36
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Problem

Given no input write a program or a function that outputs or returns the following string:

(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)

Rules

  • Shortest program wins.
  • Trailing whitespace allowed.
  • Trailing newlines allowed.
  • Unused parameters for functions allowed.
\$\endgroup\$
  • 5
    \$\begingroup\$ Note: this string is the concatenation of substrings of (<>.<>): "(<" + "(<>" + "(<>." + ... + "(<>.<>)" + "<>.<>)" + ">.<>)" + ... + ">)" \$\endgroup\$ – JungHwan Min Jun 20 '17 at 11:38
  • 2
    \$\begingroup\$ This would've been a little less "hardcode all the text!" if it were a bigger crowd... \$\endgroup\$ – totallyhuman Jun 20 '17 at 21:56
  • \$\begingroup\$ @totallyhuman Will think about it next time \$\endgroup\$ – LiefdeWen Jun 21 '17 at 6:15

45 Answers 45

13
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SOGL V0.12, 12 bytes

pΙ2○3V?hG]‘Γ

Explanation:

pΙ2○3V?hG]‘   Push compressed string of the first half of the string (this is the kind of thing this compressor was made for)
           Γ  mirror, swap chars & palindromize with one character overlap

Try it Here!

\$\endgroup\$
  • \$\begingroup\$ Nice to see you have an online interpreter available now :) \$\endgroup\$ – Emigna Jun 20 '17 at 11:12
  • \$\begingroup\$ How do you do this? Great work though!! \$\endgroup\$ – abhiagNitk Jun 23 '17 at 9:26
18
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05AB1E, 16 bytes

"(<>.".∞ηJÀ24£.∞

Try it online!

Explanation

"(<>."             # push this string
                   # STACK: "(<>."
      .∞           # intersected mirror
                   # STACK: "(<>.<>)"
        η          # compute prefixes
         J         # join to string
                   # STACK: "((<(<>(<>.(<>.<(<>.<>(<>.<>)"
          À        # rotate left
           24£     # take the first 24 chars
                   # STACK: "(<(<>(<>.(<>.<(<>.<>(<>."
              .∞   # intersected mirror
                   # STACK: "(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
\$\endgroup\$
16
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Python 3, 49 bytes

for i in range(11):print(end='(<>.<>)'[i-12:i+2])

Try it online!

\$\endgroup\$
15
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JS (Jsfuck), 10,614 bytes

Didn't have access to the website, had to figure this all out manually -.-

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+(+[]+[])[([![]]+[][[]])[+!+[]+[+[]]]+(!![]+[])[+[]]+(![]+[])[+!+[]]+(![]+[])[+!+[]+!+[]]+([![]]+[][[]])[+!+[]+[+[]]]+([][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]]+[])[!+[]+!+[]+!+[]]+(![]+[])[!+[]+!+[]+!+[]]]()[+!+[]+!+[]]
+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[+!+[]+!+[]]+(![]+[])[+!+[]+!+[]]])[+!+[]+(+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+[])]

Explanation:

Most of the characters are relatively easy: For the brackets im executing

   ([]["fill"]+[])[13] // '('

   ([]["fill"]+[])[14] // ')'

The full stop is gotten by making js create a scientific notation number like 1.1e+21 And taking the dot character after converting to a string

The square brackets are harder, we have to execute the function to create an italic html object string and steal the angle brackets. The main trick in this is getting the 'c' for the word italics, which requires building up another function to take the c from 'function'

There's a lot of scope for improvement here, mostly on which functions are used to get the square brackets. I also recall an easier way to get the character 'c' if italics is better, but i'd have to look though my old files to find it

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+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[+!+[]+!+[]]+(![]+[])[+!+[]+!+[]]])[+!+[]+(+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+[])])

