Print a 3D shape

Question

Write a program or a function to print the following cube (if you allow me to call it so) in different sizes:

          ^L^L^L^L^L^L^L^L^L^L^L
         //^L^L^L^L^L^L^L^L^L^L^L
        ////^L^L^L^L^L^L^L^L^L^L^L
       //////^L^L^L^L^L^L^L^L^L^L^L
      ////////^L^L^L^L^L^L^L^L^L^L^L
     //////////^L^L^L^L^L^L^L^L^L^L^L
    ////////////^L^L^L^L^L^L^L^L^L^L^L
   //////////////^L^L^L^L^L^L^L^L^L^L^L
  ////////////////^L^L^L^L^L^L^L^L^L^L^L
 //////////////////^L^L^L^L^L^L^L^L^L^L^L
////////////////////^L^L^L^L^L^L^L^L^L^L^L
\\\\\\\\\\\\\\\\\\\\ " " " " " " " " " " "
 \\\\\\\\\\\\\\\\\\ " " " " " " " " " " "
  \\\\\\\\\\\\\\\\ " " " " " " " " " " "
   \\\\\\\\\\\\\\ " " " " " " " " " " "
    \\\\\\\\\\\\ " " " " " " " " " " "
     \\\\\\\\\\ " " " " " " " " " " "
      \\\\\\\\ " " " " " " " " " " "
       \\\\\\ " " " " " " " " " " "
        \\\\ " " " " " " " " " " "
         \\ " " " " " " " " " " "
           " " " " " " " " " " "

The shape is made of pairs of characters: ^L, ", \\, //

Input:

An integer greater than 1, representing the size of the shape.
_{The one showed above has a size of 11 because each side of each face is made of 11 pairs of characters.

Note that only two faces are fully displayed.}

Output:

A cube of the given size.

Here's another example, with a size of 5:

    ^L^L^L^L^L
   //^L^L^L^L^L
  ////^L^L^L^L^L
 //////^L^L^L^L^L
////////^L^L^L^L^L
\\\\\\\\ " " " " "
 \\\\\\ " " " " "
  \\\\ " " " " "
   \\ " " " " "
     " " " " "

This is code-golf, the shortest in bytes wins.

_{Even though you are not required to do so, it would be nice if you could provide an easy way to check the output of your code}

May we choose which characters to use, as long as all 8 are unique? — Shaggy, Jan 23, 2021 at 20:22
@shaggy I think it's better to stick with the original characters. It's true that the algorithm would be quite the same even using other characters, but you may lose the "3D illusion". So then it wouldn't be an algorithm to print a 3D shape anymore — anotherOne, Jan 23, 2021 at 20:42

xnor · Accepted Answer · 2021-01-23 20:22:47Z

9

Python 2, 81 bytes

k=n=input()
while 1:k-=1;i=k^k/n;print' '*i+(n+~i)*2*'\/'[k/n-1]+'^ L"'[k<0::2]*n

Try it online!

Reuses some ideas from my answer on Draw an ASCII hexagon of side length n, including using the index k/n-1 both to select between / and \ in an order where the \ doesn't need escaping, as well as to terminate the loop with error at the end.

edited Jan 23, 2021 at 20:22

answered Jan 23, 2021 at 20:10

xnor

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Razetime · Accepted Answer · 2021-01-31 12:32:00Z

Canvas, ²⁷ 22 bytes

｛╷２×/×^L⁸×＋］：↕vL∙ "╋∔ｒ

Try it here!

Explanation

{|2×/×^L⁸×+]:↕vL∙ "╋∔r
{          ]           map each i from 1 to n:
 |2×                   decrement and double i
    /×                 repeat "/" that many times
      ^L⁸×             repeat "^L" n times
          +            add that to get one line
            :          copy the top half
             ↕         mirror vertically (mirrors characters)
              vL∙" ╋   replace 'vL' with '" '
                    ∔  and add vertically
                     r center the whole art

Level River St · Accepted Answer · 2021-01-24 11:11:27Z

5

Ruby, 78 bytes

->n{puts (0...n).map{|i|puts ' '*(j=n+~i)+?/*i*2+'^L'*n;' '*i+?\\*j*2+' "'*n}}

Builds an array for the bottom half while printing the top half, then dumps the bottom half at the end. I also tried recursion but it was longer.

