Sierpinski Carpets

Question

Who doesn't love a good fractal? The Sierpinski Carpet is a classic example of a fractal.

To complete this task, you will be required to generate a carpet of type \$n\$ and print the resulting image to the stdout (see example below for formatting) \$n\$, representing the level carpet. Steps can be found on this Wikipedia article.

This value will be taken from stdin or equivalent.

For example, an input of 4 would produce a level 4 carpet:

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
■   ■ ■   ■ ■   ■ ■   ■ ■   ■ ■   ■ ■   ■ ■   ■ ■   ■
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
■ ■ ■       ■ ■ ■ ■ ■ ■       ■ ■ ■ ■ ■ ■       ■ ■ ■
■   ■       ■   ■ ■   ■       ■   ■ ■   ■       ■   ■
■ ■ ■       ■ ■ ■ ■ ■ ■       ■ ■ ■ ■ ■ ■       ■ ■ ■ 
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
■   ■ ■   ■ ■   ■ ■   ■ ■   ■ ■   ■ ■   ■ ■   ■ ■   ■
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
■ ■ ■ ■ ■ ■ ■ ■ ■                   ■ ■ ■ ■ ■ ■ ■ ■ ■
■   ■ ■   ■ ■   ■                   ■   ■ ■   ■ ■   ■
■ ■ ■ ■ ■ ■ ■ ■ ■                   ■ ■ ■ ■ ■ ■ ■ ■ ■
■ ■ ■       ■ ■ ■                   ■ ■ ■       ■ ■ ■
■   ■       ■   ■                   ■   ■       ■   ■
■ ■ ■       ■ ■ ■                   ■ ■ ■       ■ ■ ■ 
■ ■ ■ ■ ■ ■ ■ ■ ■                   ■ ■ ■ ■ ■ ■ ■ ■ ■
■   ■ ■   ■ ■   ■                   ■   ■ ■   ■ ■   ■
■ ■ ■ ■ ■ ■ ■ ■ ■                   ■ ■ ■ ■ ■ ■ ■ ■ ■ 
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
■   ■ ■   ■ ■   ■ ■   ■ ■   ■ ■   ■ ■   ■ ■   ■ ■   ■
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
■ ■ ■       ■ ■ ■ ■ ■ ■       ■ ■ ■ ■ ■ ■       ■ ■ ■
■   ■       ■   ■ ■   ■       ■   ■ ■   ■       ■   ■
■ ■ ■       ■ ■ ■ ■ ■ ■       ■ ■ ■ ■ ■ ■       ■ ■ ■ 
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
■   ■ ■   ■ ■   ■ ■   ■ ■   ■ ■   ■ ■   ■ ■   ■ ■   ■
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

An input of 3 would produce a level 3 carpet:

■ ■ ■ ■ ■ ■ ■ ■ ■
■   ■ ■   ■ ■   ■
■ ■ ■ ■ ■ ■ ■ ■ ■
■ ■ ■       ■ ■ ■
■   ■       ■   ■
■ ■ ■       ■ ■ ■
■ ■ ■ ■ ■ ■ ■ ■ ■
■   ■ ■   ■ ■   ■
■ ■ ■ ■ ■ ■ ■ ■ ■

An input of 2 would produce a level 2 carpet:

■ ■ ■
■   ■ 
■ ■ ■

And a input of 1 would produce a level 1 carpet (just a square):

■

Note the spaces between the columns.

The idea is to use the ■ character for blocks and space for gaps

The line starts with the ■ character.

As always, the smallest solution wins by character count (due to non-standard characters).

Alternatively, # can be used instead of the ■ character in languages that do not support Unicode.

Nice question, but do they have to be those characters? those aren't ASCII, which rules out a lot of languages. Can they be just space and X or O? I actually think your image would look better with spaces instead of empty squares. — Level River St, Commented Oct 21, 2014 at 17:32
What if your language does not support unicode output, is there also an ASCII alternative? — flawr, Commented Oct 21, 2014 at 17:35

Martin Ender · Accepted Answer · 2014-10-22 20:51:47Z

CJam, 38 37 31 30 28 characters

Oh well, we're counting by characters, so let's do some Unicode unpacking:

"B胷맋풽巓뱖ᮨ㣙¬䙧੥墱륋청"2G#b129b:c~

Test it here. Just put the desired level into the input field.

Explanation

After base conversion, this is

3li(#,{3b1f&2b}%_f{f{&S9632c?S}N}

which is the same as the following, just with the Unicode character written as 9632c:

3li(#,{3b1f&2b}%_f{f{&S'■?S}N}

This code is based on the following observation: if we look at the coordinates (x,y) of each cell, then we get an empty cell, whenever both x and y have a 1 at the same position in their base-3 representation. If you think about it, the small-scale repeating pattern is the significant base-3 digit, then the next more significant digit governs the next larger-scale repetition and so on.

3                                      "Push a 3 on the stack.";
 li(                                   "Read input, convert to integer, decrement.";
    #                                  "Raise to that power. This yields the dimensions.";
     ,                                 "Turn into a range array.";
      {       }%                       "Map the block onto the array.";
       3b                              "Convert to base 3.";
         1f&                           "Bit-wise AND each digit with 1.";
            2b                         "Convert to base 2.";
                _                      "Duplicate this list.";
                 f{          }         "Map this block onto one list, with the second list
                                        as an additional parameter.";
                   f{      }           "Map this block onto the second list, with the first
                                        list's current element as an additional parameter.";
                                       "I.e. this iterates over all coordinate pairs.";
                     &                 "Bitwise AND to check that the base-3 representations
                                        had a 1 in the same position.";
                      S'■?             "Select the right character.";
                          S            "Push a space.";
                            N          "Push a newline";

The contents of the resulting array are printed automatically.

