10

编写最短的程序或函数来生成这1000个数字或以它们开头的序列（0或1索引）。

[0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 
 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 
 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 
 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 
 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 
 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 
 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 
 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 
 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 
 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 
 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 
 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 
 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 
 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 
 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 
 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 
 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 
 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 
 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 
 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 
 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 
 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 
 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 
 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 
 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 
 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 
 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 
 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 
 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 
 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 
 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 
 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 
 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 
 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 
 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 
 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 
 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 
 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 
 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 
 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 
 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 
 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0]

code-golf sequence kolmogorov-complexity

— 约翰·曼瓜尔
source

这是我第一次发布代码难题。如果您有任何样式上的改进。让我知道。

— 约翰·曼格

7

约翰，您好，欢迎来到PPCG！这里的挑战需要有一个客观的胜利条件（通常是代码高尔夫）。我们还建议您在发布之前通过沙盒运行所有挑战。

3

由于此问题的目标似乎是找到序列，因此我建议要求最短的代码来正确生成这前1000个元素。

@助记符听起来不错。我的代码已经很短了，我在问是否有更短的代码。随时编辑:-)，或者我可以移至沙盒

— 约翰·曼格

我忘记了谁曾挑战过。但是“找到模式”非常受欢迎。我隐约记得有人在50分钟内破解了它；但即使在那之后，人们仍继续回答。

— 魔术章鱼缸

17

果冻，11 10 字节

@Dennis节省了1个字节

ȷḶ×⁽q£:ȷ5Ḃ

在线尝试！

怎么样？

我首先注意到该模式在长度4和长度3的运行之间交替，每隔几个运行就跳过长度4的步骤。这导致我寻找一个可以划分为当前索引的数字，然后采用mod 2并下限（即检索最低有效位）以给出序列中该索引处的位。经过多次试验和错误，我发现3.41845确实可以做到这一点，但是乘以它的近似倒数（.29253）会短一个字节。

ȷḶ×⁽q£:ȷ5Ḃ    Main link. Arguments: none
ȷ             Yield 1e3, i.e. 1000.
 Ḷ            Lowered range; yield [0, 1, 2, ..., 999].
  ×⁽q£        Multiply each item by 29253.
      :ȷ5     Floor-divide each item by 1e5, i.e. 100000.
         Ḃ    Take each item mod 2.

— ETH生产
source

啊，你找到了

— 乔纳森·艾伦，

[0 ... 999]次分别乘以0.2925，mod 2和发言权（我先发言，然后发言mod 2，但等效）

— Jonathan Allan

6

嗯，这是很滑稽的，期望有更复杂的东西。

— Nit

@JonathanAllan我最初只是尝试，Ḃ但显然那只是mod 2，而不是最低位，所以我加了Ḟ修复它。现在交换

— ETHproductions'Apr 18'18

1

ȷḶ×⁽q£:ȷ5Ḃ有效，为10个字节。

— 丹尼斯

3

Dyalog APL，99 83 82字节

a←{⍵/0 1}¨(↓3 2⍴4 3 3)
{a⊢←↓⍉↑a{⍺∘{⍵/⊂⍺}¨⍵}¨↓3 3⍴⍵}¨(9/5)∘⊤¨1386531 496098
1000⍴∊a

在线尝试！

绝对不是预期的解决方案，因为它仍然具有大量的硬编码数据，但这只是一个开始。

— 宰马
source

3

Ruby，34 29 26 22字节

$.+=184while p$./629%2

在线尝试！

快速说明：之所以有效，是因为魔术数为629。我注意到该序列在第629个元素之后开始重复，因此我尝试仅使用整数数学来“改善”一些现有答案。我发现另一个“魔术数字”（0.29253）实际上是184/629。

— 国标
source

2

果冻，31 个字节

鉴于这种模式，可能还有更短的方法...

