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我正在构建一个程序来恢复句子的括号，以使它们成为格式正确的公式（句子逻辑中的 WFF）。例如，

    - 这句话a是一个WFF。
    - 该句子a > b只有一种方法可以恢复括号以使其成为 WFF，即(a > b).
    - 该句子a > b > c有两种方法可以恢复括号以使其成为 WFF - 要么((a > b) > c)要么(a > (b > c))。
ETC...

该算法有一个迭代和递归元素

# returns index of wff
def findConnective(wff, indexes):
    if len(wff) == None:
        return -1
    if (len(wff) <= 1):
        return -1                                   # it's an atomic

    for i in range(len(wff)):                       # looping through all chars in wff
        if set([i]) & set(indexes):                     # if operator has already been used
            continue
        else:                                           # if operator has not been usedl
            for j in range(len(connectives)):           # looping through all of the connectives
                if wff[i] == connectives[j]:            # if the wff contains the connective
                    indexes.append(i)                   # keeps track of which operators have already been used
                    return i

# returns what's on left of operator
def createLeft(wff, opIndex):
    if opIndex == -1:
        return wff          # return the atomic
    else:
        return wff[:opIndex]

# returns what's on right of operator
def createRight(wff, opIndex):
    if opIndex == -1:
        return wff          # return the atomic
    else:
        return wff[opIndex+1:]

# 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"))

我的输出(a>(b>c))应该是(a>(b>c))AND ((a>b)>c)。这是因为递归函数内部的循环永远不会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)

如何使函数通过所有运算符（也就是为每个函数调用执行整个循环）

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1 回答 1

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尽管循环中return是特定问题，但我相信总体问题是您使问题变得比需要的更难。您在处理 时也不一致，有时它是字符的集合，有时是单个字符，。这是我对您的代码的简化：forrec()connectivesrange(len(connectives))wff[i] == connectives[j]

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"))

输出

% python3 test.py
['(a>(b>c))', '((a>b)>c)']
%
于 2019-06-03T16:27:11.143 回答