python - 如何构建trie树来解决这个解析算法

Question

我正在尝试使用 trie 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

我发现这个问题的关键点是构建一个 trie 树。我试着写:

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

但卡在了 else 部分......还没有弄清楚如何编写这棵特里树。有什么线索吗？

Answer 1

您可能想使用 Python 中的多个解析器库。 我用过LARK parser .他们对各种 python 解析器进行了比较。

在大学期间，我用 C 语言实现了一个 LALR(1) 解析器。我想它用处不大。我在 python here 中找到了一个有用的实现，如果你想再次编写整个解析器。我还没有测试该代码的工作情况。

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)

希望对您有所帮助!