这是每周挑战2。主题：翻译

编写一个程序或函数，该程序或函数在Prelude中获取程序的源代码，并在Befunge-93中输出等效程序的代码。为了使程序等效，对于任何给定的输入，它都应产生与Prelude程序相同的输出，并且仅当Prelude程序停止时才停止。

输入语言：前奏

Python解释器：

#!/usr/bin/python

import sys

NUMERIC_OUTPUT = True
NUMERIC_INPUT = True

try:
    filename = sys.argv[1]
except:
    print "Usage:", sys.argv[0], "<filename>"
    raise SystemExit

try:
    inFile = file(filename)
except:
    print "Error when opening", filename
    raise SystemExit

# code is kept as a list of voices, each voice a string

code = []
firstLine = True
eof = False
while not eof:
    # read batch of lines
    batch = []
    while 1:
        line = inFile.readline()
        if line == '': eof = True
        if line == '' or line.rstrip() == '*':
            break
        batch.append(line.rstrip())
    maxLen = max([len(b) for b in batch])
    batch = [b + ' '*(maxLen - len(b)) for b in batch]
    if firstLine:
        code = batch
        firstLine = False
    else:
        if len(batch) != len(code):
            print "Error in the program: number of voices changes"
            raise SystemExit
        for i in range(len(batch)):
            code[i] += batch[i]


class Stack:
    def __init__(self):
        self.data = []
    def push(self, value):
        if self.data or value != 0: self.data.append(value)
    def drop(self):
        if self.data: self.data.pop()
    def top(self):
        if self.data: return self.data[-1]
        return 0
    def pop(self):
        value = self.top()
        self.drop()
        return value




numVoices = len(code)
numInstructions = len(code[0])
stacks = [Stack() for x in range(numVoices)]
topValues = [0 for x in range(numVoices)]
# establish loop couplings
loopStack = []
loops = {}
for cp in range(numInstructions):
    curr = [voice[cp] for voice in code]
    if curr.count('(') + curr.count(')') > 1:
        print "Error in the program: More than one bracket; position", cp
        raise SystemExit
    if '(' in curr:
        loopStack.append((cp, curr.index('(')))
    if ')' in curr:
        if not loopStack:
            print "Error in the program: extraneous closing bracket; position", cp
            raise SystemExit
        openingPosition, openingVoice = loopStack.pop()
        loops[openingPosition] = cp
        loops[cp] = openingPosition, openingVoice

if loopStack:
    print "Error in the program: not enough closing brackets"
    raise SystemExit


# now, actually execute the program
cp = 0 # code pointer
while cp < numInstructions:
    # technically we're supposed to shuffle our voices to make sure to perform IO
    # in random order, but screw that for now
    next_cp = cp+1 # can be modified by ( )
    for voice in range(numVoices):
        i = code[voice][cp] # current instruction
        if i == '^':
            stacks[voice].push(topValues[(voice-1) % numVoices])
        elif i == 'v' or i == 'V':
            stacks[voice].push(topValues[(voice+1) % numVoices])
        elif i == '+':
            stacks[voice].push(stacks[voice].pop() + stacks[voice].pop())
        elif i == '-':
            b = stacks[voice].pop()
            a = stacks[voice].pop()
            stacks[voice].push(a-b)
        elif i == '#':
            stacks[voice].drop()
        elif i == '?':
            if NUMERIC_INPUT:
                try: num = int(raw_input())
                except ValueError: num = 0
                stacks[voice].push(num)
            else:
                char = sys.stdin.read(1)
                if not char: char = '\0'
                stacks[voice].push(ord(char))
        elif i == '!':
            if NUMERIC_OUTPUT:
                print stacks[voice].pop()
            else:
                sys.stdout.write(chr(stacks[voice].pop()))
        elif i == '(':
            if stacks[voice].top() == 0:
                next_cp = loops[cp] + 1
        elif i == ')':
            openingPosition, openingVoice = loops[cp]
            if topValues[openingVoice] != 0:
                next_cp = openingPosition + 1
        elif i in '0123456789':
            stacks[voice].push(int(i))
    topValues = [stacks[i].top() for i in range(numVoices)]
    cp = next_cp

