Find the minimal dfa that accepts L(a∗bb) ∪ L(ab∗ba).

Sure, let's break this down step by step.

1. First, let's understand the language L(a∗bb) ∪ L(ab∗ba). This language accepts all strings that either end with 'bb' after any number of 'a's or strings that start with 'ab', end with 'ba' and have any number of 'b's in between.

2. Now, let's construct the minimal DFA for L(a∗bb):

   • We start with state q0 which is the initial state.
   • If we get an 'a', we stay in state q0 because we can have any number of 'a's before 'bb'.
   • If we get a 'b', we move to state q1.
   • If we get another 'b', we move to state q2 which is the final state.
   • If we get an 'a' after a 'b', we go back to state q0 because the string must end with 'bb'.
   • If we get a 'b' in state q2, we stay in state q2 because the string can end with more than two 'b's.

3. Now, let's construct the minimal DFA for L(ab∗ba):

   • We start with state p0 which is the initial state.
   • If we get an 'a', we move to state p1.
   • If we get a 'b', we move to state p2.
   • If we get another 'b', we stay in state p2 because we can have any number of 'b's.
   • If we get an 'a', we move to state p3 which is the final state.
   • If we get a 'b' in state p3, we go back to state p2 because the string must end with 'ba'.
   • If we get an 'a' in state p2, we stay in state p2 because the string can have 'a's in the middle.

4. Now, we need to combine these two DFAs into one that accepts L(a∗bb) ∪ L(ab∗ba). We can do this by creating a new initial state and adding epsilon transitions to the initial states of the two DFAs. The final state is the union of the final states of the two DFAs.

5. Finally, we need to minimize this DFA. However, in this case, the DFA is already minimal because there are no equivalent states that can be merged.

So, the minimal DFA that accepts L(a∗bb) ∪ L(ab∗ba) has 6 states and 8 transitions.

To find the minimal DFA that accepts L(a∗bb) ∪ L(ab∗ba), we first need to understand what these languages represent.

1. L(a∗bb) represents the language of all strings that contain zero or more 'a' followed by exactly two 'b'.
2. L(ab∗ba) represents the language of all strings that start with 'a', followed by zero or more 'b', and ending with 'ba'.

Now, let's construct the DFA for each language separately:

1. DFA for L(a∗bb):

   • States: q0, q1, q2, q3 (q0 is the start state and q3 is the final state)
   • Transitions:
     • q0 on 'a' -> q0
     • q0 on 'b' -> q1
     • q1 on 'b' -> q3
     • q1 on 'a' -> q2
     • q2 on 'a' -> q2
     • q2 on 'b' -> q2
     • q3 on 'a' -> q2
     • q3 on 'b' -> q2

2. DFA for L(ab∗ba):

   • States: p0, p1, p2, p3 (p0 is the start state and p3 is the final state)
   • Transitions:
     • p0 on 'a' -> p1
     • p0 on 'b' -> p4
     • p1 on 'a' -> p4
     • p1 on 'b' -> p2
     • p2 on 'a' -> p3
     • p2 on 'b' -> p2
     • p3 on 'a' -> p4
     • p3 on 'b' -> p4
     • p4 on 'a' -> p4
     • p4 on 'b' -> p4

Now, we need to construct a DFA that accepts the union of these two languages. This can be done by creating a new start state that transitions to the start states of the two DFAs on 'a' and 'b'. The final states are the final states of the two DFAs.

3. DFA for L(a∗bb) ∪ L(ab∗ba):

   • States: s0, q0, q1, q2, q3, p0, p1, p2, p3, p4 (s0 is the new start state and q3, p3 are the final states)
   • Transitions:
     • s0 on 'a' -> q0, p0
     • s0 on 'b' -> q1, p4
     • (other transitions are the same as the individual DFAs)

Finally, we minimize this DFA by merging equivalent states. In this case, the DFA is already minimal as there are no equivalent states. So, the minimal DFA that accepts L(a∗bb) ∪ L(ab∗ba) is the one we constructed in step 3.