Lặp lại lặp đi lặp lại

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
nguồn

Bạn có biết có cách nào tốt hơn nén cơ sở không?

— Erik the Outgolfer 24/07/17

@EriktheOutgolfer Chắc chắn, vì điều này được thực hiện như một tai nạn 15 byte: pA big part of the challenge is to find the shortest way to generate it.

— dzaima

Vâng, đó là lý do tôi xóa câu trả lời của mình ... cảm thấy không thực sự xứng đáng. Oh, và những gì bạn trích dẫn là lý do tại sao tôi hỏi.

— Erik the Outgolfer 24/07/17

1

Đây dường như là một chuỗi nguyên thủy, vì vậy nó không phải là sự lặp lại tầm thường của bất cứ điều gì không may :(

— HyperNeutrino

1

@Sanchises Vâng, sau đó vẫn có khả năng tái tạo một cái gì đó thậm chí còn golf. ;)

— Erik the Outgolfer

35

Japt , 35 33 29 26 25 24 byte

'1
#˜5s £=r1,26¤²²sX
s#ž

Giải trình

'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.

Đây là cách tôi có được điều này. Bắt đầu với trình tự ban đầu:

01101011010110100110101101011010110101101011010011010110101101001101011010110101101011...

Sau một vài lần loay hoay, tôi nhận ra rằng chuỗi bao gồm toàn bộ các bản sao của 0và 11010. Thay thế mỗi trường hợp của 11010với 2cho này:

02220222222022202222220222022222202220002220222222022202222220222022222202220002220222...

Oh hey, bây giờ nó hoàn toàn được làm bằng 0và 222. Hãy thay đổi từng 222thành một 2:

02022020220202202000202202022020220200020220202202022020020220202202022020002022020220...

Trình tự này trông rất nhiều như ban đầu. Dường như có rất nhiều 22020s trong này, vì vậy tôi đã thay thế từng cái 3và nhận cái này:

02033300203330020333020333002033300203330203330020333000203330203330020333002033302033...

Bây giờ mỗi trường hợp 3được chứa trong a 020333. Thay thế từng cái một lần nữa bằng một cái duy nhất 3cho điều này:

30303303033030033030330303303003303033030330303303033030330300330303303033030033030330...

... Và chúng tôi trở lại mô hình ban đầu một lần nữa. Một lần nữa, thay thế 33030bằng 4:

30304404440444444044404444440444044444404440

Uh-oh, lúc đầu có một số ký tự đi lạc. Chúng ta có thể khắc phục điều này bằng cách thêm một phần bổ sung 3vào đầu trước khi ngưng tụ (đây là nơi có thêm 158 ký tự ở cuối). Sau đó, chúng tôi ngưng tụ mỗi lần chạy 3 4giây thành một lần 4như trước:

4040440404404044040

Aaaaand một lần nữa, chúng tôi trở lại mô hình ban đầu. Một lần cuối cùng, chúng ta hãy bản cô đặc của 44040để 5:

Hy vọng rằng bước cuối cùng này không quá khó hiểu. Nó giống như bước mà chúng tôi nén 020333s thành 3s, ngoại trừ lần này, chúng tôi nén 4040555s thành 5s:

Và ở đó bạn có chuỗi nén hoàn toàn. Mã chỉ đơn giản là đảo ngược quá trình nén lặp lại để lấy lại chuỗi gốc. Vì tất cả các lần lặp nén gần như giống hệt nhau, nên không cần phải tổ chức lại chuỗi đúng cách.

— Sản phẩm sinh học
nguồn

Như bạn có thể thấy ở đây tôi đã sử dụng Jelly's /(giảm) ... Japt có giảm / gấp không? Nếu vậy, điều đó có thể tiết kiệm byte.

— Erik the Outgolfer

@EriktheOutgolfer Japt có giảm, nhưng điều đó sẽ yêu cầu thay đổi £=ít nhất r@nếu không phải là một cái gì đó lâu hơn. (Thành thật mà nói, £=không phải là một mức giá cao để trả cho một chức năng giảm độc đáo)

— Sản phẩm ETH

Ý bạn là r@<space>sao? Tất nhiên sau đó nó sẽ là một cái gì đó giống như [[1], 1, 5, 2, 5]thay vì [1, 5, 2, 5]sử dụng giảm.

— Erik the Outgolfer

1

Sẽ là chính xác để nói rằng bạn chỉ cần thiết kế ngược lại một bản mở rộng fractal?

— Patrick Roberts

1

@PatrickRoberts Yep, khá nhiều.

— Sản phẩm ETH

14

JavaScript (ES6), 130 byte

Nén xung phi tuyến nhiều màu bằng cách khuếch đại tham số quang liên tiếp trong các cấu trúc khớp pha.

^{OK, không hẳn. Chỉ cần thay thế.}

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

Bản giới thiệu

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>

Làm sao?

