    /__\    /__\
/\  /\  /\  /\  /\  /\  /\  /\
/__\/__\/__\/__\/__\/__\/__\/__\
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+1 for the nice ASCII art – w0lf Jun 11 '12 at 9:32

QBasic 151 Characters

As an example, here is how it can be done in QBasic.

SCREEN 9
H=.5
P=300
FOR I=1 TO 9^6
N=RND
IF N > 2/3 THEN
X=H+X*H:Y=Y*H
ELSEIF N > 1/3 THEN
X=H^2+X*H:Y=H+Y*H
ELSE
X=X*H:Y=Y*H
END IF
PSET(P-X*P,P-Y*P)
NEXT

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Could you describe the measure under which this program is 129 characters? I get 151 if I strip out all the probably-unnecessary whitespace. (I'm not familiar with QBasic.) – Kevin Reid Jun 9 '12 at 12:44
I stripped out all the whitespace for my count. I guess I could count only non-essential whitespace. I'm not sure what the "official" rule is for code golf. – Kibbee Jun 10 '12 at 0:41
You should count the actual number of characters, including whitespace, in a program which runs and produces the correct output. Naturally you'll want to have no unnecessary whitespace. – Kevin Reid Jun 10 '12 at 0:43
Corrected my character count. – Kibbee Jun 12 '12 at 12:18

Python (42)

I originally wanted to post a few suggestions on boothbys solution (who actually uses rule 18 :), but i didn't have enough reputation to comment, so i made it into another answer. Since he changed his approach, i added some explanation. My suggestions would have been:

1. use '%d'*64%tuple(x) instead of ''.join(map(str,x)
2. shift in zeros instead of wraping the list around

which would have led to the following code (93 characters):

x=[0]*63
x[31]=1
exec"print'%d'*63%tuple(x);x=[a^b for a,b in zip(x[1:]+[0],[0]+x[:-1])];"*32

But i optimzed further, first by using a longint instead of an integer array and just printing the binary representation (75 characters):

x=2**31
exec"print'%d'*63%tuple(1&x>>i for i in range(63));x=x<<1^x>>1;"*32

And finally by printing the octal representation, which is already supported by printf interpolation (42 characters):

x=8**31
exec"print'%063o'%x;x=x*8^x/8;"*32

All of them will print:

000000000000000000000000000000010000000000000000000000000000000
000000000000000000000000000000101000000000000000000000000000000
000000000000000000000000000001000100000000000000000000000000000
000000000000000000000000000010101010000000000000000000000000000
000000000000000000000000000100000001000000000000000000000000000
000000000000000000000000001010000010100000000000000000000000000
000000000000000000000000010001000100010000000000000000000000000
000000000000000000000000101010101010101000000000000000000000000
000000000000000000000001000000000000000100000000000000000000000
000000000000000000000010100000000000001010000000000000000000000
000000000000000000000100010000000000010001000000000000000000000
000000000000000000001010101000000000101010100000000000000000000
000000000000000000010000000100000001000000010000000000000000000
000000000000000000101000001010000010100000101000000000000000000
000000000000000001000100010001000100010001000100000000000000000
000000000000000010101010101010101010101010101010000000000000000
000000000000000100000000000000000000000000000001000000000000000
000000000000001010000000000000000000000000000010100000000000000
000000000000010001000000000000000000000000000100010000000000000
000000000000101010100000000000000000000000001010101000000000000
000000000001000000010000000000000000000000010000000100000000000
000000000010100000101000000000000000000000101000001010000000000
000000000100010001000100000000000000000001000100010001000000000
000000001010101010101010000000000000000010101010101010100000000
000000010000000000000001000000000000000100000000000000010000000
000000101000000000000010100000000000001010000000000000101000000
000001000100000000000100010000000000010001000000000001000100000
000010101010000000001010101000000000101010100000000010101010000
000100000001000000010000000100000001000000010000000100000001000
001010000010100000101000001010000010100000101000001010000010100
010001000100010001000100010001000100010001000100010001000100010
101010101010101010101010101010101010101010101010101010101010101

Of course there is also a graphical solution (131 characters):

from PIL.Image import*
from struct import*
a=''
x=2**31
exec"a+=pack('>Q',x);x=x*2^x/2;"*32
fromstring('1',(64,32),a).save('s.png')