Thanks to Dennis for shaving off three bytes.

@Dennis I keep forgetting that & produces a truthy/falsy value (or even that it exists). Might be worth a post in the CJam tips for the canonical way to test whether an array contains some element. — Martin Ender, Commented Oct 21, 2014 at 22:58
@Ypnypn Nope. It looks longer but that's because the Unicode characters aren't monospaces. The unpacked code has 30 characters. — Martin Ender, Commented Oct 23, 2014 at 21:43

flawr · Accepted Answer · 2014-10-23 19:24:13Z

Matlab (113)(110)(99)(85)

You can try it here (You'll have to replace input('') with your desired input.)

Now 99 thanks to feersum! And now down to 85 thanks to RTL!

Golfed:

a=ones(3);a(5)=0;c=1;for i=2:input('');c=kron(c,a);end;disp(char(kron(c,[1,0])*3+32))

Ungolfed:

a=ones(3);a(5)=0;c=1;  %creating the template / anchor
for i=2:input('');
    c=kron(c,a);       %recursive iterations
end;
disp(char(kron(c,[1,0])*3+32))

~~d=[c,c]*0; %this is all just for adding the additional spaces d(:,1:2:end)=c; disp(char(d*3+32)); %converting to spaces (32) and # (35)~~

Explanation:

I am abusing the kronecker product for this task. (It is a special product defined for two each arbitrary sized matrices. Example:

A = [1,2] is a 3x2 matrix, B is a nxm matrix.
    [3,4]
    [5,6]

Then

kron(A,B) = [1*B , 2*B] is a 2n x 2m matrix.
            [3*B , 4*B]
            [5*B , 6*B]

So heres an example for n=5 (In the old counting method it's 4);

# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # 
#   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # 
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # 
# # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # 
#   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # 
# # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # 
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # 
#   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # 
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # 
# # # # # # # # #                   # # # # # # # # # # # # # # # # # #                   # # # # # # # # # # # # # # # # # #                   # # # # # # # # # 
#   # #   # #   #                   #   # #   # #   # #   # #   # #   #                   #   # #   # #   # #   # #   # #   #                   #   # #   # #   # 
# # # # # # # # #                   # # # # # # # # # # # # # # # # # #                   # # # # # # # # # # # # # # # # # #                   # # # # # # # # # 
# # #       # # #                   # # #       # # # # # #       # # #                   # # #       # # # # # #       # # #                   # # #       # # # 
#   #       #   #                   #   #       #   # #   #       #   #                   #   #       #   # #   #       #   #                   #   #       #   # 
# # #       # # #                   # # #       # # # # # #       # # #                   # # #       # # # # # #       # # #                   # # #       # # # 
# # # # # # # # #                   # # # # # # # # # # # # # # # # # #                   # # # # # # # # # # # # # # # # # #                   # # # # # # # # # 
#   # #   # #   #                   #   # #   # #   # #   # #   # #   #                   #   # #   # #   # #   # #   # #   #                   #   # #   # #   # 
# # # # # # # # #                   # # # # # # # # # # # # # # # # # #                   # # # # # # # # # # # # # # # # # #                   # # # # # # # # # 
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # 
#   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # 
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # 
# # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # 
#   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # 
# # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # 
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # 
#   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # 
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # 
# # # # # # # # # # # # # # # # # # # # # # # # # # #                                                       # # # # # # # # # # # # # # # # # # # # # # # # # # # 
#   # #   # #   # #   # #   # #   # #   # #   # #   #                                                       #   # #   # #   # #   # #   # #   # #   # #   # #   # 
# # # # # # # # # # # # # # # # # # # # # # # # # # #                                                       # # # # # # # # # # # # # # # # # # # # # # # # # # # 
# # #       # # # # # #       # # # # # #       # # #                                                       # # #       # # # # # #       # # # # # #       # # # 
#   #       #   # #   #       #   # #   #       #   #                                                       #   #       #   # #   #       #   # #   #       #   # 
# # #       # # # # # #       # # # # # #       # # #                                                       # # #       # # # # # #       # # # # # #       # # # 
# # # # # # # # # # # # # # # # # # # # # # # # # # #                                                       # # # # # # # # # # # # # # # # # # # # # # # # # # # 
#   # #   # #   # #   # #   # #   # #   # #   # #   #                                                       #   # #   # #   # #   # #   # #   # #   # #   # #   # 
# # # # # # # # # # # # # # # # # # # # # # # # # # #                                                       # # # # # # # # # # # # # # # # # # # # # # # # # # # 
# # # # # # # # #                   # # # # # # # # #                                                       # # # # # # # # #                   # # # # # # # # # 
#   # #   # #   #                   #   # #   # #   #                                                       #   # #   # #   #                   #   # #   # #   # 
# # # # # # # # #                   # # # # # # # # #                                                       # # # # # # # # #                   # # # # # # # # # 
# # #       # # #                   # # #       # # #                                                       # # #       # # #                   # # #       # # # 
#   #       #   #                   #   #       #   #                                                       #   #       #   #                   #   #       #   # 
# # #       # # #                   # # #       # # #                                                       # # #       # # #                   # # #       # # # 
# # # # # # # # #                   # # # # # # # # #                                                       # # # # # # # # #                   # # # # # # # # # 
#   # #   # #   #                   #   # #   # #   #                                                       #   # #   # #   #                   #   # #   # #   # 
# # # # # # # # #                   # # # # # # # # #                                                       # # # # # # # # #                   # # # # # # # # # 
# # # # # # # # # # # # # # # # # # # # # # # # # # #                                                       # # # # # # # # # # # # # # # # # # # # # # # # # # # 
#   # #   # #   # #   # #   # #   # #   # #   # #   #                                                       #   # #   # #   # #   # #   # #   # #   # #   # #   # 
# # # # # # # # # # # # # # # # # # # # # # # # # # #                                                       # # # # # # # # # # # # # # # # # # # # # # # # # # # 
# # #       # # # # # #       # # # # # #       # # #                                                       # # #       # # # # # #       # # # # # #       # # # 
#   #       #   # #   #       #   # #   #       #   #                                                       #   #       #   # #   #       #   # #   #       #   # 
# # #       # # # # # #       # # # # # #       # # #                                                       # # #       # # # # # #       # # # # # #       # # # 
# # # # # # # # # # # # # # # # # # # # # # # # # # #                                                       # # # # # # # # # # # # # # # # # # # # # # # # # # # 
#   # #   # #   # #   # #   # #   # #   # #   # #   #                                                       #   # #   # #   # #   # #   # #   # #   # #   # #   # 
# # # # # # # # # # # # # # # # # # # # # # # # # # #                                                       # # # # # # # # # # # # # # # # # # # # # # # # # # # 
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # 
#   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # 
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # 
# # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # 
#   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # 
# # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # 
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # 
#   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # 
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # 
# # # # # # # # #                   # # # # # # # # # # # # # # # # # #                   # # # # # # # # # # # # # # # # # #                   # # # # # # # # # 
#   # #   # #   #                   #   # #   # #   # #   # #   # #   #                   #   # #   # #   # #   # #   # #   #                   #   # #   # #   # 
# # # # # # # # #                   # # # # # # # # # # # # # # # # # #                   # # # # # # # # # # # # # # # # # #                   # # # # # # # # # 
# # #       # # #                   # # #       # # # # # #       # # #                   # # #       # # # # # #       # # #                   # # #       # # # 
#   #       #   #                   #   #       #   # #   #       #   #                   #   #       #   # #   #       #   #                   #   #       #   # 
# # #       # # #                   # # #       # # # # # #       # # #                   # # #       # # # # # #       # # #                   # # #       # # # 
# # # # # # # # #                   # # # # # # # # # # # # # # # # # #                   # # # # # # # # # # # # # # # # # #                   # # # # # # # # # 
#   # #   # #   #                   #   # #   # #   # #   # #   # #   #                   #   # #   # #   # #   # #   # #   #                   #   # #   # #   # 
# # # # # # # # #                   # # # # # # # # # # # # # # # # # #                   # # # # # # # # # # # # # # # # # #                   # # # # # # # # # 
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # 
#   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # 
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # 
# # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # 
#   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # #   #       #   # 
# # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # # # #       # # # 
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # 
#   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # #   # 
# # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # # #

