Repetir repetições 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
fonte

Você sabe se existem maneiras melhores do que a compactação básica?

Erik the Outgolfer

@EriktheOutgolfer Definitivamente, pois isso foi feito como um acidente de 15 bytes: pA big part of the challenge is to find the shortest way to generate it.

dzaima

Sim, foi por isso que excluí minha resposta ... realmente não valia a pena. Ah, e o que você citou é por que eu perguntei.

Erik the Outgolfer

11

Parece ser uma sequência primitiva, por isso, infelizmente, não é uma repetição trivial de nada :(

HyperNeutrino

11

@ Sancises Bem, ainda há a possibilidade de recriar algo ainda mais golfista. ;)

Erik the Outgolfer

35

Japt , 35 33 29 26 25 24 bytes

'1
#˜5s £=r1,26¤²²sX
s#ž

Explicação

'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.

Aqui está como eu consegui isso. Comece com a sequência original:

01101011010110100110101101011010110101101011010011010110101101001101011010110101101011...

Depois de algumas brincadeiras, percebi que a sequência consistia inteiramente de cópias de 0e 11010. Substituir cada instância de 11010com 2deu isso:

02220222222022202222220222022222202220002220222222022202222220222022222202220002220222...

Oh ei, agora é totalmente feito de 0e 222. Vamos mudar cada 222um para um único 2:

02022020220202202000202202022020220200020220202202022020020220202202022020002022020220...

Essa sequência se parece muito com o original. Parece haver muitos 22020s nisso, então substituí cada um por 3e obtive o seguinte:

02033300203330020333020333002033300203330203330020333000203330203330020333002033302033...

Agora cada instância de 3está contida em um 020333. Substituir cada uma delas novamente por uma única 3fornece:

30303303033030033030330303303003303033030330303303033030330300330303303033030033030330...

... e voltamos ao padrão original novamente. Mais uma vez, substituindo 33030por 4:

30304404440444444044404444440444044444404440

Uh-oh, há alguns caracteres perdidos no começo. Podemos consertar isso adicionando um extra 3no início antes da condensação (é daí que vêm os 158 caracteres extras no final). Em seguida, condensamos cada série de 3 4s em uma única 4, como antes:

4040440404404044040

Mais uma vez, estamos de volta ao padrão original. Uma última vez, vamos condensar cópias de 44040para 5:

Espero que este passo final não seja muito difícil de entender. É como a etapa em que comprimimos os 020333s em 3s, exceto que desta vez estamos comprimindo 4040555s em 5s:

E aí você tem a string totalmente compactada. O código simplesmente inverte a compactação repetida para recuperar a string original. Como todas as iterações de compactação eram quase idênticas, não é preciso muito para reorganizar a string corretamente.

ETHproductions
fonte

Como você pode ver aqui, eu usei o Jelly's /(reduzir) ... o Japt também reduz / dobra? Nesse caso, isso pode salvar bytes.

Erik the Outgolfer

@EriktheOutgolfer Japt tem redução, mas isso exigiria mudar £=para pelo menos r@se não algo mais. (Honestamente, £=não é um preço alto a pagar por uma função de reduzir não convencional)

ETHproductions

Você quis dizer r@<space>? Claro, então seria algo como, em [[1], 1, 5, 2, 5]vez de [1, 5, 2, 5]usar reduzir.

Erik the Outgolfer

11

Seria correto dizer que você acabou de fazer engenharia reversa de uma expansão fractal?

Patrick Roberts

11

@PatrickRoberts Sim, basicamente.

ETHproductions

14

JavaScript (ES6), 130 bytes

Compressão de pulso não linear multicolor por amplificação paramétrica óptica consecutiva em estruturas pareadas de fase quase-fase.

^{OK, não exatamente. Apenas substituições.}

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

Demo

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>

Quão?

Definitivamente, essa não é a melhor maneira de compactar essa sequência, mas pensei em publicá-la de qualquer maneira, já que tive o trabalho de otimizá-la (ligeiramente) com uma função recursiva.

