Deterministic Finite Automata for {0,1} Every5-Two0

Question

Question:

Define a DFA that accepts all strings over {0,1} such that every block of five consecutive positions contains at least two 0s. Please read the question carefully. Ask yourselves: Does this allow e (epsilon (empty string)) to be accepted? How about 0101? Such English descriptions are found in various books, and I want to make sure you know how to read and interpret.

Instructor Hint: "The "blocks of 5" DFA can be programmatically generated without much trouble. I did it both ways (by hand and programmatically). Because I'm good with Emacs and Keyboard Macros, I could do even the 'by hand' mechanically and quite fast. But programmatic is less error-prone and compact."

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I'm drawing this thing out, and I think I'm doing it wrong, as it is getting out of control.

My sketch of the DFA before I make it in python:

However, this isn't right, because index 2, 3, 4, 5, and 6 constitute a block of five consecutive positions, so I need to account for at least two zeroes in that. Oh, great, and I have been thinking it needs two 1s, not two 0s. Am I going about this the entirely wrong way? Because the way I'm thinking, this is going to have a huge amount of states.

(goes back to drawing this large DFA)

Answer 1

If you want to do it the old school way:

def check(s):
    buffer = s[:5]
    i = 5
    count0, count1 = 0, 0
    while i < len(s):
        if len(buffer) == 5:
            first = buffer[0]
            if first == '0':
                count0 -= 1
            else:
                count1 -= 1
            buffer = buffer[1:]
        buffer += s[i]
        if buffer[-1] == '0':
            count0 += 1
        else:
            count1 += 1
        if count0 < 2:
            return "REJECT"
        i += 1
    if buffer.count('0') >= 2:
        return "ACCEPT"
    else:
        return "REJECT"

A slightly smarter way:

def check(s):
    return all(ss.count('0')>=2 for ss in (s[i:i+5] for i in xrange(len(s)-4)))

The verbose code of the above method:

def check(s):
    subs = (s[i:i+5] for i in xrange(len(s)-4))
    for sub in subs:
        if sub.count('0') < 2:
            return "REJECT"
    return "ACCEPT"

Haven't tested this code, but it should likely work. Your professor probably wants the third method.

Hope this helps