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.

The background of octothorpes, the X in the middle if input is odd and each line in the big X is as long as input, not double the size. I think it's not a dupe. — totallyhuman, Aug 30 '17 at 22:51
The output for even inputs makes this not a dupe (along with the other comments) — Stephen, Aug 30 '17 at 23:19
I believe this to still be a duplicate. Here is a trivial change made to the MATL answer on the duplicate question. NB: I don't even know MATL. — H.PWiz, Aug 31 '17 at 0:16
No, but it works for all that indexed into a string that looked something like "\/X ", e.g This — H.PWiz, Aug 31 '17 at 0:59

Jonathan Allan · Accepted Answer · 2017-08-30 22:20:08Z

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

\$\begingroup\$ Clever approach! \$\endgroup\$ – Erik the Outgolfer Sep 25 '17 at 16:28 — Erik the Outgolfer, Sep 25 '17 at 16:28

Erik the Outgolfer · Accepted Answer · 2017-09-26 09:38:48Z

4

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.

edited Sep 26 '17 at 9:38

answered Sep 25 '17 at 16:46

Erik the Outgolfer

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\$\begingroup\$ Your TIO link has a footer artifact. \$\endgroup\$ – Jonathan Frech Sep 25 '17 at 16:51
\$\begingroup\$ @JonathanFrech I realized a version of mine was invalid and quickly rolled back... \$\endgroup\$ – Erik the Outgolfer Sep 25 '17 at 16:59
\$\begingroup\$ I noticed... :P \$\endgroup\$ – Jonathan Frech Sep 25 '17 at 17:00
\$\begingroup\$ Is not n/2.-.5 equivalent to ~-n/2.? \$\endgroup\$ – Jonathan Frech Sep 25 '17 at 17:05
\$\begingroup\$ @JonathanFrech missed that \$\endgroup\$ – Erik the Outgolfer Sep 25 '17 at 17:06

| show 4 more comments

G B · Accepted Answer · 2017-09-26 06:44:00Z

2

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!

edited Sep 26 '17 at 6:44

answered Sep 25 '17 at 15:03

G B

8,77614 silver badges30 bronze badges

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dzaima · Accepted Answer · 2017-08-30 22:20:37Z

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
              ╬¡  [else] palindromize with 0 overlap

Giuseppe · Accepted Answer · 2017-09-25 14:50:47Z

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 "".

totallyhuman · Accepted Answer · 2017-12-01 11:59:57Z

1

Charcoal, 21 bytes

ＵＢ#Ｎν↗∕ν²×X﹪ν²‖ＢＯL﹪ν²

Try it online! Verbose version.

edited Dec 1 '17 at 11:59

answered Aug 30 '17 at 22:04

totallyhuman

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\$\begingroup\$ There's a neat trick to save two bytes by using multidirectional mirrors. In your case, either L or ¬ would work, but you would have to draw your line in a different direction of course. \$\endgroup\$ – Neil Oct 27 '17 at 11:57
\$\begingroup\$ Note that the deverbosifier allows you to omit the else, so what you're seeing there is "if v is odd, draw an X and reflect with overlap (so that the X is not itself reflected) otherwise reflect without overlap". \$\endgroup\$ – Neil Oct 27 '17 at 11:59
\$\begingroup\$ You can actually save a byte by using ReflectButterflyOverlap which takes a second parameter which is the amount of overlap. If you give it an overlap of 0 then you get a ReflectMirror and if you give it an overlap of 1 then you get a ReflectButterfly. \$\endgroup\$ – Neil Oct 27 '17 at 12:01
1

\$\begingroup\$ Try it online! \$\endgroup\$ – Neil Nov 28 '17 at 11:12

add a comment |

Giuseppe · Accepted Answer · 2017-12-01 21:06:53Z

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.

Wolff · Accepted Answer · 2017-12-21 00:00:11Z

1

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)
  }
}

answered Dec 21 '17 at 0:00

Wolff

1112 bronze badges

\$\begingroup\$ You can save a few bytes by using a return statement instead of logging it. \$\endgroup\$ – Nissa Dec 21 '17 at 0:22
\$\begingroup\$ Thanks @StephenLeppik, but I thought I had to output the result? Would I have to include "\n" after each line? \$\endgroup\$ – Wolff Dec 21 '17 at 22:39
\$\begingroup\$ You do have to output it, but on this site your return value is valid output. \$\endgroup\$ – Nissa Dec 21 '17 at 22:51

add a comment |

DJMcMayhem · Accepted Answer · 2017-08-30 22:18:26Z

0

V, 24 bytes

Àé#ÀÄòr\jlò|òr/klòÇÜ/ãrX

Try it online!

answered Aug 30 '17 at 22:18

DJMcMayhem♦

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J42161217 · Accepted Answer · 2017-08-30 22:31:59Z

0

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)&

answered Aug 30 '17 at 22:31

J42161217

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Mr. Xcoder · Accepted Answer · 2017-08-30 22:42:10Z

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.

Rod · Accepted Answer · 2017-08-30 22:43:31Z

0

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

edited Aug 30 '17 at 22:43

answered Aug 30 '17 at 22:31

Rod

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Xcali · Accepted Answer · 2017-12-01 17:49:38Z

0

Perl 5, 81 bytes

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

Try it online!

answered Dec 1 '17 at 17:49

Xcali

6,9251 gold badge7 silver badges24 bronze badges

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hakr14 · Accepted Answer · 2018-02-16 17:31:47Z

0

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...}}

answered Feb 16 '18 at 17:31

hakr14

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\$\begingroup\$ You don't need the starting ：： as the stack (practically) contains the inputs repeated forever. And yeah, I need to add a 1 byte built-in for １１╋ \$\endgroup\$ – dzaima Apr 7 '18 at 12:22

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DanTheMan · Accepted Answer · 2018-02-16 19:35:28Z

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.

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Draw diagonal positions of me squared

Output diagonal positions of me squared

Input

Output

Examples:

15 Answers 15

Jelly, 14 bytes

How?

Python 2, 83 bytes

Ruby, 73 61 59 bytes

SOGL V0.12, 16 bytes

R, 68 bytes

Charcoal, 21 bytes

MATL, 14 bytes

Javascript, 103 bytes

V, 24 bytes

Mathematica, 114 bytes

Python 2, 133 bytes

Python 2, 122 bytes

Perl 5, 81 bytes

Canvas, 15 bytes

Mathematica, 99 bytes

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Input

Output

Examples:

15 Answers 15

Jelly, 14 bytes

How?

Python 2, 83 bytes

Ruby, 73 61 59 bytes

SOGL V0.12, 16 bytes

R, 68 bytes

Charcoal, 21 bytes

MATL, 14 bytes

Javascript, 103 bytes

V, 24 bytes

Mathematica, 114 bytes

Python 2, 133 bytes

Python 2, 122 bytes

Perl 5, 81 bytes

Canvas, 15 bytes

Mathematica, 99 bytes

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