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CSC 236            Sample Solutions for Week 13 Tutorial          Fall 2007
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A.  Construct a NFA equivalent to RE: (b+d)* (a(e + b(b+d)*))*.

    Q:  First thing to do?
    A:  Eliminate 'e' and '{}' from RE: (b+d)* (a + ab(b+d)*)*.

    Q:  Next?
    A:  Start with:

                     (b+d)* (a + ab(b+d)*)*
            --> q_0 ----------------------->(q_1)

    Q:  Next?
    A:  Remove operators one by one.

    First, concatenation:

                 (b+d)*       (a + ab(b+d)*)*
        --> q_0 -------> q_2 ---------------->(q_1)

    Next, do (b+d)*:

            b+d               (a + ab(b+d)*)*
        --> q_0          q_2 ---------------->(q_1)
               \_____________________________/|
                       (a + ab(b+d)*)*

            b+d  (a + ab(b+d)*)*
        --> q_0 ---------------->(q_1)

    (because q_2 has no more incoming transitions).

    Next, remove the next '*': it's not the only transition out of q_0
    (because of loop), so:

            b+d  a + ab(b+d)*
        -->(q_0)------------->(q_2) a + ab(b+d)*    (q_1)

    Drop q_1 (no more incoming transitions), and replace each '+' with two
    transitions:

                    a
            b,d --------->  a
        -->(q_0)          (q_2) ab(b+d)*
                --------->
                 ab(b+d)*

    Break up concatenations:

            b,d   a    a  (b+d)*
        -->(q_0)---->(q_2)<---- q_6
             |        /|  \_    /|
            a|   (b+d)*|   a\_  b|
            \|         |      \| |
            q_3 ----> q_4       q_5
                  b

    Remove '*' from the transition q_4 -> q_2:

            b,d   a    a  (b+d)*
        -->(q_0)---->(q_2)<---- q_6
             |        /|  \_    /|
            a|        a|   a\_  b|
            \|         |      \| |
            q_3 ---->(q_4)----> q_5
                  b   b,d   a

    Remove '*' from the transition q_6 -> q_2:

            b,d   a    a    a   b,d
        -->(q_0)---->(q_2)<----(q_6)
             |        /|  \_    /||
            a|        a|   a\_  b||a
            \|         |      \| ||/
            q_3 ---->(q_4)----> q_5
                  b   b,d   a

    Done!
    Compare this with the equivalent DFA:

                 a
            b,d --->
        -->(q_0)    (q_1) a
                <---
                  b

    Moral:  Sometimes, a little thinking is better than a lot of mechanical
    computation! :)

B.  Review!  Bring any question about the course material (mostly for the
    second half).