\$\endgroup\$
  • 3
    \$\begingroup\$ This looks overly-long even for JSFuck but +1 for the sheet effort of doing this manually. \$\endgroup\$ – F1Krazy Jun 20 '17 at 16:40
  • \$\begingroup\$ I call BS on doing it manually. You had to have done something to automate generating that. \$\endgroup\$ – Patrick Roberts Jun 20 '17 at 22:57
  • \$\begingroup\$ 10660 bytes actually. \$\endgroup\$ – CalculatorFeline Jun 21 '17 at 0:51
  • 6
    \$\begingroup\$ @PatrickRoberts I bet he figured it out manually and used the powers of ctrlc and v \$\endgroup\$ – bleh Jun 21 '17 at 2:21
  • 1
    \$\begingroup\$ @PatrickRoberts It's not a super difficult program, its just a case of understanding how to build up each character. As a basic example, the way to get a 't' character is to make an empty array '[]', cast to a number to make the number 0: '+[]' and then apply the not operator to it to cast to a boolean true '!+[]'. Then cast to a string, by adding an empty array again and putting it in brackets to give us the string "true" '(!+[]+[])' and finally take the zeroth letter with '(!+[]+[])[+[]]' \$\endgroup\$ – Rugnir Jun 21 '17 at 8:59
12
\$\begingroup\$

C# (Mono), 54 52 bytes

s=>"(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"

Try it online!

-2 bytes thanks to Kevin Cruijssen

RIP C#

\$\endgroup\$
  • \$\begingroup\$ I know the feeling (Java). But, you can save two bytes. The trailing ; isn't counted for lambda-answers. And () can be replaced with a single character (unused null object). Relevant meta post for the second. \$\endgroup\$ – Kevin Cruijssen Jun 20 '17 at 11:08
  • 1
    \$\begingroup\$ @KevinCruijssen Yes I did specify unused function parameters allowed so s=> is perfectly fine. \$\endgroup\$ – LiefdeWen Jun 20 '17 at 11:13
  • \$\begingroup\$ -1: There was no effort done to golf the answer in any way. \$\endgroup\$ – MechMK1 yesterday
  • \$\begingroup\$ @MechMK1 that's the point, it's quite literally as golfed as possible in C#. Literal output of the string is the tersest way to output this particular string in this particular language. See Erlantz's C# answer \$\endgroup\$ – Skidsdev 8 hours ago
  • \$\begingroup\$ @Skidsdev I still consider that it's against the spirit of Code Golf, boring to write and boring to read. I mean, my vote doesn't count because I don't actively participate, so don't worry about that. It's just my 2 cents, so to say. \$\endgroup\$ – MechMK1 8 hours ago
11
\$\begingroup\$

Retina, 30 bytes


(<>.<>)
5`.
$&$'
7`.
$`$&
^.

Try it online!

\$\endgroup\$
  • 1
    \$\begingroup\$ um what even is this \$\endgroup\$ – CalculatorFeline Jun 20 '17 at 18:52
7
\$\begingroup\$

V, 28 27 25 22 bytes

3? 4? bytes thanks to @KritixiLithos

i(<>.<>)òÙxlHÄ$xGòxÍî

Try it online!

i(<>.<>)                'insert (<>.<>)
         ò        ò      'recursively
          Ù              'duplicate this (bottom line) down
           x             'delete the first character
            l            'and break when there's only one character left
             H           'go to the top of the buffer
              Ä          'duplicate that line up
               $x        'delete the last character
                 G       'and go back to the bottom of the buffer
                   x     'delete the extra ) left by the loop
                    Íî   'and remove all newlines

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  • \$\begingroup\$ You can replace hx with just X. \$\endgroup\$ – Cows quack Jun 20 '17 at 11:31
  • \$\begingroup\$ And dkHxògJ is 1 byte shorter than dkVHgJ0x. \$\endgroup\$ – Cows quack Jun 20 '17 at 11:35
  • \$\begingroup\$ Yeah DJ's Íî thing works too \$\endgroup\$ – nmjcman101 Jun 20 '17 at 11:36
  • \$\begingroup\$ I get i(<>.<>)òÙxlHÄ$xGòddÍî for 23 \$\endgroup\$ – Cows quack Jun 20 '17 at 11:40
  • \$\begingroup\$ @KritixiLithos I literally JUST switched the lX for xl too :D \$\endgroup\$ – nmjcman101 Jun 20 '17 at 11:41
7
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R, 42 bytes

cat(substring('(<>.<>)',-4:6,2:12),sep='')

Fairly simple program that takes advantage of the way substring works in R. So substring('(<>.<>)',-4:6,2:12) produces the following vector

> substring('(<>.<>)',-4:6,2:12)
 [1] "(<"      "(<>"     "(<>."    "(<>.<"   "(<>.<>"  "(<>.<>)" "<>.<>)"  ">.<>)"  
 [9] ".<>)"    "<>)"     ">)"  

cat with an empty separator outputs it to STDOUT in the required format.

