29
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Write a program or function that takes in three positive integers, W, H, and N. Print or return a W×H grid of .'s where every Nth . in normal English reading order is replaced with an X.

For example, given W = 7, H = 3, N = 3, the grid is 7 characters wide and 3 high, and every third character reading from the top left is an X:

..X..X.
.X..X..
X..X..X

Similarly, if the input is W = 10, H = 4, N = 5, the output would be:

....X....X
....X....X
....X....X
....X....X

Notes

  • "Normal English reading order" means going left to right on each line, from the top line to the bottom.
  • When N is 1 then all the .'s will become X's.
  • You may use any two distinct printable ASCII characters in place of . and X.
    • If you use space ( ) then trailing spaces are not required when the result would be visually the same. (Empty lines are still required.)
    • You may not using something else in place of the newlines that shape the grid.
  • The exact input format and order of W, H, and N is not super important. Things like [H,W,N] or N\nW,H are alright.
  • A trailing newline in the output is fine.
  • The shortest code in bytes wins!

Examples

W = 5, H = 3, N = 1
XXXXX
XXXXX
XXXXX

W = 5, H = 3, N = 2
.X.X.
X.X.X
.X.X.

W = 5, H = 3, N = 3
..X..
X..X.
.X..X

W = 5, H = 3, N = 4
...X.
..X..
.X...

W = 5, H = 3, N = 5
....X
....X
....X

W = 5, H = 3, N = 6
.....
X....
.X...

W = 5, H = 3, N = 7
.....
.X...
...X.

W = 5, H = 3, N = 15
.....
.....
....X

W = 5, H = 3, N = 16 (or more)
.....
.....
.....

W = 1, H = 1, N = 1
X

W = 1, H = 1, N = 2 (or more)
.

W = 8, H = 6, N = 2
.X.X.X.X
.X.X.X.X
.X.X.X.X
.X.X.X.X
.X.X.X.X
.X.X.X.X

W = 8, H = 6, N = 3
..X..X..
X..X..X.
.X..X..X
..X..X..
X..X..X.
.X..X..X

W = 8, H = 6, N = 4
...X...X
...X...X
...X...X
...X...X
...X...X
...X...X

W = 8, H = 6, N = 7
......X.
.....X..
....X...
...X....
..X.....
.X......

W = 8, H = 6, N = 16
........
.......X
........
.......X
........
.......X

W = 37, H = 1, N = 4
...X...X...X...X...X...X...X...X...X.

W = 1, H = 10, N = 8
.
.
.
.
.
.
.
X
.
.
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2
  • 1
    \$\begingroup\$ Am I correct to assume that the restriction "You may not using something else in place of the newlines that shape the grid" includes "You may not return an array ["..X..X.", ".X..X..", "X..X..X"] as the grid"? \$\endgroup\$ – Peter Taylor Nov 29 '15 at 8:32
  • \$\begingroup\$ @PeterTaylor Correct \$\endgroup\$ – Calvin's Hobbies Nov 29 '15 at 12:36

25 Answers 25

13
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J, 9 5 bytes

$":&1

Uses spaces and 1's and expects input in the form H W f N

Explanation:

$":&1
   &1 bonds the fixed right argument 1 to ":
 ":   formats the right argument number (1) to take up left argument (N) number of cells
      padding with spaces, resulting  in "  1"
$     reshape to H-by-W with repeating the string if necessary 

Usage:

   3 7 ($":&1) 3
  1  1 
 1  1  
1  1  1

Try it online here.

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4
  • \$\begingroup\$ Does it also truncate the array if W*H is less than N? \$\endgroup\$ – Martin Ender Nov 29 '15 at 10:26
  • \$\begingroup\$ @MartinBüttner Yes. \$\endgroup\$ – randomra Nov 29 '15 at 11:13
  • \$\begingroup\$ If the argument is ($":&1), wouldn't that count as 7 bytes? \$\endgroup\$ – Reto Koradi Nov 29 '15 at 14:55
  • 1
    \$\begingroup\$ No, the () aren't part of the function; you could write f =. $":&1 and then 3 7 f 3. \$\endgroup\$ – Lynn Nov 29 '15 at 15:51
11
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Python 2, 60 bytes

w,h,n=input()
s='%%%dd'%n%0*w*h
exec"print s[:w];s=s[w:];"*h

This prints space and 0 in place of . and X. Input is taken as a tuple in the form of w,h,n.