:D

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36: x=8**31;exec"print'%o'%x;x^=x/8;"*32 – aditsu Feb 27 '13 at 11:08

APL (51)

A←67⍴0⋄A[34]←1⋄' ○'[1+32 67⍴{~⊃⍵:⍵,∇(1⌽⍵)≠¯1⌽⍵⋄⍬}A]

Explanation:

• A←67⍴0: A is a vector of 67 zeroes
• A[34]←1: the 34th element is 1
• {...}A: starting with A, do:
• ~⊃⍵:: if the first element of the current row is zero
• ⍵,∇: add the current row to the answer, and recurse with:
• (1⌽⍵)≠¯1⌽⍵: the vector where each element is the XOR of its neighbours in the previous generation
• ⋄⍬: otherwise, we're done
• 32 67⍴: format this in a 67x32 matrix
• 1+: add one to select the right value from the character array
• ' ○'[...]: output either a space (not part of the triangle) or a circle (when it is part of the triangle)

Output:

○
○ ○
○   ○
○ ○ ○ ○
○       ○
○ ○     ○ ○
○   ○   ○   ○
○ ○ ○ ○ ○ ○ ○ ○
○               ○
○ ○             ○ ○
○   ○           ○   ○
○ ○ ○ ○         ○ ○ ○ ○
○       ○       ○       ○
○ ○     ○ ○     ○ ○     ○ ○
○   ○   ○   ○   ○   ○   ○   ○
○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○
○                               ○
○ ○                             ○ ○
○   ○                           ○   ○
○ ○ ○ ○                         ○ ○ ○ ○
○       ○                       ○       ○
○ ○     ○ ○                     ○ ○     ○ ○
○   ○   ○   ○                   ○   ○   ○   ○
○ ○ ○ ○ ○ ○ ○ ○                 ○ ○ ○ ○ ○ ○ ○ ○
○               ○               ○               ○
○ ○             ○ ○             ○ ○             ○ ○
○   ○           ○   ○           ○   ○           ○   ○
○ ○ ○ ○         ○ ○ ○ ○         ○ ○ ○ ○         ○ ○ ○ ○
○       ○       ○       ○       ○       ○       ○       ○
○ ○     ○ ○     ○ ○     ○ ○     ○ ○     ○ ○     ○ ○     ○ ○
○   ○   ○   ○   ○   ○   ○   ○   ○   ○   ○   ○   ○   ○   ○   ○
○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○
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Yikes. I expected this to be 4 characters, using binomials mod 2... (ok... maybe a little longer than that) – boothby Jun 9 '12 at 17:47

C 127119116108 65

This one uses the trick of the HTML answer of ^ i & j getting it to print pretty output would take 1 more char (you can get really ugly output by sacrificing the a^).

a=32,j;main(i){for(;++i<a;)putchar(a^i&j);++j<a&&main(puts(""));}

To make it pretty turn (32^i&j) to (32|!(i&j)) and turn it from ++i<a to ++i<=a. However wasting chars on looks seems ungolfish to me.

Ugly output:

! ! ! ! ! ! ! ! ! ! ! ! ! ! !
""  ""  ""  ""  ""  ""  ""  ""
"# !"# !"# !"# !"# !"# !"# !"#
\$\$\$\$    \$\$\$\$    \$\$\$\$    \$\$\$\$
!\$%\$% ! !\$%\$% ! !\$%\$% ! !\$%\$%
""\$\$&&  ""\$\$&&  ""\$\$&&  ""\$\$&&
"#\$%&' !"#\$%&' !"#\$%&' !"#\$%&'
((((((((        ((((((((
! ! !()()()() ! ! ! !()()()()
""  ""((**((**  ""  ""((**((**
"# !"#()*+()*+ !"# !"#()*+()*+
\$\$\$\$((((,,,,    \$\$\$\$((((,,,,
!\$%\$%()(),-,- ! !\$%\$%()(),-,-
""\$\$&&((**,,..  ""\$\$&&((**,,..
"#\$%&'()*+,-./ !"#\$%&'()*+,-./
0000000000000000
! ! ! ! ! ! !0101010101010101
""  ""  ""  ""0022002200220022
"# !"# !"# !"#0123012301230123
\$\$\$\$    \$\$\$\$0000444400004444
!\$%\$% ! !\$%\$%0101454501014545
""\$\$&&  ""\$\$&&0022446600224466
"#\$%&' !"#\$%&'0123456701234567
((((((((0000000088888888
! ! !()()()()0101010189898989
""  ""((**((**0022002288::88::
"# !"#()*+()*+0123012389:;89:;
\$\$\$\$((((,,,,000044448888<<<<
!\$%\$%()(),-,-010145458989<=<=
""\$\$&&((**,,..0022446688::<<>>
"#\$%&'()*+,-./0123456789:;<=>?

