The correct answer is
$$
\boxed{\text{C. Semi-decidable but not decidable}}
$$
There is a procedure that halts and accepts exactly the YES-instances, but no procedure that decides both YES and NO.
Detailed Proof and Explanation
To determine the decidability status of this problem, we need to prove two things:
- It is undecidable (ruling out A).
- It is semi-decidable (which confirms C and rules out B and D).
Part 1: Proof of Undecidability (via Reduction)
We can prove that this problem is undecidable by reducing the Halting Problem
$$
HALT_{TM}
=
\left\{
\langle M,w\rangle
\mid
M \text{ halts on input } w
\right\}
$$
to it.
Recall that
$$
HALT_{TM}
$$
is a known undecidable problem.
The Reduction Construction
Suppose we have a decider
$$
\mathcal{D}
$$
for the "revisit starting cell" problem.
We can construct a Turing machine
$$
M'
$$
that simulates $M$ on $w$ in such a way that $M'$ revisits cell $0$ if and only if $M$ halts on $w$.
Construction of $M'$
- Shift the tape right:
On input $x$, $M'$ immediately writes a special boundary marker, say $\$$, at cell $1$, moves the head to cell $2$, and shifts the input $x$ to the right.
2. Confine the simulation:
$M'$ simulates the computation of $M$ on $w$ strictly on the tape cells to the right of cell $1$, i.e., cells
$$
2,3,4,\ldots
$$
The construction ensures that the simulation never causes $M'$ to revisit cell $0$.
3. Trigger a revisit on halting:
If the simulation of $M$ on $w$ halts, $M'$ enters a special state and moves the tape head left until it reaches the starting cell $0$.
Analysis of $M'$'s behavior
If $M$ halts on $w$, then
$$
M \text{ halts on } w
\implies
M' \text{ eventually reaches cell } 0.
$$
Therefore, $M'$ revisits its starting cell.
On the other hand, if $M$ does not halt on $w$, then
$$
M \text{ does not halt on } w
\implies
M' \text{ never enters the special state}.
$$
Hence, $M'$ never revisits cell $0$.
Therefore,
$$
\langle M,w\rangle \in HALT_{TM}
\iff
\langle M'\rangle \in RSC
$$
where $RSC$ denotes the problem
$$
RSC
=
\left\{
\langle M\rangle
\mid
M \text{ revisits its starting tape cell}
\right\}.
$$
Thus, if we had a decider $\mathcal{D}$ for $RSC$, we could decide $HALT_{TM}$, which is impossible.
Hence,
$$
\boxed{RSC\text{ is undecidable}.}
$$
Part 2: Proof of Semi-Decidability
A language is semi-decidable (recursively enumerable) if there exists a Turing machine that halts and accepts whenever the answer is YES, although it may run forever on NO-instances.
We can easily construct a recognizer $R$ for $RSC$.
Recognizer $R$
Given an input
$$
\langle M,w\rangle,
$$
$R$ performs the following steps:
1. Simulate the computation of $M$ on $w$ step-by-step.
2. Keep track of the current position of $M$'s tape head.
3. If the head moves away from cell $0$ and later returns to cell $0$, then $R$ halts and accepts.
4. If the head never returns to cell $0$, the simulation continues forever.
For a YES-instance, there exists some finite time $t$ such that
$$
\text{HeadPosition}(t)=0
$$
after the head has previously left cell $0$.
Since $R$ simulates the computation step-by-step, it will eventually reach that step $t$ and detect the revisit.
Therefore,
$$
\langle M,w\rangle \in RSC
\implies
R \text{ eventually accepts}.
$$
For a NO-instance, the head never revisits cell $0$, so $R$ may run forever.
Thus,
$$
RSC \text{ is semi-decidable}.
$$
Combining both results,
$$
\boxed{
RSC\text{ is semi-decidable but not decidable}
}
$$
Therefore, the correct answer is
$$
\boxed{\text{C}}
$$
