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

Vim, 59, 57, 48 bytes/keystrokes

$:let @q=float2nr(sqrt(col('.')))."|li<C-v><cr><C-v><esc>@q"<cr>@q

Since V is backwards compatible, you can Try it online!

I randomly received an upvote on this answer, so I looked over it again. My vim-golfing skills have greatly increased over the last 7 months, so I saw that this answer was very poorly golfed. This one is much better.

Answer 2

Brainfuck, 116 112 bytes

>>>>,[[<]<<+>>>[>],]<[<]<+<[>>+<[-<-<+>>]<<++[->>+<<]>]>[-]>>[<[->.[-]<[->+<]<+[->+<]>>]++++++++++.[-]<[->+<]>>]

Try it online!

Safe in flavours of BF that does not mask the cells with 256, does not support null bytes.

Remove the initial right arrows if the flavour supports negative memory for 4 bytes saved.

Explanation

The program is divided into 3 stages:

Stage 1: >>>>,[[<]<<+>>>[>],]<[<]
Stage 2: <+<[>>+<[-<-<+>>]<<++[->>+<<]>]>[-]>>
Stage 3: [<[->.[-]<[->+<]<+[->+<]>>]++++++++++.[-]<[->+<]>>]

Stage 1

In this stage, we put all the characters onto the tape, while keeping count of the number of characters.

This is the tape for the input abcdefghi after this tape:

000 009 000 000 095 096 097 098 099 100 101 102 103
             ^

The 009 is the count.

For each character, we move the the first zero on the left [<] and then add one to the count <<+>>>, and then move to the rightmost zero [>] to get ready for the next character.

Stage 2

This stage does the square root of the length stored in the second cell.

It keeps subtracting by 1, 3, 5, 7, ... until the number reaches zero, while keeping check of the number of iterations.

It works because square numbers can be expressed as 1 + 3 + 5 + ....

Stage 3

Denote the square root of the length found above as n.

This stage outputs n characters at a time, and then output a newline, until the tape is cleared.

Answer 3

Python 2, 55 bytes

s=input()
n=int(len(s)**.5)
while s:print s[:n];s=s[n:]

Answer 4

05AB1E, 5 bytes

Dgtô«

Try it online!

D    duplicate a (implicit input)
g    length of a
t    square root of a
ô    push a split in pieces of b
«    join by newlines (implicit output)

Answer 5

MATL, 6 bytes

tnX^e!

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Explanation

t     % Take input implicitly. Push another copy
n     % Get number of elements of the copy
X^    % Take square root
e     % Reshape the input into that number of rows, in column-major order
      % (which means: down, then across)
!     % Transpose so that text reads horizontally. Implicitly display

Answer 6

Jelly, 8 7 bytes

sLƽ$j⁷

Saved a byte thanks to @Dennis.

Try it online.

Explanation

sLƽ$j⁷  Input: string S
    $    Monadic chain
 L         Get the length of S
  ƽ       Take the integer square root of it, call it n
s        Split S into chunks of size n
     j⁷  Join using newline

Answer 7

JavaScript (ES7), 49 bytes

s=>s.match(eval(`/.{${s.length**.5}}/g`)).join`
`

44 bytes in Firefox Nightly 43-46 only (** was introduced some time between Firefox Nightly 42 and 43 and g as a separate parameter was removed some time between Firefox Nightly 46 and 47):

s=>s.match(`.{${s.length**.5}}`,`g`).join`
`

Answer 8

J, 9 bytes

$~,~@%:@#

This is a monadic hook over the input string:

$~ ,~@%:@#

The right tine is a series of compositions:

,~ @ %: @ #

The left is a shaping verb, switched such that it works in the hook format.

Here are some intermediate results:

   # 'hiya'
4
   %:@# 'hiya'
2
   ,~@%:@# 'hiya'
2 2

In words:

   size =: #
   sqrt =: %:
   dup =: ,~
   on =: @
   shape =: $~
   block =: shape dup on sqrt on size
   block 'Hello, World! :)'
Hell
o, W
orld
! :)

Answer 9

C, 64 bytes

Call f() with the string to square.

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

Try it on ideone.

Answer 10

Perl, 23 + 4 (-pF flags) = 27 bytes

-2 bytes thanks to @DomHastings
-1 bytes thanks to @DomHastings

$==sqrt@F;s/.{$=}/$&
/g

Try it online!

