Répéter les répétitions répétitives

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
la source

Savez-vous s'il existe de meilleurs moyens que la compression de base?

Erik the Outgolfer le

@EriktheOutgolfer Certainement, car cela a été fait comme un accident de 15 octets: pA big part of the challenge is to find the shortest way to generate it.

dzaima

Oui, c'est la raison pour laquelle j'ai supprimé ma réponse ... je ne me sentais pas vraiment la peine. Oh, et ce que vous avez cité, c'est pourquoi j'ai demandé.

Erik the Outgolfer

1

Cela semble être une séquence primitive, donc ce n'est malheureusement pas une répétition triviale de quoi que ce soit :(

HyperNeutrino

1

@Sanchises Eh bien, il y a encore la possibilité de recréer quelque chose de plus golfeur. ;)

Erik the Outgolfer

35

Japt , 35 33 29 26 25 24 octets

'1
#˜5s £=r1,26¤²²sX
s#ž

Explication

'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.

Voici comment j'ai obtenu ça. Commencez avec la séquence d'origine:

01101011010110100110101101011010110101101011010011010110101101001101011010110101101011...

Après quelques déconner, j'ai réalisé que la séquence se composait entièrement de copies de 0et 11010. Remplacer chaque instance de 11010par a 2donné ceci:

02220222222022202222220222022222202220002220222222022202222220222022222202220002220222...

Oh hé, maintenant c'est entièrement fait de 0et 222. Changeons chacun 222en un seul 2:

02022020220202202000202202022020220200020220202202022020020220202202022020002022020220...

Cette séquence ressemble énormément à l'original. Il semble y avoir beaucoup de 22020s, donc j'ai remplacé chacun par 3et j'ai obtenu ceci:

02033300203330020333020333002033300203330203330020333000203330203330020333002033302033...

Maintenant, chaque instance de 3est contenue dans a 020333. Remplacer chacun de ceux-ci à nouveau par un seul 3donne ceci:

30303303033030033030330303303003303033030330303303033030330300330303303033030033030330...

... et nous revenons au modèle d'origine. Encore une fois, en remplaçant 33030par 4:

30304404440444444044404444440444044444404440

Uh-oh, il y a des caractères errants au début. Nous pouvons résoudre ce problème en ajoutant un supplément3 au début avant de condenser (c'est de là que viennent les 158 caractères supplémentaires à la fin). Ensuite, nous condensons chaque cycle de 3 4s en un seul 4, comme précédemment:

4040440404404044040

Aaaaand encore une fois, nous revenons au modèle d'origine. Une dernière fois, condensons des copies de44040 à 5:

Espérons que cette dernière étape ne soit pas trop difficile à comprendre. C'est comme l'étape où nous avons compressé le 020333s en 3s, sauf que cette fois, nous compressons4040555 s en 5s:

Et là, vous avez la chaîne entièrement compressée. Le code inverse simplement la compression répétée pour récupérer la chaîne d'origine. Étant donné que toutes les itérations de compression étaient presque identiques, il ne faut pas grand-chose pour réorganiser correctement la chaîne.

ETHproductions
la source

Comme vous pouvez le voir ici, j'ai utilisé Jelly /(réduire) ... Japt a-t-il également réduit / plié? Si c'est le cas, cela pourrait économiser des octets.

Erik the Outgolfer

@EriktheOutgolfer Japt a une réduction, mais cela nécessiterait de passer £=à au moins r@sinon quelque chose de plus long. (Honnêtement, ce £=n'est pas un prix élevé à payer pour une fonction de réduction non conventionnelle)

ETHproductions

Tu voulais dire r@<space>? Bien sûr, ce serait quelque chose comme [[1], 1, 5, 2, 5]au lieu d' [1, 5, 2, 5]utiliser réduire.

Erik the Outgolfer

1

Serait-il exact de dire que vous venez de procéder à une ingénierie inverse d'une expansion fractale?

Patrick Roberts

1

@PatrickRoberts Oui, à peu près.

ETHproductions

14

JavaScript (ES6), 130 octets

Compression d'impulsions non linéaires multicolores par amplification optique paramétrique consécutive dans des structures adaptées en quasi-phase.

^{OK, pas tout à fait. Juste des substitutions.}

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

Démo

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>

Comment?