Đây chắc chắn không phải là cách tốt nhất để nén chuỗi này, nhưng tôi nghĩ dù sao tôi cũng đã đăng nó vì tôi gặp rắc rối khi tối ưu hóa nó (một chút) với chức năng đệ quy.

Chúng tôi bắt đầu với chuỗi s="pqh0f9bc9bc9bc900"và áp dụng liên tiếp 20 thay thế sau:

 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

Chẳng hạn, "9" sẽ mở rộng thành "ca" , "cbd" , "c0cd" , "dd0ddd" , ..., và cuối cùng là chuỗi 2381 ký tự gồm 0 và 1 giây .

— Arnauld
nguồn

Kỹ thuật thú vị, tôi chắc chắn rằng nó sẽ hoạt động tốt trên các chuỗi có ít mẫu hơn thế này.

— Sản phẩm ETH

3

Bạn đã mất tôi tại "nhiều màu"

— Patrick Roberts

6

Toán học, 128 byte

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

— J42161217
nguồn

Bạn không thể làm điều này ngắn hơn bằng cách chỉ sử dụng một StringReplace và có một danh sách các quy tắc thay thế?

— DanTheMan

vâng tất nhiên ... tôi mới bắt đầu chơi golf này ...

— J42161217

Nếu việc thay thế dễ dàng thì Retina nên làm tốt ...

— 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)

Một phần của một câu trả lời Japt của tôi. Một lời giải thích về mô hình có thể được tìm thấy ở đó.

Kiểm tra đoạn

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);

— Sản phẩm sinh học
nguồn

2

Có một chút khó hiểu ... cho đến khi bạn phát hiện ra j="1"thực sự là hai điều, một trong số chúng bị lạm dụng.

— Erik the Outgolfer

6

Thạch , 28 22 20 byte

⁽.⁷Dð26Bẋ4ṫȧ€Ẏð/ṫ159

-3 cảm ơn Jonathan Allan .

(Hơi) cảng ETHproductions tuyệt vời Japt 's câu trả lời .

Vẫn không hiệu quả như của dzaima (phiên bản SOGL cũ)!

Điều này trả về một mảng các chữ số.

— Erik là người vượt trội
nguồn

Điều này đã được kiểm tra tính hợp lệ bằng cách sử dụng Diff Checker !

— Erik the Outgolfer

Hãy để chúng tôi tiếp tục cuộc thảo luận này trong trò chuyện .

— Erik the Outgolfer

6

Võng mạc, 69 66 65 byte

Câu trả lời JavaScript (ES6) của Port @ ETHproduction: Hãy thử trực tuyến! NB hơi chậm.


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

^.{158}|¶

Một cổng của câu trả lời Mathicala của @ Jenny_mathy là 82 66 byte: Hãy thử trực tuyến!


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

Một cổng giải thích Japt của @ ETHproduction là 69 byte: Hãy thử trực tuyến!

Một cổng giải thích Japt trước đây của @ ETHproduction cũng là 69 byte: Hãy thử trực tuyến!

Một cổng của câu trả lời JavaScript của @ RickHitchcock là 74 byte: Hãy thử trực tuyến!


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

Một cổng của câu trả lời JavaScript của @ Arnauld là 85 byte: Hãy thử trực tuyến!


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

— Neil
nguồn

Tốt đẹp! bạn có thể vui lòng cập nhật của tôi? (đây là phiên bản cuối cùng của tôi) thanx!

— J42161217

@Jenny_mathy Tôi đã có thể chơi thêm một số byte để bây giờ bạn là cổng Retina hàng đầu, cảm ơn!

— Neil

3

JavaScript (ES6), 158 143 139 byte

(Đã lưu một số byte nhờ @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
nguồn

Bạn chỉ cần tạo chuỗi, vì vậy bạn có thể thả chuỗi _=>.

— DanTheMan

2

Điều này có thể ngắn hơn một chút (ít nhất 30 byte). Gợi ý số 1: chuyển từ .replace(/x/g,y)sang.split(x).join(y)

— ETHproductions

Gợi ý số 2: Sử dụng các chữ số> 1thay cho các chữ cái trong phần nén. Điều này sẽ cho phép bạn dễ dàng thay thế 9, ví dụ , sau 8đó 7, v.v.

— ETHproductions

@ETHproductions, tôi thấy cách tôi có thể tiết kiệm 1 byte bằng cách chuyển sang splitvà joinvới chuỗi mẫu: 'ifififhg0'[s='split']BTiBT[j='join']BThhgBT[s]BThBT[j]BTg0fBT(trong đó BT là backtick)

— Rick Hitchcock

Gợi ý số 3: ánh xạ qua một loạt các thay thế, ví dụ:j='string',i=9,[replacements].map(x=>j=j.split(i--).join(x)),j

— ETHproductions

2

05AB1E , 20 byte

1Ž5ûv1₂b4×y.$.:]Ƶv.$

Cảng JAPT câu trả lời ETHProductions của .

— Grimmy
nguồn

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]