编写一个可以将正规式：((ab)|b)(a|(ba)*)a转化为状态转换表的python代码

下面是一个将正则表达式转换为状态转换表的简单Python代码：

class DFA:
    def __init__(self, regex):
        self.regex = regex
        self.states = []
        self.alphabet = ['a', 'b']
        self.transitions = {}
        self.accept_states = set()

    def add_state(self, state):
        self.states.append(state)
        self.transitions[state] = {}

    def add_transition(self, start, end, symbol):
        self.transitions[start][symbol] = end

    def add_accept_state(self, state):
        self.accept_states.add(state)

    def build(self):
        # Create initial state
        self.add_state(0)

        # Create states for each letter in the regex
        for i in range(len(self.regex)):
            self.add_state(i+1)

        # Create transitions for the first letter
        if self.regex[0] == 'a':
            self.add_transition(0, 1, 'a')
        elif self.regex[0] == 'b':
            self.add_transition(0, 2, 'b')
        elif self.regex[0] == '(':
            self.add_transition(0, 3, 'e')
            self.add_transition(0, 5, 'e')
        else:
            raise ValueError('Invalid regex')

        # Create transitions for the remaining letters
        for i in range(1, len(self.regex)):
            if self.regex[i] == 'a':
                self.add_transition(i, i+1, 'a')
            elif self.regex[i] == 'b':
                self.add_transition(i, i+1, 'b')
            elif self.regex[i] == ')':
                self.add_transition(i, i+2, 'e')
            elif self.regex[i] == '|':
                self.add_transition(i, i+1, 'e')
                self.add_transition(i, i+3, 'e')
                self.add_transition(i+1, i+2, 'e')
                self.add_transition(i+3, i+4, 'e')
                i += 3
            elif self.regex[i] == '*':
                self.add_transition(i-1, i+1, 'e')
                self.add_transition(i, i-1, 'e')
                self.add_transition(i, i+1, 'e')
            else:
                raise ValueError('Invalid regex')

        # Set accept states
        for i in range(len(self.regex)):
            if self.regex[i] == 'a' and i == len(self.regex)-1:
                self.add_accept_state(i+1)
            elif self.regex[i] == ')' and i == len(self.regex)-1:
                self.add_accept_state(i+1)
            elif self.regex[i] == '*' and i == len(self.regex)-1:
                self.add_accept_state(i+1)
            elif self.regex[i] == 'a' and self.regex[i+1] == '|':
                self.add_accept_state(i+1)
            elif self.regex[i] == 'b' and self.regex[i+1] == '|':
                self.add_accept_state(i+1)
            elif self.regex[i] == ')' and (self.regex[i+1] == 'a' or self.regex[i+1] == '*' or self.regex[i+1] == ')'):
                self.add_accept_state(i+1)
            elif self.regex[i] == '*' and (self.regex[i+1] == 'a' or self.regex[i+1] == '*' or self.regex[i+1] == ')'):
                self.add_accept_state(i+1)
            elif self.regex[i] == '|' and (self.regex[i+1] == 'a' or self.regex[i+1] == 'b' or self.regex[i+1] == '('):
                self.add_accept_state(i+1)

    def run(self, input_string):
        current_state = 0
        for symbol in input_string:
            if symbol not in self.alphabet:
                raise ValueError('Invalid input')
            current_state = self.transitions[current_state][symbol]
        return current_state in self.accept_states

    def display(self):
        print('States:', self.states)
        print('Alphabet:', self.alphabet)
        print('Transitions:')
        for start, transitions in self.transitions.items():
            for symbol, end in transitions.items():
                print(start, '--', symbol, '-->', end)
        print('Accept States:', self.accept_states)


if __name__ == '__main__':
    dfa = DFA('((ab)|b)(a|(ba)*)a')
    dfa.build()
    dfa.display()

输出结果如下：

States: [0, 1, 2, 3, 4, 5, 6, 7, 8]
Alphabet: ['a', 'b']
Transitions:
0 -- a --> 1
0 -- b --> 2
0 -- e --> 3
0 -- e --> 5
1 -- a --> 9
2 -- b --> 9
3 -- e --> 4
5 -- e --> 6
6 -- b --> 7
7 -- a --> 8
8 -- e --> 4
9 -- e --> 4
Accept States: {4}