Começamos com a sequência s="pqh0f9bc9bc9bc900"e aplicamos sucessivamente as 20 substituições a seguir:

 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 exemplo, um "9" será expandido para "ca" , "cbd" , "c0cd" , "dd0ddd" , ... e, finalmente, uma cadeia de 2381 caracteres de 0 e 1.

Arnauld
fonte

Técnica interessante, tenho certeza de que funcionaria bem em strings com menos padrão do que isso.

ETHproductions

3

Você me perdeu no "multicolor"

Patrick Roberts

6

Mathematica, 128 bytes

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

J42161217
fonte

Você não poderia reduzir isso usando apenas um StringReplace e tendo uma lista de regras de substituição?

DanTheMan

sim, claro ... Eu apenas comecei a jogar isso ...

J42161217

Se a substituição é que fácil, então Retina deve fazer bem ...

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)

Um tanto quanto um porto da minha resposta japonesa. Uma explicação do padrão pode ser encontrada por lá.

Snippet de teste

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);

ETHproductions
fonte

2

É um pouco confuso ... até você descobrir j="1"duas coisas, uma delas sendo abusada.

Erik the Outgolfer

6

Geléia , 28 22 20 bytes

⁽.⁷Dð26Bẋ4ṫȧ€Ẏð/ṫ159

-3 graças a Jonathan Allan .

(Um pouco) porta de ETHproductions excelente Japt 's resposta .

Ainda não é eficiente como o dzaima (versão SOGL antiga)!

Isso retorna uma matriz de dígitos.

Erik, o Outgolfer
fonte

Isso foi verificado quanto à validade usando o Diff Checker !

Erik the Outgolfer

Vamos continuar esta discussão no chat .

Erik the Outgolfer

6

Retina, 69 66 65 bytes

Porto da resposta JavaScript (ES6) da @ ETHproduction: Experimente online! NB Um pouco lento.


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

^.{158}|¶

Uma porta da resposta Mathematica de @ Jenny_mathy era 82 66 bytes: Experimente online!


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

Uma porta da explicação do Japt da @ ETHproduction era de 69 bytes: Experimente online!

Uma porta da explicação japonesa anterior da @ ETHproduction também tinha 69 bytes: Experimente online!

Uma porta da resposta JavaScript de @ RickHitchcock era 74 bytes: Experimente online!


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

Uma porta da resposta JavaScript de @ Arnauld era de 85 bytes: Experimente online!


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

Neil
fonte

Agradável! você pode atualizar o meu? (esta é minha versão final) thanx!

J42161217

@Jenny_mathy Consegui jogar mais alguns bytes, então agora você é a principal porta Retina, obrigado!

Neil

3

JavaScript (ES6), 158 143 139 bytes

(Salvou alguns bytes graças 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
fonte

Você só precisa gerar a string, para poder soltar o _=>.

21417 DanTheMan

2

Isso pode ser um pouco menor (pelo menos 30 bytes). Dica # 1: mude de.replace(/x/g,y) para.split(x).join(y)

ETHproductions

Dica 2: use dígitos> 1no lugar das letras na compactação. Isso permitirá que você substitua facilmente 9, por exemplo , então 8, então 7etc.

ETHproductions

@ETHproductions, vejo como posso salvar 1 byte alternando para splite joincom as strings do modelo: 'ifififhg0'[s='split']BTiBT[j='join']BThhgBT[s]BThBT[j]BTg0fBT (onde a BT é crase)

Rick Hitchcock

Dica # 3: mapeie através de uma matriz de substituições, por exemploj='string',i=9,[replacements].map(x=>j=j.split(i--).join(x)),j

ETHproductions

2

05AB1E , 20 bytes

1Ž5ûv1₂b4×y.$.:]Ƶv.$

Porto da resposta JAPT da ETHProductions .

Grimmy
fonte

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]