A square of text

Question

The Challenge

Given a string, output the text in the shape of a square.

You can assume that the text will always fit in a square, and that it will never be an empty string.

You can also assume it will never have newlines.

Example

Input:
Hi, world

Output:
Hi,
 wo
rld

Test Cases

Input:
Hi, world! Hello

Output:
Hi, 
worl
d! H
ello

Input:
Lorem ipsum dolor sit amt

Output:
Lorem
 ipsu
m dol
or si
t amt

Input:
H

Output:
H

Rules

• This is code-golf, so shortest answer in bytes wins! Tiebreaker is most upvoted answer.
• Standard loopholes are forbidden.

Answer 1

Vyxal jr, 3 bytes

L√/

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  / # (Implicit input) divided into...
 √  # Square root of...
L   # Length of...
    # (implicit input)
  / # Pieces. 

Answer 2

Risky, 14 bytes

2_2+_2+__1+_1*__?*_1/}!?*_1

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Explanation

So many no-ops... 0__0

Parse tree:

             *
      +             /
2      _      _      }
   +      +      *      *
 _  _   _  _   _  _   !  _
  2  2   1  1   ?  1   ?  1

How it works:

                ?            The input string (as a list of charcodes)
              __ *_1         (no-ops)
                    /        Group into sublists of length...
                     }       square root of...
                      !      length of...
                       ?     the input string again
                        *_1  (no-ops)
2_2+_2+__1+_1                Generate the number 10 in a needlessly complicated way
             *               Join the list of lists on 10 (newline)
                             Output the resulting list of numbers as a string

Answer 3

Convex, 7 bytes

_,mQ/N*

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Fun fact:

_,mQ/\* also works on TIO due to how it works.

How have I forgotten to make a 1-char square root op?

Answer 4

Matlab, 28 bytes

@(s)reshape(s,nnz(s)^.5,[])'

reshape puts output in columns first, so we have to transpose the whole thing to get the desired output.

Answer 5

PHP, 47 bytes

<?=chunk_split($a=$argv[1],strlen($a)**.5,"
");

For multibyte strings, 66 bytes

<?=preg_replace("/.{1,".sqrt(strlen($a=$argv[1]))."}/u","$0
",$a);

Answer 6

QBasic, 60 bytes

Nothing special here... just your BASIC program.

INPUT i$
t=SQR(len(i$))
FOR x=0TO t-1
?MID$(i$,x*t+1,t)
NEXT

Answer 7

q/kdb+, 25 22 bytes

Solution:

-1{(2#6h$sqrt(#)x)#x};

Example:

q)-1{(2#6h$sqrt(#)x)#x}"Lorem ipsum dolor sit amt";
Lorem
 ipsu
m dol
or si
t amt

Explanation:

-1{(2#6h$sqrt count x)#x}; / ungolfed solution
  {                     }  / lambda function
                      #x   / take from x (input)
   (                 )     / do everything in the brackets first
              count x      / length of x
         sqrt              / calculate square-root
      6h$                  / cast to integer
    2#                     / duplicate this number, e.g. 5->(5;5)
-1                       ; / print to stdout

Answer 8

SOGL V0.12, 1 byte

√

Try it Here! - this expects the input on the stack as a function, so there it's called as a function F and then, in the next line, , pushes the input and F executes the function.

without builtins, 3 bytes

l√n

Try it Here!
push length, square root it, split input into strings of that length, implicitly output joined with newlines

Answer 9

Perl 6, 28 27 bytes

-1 byte thanks to nwellnhof

{.comb(.comb**.5+|0)>>.say}

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Answer 10

Momema, 99 bytes

03i00+1*0*0*-9i=+1**00+-4*0q00+*0-+1+*1*1 1+1*1q=*02*1o03*1p0-9*+4*00+1*03+-1*3p=*3 2+-1*2-9 10o=*2

Try it online!