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1
  • 4
    \$\begingroup\$ That's a clever string format. \$\endgroup\$ – xnor Nov 29 '15 at 4:03
7
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J, 12 bytes

$'X'_1}#&'.'

This is a dyadic function that takes the array H W as its left argument and N as its right argument. Usage:

  f =: $'X'_1}#&'.'
  3 5 f 3
..X..
X..X.
.X..X

Explanation

$'X'_1}#&'.'
         '.'  The character '.'
       #&     repeated N times
    _1}       with the last character
 'X'          replaced by 'X'
$             reshaped into an HxW array
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3
  • \$\begingroup\$ Right tool for the job? \$\endgroup\$ – Addison Crump Nov 29 '15 at 2:48
  • \$\begingroup\$ Is the use of X. really shortest? \$\endgroup\$ – lirtosiast Nov 29 '15 at 4:02
  • \$\begingroup\$ @ThomasKwa I believe so. I tried to use the numbers 0 and 1 instead, but then I had to surround the one next to _1 with parentheses, and format away the spaces between columns, and it ended up being longer. \$\endgroup\$ – Zgarb Nov 29 '15 at 4:17
5
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BBC Basic, 67 ASCII characters, tokenised filesize 43 bytes

Download interpreter at http://www.bbcbasic.co.uk/bbcwin/download.html

INPUTw,h,n:WIDTHw:PRINTLEFT$(STRING$(w*h,STRING$(n-1,".")+"X"),w*h)

BBC basic has a handy command for limiting the field width. We use STRING$ to make w*h copies of the string of n-1 periods followed by an X. Then we use LEFT$ to truncate this to w*h characters.

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4
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Minkolang 0.14, 34 30 28 22 bytes

n2-D1n$zn[z[1Rd6ZO]lO]

Check one case here and check all test cases here. Expects input like N W H.

Explanation

n                 Take number from input (N)
 2-               Subtract 2
   D              Duplicate the top of stack (which is 0 because it's empty) N-2 times
    1             Push a 1 onto the stack
n                 Take number from input (W)
 $z               Store W in the register (z)
n                 Take number from input (H)
 [                Open a for loop that repeats H times
  z[              Open a for loop that repeats W times
    1R            Rotate 1 step to the right
      d           Duplicate top of stack
       6Z         Convert number to string
         O        Output as character
          ]       Close for loop
           lO     Output a newline
             ]    Close for loop

As Minkolang's codebox is toroidal, this will wrap around to the beginning. As every n will now take in -1, this eventually crashes with an error and no further output, which is allowed.

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2
  • \$\begingroup\$ So it's easy for you to compare. (Note that it's not quite the exact same code.) \$\endgroup\$ – El'endia Starman Nov 29 '15 at 2:20
  • \$\begingroup\$ Ahead of you! :P :) \$\endgroup\$ – El'endia Starman Nov 29 '15 at 2:23
4
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CJam (16 bytes)

{1$*,:)@f%:!/N*}

Takes input on the stack in the order N W H, returns string using characters 0 and 1. Online demo

Dissection

{        e# Anonymous function. Stack: N W H
  1$*,   e# Stack: N W [0 1 ... W*H-1]
  :)     e# Stack: N W [1 2 ... W*H]
  @f%    e# Stack: W [1%N 2%N ... W*H%N]
  :!     e# Map Boolean not, taking 0 to 1 and anything else to 0
  /      e# Split into W-sized chunks (i.e. the lines of the grid)
  N*     e# Join the lines with newlines
}
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1
  • \$\begingroup\$ ;-; you beat me ;-; but good job! :D \$\endgroup\$ – anOKsquirrel Nov 29 '15 at 13:30
4
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APL, 13 bytes

{⍪,/⍕¨⍺⍴⍵=⍳⍵}

This takes H W as the left argument and N as the right argument.

