Example 2-1 There are $n$ disks of different sizes stacked on column $A$ in ascending order of their radii, as shown in Figure 2-1. Each move allows transferring only one disk to either column $B$ or column $C$, and it is not allowed to place a larger disk on top of a smaller one. The goal is to transfer all $n$ disks from column $A$ to column $C$. Please design a method and estimate how many moves are required. The problem provides three columns $A, B, C$ for use.
The typical significance of the Tower of Hanoi problem lies in first finding the algorithm, and then estimating its time complexity, specifically the number of moves required. The research on the algorithm is roughly as follows.

Please consider practical scenarios, simulate with smaller-scale problems as a "sandbox exercise," identify regular patterns, and then generalize to the general case.

As shown in Figure 2-1, first move the top disk to column B, then move the next disk to column C, and finally move the disk from column B to column C, making a total of 3 moves. The Tower of Hanoi problem with \( n=2 \) is thus solved. Next, consider the problem with \( n=3 \). Since the problem with \( n=2 \) has been solved, the first step is to move the top two disks from column A to column B, then move the last disk from column A to column C. Finally, use the method for moving two disks to transfer the two disks from column B to column C, thus solving the Tower of Hanoi problem with \( n=3 \). The problem with \( n=4 \) can be solved similarly, leading to a chain reaction, solving the problems with \( n=5, n=6, \cdots \) in sequence. Solving the Tower of Hanoi problem with \( n-1 \) disks will lead to the solution for \( n \) disks. This method is called a recursive algorithm.

Algorithm Analysis: Let \( H(n) \) represent the number of moves required for the Tower of Hanoi problem with \( n \) disks. According to the algorithm, two \( n-1 \) disk problems need to be solved: first, move the top \( n-1 \) disks from column A to column B, then move the last disk from column A to column C, and finally move the \( n-1 \) disks from column B to column C. Therefore, we have:
\[ 
H(n) = 2H(n-1) + 1, \quad H(1) = 1 \quad (2-1)
\]
\( H(1) = 1 \) is the initial condition of the recurrence relation (2-1), meaning one disk requires one move.
Thus, we get the sequence:
\[
\begin{array}{l}
H(1) = 1, \\
H(2) = 2 + 1 = 2^2 - 1, \\
H(3) = 2(2^2 - 1) + 1 = 2^3 - 1, \\
H(4) = 2(2^3 - 1) + 1 = 2^5 - 1.
\end{array}
\]

We can prove by mathematical induction that \( H(n) = 2^n - 1 \). Clearly, \( H(1) = 2^1 - 1 = 1 \) holds.
Assume
\[
H(n-1) = 2^{n-1} - 1 \text{ holds}
\]
Then
\[
H(n) = 2H(n-1) + 1 = 2(2^{n-1} - 1) + 1 = 2^n - 1
\]

If \( n = 50 \), then \( H(50) = 2^{50} - 1 \approx 1.126 \times 10^{15} \).
What does this mean? If 10 million moves can be made per second, and there are 365 days in a year, then there are
\[
N = 365 \times 24 \times 3600 = 3.1536 \times 10^7 \text{ seconds per year}
\]

Thus, solving the Tower of Hanoi problem with \( n = 50 \) disks would require
\[
T = \frac{1.126 \times 10^{15}}{3.1536 \times 10^7} = 3.5705 \text{ years}
\]