[Question]
  [
Consider a non-empty string of correctly balanced parentheses:
```
(()(()())()((())))(())
```
We can imagine that each pair of parentheses represents a ring in a collapsed [telescopic construction](http://en.wikipedia.org/wiki/Telescoping_(mechanics)). So let's extend the telescope:
```
(                )(  )
 ()(    )()(    )  ()
    ()()    (  )
             ()
```
Another way to look at it is that the parentheses at depth *n* are moved to line *n*, while keeping their horizontal position.
Your task is to take such a string of balanced parentheses and produce the extended version.
You may write a program or function, taking input via STDIN (or closest equivalent), command-line argument or function parameter, and producing output via STDOUT (or closest equivalent), return value or function (out) parameter.
You may assume that the input string is valid, i.e. consists only parentheses, which are correctly balanced.
You may print trailing spaces on each line, but not any more leading spaces than necessary. In total the lines must not be longer than twice the length of the input string. You may optionally print a single trailing newline.
## Examples
In addition to the above example, here are a few more test cases (input and output are separated by an empty line).
```
()
()
```
```
(((())))
(      )
 (    )
  (  )
   ()
```
```
()(())((()))(())()
()(  )(    )(  )()
   ()  (  )  ()
        ()
```
```
((()())()(()(())()))
(                  )
 (    )()(        )
  ()()    ()(  )()
             ()
```
**Related Challenges:**
* **[Topographic Strings](https://codegolf.stackexchange.com/q/6600/8478)**, which asks you to produce what is essentially the complement of the output in this challenge.
* **[Code Explanation Formatter](https://codegolf.stackexchange.com/q/49016/8478)**, a broad generalisation of the ideas in this challenge, posted recently by PhiNotPi. (In fact, PhiNotPi's original description of his idea was what inspired this challenge.)
## Leaderboards
Huh, this got quite a lot of participation, so here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.
To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:
```
# Language Name, N bytes
```
where `N` is the size of your submission. If you improve your score, you *can* keep old scores in the headline, by striking them through. For instance:
```
# Ruby, <s>104</s> <s>101</s> 96 bytes
```
```
function answersUrl(e){return"http://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function getAnswers(){$.ajax({url:answersUrl(page++),method:"get",dataType:"jsonp",crossDomain:true,success:function(e){answers.push.apply(answers,e.items);if(e.has_more)getAnswers();else process()}})}function shouldHaveHeading(e){var t=false;var n=e.body_markdown.split("\n");try{t|=/^#/.test(e.body_markdown);t|=["-","="].indexOf(n[1][0])>-1;t&=LANGUAGE_REG.test(e.body_markdown)}catch(r){}return t}function shouldHaveScore(e){var t=false;try{t|=SIZE_REG.test(e.body_markdown.split("\n")[0])}catch(n){}return t}function getAuthorName(e){return e.owner.display_name}function process(){answers=answers.filter(shouldHaveScore).filter(shouldHaveHeading);answers.sort(function(e,t){var n=+(e.body_markdown.split("\n")[0].match(SIZE_REG)||[Infinity])[0],r=+(t.body_markdown.split("\n")[0].match(SIZE_REG)||[Infinity])[0];return n-r});var e={};var t=0,c=0,p=-1;answers.forEach(function(n){var r=n.body_markdown.split("\n")[0];var i=$("#answer-template").html();var s=r.match(NUMBER_REG)[0];var o=(r.match(SIZE_REG)||[0])[0];var u=r.match(LANGUAGE_REG)[1];var a=getAuthorName(n);t++;c=p==o?c:t;i=i.replace("{{PLACE}}",c+".").replace("{{NAME}}",a).replace("{{LANGUAGE}}",u).replace("{{SIZE}}",o).replace("{{LINK}}",n.share_link);i=$(i);p=o;$("#answers").append(i);e[u]=e[u]||{lang:u,user:a,size:o,link:n.share_link}});var n=[];for(var r in e)if(e.hasOwnProperty(r))n.push(e[r]);n.sort(function(e,t){if(e.lang>t.lang)return 1;if(e.lang<t.lang)return-1;return 0});for(var i=0;i<n.length;++i){var s=$("#language-template").html();var r=n[i];s=s.replace("{{LANGUAGE}}",r.lang).replace("{{NAME}}",r.user).replace("{{SIZE}}",r.size).replace("{{LINK}}",r.link);s=$(s);$("#languages").append(s)}}var QUESTION_ID=49042;var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";var answers=[],page=1;getAnswers();var SIZE_REG=/\d+(?=[^\d&]*(?:&lt;(?:s&gt;[^&]*&lt;\/s&gt;|[^&]+&gt;)[^\d&]*)*$)/;var NUMBER_REG=/\d+/;var LANGUAGE_REG=/^#*\s*([^,]+)/
```
```
body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}
```
```
