Una multitud de miradas en blanco


36

Problema

Sin entrada, escriba un programa o una función que genere o devuelva la siguiente cadena:

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

Reglas

  • El programa más corto gana.
  • Espacio en blanco al final permitido.
  • Nuevas líneas finales permitidas.
  • Parámetros no utilizados para funciones permitidas.

55
Nota: esta cadena es la concatenación de subcadenas de (<>.<>):"(<" + "(<>" + "(<>." + ... + "(<>.<>)" + "<>.<>)" + ">.<>)" + ... + ">)"
JungHwan Min

2
Esto hubiera sido un poco menos "codificar todo el texto". si se tratara de una multitud más grande ...
Totalmente humano

@totallyhuman lo pensará la próxima vez
LiefdeWen

Respuestas:


13

SOGL V0.12 , 12 bytes

pΙ2○3V?hG]‘Γ

Explicación:

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

Pruébalo aquí!


Es bueno ver que tienes un intérprete en línea disponible ahora :)
Emigna

¿Cómo haces esto? ¡Buen trabajo!
abhiagNitk

18

05AB1E , 16 bytes

"(<>.".∞ηJÀ24£.∞

Pruébalo en línea!

Explicación

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


15

JS (Jsfuck), 10,614 bytes

No tenía acceso al sitio web, tuve que resolver todo esto manualmente.

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Explicación:

La mayoría de los caracteres son relativamente fáciles: para los corchetes que estoy ejecutando

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

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

El punto final se obtiene haciendo que js cree un número de notación científica como 1.1e + 21 Y tomando el carácter de punto después de convertirlo en una cadena

Los corchetes son más difíciles, tenemos que ejecutar la función para crear una cadena de objetos html en cursiva y robar los corchetes angulares. El truco principal en esto es obtener la 'c' para la palabra cursiva, que requiere construir otra función para tomar la c de 'función'

Aquí hay muchas posibilidades de mejora, principalmente en qué funciones se utilizan para obtener los corchetes. También recuerdo una forma más fácil de obtener el carácter 'c' si la cursiva es mejor, pero tendría que revisar mis archivos antiguos para encontrarlo

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3
Esto parece demasiado largo incluso para JSFuck pero +1 para el esfuerzo de la hoja de hacer esto manualmente.
F1Krazy

Llamo a BS al hacerlo manualmente. Tenías que haber hecho algo para automatizar la generación de eso.
Patrick Roberts

10660 bytes en realidad.
CalculatorFeline

66
@PatrickRoberts Apuesto a que lo descubrió manualmente y usó los poderes de ctrlc y v
bleh

1
@PatrickRoberts No es un programa súper difícil, solo se trata de entender cómo construir cada personaje. Como ejemplo básico, la forma de obtener un carácter 't' es hacer una matriz vacía '[]', convertir a un número para hacer el número 0: '+ []' y luego aplicarle el operador no para emitir a un verdadero booleano '! + []'. Luego se convierte en una cadena, agregando una matriz vacía nuevamente y poniéndola entre paréntesis para darnos la cadena "verdadero" '(! + [] + [])' Y finalmente tomar la letra cero con '(! + [] + []) [+ []] '
Rugnir

12

C # (Mono) , 54 52 bytes

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

Pruébalo en línea!

-2 bytes gracias a Kevin Cruijssen

RIP C #


Conozco el sentimiento (Java). Pero, puede guardar dos bytes. El final ;no se cuenta para las respuestas lambda. Y ()se puede reemplazar con un solo carácter ( nullobjeto no utilizado ). Meta publicación relevante para el segundo.
Kevin Cruijssen

1
@KevinCruijssen Sí, especifiqué los parámetros de función no utilizados permitidos, por lo que s=>está perfectamente bien.
LiefdeWen

-1: No se hizo ningún esfuerzo para mostrar la respuesta de ninguna manera.
MechMK1

@ MechMK1 ese es el punto, es literalmente lo más golfizado posible en C #. La salida literal de la cadena es la mejor manera de generar esta cadena en particular en este idioma en particular. Ver la respuesta C # de
Erlantz

@Skidsdev Todavía considero que va en contra del espíritu de Code Golf, aburrido de escribir y aburrido de leer. Quiero decir, mi voto no cuenta porque no participo activamente, así que no te preocupes por eso. Son solo mis 2 centavos, por así decirlo.
MechMK1


7

V , 28 27 25 22 bytes

3? 4? bytes gracias a @KritixiLithos

i(<>.<>)òÙxlHÄ$xGòxÍî

Pruébalo en línea!

