Example 5 Suppose we have a $4 \times 4$ grid where each cell is colored differently. Each cell is filled with either 0 or 1, such that the product of the numbers in any two adjacent cells is 0. How many different ways are there to fill the grid with numbers?

(2007, Korean Mathematical Olympiad)
【Analysis】Since in any two adjacent cells there is at least one 0, when a cell is filled with 1, all its neighboring cells must be filled with 0. The solution can start from a cell filled with 1.

First, consider which position, when filled with 1, can produce as many 0s as possible. Since the cells in the middle have the most neighbors, we can discuss the number of 1s in the four middle cells.

If there is no 1 in the four middle cells, then the surrounding 12 cells form a circle, and in any two adjacent cells, there is at least one 0. Therefore, we first solve the following local problem.

Lemma 1 If $n$ cells form a circle, and each cell is filled with 0 or 1, such that the product of the numbers in any two adjacent cells is 0, then the number of different filling methods is
$$
b_{n}=F_{n+1}+F_{n-1} \text {. }
$$

The proof can use reduction by recursion, fixing one of the cells. If this cell is filled with 0, then the remaining $n-1$ cells do not form a circle (separated by the cell filled with 0), but form a row. Therefore, to prove Lemma 1, we need to first prove Lemma 2.

Lemma 2 If $n$ cells form a row, and each cell is filled with 0 or 1, such that the product of the numbers in any two adjacent cells is 0, then the number of different filling methods is $a_{n}=F_{n+2}$, where $F_{1}=F_{2}=1, F_{n}=F_{n-1}+F_{n-2}$ is the Fibonacci sequence.

Proof of Lemma 2 Consider the first cell. If this cell is filled with 0, then the remaining $n-1$ cells have $a_{n-1}$ filling methods; if this cell is filled with 1, then its neighboring cell must be filled with 0, and the remaining $n-2$ cells have $a_{n-2}$ filling methods. Therefore,
$$
\begin{array}{l}
a_{n}=a_{n-1}+a_{n-2} . \\
\text { Also, } a_{1}=2=a_{3}, a_{2}=3=a_{5}, \text { thus, } \\
a_{n}=F_{n+2} .
\end{array}
$$

Lemma 2 is proved.
Proof of Lemma 1 Fix one of the cells. If this cell is filled with 0, then the remaining $n-1$ cells form a row, with $a_{n-1}$ filling methods; if this cell is filled with 1, then its two neighboring cells must be filled with 0, and the remaining $n-3$ cells form a row, with $a_{n-3}$ filling methods.
Thus, $b_{n}=a_{n-1}+a_{n-3}=F_{n+1}+F_{n-1}$.
Lemma 1 is proved.
Consider three cases.
(1) If there is no 1 in the four middle cells, then the surrounding 12 cells form a circle, and in any two adjacent cells, there is at least one 0. By Lemma 1, there are
$$
b_{12}=F_{13}+F_{11}=233+89=322
$$

filling methods.
(2) If there is one 1 and three 0s in the four middle cells, then there are 4 ways to fill the 1, and the filling method for 0 is unique. For any of these filling methods, the two neighboring cells of the cell filled with 1 can only be filled with 0, and the common neighboring cell between these two newly filled 0s has 2 filling methods. The remaining 9 cells form a row, and by Lemma 2, there are $a_{9}=F_{11}=89$ filling methods.
Thus, there are $4 \times 2 \times 89=712$ filling methods.
(3) If there are two 1s and two 0s in the four middle cells, then there are 2 ways to fill the two 1s (filling diagonally), and for any of these filling methods, the two cells filled with 1 each have two neighboring cells not filled, which can only be filled with 0, and the common neighboring cell between these two newly filled 0s can be filled with 0 or 1, with 2 filling methods. The remaining six cells form two rows of length 3, and in any two adjacent cells, there is at least one 0. By Lemma 2, each row has $a_{3}=F_{5}=5$ filling methods.
Thus, there are $2^{2} \times 5^{2} \times 2=200$ filling methods.
In total, there are $322+712+200=1234$ filling methods.