<script src=https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js></script><link rel=stylesheet type=text/css href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"><div id=answer-list><h2>Leaderboard</h2><table class=answer-list><thead><tr><td></td><td>Author<td>Language<td>Size<tbody id=answers></table></div><div id=language-list><h2>Winners by Language</h2><table class=language-list><thead><tr><td>Language<td>User<td>Score<tbody id=languages></table></div><table style=display:none><tbody id=answer-template><tr><td>{{PLACE}}</td><td>{{NAME}}<td>{{LANGUAGE}}<td>{{SIZE}}<td><a href={{LINK}}>Link</a></table><table style=display:none><tbody id=language-template><tr><td>{{LANGUAGE}}<td>{{NAME}}<td>{{SIZE}}<td><a href={{LINK}}>Link</a></table>
```
  
[Answer]
# x86 machine code, ~~39~~ ~~34~~ ~~33~~ ~~30~~ 29 bytes
```
00000000  68 c3 b8 07 31 ff be 82  00 b3 a0 ad 4e 3c 28 7c  |h...1.......N<(||
00000010  f0 77 05 ab 01 df eb f3  29 df ab eb ee           |.w......)....|
0000001d
```
x86 assembly for DOS, with some tricks:
```
    org 100h
section .text
start:
    ; point the segment ES to video memory
    ; (c3 is chosen so that it doubles as a "ret")
    push 0b8c3h
    pop es
    ; di: output pointer to video memory
    xor di,di
    ; si: input pointer from the command line
    mov si,82h
    ; one row=160 bytes (assume bh=0, as should be)
    mov bl,160
lop:
    ; read & increment si (assume direction flag clean)
    ; we read a whole word, so that later we have something nonzero to
    ; put into character attributes
    lodsw
    ; we read 2 bytes, go back 1
    dec si
    ; check what we read
    cmp al,'('
    ; less than `(`: we got the final `\n` - quit
    ; (we jump mid-instruction to get a c3 i.e. a ret)
    jl start+1
    ; more than `(`: assume we got a `)`
    ja closed
    ; write a whole word (char+attrs), so we end
    ; one position on the right
    stosw
    ; move down
    add di,bx
    ; rinse & repeat
    jmp lop
closed:
    ; move up
    sub di,bx
    ; as above
    stosw
    jmp lop
```
*Limitations*:
* it always prints starting at the bottom of the screen, without erasing first; a `cls` before running is almost mandatory;
* the colors are ugly; that's the consequence of recycling the next character as color attributes to save two bytes here and there;
* the code assumes `bh=0` and the direction flag clear on start, both undocumented; OTOH, `bx` is explicitly set to zero in all DOS variants I saw (DosBox, MS-DOS 2, FreeDOS), and everywhere I tested the flags were already OK.
![enter image description here](https://i.stack.imgur.com/hwpmW.png)
[Answer]
## J, ~~32~~ 28 bytes
This was a fun one.
```
0|:')(('&(i.-<:@+/\@i:){."0]
```
### Explanation
This is how this solution works, including an explanation of how it has been golfed.
```
   NB. Let a be a test case
   a =. '((()())()(()(())()))'
   NB. level alterations
   _1 + ').(' i. a
1 1 1 _1 1 _1 _1 1 _1 1 1 _1 1 1 _1 _1 1 _1 _1 _1
   NB. absolute levels
   +/\ _1 + ').(' i. a
1 2 3 2 3 2 1 2 1 2 3 2 3 4 3 2 3 2 1 0
   NB. adjusted levels
   (+/\ _1 + ').(' i. a) - ')(' i. a
0 1 2 2 2 2 1 1 1 1 2 2 2 3 3 2 2 2 1 0
   NB. take level from end of each item of a and transpose
   |: a {."0~ _1 - (+/\ _1 + ').(' i. a) - ')(' i. a
(                  )
 (    )()(        ) 
  ()()    ()(  )()  
             ()     
   NB. code as a tacit verb
   [: |: ] {."0~ _1 - ([: +/\ _1 + ').(' i. ]) - ')(' i. ]
   NB. subtractions pulled into the prefix insert
   [: |: ] {."0~ (')(' i. ]) - [: <:@+/\ ').(' i. ]
   NB. i: instead of i. so we can use the same string constant
   [: |: ] {."0~ (')((' i. ]) - [: <:@+/\ ')((' i: ]
   NB. get rid of the caps
   0 |: ] {."0~ (')((' i. ]) - ')((' <:@+/\@i: ]
   NB. join the two usages of ')((' into a single dyadic phrase
   0 |: ] {."0~ ')((' (i. - <:@+/\@i:) ]
   NB. bond ')((' and flip arguments to {."0
   0 |: ')(('&(i. - <:@+/\@i:) {."0 ]
```
[Answer]
# C, 150 bytes
```
t;f(char*c){char l=strlen(c)+1,o[l*l],*A=o,m=0;for(t=1;t<l*l;t++)o[t-1]=t%l?32:10;for(t=-1;*c;c++)A++[l*(*c-41?++t>m?m=t:t:t--)]=*c;A[m*l]=0;puts(o);}
```
This was *crazy* fun to golf. I'm still not convinced I'm done with it.
We define a single function, `f`, that takes the string as input and outputs to stdout.
Let's go through the code, line by line:
```
/* t will represent the current depth of a parentheses. It must be an int. */
t;
f(char*c){
    //Our variables:
    char l=strlen(c)+1,    //The length of each row of output, including newlines
         o[l*l],           //The output string. It's way larger than it needs to be.