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


Se puede reemplazar hxcon solo X.
Kritixi Lithos

Y dkHxògJes 1 byte más corto que dkVHgJ0x.
Kritixi Lithos

Sí DJ Íîque trabaja demasiado
nmjcman101

Me sale i(<>.<>)òÙxlHÄ$xGòddÍîpor 23
Kritixi Lithos

@KritixiLithos Literalmente SOLO cambié el lXpor xltambién: D
nmjcman101

7

R , 42 bytes

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

Programa bastante simple que aprovecha la forma en que funciona la subcadena en R. Entonces substring('(<>.<>)',-4:6,2:12)produce el siguiente vector

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

cat con un separador vacío lo envía a STDOUT en el formato requerido.

Pruébalo en línea!



5

TrumpScript, 70 bytes

Lo sé, es una solución aburrida.

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

Bien hecho, no obstante.
Adam

4

Brachylog , 14 bytes

"(<>.<>)"aᶠckb

Pruébalo en línea!

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

A partir de ahora, la cara completa parece duplicarse, pero puede solucionarlo simplemente cambiando el a .
Cadena no relacionada

3

C (gcc) , 59 57 bytes

2 4 bytes menos que una puts()solución simple . Seguramente habrá una solución recursiva elegante, pero hasta ahora la sobrecarga se vuelve demasiado grande en cada intento.

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

Pruébalo en línea!


3

Brainfuck, 198 175 167 Bytes

Nunca he hecho un codegolf, así que este es el primero. La retroalimentación es muy apreciada.

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

Pruébalo en línea!

Fui por la solución más obvia en mi opinión. Primero, configuro las celdas 1-5 a una de las letras "(). <>". Luego solo voy a las celdas correctas y saco el personaje.

ACTUALIZACIÓN: Cambié el orden en que aparecen los caracteres en la "cinta" haciendo que el programa sea más corto y eficiente.

ACTUALIZACIÓN 2: solo volví a visitar mi publicación después de un tiempo y me di cuenta de que, usando una secuencia más corta para configurar las celdas, podía guardar algunos bytes.


Bienvenido a PPCG! :)
Shaggy

La primera respuesta agradable, por lo general, las respuestas que no son evidentes de inmediato cómo funcionan necesitan una explicación, pero este desafío es bastante trivial, por lo que no es realmente necesario. Pero si haces un brainfuck de 1200 bytes en el futuro, solo una simple explicación es muy útil.
LiefdeWen

Me aseguraré de hacer eso en el futuro.
Polvo

Ok, agregué una descripción corta (y un poco mala) de mi código.
Polvo el

2

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>');

Utiliza el mismo método que la respuesta Python de Rod .



2

Carbón de leña , 20 bytes

A(<>ι(<ιι.ι.<ι.ιι.‖B

Pruébalo en línea!

Gracias a Destructible Lemon por notar un patrón (-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

Lo que ‖Bhace es básicamente palindromizar visualmente el lienzo.


Explicación por favor.
CalculatorFeline

@CalculatorFeline Se agregó AST tal como lo devolvió -a.
Erik the Outgolfer

Huh Eso fue más útil de lo que pensaba.
CalculatorFeline

Creo que esto sería más claro incluso si acaba de agregar que la impresión está implícita con los objetos
Destructible Lemon

@DestructibleLemon Pero el AST ya tiene el Prints allí, no estoy seguro de lo que quieres decir.
Erik the Outgolfer

2

C (gcc) , 61 bytes

Sé que esto es lamentable, pero es mucho más corto que mi otra solución ...

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

C (gcc) , 103 bytes

Aquí hay una versión en la que trato de ser un poco complicado ...

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 bytes

Una versión recursiva que es completamente complicada ...