ĖŒṙḂ
“ṁ⁽⁺ḄæI’BḤ+3żḂ$ẎÇo2Ç+3Çḣȷ¬

在线尝试！

怎么样？

利用重复的游程长度结构，该结构在三个深度处都是明显的。

ĖŒṙḂ - Link 1, make runs of bits: list of lengths    e.g. [5,3,5,3,3]
Ė    - enumerate                      [[1,5],[2,3],[3,5],[4,3],[5,3]]
 Œṙ  - run-length decode      [1,1,1,1,1,2,2,2,3,3,3,3,3,4,4,4,5,5,5]
   Ḃ - bit (modulo by 2)      [1,1,1,1,1,0,0,0,1,1,1,1,1,0,0,0,1,1,1]

“ṁ⁽⁺ḄæI’BḤ+3żḂ$ẎÇo2Ç+3Çḣȷ¬ - Main link: no arguments
“ṁ⁽⁺ḄæI’                   - literal 234931870193324
        B                  - to binary = [1,1,0,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,0,1,1,0,0]
         Ḥ                 - double    = [2,2,0,2,0,2,0,2,2,0,2,0,2,0,2,2,0,2,0,2,0,2,2,0,2,0,2,0,2,0,2,2,0,2,0,2,0,2,2,0,2,0,2,0,2,2,0,0]
          +3               - add three = [5,5,3,5,3,5,3,5,5,3,5,3,5,3,5,5,3,5,3,5,3,5,5,3,5,3,5,3,5,3,5,5,3,5,3,5,3,5,5,3,5,3,5,3,5,5,3,3]
              $            - last two links as a monad:
             Ḃ             -   bit     = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]
            ż              -   zip     = [[5,1],[5,1],[3,1],[5,1],[3,1],[5,1],[3,1],[5,1],[5,1],[3,1],[5,1],[3,1],[5,1],[3,1],[5,1],[5,1],[3,1],[5,1],[3,1],[5,1],[3,1],[5,1],[5,1],[3,1],[5,1],[3,1],[5,1],[3,1],[5,1],[3,1],[5,1],[5,1],[3,1],[5,1],[3,1],[5,1],[3,1],[5,1],[5,1],[3,1],[5,1],[3,1],[5,1],[3,1],[5,1],[5,1],[3,1],[3,1]]
               Ẏ           - tighten   = [5,1,5,1,3,1,5,1,3,1,5,1,3,1,5,1,5,1,3,1,5,1,3,1,5,1,3,1,5,1,5,1,3,1,5,1,3,1,5,1,3,1,5,1,5,1,3,1,5,1,3,1,5,1,3,1,5,1,3,1,5,1,5,1,3,1,5,1,3,1,5,1,3,1,5,1,5,1,3,1,5,1,3,1,5,1,3,1,5,1,5,1,3,1,3,1]
                Ç          - call the last Link (1) as a monad
                           -           = [1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0]
                 o2        - OR 2      = [1,1,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,1,1,2]
                   Ç       - Link 1... = [1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0]
                    +3     - add three = [4,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,4,3,3,4,3,4,3,3,4,3,4,3,3]
                      Ç    - Link 1... = [1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0]
                        ȷ  - literal 1000
                       ḣ   - head      = [1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1]
                         ¬ - NOT       = [0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,0]