         *A=o,             //We need another pointer to keep track of things.
         m=0;              //The maximum depth recorded thus far.
    for(t=1;t<l*l;t++)     //For each character in our output...
        o[t-1]=t%l?32:10;  //If it's at the end of a line, make it '\n'. Else, ' '.
    for(t=-1;*c;c++)       //While we have an input string...
        //Perhaps a personal record for ugliest warning-less line...
        A++[l*(*c-41?++t>m?m=t:t:t--)]=*c;
    /* 
        A breakdown:
        A++        --> Go to the next *column* of output, after writing. 
                   --> There will only ever be one parentheses per output column.
        [l*(...)]  --> A[l*X] is the character in the current column at depth X.
        (*c-41?    --> If the character is a '('...    
        ++t>m?     --> Increment t *before* we write it. If this is a new highest depth
        m=t:       --> Set m to t, and set the whole expression to t.
        t:         --> If it's not a new highest depth, don't set m.
        t--)       --> If the character was a ')', decrement t *after* we write it.
        =*c        --> Write our output character to whatever the input read.
    */    
    
    A[m*l]=0; //The last character of the maximum-depth line should be null terminated.
    puts(o);  //Output!
}
```
I'll answer any questions you may have!
Try a test program [online](http://ideone.com/v5nyZ5)!
[Answer]
# Retina + Bash, 27 bytes (14 + 10 + 3 = 27)
This makes use of ANSI Escapes:
```
\(
(\e[B
\)
\e[A)
```
Equivalent to `sed -e "s/(/(\\\e[B/g;s/)/\\\e[A)/g"`.
The `\e[B` escape code means move cursor down one row, and the `\e[A` means move cursor up one row, so this solution simply inserts those codes after and before the start and end of each nested pair of parentheses. Input is passed through STDIN.
You'll have to call it as `printf $(Retina ...)` to see the output correctly.
## Output
```
(((())))
(\e[B(\e[B(\e[B(\e[B\e[A)\e[A)\e[A)\e[A)
^C
amans:~ a$ printf "(\e[B(\e[B(\e[B(\e[B\e[A)\e[A)\e[A)\e[A)"
(      )amans:~ a$ 
 (    )
  (  )
   ()
((()())()(()(())()))
(\e[B(\e[B(\e[B\e[A)(\e[B\e[A)\e[A)(\e[B\e[A)(\e[B(\e[B\e[A)(\e[B(\e[B\e[A)\e[A)(\e[B\e[A)\e[A)\e[A)
^C
amans:~ a$ printf "(\e[B(\e[B(\e[B\e[A)(\e[B\e[A)\e[A)(\e[B\e[A)(\e[B(\e[B\e[A)(\e[B(\e[B\e[A)\e[A)(\e[B\e[A)\e[A)\e[A)"
(                  )amans:~ a$ 
 (    )()(        )
  ()()    ()(  )()
             ()
```
[Answer]
# TI-BASIC, ~~69 60 56~~ 55 bytes
This is for the TI-83+/84+ family of calculators, although it was written on an 84+ C Silver Edition.
The program shows up as larger on-calc due to VAT+size info being included. Also, there are more than 56 characters here; the reason it's 56 bytes is because all commands that are more than one character are compressed down to tokens that are either one or two bytes in size.
```
Input Str1
1→B
For(A,1,length(Str1
sub(Str1,A,1→Str2
Ans="(
Output(B+Ans,A,Str2
B-1+2Ans→B
End
```
Shaved off another byte thanks to [thomas-kwa](http://stackexchange.com/users/5456847/thomas-kwa)! (also from him was the jump from 60 to 56.)
[Answer]
# CJam, ~~17~~ ~~16~~ 15 bytes
```
0000000: 72 3a 69 22 28 0b 20 9b 41 29 22 53 2f 66 3d     r:i"(. .A)"S/f=
```
The above is a reversible xxd dump, since the source code contains the unprintable characters VT (0x0b) and CSI (0x9b).
Like [this answer](https://codegolf.stackexchange.com/a/49118), it uses [ANSI escape sequences](https://en.wikipedia.org/wiki/ANSI_escape_code), but it also uses vertical tabs and it prints the control characters directly to avoid using **printf**.
This requires a supporting video text terminal, which includes most non-Windows terminal emulators.
### Test run
We have to set the shell variable *LANG* and the terminal emulator's encoding to ISO 8859-1. The former is achieved by executing
```
$ LANGsave="$LANG"
$ LANG=en_US
```
Also, before executing the actual code, we'll disable the prompt and clear the screen.
```
$ PS1save="$PS1"
$ unset PS1
$ clear
```
This makes sure the output is shown properly.
```
echo -n '()(())((()))(())()' | cjam <(base64 -d <<< cjppIigLIJtBKSJTL2Y9)
()(  )(    )(  )()
   ()  (  )  ()
        ()
```
To restore *LANG* and the prompt, execute this:
```
$ LANG="$LANGsave"
$ PS1="$PS1save"
```
### How it works
We insert a vertical tab after each **(** to move the cursor down and the byte sequence **9b 41** (`"\x9bA"`) before each **)** to move the cursor up.
```
r         e# Read a whitespace-separated token from STDIN.
:i        e# Replace each character by its code point.
          e#   '(' -> 40, ')' -> 41
"(. .A)"  e# Push the string "(\v \x9bA)".
