Rectangular pseudo-fractal

Question

Goal

The program's goal is to draw an ASCII art rectangle repeatedly doubled in size, alternating horizontally and vertically. Each time the rectangle doubles in size, the extra area is represented by a different character and the previous areas remain unchanged. The two smallest sections contain one character each and may be in any corner.

The program accepts a single integer as input, defining the number of sections the full rectangle contains.

No other external resources or inputs are allowed.

Sample input and output

10

ABDDFFFFHHHHHHHHJJJJJJJJJJJJJJJJ
CCDDFFFFHHHHHHHHJJJJJJJJJJJJJJJJ
EEEEFFFFHHHHHHHHJJJJJJJJJJJJJJJJ
EEEEFFFFHHHHHHHHJJJJJJJJJJJJJJJJ
GGGGGGGGHHHHHHHHJJJJJJJJJJJJJJJJ
GGGGGGGGHHHHHHHHJJJJJJJJJJJJJJJJ
GGGGGGGGHHHHHHHHJJJJJJJJJJJJJJJJ
GGGGGGGGHHHHHHHHJJJJJJJJJJJJJJJJ
IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ
IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ
IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ
IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ
IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ
IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ
IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ
IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ

Selection criteria

The shortest code in a week wins the points.

@marinus What is special about powers of 4? Maybe you misunderstood the example? — Kendall Frey, Jan 31, 2014 at 1:28
@KendallFrey, no, there just needs to be n unique printable characters. — Hand-E-Food, Jan 31, 2014 at 2:48

Tobia · Accepted Answer · 2014-02-03 20:32:01Z

APL, 25 chars/bytes*

{⍉⍣⍵⊃{a,⍺⍴⍨⍴a←⍉⍪⍵}/⌽⍵↑⎕A}

Exploded view

{                   ⍵↑⎕A}   ⍝ take the first ⍵ letters
    ⊃{           }/⌽        ⍝ fold over them, using the first one as initial accum. value
            a←⍉⍪⍵           ⍝    ensure the accum. is a table, transpose it and call it 'a'
        ⍺⍴⍨⍴                ⍝    make a table as large as 'a' filled with the next letter
      a,                    ⍝    append it to the right of 'a' and loop as new accumulator
 ⍉⍣⍵                        ⍝ transpose the result as many times as the original ⍵ number

Examples

      {⍉⍣⍵⊃{a,⍺⍴⍨⍴a←⍉⍪⍵}/⌽⍵↑⎕A}¨⍳8
A AB  AB  ABDD  ABDD  ABDDFFFF  ABDDFFFF  ABDDFFFFHHHHHHHH
      CC  CCDD  CCDD  CCDDFFFF  CCDDFFFF  CCDDFFFFHHHHHHHH
                EEEE  EEEEFFFF  EEEEFFFF  EEEEFFFFHHHHHHHH
                EEEE  EEEEFFFF  EEEEFFFF  EEEEFFFFHHHHHHHH
                                GGGGGGGG  GGGGGGGGHHHHHHHH
                                GGGGGGGG  GGGGGGGGHHHHHHHH
                                GGGGGGGG  GGGGGGGGHHHHHHHH
                                GGGGGGGG  GGGGGGGGHHHHHHHH

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
^{*: APL can be written in its own (legacy) single-byte charset that maps APL symbols to the upper 128 byte values. Therefore, for the purpose of scoring, a program of N chars that only uses ASCII characters and APL symbols can be considered to be N bytes long.}

