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如何建立特里树来解决这个解析算法

[英]How to build trie tree to solve this parse algorithm

我正在尝试使用特里树来解决这个问题:

Symbol string generator consists of two parts, a set of the start symbol and a set of rules of generation.
For example:
Start symbol: ['S'], Rules of generation: ["S → abc", "S → aA", "A → b", "A → c"]
Then, symbolic string abc can be generated because S → abc. 
Symbolic string ab can be generated because S → aA → ab.
Symbolic string abc can be generated because S → aA → ac.
Now, give you a symbolic string generator and a symbolic string, and you need to return True if the symbolic string can be generated, False otherwise

Example
Given generator = ["S -> abcd", "S -> Ad", "A -> ab", "A -> c"], startSymbol = S, symbolString = “abd”, return True.

explanation:
S → Ad → abd

Given generator = ["S → abc", "S → aA", "A → b", "A → c"], startSymbol = S, symbolString = “a”, return False

我发现此问题的关键是建立特里树。 我试图写:

def build_trie(values): #value is like ['abc', 'Ad'...]
    root = {}
    for word in values:
        current = root
        is_end = False
        for c in word:
            if 'A' <= c <= 'Z':
                vals = m[c] #m is a mapping of {'S': ['abc', 'Ad'], ...}
                rs = build_trie(vals)
                for k in rs:
                    if k not in current:
                        current[k] = rs[k]
                    else:
                        # stuck here...
                        pass

                        # temp = collections.defaultdict(dict)
                        # for d in (current[k], rs[k]):
                        #     for k, v in d.items():
                        #         if k in temp and k != '__end__':
                        #             temp[k].update(v)
                        #         else:
                        #             temp[k] = v
                        # # current[k].update(rs[k])
                        # current[k] = temp[k]
                is_end = True
            else:
                current = current.setdefault(c, {})
                is_end = False
        if not is_end:
            current['__end__'] = '__end__'
    return root

但是卡在其他部分上...还没有弄清楚如何编写这棵特里树。 有什么线索吗?

您可能要使用python中的多个解析器库。 我使用了LARK解析器 他们给出了各种python解析器的比较。

在大学期间,我在C语言中实现了LALR(1)解析器。我想它的用处会更少。 我发现了一个Python有用的实现在这里 ,如果你想重新写入整个分析器。 我还没有测试该代码的工作原理。

对于给定的语法,我使用LARK编写了一个验证器,如下所示。

from lark import Lark
import sys

grammar = """
        start: "abcd"
         | A "d"
        A: "ab"
         | "c"
        """

parser = Lark(grammar)

def check_grammer(word):
    try:
            parser.parse(word)
            return True
    except Exception as exception:
            print exception
            return False



word = sys.argv[1]
print check_grammer(word)

希望能帮助到你!

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