I actually kind of like how it looks. But if you insist on it being pretty you can dock four chars. Pretty Output:

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !
!!  !!  !!  !!  !!  !!  !!  !
!   !   !   !   !   !   !   !
!!    !!!!    !!!!    !!!!    !
!     ! !     ! !     ! !     !
!!      !!      !!      !
!       !       !       !
!!!!!!        !!!!!!!!        !
! ! !         ! ! ! !         !
!!          !!  !!          !
!           !   !           !
!!            !!!!            !
!             ! !             !
!!              !
!               !
!!!!!!!!!!!!!!                !
! ! ! ! ! ! !                 !
!!  !!  !!                  !
!   !   !                   !
!!    !!!!                    !
!     ! !                     !
!!                      !
!                       !
!!!!!!                        !
! ! !                         !
!!                          !
!                           !
!!                            !
!                             !
!
!

Leaving up the older 108 char, cellular automata version.

j,d[99][99];main(i){d[0][31]=3;for(;i<64;)d[j+1][i]=putchar(32|d[j][i+2]^d[j][i++]);++j<32&&main(puts(""));}

So I don't think I'm going to get it much shorter than this so I'll explain the code. I'll leave this explanation up, as some of the tricks could be useful.

j,d[99][99]; // these init as 0
main(i){ //starts at 1 (argc)
d[0][48]=3; //seed the automata (3 gives us # instead of !)
for(;i<98;) // print a row
d[j+1][i]=putchar(32|d[j][i+2]]^d[j][i++]);
//relies on undefined behavoir. Works on ubuntu with gcc ix864
//does the automata rule. 32 + (bitwise or can serve as + if you know
//that (a|b)==(a^b)), putchar returns the char it prints
++j<32&&main(puts(""));
// repeat 32 times
// puts("") prints a newline and returns 1, which is nice
}

Some output

# #
#   #
# # # #
#       #
# #     # #
#   #   #   #
# # # # # # # #
#               #
# #             # #
#   #           #   #
# # # #         # # # #
#       #       #       #
# #     # #     # #     # #
#   #   #   #   #   #   #   #
# # # # # # # # # # # # # # # #
#                               #
# #                             # #
#   #                           #   #
# # # #                         # # # #
#       #                       #       #
# #     # #                     # #     # #
#   #   #   #                   #   #   #   #
# # # # # # # #                 # # # # # # # #
#               #               #               #
# #             # #             # #             # #
#   #           #   #           #   #           #   #
# # # #         # # # #         # # # #         # # # #
#       #       #       #       #       #       #       #
# #     # #     # #     # #     # #     # #     # #     # #
#   #   #   #   #   #   #   #   #   #   #   #   #   #   #   #
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #
-
This does not appear to be a Sierpinski triangle; it splits into three sub-triangles (proceeding downward) rather than two, and it can be seen that this produces no large central empty triangle. – Kevin Reid Jun 9 '12 at 13:10
That's because I used the wrong rule :O. Fixed, and shaved a couple of chars. – walpen Jun 9 '12 at 13:59

8086 Machine code - 30 bytes.

NOTE: This is not my code and should not be accepted as an answer. I found this while working on a different CG problem to emulate an 8086 CPU. The included text file credits David Stafford, but that's the best I could come up with.

I'm posting this because it's clever, short, and I thought you'd want to see it.