Thanks, I didn't notice that the requirements changed. I also updated the code (I just had to change the starting value of the for loop.) — flawr, Commented Oct 21, 2014 at 18:32
Instead of d=zeros(size([c,c])) you should use d=[c c]*0. — feersum, Commented Oct 21, 2014 at 19:48
nice, a suggestion: the variable d can be avoided by abusing kron once more: disp(char(kron(c,[1,0])*3+32));, — RTL, Commented Oct 23, 2014 at 10:28

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

6

Haskell, 114 108

s 1=["# "]
s n=k j++k(map(>>" ")j)++k j where j=s$n-1;k i=j%i%j
(%)=zipWith(++)
main=interact$unlines.s.read

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 21, 2014 at 19:47

proud haskeller

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nutki · Accepted Answer · 2014-10-23 08:55:32Z

Perl 5: 68 characters

n on stdin.

#!/usr/bin/perl -l
//,print map/2/?"  ":"■ ",map$_+$',@,for@,=grep!/[24-9]/,0..3x<>/10

A decimal coded ternary representation of the list of coordinates is first generated in @, (name chosen so there is no need for space between @, and for) using digits 0, 1, and 3. Now adding x and y coordinates in the decimal domain will have a 2 in the result if and only if there were two ones at matching positions. // is used to propagate the default variable $_ from the outer loop (for) to the postmatch variable $' when it is shadowed by the default map variable in the inner loop.

Wrzlprmft · Accepted Answer · 2014-10-22 14:09:42Z

5

Python – 100

r=range(3**~-input())
for i in r:print" ".join("# "[any(i/3**k%3==j/3**k%3==1for k in r)]for j in r)

edited Oct 22, 2014 at 14:09

answered Oct 22, 2014 at 14:04

Wrzlprmft

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Matthew Butterick · Accepted Answer · 2014-10-23 15:09:47Z

Racket 230 229 225 220

Not Racket's finest hour for golfing.

