Making Squared Words

Question

The Challenge

Your task is to create a program that takes any given string input, and outputs the input in a squared format. Empty strings should return an empty string.

Examples

Given the input:

golf

Your program should output:

golf
o  l
l  o
flog

Input:

123

Output:

123
2 2
321

Input:

a

Output:

a

Input:

Hello, world!

Output (notice the space between the , and w - the gap is not just a newline):

Hello, world!
e           d
l           l
l           r
o           o
,           w

w           ,
o           o
r           l
l           l
d           e
!dlrow ,olleH

Scoring

This is code-golf, so the shortest answer in each language wins.

Answer 1

Charcoal, 7 5 bytes

θ‖Ｏ↙↘

Try it online! Link is to verbose version of code. Edit: Saved 2 bytes thanks to @CarlosAlejo. Explanation:

θ       Print the input string, making the top row
 ‖Ｏ     Reflect with overlap...
   ↙    ... down and left, to create the left side
    ↘   ... down and right, to create the bottom and right sides

(Multiple directions to the Reflect command run consecutively rather than simultaneously.)

Answer 2

MATL, 20 16 11 bytes

otYTO6Lt&(c

Try it at MATL online!

EDIT: The code works in release 20.2.1, which predates the challenge. The link uses that release. (In 20.2.2 the code would be shorter, but it postdates the challenge).

Explanation

o     % Implicitly input a string. Convert chars to ASCII codes
t     % Duplicate
YT    % 2-input Toeplitz matrix
O     % Push 0
6L    % Push predefined literal [2, -1+1j]. When used as an index, this is
      % interpreted as 2:end-1 (indexing is 1-based)
t     % Duplicate
&(    % 4-input assignment indexing. This writes 0 at the square formed by
      % rows 2:end-1 and columns 2:end-1 
c     % Convert to char. Char 0 is shown as space. Implicitly display

Answer 3

Jelly,  29 22  17 bytes

Charcoal will trounce+d this score...

J⁶ẋa0,1¦"ṚṚ;$ṖŒḌY

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

Try it online!

How?

J⁶ẋa0,1¦"ṚṚ;$ṖŒḌY - Link: list of characters, w     e.g. "whole"
 ⁶                - literal space character              ' '
J                 - range(length(w))                     [1,2,3,4,5]
  ẋ               - repeat                               [" ","  ","   ","    ","     "]
         Ṛ        - reverse w                            "elohw"
        "         - zip with:
       ¦          -   sparse application of:
   a              -     and (vectorises)
    0,1           -     to indexes: [0,1] (last and 1st) ["e","ll","o o","h  h","w   w"]
            $     - last two links as a monad:
          Ṛ       -   reverse                            ["w   w","h  h","o o","ll","e"]
           ;      -   concatenate                        ["w   w","h  h","o o","ll","e","e","ll","o o","h  h","w   w"]
             Ṗ    - pop (remove last entry)              ["w   w","h  h","o o","ll","e","e","ll","o o","h  h"]
              ŒḌ  - reconstruct matrix from diagonals    ["whole","h   l","o   o","l   h","elohw"]
                Y - join with newlines                   "whole\nh   l\no   o\nl   h\nelohw"
                  - if running as a full program implicit print

Answer 4

Haskell, 84 78 bytes

f s@(_:x)|_:y<-r x=s:[a:(y>>" ")++[b]|(a,b)<-zip x y]++[r s];f s=[s]
r=reverse

Try it online! Usage: f "test". Returns a list of lines.

Edit: -6 bytes thanks to dianne!

Answer 5

C, 109 bytes

i,j;f(char*s){for(i=j=printf("%s\n",s)-2;1[++s];)printf("%c%*c\n",*s,i,s[j-=2]);for(;*s*~i--;)putchar(*s--);}

Try it online!

Noteworthy tricks:

• Instead of wasting bytes on strlen, we simply grab the length of the string while simultaneously printing the first line:

  i=j=printf("%s\n",s)-2
  

  This works because printf returns the number of bytes written.