Timwi · Accepted Answer · 2014-02-02 17:04:02Z

9

GolfScript, 30 characters

~(,[0`]{{[49+]1$,*+}+%zip}@/n*

Example (run online):

edited Feb 2, 2014 at 17:04

Timwi

12.9k3 gold badges44 silver badges66 bronze badges

answered Jan 31, 2014 at 6:47

Howard

23.5k2 gold badges43 silver badges81 bronze badges

\$\begingroup\$ This produces the wrong output for even numbers, such as the one in the question... \$\endgroup\$
– Timwi
Feb 1, 2014 at 15:24
\$\begingroup\$ @Timwi I just tested it an it works for me. The output is transposed but orientation wasn't specified in the question. \$\endgroup\$
– Howard
Feb 2, 2014 at 11:28
\$\begingroup\$ Alright, I guess I was too strict then :) \$\endgroup\$
– Timwi
Feb 2, 2014 at 17:03
\$\begingroup\$ @Howard Hm, that's how I understand "and the previous areas remain unchanged". He says the first two characters may be in any corner, but he doesn't say the orientation may change. \$\endgroup\$
– Martin Ender
Jan 30, 2015 at 13:49

Add a comment |

score 7 · Accepted Answer · 2014-02-02 22:44:39Z

Python 2.7 - 85 103

This uses the zip(*s) syntax to continually transpose the list. Big thanks to Daniel for his tip that shaved 12 characters! Then shaved a few more by using numbers instead of letters.

s=[]
for i in range(input()):x=1<<i/2;s=zip(*s+[chr(65+i)*x]*x)
for i in s:print''.join(i)

Also, this uses 1<<x rather than 2**x as bit shift has lower(?) precedence. Observe:

>>> 1<<(2*3)
64
>>> 1<<2*3
64
>>> 2**2*3
12
>>> 2**(2*3)
64

And some output:

10
01335555777777779999999999999999
22335555777777779999999999999999
44445555777777779999999999999999
44445555777777779999999999999999
66666666777777779999999999999999
66666666777777779999999999999999
66666666777777779999999999999999
66666666777777779999999999999999
88888888888888889999999999999999
88888888888888889999999999999999
88888888888888889999999999999999
88888888888888889999999999999999
88888888888888889999999999999999
88888888888888889999999999999999
88888888888888889999999999999999
88888888888888889999999999999999

Nice. You can shorten it a bit with for i in s:print''.join(i). — Daniel Lubarov, Feb 2, 2014 at 9:03

Paul Prestidge · Accepted Answer · 2014-01-31 04:19:09Z

Ruby, 88

Reads N from standard input.

s=[?A]
66.upto(64+gets.to_i){|i|x=i.chr*y=s.size;i%2<1?s.map!{|r|r+x}:s+=[x*2]*y}
puts s

Example Usage for N=8:

echo 8 | rectangular-pseudo-fractal.rb

Output:

ABDDFFFFHHHHHHHH
CCDDFFFFHHHHHHHH
EEEEFFFFHHHHHHHH
EEEEFFFFHHHHHHHH
GGGGGGGGHHHHHHHH
GGGGGGGGHHHHHHHH
GGGGGGGGHHHHHHHH
GGGGGGGGHHHHHHHH

N=10

echo 10 | rectangular-pseudo-fractal.rb

Output:

ABDDFFFFHHHHHHHHJJJJJJJJJJJJJJJJ
CCDDFFFFHHHHHHHHJJJJJJJJJJJJJJJJ
EEEEFFFFHHHHHHHHJJJJJJJJJJJJJJJJ
EEEEFFFFHHHHHHHHJJJJJJJJJJJJJJJJ
GGGGGGGGHHHHHHHHJJJJJJJJJJJJJJJJ
GGGGGGGGHHHHHHHHJJJJJJJJJJJJJJJJ
GGGGGGGGHHHHHHHHJJJJJJJJJJJJJJJJ
GGGGGGGGHHHHHHHHJJJJJJJJJJJJJJJJ
IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ
IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ
IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ
IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ
IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ
IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ
IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ
IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ

@LegoStormtroopr added some examples, it's the exact same format as the question though. — Paul Prestidge, Jan 31, 2014 at 4:00

barbermot · Accepted Answer · 2014-02-01 13:07:48Z

4

J, 57 43

(,`,.@.(=/@$@[)$${&a.@(66+2&^.@#@,)^:)1$'A'

Examples:

5 (,`,.@.(=/@$@[)$${&a.@(66+2&^.@#@,)^:)1$'A'
ABDDFFFF
CCDDFFFF
EEEEFFFF
EEEEFFFF

7 (,`,.@.(=/@$@[)$${&a.@(66+2&^.@#@,)^:)1$'A'
ABDDFFFFHHHHHHHH
CCDDFFFFHHHHHHHH
EEEEFFFFHHHHHHHH
EEEEFFFFHHHHHHHH
GGGGGGGGHHHHHHHH
GGGGGGGGHHHHHHHH
GGGGGGGGHHHHHHHH
GGGGGGGGHHHHHHHH

edited Feb 1, 2014 at 13:07

answered Jan 31, 2014 at 9:36

barbermot

1413 bronze badges

\$\begingroup\$ C and D both extend horizontally. They should alternate horizontally and vertically. \$\endgroup\$
– Hand-E-Food
Feb 1, 2014 at 11:46
\$\begingroup\$ @Hand-E-Food you're right. Thanks for pointing that out. I've fixed code (and post). \$\endgroup\$
– barbermot
Feb 1, 2014 at 13:08

Add a comment |

knedlsepp · Accepted Answer · 2015-01-30 21:08:44Z

4

MATLAB, 86 Characters

My shortest try in MATLAB, pimped by @flawr (twice!):

function M=f(n)
M='';
if n
M=cat(mod(n,2)+1,f(n-1),64+n*ones(2.^fix(n/2-[.5,1])));
end

Example output:

>> disp(f(7))
ACEEGGGG
BCEEGGGG
DDEEGGGG
DDEEGGGG
FFFFGGGG
FFFFGGGG
FFFFGGGG
FFFFGGGG

edited Jan 30, 2015 at 21:08

answered Feb 2, 2014 at 22:21

knedlsepp

2861 silver badge7 bronze badges

\$\begingroup\$ This will save you some bytes: function M=f(n) M=''; if n M=cat(mod(n,2)+1,f(n-1),64+n*ones(2.^fix([n-1,n-2]/2))); end \$\endgroup\$
– flawr
Jan 30, 2015 at 20:41
\$\begingroup\$ @flawr: Oh! Obviously! \$\endgroup\$
– knedlsepp
Jan 30, 2015 at 21:02
\$\begingroup\$ Save another byte by replacing the argument of fix with fix(n/2-[.5,1]) PS: Really nice solution with cat, did not know about this use where you can choose the dimension=) \$\endgroup\$
– flawr
Jan 30, 2015 at 21:05
\$\begingroup\$ @flawr: It seems to me I'm quite wasteful. ;-) \$\endgroup\$
– knedlsepp
Jan 30, 2015 at 21:06
\$\begingroup\$ I just noticed you're new here, so welcome on codegolf.SE, it's nice to have a.) some more matlab junkies, b.) german speakers here (I assume)! \$\endgroup\$
– flawr
Jan 30, 2015 at 21:36

| Show 2 more comments

nyi · Accepted Answer · 2014-01-31 19:16:27Z

q [73 chars]

{"c"$64+{n:x 0;m:x 1;if[2>n;m:(),m];(o;$[n-2*n div 2;,';,][m;(#m;#m 0)#o:n+1])}/[x-1;(1;1)]1}

example

10
"ABDDFFFFHHHHHHHHJJJJJJJJJJJJJJJJ"
"CCDDFFFFHHHHHHHHJJJJJJJJJJJJJJJJ"
"EEEEFFFFHHHHHHHHJJJJJJJJJJJJJJJJ"
"EEEEFFFFHHHHHHHHJJJJJJJJJJJJJJJJ"
"GGGGGGGGHHHHHHHHJJJJJJJJJJJJJJJJ"
"GGGGGGGGHHHHHHHHJJJJJJJJJJJJJJJJ"
"GGGGGGGGHHHHHHHHJJJJJJJJJJJJJJJJ"
"GGGGGGGGHHHHHHHHJJJJJJJJJJJJJJJJ"
"IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ"
"IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ"
"IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ"
"IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ"
"IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ"
"IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ"
"IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ"
"IIIIIIIIIIIIIIIIJJJJJJJJJJJJJJJJ"

3
"AB"
"CC"

6
"ABDDFFFF"
"CCDDFFFF"
"EEEEFFFF"
"EEEEFFFF"

Timwi · Accepted Answer · 2014-02-01 16:54:26Z

Sclipting, 59 characters

❶塊갠分감⓶左貶終辦감標가⓺貶⓹開上❶❶貶雙是不⒉갠乘⒉終가①上뀀❷②갠分小是增終❸⓷另要감右⓶갠加⓶終丟字⓶終丟겠終

(This program could be quite a bit shorter if I had instructions for base-2 logarithm, but I don’t, so I do it manually with a loop.)