S/        e# Split at spaces into ["(\v" "\x9bA)"].
f=        e# Select the corresponding chunks.
          e# Since arrays wrap around in CJam, ["(\v" "\x9bA)"]40= and 
          e# ["(\v" "\x9bA)"]41= select the first and second chunk, respectively.
```
[Answer]
# Python 2, 115 bytes
```
def f(L,n=0,O=()):
 for c in L:n-=c>"(";O+=" "*n+c,;n+=c<")"
 for r in map(None,*O):print"".join(c or" "for c in r)
```
Call like `f("((()())()(()(())()))")`, and output is to STDOUT.
## Explanation
We start with `n = 0`. For each char in the input line:
* If the char is `(`, we prepend `n` spaces then increment `n`
* If the char is `)`, we decrement `n` then prepend `n` spaces
The result is then zipped and printed. Note that Python's `zip` zips to match the length of the *shortest* element, e.g.
```
>>> zip([1, 2], [3, 4], [5, 6, 7])
[(1, 3, 5), (2, 4, 6)]
```
Usually one would use `itertools.zip_longest` (`izip_longest`) if they wanted `zip` to pad to the length of the *longest* element.
```
>>> import itertools
>>> list(itertools.izip_longest([1, 2], [3, 4], [5, 6, 7]))
[(1, 3, 5), (2, 4, 6), (None, None, 7)]
```
But in Python 2, this behaviour can be simulated by mapping `None`:
```
>>> map(None, [1, 2], [3, 4], [5, 6, 7])
[(1, 3, 5), (2, 4, 6), (None, None, 7)]
```
---
## Python 3, 115 bytes
```
L,d,*O=input(),0
for i,c in enumerate(L):b=c>"(";O+="",;O[d-b]=O[d-b].ljust(i)+c;d-=b*2-1
for l in O:l and print(l)
```
No zipping, just padding appropriately with `ljust`. This one seems to have some golfing potential.
[Answer]
# C, ~~58 53 52 51~~ 49 bytes
Makes use of ANSI escape sequences to move the cursor position.
```
f(char*s){while(*s)printf(*s++&1?"\e[A)":"(\v");}
```
If not using gcc or another compiler that supports `\e` then it can be replaced with `\x1B` for a total of 2 extra bytes. `\e[A` moves the cursor up one row and `\e[B` moves the cursor down one row. It's not necessary to use `\e[B` to move down one row as it's two bytes shorter to use the ASCII vertical tab character `0xB` or `\v`.
The input string is assumed, from the question, to consist of only (balanced) parentheses, so checking the parity of the character, with `&1`, is enough to distinguish between `(` and `)`.
[Answer]
# Pip, 53 bytes
[**Pip**](https://github.com/dloscutoff/pip) is a code-golf language of my invention. The first version was published on Saturday, so I can officially take it for a spin! The solution below isn't terribly competitive as golfing languages go, but that's partly because I haven't implemented things like zip and max yet.
```
z:{aEQ'(?++v--v+1}MaW(o:{z@++v=i?as}Ma)RMs{Pov:-1++i}
```
Expects the string of parentheses as a command-line argument.
"Ungolfed" version:
```
z:{
   a EQ '( ?
    ++v
    --v+1
  } M a
W (o:{
      z @ ++v = i ?
       a
       s
     } M a
  ) RM s
{
 P o
 v:-1
 ++i
}
```
**Explanation:**
Unlike most golfing languages, Pip is imperative with infix operators, so the syntax is somewhat closer to C and its derivatives. It also borrows ideas from functional and array-based programming. See the repository for further documentation.
The program first generates a list of depths (storing it in `z`) by mapping a function to the input string `a`. The global variable `v` tracks the current level. (Variables `a-g` in Pip are function-local variables, but `h-z` are global. `v` is handy because it's preinitialized to -1.)
Next, we use a `W`hile loop to generate and print each line, until the line generated would consist of all spaces. `v` is now used for columns, and `i` for rows. The `{z@++v=i?as}` function, repeatedly mapped to the original input string, tests whether the current line `i` matches the line the current parenthesis is supposed to be on (as stored in the `z` list). If so, use the parenthesis (`a`); if not, use `s` (preinitialized to space). The end result is that on each iteration, `o` gets assigned a list of characters equivalent to the next line of the output.
To test whether we should continue looping, we check if `o` with all the spaces `RM`'d is empty. If not, print it (which by default concatenates everything together as in CJam), reset the column number to -1, and increment the row number.
(Fun fact: I had a 51-byte solution at first... which didn't work because it turned up a bug in the interpreter.)
