Problem

Przy braku danych wejściowych napisz program lub funkcję, która wyświetla lub zwraca następujący ciąg:

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

Zasady

• Najkrótszy program wygrywa.
• Dozwolone końcowe znaki odstępu.
• Końcowe znaki nowej linii są dozwolone.
• Nieużywane parametry funkcji są dozwolone.

SOGL V0.12 , 12 bajtów

pΙ2○3V?hG]‘Γ

Wyjaśnienie:

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

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05AB1E , 16 bajtów

"(<>.".∞ηJÀ24£.∞

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Wyjaśnienie

"(<>."             # 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: "(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"

Python 3 , 49 bajtów

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

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JS (Jsfuck), 10 614 bajtów

Nie miałem dostępu do strony, musiałem to wszystko rozwiązać ręcznie. -

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Wyjaśnienie:

Większość znaków jest stosunkowo łatwa: w nawiasach wykonuję

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

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

Kropkę uzyskuje się, gdy js tworzy numer notacji naukowej, taki jak 1.1e + 21, i przyjmuje znak kropki po konwersji na ciąg

Nawiasy kwadratowe są trudniejsze, musimy wykonać funkcję, aby utworzyć kursywą ciąg znaków html i ukraść nawiasy kątowe. Główną sztuczką w tym przypadku jest uzyskanie litery „c” dla słowa kursywą, co wymaga zbudowania innej funkcji, aby pobrać c z „funkcji”

Istnieje wiele możliwości ulepszeń, głównie tych, które służą do uzyskania nawiasów kwadratowych. Pamiętam też łatwiejszy sposób na uzyskanie znaku „c”, jeśli kursywa jest lepsza, ale musiałbym przejrzeć moje stare pliki, aby je znaleźć

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C # (mono) , 54 52 bajtów

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

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-2 bajty dzięki Kevin Cruijssen

RIP C #

Siatkówka , 30 bajtów

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

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V , 28 27 25 22 bajtów

3? 4? bajty dzięki @KritixiLithos

i(<>.<>)òÙxlHÄ$xGòxÍî

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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

R , 42 bajty

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

Dość prosty program, który wykorzystuje sposób działania podciągów w R. Tak więc substring('(<>.<>)',-4:6,2:12)produkuje następujący wektor

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

cat z pustym separatorem wysyła go do STDOUT w wymaganym formacie.

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Węgiel drzewny, 18 17 bajtów

Ｆ⁶…(<>.<>)⁺²ι‖ＢＯ⁷

Wypróbuj online! Link jest do pełnej wersji kodu. Korzysta z nowego operatora plastra.

TrumpScript, 70 bajtów

Wiem, że to nudne rozwiązanie.

say "(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
America is great

Brachylog , 14 bajtów

"(<>.<>)"aᶠckb

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"(<>.<>)"           #   string
         aᶠ         #       get all Affixes (pre- and suffixes)
           c        #       Concatenate
            kb      #       remove last (Knife) and first (Behead)

C (gcc) , 59 57 bajtów

2 4 bajty mniej niż proste puts()rozwiązanie. Na pewno będzie jakieś eleganckie rozwiązanie rekurencyjne, ale do tej pory koszty ogólne stają się zbyt duże przy każdej próbie.

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

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Brainfuck, 198 175 167 bajtów

Nigdy nie robiłem codegolfa, więc to mój pierwszy. Informacje zwrotne są bardzo mile widziane.

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

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Moim zdaniem wybrałem najbardziej oczywiste rozwiązanie. Najpierw ustawiam komórki 1-5 na jedną z liter „(). <>”. Następnie przechodzę do właściwych komórek i wypisuję postać.

AKTUALIZACJA: Zmieniłem kolejność, w jakiej znaki pojawiają się na „taśmie”, dzięki czemu program jest krótszy i wydajniejszy.

AKTUALIZACJA 2: Właśnie wróciłem do mojego posta po chwili i zdałem sobie sprawę, że używając krótszej sekwencji do konfigurowania komórek, mógłbym zaoszczędzić trochę bajtów.

JavaScript, 47 bajtów

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>');

Używa tej samej metody, co odpowiedź Rodona w Pythonie .

Mathematica 41 bajtów

""<>"(<>.<>)"~StringDrop~i~Table~{i,-5,5}

Węgiel drzewny , 20 bajtów

Ａ(<>ι(<ιι.ι.<ι.ιι.‖Ｂ

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Dzięki Destructible Lemon za zauważenie wzoru (-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

To, ‖Ｂco właściwie robi, to wizualnie palindromize płótna.