• For the middle lines, we need to loop over the string but exclude both the first and last character. This is achieved with the condition

  1[++s]
  

  (which is shorter than (++s)[1]), which skips the first character due to the ++'s being in the condition and skips the last one by stopping when the character past the current character is '\0' (rather than stopping at '\0').

• In the body of the first loop,

  printf("%c%*c\n",*s,i,s[j-=2])
  

  we print the current character, then the appropriate "mirrored" character (keeping track with j, which does go into the negatives, resulting in the odd situation of indexing into a string with a negative number) padded to a length of i with spaces (where i is conveniently strlen(s) - 1).

• The reversed printing on the last line is pretty straightforward; the only trick is the *s*~i--, which is the shortest way to get i+2 iterations of the loop body (which doesn't depend on i; all i is used for is to count). The funky *s* part makes sure the loop doesn't run if *s is '\0', which happens on length-1 input.

Answer 6

Octave, 40 bytes

@(s,t=toeplitz(s),u=t(x=2:end-1,x)=32)t;

Try it online!

It is my answer but posted after @Luis MATL answer

Answer 7

Python 2, 89 81 88 86 bytes

i=input();n=k=len(i)-2
print i
exec'print i[~k]+" "*n+i[k];k-=1;'*n
if~k:print i[::-1]

Try it online!

Answer 8

R, 113 bytes

function(s){n=length(s<-strsplit(s,'')[[1]])
m=matrix(' ',n,n)
m[n,n:1]=m[,1]=m[1,]=m[n:1,n]=s
write(m,'',n,,'')}

Try it online!

Answer 9

05AB1E, 17 16 15 19 bytes

ÂDÂø¦¨Dgú€Ás)˜»Igi¨

Try it online!

Explanation

Example with input = golf

ÂDÂ                  # bifurcate, duplicate, bifurcate
                     # STACK: 'golf', 'flog', 'flog', 'golf'
   ø                 # zip the top 2 items
                     # STACK: 'golf', 'flog', ['fg', 'lo', 'ol', 'gf']
    ¦¨               # remove the first and last element
                     # STACK: 'golf', 'flog', ['lo', 'ol']
      Dg             # get the length of the list
                     # STACK: 'golf', 'flog', ['lo', 'ol'], 2
        ú            # prepend that many spaces to each element
                     # STACK: 'golf', 'flog', ['  lo', '  ol']
         €Á          # rotate each right
                     # STACK: 'golf', 'flog', ['o  l', 'l  o']
           s         # swap the top 2 items
                     # STACK: 'golf', ['o  l', 'l  o'], 'flog'
            )˜       # wrap in a flattened list
                     # STACK: ['golf', 'o  l', 'l  o', 'flog']
              »      # join on newline
               Igi¨  # if input is length 1, remove last char

The fix for 1-letter input was quite expensive.
I feel like a different approach might be better now.

Answer 10

Python 2, 99 88 bytes

-4 bytes thanks to musicman523.

lambda s:s[1:]and[s]+[s[i]+' '*(len(s)-2)+s[~i]for i in range(1,len(s)-1)]+[s[::-1]]or s

Try it online!

Returns a list of strings.

Answer 11

Swift 3, 215 199 bytes

let s=readLine()!,c=s.characters,r:[Character]=c.reversed(),b=c.count
print(s)
if b>1{for i in 0..<b-2{print("\(r.reversed()[i+1])\(String.init(repeating:" ",count:b-2))\(r[i+1])")};print(String(r))}

Try it online

Answer 12

APL, 58 bytes

{(' ',⍵)[x+(x←∘.{((c-1)=⍺⌈⍵)∨0=⍺⌊⍵}⍨r)×o⌊⌽⊖o←∘.⌈⍨r←⍳c←≢⍵]}

With ⎕IO←0.

Try it online!