— 乔纳森·艾伦
source

我从未见过果冻！

— 约翰·曼格

欢迎使用PPCG :)-这是我们的主持人之一丹尼斯（Dennis）编写的一种高尔夫语言。单击标题为git-hub的页面，其中有一个Wiki。

— 乔纳森·艾伦

我保证会更好地提出问题。我看到有一个沙箱和一些标准格式。

— 约翰·曼格

当我开始时，这几乎就是我的方法。

— 硕果累累'18

@EsolangingFruit我以为正在做重复动作可能只是一小部分...似乎是117/400！

— 乔纳森·艾伦

2

Java 8，75 64 62字节

v->{for(int i=0;i<1e3;)System.out.print((int)(i++*.29253)%2);}

打印整个序列而无需定界符以保存字节，因为无论如何它们都将是0and 1。

@ETHproductions的端口Jelly回答，因为我怀疑我能找到更短的东西。

在线尝试。

说明：

v->{                     // Method with empty unused parameter and no return-type
  for(int i=0;i<1e3;)    //  Loop `i` in range [0,1000)
    System.out.print(    //   Print:
      (int)(i++*.29253)  //    `i` multiplied with 0.29253,
                         //    and then truncated of their decimal values by casting to int
      %2);}              //    Modulo-2 to result in either 0 or 1

旧答案返回结果数组（75字节）：

v->{int i=1000,r[]=new int[i];for(;i-->0;)r[i]=(int)(i*.29253)%2;return r;}

在线尝试。

说明：

v->{                   // Method with empty unused parameter and integer-array return-type
  int i=1000,          //  Index `i`, starting at 1000
      r[]=new int[i];  //  Result-array of size 1000
  for(;i-->0;)         //  Loop `i` in range (1000,0]
    r[i]=              //   Set the item in the array at index `i` to:
      (int)(i*.29253)  //    `i` multiplied with 0.29253,
                       //    and then truncated of their decimal values by casting to int
      %2;              //    Modulo-2 to result in either 0 or 1
  return r;}           //  Return the resulting integer-array

— 凯文·克鲁伊森
source

1

JavaScript（Node.js），41 33字节，端口

感谢Rick Hitchcock提供4个以上的字节

f=i=>i>999?'':(i*.29253&1)+f(-~i)

在线尝试！

JavaScript（Node.js），121字节，原始

_=>'8888y888'[s='replace'](/8/g,'aaa3yyy')[s](/y/g,'aaa3aa3')[s](/a/g,34)[s](/./g,t=>(_=+!+_+[]).repeat(t)).slice(3,1003)

在线尝试！

— 平方米
source

使用递归节省4个字节： f=(i=0)=>i<1e3?(i*.29253&1)+f(i+1):''

— Rick Hitchcock

1

Stax，13 11 字节

í?♫~╘äqx-G▄

在staxlang.xyz上运行并调试它！

@ETHproductions的Jelly答案的 Stax端口（在修改之前），通过@recursive进行一些修改以节省两个字节。

— 库尔德拉西斯·纳巴里亚
source

您可以将分母加倍，保留小数部分，然后四舍五入到最接近的整数，而不使用模数。如果我没记错的话，它还会给出11

— 递归的结果

1

Wolfram语言（Mathematica），96字节

我搜索了一个元胞自动机，当您将数据划分为长度7并保留每三行时，它会查看左侧的4个邻居并产生在数据中看到的向左行走的模式。

这个细胞自动机将运行29代，每一个都进行三次，完全匹配字符1至629的序列。但是，该序列在第630个字符处开始重复，而不是继续观察到的模式，因此需要额外的代码来处理截断的模式。我两次生成主模式，以获得1258个字符。

Most@Flatten[{#,#,#}&/@CellularAutomaton[{271,2,-{{4},{3},{2},{1}}},{0,0,0,0,1,1,1},29]]~Table~2

没有这种故障，我们可以用较短的74个字节来完成。47是获得1000个字符所需的世代数（实际上是1008 = 48 * 7 * 3）

{#,#,#}&/@CellularAutomaton[{271,2,-{{4},{3},{2},{1}}},{0,0,0,0,1,1,1},47]

在线尝试！

— 凯莉·洛德（Kelly Lowder）
source

1

Z80Golf，27个字节

00000000: 018d 2b7b 1f1f e601 f630 ff09 3001 1313  ..+{.....0..0...
00000010: 7bfe 9220 ee7a fe04 20e9 76              {.. .z.. .v

在线尝试！

从下面的C代码翻译：

for (n = 0; n >> 16 != 1170; n += 11149 + 65536)
    putchar('0'|n>>18&1);

