Can All States Be Final In A Deterministic Pushdown Automata?

Question

When constructing deterministic pushdown automata, can every state be a final state?

I'm having trouble, specifically, with constructing a DPDA that accepts the following language:

L = { 0ⁿ 1^m | n ≥ m }

My approach is to make the initial state a final state that can push 0's to the stack for input 0's then transition to another final state that can pop 0's from the stack for input 1's. I believe this solution is correct, but it seems unusual for every state to be a final state, and I want to verify that my approach is valid.

Is this the correct approach? Can I have both states as final states? Here is the exact transition function δ for my DPDA.

δ(q₀, 0, Z₀) = { (q₀, 0 Z₀) }

δ(q₀, 0, 0) = { (q₀, 0 0) }

δ(q₀, 1, 0) = { (q₁, ε) }

δ(q₁, 1, 0) = { (q₁, ε) }

Answer 1

Of course, every state can be final in a deterministic pushdown automaton. Your approach seems correct to me. Depending on your definition on determinism it might be necessary to also add a transition that deals with the case where you read a 0 in state q_1 in order to have a total transition function (but this depends on what deterministic in your context is defined to be.)