Turing Machine Encoding and Complement of the Diagonal Language In trying to understand how Turing Machine encoding works, there is a question that I have come across that goes something like this:

Given that the diagonal language, $L_d = \{ w_i \mid w_i \space is \space not \space accepted \space by \space M_i \}$, what is the first string of the encoding of the complement of $L_d$ which is $\bar{L_d}$?

I am assuming also that this is operating with the alphabet of $\Sigma = \{0, 1\}$. Is there any way that it is even possible? I am confused also since $L_d$ is an undecidable language how this is possible?

Also by encoding I think that the question is referring to binary encoding with a tape alphabet $\{0, 1, B\}$ with Turing machine $M$ defined as:

$M = &lt; Q, \{0, 1\}, \{0, 1, B\},\delta,q_1,B,\{q_2\} &gt;$

with moves defined by:

$\delta(q_i,X_j) = (q_k,X_n,D_m) \rightarrow 0^i10^j10^k10^n10^m $

where $X_1 = 0, X_2 = 1, X_3 = B$
and Direction is given by: $D_1 = L, D_2 = R$,

and the Binary encoding of $M$ as $&lt;M&gt;$: $111 \space move_1 \space 11 \space move_2 \space 11 \space ... \space 11 \space move_{last} \space 111$

Update: If my understanding is correct, the correct encoding should represent a Turing machine which accepts every string since originally we do not accept any string? How would one go about formulating an encoding for such a Turing machine? Does it have to be a unique encoding, or does it just imply that any Turing machine that accepts every string is valid?
The following hint repeats my comments and assumes that the question asks for the first encoding of a Turing machine which is in $\overline{L_d}$.

Hint: Make a list of the first few encodings of Turing machines, in lexicographic order. For each one, check whether it's in $\overline{L_d}$. The first one which is in $\overline{L_d}$ is the answer.

By the way, how does your encoding handle termination? Does it mark accepting/rejecting states in some way, or does it use another convention?