Prelude程序由多个“声音”组成，这些声音同时执行指令。每个语音的说明在单独的一行上。每个声音都有一个单独的堆栈，并使用无限数量的零进行初始化。执行从最左边的列开始，并在每个刻度线前向右移一列，除非受)或(指令的影响。当到达最后一列时，程序终止。

此挑战的前奏规格：

Digits 0-9      Push onto the stack a number from 0 to 9. Only single-digit
                    numeric literals can be used.
^               Push onto the stack the top value of the stack of the above 
                    voice.
v               Push onto the stack the top value of the stack of the below 
                    voice.
#               Remove the top value from the stack.
+               Pop the top two integers from the stack and push their sum.
-               Pop the top two integers from the stack, subtract the topmost 
                    from the second, and push the result.
(               If the top of the stack is 0, jump to the column after the 
                    matching `)` after the current column executes.
)               If the top of the stack is not 0, jump to the column after 
                    the matching `(` after the current column executes.
?               Read an integer from STDIN.
!               Pop one value from the stack and print it to STDOUT as an
                    integer.
<space>         No-op

笔记

• v并^循环执行，因此v底部声音上的会复制顶部声音的stack元素，顶部声音^上的会从底部声音复制。推论：无论v和^复制堆栈的顶部在一个单一的语音程序。
• A (及其匹配项)可能位于不同的行上。但是，a )将始终查看(放置相应语音的语音堆栈，而不是其)本身放置的语音堆栈。
• ^和v指令产生的值将在同一列中的任何其他操作完成之前对存在的值进行操作。
• ?并且其!操作与esolangs.org上的规范不同，因此请务必使用本文中提供的稍有改动的解释器进行测试。

输入保证有：

• 匹配括号
• 一栏中最多只能有一个括号
• 每行字符数相同
• 至少一行
• 没有超过一个I / O（!或?）指令的列
• 每个语音指令后一个换行字符
• 除上述字符外，没有其他字符

输出语言：Befunge-93

Befunge是一种基于堆栈的语言，其程序计数器（PC；指向当前指令的指针）在二维网格上自由移动。它从左上角开始，向右移动。运动场是环形的，即PC运动环绕两个边缘。Befunge还具有一个堆栈，该堆栈被初始化为无数个零。Befunge具有以下操作：

Digits 0-9      Push onto the stack a number from 0 to 9. Only single-digit
                        numeric literals can be used.
    +               Pop the top two integers from the stack and push their sum.
    -               Pop the top two integers from the stack, subtract the topmost 
                        from the second, and push the result.
    *               Pop the top two integers from the stack and push their product.
    /               Pop the top two integers from the stack, divide the second by
                        the topmost, and push the result. Rounds down.
    %               Pop the top two integers from the stack, divide the second by
                        the topmost, and push the remainder.
    !               Pop a value, push 1 if it's zero, push 0 otherwise.
    `               Pop the top two integers from the stack, push 1 if the second
                        is greater than the topmost, push 0 otherwise.
    >               Set PC direction to right.
    <               Set PC direction to left.
    ^               Set PC direction to up.
    v               Set PC direction to down.
    ?               Set PC direction randomly.
    _               Pop a value, if it's 0 set PC direction to right, left otherwise.
    |               Pop a value, if it's 0 set PC direction to down, up otherwise.
    "               Toggle string mode. While in string mode, push each character's
                        ASCII value instead of executing it.
    :               Duplicate top stack element.
    \               Swap top two stack elements.
    $               Pop and discard top stack element.
    .               Pop a value and output it as an integer. Whitespace not included.
    ,               Pop a value and output the corresponding ASCII character.
    #               Jump over the next instruction.
    g               Pop y, pop x, push the character at coordinate (x,y).
    p               Pop y, pop x, pop v, set character at coordinate (x,y) to v.
    &               Read an integer from STDIN and push it.
    ~               Read a character from STDIN and push it ASCII value.
    @               End program.

您可以假定Befunge-93编译器/解释器具有以下特征：

• 整数是无限精度的。
• 它允许任何大小的网格。
• 网格坐标（for g和p）从0开始。

计分

为了防止仅在Befunge中产生Prelude解释器并将Prelude源硬编码到其中的提交，目标是最大程度地减少生成的Befunge源代码的大小。

下面提供了一些Prelude程序。您的翻译器将在所有这些文件上运行。您的分数是Befunge程序的大小之和，只要它们均有效。

您的翻译器不应针对这些测试用例进行专门优化（例如，通过为它们手动编写手写的Befunge程序）。如果我怀疑这样做有任何答案，我保留更改输入或创建其他输入的权利。

输入样例

打印n-1到0：

?(1-^!)

逻辑与：

?  (0)  
 ?(0  ) 
    1  !

逻辑或：

 ?   (0) 
? (0)    
   1  1 !

检查非负数输入（即模2）的奇偶校验：

?(1-)   
 ^  v   
 v1-^^-!

平方输入：

 ^    
  ^+ !
?(1-) 

打印第n个斐波那契数，其中n = 00 n = 1对应于1：

0 v+v!
1   ^ 
?(1-) 

签名：

  1) v #  - !
 vv (##^v^+) 
?(#   ^   ## 

非负输入除法：

1 (#  1) v #  - 1+)   
     vv (##^v^+)      
?  v-(0 # ^   #       
 ?                    
   1+              1-!

当然，即使未指定示例程序的负数行为，您的程序在所有情况下都必须表现出相同的行为。

最后，您的翻译器不应过长：

• 它必须包含在Stack Exchange帖子中
• 它应在典型的台式计算机上在10分钟内处理样本输入。

请注意，Prelude或Befunge的数字输入是可选的负号，后跟一个或多个十进制数字，后跟换行符。其他输入是未定义的行为。

您可以用任何语言编写翻译器。最短的翻译Befunge代码获胜。

排行榜

• Sp3000：16430字节

Python 3，稍后会得分

from collections import defaultdict
from functools import lru_cache
import sys

NUMERIC_OUTPUT = True

@lru_cache(maxsize=1024)
def to_befunge_num(n):
    # Convert number to Befunge number, using base 9 encoding (non-optimal,
    # but something simple is good for now)

    assert isinstance(n, int) and n >= 0

    if n == 0:
        return "0"

    digits = []

    while n:
        digits.append(n%9)
        n //= 9

    output = [str(digits.pop())]

    while digits:
        output.append("9*")
        d = digits.pop()

        if d:
            output.append(str(d))
            output.append("+")

    output = "".join(output)

    if output.startswith("19*"):
        return "9" + output[3:]

    return output

def translate(program_str):
    if program_str.count("(") != program_str.count(")"):
        exit("Error: number of opening and closing parentheses do not match")

    program = program_str.splitlines()
    row_len = max(len(row) for row in program)
    program = [row.ljust(row_len) for row in program]
    num_stacks = len(program)


    loop_offset = 3
    stack_len_offset = program_str.count("(")*2 + loop_offset
    stack_offset = stack_len_offset + 1
    output = [[1, ["v"]], [1, [">"]]] # (len, [strings]) for each row
    max_len = 1 # Maximum row length so far

    HEADER_ROW = 0
    MAIN_ROW = 1
    FOOTER_ROW = 2
    # Then stack lengths, then loop rows, then stacks

    # Match closing parens with opening parens
    loop_map = {} # {column: (loop num, stack number to check, is_start)}
    loop_stack = []
    loop_num = 0

    for col in range(row_len):
        col_str = "".join(program[stack][col] for stack in range(num_stacks))

        if col_str.count("(") + col_str.count(")") >= 2:
            exit("Error: more than one parenthesis in a column")

        if "(" in col_str:
            stack_num = col_str.index("(")

            loop_map[col] = (loop_num, stack_num, True)
            loop_stack.append((loop_num, stack_num, False))
            loop_num += 1

        elif ")" in col_str:
            if loop_stack:
                loop_map[col] = loop_stack.pop()

            else:
                exit("Error: mismatched parentheses")


    def pad_max(row):
        nonlocal max_len, output

        while len(output) - 1 < row:
            output.append([0, []])

        if output[row][0] < max_len:
            output[row][1].append(" "*(max_len - output[row][0]))
            output[row][0] = max_len


    def write(string, row):
        nonlocal max_len, output

        output[row][1].append(string)
        output[row][0] += len(string)

        max_len = max(output[row][0], max_len)


    def stack_len(stack, put=False):
        return (to_befunge_num(stack) + # x
                str(stack_len_offset) + # y
                "gp"[put])


    def get(stack, offset=0):
        assert offset in [0, 1] # 1 needed for 2-arity ops

        # Check stack length
        write(stack_len(stack) + "1-"*(offset == 1) + ":0`", MAIN_ROW)

        pad_max(HEADER_ROW)
        pad_max(MAIN_ROW)
        pad_max(FOOTER_ROW)

        write(">" + to_befunge_num(stack + stack_offset) + "g", HEADER_ROW)
        write("|", MAIN_ROW)
        write(">$0", FOOTER_ROW)

        pad_max(HEADER_ROW)
        pad_max(MAIN_ROW)
        pad_max(FOOTER_ROW)

        write("v", HEADER_ROW)
        write(">", MAIN_ROW)
        write("^", FOOTER_ROW)


    def put(stack, value=""):
        put_inst = (value +
                    stack_len(stack) +
                    to_befunge_num(stack + stack_offset) +
                    "p")

        post_insts.append(put_inst)


    def pop(stack):
        put(stack, "0")


    def inc_stack_len(stack):
        post_insts.append(stack_len(stack) + "1+")
        post_insts.append(stack_len(stack, put=True))


    def dec_stack_len(stack):
        post_insts.append(stack_len(stack) + ":0`-") # Ensure nonnegativity
        post_insts.append(stack_len(stack, put=True))


    # Technically not necessary to initialise stack lengths per spec, but it makes it
    # more portable and easier to test against other Befunge interpreters

    for stack in range(num_stacks):
        write("0" + stack_len(stack, put=True), MAIN_ROW)

    for col in range(row_len):
        post_insts_all = []

        loop_start = False
        loop_end = False

        if col in loop_map:
            if loop_map[col][2]:
                loop_start = True
            else:
                loop_end = True

        if loop_start:
            loop_row = loop_offset + 2*loop_map[col][0]
            get(loop_map[col][1])

        elif loop_end:
            get(loop_map[col][1])
            write("!", MAIN_ROW)


        for stack in range(num_stacks-1, -1, -1):
            char = program[stack][col]
            post_insts = [] # Executed after the gets in reverse order, i.e. last added first

            if char in " ()":
                continue

            # Pre-inc, post-dec
            elif char.isdigit():
                inc_stack_len(stack)
                put(stack, char)

            elif char == "?":
                inc_stack_len(stack)
                put(stack, "&")

            elif char == "!":
                get(stack)
                post_insts.append(".91+," if NUMERIC_OUTPUT else ",")
                pop(stack)
                dec_stack_len(stack)

            elif char == "#":
                pop(stack)
                dec_stack_len(stack)

            elif char in "+-":
                get(stack, 1)
                get(stack)
                post_insts.append(char)
                pop(stack) # This one first in case of ! or 1!
                post_insts.append(stack_len(stack) + ":1`-:1\\`+") # Ensure >= 1
                post_insts.append(stack_len(stack, put=True))
                put(stack)                

            elif char in "^v":
                offset = -1 if char == "^" else 1

                get((stack + offset) % num_stacks)
                inc_stack_len(stack)
                put(stack)

            else:
                exit("Error: invalid character " + char)

            post_insts_all.append(post_insts)


        while post_insts_all:
            write("".join(post_insts_all.pop()), MAIN_ROW)

        if loop_start or loop_end:
            loop_row = loop_offset + 2*loop_map[col][0]