拆卸：

  ld bc, 11149
loop:
  ld a, e
  rra
  rra
  and 1
  or '0'
  rst $38           ; putchar
  add hl, bc        ; Add 11149 to n = DEHL.
  jr nc, just_one   ; Add 65536 to n, possibly with carry from low 16 bits.
  inc de
just_one:
  inc de
  ld a, e
  cp 1170 & 255
  jr nz, loop
  ld a, d
  cp 1170 >> 8
  jr nz, loop
  halt

这实质上是一个定点算术方法：（11149 + 65536）/ 2 ¹⁸ ≈0.29253，通过其他的答案中使用的常数。

— 林恩
source

0

J，17个字节

2|<.0.29253*i.1e3

ETHproduction的Jelly答案的AJ端口。

在线尝试！

— 加伦·伊万诺夫（Galen Ivanov）
source

0

Japt，13个字节

A³Ç*.29#ý f u
A³             // Given 10³,
  Ç            // map over it as a range, returning the given number
   *.29253     // times the constant,
           f u // floored and mod-2.

ETHproduction 的Jelly答案的 Japt版本。
修正了奥利弗（Oliver）的错误。

在这里尝试。

— 尼特
source

0

木炭，13字节

Ｅφ§01×·²⁹²⁵³ι

在线尝试！链接是详细版本的代码。说明：

 φ              Predefined variable 1000
Ｅ               Map over implicit range
            ι   Current value
      ·²⁹²⁵³    Literal constant `0.29253`
     ×          Multiply
   01           Literal string `01`
  §             Cyclically index
                Implicitly print each result on its own line

感谢@ ASCII-only，它允许索引接受转换为整数的浮点数（在这种情况下，该浮点数将自动减少为模2）。

— 尼尔
source

0

C，55 53 52字节

f(i,j){for(i=0;j=.29253*i,i++-1e3;)putchar(j%2+48);}

凯文·克鲁伊森（Kevin Cruijssen）的Java 回答端口。在这里在线尝试。

感谢vazt打了2个字节的高尔夫球，并感谢Jonathan Frech 再打了一个高尔夫球。

非高尔夫版本：

f(i, j) { // function taking two dummy arguments (implicitly int) and implicitly returning an unused int
    for(i = 0; j = .29253*i, i++ - 1e3; ) //  loop 1000 times, multiply i with 0.29253, truncating to an integer
        putchar(j % 2 + 48);  // modulo the truncated integer by 2, yielding 0 or 1, then convert to ASCII (48 is ASCII code for '0') and print
}

— OOBalance
source

i因为它是全局的，所以被初始化为0，因此您可以i=0从for循环初始化程序中删除for以节省3个字节。另外，如果您引入了第二个变量（作为的参数f()）并分配i++*.29253给它，则可以避免强制转换并另外保存2个字节：i;f(j){for(;i<1e3;)printf("%d",(j=i++*.29253)%2);} 在线尝试！

— vazt

@vazt是的，i在一开始被初始化为0，但是如果我们要多次调用此函数，那还不够。使用j以避免投是一个伟大的高尔夫，谢谢。

— OOBalance

52个字节。

— 乔纳森·弗雷奇

0

///，63字节

/b/000111//A/b1b//B/b0b//C/0BA1//X/CACACACA0bCBCBCB/0bXXCBXCAC0

在线尝试！

— 暴民埃里克
source

在1和0的数字序列中找到模式

果冻，11 10 字节

怎么样？

Dyalog APL，99 83 82字节

Ruby，34 29 26 22字节

果冻，31 个字节

怎么样？

Java 8，75 64 62字节

JavaScript（Node.js），41 33字节，端口

JavaScript（Node.js），121字节，原始

Stax，13 11 字节

Wolfram语言（Mathematica），96字节

Z80Golf，27个字节

J，17个字节

Japt，13个字节

木炭，13字节

C，55 53 52字节

///，63字节