Is my Turing Machine (Transition Function) correct for finding if a string is of even or odd length?

Question

I've been asked to create a formal Turing Machine (by means of a transition function) to which takes a string $a^n \in$ {a}* and decides whether it is an even or odd length.

What I have made is the following

({E, O, H}, {a, _, >}, P, E, {H})

Where P is

State, Letter $\to$ P(State/Letter)

E, a $\to$ (O, _)

E, _ $\to$ (E, $\to$)

E, > $\to$ (E, $\to$)

O, a $\to$ (E, _)

O, _ $\to$ (O, $\to$)

O, > $\to$ (O, $\to$)

I'm fairly sure I've got the right idea but I don't know how exactly I am meant to finish. How do I 'finish' and where in the input do I start reading from?

Answer 1

Try to put more structure here: The states $\{E, O, H_E, H_O\}$ are meant to denote even, odd and halted even, odd states. The Alphabet may only consist of $\{a, \Box\}$ i.e. the letter and the blank. The algorithm we think up is

switch (state)
case even:
    if letter == a then
        state = odd
        goRight
    else
        state = halt_even
    end if
case odd:
    if letter == a then
        state = even
        goRight
    else
        state = halt_odd
    end if
end switch

Now we can say the transition function is given by $$(s,l) \mapsto \delta(s, l) = (s, l, d)\\ (E, \Box) \mapsto (H_E, -, -) \\ (E, a) \mapsto (O, -, \rightarrow) \\ (O, \Box) \mapsto (H_O, -, -) \\ (O, a) \mapsto (E, -, \rightarrow)$$

The halting states are $H_E, H_O$. Our TM is then given by $$TM = (\{E, O, H_E, H_O\}, \{a, \Box\}, \delta, E, \{H_E, H_O\})$$ We consider two acceptance rules:

1. $TM$ accepts $a^{2n}$ iff the halting state is $H_E$ and
2. $TM$ accepts $a^{2n+1}$ iff the halting state is $H_O$.