Expanations : computes the square root (lets call it S for the explanation) of the size of the input (it will be always be an integer) (@F is used in scalar context, thus returning its size), then add a newline after each bloc of S characters.

Answer 11

05AB1E, 8 6 bytes

Thanks to @quartata for letting me know about the square-root function

Dgtô¶ý

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Explanation

D     Implicit input. Duplicate
g     Number of elements
t     Square root
ô     Split into chunks of that length
¶     Push newline character
ý     Join list by newlines. Implicit display

Answer 12

Dyalog APL, 10 bytes

⊢⍴⍨2⍴.5*⍨≢

Explanation:

         ≢   length of the argument   
     .5*⍨    square root 
   2⍴        reshape that to a length-2 vector
⊢⍴⍨          reshape the input by that vector

Tests:

      (⊢⍴⍨2⍴.5*⍨≢)'Hi, world'
Hi,
 wo
rld
      (⊢⍴⍨2⍴.5*⍨≢)'Hi, world! Hello'
Hi, 
worl
d! H
ello
      (⊢⍴⍨2⍴.5*⍨≢)'Lorem ipsum dolor sit amt'
Lorem
 ipsu
m dol
or si
t amt
      (⊢⍴⍨2⍴.5*⍨≢) 'H'
H

Answer 13

Python, 94 75 71 65 63 bytes

import re;lambda r:"\n".join(re.findall("."*int(len(r)**.5),r))

Old version:

lambda r:"\n".join(map("".join,zip(*[iter(r)]*int(len(r)**.5))))

Answer 14

zsh, 36 33 bytes

fold -`tr -d .<<<$[$#1**.5]`<<<$1

Try it online!

Takes input as a command line argument, outputs to STDOUT.

                   $#1             # get the length of the input string
                 $[   **.5]        # take it to the .5 power (sqrt)
              <<<                  # and pass the result to
       tr -d .                     # tr, to delete the trailing decimal point
                                   #    (sqrt(9) is 3. instead of 3)
     -`                    `       # give the result as a command line flag to
fold                               # the fold util, which wraps at nth column
                            <<<$1  # pass the input as input to fold

Answer 15

Cheddar, 27 bytes

s->s.chunk(s.len**.5).vfuse

I added the .chunk function a while ago but I removed it in the transition to the new stdlib format and forgot to re-add it. Cheddar has a dedicated sqrt operator but **.5 is shorter

Try it online!

Explanation

s ->              // Function with argument s
    s.chunk(      // Chunk it into pieces of size...
      s.len ** .5 // Square root of length.
    ).vfuse       // Vertical-fuse. Join on newlines

Answer 16

CJam, 8 bytes

l_,mQ/N*

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Explanation

l     e# Read line from input
_,    e# Duplicate. Get length 
mQ    e# Integer square root
/     e# Split into pieces of that size
N*    e# Join by newline. Implicitly display

Answer 17

Pyth, 8 bytes

jcs@lQ2Q

Try it online

How it works

    lQ     length of input
   @  2    square root
  s        floor
 c     Q   chop input into that many equal pieces
j          join on newline

Answer 18

Bash + GNU utilities, 25

fold -`dc -e${#1}vp`<<<$1

Not so different to @Doorknob's answer, but dc is a shorter way to get the square root.

Answer 19

𝔼𝕊𝕄𝕚𝕟, 11 chars / 14 bytes

ѨĊ(ï,√ ïꝈ⸩Ė⬮

Try it here (ES6 browsers only).

Generated using this code (run in the interpreter's browser console):

c.value=`Ѩ${alias(_,'chunk')}(ï,√ ïꝈ⸩${alias(Array.prototype,'mjoin')}⬮`

Answer 20

Brainfuck, 83 bytes

,[>+[>+<-],]
>
[
  >>[<+<-->>-]
  +<[>+<-]
  <-
]
<<
[
  [<]
  >.,
  >[>]
  >>+>-[<]
  <[[>+<-]++++++++++.,<<]
  <
]

Try it online!

This uses the same idea as Leaky Nun's answer. He asked for help golfing it in chat, then suggested that I add this as a new answer. (Actually what I wrote in chat was an 84-byte solution very similar to this.)

