9.5. The cells of a chessboard are painted in 3 colors - white, gray, and black - in such a way that adjacent cells, sharing a side, must differ in color, but a sharp change in color (i.e., a white cell adjacent to a black cell) is prohibited. Find the number of such colorings of the chessboard (colorings that coincide when the board is rotated by 90 and 180 degrees are considered different).
Solution. To avoid confusion with the natural coloring of the chessboard, let's choose three colors: red, blue, and gray. Suppose the field a1 is painted gray. Then, according to our coloring rules, all black fields of the board (recall that a1 is a black field) automatically become gray, while the color of the remaining thirty-two (white) fields has only one restriction: it must be red or blue. This gives us $2^{32}$ coloring options. If the field $\mathbf{a} 1$ is painted red or blue, then all white fields of the chessboard automatically become gray under any coloring according to our rules, and the color of any of the remaining thirty-two (black) fields can be red or blue. Another $2^{32}$ options. In total, $2 \cdot 2^{32}=2^{33}$.

Criteria.

$(+/ 2)$ written or drawn, which cells should be painted gray (not in some particular cases, but in general),

(+.) The solution is correct and complete, with no significant omissions except for the following:

- half the cases are missed due to the second case of gray cell placement not being considered, the answer is twice as small as the correct one, or the answer is twice as large as the correct one because instead of two ways of placing gray cells, four were considered, or
- the solution assumes that all three colors must be used in the coloring, resulting in an answer that is 2 less than the correct one.

$(+)$ Correct answer and correct solution.