I want to find a non-deterministic 2-tape Turing machine, that accepts the language L over $\Sigma=\{0,1\}$ in $n$ steps, with input of length $n$, $L=\{x1y \mid |y|=2|x|>0\}$.
Should the Turing machine do the following?
Each time that the machine reads 1 it should check if the length of the subword before 1 is equal to the half of the length of the subword after 1.
How can this be done by a non-deterministic 2-tape Turing machine? Could you give me a hint?
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EDIT:
The idea is the following:
We have to guess the position of $1$. In the state $z_1$ we copy the input of the first tape, till this $1$, to the second one. Then after having reached $1$, the head of the first tape will make $|x|$ to the right. If the input ends there, the input is accepted. If not, the input is rejected.
Is this correct?
Is the transition function then the following?
$(z_0, 1, \square)\mapsto (z_0,1,1,R,R) \mid (z_1, X,X,N,N)$
$(z_0, 0, \square)\mapsto (z_0,0,0,R,R)$
$(z_1, a, \square)\mapsto (z_1, a,a,R,R)$
$(z_1, X,X)\mapsto (z_2, a, \square, R, N)$
$(z_2, a, \square)\mapsto (z_2, a, \square, R, N)$
$(z_2,\square, \square )\mapsto (z_3, \square, \square, N, N)$