做一个随机的鼓环

Question

随机生成的鼓环听起来好吗？

drum循环是5\times 32矩阵A，由1s和0s组成

1. A_{1,1}=A_{1,17}=A_{2,9}=A_{2,25}=1，
2. 对于每一个i，i第四行都有完全不同的1s，其中f(1)=3, f(2)=2, f(3)=6, f(4)=8, f(5)=5，
3. 对于每个j\equiv 1\bmod 4，jth列正好有一个1，并且
4. 对于每个j\not\equiv 1\bmod 4，j第四列最多只有一个1。

在这里，行为五个鼓中的每一个提供指令，每一列代表一个节拍。特别是，第一行对应于踢，而第二行对应于圈套。

编写最短的代码，从一组鼓循环中均匀地绘制矩阵。

可以随意使用您的代码来绘制一个随机的鼓循环，在本网站上实现鼓循环，然后在您的文章中发布一个指向您的创建的链接。(请注意，我们只使用这个在线节拍器的前5行。)

Edit 1:我们使用的是基于1的索引:行索引是\{1,\ldots,5\}，列索引是\{1,\ldots,32\}。

Edit 2:从当前提交的文件来看，我最初问题的答案似乎是“是的”：随机生成的鼓环听起来不错。下面是指向当前示例的链接：

• @Arnauld随机鼓环
• 随机鼓环
• @KevinCruijssen随机鼓环

Answer 1

05AB1E，65 字节数

31Ý©8ˆ24ˆ40¯ǝ0ˆ16ˆ®¯KΩˆ48¯¦¦ǝ•Wk¤]•3ôεćи}˜.ržw8и«®¯K.r.¡4Ö}é˜ǝb€¦

输出作为字符串的转置列表。

在网上试试。 (页脚漂亮-将其打印为预期的行/列)。请随意删除页脚以查看实际结果。)

解释：

31Ý                # Push a list in the range [0,31]
   ©               # Store it in variable `®` (without popping)
8ˆ                 # Add 8 to the global array
  24ˆ              # Add 24 to the global array as well
     40¯ǝ          # Replace the values at the indices of the global array with 40
0ˆ                 # Add 0 to the global array
  16ˆ              # Add 16 to the global array as well
     ®             # Push the [0,31] list from variable `®` again
      ¯K           # Remove all values of the global array (the [0,8,16,24])
        Ω          # Pop and push a random index from this list
         ˆ         # Add it to the global array as well
            ¯¦¦    # Push the global array with the first two (8,24) removed
          48   ǝ   # Replace the values at the remaining three indices with 48
•Wk¤]•             # Push compressed integer 533636834
      3ô           # Split it into parts of size 3: [533,636,834]
        ε          # Map over each integer
         ć         #  Extract head; pop and push remainder and head
          и        #  Repeat the remainder the head amount of times
                   #   i.e. 533 becomes [33,33,33,33,33]
        }˜         # After the map: flatten the list of list of integers
          .r       # And randomly shuffle it
            žw8и   # Then push a list of 8 32s
                «  # And merge it to the end of the list
®                  # Push the [0,31] list from variable `®` again
 ¯K                # Remove all indices stored in the global array
                   # (the [0,8,16,24] and one random index)
   .r              # Randomly shuffle these remaining indices
     .¡            # Group the remaining indices by:
       4Ö          #  Where this (0-based) index is divisible by 4
                   #  (which would be i % 4 == 1 for the corresponding 1-based index)
     }é            # After the group by, sort the two inner lists by length,
                   # so the indices divisible by 4 are before the other indices
       ˜           # And flatten it to a single list
ǝ                  # Then insert the shuffled integers at those shuffled indices
 b                 # Convert each integer to a binary string
                   # (32=100000;33=100001;34=100010;36=100100;40=101000;48=110000)
  €¦               # And then remove the leading "1" from each
                   # (after which the resulting list is output implicitly)

看这个05AB1E我的尖端(部分)如何压缩大整数？)来理解为什么•Wk¤]•是533636834。

Answer 2

果冻，43字节

32s8Zs4ZḢ;FẊḣ⁴ƲƊḢŒœ;FẊ$ƲFṙ-“¥©€Þı‘œṖ;¹Ṭz0ZṖ

在网上试试！

返回代表鼓循环的1s和0的32x5矩阵的niladic链接。页脚只需将其连在一起并用换行符分隔。

解释

32                                          | 32
  s8                                        | Split into groups of 8
    Z                                       | Transpose
     s4                                     | Split into groups of 4
       Z                                    | Transpose
               Ɗ                            | Following as a monad:
        Ḣ                                   | - Head
         ;    Ʋ                             | - Concatenate to following as a monad:
          F                                 |   - Flatten
           Ẋ                                |   - Shuffle list
            ḣ⁴                              |   - First 16
                       Ʋ                    | Following as a monad:
                Ḣ                           | - Head
                 Œœ                         | - Odd/even indexed items
                   ;  $                     | - Concatenate to:
                    F                       |   - Remainder of list flattened
                     Ẋ                      |   - Shuffled
                        F                   | Flatten
                         ṙ-                 | Rotate right 1 item
                           “¥©€Þı‘œṖ        | Split list before positions 4, 6, 12, 20, 25
                                    ;¹      | Concatenate to 32 (because 32 was the beginning of this niladic link)
                                      Ṭ     | Convert from lists of numeric indices to logical lists with 1s at those positions
                                       z0   | Transpose with 0 as filler
                                         Z  | Transpose
                                          Ṗ | Remove last (the sixth list introduced with the 32 above)

Answer 3

Python 3，180个字节

from random import*
r=range
s=sample
a=s(s([x for x in r(32)if x%4],16)+[x*8+4 for x in r(4)],20)
print([[i in t for i in r(32)]for t in[[0,16,a[0]],[8,24],a[1:7],a[7:15],a[15:]]])

在网上试试！

打印表示鼓循环的逻辑矩阵。