Explanation

                 # -- Read the input onto the tape.
                 # -- Set aside the first 4 cells as variables.
0   3            # p = 3
i   0            # do {
0   +1*0         #   p = p+1
*0  *-9          #   tape[p] = getchar()
i   =+1**0       # } while (tape[p] != -1)
0   +-4*0        # p = p - 4
                 # -- Calculate the square root of the input length.
q   0            # do {
0   +*0-+1+*1*1  #   p = p - (1 + r + r)
1   +1*1         #   r = r + 1
q   =*0          # } while (p)
                 # -- Print the formatted output.
2   *1           # i = r
o   0            # do {
3   *1           #   j = r
p   0            #   do {
-9  *+4*0        #     putchar(tape[p + 4])
0   +1*0         #     p = p + 1
3   +-1*3        #     j = j - 1
p   =*3          #   } while (j)
2   +-1*2        #   i = i - 1
-9  10           #   putchar(10)
o   =*2          # } while (i)

Answer 11

Python 3.8, 68 bytes

a=int(len(s:=input())**.5)
for i in range(0,a**2,a):print(s[i:][:a])

Not the shortest Python answer, but also not the longest. Goes through the string by intervals of the square root of the length of the string. I saved a byte by using Python 3.8's new-fangled assignment expressions.

Answer 12

Kotlin, 57 bytes

fun String.s()=chunked(sqrt(length+.0).toInt(),::println)

Explanation:

fun String.s()=   # Extension method on the string, so we can avoid declaring a param
chunked           # Actually this method does all the heavy lifting, it splits the string into a list of size N strings
(sqrt(length+.0)  # Kotlin's sqrt only takes double and produces double :/ 
.toInt(),         # So we need to cast the result for chunked in an ugly way :/
::println)        # But hey, applying a transform to print is actually included for free!

Answer 13

Python 3, 78 bytes

def f(s):
	r=int(len(s)**.5)
	return'\n'.join([s[i*r:i*r+r]for i in range(r)])

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Answer 14

TI-Basic, 33 Bytes

Input Str1
length(Str1→Z
For(X,1,Z,√(Z
Disp sub(Str1,X,√(Z
End

This is not a particularly interesting answer, due to the limitations of TI-Basic, but I like the language too much to resist. Note that TI-Basic is tokenized, so byte count for the program is not the same as character count.

Answer 15

Stax, 3 bytes

:Jm

Run and debug it at staxlang.xyz!

Very simple. :J squarifies an array. That m is necessary because Stax will print a list of strings on one line.

Without the builtin, six bytes

c%|q/m

Run and debug it at staxlang.xyz!

Copy the string (c) and take its length (%). Square root that length (|q), and split the string into substrings of that (square-rooted) length (/).

Answer 16

Zsh, 54 51 bytes

-3 by ^= instead of =0^().

((n=$#1,l^=n**0.5))
eval '<<<${1:'{0..$n..$l}':$l}'

Try it online! Try it online!

Shoutouts to eval jank.

((n=$#1,l^=n**0.5))              # the xor with 0 casts n**0.5 to an integer
eval '<<<${1:'           ':$l}'  # when eval'd, prints a subtring of length l
              {0..$n..$l}        # brace expansion: {0..4..16} => 0 4 8 12 16

I left in set -x in the TIO link, scroll down to the debug section to see exactly what happens with the brace expansion.

Answer 17

Silicon, 6 bytes

I added some commands right after I noticed this challenge.

iLqnåê

Explanation:

iLqnåê

i          Input
  q        Square root of...
 L         Length
   n       Convert to integer
    å      Split into chunks of n
     ê     Join at newlines 

Answer 18

Vyxal j, 1 byte

²

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Yes, there's a built in for it.

"But there's already another vyxal answer!" This is using 2.16.0 while the other one is using <2.4.0

Answer 19

ListSharp, 157 bytes

STRG a=READ[<here>+"\\a.txt"]
NUMB b=(int)Math.Sqrt(a LENGTH)
[FOREACH NUMB IN 0 TO b-1 AS x]
{
STRG t=GETRANGE a FROM [x*b+1] TO [x*b+b]
ROWS f=f+t
}
SHOW=f

reads text from local file a.txt , im using embedded c# code and other tricks to make this work. feel free to ask me questions about this code or listsharp in general

Answer 20

C, 64 bytes

f(s,m,k){for(m=k=sqrt(strlen(s));k--;s+=m)printf("%.*s\n",m,s);}

Inspired by the other C answer. Only works on platforms where int has the same size as char* (e.g. 32-bit Windows).