[Answer]
# R, ~~151~~ 127 characters
```
S=strsplit(scan(,""),"")[[1]];C=cumsum;D=c(C(S=="("),0)-c(0,C(S==")"));for(j in 1:max(D)){X=S;X[D!=j]=' ';cat(X,sep='',fill=T)}
```
With indents and newlines:
```
S=strsplit(scan(,""),"")[[1]]
C=cumsum
D=c(C(S=="("),0)-c(0,C(S==")"))
for(j in 1:max(D)){
    X=S
    X[D!=j]=' '
    cat(X,sep='',fill=T)
    }
```
Usage:
```
> S=strsplit(scan(,""),"")[[1]];C=cumsum;D=c(C(S=="("),0)-c(0,C(S==")"));for(j in 1:max(D)){X=S;X[D!=j]=' ';cat(X,sep='',fill=T)}
1: ()(())((()))(())()
2: 
Read 1 item
()(  )(    )(  )()
   ()  (  )  ()   
        ()        
> S=strsplit(scan(,""),"")[[1]];C=cumsum;D=c(C(S=="("),0)-c(0,C(S==")"));for(j in 1:max(D)){X=S;X[D!=j]=' ';cat(X,sep='',fill=T)}
1: ((()())()(()(())()))
2: 
Read 1 item
(                  )
 (    )()(        ) 
  ()()    ()(  )()  
             ()     
```
It reads the string as stdin, splits it as a vector of single characters, computes the cumulative sum of `(` and `)`, substracts the former with the latter (with a lag) thus computing the "level" of each parentheses. It then prints to stdout, for each level, either the corresponding parentheses or a space.
Thanks to @MickyT for helping me shortening it considerably!
[Answer]
# Pyth, 31 bytes
```
VzJs.e?YqN-/<zk\(/<zhk\)dzI-JdJ
```
[Try it online.](https://pyth.herokuapp.com/?code=VzJs.e%3FYqN-%2F%3Czk%5C(%2F%3Czhk%5C)dzI-JdJ&input=(()(()())()((())))(())&debug=0)
`-/<zk\(/<zhk\)`: Finds the appropriate level for the current character position.
`?YqN-/<zk\(/<zhk\)d`: A space if the appropriate level is not the current level, current character otherwise.
`Js.e?YqN-/<zk\(/<zhk\)dz`: Generate the string, save it to `J`.
`I-JdJ`: If `J` is not all spaces, print it out.
`Vz`: Loop `z` times.
[Answer]
# GNU Bash + coreutils + indent, 135
```
eval paste "`tr '()' {}|indent -nut -i1 -nbap|sed 's/.*/<(fold -1<<<"&")/'|tr '
' \ `"|expand -t2|sed 'y/{}/()/;s/\(.\) /\1/g;s/ \+$//'
```
Input/output via STDIN/STDOUT:
```
$ ./telescopic.sh <<< "(()(()())()((())))(())"
(                )(  )
 ()(    )()(    )  ()
    ()()    (  )
             ()
$ 
```
`indent` does most of the heavy lifting, but needs to work with braces instead of parens. The rest is [modification of this answer](https://codegolf.stackexchange.com/questions/17305/transpose-a-page-of-text/24501#24501) to transpose the output of `indent`.
[Answer]
# Python 2, 92
```
def f(s,i=0,z=''):
 for x in s:b=x>'(';z+=[' ',x][i==b];i-=2*b-1
 if'('in z:print z;f(s,i-1)
```
Prints line by line. For a given line number `i` (actually, its negation), goes through the input string `s`, and makes a new string `z` that only contains the characters of `s` at depth `i`. This is done by incrementing or decrementing `i` to track the current depth, and adding the current characters when `i` is `0` adjusted for paren type, and otherwise adding a space.
Then, prints and recurses to the next `i` unless the current line was all spaces. Note that since the parens are balanced, the `i` after a loop is the same as at the start.
Python 3 would be same except for a character for `print(z)`.
[Answer]
# cheating :(  ~~Retina + TeX, N bytes~~ cheating :(
This only works if you render(?) the output using MathJax or some other TeX, which is currently disabled for this SE :(
```
\(
({
\)
})
\{\(
_{(
```
Each line should be in a different file, but you can test it by using `Retina -e "\(" -e "({" -e "\)" -e "})" -e "\{\(" -e "_{("` (or the equivalent sed command `sed -e "s/(/({/g;s/)/})/g;s/{(/_{(/g"`). Input is passed through STDIN.
This works by enclosing the contents of each pair of parentheses in braces, and then subscripting all the items inside them.
## Output
```
(((())))
(_{(_{(_{({})})})})
()(())((()))(())()
({})(_{({})})(_{(_{({})})})(_{({})})({})
((()())()(()(())()))
(_{(_{({})({})})({})(_{({})(_{({})})({})})})
```
![TeX Output](https://i.stack.imgur.com/inv2p.png)
[Answer]
# Java, 232 226 224 222 bytes
Golfed version:
```
int i,j,k,l,m,a[];void f(String s){a=new int[s.length()];j=a.length;for(k=0;k<j;){a[k]=s.charAt(k++)<41?i++:--i;m=m<i?i:m;}for(k=0;k<m;k++)for(l=0;l<j;)System.out.print(k==a[l++]?i++%2<1?'(':l==j?")\n":')':l==j?'\n':' ');}
```
Long version:
```
int i, j, k, l, m, a[];
void f(String s) {
    a = new int[s.length()];
    j = a.length;
    for (k = 0; k < j;) {
        a[k] = s.charAt(k++) < 41 ? i++ : --i;
        m = m < i ? i : m;
    }
    for (k = 0; k < m; k++)
        for (l = 0; l < j;)
            System.out.print(k == a[l++] ? (i++ % 2 < 1 ? '(' : (l == j ? ")\n" : ')')) : (l == j ? '\n':' '));
}
```
The input string is analyzed first, looking for "(" and ")" to add/subtract a counter and store its value determining how far down the parentheses should go in an array while also keeping track of how deep the deepest one goes. Then the array is analyzed; the parentheses with lesser values are printed first, and will continue printing line by line until the maximum is reached.
