9.4. In a regular 100-gon, each vertex is covered by a cloth. It is known that exactly one of the cloths hides a coin. The following action is called an operation: choose any four cloths to check if there is a coin underneath, and after each operation, the cloths are returned to their original positions, while the coin is secretly moved to the cloth under an adjacent vertex. Find the minimum number of operations required to definitely find the coin.
9.4.33 times.

Assume that a regular 100-gon is placed on a rotatable horizontal circular table, and the vector from the center to one vertex points due north. The initial positions of the vertices of the polygon are defined from due north in a counterclockwise direction as $0, 1, \cdots, 99$. Each operation and coin transfer will cause the table to rotate automatically by $\frac{2 \pi}{100}$ in a clockwise direction. This means the coin's position either remains unchanged or moves two positions clockwise. Therefore, the parity of the coin's position remains unchanged after each automatic rotation of the table.

If a position cannot have a coin at a certain moment, it is called "empty." Otherwise, it is called "non-empty."
Clearly, at the initial moment, the number of empty positions is 0.
Thus, each operation can increase the number of empty positions by at most 4.

Consecutive odd (even) empty positions form an odd (even) empty interval.

If an odd (even) empty interval is not a proper subset of another odd (even) empty interval, it is called "maximal."
Due to the transfer of the coin, after the table rotates,

the counterclockwise endpoint of each maximal odd (even) empty interval will become non-empty. This indicates that during the entire process of finding the coin, the transfer of the coin will not increase the number of non-empty positions only in the following situation: when all odd (even) positions are empty, and all even (odd) positions are non-empty (this situation can occur at most once). In all other cases, the number of empty positions before this operation is at most 3 more than the number of empty positions before the previous operation. Thus, after 32 operations, the number of empty positions is at most $31 \times 3 + 4 = 97$, which does not guarantee finding the coin.

Next, we show that the coin can be found in at most 33 operations.
First, check positions $0, 2, 4, 6$.
If the coin is found, the problem is solved;
If not, then after the table rotates, positions $0, 2, 4$ are empty (assuming the coin is not found in the first 32 operations).

Generally, if after the table rotates, it is known that $0, 2, \cdots, 2s$ are empty positions, then the next operation checks $2s+2, 2s+4, 2s+6, 2s+8$.

After the coin transfer and table rotation, $0, 2, \cdots, 2s+6$ are empty positions. Thus, after 16 operations and table rotations, 48 empty positions are obtained, i.e., $0, 2, \cdots, 94$.
The 17th operation checks $96, 98, 1, 3$.
After the table rotates, 51 empty positions are obtained, i.e., $0, 1, 2, 4, 6, \cdots, 98$. Next, check $3, 5, 7, 9; 9, 11, 13, 15; \cdots$.

After the 32nd operation and table rotation, 96 empty positions are obtained, i.e., $0, 1, 2, \cdots, 91, 92, 94, 96, 98$.

The 33rd operation checks the last four positions $93, 95, 97, 99$, and the coin will certainly be found.