Answer 21

Lua, 68 Bytes

o=io.read()for s in o:gmatch(("."):rep(math.sqrt(#o)))do print(s)end

Simple and dirty, just how I like it. I use o=io.read() instead of function(o) Which works out as the same length, to save from having to add an extra "end" at the end.

Pretty simple in functionality, it just uses gmatch to loop through the string in chunks of math.sqrt(#o), Which is the square root of the length of the string. Sadly, Lua patterns don't offer the regex functionality of {} to denote a specific count, so I had to do :rep() instead.

Answer 22

Racket 102 bytes

(λ(s)(let*((l(string-length s))(q(sqrt l)))(for((i(range 0 l q)))(displayln(substring s i(+ i q))))))

More readable form:

(define (f s)
  (let* ((l (string-length s))
         (q (sqrt l)))
    (for((i(range 0 l q)))
      (displayln (substring s i(+ i q))))))

Testing:

(f "Lorem ipsum dolor sit amt")
Lorem
 ipsu
m dol
or si
t amt

Answer 23

QBIC, 41 bytes

;c=sqr(len(A))[1,c|?mid$$|(A,q,c) q=c*a+1

Unfortunately, it doesn't work on text parameters with spaces in it. I'll put it on the QBIC todo-list.

Explanation:

;           gets the input as string A$
sqr(len(A)) calls the QBasic square root function, with the length of A$ as argument
[1,c...     Loops from 1 to the height of the square
?mid$$|(..) Prints the middel part of A$. The '$' has a reserved meaning in QBIC and needs to be escaped
q=c*a+1     Sets the starting point of the next mid$
<implicit>  NEXT

Note that a more modern version of QBIC can solve this in 27 bytes:

_L;|g=sqr(a)[1,a,g|?_sA,b,g

Improvements are: inline cmd-arg-grabbing (;), implementation of len (_L or _l) and substring (_s). Also, I've improved conditions and workings of the FOR-loop itself.

Answer 24

K (oK), 14 bytes

Solution:

`0:#[2#%#x]x:;

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Example:

`0:#[2#%#x]x:"Lorem ipsum dolor sit amt";
Lorem
ipsu
m dol
or si
t amt

Explanation:

Similar to my q/kdb+ solution, but a little shorter. Use # to reshape string into a square:

`0:#[2#%#x]x:; / the solution
`0:          ; / print to stdout and swallow return value
           x:  / save input as x
   #[     ]    / reshape
        #x     / count x
       %       / sqrt
     2#        / duplicate

Answer 25

Ly, 31 bytes

irysp>l12/^s<[ol,s!['
o>s<1$]p]

Try it online!

Try a square version online, for fun

Answer 26

JavaScript (Node.js), 54 bytes

s=>[...s].map((c,i)=>++i%s.length**.5?c:c+`\n`).join``

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Answer 27

Python 3, 76 bytes

I understand the length of my answer, but I didn't want to copy off the other people that did python, so I took the bit longer route (by like an extra 20 some bytes).

i=input()
for c in range(len(i)):print(i[c],end="\n"*((c+1)%len(i)**0.5==0))

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So rather than cut my string after I use the part I want, I keep printing characters until I get to the edge of the square, then I add a newline and keep spewing characters and doing that till we get to the end.

Answer 28

C#, 151 Bytes

static string T(string c){string s="";int a=(int)Math.Sqrt(c.Length);for(int i=0;i<c.Length;i++){s=s+c.Substring(0,a)+"\n";c=c.Remove(0,a);}return s;}

Answer 29

Japt -R, 4 bytes

òUÊq

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Explanation:

òUÊq
ò        // Split the input into slices of length:
   q     //   Square root of 
 UÊ      //   Length of the input
-R       // Join with newlines

Answer 30

Japt -R, 7 6 4 bytes

òUʬ

Try it

òUʬ     :Implicit input of string U
ò        :Partition into chunks of length
 UÊ      :  Length of U
   ¬     :  Square root
         :Implicitly join with newlines and output