I'll probably find ways to golf this further later.
[Answer]
# Javascript/ES6, 97 chars
```
f=s=>{for(n in s){m=o=d='';for(c of s)o+=n==(c<')'?d++:--d)?c:' ',m=m<d?d:m;n<m&&console.log(o)}}
```
## Usage
```
f("(()(()())()((())))(())")
```
## Explanation
```
fn=str=>{                          // accepts string of parenthesis
  for(line in str){                // repeat process n times where n = str.length
    max=output=depth='';           // max: max depth, output: what to print, depth: current depth
    for(char of str)               // iterate over chars of str
      output+=
        line==(char<')'?depth++:--depth)? // update depth, if line is equal to current depth
        char:' ',                  // append either '(', ')', or ' '
        max=max<depth?depth:max;   // update max depth
    line<max&&console.log(output)  // print if current line is less than max depth
  }
}
```
[Answer]
# CJam, ~~43 41~~ 36 bytes
Not too golfed (I think), but here goes my first attempt:
```
l:L,{)L<)_')=@~zS*\+}%_$0=,f{Se]}zN*
```
**How it works**
I am using the very handy fact that `)` and `(` in CJam mean increment and decrement respectively. Thus, I simply evaluate the brackets to get the depth.
```
l:L,{)L<)_')=@~zS*\+}%_$0=,f{Se]}zN*
l:L,{                    }%                "Store input line in L and iterate over [0,L)";
     )L<                                   "substr(L, 0, iterator + 1)";
        )                                  "Slice off the last character to stack";
         _')=                              "Put 0 on stack if the sliced character is (,
                                            else 1 if sliced character is )";
             @~                            "bring forth the remaining
                                            brackets after slicing and evaluate them";
               zS*                         "Stack has negative depth number, take absolute
                                            value and get that many spaces";
                  \+                       "Prepend to the sliced character";
                      _$0=,                "Get the maximum depth of brackets";
                           f{Se]}          "Pad enough spaces after each string to match
                                            the length of each part";
                                 zN*       "Transpose and join with new lines";
```
[Try it online here](http://cjam.aditsu.net/#code=l%3AL%2C%7B)L%3C)_')%3D%40~zS*%5C%2B%7D%25_%240%3D%2Cf%7BSe%5D%7DzN*&input=(()(()())()((())))(()))
[Answer]
# Octave, 85 chars
```
function r=p(s)i=j=0;for b=s k=b==40;k&&++j;t(j,++i)=9-k;k||--j;r=char(t+32);end;end
```
It's an optimization of the naïve approach, which is actually pretty natural for Matlab and Octave:
```
function r=p(s)
i=j=1;
for b=s
 if b=='(' t(++j,i++)='(' else t(j--,i++)=')' end; end; t(~t)=' '; r=char(t);
end;
```
The table `t` may even not yet exist, and we may assign to any element right away, and it reshapes to the smallest dimension that is required for this element to exist which is quite convenient.
[Answer]
# Perl, ~~91~~ ~~89~~ ~~88~~ ~~84~~ ~~80~~ 79 bytes
```
$t=<>;{$_=$t;s/\((?{$l++})|.(?{--$l})/$^R==$c?$&:$"/ge;print,++$c,redo if/\S/}
```
* $t is the input string.
* $c is the depth we want to print on the current line.
* $l is the depth we are at after encountering a paren.
* $l is updated in [regex embedded code blocks](http://perldoc.perl.org/perlre.html#(%3F%7B-code-%7D)).
* $^R is the result of the most recent code block.
[Answer]
# Haskell, 154 bytes
```
f h('(':s)=h:f(h+1)s;f h(')':s)=(h-1):f(h-1)s;f _ _=[]
main=interact$ \s->unlines[[if i==h then c else ' '|(c,i)<-zip s l]|let l=f 0 s,h<-[0..maximum l]]
```
same idea as the other Haskell solution, but somewhat shorter. - Usage:
```
echo  '(((())())(()))' | runghc Golf.hs
```
[Answer]
# [Brain-Flak](https://github.com/Flakheads/BrainHack), 338 bytes
```
([]){(({})[()]<([{}][])([][()]){({}[()]<(({}<(({})(())){({}[()]<([{}]())>)}>{})<({}{})((){[()](<{}<>((((()()()()){}){}){})<>>)}{}){{}(({})<>)(<>)}{}(({})(())){({}[()]<([{}])>)}{}>{}<({}<(([])<{{}({}<>)<>([])}{}<>>)<>>)<>{({}[()]<({}<>)<>>)}{}<>>)>)}{}{}<>((()()()()()){})<>>)}{}{{}}<>{{}{}(({})[((((()()()){}())){}{}){}]())}{}{({}<>)<>}<>
```
[Try it online!](https://tio.run/##dU85DgMhDPzLVp4iP7As5R2IYpMmUaQt0lq8nYwNaLcJl8wceHh89/fx2p@f3qVUuIg3FEFVKd4qIcJxJ@VtECzygAhwwcNAxNCMpBIfGg9alC6TGBgTZMdSoydKb/muGkQT@tcGyVoEyTDMqeGOHrTHvUVtGPt0T4UtPosZDddoU8NXSbqvMOX8AlWZLKPn10O/OvDsvW9THELurd/uPw "Brain-Flak (BrainHack) – Try It Online")
## Idea
The idea for this program is that we make a number of passes over the string. Each time we make a pass we keep track of a counter, we decrement the counter if we encounter a `(` and increment it if we encounter a `)`. If `(` brings it to -1 we copy it to the offstack and if `)` brings it to zero we copy it to the offstack. Otherwise we just copy a space. We do this as many times as the length of the string starting the counter at 0 the first time and decreasing it each loop. This way we catch the different layers in order. When we are done we just trim characters until we reach a `)`.