Try it online!

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7
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Charcoal, 18 17 bytes

F⁶…(<>.<>)⁺²ι‖BO⁷

Try it online! Link is to verbose version of code. Uses the new Slice operator.

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5
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TrumpScript, 70 Bytes

I know, it's a boring solution.

say "(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
America is great
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  • \$\begingroup\$ Well done, nonetheless. \$\endgroup\$ – Adam Sep 2 '18 at 5:17
4
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Brachylog, 14 bytes

"(<>.<>)"aᶠckb

Try it online!

"(<>.<>)"           #   string
         aᶠ         #       get all Affixes (pre- and suffixes)
           c        #       Concatenate
            kb      #       remove last (Knife) and first (Behead)
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  • \$\begingroup\$ As of now the full face seems to get duplicated, but you can fix it by just changing the into a . \$\endgroup\$ – Unrelated String 2 days ago
3
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C (gcc), 59 57 bytes

2 4 bytes less than a simple puts() solution. There is sure to be some elegant recursive solution, but so far the overhead becomes too large at every attempt.

f(i){for(i=1;i++<12;)printf("%.*s",i,"(<>.<>)"+i/7*i%7);}

Try it online!

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3
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Brainfuck, 198 175 167 Bytes

I have never done a codegolf, so this is my first one. Feedback is highly appreciated.

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

Try it online!

I went for the most obvious solution in my opinion. First, I set cells 1-5 to one of the letters "().<>". Then I just go to the right cells and output the character.

UPDATE: I changed the order in which the characters appear on the "tape" making the program way shorter and more efficient.

UPDATE 2: Just revisited my post after a while and realized, that using a shorter sequence for setting up the cells I could save some bytes.

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  • \$\begingroup\$ Welcome to PPCG! :) \$\endgroup\$ – Shaggy Jan 25 '18 at 9:33
  • \$\begingroup\$ Nice, first answer, usually answers that are not immediately apparent how they work need an explanation but this challenge is pretty trivial so not really needed. But if you do a 1200 bytes brainfuck in the future just a simple explanation goes a long way. \$\endgroup\$ – LiefdeWen Jan 25 '18 at 9:44
  • \$\begingroup\$ I will be sure to do that in the future. \$\endgroup\$ – Dust Jan 25 '18 at 9:45
  • \$\begingroup\$ Ok, I added a short (and kinda bad) description of my code. \$\endgroup\$ – Dust Jan 25 '18 at 9:51
2
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JavaScript, 47 bytes

g=(i=2)=>i>12?'':'(<>.<>)'.slice(i-14,i)+g(i+1)
f=(i=11)=>i?f(i-1)+'(<>.<>)'.slice(i-13,i+1):''

// Both of the above work and are the same size

document.write('<pre>Actual g(): ' + g() + '\nActual f(): ' + f() + '\nExpected:   (<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)</pre>');

Uses the same method as Rod's Python answer.

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2
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Mathematica 41 bytes

""<>"(<>.<>)"~StringDrop~i~Table~{i,-5,5}
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2
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Charcoal, 20 bytes

A(<>ι(<ιι.ι.<ι.ιι.‖B

Try it online!

Thanks to Destructible Lemon for noticing a pattern (-4).

AST:

Program
├A: Assign
│├'(<>': String '(<>'
│└ι: Identifier ι
├Print
│└'(<': String '(<'
├Print
│└ι: Identifier ι
├Print
│└ι: Identifier ι
├Print
│└'.': String '.'
├Print
│└ι: Identifier ι
├Print
│└'.<': String '.<'
├Print
│└ι: Identifier ι
├Print
│└'.': String '.'
├Print
│└ι: Identifier ι
├Print
│└ι: Identifier ι
├Print
│└'.': String '.'
└‖B: Reflect butterfly
 └Multidirectional

What ‖B does is basically visually palindromize canvas.

