# Can a DFA have multiple of the same state?

I need to create a DFA for a sequential order of states e.g. A -> B -> C -> B -> A
and so on, where 'A' is the start and finish state, 1 is a transition to the next state and 0 just loops back to the current state. 1 and 0 are the only inputs

The only way I've found I can do it in this order is by having 2 B states, Is this ok or is there another way to do this?

• I don't understand what you are trying to do. What does it mean to "create a DFA for a sequential order of states"? Can you edit your question to specify your requirements more clearly, and motivate them (why you want them to hold)? It might help to also give an example of a sample input and the desired DFA you want it to produce. For the example you give, I suggest you try drawing the DFA you have in mind, show us a picture of it, check whether you are satisfied with it, and tell us why you are or aren't satisfied with it.
– D.W.
Feb 21 at 19:35

Ask yourself this: what do you mean by the same state? In particular, why do your two $$B$$ states need to be the same state and what value does that give you over treating them as two separate states?
For a more thorough answer: consider the formal definition of the DFA, which includes the transition function $$\delta(Q \times \Sigma) \rightarrow Q$$ ($$Q$$ being the set of states, $$\Sigma$$ being the set of input symbols). The notion of there being "duplicate" states is already invalidated by $$Q$$ being a set of states, but since the actual behavior of the automaton emerges from the state transition function, let's consider the implication of there being duplicate states regarding that.
Let's call the two states you've labeled $$B$$ by the names $$b_1$$ and $$b_2$$. Since you want them to be the same state, $$b_1 = b_2$$. Since $$\delta$$ is a function, for any $$\sigma \in \Sigma$$ it must therefore hold that $$\delta(b_1, \sigma) = \delta(b_2, \sigma)$$. In English: if $$b_1$$ and $$b_2$$ are the same state they cannot behave differently between each other wrt. the state transition function. Therefore your $$B$$ states must be separate states: if $$\delta(B, 1) = C$$ at one point, it will be that, always.
A DFA must have at most a single transition on a given input. In your example, being in state $$B$$ and reading a 1 you have two choices for the resulting state: go to $$A$$ or $$C$$. You could, though have a nondeterministic FA that would achieve this. The resulting language would consist of all strings of $$1$$s of even length.