Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 104 / 4

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.

If the Turing machine halts on $w$, follow its computation from $h{q}_{1}wh$. A transition at an already represented cell is one of the corresponding relations. At a boundary, first insert the required blank using a padding relation. Thus the computation gives a finite equality derivation to $hu{q}_{0}vh$. Replace $q_0$ by $q$, erase the symbols of $u,v$ using $s_{j}q=q=q{s}_j$, and erase both boundary markers. This proves $h{q}_{1}wh=q$ in the semigroup.

For the converse, equality in a presented semigroup means a finite sequence of contextual replacements using defining relations in either direction. In any derivation from $h{q}_{1}wh$ to $q$, consider the first occurrence of the special symbol $q$. Before it occurs, the erasure relations cannot be used in either direction. Each word therefore still has exactly two markers $h$, exactly one state letter, and the form $hu{q}_{i}vh$. Both directions of a transition relation join configurations linked by one actual machine step; both directions of a boundary-padding relation merely change how much blank tape is represented.

The first occurrence of $q$ must come from $q_0=q$ in the forward direction. Hence the initial configuration is connected, by transitions in either direction and harmless padding, to a configuration in state $q_0$. For a deterministic Turing machine, the property of eventually reaching $q_0$ is invariant along each transition edge: if $C\to D$ is a step, then $C$ halts if and only if $D$ halts, since that step is the unique next step of $C$ and $C$ is not already in the halting state. The same property is unchanged by padding. A finite path of these edges to $q_0$ therefore shows that the initial configuration halts.

ConsequentlyThe invariance argument is necessary because equality in a semigroup presentation allows reversed transitions as well as forward computation steps.