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  • \$\begingroup\$ Explanation please. \$\endgroup\$ – CalculatorFeline Jun 20 '17 at 18:56
  • \$\begingroup\$ @CalculatorFeline Added AST as returned by -a. \$\endgroup\$ – Erik the Outgolfer Jun 20 '17 at 18:59
  • \$\begingroup\$ Huh. That was more useful than I thought. \$\endgroup\$ – CalculatorFeline Jun 20 '17 at 19:05
  • \$\begingroup\$ I think this would be clearer even if you just added that print is implicit with objects \$\endgroup\$ – Destructible Lemon Jun 22 '17 at 2:32
  • \$\begingroup\$ @DestructibleLemon But the AST already has the Prints in there, not sure what you mean. \$\endgroup\$ – Erik the Outgolfer Jun 22 '17 at 8:56
2
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C (gcc), 61 bytes

I know this is lame but it's much shorter than my other solution...

f(){puts("(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)");}

C (gcc), 103 bytes

Here's a version where I'm trying to be somewhat tricky...

char*e="(<>.<>)<>.<>)>.<>).<>)<>)>)";f(i,n){for(n=2;n<8;n++)for(i=0;i<n;i++)putchar(e[i]);puts(&e[i]);}

C (gcc), 117 115 bytes

A recursive version that's fully tricky...

char*e="(<>.<>)";main(i,j){if(i<6){for(++i,j=0;j<i;putchar(e[j++]));main(i);}else if(i<12)printf(e+i++-6),main(i);}

Try it online!

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  • 1
    \$\begingroup\$ Please also share your none-hardcoded solution. Despite being longer I'm sure it's much more interesting than this trivial one. \$\endgroup\$ – Laikoni Jun 20 '17 at 17:45
2
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A crowd of blank stares watching Batman in honor of Adam West...

C# (.NET Core), 393 bytes

s=>{string b="NanananaNanaBATMAN!NanaBATMAN!NANANanaBATMAN!NANAnanaNanaBATMAN!NANABATMAN!NanaBATMAN!NANABATMAN!naNABATMAN!NANABATMAN!naNANAnaNANABATMAN!naNANANABATMAN!naNABATMAN!naNANAnanaNA",r="",t;for(int i=0;i<176;)if(b[i]=='B'){r+="<>";i+=7;}else{t=b.Substring(i,4);if(t=="Nana")r+="(";else if(t=="nana")r+="<";else if(t=="NAna")r+=">";else if(t=="NANA")r+=".";else r+=")";i+=4;}return r;}

Try it online!

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2
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q, 29 bytes

raze"(<>.<>)"{(y-5)_x}/:(!)11

-2 bytes thanks to streetster

EDIT for explanation:

Language is right to left interpreted.

How it works

raze"(<>.<>)"{(y-5)_x}/:(!)11
                        (!)11 /yields indices [0-10], right parameter of function (y)
                      /:      /each right: loop right parameter of a dyadic function
             {       }        /function
    "(<>.<>)"                 /left parameter of function (x)
              (y-5)           /subtract 5 from y (indices)
                   _          /remove first y chars from x, or last y chars if negative
raze                          /flatten string output to produce the final string
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  • 4
    \$\begingroup\$ An explanation here would be great. \$\endgroup\$ – Gryphon Jun 22 '17 at 10:41
  • 1
    \$\begingroup\$ You can remove the space before the _x for 1 byte, and change til 11 to (!)11 for another 1 byte saving = 2 bytes saved :) \$\endgroup\$ – streetster Jun 23 '17 at 0:00
  • \$\begingroup\$ @streetster great stuff :) \$\endgroup\$ – B.Wong Jun 23 '17 at 0:11
2
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Haskell, 55 45 bytes

map take[2..7]++map drop[1..5]>>=($"(<>.<>)")

Try it online!

First Haskell answer to beat the hardcoded solution!

Explanation

The way this answer works is by building a list of functions to be applied to the string (<>.<>). First we build the left and center with

map take[2..7]

which gives us all the prefixes from size two to seven. Then we build the right with

map drop[1..5]

which gives us all the suffixes from size six to two.

Once we have the list of functions we use a monadic bind (>>=) which is just concatMap but shorter. The function we concatmap with is ($"(<>.<>)") which applies the input to the string (<>.<>).

This makes the string.

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1
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Jelly, 22 bytes

1 byte thanks to Kritixi Lithos.

U;\UṚṖ
“(<>.<>)”;\ḊṖ,Ç

Try it online!