Try it online!

edited Jan 24, 2021 at 11:11

answered Jan 24, 2021 at 10:45

Level River St

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3

\$\begingroup\$ you can remove the space between puts and ' ' \$\endgroup\$
– Razetime
Jan 24, 2021 at 14:42

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Alex bries · Accepted Answer · 2021-02-10 09:54:29Z

5

JavaScript (Node.js), 121 bytes

s=>(j=n=>' '[b="repeat"]((a=n>=s)?n-s:s+~n)+(a?"//":"\\\\")[b](a?s*2-n:n+1)+(a?"^L":' "')[b](s)+'\n'+(~n?j(--n):''))(s*2)

Try it online!

edited Feb 10, 2021 at 9:54

answered Jan 23, 2021 at 21:17

Alex bries

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coltim · Accepted Answer · 2021-01-24 02:55:57Z

K (ngn/k), 69 bytes

{` 0:,/',/$(`" "`"//"`"^L";`" "`"\\\\"`" \"")@'&''(t;|t:+(|!x;!x;x))}

Try it online!

Not a great fit for the language, but an interesting exercise...

(t;|t:+(|!x;!x;x)) build a nested list containing the number of copies of each type of symbol to take. (i.e. spaces from x..0,0..x, slashes from 0..x,x..0, and x copies of ^L and ").
&'' convert to indices, e.g. 3 1 5 => 0 0 0 1 2 2 2 2 2
(`" "`"//"`"^L";`" "`"\\\\"`" \"") build another nested list (of the same shape), this time containing symbol versions of the output characters. This avoids having to enlist the individual strings and simplifies the list notation a bit. Since they contain special characters (as opposed to being alphanumeric), the contents need to be enclosed in "'s, with " and \ also needing to be escaped.
(...)@'(...) retrieve the correct number of copies of each type of symbol
,/',/$ stringify the symbols, and raze them into the lines of the output
` 0: print them (this elides the wrapping "s and \ escapes otherwise present)

ovs · Accepted Answer · 2021-01-25 08:06:09Z

05AB1E, 28 26 bytes

-2 bytes thanks to Kevin Cruijssen!

Without this bug, »∊¶¡ could be just ∊.

L<·'/×»∊¶¡„^L„ "‚I×Iδи˜ø.c

Try it online!

Commented:

L<·'/×»∊¶¡      # generate the left part
L               # push the range [1..n]
 <              # decrement to get [0..n-1]
  ·             # double each value: [0, 2, ..., 2*(n-1)]
   '/           # push the string '/'
     ×          # for each number in the list, repeat the string this many times
      »∊¶¡      # vertically mirror to get lower left half

„^L„ "‚I×Iδи˜   # generate the right part
„^L             # push the string '^L'
   „ "          # push the string ' "'
      ‚         # pair into a list: ['^L', ' "']
       I×       # repeat each string input times 
         Iδи    # for each string create a list containing the string input times
            ˜   # flatten into a single list

 ø              # zip both parts to get the rows
   .c           # join each row, centralize and join by newlines

Try it with step-by-step output!

Some slightly longer alternatives:

L<·'/×»∊¶¡εNI@i„ "ë„^L}I×«}.c
L<·'/×»∊¶¡ε„^LNI@i„ "}I×yì}.c
L<·Â«εNI@i'\×„ "ë'/×„^L}I×«}.c

‚øJ.c can be ø.c for -2, because the ø uses two arguments implicitly with a string-list in the new 05AB1E version (unlike the legacy, where it would transpose a string-list as if it were a character-matrix), and the .c joins implicitly. — Kevin Cruijssen, Jan 25, 2021 at 7:37

mazzy · Accepted Answer · 2021-01-26 14:16:50Z

4

PowerShell, 88 bytes

param($n)$n..1|%{($l=' '*--$_)+'//'*($j=$n-$_-1)+'^L'*$n
$b=,($l+'\\'*$j+' "'*$n)+$b}
$b

Try it online!

PowerShell, 104 102 92 bytes, Alternative

param($n),{$l='//';'^L';$i++}*$n+,{$l='\\';--$i;' "'}*$n|%{.$_}|%{' '*($n-$i-1)+$l*$i+$_*$n}

Try it online!

The script generates the array of scriptblocks and executes its in the same context. Less golfed:

param($n)filter g{' '*($n-$i-1)+$l*$i+$_*$n}
,{$l='//';'^L'|g;$i++}*$n+,{$l='\\';--$i;' "'|g}*$n|%{.$_}

edited Jan 26, 2021 at 14:16

answered Jan 26, 2021 at 10:33

mazzy

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Kevin Cruijssen · Accepted Answer · 2021-01-26 15:23:44Z

Java 11, 148 136 bytes

n->{String r="",t=r;for(int i=n;i-->0;t+=" ")r=t+"//".repeat(i)+"^L".repeat(n)+"\n"+r+t+"\\\\".repeat(i)+" \"".repeat(n)+"\n";return r;}

-12 bytes thanks to @OlivierGrégoire.

Try it online.

Explanation:

n->{                        // Method with integer parameter and String return-type
  String r="",              //  Result-String, starting empty
         t=r;               //  Temp-string, starting empty as well
  for(int i=n;i-->0         //  Loop `i` in the range [input, 0):
      ;                     //    After every iteration:
       t+=" ")              //     Append a space to `t`
    r=...+r+...;            //   Change the result-String to:
                            //    Prepend to the result-String:
      t                     //     The spaces `t`
      +"//".repeat(i)       //     plus '//' repeated `i-1` amount of times
      +"^L".repeat(n)       //     plus '^L' repeated the input amount of times
      +"\n"                 //     plus a newline
                            //    Append to the result-String:
      t                     //     The spaces `t` again
      +"\\\\".repeat(i)     //     plus '\\' repeated `i-1` amount of times
      +" \"".repeat(n)      //     plus ' "' repeated the input amount of times
      +"\n";                //     plus a newline again
  return r;}                //  After the loop, return the result-String

136 bytes, mostly changing t=" ".repeat(...) with t+=" " and rearranging the rest a bit. — Olivier Grégoire, Jan 26, 2021 at 14:22

Neil · Accepted Answer · 2021-01-23 18:06:09Z

3

Charcoal, 42 32 bytes

ＮθＧ⌈⊖θ\↓‖Ｃ‖Ｍ↑Ｆθ«Ｐ×^Lθ↘»Ｆθ«Ｐ×" θ↙

Try it online! Link is to verbose version of code. Explanation:

Ｎθ

Input the size.

Ｇ⌈⊖θ\↓

Print a small triangle of \s.

‖Ｃ‖Ｍ↑

Reflect it twice, once horizontally to complete the bottom left block of \s, once vertically with mirroring to produce the top left block of /s.

Ｆθ«Ｐ×^Lθ↘»

Print the top right block of ^s and Ls.

Ｆθ«Ｐ×" θ↙

Print the bottom right block of "s and spaces.

edited Jan 23, 2021 at 18:06

answered Jan 23, 2021 at 17:54

Neil

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\$\begingroup\$ Doesn't the vertical reflection change the slashes in backslashes? If so, wow! \$\endgroup\$
– anotherOne
Jan 23, 2021 at 18:02
2

\$\begingroup\$ @Davide It depends on whether you use ‖Ｃ or ‖Ｍ, see my updated answer. \$\endgroup\$
– Neil
Jan 23, 2021 at 18:06
\$\begingroup\$ Oh that's powerful \$\endgroup\$
– anotherOne
Jan 23, 2021 at 18:13

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Danis · Accepted Answer · 2021-01-24 16:36:52Z

3

Python 3.8 (pre-release), ^{97 93} 92 bytes

lambda a:sum(zip(*((' '*(x:=a+~i)+'//'*i+'^L'*a,' '*i+r'\\'*x+' "'*a)for i in range(a))),())

Try it online!