It makes use of overlapping opcodes to pack more instructions in a smaller space. Amazingly clever. Here is the machine code:

B0 13 CD 10 B3 03 BE A0 A0 8E DE B9 8B 0C 32 28 88 AC C2 FE 4E 75 F5 CD 16 87 C3 CD 10 C3

A straight-up decode looks like this:

0100: B0 13              mov AL, 13h
0102: CD 10              int 10h
0104: B3 03              mov BL, 3h
0106: BE A0 A0           mov SI, A0A0h
0109: 8E DE              mov  DS, SI
010B: B9 8B 0C           mov CX, C8Bh
010E: 32 28              xor  CH, [BX+SI]
0110: 88 AC C2 FE        mov  [SI+FEC2h], CH
0114: 4E                 dec SI
0115: 75 F5              jne/jnz -11

When run, when the jump at 0x0115 happens, notice it jumps back to 0x010C, right into the middle of a previous instruction:

0100: B0 13              mov AL, 13h
0102: CD 10              int 10h
0104: B3 03              mov BL, 3h
0106: BE A0 A0           mov SI, A0A0h
0109: 8E DE              mov  DS, SI
010B: B9 8B 0C           mov CX, C8Bh
010E: 32 28              xor  CH, [BX+SI]
0110: 88 AC C2 FE        mov  [SI+FEC2h], CH
0114: 4E                 dec SI
0115: 75 F5              jne/jnz -11
010C: 8B 0C              mov  CX, [SI]
010E: 32 28              xor  CH, [BX+SI]
0110: 88 AC C2 FE        mov  [SI+FEC2h], CH
0114: 4E                 dec SI
0115: 75 F5              jne/jnz -11
010C: 8B 0C              mov  CX, [SI]

Brilliant! Hope you guys don't mind me sharing this. I know it's not an answer per se, but it's of interest to the challenge.

Here it is in action:

-

PostScript, 120 chars

-7 -4 moveto
14 0 rlineto
7{true upath dup
2{120 rotate uappend}repeat[2 0 0 2 7 4]concat}repeat
matrix setmatrix
stroke

Ghostscript output:

This is drawing the figure by recursively tripling what's already drawn.

First step is drawing a line. The line is saved as a userpath, then the userpath is added two more times after rotating 120 degrees each time. [2 0 0 2 7 4]concat moves the "rotation point" to the center of the next big white "center triangle" that is to be enclosed by replications of the triangle we already have. Here, we're back to step 1 (creating a upath that's tripled by rotation).

The number of iterations is controlled by the first number in line 3.

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+1 Very nice. I had no idea upath could be used like that. – luser droog Jul 20 '12 at 6:28
Hey, you've got the rep to add that image now! – luser droog Jan 1 '13 at 7:01
@luserdroog: That's right (even partly thanks to you)! – Thomas W. Jan 1 '13 at 15:21

J (9 characters)

Easily the ugliest, you really need to squint to see the output ;)

2|!/~i.32

produces the output

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1
0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1
0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1
0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1
0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1
0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1
0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1
0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1
0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

of course you can display it graphically:

viewmat 2|!/~i.32

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how the... what? – acolyte Apr 3 '13 at 17:16
The code exploits the property of Pascal's triangle that if you colour all the odd (even) numbers black (white), you end up with the Sierpinski triangle. (see this image). i.32 generates the list 0 1 2 ... 31. Then !/~ computes the binomial coefficients of each element in the list against itself, i.e. produces a 32 x 32 matrix which has Pascal's triangle embedded in it. Then 2| is simply each element in this matrix mod 2, producing Sierpinski's triangle. – Mark Allen Apr 7 '13 at 17:55

APL, 37 32 (28 23)

Upright triangle (37 32-char)

({((-1⌷⍴⍵)⌽⍵,∊⍵)⍪⍵,⍵}⍣⎕)1 2⍴'/\'

Explanation

• 1 2⍴'/\': Create a 1×2 character matrix /\
• {((-1⌷⍴⍵)⌽⍵,∊⍵)⍪⍵,⍵}: A function that pad the right argument on both sides with blanks to create a matrix double as wide, then laminates the right argument itself doubled onto the bottom.
E.g. /\ would become
/\
/\/\
• ⍣⎕: Recur the function (user input) times.

