10. (5 points) Arrange 6 balls in a row, where balls 1, 2, and 3 are black, and balls 4, 5, and 6 are white, as shown in Figure 1. If balls 2 and 5 are swapped, the 6 balls will be arranged in an alternating black and white pattern, as shown in Figure 2. Now, there are 20 balls arranged in a row in sequence, with balls 1 to 10 being black and balls 11 to 20 being white. If we want to arrange these 20 balls in an alternating black and white pattern, what is the minimum number of swaps required? $\qquad$
【Analysis】From the problem, we know that 6 balls (3 black, 3 white) only need to swap the middle two balls once to form a black and white alternating arrangement, i.e., the middle 2 balls remain unchanged, and the remaining 4 balls need to be swapped once. Therefore, these 4 balls can be considered as one cycle. 
If we want to arrange 20 balls (10 black, 10 white) in a black and white alternating pattern, apart from the middle 2 balls, there are \(20-2=18\) balls left. \(18 \div 4=4\) (times) \(\cdots 2\) (balls), so we know that the minimum number of swaps required is \(4+1=5\) times; according to this, we can solve the problem.
【Solution】Solution: 6 balls (3 black, 3 white), the middle 2 balls remain unchanged, and the remaining 4 balls need to be swapped once,
$$
\begin{array}{l}
20-2=18 \text { (balls) } \\
18 \div 4=4 \text { (times) } \cdots 2 \text { (balls) } \\
4+1=5 \text { (times) }
\end{array}
$$

Answer: The minimum number of swaps required to arrange 20 balls in a black and white alternating pattern is 5 times.
The answer is: 5.
【Comment】The key to solving this problem is to understand that for every 6 balls (3 black, 3 white), only the middle two balls need to be swapped once to form a black and white alternating arrangement, i.e., to consider these 4 balls as one cycle, and solve it as a periodic problem.