14. The figure below shows a switch array, where each switch has only two states: "on" and "off". Initially, all switches are off. Pressing a switch once will change the state of the switch itself and all adjacent switches. For example, pressing $(2,2)$ will change the states of $(1,2),(2,1),(2,2),(2,3),(3,2)$. If each switch can be pressed at most once, and the total number of presses is minimized:
\begin{tabular}{|l|l|l|l|}
\hline$(1,1)$ & $(1,2)$ & $(1,3)$ & $(1,4)$ \\
\hline$(2,1)$ & $(2,2)$ & $(2,3)$ & $(2,4)$ \\
\hline$(3,1)$ & $(3,2)$ & $(3,3)$ & $(3,4)$ \\
\hline$(4,1)$ & $(4,2)$ & $(4,3)$ & $(4,4)$ \\
\hline
\end{tabular}
(1) If the final state is exactly that the switches at $(2,1),(3,2),(4,3),(1,2),(2,3),(3,4)$ are on, how many times at least do you need to press the switches? Please provide a proof;
(2) If all switches need to be off, how many times at least do you need to press the switches? Please directly write the result and a solution.
(1) 6 times are enough, as shown in the figure below.

The proof that the minimum number of operations required is 6 is as follows:
Consider the diagonal $(1,1),(2,2),(3,3),(4,4)$, denoted as region $A$. The switches in region $A$ change an odd number of times only when the switches inside region $A$ are pressed. Since the switches in region $A$ are all closed in the final state, the switches in region $A$ must have been pressed an even number of times. Let $S(X)$ denote the total number of times the switches in region $X$ are pressed.
$1^{\circ} S(A)=4$.
After operating the switches in $A$, the switch at $(2,1)$ is off, so the lower-left corner of $A$ must be operated on at least once. Similarly, the upper-right corner of $A$ must also be operated on at least once, so the total number of operations is at least 6.
$$
2^{\circ} S(A)=2 \text{. }
$$
(i) As shown in Figure $a$, after operating the switches in $A$, the switch at $(2,1)$ is off, so $S(B)$ is odd, i.e., $S(B) \geqslant 1$. Similarly, the switches in regions $C, D, E$ must each be pressed at least once, so the total number of operations is at least 6 times;
(ii) As shown in Figure $b$, after operating the switches in $A$, the switches at $(2,3),(3,2)$ are off, so the total number of operations is at least 6 times;
(iii) As shown in Figure $c$, after operating the switches in $A$, the switches at $(2,1),(1,2),(4,3),(3,4)$ are on. If the switches in regions $B, E$ are each pressed an even number of times, then the switch at $(1,1)$ will be on, which does not meet the requirement. Therefore, $S(B), S(E)$ cannot both be 0. Suppose $S(E)=2$. After operating the switches in $E$ (as shown in Figure $d$), the switches at $(1,4),(3,3)$ are on. Since the final state of $(1,4),(3,3)$ is off, $S(F), S(H)$ must both be odd. Therefore, the total number of operations $\geqslant S(A)+S(E)+S(H)+S(F) \geqslant 2+2+1+1=6$; $3^{\circ} S(A)=0$.
To minimize the number of operations, by symmetry, the number of operations in the upper-right corner of region $A$ must be minimized. In fact, the operations that ensure the upper-right corner meets the conditions, when mirrored along the diagonal (region $A$), will also meet the requirements. The three cells in the upper-right corner are denoted as region $G$, with the number of operations being $x, y, z$, as shown in the figure. Note that the switch at $(1,4)$ is off in the final state, so $x+y+z$ is even.
The switches at $(2,1),(3,2),(4,3)$ are denoted as region $J$.
(i) If $x+y+z=0$, then since the switches in region $J$ are all on, each of these switches must be pressed once, requiring at least $2 \times 3=6$ operations;
(ii) If $x+y+z=2$, then at least one of $x+y, y+z$ is odd. Suppose $x+y$ is odd, then after operating the switches in regions $A$ and $G$, the switch at $(1,3)$ is on. To turn it off, $S(J) \geqslant 1$. Therefore, the number of operations in the upper-right corner is at least $1+2=3$, and the total number of operations is at least 6. In summary, the total number of operations is at least 6.
(2) 8 times, a possible solution is: