Draw a big slash X

Question

Given a non-negative integer, print out an X that is that big. X is the output for input 0, and you will add slashes equal to the input in each direction to extend the X for larger inputs.

Test Cases

0

X

1

\ /
 X
/ \

2

\   /
 \ /
  X
 / \
/   \

...

10

\                   /
 \                 /
  \               /
   \             /
    \           /
     \         /
      \       /
       \     /
        \   /
         \ /
          X
         / \
        /   \
       /     \
      /       \
     /         \
    /           \
   /             \
  /               \
 /                 \
/                   \

Rules

You may either print the output, or return a string or list of string from a function. A trailing newline, as well as extra interior whitespace that does not affect appear, is allowed.

This is code-golf, so shortest answer in bytes wins!

Answer 1

Perl 5 -n, 59 bytes

map{//;say map$_|$'?$_-$'?$_+$'?$":'/':'\\':X,@a}@a=-$_..$_

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Uses the same logic as my ASCII sun answer here.

It treats the output as an X-Y grid from -n to +n where n is the input value. If X==-Y, output a slash. If X==Y, output a backslash. Put an X at the origin. Everything else is a space.

Answer 2

YASEPL, 114 111 bytes

=i!m'!n$!u$m]2!f$47»!b$92»!l$+`1!o;" ",i~#b$u-i*-;" ",o~>f!i+n}l,m-`2!n[3--!o;" ",m~>"X"!m]3=m-!l$!b)f!f$92»|`3

explanation (poor)

=i!m'!n$!u$m]2!f$47»!b$92»!l$+`1!o;" ",i~#b$u-i*-;" ",o~>f!i+n}l,m-`2!n[3--!o;" ",m~>"X"!m]3=m-!l$!b)f!f$92»|`3         packed
=i                                                                                                                      increment variable
  !m'                                                                                                                   input 
     !n$                                                                                                                N = 1
        !u$m                                                                                                            U = M
            ]2                                                     `2                                                   If U != 0 ...
              !f$47»!b$92»                                                                                                F = "/", B = "\"
                          !l$+                                                                                            set L to 2
                              `1                          !i  }l,m                                                        while (L == 2 and I < M) xor (L == 1 and I > M) ...
                                !o;" ",i~                                                                                   write a space I amount of times (inline)
                                         #b                                                                                 write B (inline)
                                           $u-i*-;" ",o~                                                                    write a space [U - I] * 2 - 1 amount of times (inline)
                                                        >f                                                                  print F with newline at end
                                                            +n                                                              add N to I and loop.
                                                                  -                                                     decrement I
                                                                     !n[3                                    `3         If N == 1 ...
                                                                         --                                               subtract 2 from N
                                                                           !o;" ",m~>"X"                                  write a space M amount of times inline and an X with newline
                                                                                        !m]3                 `3           if M != 0 ...
                                                                                            =m-                             set M to -1
                                                                                               !l$                          set L to 1
                                                                                                  !b)f!f$92»                swap B and F
                                                                                                            |               loop back to 1 and  print other half

Answer 3

K (ngn/k), 19 18 bytes

-1 byte from @att

"X\\/"@2/~|:\=1+2*

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Thanks to @tubular.

• 1+2* multiply the (implicit) input by 2, then add 1
• = generate an identity matrix of the above size
• |:\ return a list containing the identity matrix and its reverse (i.e. the antidiagonal)
• ~ "not" the above, i.e. convert 1's to 0's and 0's to 1's
• 2/ convert to base-2
• "X\\/"@ index into a string containing the appropriate characters

Answer 4

Perl 5, 110 + 1 = 111 bytes

Uses -n flag.

$x=$_;$,="\n";push@x,$"x($x-$_)."\\".$"x(2*--$_+1)."/"while$_>0;say@x,$"x$x."x";for(reverse@x){y!\\/!/\\!;say}

Answer 5

QBIC, 90 bytes

~b=0|?@X`\[0,:-1|X=space$(a)┘Z=Z+X+@\`+space$((b-a)*2-1)+@/`+X+@┘`]Z=Z+space$(b)+A+_fZ

How this monstrosity works, is left as an excercise for the reader...

Sample output:

Command line: 3
\     /
 \   / 
  \ /  
   X
  / \  
 /   \ 
/     \

Answer 6

Visual Basic.Net, 454 450 Bytes

Option Strict Off
module m
sub main(a As String())
dim v=Convert.toInt32(a(0))
for i as Integer=v to 1 step -1
for j as Object=1 to v-i
w(" ")
next
w("\")
for j as Object=1 to i*2-1
w(" ")
next
console.writeline("/")
next
console.writeline(new String(" ",v)&"X")
for i as Object=1 to v
for j as Object=1 to v-i
w(" ")
next
w("/")
for j as Object=1 to i*2-1
w(" ")
next
console.writeline("\")
next
end sub
sub w(s)
console.write(s)
end Sub
end module

not sure whether making a func for writeline will save some bytes thanks to Stephen S for pointing at as ... removal also changed integer into object last edit changed the first one back

Answer 7

05AB1E, 22 bytes

F'\IN-úR.∞})Âí'Xs)˜.c»

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Explanation

F                        # for N in [0...input-1] do
 '\                      # push a backslash
   IN-ú                  # prepend input-N spaces
       R                 # reverse
        .∞               # mirror
          }              # end loop
           )             # wrap stack in a list
            Â            # bifurcate
             í           # reverse each item
              'Xs        # push an "X" between the 2 lists on the stack
                 )˜      # wrap in flattened list
                   .c    # pad lines to equal length
                     »   # join on newlines

Alternative 22 byte solution

F'\N·>ú'/ì})Âí'X¸«ì.c»

Answer 8

Pyke, 14 bytes

\XQV.X \   /\/

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\X             - "X"
  QV           - repeat input times:
    .X \   /\/ -  surround(^, all=" ", 
                              tl="\", 
                              left=" ",  
                              right=" ",  
                              lower=" ",  
                              tr="/",  
                              br="\",  
                              bl="/")

Answer 9

tcl, 134

proc P {x s b} {time {puts [format %[incr ::i $x]s%[expr ($::n-$::i)*2+2]s $s $b]} $::n}
P 1 \\ /
puts [format %[incr i]s X]
P -1 / \\

demo

Set n on the first line.

