ASCII Brainfuck

Question

给出了使用Brainfuck命令编写的随机代码。您的程序应该将该代码转换为以下ascii技巧:每个Brainfuck命令表示为5x9数组的A和B符号。所有命令都与5x3数组的A符号连接。符号A和B你选择自己。

例如，当输入为><+,.[]时，程序可能会输出：

==                 ==       =                                 = = =       = = =  
    ==         ==           =                                 =               =  
       ==   ==          = = = = =                             =               =  
    ==         ==           =           =           =         =               =  
==                 ==       =          =                      = = =       = = =  

或

@@                 @@       @                                 @ @ @       @ @ @  
    @@         @@           @                                 @               @  
       @@   @@          @ @ @ @ @                             @               @  
    @@         @@           @           @           @         @               @  
@@                 @@       @          @                      @ @ @       @ @ @  

甚至

..ooooooooooooooooo..ooooooo.ooooooooooooooooooooooooooooooooo.o.o.ooooooo.o.o.oo
oooo..ooooooooo..ooooooooooo.ooooooooooooooooooooooooooooooooo.ooooooooooooooo.oo
ooooooo..ooo..oooooooooo.o.o.o.o.ooooooooooooooooooooooooooooo.ooooooooooooooo.oo
oooo..ooooooooo..ooooooooooo.ooooooooooo.ooooooooooo.ooooooooo.ooooooooooooooo.oo
..ooooooooooooooooo..ooooooo.oooooooooo.oooooooooooooooooooooo.o.o.ooooooo.o.o.oo

等等，尾随空格和A符号是允许的。

保证输入仅由符号> < + , . [ ]组成。

这是一个代码-高尔夫挑战，所以最短的解决方案获胜。

Answer 1

果冻，57 字节数

“ṾÆxÑ3øḤC⁵ịJYSLẇɱ⁼’Bs3ŒBJṫ⁴Ɗ¦z1m€0Ṛs€9
⁽HṠ,17ḥⱮịⱮ¢j€⁶ẋ3¤Y

接受字符串作为输出结果的第一个程序参数的完整程序，使用0表示墨迹，而1不使用墨水。

在网上试试！ (页脚将1s替换为空格，以便于检查)。

怎么做？

“...’Bs3ŒBJṫ⁴Ɗ¦z1m€0Ṛs€9 - Link 1: no arguments
“...’                    - 43552819895870853685529620449548093603391
     B                   - in binary
      s3                 - split into chunks of length 3
                           -> columns of the bottom half of 5 of the 7 characters
                              the '[' and '>' are created using reflection later
                              the comma is reversed too, we'll use its reflection
              ¦          - sparse application...
             Ɗ           - ...to indices: last three links as a monad:
          J              -   range of length
            ⁴            -   16
           ṫ             -   tail (covers the columns containing ',' and '.')
        ŒB               - ...apply: bounce ("something" -> "somethingnihtemos")
               z1        - transpose with filler 1 (fills where we didn't reflect)
                 m€0     - reflect each (constructing our ']', '>', and ',')
                    Ṛ    - reverse
                           -> a list of the lines of a representation (without
                              any spacing) of ",.+<][>+.," where the first
                              comma is reversed.
                     s€9 - split each line into chunks of length 9

⁽HṠ,17ḥⱮịⱮ¢j€⁶ẋ3¤Y - Main Link: list of characters, C
⁽HṠ,17             - [19206,17]
       Ɱ           - map across C with:
      ḥ            -   hash with salt 19206 and domain [1,2,...,17]
                       -> maps: ".[,+<]>" to 2,6,10,13,14,15,17
          ¢        - call Link 1 as a nilad
         Ɱ         - map across the lines of the result with:
        ị          -   index into
                       -> 1-indexed and modular, so e.g. 17 gets the 7th item,
                          the part of the line for a '>'.
                ¤  - nilad followed by links as a nilad:
             ⁶     -   space character
              ẋ3   -   repeated three times
           j€      - join each list of line parts with three spaces
                 Y - join with newline characters
                   - implicit, smashing print

Answer 2

Python3.8 (预发行版)，257个字节

lambda s:'\n'.join(list(map(lambda i:print("111".join([list(map(lambda s:str(s)[3:],map(bin,[0x9FFF9EFFFFFF57AB,0xF9F9FEFFFFFF7FFB,0xFF0FEAAFFFFF7FFB,0xF9F9FEFF7FBF7FFB,0x9FFF9EFEFFFF57AB])))[i][(p:="><+,.[]".index(c))*9:(p+1)*9]for c in s])),[0,1,2,3,4])))

我不知道为什么它给我一个TypeError，但它有效:)

在网上试试！