Example output

/\
/\/\
/\  /\
/\/\/\/\
/\      /\
/\/\    /\/\
/\  /\  /\  /\
/\/\/\/\/\/\/\/\
/\              /\
/\/\            /\/\
/\  /\          /\  /\
/\/\/\/\        /\/\/\/\
/\      /\      /\      /\
/\/\    /\/\    /\/\    /\/\
/\  /\  /\  /\  /\  /\  /\  /\
/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\

Skewed Triangle (28 23-char)

({(⍵,∊⍵)⍪⍵,⍵}⍣⎕)1 1⍴'○'

Explaination

• 1 1⍴'○': Create a 1×1 character matrix
• {(⍵,∊⍵)⍪⍵,⍵}: A function that pad the right argument on the right with blanks to create a matrix double as wide, then laminates the right argument itself doubled onto the bottom.
E.g. would become
○
○○
• ⍣⎕: Recur the function (user input) times.

Example output

○
○○
○ ○
○○○○
○   ○
○○  ○○
○ ○ ○ ○
○○○○○○○○
○       ○
○○      ○○
○ ○     ○ ○
○○○○    ○○○○
○   ○   ○   ○
○○  ○○  ○○  ○○
○ ○ ○ ○ ○ ○ ○ ○
○○○○○○○○○○○○○○○○
-

Logo, 75 characters

59 characters for just the first function, the second one calls the first with the size and the depth/number of iterations. So you could just call the first function from the interpreter with the command: e 99 5, or whatever size you want to output

to e :s :l
if :l>0[repeat 3[e :s/2 :l-1 fd :s rt 120]]
end
to f
e 99 5
end
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+1 I've read about Logo. What interpreter are you using? ... Logo might be a natural fit for my l-system challenge. – luser droog Feb 28 '13 at 2:50
If you just remove the to f and end around e 99 5, you have a complete runnable program in fewer characters. Also, in UCBLogo (though not other versions) you can lose the colons on the variables to save more chars. – Mark Reed Jan 21 '14 at 3:38

matlab 56

v=[1;-1;j];plot(filter(1,[1,-.5],v(randi(3,1,1e4))),'.')

-

Python (75)

I'm two years late to the party, but I'm surprised that nobody has taken this approach yet

from pylab import*
x=[[1,1],[1,0]]
for i in'123':x=kron(x,x)
imsave('a',x)

Uses the Kronecker product to replace a matrix by multiple copies of itself.

I could save two chars by using x=kron(x,x);x=kron(x,x) in line three to get a 16x16 pixel image with three visible levels or add another char to the iterator and end up with a 2^16 x 2^16 = 4.3 Gigapixel image and 15 triangle levels.

-

Mathematica 67

ListPlot@NestList[(#+RandomChoice@{{0,0},{2,0},{1,2}})/2&,{0,0},8!]

Mathematica 92

Graphics@Polygon@Nest[Join@@(Mean/@#&/@#~Tuples~2~Partition~3&/@#)&,{{{0,0},{2,0},{1,1}}},3]

-

Python (90 chars)

from turtle import*
def l():left(60)
def r():right(60)
def f():forward(1)
def L(n):
if n:n-=1;R(n);l();L(n);l();R(n)
else:f()
def R(n):
if n:n-=1;L(n);r();R(n);r();L(n)
else:f()
l();L(8)

Draw fractal line filling Sierpinsky Triangle

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Postscript, 205 203

[48(0-1+0+1-0)49(11)43(+)45(-)/s{dup
0 eq{exch{[48{1 0 rlineto}49 1 index
43{240 rotate}45{120 rotate}>>exch
exch 1 sub s}forall}ifelse 1 add}>>begin
9 9 moveto(0-1-1)9 s fill

Rewrite using strings and recursion ends up at exactly the same count. But the depth-limitations of the macro-approach are overcome.

Edit: fill is shorter than stroke.

Indented and commented.

%!
[   % begin dictionary
48(0-1+0+1-0) % 0
49(11)        % 1
43(+)         % +
45(-)         % -
/s{ % string recursion-level
dup 0 eq{ % level=0
exch{  % iterate through string
[
48{1 0 rlineto} % 0
49 1 index      % 1
43{240 rotate}  % +
45{120 rotate}  % -
>>exch get exec % interpret turtle command
}forall
}{ % level>0
exch{  % iterate through string
1 sub s    % recurse with level-1
}forall
}ifelse
}
>>begin
9 9 moveto(0-1-1)9 s fill % execute and fill

Adding 0 setlinewidth gives a better impression of how deep this one goes.