Output is a tuple of lines.

-4 by @Seb

-1 by @Danis

edited Jan 24, 2021 at 16:36

Danis

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answered Jan 23, 2021 at 19:04

pxeger

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\$\begingroup\$ You can save a byte by changing '\\\\' to r'\\', and 3 more by using Python 3.8 for its walrus operator \$\endgroup\$
– Seb
Jan 24, 2021 at 3:51
\$\begingroup\$ a-i-1 --> a+~1 save 1 byte \$\endgroup\$
– Danis
Jan 24, 2021 at 11:35

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Noodle9 · Accepted Answer · 2021-01-25 09:03:42Z

C (gcc), ^{160 135} 130 bytes

Saved 5 bytes thanks to ceilingcat!!!

p(n,c){for(;n--;)printf(L"\n//\\\\L^ \"  "+c+n%2);}i;t;f(n){for(t=i=1;i;i+=t,t=i>n?i=n,-1:t)p(n-i,9),p(!p(2*n,6-t),p(2*~-i,2-t));}

Try it online!

Explanation (before some golfs)

d;p(n,c){                             // p is a helper function that prints     
                                      // n/2 pairs of chars  
  for(d=0;n--;d^=1)                   // loop n times, flipping d between  
                                      //   0 and 1 each time  
    printf(L"                         // print one char of a pair of 
                                      //   chars based on c (then select the  
                                      //   other one on the next loop as d 
                                      //   flips between 0 and 1):  
             //                       //   c==0: '//'  
               \\\\                   //   c==2: '\\'  
                   ^L                 //   c==4: '^L'  
                      \"              //   c==6: '" '  
                                      //   c==7: any number of spaces
                          \n"         //   c==9: p(1,9) prints a newline  
             +c+d);                   // use c and d to select char
}                                     // returns 0 
i;t;f(n){                             // f is the main function    
  for(t=i=1;                          // main loop i goes from 1 to n and  
                                      // back down to 1 again  
    i;                                // loop until i is 0  
    i+=t,                             // bump i up for 1st half and then  
                                      // bump i down for 2nd half  
         t=i>n?i=n,-1:t)              // flip t from 1 to -1 half way thru  
               p(n-i,7),              // print leading spaces   
               p(2*~-i,1-t),          // print left side slashes   
               p(2*n,5-t),            // print right side of cube   
               p(1,9);                // print newline
}                                     //

This is shorter than the one I made before to post the challenge. Nice usage of printf! I need to learn that — anotherOne, Jan 24, 2021 at 12:33
How does it work this L in printf, before the string? I tried to find out, but I get only results about the L in %L — anotherOne, Jan 24, 2021 at 16:15
@Davide So the way it works is the 8-bit chars are zero expanded when converted to 32-bit wchars. printf sees the zeros as end of strings so only prints the single char.that's pointed to by the pointer after the int offset is added. — Noodle9, Jan 24, 2021 at 22:47

Jonathan Allan · Accepted Answer · 2021-01-25 19:09:03Z

Jelly, 30 bytes

ḶṚżḤƊØ.xⱮ;€3,5ẋ$;ṚḤ$$ị“/\^"L ”

A monadic Link accepting a positive integer which yields a list of lists of characters (a list of lines).

Try it online! (The footer calls the Link and joins the resulting lines with newline characters.)

How?