Golfed:

(define(s n)(letrec([t(λ(x y)(if(or(= x 0)(= y 0))"■"(if(=(modulo x 3)(modulo y 3)1)" "(t(floor(/ x 3))(floor(/ y 3))))))][i(expt 3(- n 1))])(for-each displayln(for/list([r i])(string-join(for/list([c i])(t r c))" ")))))

Ungolfed:

(define (s n)
  (letrec ([t (λ (x y)
                (if (or (= x 0) (= y 0)) 
                    "■"
                    (if (= (modulo x 3) (modulo y 3) 1)
                        " "
                        (t (floor (/ x 3)) (floor (/ y 3))))))]
           [i (expt 3 (- n 1))])
    (for-each displayln 
              (for/list ([r i])
                (string-join 
                 (for/list ([c i]) 
                   (t r c)) " ")))))

nutki · Accepted Answer · 2014-10-23 15:43:07Z

C: 123 118 111 104 characters

Based on a similar idea as my perl solution. After adding some spaces:

m=0x55555555;
x;
main(n){
  scanf("%d",&n);
  n=1<<2*--n;
  for(x=n*n;x--;)
    printf(x&x/2&m?"":"%c%c",x&x/n&m?32:35,x&n-1?32:10);
}

Uses ternary system coding each digit with 2 bits. Illegal values (having two ones in odd-even position) are filtered with x & (x>>1) & 0b01010101. Both coordinates are stored in one value, so checking the pixel color is down to x & (x >> 2 * n) & 0b01010101. n is stored as a power of 2 for convenience.

Edit

Replaced define with a simple constant m.

Edit 2

0x5555555 mask can be represented with (1LL<<32)/3, but we only need n of those bits so n/3 is sufficient.

x;
main(n){
  scanf("%d",&n);
  n=1<<2*--n;
  for(x=n*n;x--;)
    printf(x&x/2&n*n/3?"":"%c%c",x&x/n&n/3?32:35,x&n-1?32:10);
}

Edit 3

Minor tweaks. One 2 char gain relying on scanf being executed before the loading value of n for the execution of --n. The eol can only follow #, duh.

x;
main(n){
  n=scanf("%d",&n)<<2*--n;
  for(x=n*n;x--;)
    x&x/2&n*n/3||printf(x&x/n&n/3?"  ":x&n-1?"# ":"#\n");
}

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

4

Mathematica, 71 bytes

Grid@Nest[ArrayFlatten@ArrayPad[{{0}},1,{{#}}]&,1,#]/.{0->"",1->"■"}&

input

3

output

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jun 13, 2017 at 22:15

ZaMoC

24.9k2 gold badges30 silver badges81 bronze badges

\$\begingroup\$ I followed you like a total creeper. +1 ;). \$\endgroup\$
– Magic Octopus Urn
Commented Jun 13, 2017 at 22:17
\$\begingroup\$ Do you think you can shorten this with SubstitutionSystem ? \$\endgroup\$
– Vitaliy Kaurov
Commented Sep 26, 2017 at 16:57

Add a comment |

Compass · Accepted Answer · 2014-10-21 18:46:19Z

HTML/JavaScript, 205 Chars

Obfuscatweet, 205 Chars

document.write(unescape(escape('🁳𨱲𪑰𭀾𬰽𙰦𫡢𬱰𞰧𞱮🐴𞱭👍𨑴𪀮𬁯𭰨𜰬𫠭𜐩𞱦𫱲𚁩🐰𞱩🁭𞱩𚰫𛁤𚀧🁢𬠾𙰩𚑻𩡯𬠨𪠽𜀻𪠼𫐻𚑩𩠨𨰨𪐬𪠫𚰩𚑤𚀧𘰧𚐻𩑬𬱥𘁤𚁳𚑽𩡵𫡣𭁩𫱮𘁣𚁸𛁹𚑻𭱨𪑬𩐨𮁼𯁹𚑻𪑦𚁸𙐳🐽𜐦𙡹𙐳🐽𜐩𬡥𭁵𬡮𘀰𞱸👦𚁸𚐻𮐽𩠨𮐩𯑲𩑴𭑲𫠠𜑽𩡵𫡣𭁩𫱮𘁦𚁡𚑻𬡥𭁵𬡮𘁍𨑴𪀮𩡬𫱯𬠨𨐯𜰩𯑦𭑮𨱴𪑯𫠠𩀨𨐩𮱤𫱣𭑭𩑮𭀮𭱲𪑴𩐨𬰫𨐩𯐼𛱳𨱲𪑰𭀾🁳𭁹𫁥🠪𮱦𫱮𭀭𩡡𫑩𫁹𞠢𠱯𭑲𪑥𬠢').replace(/uD./g,'')))

HTML/JS, 298 Chars

Due to how HTML plays with whitespace, a few characters had to be dedicated to the nbsp char. In addition, the default font of most browsers is not Courier, so I had to set it to that, too. About 20 characters worth of styling. If this requires a direct input method, I can add it, but changing the input currently is setting n to a different value.

Demo

<script>s='&nbsp;';n=4;m=Math.pow(3,n-1);for(i=0;i<m;i++,d('<br>')){for(j=0;j<m;)if(c(i,j++))d('#');else d(s)}function c(x,y){while(x||y){if(x%3==1&&y%3==1)return 0;x=f(x);y=f(y)}return 1}function f(a){return Math.floor(a/3)}function d(a){document.write(s+a)}</script><style>*{font-family:"Courier"

Readable HTML/JS

<script>
    s='&nbsp;';
    n=4;
    m=Math.pow(3,n-1);
    for(i=0;i<m;i++,d('<br>')){
        for(j=0;j<m;)
            if(c(i,j++))
                d('#');
            else d(s)
    }
    function c(x,y){
        while(x||y){
            if(x%3==1&&y%3==1)
                return 0;
            x=f(x);
            y=f(y)
        }
        return 1
    }
    function f(a){
        return Math.floor(a/3)
    }
    function d(a){
        document.write(s+a)
        }
</script>
<style>
*{font-family:"Courier"

Dennis · Accepted Answer · 2014-10-21 23:07:05Z

CJam, 38 35 32 31 characters

"■ ""՛ୗ䁎뽔휼ꆩ闳⾿➺⥧槲㩡郊"6e4b128b:c~

Try it online.

