Construct a Turing Machine that recognizes the set $\{0^{2n}1^n|n\geq 0\}$ The full question is: Construct a Turing machine that recognizes the set $\{0^{2n}1^n \mid n\geq 0\}$.  The Turing machine starts with the input on the tape and the head over the leftmost symbol of the input.  The symbols to the left of the input are blanks out to infinity, as are the symbols to the right of the input.  If the string is accepted, the machine should halt with 1 on the tape (the rest of the tape should be blank).  If the string is rejected, the machine should halt with 0 on the tape.

Yes, this is a homework question, but I think I just need some help understanding the problem. We have not gone very in depth into Turing Machines, and he suggested we should just write the answer graphically like a  Deterministic Finite State Machine with read, write, and direction in the transitions.  

I am getting stuck because I can only think about it in terms of 3 different paths (being 001, 100, 010), but I know this is most likely not correct, and I do not know how to recognize if the string is not in the set. I understand that the machine know the string has ended when it reads a blank, and then it will move right one and write either a 1 or 0 depending on if the string is accepted or not, but I can only think about it as 3 different paths. 

Help would be appreciated! Thanks!
Here is a Turing machine on the online simulator I found here.

name: Recongnize 0^2n 1^n
init: s0
accept: acc

s0, 0
s1, _, &gt;

s1, 0
s2, _, &gt;

s2, 0
s2, 0, &gt;

s2, 1
s2, 1, &gt;

s2, _
s3, _, &lt;

s3, 1
s4, _, &lt;

s4, 0
s4, 0, &lt;

s4, 1
s4, 1, &lt;

s4, _
s0, _, &gt;

s0, _
acc, _, -


The idea is to delete the two zeroes on the left, got to the end on the right, delete one, go to the way to the left, repeat. Accept only when after deleting 1 on the right the tape is empty.