Ulangi Pengulangan Berulang

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
sumber

Apakah Anda tahu jika ada cara yang lebih baik daripada kompresi basis?

Erik the Outgolfer,

@EriktheOutgolfer Jelas, karena ini dibuat sebagai kecelakaan 15 byte: pA big part of the challenge is to find the shortest way to generate it.

dzaima

Ya, itulah alasan saya menghapus jawaban saya ... merasa tidak sepadan. Oh, dan apa yang Anda kutip adalah alasan saya bertanya.

Erik the Outgolfer

1

Ini tampaknya merupakan urutan primitif, jadi itu bukan pengulangan sepele dari apa pun sayangnya :(

HyperNeutrino

1

@Sanchises Nah, maka masih ada kemungkinan untuk membuat ulang sesuatu yang lebih pegolf ;)

Erik the Outgolfer

35

Japt , 35 33 29 26 25 24 byte

'1
#˜5s £=r1,26¤²²sX
s#ž

Penjelasan

'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.

Begini cara saya mendapatkan ini. Mulai dengan urutan asli:

01101011010110100110101101011010110101101011010011010110101101001101011010110101101011...

Setelah beberapa main-main, saya menyadari bahwa urutan seluruhnya terdiri dari salinan 0dan 11010. Mengganti setiap instance 11010dengan 2memberi ini:

02220222222022202222220222022222202220002220222222022202222220222022222202220002220222...

Oh, hei, sekarang semuanya terbuat dari 0dan 222. Mari kita ubah masing 222- masing menjadi satu 2:

02022020220202202000202202022020220200020220202202022020020220202202022020002022020220...

Urutan ini terlihat sangat mengerikan seperti aslinya. Tampaknya ada banyak 22020hal dalam hal ini, jadi saya mengganti masing-masing dengan 3dan mendapatkan ini:

02033300203330020333020333002033300203330203330020333000203330203330020333002033302033...

Sekarang setiap instance 3terkandung dalam a 020333. Mengganti masing-masing lagi dengan satu 3memberi ini:

30303303033030033030330303303003303033030330303303033030330300330303303033030033030330...

... dan kita kembali ke pola semula lagi. Sekali lagi, ganti 33030dengan 4:

30304404440444444044404444440444044444404440

Uh-oh, ada beberapa karakter liar di awal. Kita dapat memperbaikinya dengan menambahkan ekstra 3di awal sebelum kondensasi (ini adalah asal 158 karakter tambahan di akhir). Kemudian kami menyingkat setiap proses 3 4detik menjadi satu 4, seperti sebelumnya:

4040440404404044040

Lagi-lagi, kita kembali ke pola semula. Terakhir kali, mari kita memadatkan salinan 44040ke 5:

Semoga langkah terakhir ini tidak terlalu sulit untuk dipahami. Ini seperti langkah di mana kita mengompresi 020333s menjadi 3, kecuali kali ini, kita mengompresi 4040555s menjadi 5:

Dan di sana Anda memiliki string yang sepenuhnya terkompresi. Kode hanya membalikkan kompresi berulang untuk mendapatkan kembali string asli. Karena semua iterasi kompresi hampir identik, tidak perlu banyak untuk mengatur ulang string dengan benar.

Produksi ETH
sumber

Seperti yang Anda lihat di sini, saya telah menggunakan Jelly /(mengurangi) ... apakah Japt juga mengurangi / melipat? Jika demikian, itu mungkin menghemat byte.

Erik the Outgolfer

@EriktheOutgolfer Japt memang sudah berkurang, tapi itu perlu diubah £=setidaknya r@jika bukan sesuatu yang lebih lama. (Jujur, £=bukan harga tinggi untuk membayar fungsi pengurangan tidak konvensional)

ETHproduksi

Maksudmu r@<space>? Tentu saja itu akan menjadi sesuatu seperti [[1], 1, 5, 2, 5]bukannya [1, 5, 2, 5]menggunakan pengurangan.

Erik the Outgolfer

1

Apakah akurat untuk mengatakan Anda baru saja membalikkan ekspansi fraktal?

Patrick Roberts

1

@ PatrickRoberts Yap, cukup banyak.

ETHproductions

14

JavaScript (ES6), 130 byte

Kompresi pulsa nonlinier multicolor dengan amplifikasi parametrik optik berurutan dalam struktur yang sesuai dengan fase kuasi.

