用python实现上述要求

好的，我可以给你提供一个用 Python 实现上述要求的代码示例。这个示例代码实现了文法的 first 集、follow 集和 select 集的计算，以及判断是否为 LL(1) 文法，如果不是则尝试消除左递归并转化为 LL(1) 文法。

# 定义文法，这里以 CFG 为例
grammar = {
    'S': ['aB', 'bA'],
    'A': ['aS', 'bAA', 'c'],
    'B': ['bS', 'aBB', 'd']
}

# 计算每个非终结符的 first 集
def compute_first(grammar):
    first = {}
    for nonterminal in grammar.keys():
        first[nonterminal] = set()
    for nonterminal in grammar.keys():
        for production in grammar[nonterminal]:
            if production[0].isupper():
                first[nonterminal] |= first[production[0]]
            else:
                first[nonterminal].add(production[0])
    return first

# 计算每个非终结符的 follow 集
def compute_follow(grammar, first):
    follow = {}
    for nonterminal in grammar.keys():
        follow[nonterminal] = set()
    follow['S'].add('$')
    for nonterminal in grammar.keys():
        for production in grammar[nonterminal]:
            for i in range(len(production)):
                if production[i].isupper():
                    if i == len(production) - 1:
                        follow[production[i]] |= follow[nonterminal]
                    else:
                        j = i + 1
                        while j < len(production):
                            if production[j].isupper():
                                follow[production[i]] |= first[production[j]]
                                if 'ε' not in first[production[j]]:
                                    break
                            else:
                                follow[production[i]].add(production[j])
                                break
                            j += 1
    return follow

# 计算每个产生式的 select 集
def compute_select(grammar, first, follow):
    select = {}
    for nonterminal in grammar.keys():
        for production in grammar[nonterminal]:
            select[(nonterminal, production)] = set()
            if 'ε' in first[production[0]]:
                select[(nonterminal, production)] |= follow[nonterminal]
            for i in range(len(production)):
                if production[i].isupper():
                    select[(nonterminal, production)] |= first[production[i]]
                    if 'ε' not in first[production[i]]:
                        break
                else:
                    select[(nonterminal, production)].add(production[i])
                    break
    return select

# 判断是否为 LL(1) 文法
def is_LL1_grammar(grammar, select):
    for nonterminal in grammar.keys():
        for i in range(len(grammar[nonterminal])):
            for j in range(i + 1, len(grammar[nonterminal])):
                if len(select[(nonterminal, grammar[nonterminal][i])] & select[(nonterminal, grammar[nonterminal][j])]) > 0:
                    return False
    return True

# 消除左递归
def eliminate_left_recursion(grammar):
    new_grammar = {}
    nonterminals = list(grammar.keys())
    for i in range(len(nonterminals)):
        new_grammar[nonterminals[i]] = []
        for j in range(i):
            for production in grammar[nonterminals[i]]:
                if production.startswith(nonterminals[j]):
                    for new_production in new_grammar[nonterminals[j]]:
                        new_grammar[nonterminals[i]].append(production.replace(nonterminals[j], new_production))
                else:
                    new_grammar[nonterminals[i]].append(production)
        new_productions = []
        for production in new_grammar[nonterminals[i]]:
            if production.startswith(nonterminals[i]):
                new_nonterminal = nonterminals[i] + "'"
                new_productions.append(production[1:] + new_nonterminal)
                new_productions.append('ε')
                if new_nonterminal not in new_grammar:
                    new_grammar[new_nonterminal] = []
                for production2 in grammar[nonterminals[i]]:
                    new_grammar[new_nonterminal].append(production2[1:] + new_nonterminal)
            else:
                new_productions.append(production)
        new_grammar[nonterminals[i]] = new_productions
    return new_grammar

# 转化为 LL(1) 文法
def to_LL1_grammar(grammar):
    new_grammar = grammar
    while True:
        first = compute_first(new_grammar)
        follow = compute_follow(new_grammar, first)
        select = compute_select(new_grammar, first, follow)
        if is_LL1_grammar(new_grammar, select):
            return new_grammar, select
        else:
            new_grammar = eliminate_left_recursion(new_grammar)

# 测试代码
if __name__ == '__main__':
    first = compute_first(grammar)
    follow = compute_follow(grammar, first)
    select = compute_select(grammar, first, follow)
    print('First集：', first)
    print('Follow集：', follow)
    print('Select集：', select)
    if is_LL1_grammar(grammar, select):
        print('该文法是LL(1)文法')
    else:
        new_grammar = eliminate_left_recursion(grammar)
        new_grammar, select = to_LL1_grammar(new_grammar)
        print('消除左递归并转化为LL(1)文法：', new_grammar)
        print('新文法的Select集：', select)

以上是一个简单的实现示例，可以帮助你完成上述要求。