CS250      Lab2 - Nondeterministic Finite Automata

Objectives:

• Understand the concept of nondeterminism
• See that NFAs and DFAs are equivalent

1. Download the lab2 files and start JFLAP 6.0.  Open file file ex2.1.  If you go to Test- Highlight Nondeterminism you will see that in state q0 we have a choice of either going to state q1 or state q3 on an input of a. 
   
   Use the step by step run to test the input aa.  After one step we can be in either q1 or q3.  After one more step we can be in either q2 or q4.  Since q2 is an accepting state, the input aa is accepted.  Notice that it does not matter that the other choice is not accepting.  An NFA accepts if there EXISTS a choice which leads to an accepting state.
   
   Try the step by step again with the input aaa.  Does it accept?
   Try the step by step again with the input aaaa.  Does it accept?
    

   Clearly and succinctly describe the language that is accepted by the NFA in ex2.1

    

2. Open file ex2.2.  This NFA has a lambda transition.  That means that there is a choice to move to that state without consuming any input.  Test this NFA on a number of input strings to see how it behaves.
    

   Give 3 inputs in the alphabet {a,b} which are accepted by ex2.2 and 3 inputs which are rejected.

    

3. Open file ex2.3.  This NFA has a single transition that can either consume two a's or three a's.  We want an FA which recognizes the language L of any string of a's of some length divisible by 2 or 3. Does this FA achieve this?
    

   Produce at least 4 strings which either are in the language L but are not accepted, or are not in L and are accepted.  Modify the FA so it actually recognizes the language L.  Simulate those same strings again to verify your modification.

    

4. Any language that can be accepted by an NFA can also be accepted by a DFA.  The algorithm for converting an NFA to an equivalent DFA is given on p59 of your text but JFLAP can assist you with the process. 
   
   Go back to ex2.2.  In the editor window, select Convert to DFA.  This will bring up a new split screen with the NFA on the left.  If it is too tiny, simply drag the vertical line between the screens to make it larger.  On the right we will build the DFA.  You are given an initial state.  The subscripts of 0,1 on the state indicate that initially we can be in either state 0 or state 1 of our NFA without consuming any input. Do you see why this is so?
   
   To add the next state click on the Expand on Terminal tool and click and hold on q0 and drag to a location for the next state.  You will be asked to enter a terminal and type a.  You will then be asked to enter what states can be reached from either 0 or 1 of the NFA on an input of a.  ----- The answer is q2 and q3.  Do you see why?  Type 2,3 and JFLAP will add that state.
   
   Continue the process of adding new states.  If you press the Done? button, JFLAP will inform you of what you have left to do, if anything.
   
   Test out the DFA to convince yourself that it accepts the same language.

    

5. An NFA recognizes language L if for each string w in L there exists of a path labeled w to an accepting state. How would we define the complement of L?  Exercises 5 and 6 on p62 of your text attempt to do this, although only one of them is correct.  You may have some trouble reading the mathematical notation but exercise 5 says that none of the paths labeled w lead to accepting states. Exercise 6 says that there exists a path labeled w which leads to a non-accepting state.  Read the notation carefully so that you understand what it is saying. 
    

   For each of exercise 5 and 6 on p62, either explain why it correctly describes the complement of L or give a counterexample (some NFA M whose complement does not meet the definition). Be sure to carefully explain your answers in each case.