我一直在做Udacity CS262，对于检测歧义问题，我不确定我的解决方案是否正确，也不确定"官方"解决方案是否准确。
问题的简要描述：编写一个函数isambig（grammar，start，string），它采用有限上下文无关语法（编码为python字典）、语法的起始符号和字符串。如果有两个解析树指向字符串，那么语法是模糊的（或者至少这是我对模糊性的理解——如果我错了，请纠正我）。如果语法不明确，则返回True。否则返回False。
测试用例：
grammar1 = [
("S", [ "P", ]),
("S", [ "a", "Q", ]) ,
("P", [ "a", "T"]),
("P", [ "c" ]),
("Q", [ "b" ]),
("T", [ "b" ]),
]
print isambig(grammar1, "S", ["a", "b"]) == True
print isambig(grammar1, "S", ["c"]) == False
grammar2 = [
("A", [ "B", ]),
("B", [ "C", ]),
("C", [ "D", ]),
("D", [ "E", ]),
("E", [ "F", ]),
("E", [ "G", ]),
("E", [ "x", "H", ]),
("F", [ "x", "H"]),
("G", [ "x", "H"]),
("H", [ "y", ]),
]
print isambig(grammar2, "A", ["x", "y"]) == True
print isambig(grammar2, "E", ["y"]) == False
grammar3 = [ # Rivers in Kenya
("A", [ "B", "C"]),
("A", [ "D", ]),
("B", [ "Dawa", ]),
("C", [ "Gucha", ]),
("D", [ "B", "Gucha"]),
("A", [ "E", "Mbagathi"]),
("A", [ "F", "Nairobi"]),
("E", [ "Tsavo" ]),
("F", [ "Dawa", "Gucha" ])
]
print isambig(grammar3, "A", ["Dawa", "Gucha"]) == True
print isambig(grammar3, "A", ["Dawa", "Gucha", "Nairobi"]) == False
print isambig(grammar3, "A", ["Tsavo"]) == False
我添加了自己的测试用例。我不确定这是否正确，但我只能看到可能的一个解析树导致字符串"a b"，因此该字符串不会证明语法不明确。我不认为语法含糊不清。
grammar4 = [ # Simple test case
("S", [ "P", "Q"]),
("P", [ "a", ]),
("Q", [ "b", ]),
]
print isambig(grammar4, "S", ["a", "b"]) == False
以下是"官方"程序：
def expand(tokens_and_derivation, grammar):
(tokens,derivation) = tokens_and_derivation
for token_pos in range(len(tokens)):
for rule_index in range(len(grammar)):
rule = grammar[rule_index]
if tokens[token_pos] == rule[0]:
yield ((tokens[0:token_pos] + rule[1] + tokens[token_pos+1:]), derivation + [rule_index])
def isambig(grammar, start, utterance):
enumerated = [([start], [])]
while True:
new_enumerated = enumerated
for u in enumerated:
for i in expand(u,grammar):
if not i in new_enumerated:
new_enumerated = new_enumerated + [i]
if new_enumerated != enumerated:
enumerated = new_enumerated
else:
break
result = [xrange for xrange in enumerated if xrange[0] == utterance]
print result
return len(result) > 1
这是我自己的，更长的程序：
def expand(grammar, symbol):
result = []
for rule in grammar:
if rule[0] == symbol:
result.append(rule[1])
return result
def expand_first_nonterminal(grammar, string):
result = []
for i in xrange(len(string)):
if isterminal(grammar, string[i]) == False:
for j in expand(grammar, string[i]):
result.append(string[:i]+j+string[i+1:])
return result
return None
def full_expand_string(grammar,string, result):
for i in expand_first_nonterminal(grammar,string):
if allterminals(grammar,i):
result.append(i)
else:
full_expand_string(grammar,i,result)
def isterminal(grammar,symbol):
for rule in grammar:
if rule[0] == symbol:
return False
return True
def allterminals(grammar,string):
for symbol in string:
if isterminal(grammar,symbol) == False:
return False
return True
def returnall(grammar, start):
result = []
for rule in grammar:
if rule[0] == start:
if allterminals(grammar,rule[1]):
return rule[1]
else:
full_expand_string(grammar, rule[1], result)
return result
def isambig(grammar, start, utterance):
count = 0
for i in returnall(grammar,start):
if i == utterance:
count+=1
if count > 1:
return True
else:
return False
现在，我的程序通过了所有的测试用例，包括我添加的那个（grammar4），但官方解决方案通过了除了我添加的测试用例之外的所有测试用例。在我看来，要么测试用例是错误的，要么官方解决方案是错误的。
官方的解决方案正确吗？我的解决方案正确吗？
对我来说，grammar4似乎并不含糊。只有一个解析树：
S -> PQ
P -> a
Q -> b
S
|
___|____
P Q
| |
a b
然而，官方程序表示，由于它使用了规则，因此它是模棱两可的P -> a和Q -> b连续：
[(['a', 'b'], [0, 1, 2]), (['a', 'b'], [0, 2, 1])]
（现在有两个规则序列0,1,2和0,2,1。）
因此，"官方"程序似乎错误地检测到grammar4是模糊的。
更新：我查看了您的代码，做了一些测试，除了不处理递归（官方版本也不处理递归），你的程序似乎正确地区分了歧义和毫不含糊的
简单测试：
grammar5 = [
("S", ["A", "B"]),
("S", ["B", "A"]),
("A", ["a"]),
("B", ["a"]),
]
print(isambig(grammar5, "S", ["a", "a"]))
S -> AB
S -> BA
A -> a
B -> a
S
|
___|____
A B
| |
a a
S
|
___|____
B A
| |
a a
您的版本返回"不明确"（"官方"版本也是如此。）
如果删除("S", ["B", "A"])，则您的版本正确切换到"不含糊"，而另一个版本仍然返回"含糊"（我们回到grammar4案例。）
也许其他人（比我更有经验）可以插话。
更新2:Ira Baxter提到上下文无关语法是模棱两可的。
另请参阅如何证明一种上下文无关的语言是模糊的、不可判定的？