Repetir repeticiones repetitivas

function check() {
  if (chr0.value == "") chr0.value = "0"
  if (chr1.value == "") chr1.value = "1"
  inp.value = inp.value.replace(new RegExp("[^"+JSON.stringify(chr0.value+chr1.value)+"]", "g"), "")
  if (inp.value == correct)
    output.innerHTML = "Matches"
  else
    output.innerHTML = "Doesn't match"
}
correct = "0110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000011010110101101001101011010110101101011010110100110101101011010011010110101101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1011010110101101001101011010110100110101101011010110101101011010011010110101101000011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010000"

<input id="inp" placeholder="input your result here"/><br>
<input id="chr0" placeholder="character for 0 (def=0)"/>
<input id="chr1" placeholder="character for 1 (def=1)"/><br>
<button id="chk" onclick="check()">check</button>
<div id=output></div>

code-golf kolmogorov-complexity dzaima
fuente

¿Sabes si hay mejores formas que la compresión base?

Erik the Outgolfer

@EriktheOutgolfer Definitivamente, ya que esto se hizo como un accidente de 15 bytes: pA big part of the challenge is to find the shortest way to generate it.

dzaima

Sí, esa es la razón por la que borré mi respuesta ... sentí que realmente no valía la pena. Ah, y lo que citó es por qué pregunté.

Erik the Outgolfer

1

Esto parece ser una secuencia primitiva, por lo que desafortunadamente no es una repetición trivial de nada :(

HyperNeutrino

1

@Sanchises Bueno, todavía existe la posibilidad de recrear algo aún más golfista. ;)

Erik the Outgolfer

35

Japt , 35 33 29 26 25 24 bytes

'1
#˜5s £=r1,26¤²²sX
s#ž

Explicación

'1   First line: Set variable U to this value.
'1   Yield the string "1". (This part gets simpler every time I golf this answer...)

1525s £=r1,26¤²²sX   Second line: Set variable V to this value (though that's not important).
1525s                Yield the string "1525".
      £              For each char X in this string:
           26¤         Yield the binary representation of 26. "11010"
              ²²       Double it twice.                       "11010110101101011010"
                sX     Slice off the first X chars. This gives "1010110101101011010",
                       "110101101011010", "010110101101011010", and "110101101011010".
       =               Set U to
        r1,              U with each instance of 1 replaced with the string calculated above.

s158   Last line: output the result of this line.
s158   Take U and slice off the first 158 characters.

Así es como obtuve esto. Comience con la secuencia original:

01101011010110100110101101011010110101101011010011010110101101001101011010110101101011...

Después de jugar un poco, me di cuenta de que la secuencia consistía completamente en copias de 0y 11010. Reemplazar cada instancia de 11010con 2dio esto:

02220222222022202222220222022222202220002220222222022202222220222022222202220002220222...

Oh, oye, ahora está completamente hecho de 0y 222. Cambiemos cada uno 222a un solo 2:

02022020220202202000202202022020220200020220202202022020020220202202022020002022020220...

Esta secuencia se parece muchísimo a la original. Parece que hay muchos 22020s en esto, así que reemplacé cada uno 3y obtuve esto:

02033300203330020333020333002033300203330203330020333000203330203330020333002033302033...

Ahora cada instancia de 3está contenida en a 020333. Reemplazar cada uno de estos nuevamente con un solo 3da esto:

30303303033030033030330303303003303033030330303303033030330300330303303033030033030330...

... y volvemos al patrón original nuevamente. Una vez más, reemplazando 33030con 4:

30304404440444444044404444440444044444404440

Uh-oh, hay algunos caracteres extraviados al principio. Podemos arreglar esto agregando un extra 3al principio antes de condensar (de aquí es de donde provienen los 158 caracteres extra al final). Luego condensamos cada serie de 3 4s en una sola 4, como antes:

4040440404404044040

Aaaa y una vez más, volvemos al patrón original. Una última vez, condensemos copias de 44040a 5:

Esperemos que este paso final no sea demasiado difícil de entender. Es como el paso donde comprimimos la 020333s en 3s, excepto que esta vez, estamos comprimiendo 4040555s en 5s:

Y ahí tienes la cadena completamente comprimida. El código simplemente invierte la compresión repetida para recuperar la cadena original. Como todas las iteraciones de compresión fueron casi idénticas, no se necesita mucho para reorganizar la cadena correctamente.

