循环从不在递归函数内完成

Question

我正在构建一个程序来将括号恢复为句子，使它们成为格式良好的公式（句子逻辑中的WFF）。 例如，

- 句子a是WFF。

- 句子a > b只有1种方法可以修复括号，使其成为WFF，即(a > b) 。

- 句子a > b > c有2种方法可以修复括号以使其成为WFF - ((a > b) > c)或(a > (b > c)) 。

等等...

这个算法有一个迭代和递归元素

# returns number of connectives
def numConnectives(wff):
    count = 0
    for c in wff:
        if c == connectives:
            count += 1
    return count

def rec(wff):
    result = []
    ind = []                            # list storing indexes of connectives used
    if len(wff) == 1:
        return wff
    else:
        for i in range(numConnectives(wff)):
            opIndex = findConnective(wff, ind)          # index where the operator is at

            right   = createRight(wff, opIndex)     # right formula
                                                    # the first time it goes through, right is b>c
                                                    # then right is c
            left    = createLeft(wff, opIndex)      # left formula
                                                    # left is a
                                                    # then it is b
            return "(" + rec(left) + wff[opIndex] + rec(right) + ")"

 print(rec("a>b>c"))

 def rec(wff): result = [] ind = [] # list storing indexes of connectives used if len(wff) == 1: return wff else: for i in range(numConnectives(wff)): opIndex = findConnective(wff, ind) # index where the operator is at right = createRight(wff, opIndex) # right formula # the first time it goes through, right is b>c # then right is c left = createLeft(wff, opIndex) # left formula # left is a # then it is b return "(" + rec(left) + wff[opIndex] + rec(right) + ")" 

  print(rec("a>b>c")) 

我的输出是(a>(b>c)) ，它应该是(a>(b>c)) AND ((a>b)>c) 。 发生这种情况是因为递归函数内部的循环从不选择第二个运算符来执行递归调用。 当return语句在for循环之外时，输出为((a>b)>c)

我如何使它功能遍历所有操作符（也就是每个函数调用执行整个循环）

Answer 1

尽管rec() for循环的return是特定问题，但我认为整体问题是你使问题变得比它需要的更难。 你还在你的处理是不一致的connectives ，有时它的字符，集合range(len(connectives))有时它的单个字符， wff[i] == connectives[j] 这是我对代码的简化：

connectives = {'>'}

def findConnectives(wff):
    ''' returns index of wff '''

    if wff is None or len(wff) <= 1:
        yield -1  # it's an atomic
    else:
        for i, character in enumerate(wff):  # looping through all chars in wff
            if character in connectives:  # if the wff contains the connective
                yield i

def createLeft(wff, opIndex):

    ''' returns what's on left of operator '''

    return wff[:opIndex]

def createRight(wff, opIndex):

    ''' returns what's on right of operator '''

    return wff[opIndex + 1:]

def rec(wff):
    if len(wff) == 1:
        return [wff]

    result = []

    for opIndex in findConnectives(wff):
        if opIndex == -1:
            break

        left = createLeft(wff, opIndex) # left formula

        right = createRight(wff, opIndex)  # right formula

        for left_hand in rec(left):
            for right_hand in rec(right):
                result.append("(" + left_hand + wff[opIndex] + right_hand + ")")

    return result

print(rec("a>b>c"))

OUTPUT

% python3 test.py
['(a>(b>c))', '((a>b)>c)']
%