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:$M = < Q, \{0, 1\}, \{0, 1, B\},\delta,q_1,B,\{q_2\} >$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 $<M>$: $111 \space move_1 \space 11 \space move_2 \space 11 \space ... \space 11 \space move_{last} \space 111$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?
Turing Machine Encoding and Complement of the Diagonal Language
computability;turing machines