ETHproducciones
fuente

Como puede ver aquí , he usado Jelly's /(reduce) ... ¿Japt también tiene reduce / fold? Si es así, eso podría ahorrar bytes.

Erik the Outgolfer

@EriktheOutgolfer Japt tiene una reducción, pero eso requeriría cambiar £=al menos a r@algo más. (Honestamente, £=no es un precio alto a pagar por una función de reducción no convencional)

ETHproductions

Querías decir r@<space>? Por supuesto, entonces sería algo así como en [[1], 1, 5, 2, 5]lugar de [1, 5, 2, 5]usar reducir.

Erik the Outgolfer

1

¿Sería exacto decir que simplemente hiciste ingeniería inversa para una expansión fractal?

Patrick Roberts

1

@PatrickRoberts Sí, más o menos.

ETHproductions

14

JavaScript (ES6), 130 bytes

Compresión de pulso no lineal multicolor por amplificación paramétrica óptica consecutiva en estructuras coincidentes cuasi-fase.

^{OK, no del todo. Solo sustituciones.}

f=(n=0,s='pqh0f9bc9bc9bc900')=>n<20?f(n+1,s.split((n+9).toString(36)).join('cb0dgfgnjkkoppo0qs10'[n]+'adcdef0hi0klmnqpqrs1'[n])):s

Manifestación

f=(n=0,s='pqh0f9bc9bc9bc900')=>n<20?f(n+1,s.split((n+9).toString(36)).join('cb0dgfgnjkkoppo0qs10'[n]+'adcdef0hi0klmnqpqrs1'[n])):s

o.innerHTML = f()

<pre id=o></pre>

¿Cómo?

Definitivamente, esta no es la mejor manera de comprimir esta cadena, pero pensé en publicarla de todos modos ya que me tomé la molestia de optimizarla (ligeramente) con una función recursiva.

Comenzamos con la cadena s="pqh0f9bc9bc9bc900"y aplicamos sucesivamente las siguientes 20 sustituciones:

 9  | a  | b  | c  | d  | e  | f  | g  | h  | i  | j  | k  | l  | m  | n  | o  | p  | q  | r  | s
----+----+----+----+----+----+----+----+----+----+----+----+----+----+----+----+----+----+----+----
 ca | bd | 0c | dd | ge | ff | g0 | nh | ji | k0 | kk | ol | pm | pn | oq | 0p | qq | sr | 1s | 01

Por ejemplo, un "9" se expandirá a "ca" , "cbd" , "c0cd" , "dd0ddd" , ..., y finalmente una cadena de 2381 caracteres de 0 y 1.

Arnauld
fuente

Técnica interesante, estoy seguro de que funcionaría bien en cadenas con menos patrones que este.

ETHproductions

3

Me perdiste en "multicolor"

Patrick Roberts

6

Mathematica, 128 bytes

Fold[StringReplace,"AAAY0PY0",{{"A"->"Y0PY0PYP"},{"Y"->"P0P0PP0P0PP0P0"},{"P"->"QQQXXX00"},{"Q"->"XXX0XXXXXX0"},{"X"->"01101"}}]

J42161217
fuente

¿No podría acortar esto usando solo un StringReplace y tener una lista de reglas de reemplazo?

DanTheMan

sí, por supuesto ... Acabo de empezar a jugar al golf esta abajo ...

J42161217

Si el reemplazo es así de fácil, entonces Retina debería funcionar bien ...

Neil

6

JavaScript (ES6), 85 80 bytes

[j="1",5,2,5].map(x=>j=j.replace(/1/g,"11010".repeat(4).slice(x)))&&j.slice(158)

Un poco de un puerto de mi respuesta Japt. Una explicación del patrón se puede encontrar allí.

Fragmento de prueba

var q = [j="1",5,2,5].map(x=>j=j.replace(/1/g,"11010".repeat(4).slice(x)))&&j.slice(158);

var o = "0110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000011010110101101001101011010110101101011010110100110101101011010011010110101101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1011010110101101001101011010110100110101101011010110101101011010011010110101101000011010110101101001101011010110101101011010110100110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010000110101101011010011010110101101011010110101101001101011010110100110101101011010110101101011010011010110101101001101011010110101101011010110100110101101011010000";

console.log(q === o);

ETHproducciones
fuente

2

Es un poco confuso ... hasta que descubres que en j="1"realidad hay dos cosas, una de ellas es de abuso de tipo.