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);}

Pruébalo en línea!


1
Por favor también comparta su solución no codificada. A pesar de ser más largo, estoy seguro de que es mucho más interesante que este trivial.
Laikoni

2

Una multitud de miradas en blanco observando a Batman en honor a 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;}

Pruébalo en línea!


2

q , 29 bytes

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

-2 bytes gracias a streetster

EDITAR para la explicación:

El lenguaje se interpreta de derecha a izquierda.

Cómo funciona

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

44
Una explicación aquí sería genial.
Gryphon - Restablece a Monica el

1
Se puede quitar el espacio antes de la _xde 1 byte, y el cambio til 11a (!)11por otros 1 byte = ahorro de 2 bytes salvado :)
streetster

@streetster great stuff :)
B.Wong

2

Haskell , 55 45 bytes

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

Pruébalo en línea!

¡La primera respuesta de Haskell para superar la solución codificada!

Explicación

La forma en que funciona esta respuesta es creando una lista de funciones que se aplicarán a la cadena (<>.<>). Primero construimos la izquierda y el centro con

map take[2..7]

lo que nos da todos los prefijos del tamaño dos al siete. Luego construimos el derecho con

map drop[1..5]

lo que nos da todos los sufijos del tamaño seis al dos.

Una vez que tenemos la lista de funciones, usamos un enlace monádico ( >>=) que es solo concatMappero más corto. La función con la que concatmap es la ($"(<>.<>)")que aplica la entrada a la cadena (<>.<>).

Esto hace que la cadena.



1

Java 8, 52 bytes

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

Aburrido, pero no hay ninguna forma de acortar esto en Java ... Solo inicializar una cadena temporal ya tiene 11 bytes ... ( String t="something";), y usar substringun par de veces ciertamente cuesta demasiados bytes ...

La alternativa más corta a un retorno literal es probablemente esto (58 bytes):

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

Pruébalo aquí


También funciona en JavaScript, utilizando una función de "flecha gruesa" ( =>) si desea agregarlo.
Shaggy

@Shaggy Nah, puedes mantener tu respuesta de JavaScript si quieres. Lo mismo se aplica a C # que también ya tiene una respuesta separada.
Kevin Cruijssen

1

Python 2, 57 bytes

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

Digo que parece una pandilla de búhos asomándose detrás de su líder. Solo digo.



1

cerebro , 245 210 bytes

Golf en progreso.

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

Pruébalo en línea!


1

Haskell, 95 bytes

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

Esto es mucho más largo que los 49 bytes necesarios para un literal de cadena con la salida, pero lo mejor que puedo hacer para utilizar la estructura . Como de costumbre, me encantan las flechas, y inits &&& tailsproduce una tupla de la lista de subcadenas iniciales y la lista de subcadenas finales de la entrada. Luego, esos dos elementos de tupla se unen en una lista pasando la tupla a ++, y esa lista se concatenvasa en una cadena grande. Las llamadas taily initevitan duplicar el (<>.<>)en el medio (uno generado por inits, el otro por tails) y descartar el paréntesis no deseado desde el principio y el final, teniendo en cuenta solo las subcadenas de longitud 2 o más.


¿Puedes por favor agregar una explicación?
CalculatorFeline

@CalculatorFeline Done
Bergi

1

Jalea , 18 bytes

“(<>.<>)”ḣJœ|ṫJ$ḊṖ

Pruébalo en línea!

Cómo funciona

“(<>.<>)”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., ")".

1

C (clang) , 62 61 bytes

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

Esto termina solo después de que se agota el tiempo de espera en TIO porque la escritura en la entrada estándar falla, pero lo hará terminan en un terminal. El programa se basa en un orden específico de evaluación (comportamiento indefinido) y no funciona con, por ejemplo, gcc.

¡Gracias a @Steadybox por una idea que salvó un byte!

Pruébalo en línea!

Verificación

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

Versión alternativa, 62 bytes.

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

A costa de un byte más, la solución se vuelve mucho más portátil.

Pruébalo en línea!


1
60 bytes como función , pero probablemente no sea confiable.
Steadybox


@ Dennis Sí, (<falta el allí.
Erik the Outgolfer


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