Example run

$ cjam <(echo '"■ ""՛ୗ䁎뽔휼ꆩ闳⾿➺⥧槲㩡郊"6e4b128b:c~') <<< 3; echo
■ ■ ■ ■ ■ ■ ■ ■ ■ 
■   ■ ■   ■ ■   ■ 
■ ■ ■ ■ ■ ■ ■ ■ ■ 
■ ■ ■       ■ ■ ■ 
■   ■       ■   ■ 
■ ■ ■       ■ ■ ■ 
■ ■ ■ ■ ■ ■ ■ ■ ■ 
■   ■ ■   ■ ■   ■ 
■ ■ ■ ■ ■ ■ ■ ■ ■

How it works

"՛ୗ䁎뽔휼ꆩ闳⾿➺⥧槲㩡郊"6e4b128b:c~

converts the Unicode string from base 60,000 (6e4b) to base 128 (128b), casts to Character (:c) and evaluates the resulting string (~).

As a result, the following code gets executed:

"■ "                                " C := '■ '                                           ";
    li(                             " J := int(input()) - 1                               ";
       {                      }*    " for I in range(J):                                  ";
        z]A*                        "   T := [zip(C)] * 10                                ";
            ):,                     "   U := pop(T)                                       ";
               Sf*                  "   V := [ ' ' * len(u) : u ∊ U ]                     ";
                  4\t               "   U[4] := V                                         ";
                     3/{:+z~}%      "   C := { c : c ∊ zip(concat(u)), u ∊ U.split(3) }   ";
                              zN*  " print '\n'.join(zip(C))                              ";

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

Python 3 – 116 113 characters

EDIT: Well, used the trick I don't like too much myself and compressed the code by 3 bytes. Dunno if that's the best possible way, but I'll go with it.

exec(bytes('ⱴ㵬❛離₠崧氬浡摢⁡㩺癥污稨✫⁫潦⁲⁫湩琠❝਩潦⁲⁩湩爠湡敧椨瑮椨灮瑵⤨⴩⤱爺瀬氽✨㍛✪Ⱙ⡬嬢⭫‧✠⠪⨳椪⬩⤢琻爽瀫爫昊牯椠椠⁮㩴牰湩⡴⥩','utf-16')[2:].decode('utf-8'))

It may contain some unprintable characters, so here's a printable version of the string.

'ⱴ㵬❛離₠崧氬浡摢\u2061㩺癥污稨✫\u206b潦\u2072\u206b湩琠❝\u0a29潦\u2072\u2069湩爠湡敧椨瑮椨灮瑵⤨\u2d29⤱爺瀬氽✨㍛✪Ⱙ⡬嬢\u2b6b‧✠⠪⨳椪⬩⤢琻爽瀫爫昊牯椠椠\u206e㩴牰湩⡴⥩'

Not too great, but at least beats some languages.

What it expands to:

t=['■ ']
for i in range(int(input())-1):r,p=[k*3for k in t],[k+'  '*(3**i)+k for k in t];t=r+p+r
for i in t:print(i)

Somewhat ungolfed in case someone can't see how it works:

t=['■ ']                                     # iteration 1
for i in range(int(input()) - 1):            # do n-1 more iterations
    r = [k * 3 for k in t]                   # first & last rows are the last carpet x3
    p = [k + '  ' * (3 ** i) + k for k in t] # middle row: last carpet, space, last carpet
    t = r + p + r                            # and put them to the new carpet
for i in t:                                  # print final iteration
    print(i)

\$\begingroup\$ Wow python, that's impressive \$\endgroup\$
– Paul Clavier
Commented Oct 22, 2014 at 11:13 — Paul Clavier, Commented Oct 22, 2014 at 11:13

Paul Clavier · Accepted Answer · 2014-10-21 18:03:06Z

2

CJam, 76 characters

3ri(#:M_*,{_M/:I\M%:J;;{I3%1=J3%1=&0X?:X;I3/:I0>J3/:J0>|}gX'■S?1:X;}%M/Sf*N*

This is a direct translation of the formula given here

Try it here

edited Oct 21, 2014 at 18:03

Paul Clavier

3653 silver badges5 bronze badges

answered Oct 21, 2014 at 17:47

Optimizer

26.6k7 gold badges66 silver badges142 bronze badges

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

Bash+coreutils, 105 unicode characters

Since we're counting characters and not bytes:

eval `iconv -tunicode<<<潦⡲眨㌽⨪␨ⴱ⤱琬〽琻眼眪琻⬫⤩笻戠ꂖ昻牯⠨㵸╴ⱷ㵹⽴㭷㹸簰祼〾砻㴯ⰳ⽹㌽⤩笻⠠砨㌥ㄭ籼╹ⴳ⤱簩扼尽簠硼〽紻瀻楲瑮⁦␢⁢㬢⠨╴⵷⭷⤱簩敼档㭯੽|cut -b3-`

In its decoded form:

Pure Bash, 143 bytes

for((w=3**($1-1),t=0;t<w*w;t++));{
b=■
for((x=t%w,y=t/w;x>0||y>0;x/=3,y/=3));{
((x%3-1||y%3-1))||b=\ ||x=0
}
printf "$b "
((t%w-w+1))||echo
}