Annotated code

n is the input.

❶ | n n

f = i => (1 << (i/2)) - 1;
塊갠分감⓶左貶終 | n n f

w = f(n);
辦 | n w f

d = 1;
감 | n w f d

s = "";
標 | n w f d M [s]

for (y in [0..f(n-1)])
가⓺貶⓹開上 | w d M [s] y

    if ((y & (y-1)) == 0) d *= 2;
    ❶❶貶雙是不⒉갠乘⒉終 | w d M [s] y

    for (x in [0..w])
    가①上 | w d M [s] y x

        c = 64; // '@'
        뀀 | w d M [s] y x c

        if (x < d/2) c++;
        ❷②갠分小是增終 | w d M [s] y x c

        a = x | y;
        ❸⓷另 | w d M [s] y c a

        while (a > 0) { a >>= 1; c += 2; }
        要감右⓶갠加⓶終丟 | w d M [s] y c

        s += (char) c;
        字⓶ | w d M [s] y
    終丟 | w d M [s]

    s += "\n"
    겠 | w d M [s]
終

Output

For n = 6:

ABDDFFFF
CCDDFFFF
EEEEFFFF
EEEEFFFF

Of course you can change 뀀 (@) to any other base character, e.g. with 글 (space) and n = 7:

!"$$&&&&
##$$&&&&
%%%%&&&&
%%%%&&&&
''''''''
''''''''
''''''''
''''''''

The highest number that doesn’t make the program longer is 믰 (= 255), which gives us (n = 8 this time):

Āāăăąąąąćććććććć
ĂĂăăąąąąćććććććć
ĄĄĄĄąąąąćććććććć
ĄĄĄĄąąąąćććććććć
ĆĆĆĆĆĆĆĆćććććććć
ĆĆĆĆĆĆĆĆćććććććć
ĆĆĆĆĆĆĆĆćććććććć
ĆĆĆĆĆĆĆĆćććććććć

If we make the program 1 character longer, e.g. use 냟및 (= \u4DFF) and n = 9, we get:

一丁七七丅丅丅丅万万万万万万万万
丂丂七七丅丅丅丅万万万万万万万万
丄丄丄丄丅丅丅丅万万万万万万万万
丄丄丄丄丅丅丅丅万万万万万万万万
丆丆丆丆丆丆丆丆万万万万万万万万
丆丆丆丆丆丆丆丆万万万万万万万万
丆丆丆丆丆丆丆丆万万万万万万万万
丆丆丆丆丆丆丆丆万万万万万万万万
丈丈丈丈丈丈丈丈丈丈丈丈丈丈丈丈
丈丈丈丈丈丈丈丈丈丈丈丈丈丈丈丈
丈丈丈丈丈丈丈丈丈丈丈丈丈丈丈丈
丈丈丈丈丈丈丈丈丈丈丈丈丈丈丈丈
丈丈丈丈丈丈丈丈丈丈丈丈丈丈丈丈
丈丈丈丈丈丈丈丈丈丈丈丈丈丈丈丈
丈丈丈丈丈丈丈丈丈丈丈丈丈丈丈丈
丈丈丈丈丈丈丈丈丈丈丈丈丈丈丈丈

Hand-E-Food · Accepted Answer · 2014-02-04 00:15:23Z

C#, 239 185 182 180 bytes

C# has nothing on the less verbose languages.

