1. CSC312
Automata Theory
Lecture # 24
Chapter # 11 by Cohen
Decidability

2. Decidability Problems
Method-2: (Blue paint method)
To examine whether a certain FA accepts any word, it is required to seek paths from initial to final state. Following is the procedure to find such paths.
Step-1: Paint the start state Blue.
Step-2: From every blue state follow every edge that leads out of it and paint the destination state blue, then delete these edges from the machine.
Step-3: Repeat step-2 until no new state is painted blue, then stop.
Step-4: When the procedure stops, if any of the final states are painted blue, then the machine accepts some words, otherwise not.

3. Decidability Problems
Method-3:
If the FA has N state, then test all words fewer than N letters by running them on the FA. If the FA accepts none of them, then it accepts no word at all.
Example:

4. Decidability Problems
Finiteness:
How can we decide whether an FA, or RE, defines a finite or infinite language?
For a RE it is easy.
If the RE contains closure operator then it may define an infinite language, with the exception of *, as we know that * = 
Example:
(+a)(*+)* defines a finite language. While
(*+a)*(*+)* defines an infinite language.

5. If we want to decide this question for an FA, we could convert it to a regular expression OR we can use the following theorem.
Theorem 19:
Let F be an FA with N states, then
If F accepts an input string w such that
N  length(w)  2N
then F accepts an infinite language OR
2. If F accepts infinitely many words, then F accepts some word w such that
Practice questions Exercise Ch # 11 ,Q. No. 1-13

6. Remarks:
For a machine with N number of states, the total number of strings to be tested, defined over an alphabet  of m letters, is mN + mN+1 + … m2N-1. After testing all theses strings on the machine, if any one is accepted then the machine accepts infinite language otherwise not e.g. for a machine of 3 states and alphabet of two letters, = 56 strings are to be tested.