How?

c←≢⍵ - length of the string

r←⍳ - range

o←∘.⌈⍨ - outer product with minimum

123
223
333

o⌊⌽⊖ - minimalize with itself turned 180^o

123  ⌊  333  =  123
223  ⌊  322  =  222
333  ⌊  321  =  321

× - multiply with

x←∘....⍨r - outer product of the range with

    ((c-1)=⍺⌈⍵)∨0=⍺⌊⍵ - the frame of the matrix

111  ×  123  =  123
101  ×  222  =  202
111  ×  321  =  321

x+ - add the frame

111  +  123  =  234
101  +  202  =  303
111  +  321  =  432

(' ',⍵)[...] - get by index from the string concatenated to space

Answer 13

Mathematica, 128 bytes

(c=Column;g=Length[x=Characters@#]-1;If[g==0,#,c@{#,c@Table[""<>{x[[i]],""<>Table[" ",g-1],x[[-i]]},{i,2,g}],StringReverse@#}])&

Answer 14

C, 96 bytes

i,l;f(char*s){for(i=l=puts(s)-2;--i;)printf("%c%*c\n",s[l-i],l,s[i]);for(;l+1;)putchar(s[l--]);}

Bonus version (122 bytes):

x,y,i,l;f(char*s){for(i=l=puts(s)-1;++i<l*-~l;putchar(x==l?10:x%~-l*(y%~-l)?32:s[(x*y?l+l-2-x-y:x+y)%l]))x=i%-~l,y=i/-~l;}

Answer 15

JavaScript (ES8), 108 112 bytes

let f = 

s=>(n=s.length)<2?s:(r=[...s].reverse()).slice(1,-1).reduce((a,v,i)=>a+`
`+s[i+1].padEnd(n-1)+v,s)+`
`+r.join``

o.innerHTML = f("hello folks!")

<pre id="o"></pre>

Less golphed

s=>
   (n=s.length) < 2 ?    // record and check length
   s :                   // s has 0 or 1 char, else
   (r=[...s].reverse())  // reverse characters,
   .slice(1,-1)          // exclude 1st and last
   .reduce((a,v,i)=>     // append inner lines
      a +'\n' +s[i+1].padEnd(n-1) + v
    ,s)                  // to first line
    + '\n' +r.join("")   // append characters reversed

Thanks to Justin Mariner for saving lots of bytes, which then all got used up adding the zero or single character check needed to comply with the challenge. You get that :-(

Answer 16

Python 3, 88 bytes

w=input();p=print;l=len(w)-2
[p(w[n+1]+' '*l+w[l-n])for n in range(l)]
l+1and p(w[::-1])

Answer 17

PHP, 118 bytes

echo $s=$argv[1];$l=strlen($r=strrev($s))-1;for($i=$l-1;$l&&$i;)echo "\n".str_pad($r[$i],$l).$s[$i].(--$i?"":"\n$r");

Try it online!

echo $s = $argv[1];
$l = strlen($r = strrev($s)) - 1;

for ($i = $l - 1; $l && $i;)
    echo "\n" . str_pad($r[$i], $l) . $s[$i] . (--$i ? "" : "\n$r");

Answer 18

Lua, 104 bytes

print(s);for a=2,#s-1 do print(s:sub(a,a)..(" "):rep(#s-2)..s:sub(#s-a+1,#s-a+1)) end;print(s:reverse())

Try it online!

Answer 19

JavaScript (ES2017), 87 bytes

s=>[...s].reverse(l=s.length-1).map((c,i,z)=>i?l-i?s[i].padEnd(l)+c:z.join``:s).join`
`

ES6 version: 93 bytes

s=>[...s].reverse(l=s.length-1).map((c,i,z)=>i?l-i?s[i]+' '.repeat(l-1)+c:z.join``:s).join`
`

Less golfed

s => (
  l = s.length-1,
  [...s].reverse().map( // scan string backwards
     (c, i, z) => 
     i != 0 // check if top row
     ? l-i != 0 // check if bottom row
       ? s[i].padEnd(l) + c // any middle row
       : z.join`` // bottom row: string reversed
     :s // top row: original string
  ).join`\n`
)