Explanation:

{⍪,/⍕¨⍺⍴⍵=⍳⍵}     Dyadic function (args are ⍺ on left, ⍵ on right):
        ⍵=⍳⍵      ⍵ = (1 2 3...⍵); this is ⍵-1 0s followed by a 1
      ⍺⍴          Shape by the left argument; e.g. 5 3 gives a 5x3 array
    ⍕¨            Stringify each entry
  ,/              Join the strings in each row 
 ⍪                Make column vector of strings

Try it online: first test cases, last test case. Note that although this shows boxed output, my copy of Dyalog doesn't.

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4
  • \$\begingroup\$ Are those actually just boxes, or is the SE app not displaying the characters properly? \$\endgroup\$ – Carcigenicate Nov 29 '15 at 18:22
  • \$\begingroup\$ @Carcigenicate They're not boxes. The characters should display fine on the online link, because it has a different font. \$\endgroup\$ – lirtosiast Nov 29 '15 at 18:45
  • \$\begingroup\$ Ahh, right. I missed that. Do you have a special keyboard or are you a masochist? \$\endgroup\$ – Carcigenicate Nov 29 '15 at 18:47
  • \$\begingroup\$ @Carcigenicate On tryapl (and Dyalog student edition) you can type APL characters using backticks. `a turns into ⍺, for example. \$\endgroup\$ – lirtosiast Nov 29 '15 at 18:51
2
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CJam, 20 Bytes

q~:Z;_@*,:){Z%!}%/N*

Takes input as H W N.

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4
  • \$\begingroup\$ whoops, invalid \$\endgroup\$ – anOKsquirrel Nov 29 '15 at 2:03
  • \$\begingroup\$ fixed :D :D :D :D \$\endgroup\$ – anOKsquirrel Nov 29 '15 at 2:08
  • \$\begingroup\$ Still much longer than some of the solutions in other languages, but I got it to 18 bytes with CJam: q~_@*,@(S*'X+f=/N*, with input in order N H W. \$\endgroup\$ – Reto Koradi Nov 29 '15 at 4:53
  • 1
    \$\begingroup\$ @RetoKoradi Take another one off by replacing 'X with 0, and that'll be 17 \$\endgroup\$ – Sp3000 Nov 29 '15 at 5:08
2
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MATLAB, 61 55 54 bytes

function c=g(d,n);b=ones(d);b(n:n:end)=0;c=[b'+45,''];

Wow, I thought MATLAB would be competitive in this one, but how wrong I was!

The function creates an array of 1's of the correct dimensions, and then sets every n'th element to be 0 (MATLAB implicitly handles wrapping around the indices into 2D). We then add on 45 ('-') to this number and converted to a char array to be returned.

The questions allows any distinct two ASCII characters to be used for the grid, I am using '-' in place of 'x' to save some bytes. The input format is also not fixed, so it should be supplied as [w h],n - i.e. an array of width and height, and then n as a second parameter.


This also works with Octave and can be tried online here. The function is already set up in the linked workspace, so you can simply call for example:

g([4,5],3)

Which outputs:

..-.
.-..
-..-
..-.
.-..
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2
  • \$\begingroup\$ Save one byte: c=[b'+45,'']; \$\endgroup\$ – Stewie Griffin Nov 29 '15 at 9:21
  • \$\begingroup\$ @StewieGriffin Thanks :). For some reason when I'd tried that I didn't think it saved any bytes, I must have miscounted! \$\endgroup\$ – Tom Carpenter Nov 29 '15 at 13:43
2
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Processing, 93 bytes (Java, 104 bytes)

void f(int a,int b,int c){for(int i=0;i<a*b;i++)print((i%c>c-2?"X":".")+(i%a>a-2?"\n":""));}}

The reason I used Processing instead of Java is that you don't need to acces the pointer by tiping System.out because a local variable is directly accessible. I earned 11 bytes with this. The function doesn't return the result but prints it.

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1
  • 2
    \$\begingroup\$ You can save another by moving the increment (like i++%a...), and it looks like you left a spare } at the end you don't need, also. \$\endgroup\$ – Geobits Dec 7 '15 at 14:28
2
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Japt, 33 32 27 25 bytes

SpW-1 +Q p-~U*V/W f'.pU)·

Takes input in format W H N. Uses   and " in place of . and X, respectively. Try it online!

