Draw diagonal positions of me squared

Question

Based on the interpretation user @ThePirateBay made from the first version of my own challenge

Output diagonal positions of me squared

now I challenge you to make ASCII art that draws diagonal lines based on a square character matrix based on a given number n.

Input

A number n.

Output

A square matrix of size n^2 which outputs diagonals represented by the adequate one of the \ X /chars. All other positions are to be fulfilled with # chars.

Examples:

1
X

2
\/
/\

3
\#/
#X#
/#\

4
\##/
#\/#
#/\#
/##\

5
\###/
#\#/#
##X##
#/#\#
/###\  

There will be no accepted answer. I want to know the shortest code for each language.

Answer 1

Jelly, 14 bytes

⁼þµḤ+Ṛị“/\X#”Y

A monadic link taking a number and returning a list of characters; or a full program printing the result.

Try it online! or see [1-12] in one go.

How?

⁼þµḤ+Ṛị“/\X#”Y - Link: number, n
 þ             - outer product (implicit range build on left AND right from n) with:
⁼              -   is equal (yields a table of 0s except the main diagonal is 1s)
  µ            - monadic chain separation, call that I
   Ḥ           - double (change all the 1s to 2s)
     Ṛ         - reverse I (a table of 0s with 1s as the anti-diagonal)
    +          - add (vectorises to make a table of zeros with 2s on the diagonal,
               -      1s on the anti-diagonal and, if the meet, a 3 at the centre)
       “/\X#”  - literal list of characters ['/','\','X','#']
      ị        - index into (replaces... 1:'/'; 2:'\'; 3:'X'; and 0:'#')
             Y - join with newline characters
               - as a full program: implicit print

Answer 2

Python 2, 83 bytes

n=input()
for i in range(n):l=['#']*n;l[i],l[~i]='\X/X'[i-~i==n::2];print''.join(l)

Try it online!

-3 thanks to Jonathan Frech and G B.

Answer 3

Ruby, 73 61 59 bytes

->n{(0...n).map{|x|w=?#*n;w[x],w[~x]=?\\,n==1+x*2??X:?/;w}}

Try it online!

Answer 4

SOGL V0.12, 16 bytes

.↔╝Ζ #ŗ.2%?╬3←╬¡

Try it Here!

Finally got use out of the fact that palindromizating / or \ on the edge creates X (but this could be ~5 bytes less with a couple features I have had in mind for a while (e.g. that overlapping would overlap slashes too and palindomizating commands have the option to choose the overlap amount from the remainder of ceiling dividing))

Explanation:

.↔                push input ceiling-divided by 2 (. is required because SOGLs input is taken lazily)
  ╝               create a diagonal of that length
   Ζ #ŗ           replace spaces with hashes
       .2%?       if the input % 2 isn't 0
           ╬3       palindromize with 1 overlap
             ←      stop program
              έ  [else] palindromize with 0 overlap

Answer 5

R, 68 bytes

write(c("#","\\","/","X")[(d=diag(n<-scan()))+d[n:1,]*2+1],"",n,,"")

Try it online!

Reads n from stdin. Creates a matrix of indices (d=diag(n))+d[n:1,]*2+1 to index into the vector of characters to print, which results in a vector of characters. write writes it as a matrix to "" (stdout) with n columns and separator "".

Answer 6

Charcoal, 21 bytes

ＵＢ#Ｎν↗∕ν²×X﹪ν²‖ＢＯL﹪ν²

Try it online! Verbose version.

Answer 7

MATL, 14 bytes

'\/X#'iXytPE+)

Try it online!

Explanation: suppose the input is 3

'\/X#'               push this character string.
i                    read in input. stack is ['\/X#';3]
Xy                   push nxn identity matrix.
                     stack is ['\/X#'; [1 0 0 ; 0 1 0 ; 0 0 1]]
tP                   duplicate and flip left-right
                     stack is ['\/X#'; [1 0 0 ; 0 1 0 ; 0 0 1];
                                       [0 0 1 ; 0 1 0 ; 1 0 0]]
E+                   double TOS and add top 2 stack elements
                     stack is ['\/X#'; [1 0 2 ; 0 3 0 ; 2 0 1]]
)                    index; 1-based modular, so
                     1 -> \, 2 -> /, 3 -> X, 0 -> #
                     implicit output.

Answer 8

Javascript, 103 bytes

n=>{for(let i=n,j,s;i--;){s="";for(j=n;j--;)s+=i==j?n-j-1==j?"X":"\\":n-j-1==i?"/":"#";console.log(s)}}

Try it online!

A more readable version with spaces and breaks added:

n => {
  for (let i = n, j, s; i--;) {
    s="";
    for (j = n; j--;) 
      s += i == j
        ? n - j - 1 == j
          ? "X"
          : "\\"
        : n - j - 1 == i
          ? "/"
          : "#";
    console.log(s)
  }
}

Answer 9

Husk, 20 bytes

´Ṫ§o!"¦/X#"+=ȯD=←¹+ŀ

Try it online!

Answer 10

V, 24 bytes

Àé#ÀÄòr\jlò|òr/klòÇÜ/ãrX

Try it online!

Answer 11

Mathematica, 114 bytes

(s=#;F[x_]:=DiagonalMatrix[Table[x,s]];z=F@"\\"+Reverse@F@"/"//. 0->"#";If[OddQ@s,z[[t=⌈s/2⌉,t]]="X"];Grid@z)&

Answer 12

Python 2, 133 bytes

def f(k,h="#"):s=[h*i+"\\"+h*(k-i*2-2)+"/"+i*h for i in range(k/2)];print"\n".join(s+k%2*[k/2*h+"X"+k/2*h]+[e[::-1]for e in s[::-1]])

Try it online!

A beautiful mess done on mobile :-). If you have golfing suggestions, you are welcome to edit in yourselves (and credit you!), because I will be sleeping :P.

Answer 13

Python 2, 122 bytes

def f(n):s='\\'+'#'*(n-2)+'/';return['x'][:n]if n<2else[s]+['#'+l+'#'for l in f(n-2)]+[s[::-1]]
print'\n'.join(f(input()))

Try it online!

A recursive aproach

Answer 14

Perl 5, 81 bytes

for$i(0..($n=<>)-1){$_='#'x$n;s|.{$i}\K.|\\|;s%.(.{$i})$%/$1%;s%/%X% if!/\\/;say}

Try it online!

Answer 15

Canvas, 15 bytes

：：#×＊；＼：↔１１╋１１╋

Try it online!

^{_{could get it to 11 bytes if canvas's overlap function had a default setting but it doesn't so...}}

Answer 16

Mathematica, 99 bytes

FromCharacterCode[(2#+Reverse@#+2)&@DiagonalMatrix[45~Table~#]/.{2->35,137->88}]~StringRiffle~"\n"&

This uses a somewhat different technique from the other Mathematica solution. Also, we should probably start calling it the Wolfram Language at some point.