Ce n'est certainement pas la meilleure façon de compresser cette chaîne, mais je pensais que je la publierais quand même car je me suis donné la peine de l'optimiser (légèrement) avec une fonction récursive.

Nous commençons par la chaîne s="pqh0f9bc9bc9bc900"et appliquons successivement les 20 substitutions suivantes:

 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

Par exemple, un "9" se développera en "ca" , "cbd" , "c0cd" , "dd0ddd" , ..., et finalement une chaîne de 2381 caractères de 0 et 1.

Arnauld
la source

Technique intéressante, je suis sûr que cela fonctionnerait bien sur des cordes avec moins de motifs que cela.

ETHproductions

3

Tu m'as perdu au "multicolore"

Patrick Roberts

6

Mathematica, 128 octets

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

J42161217
la source

Ne pourriez-vous pas raccourcir cela en utilisant un seul StringReplace et en ayant une liste de règles de remplacement?

DanTheMan

oui bien sûr ... Je viens de commencer à jouer au golf ...

J42161217

Si le remplacement est aussi simple que ça, Retina devrait bien faire ...

Neil

6

JavaScript (ES6), 85 80 octets

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

Un peu un port de ma réponse Japt. Une explication du modèle peut être trouvée là-bas.

Extrait de test

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
la source

2

C'est un peu déroutant ... jusqu'à ce que vous découvriez en j="1"fait deux choses, dont l'une est abusée.

Erik the Outgolfer

6

Gelée , 28 22 20 octets

⁽.⁷Dð26Bẋ4ṫȧ€Ẏð/ṫ159

-3 merci à Jonathan Allan .

Port (quelque peu) de ETHproductions excellente Japt s » réponse .

Toujours pas aussi efficace que celui de dzaima (ancienne version SOGL)!

Cela renvoie un tableau de chiffres.

Erik le Outgolfer
la source

La validité a été vérifiée à l'aide de Diff Checker !

Erik the Outgolfer

Continuons cette discussion dans le chat .

Erik the Outgolfer

6

Rétine, 69 66 65 octets

Réponse de JavaScript (ES6) de Port of @ ETHproduction: Essayez-le en ligne! NB Un peu lent.


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

^.{158}|¶

Un port de @ Jenny_mathy Mathematica réponse était 82 66 octets: Essayez-le en ligne!


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

Un port de l'explication Japt de @ ETHproduction était de 69 octets: Essayez-le en ligne!

Un port de l'explication Japt de @ ETHproduction était également de 69 octets: Essayez-le en ligne!

Un port de la réponse JavaScript de @ RickHitchcock était de 74 octets: Essayez-le en ligne!


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

Un port de la réponse JavaScript de @ Arnauld était de 85 octets: essayez-le en ligne!


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

Neil
la source

Agréable! pouvez-vous mettre à jour le mien? (c'est ma version finale) merci!

J42161217

@Jenny_mathy J'ai pu jouer au golf sur quelques octets supplémentaires, vous êtes donc maintenant le premier port Retina, merci!

Neil

3

JavaScript (ES6), 158 143 139 octets

(Enregistré quelques octets grâce à @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
la source

Vous avez juste besoin de générer la chaîne pour pouvoir supprimer le _=>.

DanTheMan

2

Cela pourrait être un peu plus court (au moins 30 octets). Astuce # 1: passer de .replace(/x/g,y)à.split(x).join(y)

ETHproductions

Astuce # 2: Utilisez des chiffres> 1à la place des lettres dans la compression. Cela vous permettra de remplacer facilement par exemple 9, puis 8, puis 7, etc.

ETHproductions

@ETHproductions, je vois comment je peux économiser 1 octet en basculant vers splitet joinavec les chaînes de modèle: 'ifififhg0'[s='split']BTiBT[j='join']BThhgBT[s]BThBT[j]BTg0fBT(où BT est le backtick)

Rick Hitchcock

Astuce # 3: j='string',i=9,[replacements].map(x=>j=j.split(i--).join(x)),j

mappez à

2

05AB1E , 20 octets

1Ž5ûv1₂b4×y.$.:]Ƶv.$

Réponse JAPT de Port of ETHProductions .

Grimmy
la source

1

K (ngn / k) , 86 octets

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