Python中{ab，abc}*的非确定性有限自动机

Question

我一直在尝试画这个非确定性有限自动机：

NFA语言{ab，abc}*的州数不超过3个，我得出了下图中的解决方案：

NFA diagram

问题似乎在于代码逻辑，因为我的代码总是打印“拒绝”。如果有人能给出一些关于如何编写此算法的提示，我将不胜感激

print("Insert below the string: ")
string = str(input())


def state_q0(string):
    if len(string) == 0:
        print("string not accepted")
    else:
        if string[0] == "a":
            state_q1(string[1:])
        elif string[0] == "b":
            print("rejected")


def state_q1(string):
    if len(string) == 0:
        print("string not accepted")
    else:
        if string[1] == "b":
            state_q2(string[2:])
        else:
            print("rejected")


def state_q2(string):
    if len(string) == 0:
        print("string not accepted")
    else:
        if string[2] == "c":
            print("accepted -> q0")
        elif string[2] == "b":
            print("rejected")


state_q0(string)

Answer 1

您应该始终查看字符串的第一个字符，并使用除第一个字符以外的所有内容调用递归调用

在您的图表中也没有注意到，但是我假设q0是接受状态，因为它对应于(ab + abc)*

因此，以您的风格（我个人不会使用，但可以）：

def q0(s):
    if not s: return "accept"  # Accepting state.
    if s[0] == "a": return q1(s[1:])
    return "reject"

def q1(s):
    if not s: return "reject"
    # NFA, must try both edges.
    if (s[0] == "b" and q0(s[1:]) == "accept" or
        s[0] == "b" and q2(s[1:]) == "accept"):
        return "accept"
    return "reject"

def q2(s):
    if not s: return "reject"
    if s[0] == "c": return q0(s[1:])
    return "reject"

然而，我将如何对NFA进行编码：

transitions = [
    {"a": {1}},
    {"b": {0, 2}},
    {"c": {0}}
]
starting_state = 0
accepting_states = {0}

def nfa(w):
    cur_states = {starting_state}
    for c in w:
        if not cur_states: return "reject"
        cur_states = set.union(*
            (transitions[s].get(c, set()) for s in cur_states))
    return "accept" if cur_states & accepting_states else "reject"