[Answer]
# J, 46
Not as great as the other 'golfing languages', but in my defence: J is terrible with strings.
```
[:|:(((,~(' '#~]))"0)(0,2%~[:+/\2+/\1-'(('i.]))~
```
Takes the string as input for a function. There's also probably a better way to do it in J.
## Usage:
```
   f=:[:|:(((,~(' '#~]))"0)(0,2%~[:+/\2+/\1-'(('i.]))~
   f '(()(()())()((())))(())'
(                )(  )
 ()(    )()(    )  () 
    ()()    (  )      
             ()       
```
[Answer]
# Ruby, ~~119~~ ~~115~~ 114
```
->s{r=[""]*s.size
d=0
s.chars.map{|l|r.map!{|s|s+" "}
b=l>"("?1:0
d-=b
r[d][-1]=l
d+=1-b}.max.times{|i|puts r[i]}}
```
Explanation:
```
->s{r=[""]*s.size  # Take an array of strings big enough
d=0                # This will contain the current depth
s.chars.map{|l|r.map!{|s|s+" "}  # Add a new space to every array
b=l>"("?1:0       # Inc/Dec value of the depth
d-=b               # Decrement depth if we are at a closing paren
r[d][-1]=l         # Set the corresponding space to the actual open/close paren
d+=1-b             # Increment the depth if we are at a opening paren
}.max.times{|i|puts r[i]}}  # Print only the lines up to the max depth
```
[Answer]
# Java, ~~233~~ 214 bytes
```
void f(String s){int p,x,d,l=s.length();char c,m[]=new char[l*l];java.util.Arrays.fill(m,' ');p=x=0;while(x<l){d=(c=s.charAt(x))==40?p++:--p;m[d*l+x++]=c;}for(x=0;x<l*l;x++)System.out.print((x%l==0?"\n":"")+m[x]);}
```
Indented:
```
void f(String s){
    int p, x, d, l = s.length();
    char c, m[] = new char[l * l];
    java.util.Arrays.fill(m, ' ');
    p = x = 0;
    while (x < l){
        d = (c = s.charAt(x)) == 40
                ? p++
                : --p;
        m[d * l + x++] = c;
    }
    for (x = 0; x < l * l; x++)
        System.out.print((x % l == 0 ? "\n" : "") + m[x]);
}
```
I guess the final loop could be shortened, but I'll leave it as an exercise to the reader. ;-)
---
Old, 233 bytes answer:
```
void f(String s){int y=s.length(),x=0;char[][]m=new char[y][y];for(char[]q:m)java.util.Arrays.fill(q,' ');y=0;for(char c:s.toCharArray())if(c=='(')m[y++][x++]=c;else m[--y][x++]=c;for(char[]q:m)System.out.println(String.valueOf(q));}
```
Indented:
```
static void f(String s) {
    int y = s.length(), x = 0;
    char[][] m = new char[y][y];
    for(char[] q : m)
        java.util.Arrays.fill(q, ' ');
    y = 0;
    for(char c : s.toCharArray())
        if(c == '(')
            m[y++][x++] = c;
        else
            m[--y][x++] = c;
    for(char[] q : m)
        System.out.println(String.valueOf(q));
}
```
[Answer]
# C#, 195 bytes
First try at golf - yell if I did something wrong.
Alternative C# version using SetCursorPosition and working left to right taking the input as a commandline arg.
```
using System;class P{static void Main(string[] a){Action<int,int>p=Console.SetCursorPosition;int r=0,c=0;foreach(var x in a[0]){r+=x==')'?-1:0;p(c,r);Console.Write(x);r+=x=='('?1:0;p(c,r);c++;}}}
```
I thought it would be fun to adjust the write position based on the open/close paren and not full lines. Close paren moves the position up before writing; open paren moves it down after writing. Actioning SetCursorPosition saves five bytes. Moving the cursor to the next line after the output would take quite a bit extra.
```
using System;
class P
{
    static void Main(string[] a)
    {
        Action<int, int> p = Console.SetCursorPosition;
        int r = 0, c = 0;
        foreach (var x in a[0])
        {            
            r += x == ')' ? -1 : 0;
            p(c, r);
            Console.Write(x);
            r += x == '(' ? 1 : 0;
            p(c, r);
            c++;
        }
    }
}
```
[Answer]
# Batch, ~~356~~ 335 bytes
I know that there already exists a Batch solution for this challenge, but this one is golfed significantly more and seems to take a different approach. Most importantly, the other batch solution contains at least one powershell command; this solution does not.
```
@echo off
setlocal enabledelayedexpansion
set p=%1
set p=%p:(="(",%
set p=%p:)=")",%
set c=0
for %%a in (%p%)do (if ")"==%%a set/ac-=1
set d=!d!,!c!%%~a
if "("==%%a set/ac+=1&if !c! GTR !m! set m=!c!)
set/am-=1
for /l %%a in (0,1,!m!)do (for %%b in (!d!)do (set t=%%b
if "%%a"=="!t:~0,-1!" (cd|set/p=!t:~-1!)else (cd|set/p=. ))
echo.)
```
There is a backspace character (`U+0008`) on the second to last line following the dot (line 12, column 57). This isn't visible in the code posted here but is included in the byte count.
[Answer]
# Batch, 424 bytes
```
@echo off
setLocal enableDelayedExpansion
set s=%1
set a=1
:c
if defined s (set/ac+=1
set "z="
if "%s:~0,1%"=="(" (set "1=(")else (set/aa-=1
set "1=)")
for %%a in (!a!)do for /f usebackq %%b in (`powershell "'!l%%a!'".Length`)do (set/ay=!c!-%%b
for /l %%a in (1,1,!y!)do set z= !z!
set "l%%a=!l%%a!!z!!1!")
if "%s:~0,1%"=="(" set/aa+=1
if !a! GTR !l! set/al=!a!-1
set "s=%s:~1%"
goto c)
for /l %%a in (1,1,!l!)do echo !l%%a!
```
## Un-golfed:
```
@echo off
setLocal enableDelayedExpansion
set s=%1
set a=1
set c=0
set l=0
:c
if defined s (
    set /a c+=1
    set "z="
    if "%s:~0,1%"=="(" (
        set "1=("
    ) else (
        set /a a-=1
        set "1=)"
    )
    for %%a in (!a!) do for /f usebackq %%b in (`powershell "'!l%%a!'".Length`) do (
        set /a y=!c!-%%b
        for /l %%a in (1,1,!y!) do set z= !z!
        set "l%%a=!l%%a!!z!!1!"
    )
    if "%s:~0,1%"=="(" set /a a+=1
    if !a! GTR !l! set /a l=!a!-1
    set "s=%s:~1%"
    goto c
)
for /l %%a in (1,1,!l!) do echo !l%%a!
```
Example:
```
h:\>par.bat (((())())(()))
 (            )
  (      )(  )
   (  )()  ()
    ()
```
[Answer]
# C, 118 117 Bytes
Another answer in C, but mine is shorter.
```
c;d;main(m,v)int**v;{while(d++<m){char*p=v[1];while(*p)c+=*p==40,putchar(c-d?*p:32),m=c>m?c:m,c-=*p++==41;puts("");}}
```
Ungolfed version:
```
c; /* current depth */
d; /* depth to print in current row */
main(m,v)int**v;{
    while(d++<m) {
        char*p=v[1];
        while(*p){
            c+=*p==40;           /* 40 = '(' */
            putchar(c-d?*p:32); /* 32 = ' ' (space) */
            m=c>m?c:m;           /* search maximum depth */
            c-=*p++==41;         /* 41 = ')' */
        }
        puts("");
    }
}
```
And it works!
```
% ./telescope '()(())((()))(())()'
()(  )(    )(  )()
   ()  (  )  ()
        ()
% ./telescope '((()())()(()(())()))'
(                  )
 (    )()(        )
  ()()    ()(  )()
             ()
```
[Answer]
# [Haskell](https://www.haskell.org/), 107 bytes
```
h#('(':s)=h:(h+1)#s
h#(_:s)|k<-h-1=k:k#s
_#_=[]
g s=[do(c,i)<-zip s$0#s;max" "[c|i==h]|h<-[0..maximum$0#s]]
```
[Try it online!](https://tio.run/##FcpBCoMwEIXhfU8xGMEJbUS31jlJCBKsmCEmLY1CKd49TeH9m4/nbPLLtuXsBDbYDEmSG9BdeynSpdhU5PSjcqonP/iCk5hIm8sKifTjifON5ai@/IJUdyLdg/1UUOn5ZCJnTjcq3bVtUQ5H@D@MycFyBAKO@/K28w41HHHjuCRoYc2IKMukLP0A "Haskell – Try It Online")
Based on [d8d0d65b3f7cf42's Haskell answer](https://codegolf.stackexchange.com/a/49191/56433).
[Answer]
# Japt, 13 bytes
```
y_ùZ<')?°T:T´
```
[Try it](https://ethproductions.github.io/japt/?v=2.0a0&code=eV/5WjwnKT+wVDpUtA==&input=IigoKCkoKSkoKSgoKSgoKSkoKSkpIg==)
---
## Explanation
```
y_                :Transpose, pass each column Z through a function and transpose back
   Z<')           :  Is Z less than ")", i.e., is Z equal to "("?
       ?          :  If so
        °T        :   Prefix increment variable T (initially 0)
          :       :  Else
           T´     :   Postfix decrement T
  ù               :  Left pad Z with spaces to length T
```
]