Input is taken from command-line args:

Output:

$ ./sierpinskicarpet.sh 3
■ ■ ■ ■ ■ ■ ■ ■ ■ 
■   ■ ■   ■ ■   ■ 
■ ■ ■ ■ ■ ■ ■ ■ ■ 
■ ■ ■       ■ ■ ■ 
■   ■       ■   ■ 
■ ■ ■       ■ ■ ■ 
■ ■ ■ ■ ■ ■ ■ ■ ■ 
■   ■ ■   ■ ■   ■ 
■ ■ ■ ■ ■ ■ ■ ■ ■ 
$

Allbeert · Accepted Answer · 2014-10-23 15:25:46Z

2

C - 154

Input from stdin. I tried to find a good way to avoid an extra function, but couldn't come up with any.

The character count includes only necessary spaces.

p(l,x,y){
    return l?x/l&y/l&1||p(l/3,x%l,y%l):0;
}
x,y;
main(v){
    scanf("%d",&v);
    v=pow(3,--v);
    for(;y<v;y++,puts(""))
        for(x=0;x<v;)
            printf("%c ",p(v,x++,y)?32:35);
}

edited Oct 23, 2014 at 15:25

answered Oct 21, 2014 at 20:33

Allbeert

5192 silver badges5 bronze badges

\$\begingroup\$ 143 bytes \$\endgroup\$
– ceilingcat
Commented Jun 9, 2020 at 2:21

Add a comment |

xnor · Accepted Answer · 2017-09-26 10:57:34Z

2

Python 2, 91 bytes

r=input()-1
for i in range(3**r):x,s='# ','  ';exec"x+=[x,s][i%3%2]+x;s*=3;i/=3;"*r;print x

Try it online!

answered Sep 26, 2017 at 10:57

xnor

147k26 gold badges279 silver badges652 bronze badges

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Jonathan Allan · Accepted Answer · 2020-05-24 18:55:38Z

Jelly, 17 bytes

’3*Ḷb3ḂḄ&þ`¬ị⁾# G

Try it online!

How?

’3*Ḷb3ḂḄ&þ`¬ị⁾# G - Link: integer, n
’                 - decrement -> n-1
 3*               - three raised to (that)
   Ḷ              - lowered range -> [0,1,...,3^(n-1)-1]
    b3            - convert to base three (vectorises)
      Ḃ           - least significant bit (vectorises) -> i.e. make 2 trits become 0s
       Ḅ          - convert from base two (vectorises)
          `       - use that as both arguments of:
         þ        -   outer product with:
        &         -     bitwise AND
           ¬      - logical NOT (vectorises)
            ị     - index into:
             ⁾#   -   list of characters ['#', ' ']
                G - format as a grid

zgrep · Accepted Answer · 2021-09-21 22:29:30Z

2

oK, 40 45 bytes

`0:" "/'" #"{(x-1)(,'//3 3#111101111b*9#,)/1}

Try it online.

It starts with 1, and then draws it in a grid (1 1 1;1 0 1;1 1 1), which it then draws in a grid the same way, etc. repeated the necessary number of times.

edited Sep 21, 2021 at 22:29

answered Jun 14, 2017 at 0:10

zgrep

1,6318 silver badges9 bronze badges

\$\begingroup\$ Ah, sorry about that! It should meet the spec now (at the cost of 5 more bytes). \$\endgroup\$
– zgrep
Commented Sep 21, 2021 at 22:29

Add a comment |

lynn · Accepted Answer · 2021-11-03 01:51:33Z

2

Haskell, 77 characters

f n|a<-mapM(:[0,1])[2..n]=[do x<-a;"■ "!!(0^product(zipWith(+)x y)):" "|y<-a]

Try it online!

answered Nov 3, 2021 at 1:51

lynn

69.2k11 gold badges133 silver badges283 bronze badges

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Gedrox · Accepted Answer · 2014-10-22 10:42:22Z

1

PHP, 194 characters

The n received as first argument in command line.

<?php
function d(&$a,$n,$e,$x,$y){if(--$n)for(;$i<9;)$p=pow(3,$n)|d($a,$n,$e|$i==4,$x+$p*$j,$y+$p*($i-$j)/3)|$j=++$i%3;else$a[$x][$y]=$e?" ":■;}@d($a,$argv[1]);foreach($a as$s)echo join($s),"
";

Readable

<?php

function draw(&$array, $n, $empty, $x, $y) {
    $n--;
    if ($n != 0) {
        for ($i = 0; $i < 9; $i++) {
            $j = $i % 3;
            $p = pow(3, $n);
            draw($array, $n, $empty || $i == 4, $x + $p * $j, $y + $p * ($i - $j) / 3);
        }
    } else {
        $array[$x][$y] = $empty ? " " : "#";
    }
}

$array = array();
draw($array, $argv[1], false, 0, 0);
foreach ($array as $line) {
    echo join($line), "\n";
}

edited Oct 22, 2014 at 10:42

answered Oct 21, 2014 at 19:12

Gedrox

1113 bronze badges

1

\$\begingroup\$ If you use the char ■ instead of #, you can save 2 chars by removing the quotes. \$\endgroup\$
– Ismael Miguel
Commented Oct 22, 2014 at 8:48

Add a comment |

Mathill · Accepted Answer · 2014-10-23 00:31:03Z

Scala 230 characters

Golfed code:

object T extends App {def m=math.pow(3,args(0).toInt-1).toInt-1;def p=print _;val x=0 to m;x.map{n=>x.map{j=>if(q(n,j)==1)p(" #");else p("  ")};p("\n");};def q(n:Int,j:Int):Int={if(n>0|j>0)if((n%3&j%3)==1)0 else q(n/3,j/3)else 1}}

Ungolfed code:

object T extends App {
  def m = math.pow(3, args(0).toInt - 1).toInt - 1;

  def p = print _;
  val x = 0 to m;
  x.map { n => x.map { j => if (q(n, j) == 1) p(" #"); else p("  ")}; p("\n");};

  def q(n: Int, j: Int): Int = {
    if (n > 0 | j > 0) if ((n % 3 & j % 3) == 1) 0 else q(n / 3, j / 3) else 1
  }
}

Only necessary spaces are included.

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

1

SOGL V0.12, 27 bytes

 ■¹.H{³³┼┼≥: ■@ŗ;┼┼⁴++}{@∑P

Try it Here!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 24, 2017 at 13:13

dzaima

20.3k2 gold badges41 silver badges75 bronze badges

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Ayb4btu · Accepted Answer · 2017-10-19 09:37:52Z

1

C# (.NET Core), 162 bytes

n=>{var f="";int s=(int)Math.Pow(3,n),z=0;while(z<s*s){int x=z%s,y=z/s,h=1;for(;x>0|y>0;y/=3,x/=3)if(y%3==1&x%3==1)h=0;f+=" #"[h]+(z++%s<s-1?" ":"\n");}return f;}

Try it online!

Degolfed

n=>{
    var f="";
    int s=(int)Math.Pow(3,n),z=0;

    while(z<s*s)
    {
        int x=z%s, y=z/s, h=1;
        for(; x>0 | y>0; y/=3, x/=3)
            if(y%3==1 & x%3==1)
                h=0;

        f += " #"[h] + (z++%s<s-1? " " : "\n");
    }

    return f;
}

answered Oct 19, 2017 at 9:37

Ayb4btu

5711 gold badge4 silver badges7 bronze badges

\$\begingroup\$ 156 bytes \$\endgroup\$
– ceilingcat
Commented Sep 7, 2020 at 5:27

Add a comment |

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

1

Canvas, 17 16 17 characters

■；╷［⌐ ＋２×；┌∔∔；３＊＋

Try it here!

-1: Used ［ instead of ｛ to delete the first ； inside the loop.
+1: Fixed errenous behavior: ■ now corresponds to level 1, as specified in the original post.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Mar 10, 2018 at 3:15

hakr14

5,0141 gold badge11 silver badges18 bronze badges

\$\begingroup\$ It is competing, per a newer consensus. \$\endgroup\$
– Erik the Outgolfer
Commented Mar 10, 2018 at 13:58
\$\begingroup\$ @EriktheOutgolfer oh, thanks for letting me know! \$\endgroup\$
– hakr14
Commented Mar 10, 2018 at 18:39
\$\begingroup\$ And yes, in this case the score is 17, but we count solutions in bytes, not characters (Canvas has a code page). \$\endgroup\$
– Erik the Outgolfer
Commented Mar 10, 2018 at 18:40
\$\begingroup\$ @EriktheOutgolfer normally yes, but this challenge uses characters, due to ■ being multibyte. Says so toward the bottom in the original post. \$\endgroup\$
– hakr14
Commented Mar 10, 2018 at 18:45
\$\begingroup\$ You need to add one to all your inputs. 1 should return a square (i.e. what 0 returns here) \$\endgroup\$
– Paul Clavier
Commented Mar 12, 2018 at 20:14

| Show 1 more comment

DLosc · Accepted Answer · 2018-05-17 21:05:55Z

Pip `-S`, 30 characters

"■■ "@MX{{aTB3R2i}MSg}MC3**a/3

Try it online!

The basic idea: consider a coordinate grid in base 3. The holes in the carpet occur where 1) a trit in the x-coordinate is 1, and 2) the trit in the same position in the y-coordinate is also 1.

"■■ "@MX{{aTB3R2i}MSg}MC3**a/3
                                i is 0; a is 1st cmdline arg (implicit)
                        3**a/3  Width/height of the carpet
                      MC        Make a coordinate grid that size and map this function
                                to the coordinate pairs:
                    g            Take each [x y] pair
                  MS             Map this function to each coord and add the results:
         {aTB3   }                Convert to base 3
              R2i                 Replace 2's with 0's
                                  E.g. 15 -> 120 -> 100 and 16 -> 121 -> 101
                                 When we add the results, we get [15 16] -> 201
      MX{            }           Take the max of that list (0, 1, or 2)
"■■ "@                           Use that to index into this string (resulting in a space
                                 iff two 1's coincided to make a 2, or ■ otherwise)
                                Print the resulting nested list with each sublist on its
                                own line and space-delimited (implicit, -S flag)

Similar solution, same number of characters but -2 bytes: {2N({aTB3R2i}MSg)?s'■}MC3**a/3

ngn · Accepted Answer · 2020-12-17 20:09:31Z

1

K (ngn/k), 50 49 bytes

thanks @DLosc for reminding me I should output spaces between the columns

thanks @coltim for -1 byte

1@"\n"/("# "1_,/1,'+/')'a*/:\:a:+2!!1_(. 0:"")#3;

Try it online!

edited Dec 17, 2020 at 20:09

answered May 17, 2018 at 19:53

ngn

15.3k2 gold badges43 silver badges87 bronze badges

\$\begingroup\$ @DLosc thanks, fixed other strange requirements too: stdin/stdout instead of arg/result; off-by-one "level" \$\endgroup\$
– ngn
Commented May 17, 2018 at 21:50
1

\$\begingroup\$ I think you can drop a byte by using ("# "1_,/1,'+/')' instead of "# "1_',/'1,''+/'' \$\endgroup\$
– coltim
Commented Dec 17, 2020 at 15:17

Add a comment |

naffetS · Accepted Answer · 2022-06-04 19:43:00Z

Vyxal, 19 bytes

‛ #3?‹eʁ3τ∷vB:v⋏†İ⁋

Try it Online!

