这些用于检测有限语法歧义的Python程序是否正确？

Question

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，我也不确定“官方”解决方案是否正确。在

问题的简要描述：编写一个函数isambig（grammar，start，string），它接受有限上下文无关语法（编码为python字典）、语法的开始符号和字符串。如果有两个解析树指向字符串，那么语法是不明确的（或者至少这是我对歧义的理解——如果我弄错了，请纠正我）。如果语法不明确，则返回True。否则返回False。在

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

我添加了我自己的测试用例。我不确定这是否正确，但我只能看到一个可能的解析树，导致字符串“ab”，因此字符串不能证明语法不明确。我不认为语法是含糊不清的。在

^{pr2}$

以下是“官方”计划：

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，更长的计划：

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，包括我添加的测试用例（grammar4），但是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，要么官方的解决方案是错误的。在

官方的解决方案正确吗？我的解决方案正确吗？在

Answer 1

在我看来，grammar4并不含糊。 只有一个解析树：

S -> PQ
P -> a
Q -> b

    S
    |
 ___|____
P        Q
|        |
a        b

然而，官方的程序说它是模棱两可的，因为它使用了规则 P -> a和{}连续：

^{pr2}$

（现在有两个规则序列0,1,2和{}。）

因此“官方”程序似乎错误地检测到grammar4是不明确的。在

更新：我查看了您的代码，并做了一些测试，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模棱两可和 毫不含糊。在

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“暧昧”（与“官方”版本一样）

如果删除("S", ["B", "A"])，则您的版本正确 切换到“not digulary”，而另一个版本仍然返回“dimengious” （我们又回到了grammar4案件。）

也许其他人（比我更有经验）可以插话。在

更新2:Ira Baxter提到，是否 上下文无关语法是模糊的。在

另请参见How is proving a context free language to be ambiguous undecidable?