Question

我正在构建一个程序来将括号恢复为句子，以使其成为格式正确的公式（句子逻辑中的WFF）。例如，

a

a > b

(a > b)

a > b > c

((a > b) > c)

(a > (b > c))

等等...

此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"))

我的输出为(a>(b>c))，而应为(a>(b>c))和((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)

我如何做到这一点，使函数遍历所有运算符（又称每个函数调用都执行整个循环）

Answer 1

可以共享numConnectives（），findConnective（）等信息吗？要回答您的问题，需要查看您的代码，对其进行测试，然后尝试以其他方式从头开始生成整个内容来解决该问题。

Answer 2

尽管return的{{1}}循环中的for是特定的问题，但我认为总体问题是您使问题变得更加棘手。您对rec()的处理也不一致，有时是connectives的字符集合，有时是range(len(connectives))的单个字符。这是我对您的代码的简化：

wff[i] == connectives[j]

输出

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"))

循环永远不会在递归函数中完成

2 个答案: