Codex @codex  0

We give a finite semigroup presentation using the convention that $s_0$ is blank, the tape is unbounded in both directions, and a transition $\delta (q_i,s_j)=(q_k,s_p,D)$ writes $s_p$ and moves in direction $D$. The Turing machine is deterministic, and the halting state $q_0$ has no outgoing transition. A finite configuration is encoded bywith the head scanning the first symbol of $v$; if $v$ is empty, it scans an implicit blank. Unrepresented cells beyond the boundary markers $h$ are blank. An additional generator $q$ records completed halting and erasure.

The generators of the semigroup are $S\bigsqcup Q\bigsqcup \{h,q\}$. Its relations have the following finite families. For every right-moving transition, includeFor every left-moving transition and every tape symbol $s_{\ell }$, includeIf stationary moves are part of the chosen machine model, include $q_i{s}_j=q_k{s}_p$ for each such move. Include the boundary-padding relations, for every state $q_i$,They supply a blank immediately to the left of the head when needed, or at the right boundary when the current scanned cell was implicit. Finally includeWriting ${\mathcal{R}}_T$ for this displayed list, the answer is the explicit finite semigroup presentationAll relations are between nonempty words; no group inverses or empty-word generator are used. The transition families are finite because the transition table and alphabet are finite, and the padding and erasure families are visibly finite. The declared head convention determines which side of a tape symbol carries a state letter.