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建立一个特里树。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

但是卡在了其他部分......还没弄明白如何写这个特丽树。任何线索？

Answer 1

你可能想要使用python中有多个解析器库。我使用过LARK parser。他们对各种python解析器进行了比较。

在大学期间，我在C中实现了一个LALR（1）解析器。我猜它的用处不大。如果你想再次编写整个解析器，我在python here中找到了一个有用的实现。我还没有测试过该代码的工作情况。

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)

希望它有所帮助！

如何构建trie树来解决这个解析算法

1 个答案: