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For the expression (ab)*ba I came up with the following (very poorly drawn):

DFA

However, this was not the correct answer - apparently the solution requires five states. I don't understand why my DFA doesn't describe the expression. My interpretation of the expression is "ab any number of times, followed by ba". Is this correct?

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For something to be a DFA, it should have transitions from every state for every letter in the alphabet. Your alphabet is {$a, b$}, so every state should have a transition for $a$ as well as $b$, e.g., your state 2 does not have a transition for $a$. Complete your DFA and you'll get what is that 5th state that you are missing.

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Your solution is correct—but you've built an NFA, not a DFA.

The difference is that an NFA can have any number of transitions on a single symbol from a single state. Anything from zero upward.

A DFA needs to have exactly one transition for each symbol from each state. Right now you're missing several of those. The fifth state to add is a "failure state" that transitions only to itself; make all the missing transitions go to this failure state.

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