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))
。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)