CODEWITHSUNDEEP
pythonpython-3.xdfanfa
Flavio Andrade
Flavio Andrade

Reputation: 15

Non-Deterministic Finite Automata of {ab, abc}* in Python

I have been trying to draw this Non-Deterministic Finite automaton:

NFA with the number of states not exceeding 3 for the language {ab, abc}*, and I reached the solution in the picture below:

NFA diagram

The issue seems to be the code logic since my code always prints "rejected. I will appreciate it if someone can give some hints on how to code this algorithm.

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)

Upvotes: 1

Views: 1361

Answers (1)

orlp
orlp

Reputation: 117771

You should always look at the first character of the string, and call recursive calls using everything except the first character.

Also it's not noted in your diagram, but I assume that q0 is the accepting state since that corresponds to (ab + abc)*.

So in your style (which I personally wouldn't use, but ok):

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"

How I would code the NFA however is like such:

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"

Upvotes: 2

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