ḶṚżḤƊØ.xⱮ;€3,5ẋ$;ṚḤ$$ị“/\^"L ” - Link: integer, n    e.g. 3
Ḷ                              - lowered range            [0,1,2]
    Ɗ                          - last three links as a monad - f(that):
 Ṛ                             -   reverse                [2,1,0]
   Ḥ                           -   double                 [0,2,4]
  ż                            -   zip together           [[2,0],[1,2],[0,4]]
        Ɱ                      - map with:
     Ø.                        -   bits                   [0,1]
       x                       -   repeat elements        [[0,0],[0,1,1],[1,1,1,1]]
               $               - last two links as a monad - f(n):
           3,5                 -   three pair five        [3,5]
              ẋ                -   repeat (n times)       [3,5,3,5,3,5]
         ;€                    - concatenate each = x     [[0,0,3,5,3,5,3,5],[0,1,1,3,5,3,5,3,5],[1,1,1,1,3,5,3,5,3,5]]
                    $          - last two links as a monad - f(x):
                   $           -   last two links as a monad - f(x):
                 Ṛ             -     reverse              [[1,1,1,1,3,5,3,5,3,5],[0,1,1,3,5,3,5,3,5],[0,0,3,5,3,5,3,5]]
                  Ḥ            -     double               [[2,2,2,2,6,10,6,10,6,10],[0,2,2,6,10,6,10,6,10],[0,0,6,10,6,10,6,10]]
                ;              -   concatenate            [[0,0,3,5,3,5,3,5],[0,1,1,3,5,3,5,3,5],[1,1,1,1,3,5,3,5,3,5],[2,2,2,2,6,10,6,10,6,10],[0,2,2,6,10,6,10,6,10],[0,0,6,10,6,10,6,10]]
                      “/\^"L ” - list of characters = '/\^"L '
                     ị         - index into (1-indexed & modular, i.e. 10->")
                                                     -> ['  ^L^L^L',
                                                         ' //^L^L^L',
                                                         '////^L^L^L',
                                                         '\\\\ " " "',
                                                         ' \\ " " "',
                                                         '   " " "',
                                                        ]

recursive · Accepted Answer · 2021-01-26 01:24:20Z

3

Stax, 27 bytes

ü╚lTà╨ú%Ñ►F↔33j♦ü♂D▄P;L.⌂î↑

Run and debug it

answered Jan 26, 2021 at 1:24

recursive

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Kaddath · Accepted Answer · 2021-01-26 10:41:59Z

3

PHP, 134 116 bytes

for($f=str_repeat;$i<$n=$argn;)$s=($t=$f(" ",$i++).$f("//",$n-$i).$f("^L",$n)."
").$s.strtr($t,'/^L','\\ "');echo$s;

Try it online!

Yes, much shorter to build the string..

EDIT: a huge 18 bytes saved thanks to optimization from Dom Hastings

Old version kept just for the fun:

157 bytes

for($f=str_repeat;$i<2*$n=$argn;$s.='//')$i++<$n?printf("%".($n-2+$i)."s%s",$s,$f("^L",$n)."
"):printf("%".(5*$n-$i)."s",$f('\\',4*$n-2*$i).$f(' "',$n)."
");

Try it online!

Here is an infamous PHP answer based on printf. Couldn't think of an uglier looking piece of code for now, but I'll work on it (starting to think it would be better to build the whole string)!

edited Jan 26, 2021 at 10:41

answered Jan 26, 2021 at 9:39

Kaddath

2,8273 silver badges9 bronze badges

1

\$\begingroup\$ Nice! Good dedication to check the alternative as well! You can also save a few more bytes using strtr too: Try it online! \$\endgroup\$
– Dom Hastings
Jan 26, 2021 at 10:30
\$\begingroup\$ @DomHastings huge optimization! if you want to post it as your own answer be sure I'll upvote it, or else I'll edit. Yes for these kind of simple problems I like to try to think by myself first rather than just port other answers \$\endgroup\$
– Kaddath
Jan 26, 2021 at 10:36
\$\begingroup\$ Not at all, I was looking at a version but even using strtr I was at 133, so it's mostly your work! Please edit in! :) Yeah, I try to avoid looking at other approaches too, at least until I think I'm done! \$\endgroup\$
– Dom Hastings
Jan 26, 2021 at 10:39

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btwiuse · Accepted Answer · 2021-01-31 18:48:11Z

Wolfram Language (Mathematica), 172 bytes

(m=MapThread[Join,{DiamondMatrix[#][[1;;#+1,1;;#]],2BoxMatrix[#][[1;;#+1,1;;#+1]]}];Print/@StringJoin@@@(Join[m,4Reverse[m]]/.{0->" ",1->"//",2->"^L",4->"\\\\",8->" \""}))&

Try it online!

Brief explanation:

Construct the upper part m with DiamondMatrix and BoxMatrix:

m = MapThread[Join, {DiamondMatrix[n][[1 ;; n + 1, 1 ;; n]], 2BoxMatrix[n][[1 ;; n + 1, 1 ;; n + 1]]}]

Derive the whole matrix M from m:

M = Join[m, 4*Reverse@m]

you can visualize it with:

MatrixPlot[M, Frame -> None]

when n = 10, it's like this:

Replace each number from the matrix with corresponding string:

M/.{0->" ",1->"//",2->"^L",4->"\\\\",8->" \""}

Convert each row to string and print it:

Print/@StringJoin@@@%

pxeger · Accepted Answer · 2021-02-13 12:17:19Z

3

Zsh commit 2a96748, ^{123 121} 119 bytes

alias R=repeat
R "i=a=$1" {s=${(l:--i:)};R b++ s+=//;R a s+=^L;<<<$s}
R a {s=${(l:i++:)};R a-i s+='\\';R a s+=' "';<<<$s}

Try it online!

This specific version fixes a bug that would otherwise require i=a=$1 to be in quotes. TIO doesn't have that version obviously, so the linked one is two bytes longer.

edited Feb 13, 2021 at 12:17

answered Jan 23, 2021 at 18:45

pxeger

23.2k4 gold badges51 silver badges134 bronze badges

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T3RR0R · Accepted Answer · 2021-08-07 05:32:03Z

batch 317 309 301 291 284 bytes 276 bytes

Cls&@Set $=@Set 
%$%/Ah=%1*2,z=0
%$%F=@For /L %%a in (1 1 &%$%P=@^<nul set/p&%$%C= @If ,%%y LEQ %1 
%f:a=y%!h!)Do @(%$%/Aw=%1-z
%F%!w!)Do %P%=ESCC
%F%!z!)Do%C%(%P%=//)Else %P%=\\
%F%%1)Do%C%(%P%=^^L)Else %P%^"=ESCC""
%C:EQ=SS%(%$%/A z+=1)Else%C:,=not %%$%/Az-=1
Echo()

Cube.bat

Notes:

must be run from the command line using: Echo(cube.bat n|Cmd /V:on where n is an integer representing cube size and cube.bat is the filename of the file
lines 1 and3 end wth trailing spaces
requires windows 10 with virtual terminal support
ESC is a stand in for the ANSI escape character, As it can't be pasted here, the scripts raw paste data can be found here.
- Or it can be generated by opening the command prompt and typing:
  - >"%TEMP%\escape.chr" Echo(ALT+027 then pressing ENTER
- Or output using a for loop with Echo off:
  - (For /F "skip=4 DELIMS=" %a in ('Echo(prompt $E^|cmd')Do Echo(%a) >"%TEMP%\escape.chr"

Vt sequence ESCC ('cursor right') is used to get around <nul set /p's inability to output leading spaces.

example output:

xigoi · Accepted Answer · 2021-01-23 19:22:35Z

2

Jelly, 37 bytes

Ḷ,Ṛ$ạ’;,ʋ³ẋ@"ɗ€"“ “//“^L”,“ “\\“ "”ẎY

Try it online!

answered Jan 23, 2021 at 19:22

xigoi

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LuaNoob · Accepted Answer · 2021-01-24 15:38:36Z

2

Lua (LuaJIT), 122 bytes

r=n.rep;for i=1,n*2 do print(i<=n+0 and r(' ',n-i)..r("/",i*2-2)..r('^L',n)or r(' ',i-1-n)..r("\\\\",n*2-i)..r(' "',n))end

Try it online!