Ungolfed and explanation

SpW-1 +Q p-~U*V/W f'.pU)·qR
          // Implicit: U = width, V = height, W = interval
SpW-1 +Q  // Create a string of W - 1 spaces, plus a quotation mark.
p-~U*V/W  // Repeat this string ceil(U*V/W) times.
f'.pU)    // Split the resulting string into groups of U characters.
qR        // Join with newlines.
          // Implicit: output last expression

Suggestions welcome!

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2
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Vitsy, 25 23 22 21 19 Bytes

Thanks to @Sp3000 for pointing out that I don't need a duplicate and saving me 2 bytes!

Takes input as N W H. Try it online!

1}\0XrV\[V\[{DN]aO]
1                         Push 1 to the stack.
 }                        Push the backmost to the front and subtract 2.
  \0X                     Duplicate the 0 temp variable times.
     r                    Reverse the stack.
      V                   Save as final global variable.
       \[         ]       Repeat top item times.
         V\[   ]          Repeat global variable times.
            {DO           Duplicate, output, then shift over an item.
                aO        Output a newline.
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1
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K, 21 19 18 14 bytes

Takes arguments as (H W;N):

{".X"x#y=1+!y}

In action:

  f:{".X"x#y=1+!y};

  f.'((3 5;1);(3 5;2);(3 7;3);(4 10;5);(3 5;16))
(("XXXXX"
  "XXXXX"
  "XXXXX")
 (".X.X."
  "X.X.X"
  ".X.X.")
 ("..X..X."
  ".X..X.."
  "X..X..X")
 ("....X....X"
  "....X....X"
  "....X....X"
  "....X....X")
 ("....."
  "....."
  "....."))
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1
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Pyth - 19 18 17 bytes

Hope to golf it more. Takes input as N\n[W, H].

jc.[k+*dtvzN*FQhQ

Test Suite.

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1
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R, 66 bytes

function(w,h,n){x=rep(".",a<-w*h);x[1:a%%n<1]="X";matrix(x,h,w,T)}

This is a function that accepts three integers and returns a matrix of character values. To call it, assign it to a variable.

Ungolfed:

f <- function(w, h, n) {
    # Get the area of the square
    a <- w*h

    # Construct a vector of dots
    x <- rep(".", a)

    # Replace every nth entry with X
    x[1:a %% n == 0] <- "X"

    # Return a matrix constructed by row
    matrix(x, nrow = h, ncol = w, byrow = TRUE)
}
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1
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JavaScript (ES6), 65 60 bytes

(w,h,n)=>eval('for(i=r=``;i++<w*h;i%w?0:r+=`\n`)r+=i%n?0:1')

Explanation

(w,h,n)=>eval('    // use eval to remove need for return keyword
  for(
    i=             // i = current grid index
      r=``;        // r = result
    i++<w*h;       // iterate for each index of the grid
    i%w?0:r+=`\n`  // if we are at the end of a line, print a newline character
                   // note: we need to escape the newline character inside the template
  )                //       string because this is already inside a string for the eval
    r+=i%n?0:1     // add a 0 for . or 1 for X to the result
                   // implicit: return r
')

Test

W = <input type="number" id="W" value="7" /><br />
H = <input type="number" id="H" value="3" /><br />
N = <input type="number" id="N" value="3" /><br />
<button onclick="result.innerHTML=(

(w,h,n)=>eval('for(i=r=``;i++<w*h;i%w?0:r+=`\n`)r+=i%n?0:1')

)(+W.value,+H.value,+N.value)">Go</button>
<pre id="result"></pre>

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1
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Mathematica, 85 bytes

""<>(#<>"
"&/@ReplacePart["."~Table~{t=# #2},List/@Range[#3,t,#3]->"X"]~Partition~#)&

As with many other solutions, this creates a single row, then partitions it.

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1
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JavaScript (ES6), 55 bytes

(w,h,n)=>(f=i=>i++<w*h?+!(i%n)+(i%w?"":`
`)+f(i):"")(0)

Uses the IIFE f to loop to save a return statement.