using C=System.Console;
class P{
    static void Main(string[]a){
        for(int x,i,n=int.Parse(a[0]);n-->0;C.CursorTop=0)
            for(i=1<<n,x=1<<n/2+n%2;i-->0;)
                C.Write((char)(n+33)+(i%x<1?"\n":""));
    }
}

Output, characters chosen for prettiness:

!"$$&&&&((((((((****************
##$$&&&&((((((((****************
%%%%&&&&((((((((****************
%%%%&&&&((((((((****************
''''''''((((((((****************
''''''''((((((((****************
''''''''((((((((****************
''''''''((((((((****************
))))))))))))))))****************
))))))))))))))))****************
))))))))))))))))****************
))))))))))))))))****************
))))))))))))))))****************
))))))))))))))))****************
))))))))))))))))****************
))))))))))))))))****************

Not sure how you counted, but I counted 184. You can save two characters by ditching the braces from the outer for loop, making 182. — Bob, Jan 31, 2014 at 6:33

Community · Accepted Answer · 2020-06-17 09:04:33Z

PERL, 122 chars

$N=<>;$x=$r=1;do{$_=chr$a+++65;$s=$x;$o=$_ x$s;$o.=$_++x$s,$s*=2while$N+65>ord++$_;print"$o\n"x$r;$r=$x;$x*=2}while++$a<$N

with added whitespace:

$N=<>;
$x=$r=1;
do{
    $_=chr$a+++65;
    $s=$x;
    $o=$_ x$s;
    $o.=$_++x$s,$s*=2 
        while $N+65>ord++$_;
    print "$o\n"x$r;
    $r=$x;
    $x*=2
} while++$a<$N

Output:

$ echo 8 | perl pseudo-fractal.pl
ABDDFFFFHHHHHHHH
CCDDFFFFHHHHHHHH
EEEEFFFFHHHHHHHH
EEEEFFFFHHHHHHHH
GGGGGGGGHHHHHHHH
GGGGGGGGHHHHHHHH
GGGGGGGGHHHHHHHH
GGGGGGGGHHHHHHHH

Community · Accepted Answer · 2020-06-17 09:04:33Z

PERL, 94 81 chars

$N=$_;$_=$:=A;$h=1;++$i%2?s/$/$:x$h/gem:($_.=($/.$:x2x$h)x$h,$h*=2)while++$:,--$N

It constructs the fractal iteratively letter by letter, adding new rows and columns and rows and columns... Uses simple string operations to do that. Note that I am abusing standard variable instead of letter one to allow for syntax sugar (like omitting spaces - $:x2 etc.)

With added whitespace and comments:

$N=$_;
$_=$:=A;                    # $: is current letter
$h=1;

++$i%2? 
s/$/$:x$h/gem:              # every odd run - add "columns"
($_.=($/.$:x2x$h)x$h,$h*=2) # every even run - add "rows"
while++$:,--$N              # iterate over letters

Some output:

$ echo 8 | perl -p pseudo-fractal.fill.pl.5a5
ABDDFFFFHHHHHHHH
CCDDFFFFHHHHHHHH
EEEEFFFFHHHHHHHH
EEEEFFFFHHHHHHHH
GGGGGGGGHHHHHHHH
GGGGGGGGHHHHHHHH
GGGGGGGGHHHHHHHH
GGGGGGGGHHHHHHHH

Timwi · Accepted Answer · 2014-02-01 21:15:39Z

Sclipting, 45 characters

가⓶貶上倘감雙⓶壹長⓸講增字⓶復⓷是標⓷各①合終并不⓶梴❸⓶疊合終不뀐標뀐并終終⓶丟各겠終

This solution works completely differently from the other Sclipting solution. It’s much more boring, but it’s shorter...

