给定一个字符串,波兰表示法中最长的 WFF 之一的长度是多少?
Given a String, What is the Length of the One of the Longest WFF in Polish Notation?
我正在 to write Python 中 WFF 'N Proof 游戏(无意侵犯版权)中一直很受欢迎的 Count-A-WFF 部分的一个版本。好吧,人气不高
我想我已经准备好了所有内容,并且 运行 根据需要准备了最多 4 个字母的字符串。
def maximum_string(s):
if cs(s) == True:
return len(s)
elif len(s) == 2:
l1 = [cs(s[0]), cs(s[1])]
if True in l1:
return len(s) - 1
else:
return 0
elif len(s) == 3:
first = s[0] + s[1]
second = s[0] + s[2]
third = s[1] + s[2]
l1 = [cs(first), cs(second), cs(third)]
if True in l1:
return len(s) - 1
l2 = [cs(s[0]), cs(s[1]), cs(s[2])]
if True in l2:
return len(s) - 2
else:
return 0
elif len(s) == 4:
first = s[0]+s[1]+s[2]
second = s[0]+s[1]+s[3]
third = s[1]+s[2]+s[3]
fourth = s[0]+s[2]+s[3]
l1 = [cs(first), cs(second), cs(third), cs(fourth)]
if True in l1:
return 3
first = s[0] + s[1]
second = s[0] + s[2]
third = s[0] + s[3]
fourth = s[1] + s[2]
fifth = s[1] + s[3]
sixth = s[2] + s[3]
l2 = [cs(first), cs(second), cs(third), cs(fourth), cs(fifth), cs(sixth)]
if True in l2:
return 2
first = s[0]
second = s[1]
third = s[2]
fourth = s[3]
l3 = [cs(first), cs(second), cs(third), cs(fourth)]
if True in l3:
return 1
else:
return 0
def cs(string):
global length_counter, counter, letter
counter = 1
length_counter = 0
letters_left = len(string)
while letters_left != 0 and length_counter < len(string):
letter = string[length_counter]
if letter == 'C' or letter == 'A' or letter == 'K' or letter == 'E' or letter == "K":
counter += 1
elif letter == 'N':
counter += 0
else:
counter -= 1
length_counter += 1
letters_left -= 1
if counter == 0 and len(string) == length_counter:
return True
else:
return False
maximum_string 辅助函数的目的是,给定任何字符串 S,找到最长可能的 wffs 之一的长度,您可以仅从 S 的字母组成。当然,我可以继续我目前拥有的 maximum_string 辅助函数的模式最大长度为 13。但是,组合爆炸是显而易见的。那么,有没有更优雅的方法来完成最大字符串辅助函数?
实际上,我之前使用的函数之一会 return 一个字符串与波兰表示法排列的距离。因此,这比我预期的更容易修复。这就是我要找的东西:
def maximum_string(string):
global length_counter, counter, letter
counter = 1
length_counter = 0
letters_left = len(string)
while letters_left != 0 and length_counter < len(string):
letter = string[length_counter]
if letter == 'C' or letter == 'A' or letter == 'K' or letter == 'E' or letter == "K":
counter += 1
elif letter == 'N':
counter += 0
else:
counter -= 1
length_counter += 1
letters_left -= 1
if ('p' in string) or ('q' in string) or ('r' in string) or ('s' in string) or ('t' in string) or ('u' in string):
return len(string) - abs(counter)
else:
return 0
我正在 to write Python 中 WFF 'N Proof 游戏(无意侵犯版权)中一直很受欢迎的 Count-A-WFF 部分的一个版本。好吧,人气不高
我想我已经准备好了所有内容,并且 运行 根据需要准备了最多 4 个字母的字符串。
def maximum_string(s):
if cs(s) == True:
return len(s)
elif len(s) == 2:
l1 = [cs(s[0]), cs(s[1])]
if True in l1:
return len(s) - 1
else:
return 0
elif len(s) == 3:
first = s[0] + s[1]
second = s[0] + s[2]
third = s[1] + s[2]
l1 = [cs(first), cs(second), cs(third)]
if True in l1:
return len(s) - 1
l2 = [cs(s[0]), cs(s[1]), cs(s[2])]
if True in l2:
return len(s) - 2
else:
return 0
elif len(s) == 4:
first = s[0]+s[1]+s[2]
second = s[0]+s[1]+s[3]
third = s[1]+s[2]+s[3]
fourth = s[0]+s[2]+s[3]
l1 = [cs(first), cs(second), cs(third), cs(fourth)]
if True in l1:
return 3
first = s[0] + s[1]
second = s[0] + s[2]
third = s[0] + s[3]
fourth = s[1] + s[2]
fifth = s[1] + s[3]
sixth = s[2] + s[3]
l2 = [cs(first), cs(second), cs(third), cs(fourth), cs(fifth), cs(sixth)]
if True in l2:
return 2
first = s[0]
second = s[1]
third = s[2]
fourth = s[3]
l3 = [cs(first), cs(second), cs(third), cs(fourth)]
if True in l3:
return 1
else:
return 0
def cs(string):
global length_counter, counter, letter
counter = 1
length_counter = 0
letters_left = len(string)
while letters_left != 0 and length_counter < len(string):
letter = string[length_counter]
if letter == 'C' or letter == 'A' or letter == 'K' or letter == 'E' or letter == "K":
counter += 1
elif letter == 'N':
counter += 0
else:
counter -= 1
length_counter += 1
letters_left -= 1
if counter == 0 and len(string) == length_counter:
return True
else:
return False
maximum_string 辅助函数的目的是,给定任何字符串 S,找到最长可能的 wffs 之一的长度,您可以仅从 S 的字母组成。当然,我可以继续我目前拥有的 maximum_string 辅助函数的模式最大长度为 13。但是,组合爆炸是显而易见的。那么,有没有更优雅的方法来完成最大字符串辅助函数?
实际上,我之前使用的函数之一会 return 一个字符串与波兰表示法排列的距离。因此,这比我预期的更容易修复。这就是我要找的东西:
def maximum_string(string):
global length_counter, counter, letter
counter = 1
length_counter = 0
letters_left = len(string)
while letters_left != 0 and length_counter < len(string):
letter = string[length_counter]
if letter == 'C' or letter == 'A' or letter == 'K' or letter == 'E' or letter == "K":
counter += 1
elif letter == 'N':
counter += 0
else:
counter -= 1
length_counter += 1
letters_left -= 1
if ('p' in string) or ('q' in string) or ('r' in string) or ('s' in string) or ('t' in string) or ('u' in string):
return len(string) - abs(counter)
else:
return 0