            pad_max(HEADER_ROW)
            pad_max(MAIN_ROW)
            pad_max(loop_row)
            pad_max(loop_row + 1)

            write(">v", HEADER_ROW)
            write("|>", MAIN_ROW)

            if loop_start:
                write(" ^", loop_row)
                write(">", loop_row + 1)

            else:
                write("<", loop_row)
                write(" ^", loop_row + 1)


    write("@", MAIN_ROW)
    return "\n".join("".join(row) for row_len, row in output)

if __name__ == '__main__':
    if len(sys.argv) < 3:
        exit("Usage: py -3 prefunge.py <input filename> <output filename>")

    with open(sys.argv[1]) as infile:
        with open(sys.argv[2], "w") as outfile:
            outfile.write(translate(infile.read()))

运行像py -3 prefunge.py <input filename> <output filename>。

对我来说这是很慢的一周，所以我终于无聊了，无法解决这个六个月大的问题。我会问为什么没有其他人尝试过，但是我仍然感觉到调试带来的痛苦（据我所知，可能仍然存在bug）。

这个问题没有提供Befunge-93解释器，因此我使用了这个，与规范稍有不同。两个主要区别是：

• 如果程序的给定行中不存在char，则无法写入该行。这意味着您需要多次按Enter键以在末尾引入足够的换行符。如果您NaN在输出中看到s，则最可能的原因是。

• 网格单元未预初始化为零-为方便起见，我在Befunge输出中包括了一些预初始化，但是由于没有必要，在我开始评分时可以将其删除。

输出程序的核心布局是这样的：

v [header row]
> [main row]
  [footer row]
  ---
   |
   | rows for loops (2 per loop)
   |
  ---
  [stack length row]
  ---
   |
   | rows for stack space (1 per voice)
   |
  ---

堆栈空间在程序外部，因此换行符是前面的Enter-spamming注释。

核心思想是为每个语音分配一行作为其堆栈。为了维护这些堆栈，我们还有一个特殊的堆栈长度行，其中每个堆栈的长度沿该行记录在一个单元格中。该程序然后是很多gets和puts，例如，用于打印的过程是：

• 在获取单元格 y = stack_row[stack], x = stack_length[stack]
• 执行.91+,，即打印为整数，然后打印换行符
• 将上面坐标处的单元格替换为0（以模拟弹出）
• 减量 stack_length[stack]

为了执行列的同时评估，在写入任何单元之前（例如，对于打印示例，在第一步和第二步之间可能会有更多指令），读取所有必需的单元并将它们的值保留在堆栈上。

`大于，用于确保堆栈长度永远不会为负，并在堆栈为空时将其推为0。这是清晰可见的分支的来源，但是我有一个想法，将删除该分支，这应该从第一行和第三行中删除大量空白。

对于循环，由于Prelude循环可以双向跳转，因此在这样的配置中，每个循环使用两行：

       >v                     >v
(cond) |>  (program)  (cond) !|>

        ^                     <
       >                       ^

目前，这些循环占了字节的大部分，但是可以通过将它们放入带有的代码箱中来轻松地找到它们p，我对转换器能够正常工作感到满意之后，我打算这样做。

这是的一些示例输出?(1-^!)，即打印n-1到0：

v                        >6gv>v                      >6gv      >6gv                                 >6gv                   >6gv                           >6gv >v
>005p05g1+05p&05g6p05g:0`|  >|>05g1+05p105g6p05g1-:0`|  >05g:0`|  >-005g6p05g:1`-:1\`+05p05g6p05g:0`|  >05g1+05p05g6p05g:0`|  >.91+,005g6p05g:0`-05p05g:0`|  >!|>@
                         >$0^                        >$0^      >$0^                                 >$0^                   >$0^                           >$0^
                              ^                                                                                                                                <
                             >                                                                                                                                  ^

输入平方：

v                                >8gv      >8gv             >v      >6gv                                   >8gv      >8gv        >7gv      >7gv                                                            >8gv >v      >7gv
>005p015p025p25g1+25p&25g8p25g:0`|  >25g:0`|  >05g1+05p05g6p|>05g:0`|  >15g1+15p15g7p25g1+25p125g8p25g1-:0`|  >25g:0`|  >15g1-:0`|  >15g:0`|  >+015g7p15g:1`-:1\`+15p15g7p-025g8p25g:1`-:1\`+25p25g8p25g:0`|  >!|>15g:0`|  >.91+,015g7p15g:0`-15p@
                                 >$0^      >$0^                     >$0^                                   >$0^      >$0^        >$0^      >$0^                                                            >$0^         >$0^
                                                             ^                                                                                                                                                  <
                                                            >                                                                                                                                                    ^

除法（建议少量输入）：

v                                                                          >91+gv>v      >94+gv                                                         >95+gv      >95+gv        >93+gv      >93+gv                                                                    >93+gv      >93+gv               >v      >93+gv                                                     >93+gv >v      >92+gv                  >v      >92+gv                                       >92+gv                                       >91+gv                                       >93+gv                     >91+gv                       >92+gv      >92+gv        >91+gv      >91+gv                                                                                      >92+gv >v                        >91+gv      >91+gv                                     >91+gv >v                        >95+gv      >95+gv                                     >95+gv
>009p019p029p039p049p09g1+09p109g91+p29g1+29p&29g93+p39g1+39p&39g94+p09g:0`|    >|>39g:0`|    >009g91+p09g:0`-09p29g1+29p29g93+p49g1+49p149g95+p49g1-:0`|    >49g:0`|    >29g1-:0`|    >29g:0`|    >-029g93+p29g:1`-:1\`+29p29g93+p+049g95+p49g:1`-:1\`+49p49g95+p29g:0`|    >29g:0`|    >19g1+19p19g92+p|>29g:0`|    >09g1+09p109g91+p19g1+19p19g92+p29g1+29p029g93+p29g:0`|    >!|>19g:0`|    >029g93+p29g:0`-29p|>19g:0`|    >09g1+09p09g91+p019g92+p19g:0`-19p19g:0`|    >019g92+p19g:0`-19p29g1+29p29g93+p09g:0`|    >009g91+p09g:0`-09p19g1+19p19g92+p29g:0`|    >19g1+19p19g92+p09g:0`|    >19g1+19p19g92+p19g1-:0`|    >19g:0`|    >09g1-:0`|    >09g:0`|    >-009g91+p09g:1`-:1\`+09p09g91+p+019g92+p19g:1`-:1\`+19p19g92+p029g93+p29g:0`-29p19g:0`|    >!|>09g1+09p109g91+p09g1-:0`|    >09g:0`|    >+009g91+p09g:1`-:1\`+09p09g91+p09g:0`|    >!|>49g1+49p149g95+p49g1-:0`|    >49g:0`|    >-049g95+p49g:1`-:1\`+49p49g95+p49g:0`|    >.91+,049g95+p49g:0`-49p@
                                                                           >$0  ^        >$0  ^                                                         >$0  ^      >$0  ^        >$0  ^      >$0  ^                                                                    >$0  ^      >$0  ^                       >$0  ^                                                     >$0  ^         >$0  ^                          >$0  ^                                       >$0  ^                                       >$0  ^                                       >$0  ^                     >$0  ^                       >$0  ^      >$0  ^        >$0  ^      >$0  ^                                                                                      >$0  ^                           >$0  ^      >$0  ^                                     >$0  ^                           >$0  ^      >$0  ^                                     >$0  ^
                                                                                  ^                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                        <
                                                                                 >                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                          ^
                                                                                                                                                                                                                                                                                                          ^                                                                        <
                                                                                                                                                                                                                                                                                                         >                                                                          ^
                                                                                                                                                                                                                                                                                                                                                                                                                    ^                                                                                                                                                                                                                                                                                                                                              <
                                                                                                                                                                                                                                                                                                                                                                                                                   >                                                                                                                                                                                                                                                                                                                                                ^

还有一堆浮现在脑海中，像更换其他次要的优化的07p07g有:07p，但我考虑在这个时间一步:)