Ripeti ripetizioni ripetitive

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

Sai se ci sono modi migliori della compressione di base?

— Erik the Outgolfer,

@EriktheOutgolfer Sicuramente, dato che questo è stato fatto come un incidente di 15 byte: pA big part of the challenge is to find the shortest way to generate it.

— dzaima

Sì, questo è il motivo per cui ho cancellato la mia risposta ... non mi è valsa davvero la pena. Oh, e quello che hai citato è il motivo per cui ho chiesto.

— Erik the Outgolfer,

1

Questa sembra essere una sequenza primitiva, quindi purtroppo non è una banale ripetizione di nulla :(

— HyperNeutrino

1

@Sanchise Bene, allora c'è ancora la possibilità di ricreare qualcosa di ancora più golfista. ;)

— Erik the Outgolfer,

35

Japt , 35 33 29 26 25 24 byte

'1
#˜5s £=r1,26¤²²sX
s#ž

Spiegazione

'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.

Ecco come ho ottenuto questo. Inizia con la sequenza originale:

01101011010110100110101101011010110101101011010011010110101101001101011010110101101011...

Dopo alcuni scherzi, mi sono reso conto che la sequenza consisteva interamente in copie di 0e 11010. La sostituzione di ogni istanza di 11010con ha 2dato questo:

02220222222022202222220222022222202220002220222222022202222220222022222202220002220222...

Oh hey, ora è interamente fatto di 0e 222. Cambiamo ciascuno 222in un singolo 2:

02022020220202202000202202022020220200020220202202022020020220202202022020002022020220...

Questa sequenza assomiglia moltissimo all'originale. Sembra che ci siano molti messaggi 22020in questo, quindi ho sostituito ciascuno con 3e ho ottenuto questo:

02033300203330020333020333002033300203330203330020333000203330203330020333002033302033...

Ora ogni istanza di 3è contenuta in a 020333. Sostituendo di nuovo ognuno di questi con un singolo si 3ottiene questo:

30303303033030033030330303303003303033030330303303033030330300330303303033030033030330...

... e torniamo di nuovo al modello originale. Ancora una volta, sostituendo 33030con 4:

30304404440444444044404444440444044444404440

Uh-oh, ci sono alcuni caratteri vaganti all'inizio. Possiamo risolvere questo problema aggiungendo un extra 3all'inizio prima della condensazione (da qui provengono i 158 caratteri extra alla fine). Quindi condensiamo ogni serie di 3 4secondi in una singola 4, come in precedenza:

4040440404404044040

Ancora una volta, siamo tornati al modello originale. Un'ultima volta, condensiamo copie di 44040a 5:

Speriamo che questo passaggio finale non sia troppo difficile da capire. È come il passaggio in cui abbiamo compresso le 020333s in 3s, tranne che questa volta, stiamo comprimendo le 4040555s in 5s:

E lì hai la stringa completamente compressa. Il codice inverte semplicemente la compressione ripetuta per ripristinare la stringa originale. Poiché tutte le iterazioni di compressione erano quasi identiche, non ci vuole molto per riorganizzare correttamente la stringa.

— ETHproductions
fonte

Come puoi vedere qui ho usato Jelly's /(ridurre) ... Japt ha anche ridurre / piegare? In tal caso, ciò potrebbe salvare byte.

— Erik the Outgolfer,

@EriktheOutgolfer Japt ha ridotto, ma ciò richiederebbe una modifica £=almeno r@se non qualcosa di più. (Onestamente, £=non è un prezzo elevato da pagare per una funzione di riduzione non convenzionale)

— ETHproductions

Intendevi r@<space>? Ovviamente, sarebbe qualcosa di simile [[1], 1, 5, 2, 5]invece di [1, 5, 2, 5]usare il metodo di riduzione.

— Erik the Outgolfer,

1

Sarebbe accurato dire che hai appena decodificato un'espansione frattale?

— Patrick Roberts,

1

@PatrickRoberts Sì, praticamente.

— ETHproductions

14

JavaScript (ES6), 130 byte

Compressione impulsiva non lineare multicolore mediante amplificazione parametrica ottica consecutiva in strutture abbinate quasi-fase.

^{OK, non del tutto. Solo sostituzioni.}

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

dimostrazione

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>

Come?