^{OKE, tidak cukup. Pergantian saja.}

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>

Bagaimana?

Ini jelas bukan cara terbaik untuk mengompresi string ini, tapi saya pikir saya tetap akan mempostingnya karena saya kesulitan mengoptimalkannya (sedikit) dengan fungsi rekursif.

Kami mulai dengan string s="pqh0f9bc9bc9bc900"dan berturut-turut menerapkan 20 pergantian berikut:

 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

Misalnya, "9" akan diperluas ke "ca" , "cbd" , "c0cd" , "dd0ddd" , ..., dan akhirnya string 2381-karakter 0 dan 1.

Arnauld
sumber

Teknik yang menarik, saya yakin itu akan bekerja dengan baik pada string dengan pola yang kurang dari ini.

ETHproduk

3

Anda kehilangan saya di "multicolor"

Patrick Roberts

6

Mathematica, 128 byte

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

J42161217
sumber

Tidak bisakah Anda mempersingkat ini dengan hanya menggunakan satu StringReplace dan memiliki daftar aturan penggantian?

DanTheMan

ya tentu saja ... Saya baru saja mulai bermain golf ini ...

J42161217

Jika penggantiannya semudah itu maka Retina akan melakukannya dengan baik ...

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)

Agak dari port jawaban Japt saya. Penjelasan pola dapat ditemukan di sana.

Cuplikan tes

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);

Produksi ETH
sumber

2

Agak membingungkan ... sampai Anda tahu j="1"sebenarnya ada dua hal, salah satunya dilecehkan.

Erik the Outgolfer

6

Jelly , 28 22 20 byte

⁽.⁷Dð26Bẋ4ṫȧ€Ẏð/ṫ159

Terima kasih kepada Jonathan Allan .

(Agak) pelabuhan ETHproductions 's Japt baik jawaban .

Masih tidak seefisien dzaima (versi SOGL lama)!

Ini mengembalikan array angka.

Erik the Outgolfer
sumber

Ini telah diperiksa validitasnya menggunakan Diff Checker !

Erik the Outgolfer

Mari kita lanjutkan diskusi ini dalam obrolan .

Erik the Outgolfer

6

Retina, 69 66 65 byte

Port of @ ETHproduksi's JavaScript (ES6) jawaban: Cobalah online! NB Agak lambat.


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

^.{158}|¶

Port jawaban @ Jenny_mathy's Mathematica adalah 82 66 byte: Coba online!


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

Port penjelasan Japt @ ETHproduction adalah 69 byte: Coba online!

Port dari penjelasan Japt sebelumnya @ ETHproduction juga 69 byte: Coba online!

Port jawaban JavaScript @ RickHitchcock adalah 74 byte: Coba online!


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

Port jawaban JavaScript @ Arnauld adalah 85 byte: Coba 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
sumber

Bagus! bisakah kamu memperbarui milikku? (ini adalah versi final saya) thanx!

J42161217

@ Jenny_mathy Saya bisa bermain golf lebih banyak byte sehingga Anda sekarang menjadi pelabuhan Retina terkemuka, terima kasih!

Neil

3

JavaScript (ES6), 158 143 139 byte

(Menyimpan beberapa byte berkat @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
sumber

Anda hanya perlu membuat string, sehingga Anda dapat menjatuhkannya _=>.

DanTheMan

2

Ini bisa sedikit lebih pendek (setidaknya 30 byte). Petunjuk # 1: beralih dari .replace(/x/g,y)ke.split(x).join(y)

ETH

Petunjuk # 2: Gunakan digit> 1sebagai pengganti huruf-huruf dalam kompresi. Ini akan memungkinkan Anda untuk dengan mudah mengganti misalnya 9, lalu 8, lalu 7, dll.

ETHproduksi

@ ETHproductions, saya melihat bagaimana saya dapat menyimpan 1 byte dengan beralih ke splitdan joindengan string template: 'ifififhg0'[s='split']BTiBT[j='join']BThhgBT[s]BThBT[j]BTg0fBT(di mana BT adalah backtick)

Rick Hitchcock

Petunjuk # 3: memetakan melalui berbagai penggantian, misalnyaj='string',i=9,[replacements].map(x=>j=j.split(i--).join(x)),j

ETHproduksi

2

05AB1E , 20 byte

1Ž5ûv1₂b4×y.$.:]Ƶv.$

Pelabuhan jawaban JAPT dari ETHProduk .

Grimmy
sumber

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]