C (gcc) , 61 bajtów

Wiem, że to kiepskie, ale znacznie krótsze niż moje inne rozwiązanie ...

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

C (gcc) , 103 bajty

Oto wersja, w której staram się być nieco podstępny ...

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 bajtów

Wersja rekurencyjna, która jest w pełni podstępna ...

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);}

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Tłum pustych spojrzeń obserwujących Batmana na cześć Adama Westa ...

C # (.NET Core) , 393 bajty

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;}

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q , 29 bajtów

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

-2 bajty dzięki streetsterowi

EDYCJA dla wyjaśnienia:

Język jest interpretowany od prawej do lewej.

Jak to działa

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

Haskell , 55 45 bajtów

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

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Pierwsza odpowiedź firmy Haskell, która pokonała zakodowane rozwiązanie!

Wyjaśnienie

Ta odpowiedź działa w ten sposób, że tworzy listę funkcji, które zostaną zastosowane do łańcucha (<>.<>). Najpierw budujemy lewy i środkowy z

map take[2..7]

co daje nam wszystkie prefiksy od rozmiaru drugiego do siódmego. Następnie budujemy z

map drop[1..5]

co daje nam wszystkie przyrostki od rozmiaru szóstego do drugiego.

Kiedy mamy już listę funkcji, używamy monadic bind ( >>=), który jest tylko, concatMapale krótszy. Funkcja, z którą konkatmapujemy, ($"(<>.<>)")stosuje dane wejściowe do łańcucha (<>.<>).

To sprawia, że ​​ciąg.

Galaretka , 22 bajty

1 bajt dzięki Kritixi Lithos.

U;\UṚṖ
“(<>.<>)”;\ḊṖ,Ç

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Java 8, 52 bajty

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

Nudne, ale nie ma sposobu, aby skrócić ten czas w Javie. Tylko zainicjowanie temp. String ma już 11 bajtów .. ( String t="something";), a użycie substringkilku razy z pewnością kosztuje zbyt wiele bajtów ..

Najkrótsza alternatywa dla dosłownego zwrotu jest prawdopodobnie następująca (58 bajtów):

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

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Python 2, 57 bajtów

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

Mówię, że wygląda to na bandę sów wystających za przywódcę. Tylko mówię.

MATL , 28 bajtów

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

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pieprzenie mózgu , 245 210 bajtów

Gra w golfa w toku.

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

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Haskell, 95 bajtów

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

Jest to znacznie dłużej niż 49 bajtów niezbędnych do literału łańcuchowego z wyjściem, ale najlepiej, co mogłem zrobić, aby wykorzystać strukturę . Jak zwykle uwielbiam strzałki i inits &&& tailstworzy krotkę listy wiodących podłańcuchów i listy końcowych podciągów wejściowych. Następnie te dwa elementy krotkowe są łączone w jedną listę, przekazując krotkę do ++, i ta lista jest concatumieszczana w jednym dużym ciągu. tailI initrozmowy uniknąć duplikowania (<>.<>)w środku (jeden generowanego przez inits, drugi przez tails) i upuść niechcianego nawias od początku i na końcu, biorąc tylko podciągi o długości 2 lub więcej pod uwagę.

Galaretka , 18 bajtów

“(<>.<>)”ḣJœ|ṫJ$ḊṖ

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Jak to działa

“(<>.<>)”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., ")".

C (clang) , 62 61 bajtów

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

Ten kończy się dopiero po rozrządu na TIO ponieważ napisz do standardowego wejścia nie powiedzie się, ale to będzie rozwiązać w terminalu. Program opiera się na określonej kolejności oceny (niezdefiniowane zachowanie) i nie będzie współpracował np. Z gcc.

Dzięki @Steadybox za pomysł, który oszczędził bajt!

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Weryfikacja

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

Alternatywna wersja, 62 bajty

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

Kosztem jeszcze jednego bajtu rozwiązanie staje się znacznie bardziej przenośne.

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Python 2 , 55 bajtów

nie mogłem się oprzeć pokonaniu aktualnej odpowiedzi w python 2

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

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Galaretka , 19 bajtów

“(<>.<>)”µṖ¹Ƥ;¹ÐƤṖḊ

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