How?

‛ #3?‹eʁ3τ∷vB:v⋏†İ⁋
    ?‹              # Push input - 1
   3  e             # Push 3 to the power of it
       ʁ            # Get range [0, that)
        3τ          # Convert each to base 3 (ternary)
          ∷         # Modulo each by two to convert twos to zeros
           vB       # Convert each from binary to integers
             :v⋏    # Outer product bitwise AND with itself
                †   # Logical NOT for each
‛ #              İ  # Index each into " #"
                  ⁋ # Join each by spaces, and then join on newlines

Giuseppe · Accepted Answer · 2018-03-12 16:38:04Z

0

R, 92 characters

94 bytes with the special character.

write(c(" ","■")[1+Reduce("%x%",rep(list(matrix(c(1,1,1,1,0,1),3,3)),n<-scan()-1),1)],1,3^n)

Try it online!

answered Mar 12, 2018 at 16:38

Giuseppe

28.8k3 gold badges31 silver badges105 bronze badges

\$\begingroup\$ I'm just amazed people are still doing this 4 years later \$\endgroup\$
– Paul Clavier
Commented Mar 12, 2018 at 20:10
1

\$\begingroup\$ @PaulClavier it's a great challenge; you should be proud :) \$\endgroup\$
– Giuseppe
Commented Mar 12, 2018 at 20:10

Add a comment |

Razetime · Accepted Answer · 2021-09-17 15:23:53Z

0

Husk, 27 bytes

mJ' ´Ṫȯ!"□ "¬nmḋ†%2mB3ŀ`^3←

Try it online!

Translation of Jonathan Allan's program.

uses the '□' character.

edited Sep 17, 2021 at 15:23

answered Nov 6, 2020 at 15:24

Razetime

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Jonah · Accepted Answer · 2021-09-17 16:14:49Z

0

J, ³⁵ 50 bytes

' #'{~1|:@}.[:,/0,:"1]9&([:,/_3,./\0"+4}(#,:)),:@1

Try it online!

+15 thanks to DLosc for pointing out there had to be spaces between columns.

This was a quick fix to satisfy the spec and could likely be re-golfed further.

J, original answer, cleaner, 35 bytes

' #'{~]9&([:,/_3,./\0"+4}(#,:)),:@1

Try it online!

edited Sep 17, 2021 at 16:14

answered May 24, 2020 at 17:26

Jonah

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Kevin Cruijssen · Accepted Answer · 2021-11-03 11:19:38Z

05AB1E, 21 bytes

„# 3I<mÝ¨3δвÉJCDδ&Āè»

Port of @JonathanAllan's Jelly answer, so make sure to upvote him as well!

Try it online or verify all test cases.

Explanation:

    I<                # Push the input-1
   3  m               # Take 3 to the power this input-1
       Ý              # Pop and push a list in the range [0,3**(input-1)]
        ¨             # Remove the last item to lower the range to: [0,3**(input-1))
          δ           # Map over each integer:
         3 в          #  Convert it to a base-3 list
            É         # Check for each trigit if it's odd, converting the 2s to 0s
             J        # Join the inner bit-lists together to strings
              C       # Convert each from a binary-string to an integer
               D      # Using a copy of itself,
                δ     # apply double-vectorized over this list of integer:
                 &    #  Bitwise-AND the integers together
                  Ā   # Check for each integer whether it's NOT 0 (0 if 0; 1 otherwise)
„#                 è  # 0-based index it into the string "# "
                    » # Join each inner list by spaces, and then each string by newlines
                      # (after which the result is output implicitly)

Sierpinski Carpets

30 Answers 30

CJam, 38 37 31 30 28 characters

Explanation

Matlab (113)(110)(99)(85)

Haskell, 114 108

Perl 5: 68 characters

Python – 100

Racket 230 229 225 220

C: 123 118 111 104 characters

Edit

Edit 2

Edit 3

Mathematica, 71 bytes

HTML/JavaScript, 205 Chars

Obfuscatweet, 205 Chars

HTML/JS, 298 Chars

Readable HTML/JS

CJam, 38 35 32 31 characters

Example run

How it works

Python 3 – 116 113 characters

CJam, 76 characters

Bash+coreutils, 105 unicode characters

Pure Bash, 143 bytes

Output:

C - 154

Python 2, 91 bytes

Jelly, 17 bytes

How?

oK, 40 45 bytes

Haskell, 77 characters

PHP, 194 characters

Readable

SOGL V0.12, 27 bytes

C# (.NET Core), 162 bytes

Degolfed

Canvas, 17 16 17 characters

Pip -S, 30 characters

K (ngn/k), 50 49 bytes

Vyxal, 19 bytes

How?

R, 92 characters

Husk, 27 bytes

J, 35 50 bytes

J, original answer, cleaner, 35 bytes

05AB1E, 21 bytes

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