Annotated

for i in [0..n-1]
가⓶貶上
    if (i != 0)
    倘
        i &= 1
        감雙
        e = list[0].Length
        ⓶壹長
        c = ((char) (c[0] + 1)).Repeat(e)
        ⓸講增字⓶復
        if (i)
        ⓷是
            concatenate c onto every element of list
            標⓷各①合終并
        else
        不
            concatenate c.Repeat(list.Length) onto list
            ⓶梴❸⓶疊合
        終
    else (i.e., i == 0)
    不
        c = "A"
        뀐
        list = ["A"]
        標뀐并
    終
終
concatenate "\n" to every element in list
⓶丟各겠終

Teun Pronk · Accepted Answer · 2014-02-04 16:09:46Z

Delphi 348 || 449 with indent

Without indent

var inp,j,i,x: integer;s:string;L:TStringlist;begin L:=TStringList.Create;readln(s);inp:=StrToIntDef(s,4);if inp<4then inp:=4;s:='';l.Add('AB');for I:=2to inp-1do begin j:=Length(L[0]);if i mod 2=0then for x:=0to L.Count-1do L.Add(s.PadLeft(j,Chr(65+i)))else for x:=0to L.Count-1do L[x]:=L[x]+s.PadLeft(j,Chr(65+i));end;Write(L.GetText);readln;end.

With indent

var
  inp,j,i,x: integer;
  s:string;
  L:TStringlist;
begin
  L:=TStringList.Create;
  readln(s);
  inp:=StrToIntDef(s,4);
  if inp<4then inp:=4;
  s:='';
  l.Add('AB');

  for I:=2to inp-1do
  begin
    j:=Length(L[0]);
    if i mod 2=0then
      for x:=0to L.Count-1do L.Add(s.PadLeft(j,Chr(65+i)))
    else
      for x:=0to L.Count-1do
        L[x]:=L[x]+s.PadLeft(j,Chr(65+i));
  end;
  Write(L.GetText);
  readln;
end.

Martin Ender · Accepted Answer · 2015-01-30 14:02:29Z

CJam, 30 (23) bytes

CJam is a few months younger than this challenge so it's not eligible for the green checkmark.

l~(Sa1${{_,I'!+*+}%z}fI\{z}*N*

Test it here.

The OP clarified in a comment that any set of unique printable characters is allowed, so I'm just taking the printable ASCII characters from the start (with a space in the corner, ! next and so on).

If the orientation may change between even and odd inputs (which I don't think, but that's what the GolfScript submission does), I can do it in 25 bytes:

S]l~({{_,I'!+*+}%z}fIN*

The idea is really simple: start with a grid containing a space, and then N-1 times transpose it and double all lines with the next character.

For the long version, at the end I also transpose again N-1 times in order to guarantee a consistent orientation.

Razetime · Accepted Answer · 2020-11-05 16:35:09Z

0

Canvas, 22 bytes

A；╷｛┐⤢ｌ；Ｌ┘¹Ｚｊ＠┘┘＊＋｝╶［⤢

Try it here!

answered Nov 5, 2020 at 16:35

Razetime

27k3 gold badges26 silver badges71 bronze badges

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Stack Exchange Network

Rectangular pseudo-fractal

Goal

Sample input and output

Selection criteria

15 Answers 15

APL, 25 chars/bytes*

GolfScript, 30 characters

Python 2.7 - 85 103

Ruby, 88

q [73 chars]

example

Sclipting, 59 characters

Annotated code

Output

C#, 239 185 182 180 bytes

PERL, 122 chars

PERL, 94 81 chars

Sclipting, 45 characters

Annotated

Delphi 348 || 449 with indent

Without indent

With indent

CJam, 30 (23) bytes

Canvas, 22 bytes

Your Answer

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code-golf
ascii-art
or ask your own question.

Hot Network Questions

Rectangular pseudo-fractal

Goal

Sample input and output

Selection criteria

15 Answers 15

APL, 25 chars/bytes*

GolfScript, 30 characters

Python 2.7 - 85 103

Ruby, 88

q [73 chars]

example

Sclipting, 59 characters

Annotated code

Output

C#, 239 185 182 180 bytes

PERL, 122 chars

PERL, 94 81 chars

Sclipting, 45 characters

Annotated

Delphi 348 || 449 with indent

Without indent

With indent

CJam, 30 (23) bytes

Canvas, 22 bytes

Your Answer

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or ask your own question.