Questo non è sicuramente il modo migliore per comprimere questa stringa, ma ho pensato di pubblicarla comunque dato che mi sono preso la briga di ottimizzarla (leggermente) con una funzione ricorsiva.

Iniziamo con la stringa s="pqh0f9bc9bc9bc900"e successivamente applichiamo le seguenti 20 sostituzioni:

 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

Ad esempio, un "9" si espanderà in "ca" , "cbd" , "c0cd" , "dd0ddd" , ... e infine una stringa di 2381 caratteri di 0 e 1.

— Arnauld
fonte

Tecnica interessante, sono sicuro che funzionerebbe bene su stringhe con meno pattern di questo.

— ETHproductions

3

Mi hai perso a "multicolor"

— Patrick Roberts il

6

Mathematica, 128 byte

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

— J42161217
fonte

Non potresti renderlo più breve usando solo un StringReplace e avere un elenco di regole di sostituzione?

— DanTheMan,

sì certo ... ho appena iniziato a giocare a golf in questo modo ...

— J42161217

Se la sostituzione è così semplice, Retina dovrebbe fare bene ...

— Neil

6

JavaScript (ES6), 85 80 byte

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

Un po 'come una porta della mia risposta di Japt. Una spiegazione del modello può essere trovata laggiù.

Snippet di prova

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

È un po 'confuso ... fino a quando non scopri che in j="1"realtà sono due cose, una delle quali è stata abusata del tipo.

— Erik the Outgolfer,

6

Gelatina , 28 22 20 byte

⁽.⁷Dð26Bẋ4ṫȧ€Ẏð/ṫ159

-3 grazie a Jonathan Allan .

(Un po ') porto di ETHproductions eccellente Japt 's risposta .

Ancora non efficiente come quello di dzaima (vecchia versione SOGL)!

Ciò restituisce una matrice di cifre.

— Erik the Outgolfer
fonte

Questo è stato verificato per validità usando Diff Checker !

— Erik the Outgolfer,

Continuiamo questa discussione in chat .

— Erik the Outgolfer,

6

Retina, 69 66 65 byte

Port of @ ETHproduction's JavaScript (ES6) answer: Provalo online! NB Un po 'lento.


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

^.{158}|¶

Un port della risposta Mathematica di @Jenny_mathy era di 82 66 byte: provalo online!


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

Un port della spiegazione Japt di @ ETHproduction era di 69 byte: provalo online!

Un port della precedente spiegazione di Japt di @ ETHproduction era anche di 69 byte: provalo online!

Una porta della risposta JavaScript di @ RickHitchcock era di 74 byte: provala online!


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

Un port della risposta JavaScript di @ Arnauld era di 85 byte: provalo 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

Bello! puoi per favore aggiornare il mio? (questa è la mia versione finale) grazie!

— J42161217

@Jenny_mathy Sono stato in grado di giocare a golf alcuni byte in più, quindi ora sei il principale porto di Retina, grazie!

— Neil,

3

JavaScript (ES6), 158 143 139 byte

(Salvati alcuni byte grazie 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

Devi solo generare la stringa, in modo da poter eliminare _=>.

— DanTheMan,

2

Potrebbe essere un po 'più breve (almeno 30 byte). Suggerimento n. 1: passare da .replace(/x/g,y)a.split(x).join(y)

— ETHproductions

Suggerimento n. 2: utilizzare le cifre> 1al posto delle lettere nella compressione. Ciò ti consentirà di sostituire facilmente ad es9 , , quindi 8, quindi 7, ecc.

— ETHproductions

@ETHproductions, vedo come posso salvare 1 byte passando a splite joincon stringhe di template: 'ifififhg0'[s='split']BTiBT[j='join']BThhgBT[s]BThBT[j]BTg0fBT(dove BT è backtick)

— Rick Hitchcock,

Suggerimento n. 3: mappa attraverso una serie di sostituzioni, ad es.j='string',i=9,[replacements].map(x=>j=j.split(i--).join(x)),j

— ETHproductions

2

05AB1E , 20 byte

1Ž5ûv1₂b4×y.$.:]Ƶv.$

Port of ETHProductions's JAPT answer .

— Grimmy
fonte

1

K (ngn / k) , 86 byte

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