(Above code abuses the TIO's input string-type to save 1 byte, see Not abused version)

edited Jan 24, 2021 at 15:38

answered Jan 24, 2021 at 15:30

LuaNoob

3091 silver badge5 bronze badges

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Samuel Waller · Accepted Answer · 2021-03-01 18:10:38Z

2

GolfScript, 68 bytes

~:i,.-1%]zip.-1%]{{~" "*\["//""\\\\"]0=*+["^L"' "']0=i*+}%1:0;n*}%n*

Try it online!

Its really messy, but it works and I don't really know a better way to do it.

answered Mar 1, 2021 at 18:10

Samuel Waller

7415 silver badges8 bronze badges

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Axuary · Accepted Answer · 2021-05-02 01:28:59Z

2

Excel, 114 bytes

=LET(x,A1,q,SEQUENCE(2*x),b,q>x,r,IF(b,q-x-1,x-q),REPT(" ",r)&REPT(IF(b,"\\","//"),x-r)&REPT(IF(b,"""""","^L"),x))

Link to Spreadsheet

answered May 2, 2021 at 1:28

Axuary

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DLosc · Accepted Answer · 2021-06-27 01:40:32Z

Regenerate, 84 bytes

(( {#2-1}| {$~1-1})(//$3|)(^L){$~1}\n){$~1}(( $6|)(\\{#7-2}|\\{#3})( "){$~1}\n){$~1}

Takes the size as a command-line input. Try it here!

Explanation

Spaces are replaced with underscores for improved visibility.

Top half:

(( {#2-1}| {$~1-1})(//$3|)(^L){$~1}\n){$~1}
(                                    )       Group 1: one row of the top half
 (                )                           Group 2: leading spaces
  _{#2-1}                                      Either one less space than the previous row
         |_{$~1-1}                             or (input-1) spaces if this is the first row
                   (     )                    Group 3: slashes
                    //$3                       Either the previous row's slashes plus two more
                        |                      or empty string if this is the first row
                          (^L)                Group 4: ^L
                              {$~1}           Repeat group 4 (input) times
                                   \n         Newline
                                      {$~1}  Repeat group 1 (input) times

Bottom half:

(( $6|)(\\{#7-2}|\\{#3})( "){$~1}\n){$~1}
(                                  )       Group 5: one row of the bottom half
 (    )                                     Group 6: leading spaces
  _$6                                        Either the previous row's spaces plus two more
     |                                       or empty string if this is the first row
       (               )                    Group 7: backslashes
        \\{#7-2}                             Either two fewer slashes than the previous row
                |\\{#3}                      or the same number of slashes as in the last match
                                              of group 3 if this is the first row
                        (_")                Group 8: space-quote
                            {$~1}           Repeat group 8 (input) times
                                 \n         Newline
                                    {$~1}  Repeat group 5 (input) times

Dom Hastings · Accepted Answer · 2021-01-25 20:41:06Z

1

Perl 5 + `-M5.10.0 -pa`, 71 bytes

say,$\=y'/^L'\ "'r.$/.$\for map$"x("@F"-$_--)."//"x$_."^L"x"@F",1..$_}{

Try it online!

answered Jan 25, 2021 at 20:41

Dom Hastings

22.8k3 gold badges49 silver badges91 bronze badges

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MarcMush · Accepted Answer · 2021-05-01 22:29:52Z