May be I can golf it more using a recursive approach

Answer 10

R, 75 72 bytes

y=diag(b<-2*scan()+1);write(c(" ",'\\','/','X')[y+2*y[,b:1]+1],'',b,,'')

Inspired by this answer, generates an implicit matrix and writes it to stdout; reads the size from stdin. It has to build a matrix of space characters and uses sep='' b/c otherwise it has spacing issues.

diag(b)                     # generates a main diagonal of 1, rest 0
2*diag(b)[,b:1]             # the other diagonal is 2
                            # [,b:1] reverses columns
                            # [b:1,] reverses the rows; they're equivalent
diag(b)+2*diag(b)[,b:1]     # sums, so we get 1 for main diagonal
                            # 2 for other diagonal, 3 for center
diag(b)+2*diag(b)[,b:1]+1   # add 1 since R is 1-indexed
                            # the result is indices for the vector
c(' ','\\','/','X')

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Answer 11

Zsh, 83 bytes

2=${(l:$1:)}
S=$2\ $2
F()for i;(S[i]=\\;S[-i]=/;<<<$S)
F {1..$1}
<<<$2X
F {-$1..-1}

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Answer 12

sed, 112 111 bytes

s/^0/X/
/X/q
s/.*/echo {0..&}/e
s/[^ ]//g
s%.*%\\&&/%
s/ //p
:a
/^\//Q
h;s%\\ /% X%p;x
s%(\\)? (/)?%\2 \1%gp
ba

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basically a port of my sed answer to another challenge (see explanation there) which uses the engine from this answer

Answer 13

J, 25 bytes

' \/X'{~1(+2*|.)@=@i.@++:

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Arrived at independently, but very similar approach to Adam's APL answer. J has a primitive = that returns the identity matrix when given a list of distinct number, making that part even shorter.

Answer 14

Rust, 150 bytes

|n|{let w=|e|" ".repeat(e);for i in 0..n{print!("{}\\{}/
",w(i),w(2*n-2*i-1))}print!("{}X
",w(n));for i in 0..n{print!("{}/{}\\
",w(n-i-1),w(2*i+1))}}

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Answer 15

Pip -l, 22 bytes

YEYa*2+1(" /\X"y*2+Ry)

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Explanation

YEYa*2+1(" /\X"y*2+Ry)­⁡​‎‎⁡⁠⁤‏⁠‎⁡⁠⁢⁡‏⁠‎⁡⁠⁢⁢‏⁠‎⁡⁠⁢⁣‏⁠‎⁡⁠⁢⁤‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁢‏⁠‎⁡⁠⁣‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁡‏‏​⁡⁠⁡‌⁤​‎‎⁡⁠⁤⁤‏⁠‎⁡⁠⁢⁡⁡‏⁠‎⁡⁠⁢⁡⁢‏‏​⁡⁠⁡‌⁢⁡​‎‎⁡⁠⁢⁡⁣‏⁠‎⁡⁠⁢⁡⁤‏⁠‎⁡⁠⁢⁢⁡‏‏​⁡⁠⁡‌⁢⁢​‎‎⁡⁠⁣⁡‏⁠‎⁡⁠⁢⁢⁢‏‏​⁡⁠⁡‌⁢⁣​‎‎⁡⁠⁣⁢‏⁠‎⁡⁠⁣⁣‏⁠‎⁡⁠⁣⁤‏⁠‎⁡⁠⁤⁡‏⁠‎⁡⁠⁤⁢‏⁠‎⁡⁠⁤⁣‏‏​⁡⁠⁡‌­
   a*2+1                ; ‎⁡Twice the argument plus 1
 EY                     ; ‎⁢Identity matrix of that size
Y                       ; ‎⁣Yank into y
               y*2      ; ‎⁤Multiply each number in y by 2
                  +Ry   ; ‎⁢⁡Add y reversed
                        ; ‎⁢⁡We now have a 2D array containing 2s on the main diagonal,
                        ; 1s on the main antidiagonal, 3 at the intersection,
                        ; and 0s elsewhere
        (            )  ; ‎⁢⁢Use that array as a (multi-)index into
         " /\X"         ; ‎⁢⁣this string
                        ; ‎⁢⁣Join each row and output them newline-separated (-l flag)
💎

Created with the help of Luminespire.

Answer 16

Go, 199 193 bytes

import."strings"
func f(n int)(s string){S,R:=" ",Repeat
for i:=range n{s+=R(S,i)+`\`+R(S,max(0,2*(n-i)-1))+`/
`}
s+=R(S,n)+`X
`
for i:=range n{s+=R(S,n-i-1)+`/`+R(S,max(0,2*i+1))+`\
`}
return}

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• -6 by Go 1.22+ range-over-int.