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  • \$\begingroup\$ I think you can drop the µ \$\endgroup\$ – Cows quack Jun 20 '17 at 10:44
1
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Java 8, 52 bytes

x->"(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"

Boring, but there isn't any way to make this shorter in Java.. Just initializing a temp String is already 11 bytes.. (String t="something";), and using substring a couple of times certainly costs too many bytes..

Shortest alternative to a literal return is probably this (58 bytes):

x->"(<(x(x.(x.<(x.x(x.x)x.x)>.x).x)x)>)".replace("x","<>")

Try it here.

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  • \$\begingroup\$ Also works in JavaScript, using a "fat arrow" function (=>) if you want to add it. \$\endgroup\$ – Shaggy Jun 20 '17 at 11:05
  • \$\begingroup\$ @Shaggy Nah, you can keep your JavaScript answer if you want. The same applies to C# which also already has a separate answer. \$\endgroup\$ – Kevin Cruijssen Jun 20 '17 at 11:10
1
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Python 2, 57 bytes

print''.join("(<>.<>)"[max(0,n-5):n+2]for n in range(11))

I say it looks like a gang of owls peeking out behind their leader. Just saying.

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1
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MATL, 28 bytes

'(<>.<>)'XH5:"H7@-:)wh2M@+)h

Try it online!

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1
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brainfuck, 245 210 bytes

Golfing in progress.

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

Try it online!

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1
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Haskell, 95 bytes

import Data.List
import Control.Arrow
concat$tail$init$uncurry(++)$inits&&&tail.tails$"(<>.<>)"

This is far longer than the 49 bytes necessary for a string literal with the output, but the best I could do to utilise the structure. As usual, I love arrows, and inits &&& tails does produce a tuple of the list of leading substrings and the list of trailing substrings of the input. Then those two tuple elements are put together in one list by passing the tuple to ++, and that list is concatenated to one large string. The tail and init calls avoid duplicating the (<>.<>) in the middle (one generated by inits, the other by tails) and drop the unwanted parenthesis from the start and end, taking only substrings of length 2 or more into account.

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  • \$\begingroup\$ Can you please add an explanation? \$\endgroup\$ – CalculatorFeline Jun 21 '17 at 17:31
  • \$\begingroup\$ @CalculatorFeline Done \$\endgroup\$ – Bergi Jun 21 '17 at 17:41
1
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Jelly, 18 bytes

“(<>.<>)”ḣJœ|ṫJ$ḊṖ

Try it online!

How it works

“(<>.<>)”ḣJœ|ṫJ$ḊṖ  Main link. No arguments.

“(<>.<>)”           Set the argument and return value to s := "(<>.<>)".
          J         Yield all indices of s, i.e., [1, 2, 3, 4, 5, 6, 7].
         ḣ          Dyadic head; yield s's prefixes of lengths 1 to 7.
               $    Combine the two links to the left into a chain.
              J         Indices; yield [1, 2, 3, 4, 5, 6, 7].
             ṫ          Dyadic tail; yield s's postfixes of lengths 1 to 7.
           œ|       Multiset union; concatenate the results to both sides,
                    discarding the copy of s. (s is both a prefix and a postfix.)
                Ḋ   Dequeue; remove the first prefix, i.e., "(".
                 Ṗ  Pop; remove the last postfix, i.e., ")".
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1
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C (clang), 62 61 bytes

main(n){while(write(++n<13,n/7*(n-7)+"(<>.<>)",n/7?14-n:n));}

This terminates only after timing out on TIO because the write to standard input fails, but it will terminate in a terminal. The program relies on a specific order of evaluation (undefined behavior) and won't work with, e.g., gcc.

Thanks to @Steadybox for an idea that saved a byte!

Try it online!

Verification

$ cat crowd.c
main(n){while(write(++n<12,n/7*(n-7)+"(<>.<>)",n/7?14-n:n));}
$ clang -o crowd crowd.c 2> /dev/null

$ ./crowd 0> /dev/null
(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)

Alternate version, 62 bytes

main(n){while(++n<13)write(1,n/7*(n-7)+"(<>.<>)",n/7?14-n:n);}

At the cost of one more byte, the solution becomes a lot more portable.

Try it online!

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1
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Python 2, 55 bytes

couldn't resist beating the current python 2 answer

print "(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"

Try it online!

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1
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Jelly, 19 bytes

“(<>.<>)”µṖ¹Ƥ;¹ÐƤṖḊ

Try it online!

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