1

Julia, ^{79 73} 69 bytes

output is a list of lines

f(N,a=-N+1:0,b=0:N-1)=@.[" "^-a*"//"^b*"^L"^N;" "^b*"\\"^-2a*" \""^N]

Try it online!

edited May 1, 2021 at 22:29

answered Jan 27, 2021 at 14:26

MarcMush

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\$\begingroup\$ If you can print one space less on the left. But anyway yes, you did it \$\endgroup\$
– anotherOne
Jan 27, 2021 at 15:16
\$\begingroup\$ I can but it will cost me a few bytes \$\endgroup\$
– MarcMush
Jan 27, 2021 at 15:29
\$\begingroup\$ Don't worry about a few bytes, the important thing is that you managed to properly reproduce the shape in as few bytes as your language allow you to do. \$\endgroup\$
– anotherOne
Jan 27, 2021 at 17:31
1

\$\begingroup\$ @Davide done, and I even managed to shave a few more bytes \$\endgroup\$
– MarcMush
Jan 28, 2021 at 9:18

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Wasif · Accepted Answer · 2021-06-28 02:48:20Z

1

Vyxal `j`, 26 bytes

ʁd\/*m‛^L‛ ""?*ƛw?ẋ;fZvṅøĊ

Try it Online!

answered Jun 28, 2021 at 2:48

Wasif

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Shaggy · Accepted Answer · 2021-07-23 19:42:11Z

Japt `-R`, 31 bytes

Æç'Li^)i'/pXÑÃû
c¡Xd`/\\^ L"`Ãw

Try it

Æç'Li^)i'/pXÑÃû\nc¡Xd`...`Ãw     :Implicit input of integer U
Æ                                :Map each X in the range [0,U)
 ç                               :  U times repeat
  'Li^                           :    "L" prepended with "^"
      )                          :  End repeat
       i                         :  Prepend
        '/p                      :    "/" repeated
           XÑ                    :    X*2 times
             Ã                   :End map
              û                  :centre pad each element with spaces to the length of the longest
               \n                :Reassign to U
                 c               :Concatenate
                  ¡              :  Map each X in U
                   Xd`...`       :    For each pair of characters in the string, replace all occurrences of the first in X with the second
                          Ã      :  End map
                           w     :  Reverse
                                 :Implicit output joined with newlines

Print a 3D shape

Input:

Output:

26 Answers 26

Python 2, 81 bytes

Canvas, 27 22 bytes

Explanation

Ruby, 78 bytes

JavaScript (Node.js), 121 bytes

K (ngn/k), 69 bytes

05AB1E, 28 26 bytes

PowerShell, 88 bytes

PowerShell, 104 102 92 bytes, Alternative

Java 11, 148 136 bytes

Charcoal, 42 32 bytes

Python 3.8 (pre-release), 97 93 92 bytes

C (gcc), 160 135 130 bytes

Explanation (before some golfs)

Jelly, 30 bytes

How?

Stax, 27 bytes

PHP, 134 116 bytes

157 bytes

Wolfram Language (Mathematica), 172 bytes

Zsh commit 2a96748, 123 121 119 bytes

batch 317 309 301 291 284 bytes 276 bytes

Jelly, 37 bytes

Lua (LuaJIT), 122 bytes

GolfScript, 68 bytes

Excel, 114 bytes

Regenerate, 84 bytes

Explanation

Perl 5 + -M5.10.0 -pa, 71 bytes

Julia, 79 73 69 bytes

Vyxal j, 26 bytes

Japt -R, 31 bytes

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Canvas, ²⁷ 22 bytes

Python 3.8 (pre-release), ^{97 93} 92 bytes

C (gcc), ^{160 135} 130 bytes

Zsh commit 2a96748, ^{123 121} 119 bytes

Perl 5 + `-M5.10.0 -pa`, 71 bytes

Julia, ^{79 73} 69 bytes

Vyxal `j`, 26 bytes

Japt `-R`, 31 bytes

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