Erik the Outgolfer

6

Jalea , 28 22 20 bytes

⁽.⁷Dð26Bẋ4ṫȧ€Ẏð/ṫ159

-3 gracias a Jonathan Allan .

Puerto (un poco) de ETHproductions excelente Japt 's respuesta .

¡Todavía no es tan eficiente como el de dzaima (antigua versión de SOGL)!

Esto devuelve una matriz de dígitos.

Erik el Outgolfer
fuente

¡Se ha verificado su validez con Diff Checker !

Erik the Outgolfer

Continuemos esta discusión en el chat .

Erik the Outgolfer

6

Retina, 69 66 65 bytes

Respuesta de JavaScript (ES6) del puerto de @ ETHproduction: ¡ Pruébelo en línea! Nota: algo lento.


1¶1010::010::
{`1(?=.*¶(\d*):)
$1ZZZ
Z
11010
}1`\d*:

^.{158}|¶

Un puerto de la respuesta de Mathematica de @ Jenny_mathy fue 82 66 bytes: ¡ Pruébelo en línea!


AAAY0PY0
A
Y0PY0PYP
Y
ZPZPZ
Z
P0P0
P
QQQR00
Q
R0RR0
R
XXX
X
01101

Un puerto de la explicación de Japt de @ ETHproduction fue de 69 bytes: ¡ Pruébelo en línea!

Un puerto de la explicación Japt anterior de @ ETHproduction también tenía 69 bytes: ¡ Pruébelo en línea!

Un puerto de la respuesta JavaScript de @ RickHitchcock fue de 74 bytes: ¡ Pruébelo en línea!


ifififhg0
i
hhg
h
g0f
g
f0f0ff0f0ff0f0
f
e0e0e0ddd00
e
ddd0dddddd
d
01101

Un puerto de la respuesta JavaScript de @ Arnauld fue de 85 bytes: ¡ Pruébelo en línea!


pqh0g09bc9bc9bc900
9
cbd
b
0c
c
dd
d
gg0g0
g
nh
h
kkk0
k
0pppn
n
0pq
p
qq
q
s1s
s
01

Neil
fuente

¡Agradable! ¿Puedes actualizar el mío? (Esta es mi versión final) ¡Gracias!

J42161217

@ Jenny_mathy Pude jugar golf más bytes así que ahora eres el puerto líder de Retina, ¡gracias!

Neil

3

JavaScript (ES6), 158 143 139 bytes

(Guardado algunos bytes gracias a @DanTheMan).

'ifififhg0'[r='replace'](/i/g,'hhg')[r](/h/g,'g0f')[r](/g/g,'f0f0ff0f0ff0f0')[r](/f/g,'e0e0e0ddd00')[r](/e/g,'ddd0dddddd')[r](/d/g,'01101')

console.log('ifififhg0'[r='replace'](/i/g,'hhg')[r](/h/g,'g0f')[r](/g/g,'f0f0ff0f0ff0f0')[r](/f/g,'e0e0e0ddd00')[r](/e/g,'ddd0dddddd')[r](/d/g,'01101'));

Rick Hitchcock
fuente

Solo necesita generar la cadena, para poder soltar el _=>.

DanTheMan

2

Esto podría ser bastante más corto (al menos 30 bytes). Sugerencia n. ° 1: cambie de .replace(/x/g,y)a.split(x).join(y)

ETHproductions

Consejo # 2: Use dígitos> 1en lugar de las letras en la compresión. Esto le permitirá reemplazar fácilmente 9, por ejemplo , entonces 8, entonces 7, etc.

ETHproductions

@ETHproductions, veo cómo puedo ahorrar 1 byte al cambiar splity joincon cadenas de plantilla: 'ifififhg0'[s='split']BTiBT[j='join']BThhgBT[s]BThBT[j]BTg0fBT(donde BT está en retroceso)

Rick Hitchcock

Sugerencia n. ° 3: mapa a través de una serie de reemplazos, pj='string',i=9,[replacements].map(x=>j=j.split(i--).join(x)),j

ETHproductions

2

05AB1E , 20 bytes

1Ž5ûv1₂b4×y.$.:]Ƶv.$

Respuesta JAPT del puerto de ETHProductions .

Mugriento
fuente

1

K (ngn / k) , 86 bytes

@[&15558;+\(,/(,/(,/(1 5,,/(1_,/(11#5 6 6),\:5 5),\:4 5),\:9#3 2 3),\:2 2),\:1 2);:;1]