F=
s=>[...s].reverse(l=s.length-1).map((c,i,z)=>i?l-i?s[i].padEnd(l)+c:z.join``:s).join`
`

function update() {
  O.textContent = F(I.value)
}

update()

<input id=I value='Hello, world!' oninput='update()'>
<pre id=O></pre>

Answer 20

C# (.NET Core), 179 161 bytes

-18 bytes thanks to TheLethalCoder

using System.Linq;
s=>{int l=s.Length-1,i=1;var d=s.Reverse().ToArray();while(i<l)s+="\n"+s[i]+new string(' ',l-1)+d[i++];if(l>1)s+="\n"+new string(d);return s;};

Try it online!

I'm not sure about the rules, if this is needed to byte count or not:

using System.Linq;

Someone please correct me about this.

Ungolfed:

s =>
{
    int l = s.Length - 1, i = 1;
    var d = s.Reverse().ToArray();
    while (i < l)
        s += "\n" + s[i] + new string(' ', l - 1) + d[i++];
    if (l > 1)
        s += "\n" + new string(d);
    return s;
};

Answer 21

Java, 191 bytes

(s)->{PrintStream o=System.out;int l=s.lenght();o.println(s);for(int i=0;i<l;i++){o.printf("%"+"s%"+(l-1)+"s%n",s.charAt(i),s.charAt(l-1-i));for(int i=0;i<l;i++){o.print(s.charAt(l-1-i));}}};

Answer 22

Python 3, 85 bytes

Using the input for the top row :)

a=input()
b=len(a)
for i in range(b-2):print(a[1+i]+" "*(b-2)+a[-2-i])
print(a[::-1])

Answer 23

Retina, 106 bytes

..+
$&¶$&
O$^`.(?=.*$)

\G.
$&$%'¶
r`.\G
¶$%`$&
+`(.+)¶¶((.*¶)*).(.*)
¶¶$1$4$2
T`p` `(?<=.¶.).*(?=.¶.)
G`.

Try it online! Explanation:

..+
$&¶$&

If there are at least two characters, duplicate the input.

O$^`.(?=.*$)

Reverse the duplicate.

\G.
$&$%'¶
r`.\G
¶$%`$&

Turn the strings into triangles. The top triangle starts with the input and removes the first character each time, while the bottom triangle starts with the first letter of the reversed input and adds a character each time.

+`(.+)¶¶((.*¶)*).(.*)
¶¶$1$4$2

Join the triangles together, overlapping so that the last character forms the / diagonal.

T`p` `(?<=.¶.).*(?=.¶.)

Change all the characters to spaces, if they are at least one character away from the end on every side.

G`.

Delete any left-over blank lines.

Answer 24

Python 3, 106 bytes

A functional version...

w=input();p=print;l=len(w)-2
def f(k):p(w[k]+' '*l+w[-k-1]);l-k>0and f(k+1)
l<0 or f(1)or l<1or p(w[::-1])

Answer 25

Python 2, 100 bytes

lambda a:'\n'.join([a]+(len(a)>1)*([a[i]+(len(a)-2)*' '+a[~i]for i in range(1,len(a)-1)]+[a[::-1]]))

Try it online!

Answer 26

Mathematica, 138 91 bytes

(b=Outer[" "&,#,#];Do[l={{1,k},{k,1}};b=ReplacePart[b,Join[l,-l]->#[[k]]],{k,Length@#}];b)&

You can try it online with by pasting the following at the Wolfram Cloud Sandbox and clicking "evaluate cell" or hitting Shift+Enter or Numpad Enter:

(b=Outer[" "&,#,#];Do[l={{1,k},{k,1}};b=ReplacePart[b,Join[l,-l]->#[[k]]],{k,Length@#}];b)&@{"g","o","l","f","y"}//MatrixForm