# Can a Turing Machine have infinite accept states?

I'm still fairly new to Turing Machines, but I've been doing some research.

I know that a Turing Machine can have an infinite tape and that it requires a finite number of states, but does it necessarily follow that a Turing Machine can have an infinite number of accept states?

I keep seeing different layouts when formally defining Turing Machines, for example:

M = (Q, Σ, Γ, τ, s, F).

1) F ⊆ Q is the set of final or accepting states. (plural)

2) F ⊆ Q is the accept state. (singular)

So I'm just wondering which one is correct?

Any help would be greatly appreciated.

• Possible duplicate of Can a Turing machine have infinite states? – sashas Nov 23 '15 at 6:12
• You are talking about three different cases here: Turing machines with one accepting state, with several (a finite number of them) or infinite ones. Which ones do you mean? – vonbrand Nov 23 '15 at 12:26

In the standard Turing machine, the set of states $Q$ is finite $|Q| < \infty$. Therefore, it cannot have infinite final states. However, it may, or may not have multiple final states. This will not change the power of the machine (i.e., one final state is equivalent to $k$ final states).

• Sorry but why are they equivalent? I don't seem to get it. – User Not Found Sep 7 '17 at 3:43
• @UserNotFound Because once you get to any final state, the Turing machine stops and accepts the input. Since it stops, the final states are indistinguishable from each other. You can easily take any two final states F1 and F2, replace F2 with F1, and there would be no difference in how the Turing machines functioned. – Tin Man Jun 6 '18 at 21:17

Turing machines are defined to have only a finite number of states. Since every accepting state is a state, it's impossible to have infinitely many accepting states. Being allowed to have more than one of something is not the same as being allowed to have infinitely many.

As for the difference between one and many accepting states, if you have a TM with multiple accepting states, you can add a new single accepting state and make sure that from the accepting states you can get to your new state in one move. Thus one/many doesn't affect the power of the machine, but it might simplify proofs. Thus you'll see varied definitions, all equivalent.