Output for w=5, h=3, n=7:

00000
01000
00010
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1
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C#, 185 bytes

using System;class x{void a(int w,int h,int n){int c=1;for(int i=0;i<h;i++){for(int j=1;j<=w;j++){if(c%n==0){Console.Write("x");}else{Console.Write(".");}c++;}Console.WriteLine();}}}

For a more readable Reading:

using System;
class x
{
  void a(int w, int h, int n)
  {
    int c = 1;
    for (int i = 0; i < h; i++)
    {
        for (int j = 1; j <= w; j++)
        {
            if (c % n == 0)
            {
                Console.Write("x");
            }
            else
            {
                Console.Write(".");
            }
            c++;
        }
        Console.WriteLine();
     }
  }
}

Usage:

new x().a(7, 3, 3);
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0
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Julia, 50 bytes

f(w,h,n)=reshape([i%n<1?"X":"." for i=1:w*h],w,h)'

This creates a function f that accepts three integers and returns a 2-dimensional array of strings.

Ungolfed:

function f(w::Integer, h::Integer, n::Integer)
    # Construct an array of strings in reading order
    a = [i % n == 0 ? "X" : "." for i = 1:w*h]

    # Reshape this columnwise into a w×h array
    r = reshape(a, w, h)

    # Return the transpose
    return transpose(r)
end
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0
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Ruby, 67 56 bytes

->w,h,n{(1..h).map{(1..w).map{o,$.=$.%n<1?1:0,$.+=1;o}}}

Printing an array since it is accepted.

67 bytes

->w,h,n{i=1;puts (1..h).map{(1..w).map{o,i=i%n<1?1:0,i+=1;o}.join}}

Ungolfed:

-> w, h, n {
  (1..h).map {
    (1..w).map {
      o, $. = $.%n < 1 ? 1 : 0, $.+ = 1
      o
    }
  }
}

Usage:

->w,h,n{(1..h).map{(1..w).map{o,$.=$.%n<1?1:0,$.+=1;o}}}[8,6,7]
=> [[0, 0, 0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 0, 1, 0, 0], [0, 0, 0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0, 0], [0, 0, 1, 0, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0, 0, 0]]
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0
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MATLAB, 44 bytes

Note: Very different approach than the one used by Tom Carpenter.

@(x,y)char(reshape(~mod(1:prod(x),y),x)'+46)

Defines an anonymous function that accepts inputs as [W,H],N. I approached this problem using not-the-modulo-of-N for an array 1:W*H and then simply reshaping the solution to a two-dimensional array, which is then converted to a character array.

Example output for [5,3],7:

.....
./...
.../.
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0
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Common Lisp, SBCL, 94 bytes

(lambda(a b c)(dotimes(i(* a b))(format t"~:[.~;X~]~@[~%~]"(=(mod(1+ i)c)0)(=(mod(1+ i)a)0))))

Explanation

~:[.~;X~] <-- takes argument - if argument is true write ., if false write X
~@[~%~] <-- takes argument - if argument is true write newline, if not treat argument as if it was not used

(=(mod(1+ i)c)0)(=(mod(1+ i)a)0) looks pretty silly (because it's so similiar but I don't know if it can be solved, saving bytes

I use (1+ i) instead of i because dotimes starts from i=0 and I want to start from 1. It is also helpful because I can use (* a b) instead of (1+(* a b))

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0
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Japt -R, 9 bytes

Takes input as [W,H],N and uses spaces instead of .s & "s instead of Xs.

×îQùV)òUÎ

Try it

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-1
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Java, 185 183 bytes

Thanks Thomas Kwa, for saving me 2 bytes!

interface B{static void main(String[] a){int w = Byte.parseByte(a[0]);for(int i=0;i++<w*Byte.parseByte(a[1]);)System.out.print((i%Byte.parseByte(a[2])>0?".":"X")+(i%w<1?"\n":""));}}

Ungolfed (ish):

interface A {
  static void main(String[] a) {
    int w = Byte.parseByte(a[0]);
    for(
      int i = 0;
      i++ < w*Byte.parseByte(a[1]);
    )
      System.out.print((
        i%Byte.parseByte(a[2]) > 0 ? "." : "X"
        )+(
        i%w < 1 ? "\n" : ""
      ));
  }
}

Usage:

$ java B 5 3 7
.....
.X...
...X.

Maybe java will win one day :P

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1
  • \$\begingroup\$ I think you can use >0 instead of !=0, and <1 instead of ==0. \$\endgroup\$ – lirtosiast Nov 30 '15 at 0:44

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