For the sake of comparison, an extra > is needed at the beginning for brainfuck implementations that don't allow negative memory addresses.

As expected, this finds the length of the input, then takes the square root, then prints the lines accordingly. It takes advantage of perfect squares being partial sums of 1 + 3 + 5 ....

Answer 21

Brain-Flak, 110 96 bytes

([]<>){({}{}(({}[()])))}{}{({}()<(({})<{({}()<<>({}<>)>)}{}((()()()()()){})>)>)}{}{}{({}<>)<>}<>

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Second solution, 96 bytes

(([]<>)<{({}({})({}[()]))}{}>){({}(({})<{({}()<<>({}<>)>)}{}((()()()()()){})>))}{}{}{({}<>)<>}<>

Try it online!

Explanation

Here I explain the first solution, both are the same length but I like the first one because it is cooler and employs some nice tricks.

The most important part of the code is a modified square root function I wrote some time ago. The original version was

{({}[({})({}())])}{}

And this works, but we actually want two copies of the negative square root. Why? We need two copies because we are looping through the string at two levels, one to make the lines and one to count the number of lines. We want it to be negative because looping with negatives is cheaper.

To make this negative we move around the [...] so it looks like this

{({}({})({}[()]))}{}

To make two copies we change when pops occur

{({}{}(({}[()])))}{}

Now that we have that bit we can put it together with a stack height to get the first chunk of code we need.

([]<>){({}{}(({}[()])))}{}

We move to the offstack because our square root function needs two free zeros for computation, and because it makes stuff a little bit cheaper int he future in terms of stack switching.

Now we construct the main loop

{({}()<(({})<{({}()<<>({}<>)>)}{}((()()()()()){})>)>)}{}{}

This is pretty straight forward, we loop n times each time moving n items and capping it with a new line (ASCII 10).

Once the loop is done we need to reverse the order of our output so we just tack on a standard reverse construct.

{({}<>)<>}<>

Answer 22

PHP, 51 bytes

<?=join("
",str_split($x=$argv[1],strlen($x)**.5));

Answer 23

Perl 6, 38 bytes

$_=get;.put for .comb: .chars.sqrt.Int

Explanation:

$_ = get;          # get a single line of input


$_.put             # print with trailing newline

for                # every one of the following:

$_.comb:           # the input split into

$_.chars.sqrt.Int  # chunks of the appropriate size

Answer 24

Cheddar, 57 bytes

n->(m->(|>m).map(i->n.slice(i*m,i*m+m)).vfuse)(n.len**.5)

Since variables are broken, I would have to pass in variables through lambda application.

Also, it turns out that even if variables worked, it would still be shorter to use lambda application.

Usage

cheddar> (n->(m->(|>m).map(i->n.slice(i*m,i*m+m)).vfuse)(n.len**.5))("abcd")
"ab
cd"

Answer 25

Pyke, 5 bytes

l,fnJ

Try it here!

Answer 26

Java 1.7, 110 bytes

void f(String s){for(int i=-1,k=(int)Math.sqrt(s.length());++i<k;)System.out.println(s.substring(i*k,i*k+k));}

Try it! (Ideone)

I tried another approach with a function returning the result as a string, but just having to declare the string and the return statement is already more expensive (byte-count-wise) than the print statement.

Gotta love Java's verbosity... :)

Answer 27

R, 59 54 bytes

function(s)write(el(strsplit(s,'')),1,nchar(s)^.5,,'')

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Prints with a trailing newline. Surprisingly short, considering how badly R handles strings.

Answer 28

PowerShell, 56 58 61 bytes

param($i)$i-replace".{$([Math]::Sqrt($i.Length))}",'$&
'

Answer 29

Ruby -p, 40 33 30 bytes

-7 bytes from @Jordan.

gsub(/#{?.*~/$/**0.5}/){$&+$/}

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Answer 30

K (ngn/k), 11 bytes

{(2#%#x)#x}

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---

the bracketed expression returns list 5 5 for input "Lorem ipsum dolor sit amt". it reads "2 take sqrt count x", where x is the string argument. then (5 5)#x means "cut x into 5 lists each of length 5". with this we get:

  {(2#%#x)#x}"Lorem ipsum dolor sit amt"
("Lorem";" ipsu";"m dol";"or si";"t amt")