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This one's my favorite. – cjfaure Jan 17 '14 at 14:10
There's a way to make it shorter with this external lib that I wrote after the fact and cannot use. :P – luser droog Jan 21 '14 at 3:05

C, 106 chars

i,j;main(){for(;i<32;j>i/2?puts(""),j=!++i:0)
printf("%*s",j++?4:33-i+i%2*2,i/2&j^j?"":i%2?"/__\\":"/\\");}

(It still amuses me that puts("") is the shortest way to output a newline in C.)

Note that you can create larger (or smaller) gaskets by replacing the 32 in the for loop's test with a larger (smaller) power of two, as long as you also replace the 33 in the middle of the printf() with the power-of-two-plus-one.

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Common Lisp, 80 chars

(#1=dotimes(i 32)(#1#(j 32)(princ(if(logtest(- j(ash i -1))i)' 'Δ)))(terpri))

Output:

ΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔ
Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ
ΔΔ  ΔΔ  ΔΔ  ΔΔ  ΔΔ  ΔΔ  ΔΔ  ΔΔ
Δ   Δ   Δ   Δ   Δ   Δ   Δ   Δ
ΔΔΔΔ    ΔΔΔΔ    ΔΔΔΔ    ΔΔΔΔ
Δ Δ     Δ Δ     Δ Δ     Δ Δ
ΔΔ      ΔΔ      ΔΔ      ΔΔ
Δ       Δ       Δ       Δ
ΔΔΔΔΔΔΔΔ        ΔΔΔΔΔΔΔΔ
Δ Δ Δ Δ         Δ Δ Δ Δ
ΔΔ  ΔΔ          ΔΔ  ΔΔ
Δ   Δ           Δ   Δ
ΔΔΔΔ            ΔΔΔΔ
Δ Δ             Δ Δ
ΔΔ              ΔΔ
Δ               Δ
ΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔ
Δ Δ Δ Δ Δ Δ Δ Δ
ΔΔ  ΔΔ  ΔΔ  ΔΔ
Δ   Δ   Δ   Δ
ΔΔΔΔ    ΔΔΔΔ
Δ Δ     Δ Δ
ΔΔ      ΔΔ
Δ       Δ
ΔΔΔΔΔΔΔΔ
Δ Δ Δ Δ
ΔΔ  ΔΔ
Δ   Δ
ΔΔΔΔ
Δ Δ
ΔΔ
Δ
-

J (18 characters)

' *'{~(,,.~)^:9 ,1

Result

*
**
* *
****
*   *
**  **
* * * *
********
*       *
**      **
* *     * *
****    ****
*   *   *   *
**  **  **  **
* * * * * * * *
****************
*               *
**              **
* *             * *
****            ****
*   *           *   *
**  **          **  **
* * * *         * * * *
********        ********
*       *       *       *
**      **      **      **
* *     * *     * *     * *
****    ****    ****    ****
*   *   *   *   *   *   *   *
**  **  **  **  **  **  **  **
* * * * * * * * * * * * * * * *
********************************
-

Python (215 209)

Uses the Chaos Theory method of generating Sierpinski's Triangle.

import random as r,pygame as p
d=p.display
x=99;X=49;y=x,x
s=d.set_mode(y)
c=[X,X]
P=(X,0),(0,x),y
while 1:
a=r.choice(P)
for i in 0,1:c[i]=(c[i]+a[i])/2
p.draw.rect(s,[x]*3,p.Rect(c[0],c[1],2,2))
d.flip()
-

JavaScript (70 chars):

for(y=32;y--;){for(s="",x=32;x--;)s+=(x-y/2)&y?" ":"o";console.log(s)}

Using HTML guy's method. This feels like cheating, though. He gets the thread.

oo
oo
o oo
oooo
o   oo
oo  oo
o o o oo
oooooooo
o       oo
oo      oo
o o     o oo
oooo    oooo
o   o   o   oo
oo  oo  oo  oo
o o o o o o o oo
oooooooooooooooo
o               oo
oo              oo
o o             o oo
oooo            oooo
o   o           o   oo
oo  oo          oo  oo
o o o o         o o o oo
oooooooo        oooooooo
o       o       o       oo
oo      oo      oo      oo
o o     o o     o o     o oo
oooo    oooo    oooo    oooo
o   o   o   o   o   o   o   oo
oo  oo  oo  oo  oo  oo  oo  oo
o o o o o o o o o o o o o o o oo
oooooooooooooooooooooooooooooooo
-