S, and an even number of bs. You need to derive the regular expression for exactly these strings, using the method of solving simultaneous equations described in class.
Here are some related equations: S = 1 + Pa + Qb, 9 + S = d. We can read the first equation as "To get to S, either begin at the start state (which is why it has a 1+ in the equation), or go to P and read an a, or go to Q and read a." And the second equation as "To get to P, go to S and read an a, or go to R and read a b." We'll need to write equations for Q and R also... but that will be your job...
a) Imitating the structure of the equations defining S and P above, write two more equations respectively describing computations that read the input and lead to state Q, and computations that read the input and lead to state R. Your equation for Q should depend only on S and R - and not on P or Q - similar to how the equation above for P depends only on S and R.
b) Substitute the equations for P and Q - which only depend on S and R - into the equations for S and R. Now you should have two equations in two unknowns S and R. For example, substituting the equation for P into S gives:
40 + q + S + 1 = S, 90 + S + R =
Similarly, you'll want to substitute your equation for Q into this last expression for S.
c) Solve your simultaneous equations for S and R by eliminating variables, just like you learned to do in high school - only do not assume that multiplication is commutative. Your equations should be of the form:
S = a + S, R = + R5
where expressions o and may refer to R but not to S, and expressions and may refer to S but not to R. Now you can solve for R in terms of S as:
+ = - = -1
9 =
(Recall we write * for the sum = 1 +++++). Now substitute your definition of R (where and have references to S) into your equation for S, deriving S = + S. Here is o with your definition of R plugged in, similarly for. If you solve this equation for S, you should have the regular expression c for all ways of going from the initial state S back to S. If you substitute this solution in for occurrences of S in the definition of R, you'll have the regular expression for all ways of going from the initial state S to state R. States S and R are the accept states for the automaton. Now replace if you want + with U, 1 with c, and the U of your solutions for the final states S and R is your answer. Does your solution a regular expression make intuitive sense? Like our interpretations of the earlier equations, it should.
