Problem 6.3. A chip was placed on a certain cell of a $5 \times 5$ board. By moving the chip to an adjacent cell, the entire board was traversed except for one cell, and the chip returned to the starting position. The chip visited each cell, except the starting one, no more than once.

The arrows in the image show the direction the chip moved from some of the cells.

Select the cell on the picture that the chip did not enter.

![](https://cdn.mathpix.com/cropped/2024_05_06_634e5517f3bc14ef6482g-3.jpg?height=487&width=537&top_left_y=467&top_left_x=432)
Answer: $C 1$.

Solution. Since the chip has returned to the starting cell, we will consider its path to be cyclic. Note that no part of the chip's path can divide the board into two areas, each of which has more than 1 cell, because after passing this part of the path, the chip must move into one of the areas - and cannot enter the second area, which contradicts the condition.

First, let's determine where the chip could have gone from cell $B 2$:

- to cell $A 2$ it could not have moved, as from there it is impossible to go to $A 1$ (the path $B 1 \rightarrow$ $B 2 \rightarrow A 2 \rightarrow A 1$ does not continue), or to $A 3$ (there is already an arrow leading to it from another cell);
- to cell $B 3$ it could not have moved, as the path $B 1 \rightarrow B 2 \rightarrow B 3 \rightarrow A 3$ divides the board into two areas with more than one cell each.

Therefore, from $B 2$ the chip moved to $C 2$, and then, according to the arrow, to $D 2$.

Where could the chip have gone from cell $D 2$?

- to $D 3$ it could not have moved, as there is already an arrow leading to it;
- to $E 2$ it could not have moved, as the path $B 1 \rightarrow B 2 \rightarrow C 2 \rightarrow D 2 \rightarrow E 2$ divides the board into two areas with more than one cell each.

Therefore, from $D 2$ the chip moved to $D 1$. It is clear that cell $C 1$ in this case must remain unvisited, as it is cut off by the part of the path $B 1 \rightarrow B 2 \rightarrow C 2 \rightarrow D 2 \rightarrow D 1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_634e5517f3bc14ef6482g-4.jpg?height=410&width=410&top_left_y=215&top_left_x=526)

Fig. 4: to the solution of problem 6.3

The path that satisfies the condition and passes through all other cells is shown in Fig. 4. It is constructed uniquely.