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$\begingroup$

I am trying to tun the following regular language into a finite automata, but I am confused by the U (union) and I am unsure of how to approach this part of the expression.

The expression: $$ L[01((0 \cup 11)11^*)^*0] $$

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    Apologies - I have amended. The "U" is obviously for union. – 2017-01-23

1 Answers 1

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Union is easy: $$ (0 \cup 11) $$ means both $0$ and $11$ are recognized. This means there exist transitions from the state $q_i$ before the opening parenthesis to the state $q_j$ after the closing parenthesis, one via accepting $0$ and another one via accepting $11$: $$ \delta(q_i, 0) = q_j \\ \delta(q_i, 11) = q_j $$

                                             0
                                           ____
                                          /    v

                                      q_i        q_j

                                          \____^ 

                                            11
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    Thanks! This is really helpful. I now understand I can go between the two states in the union. Would you recommend a process for now translating this into a finite automata? – 2017-01-23
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    It depends a bit on what kind of automaton you work with. For example a DFA accepts symbols not words, so you would have an intermediate state $q_k$ on the lower path: $\delta(q_i, 1) = q_k, \delta(q_k, 1) = q_j$. – 2017-01-23