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krahets
2024-05-01 07:30:10 +08:00
parent 583d338530
commit a08cd961b3
64 changed files with 227 additions and 227 deletions
@@ -8,7 +8,7 @@ In both merge sorting and building binary trees, we decompose the original probl
!!! question
Given three pillars, denoted as `A`, `B`, and `C`. Initially, pillar `A` is stacked with $n$ discs, arranged in order from top to bottom from smallest to largest. Our task is to move these $n$ discs to pillar `C`, maintaining their original order (as shown below). The following rules must be followed during the disc movement process:
Given three pillars, denoted as `A`, `B`, and `C`. Initially, pillar `A` is stacked with $n$ discs, arranged in order from top to bottom from smallest to largest. Our task is to move these $n$ discs to pillar `C`, maintaining their original order (as shown in Figure 12-10). The following rules must be followed during the disc movement process:
1. A disc can only be picked up from the top of a pillar and placed on top of another pillar.
2. Only one disc can be moved at a time.
@@ -22,7 +22,7 @@ In both merge sorting and building binary trees, we decompose the original probl
### 1.   Consider the base case
As shown below, for the problem $f(1)$, i.e., when there is only one disc, we can directly move it from `A` to `C`.
As shown in Figure 12-11, for the problem $f(1)$, i.e., when there is only one disc, we can directly move it from `A` to `C`.
=== "<1>"
![Solution for a problem of size 1](hanota_problem.assets/hanota_f1_step1.png){ class="animation-figure" }
@@ -32,7 +32,7 @@ As shown below, for the problem $f(1)$, i.e., when there is only one disc, we ca
<p align="center"> Figure 12-11 &nbsp; Solution for a problem of size 1 </p>
As shown below, for the problem $f(2)$, i.e., when there are two discs, **since the smaller disc must always be above the larger disc, `B` is needed to assist in the movement**.
As shown in Figure 12-12, for the problem $f(2)$, i.e., when there are two discs, **since the smaller disc must always be above the larger disc, `B` is needed to assist in the movement**.
1. First, move the smaller disc from `A` to `B`.
2. Then move the larger disc from `A` to `C`.
@@ -58,7 +58,7 @@ The process of solving the problem $f(2)$ can be summarized as: **moving two dis
For the problem $f(3)$, i.e., when there are three discs, the situation becomes slightly more complicated.
Since we already know the solutions to $f(1)$ and $f(2)$, we can think from a divide-and-conquer perspective and **consider the two top discs on `A` as a unit**, performing the steps shown below. This way, the three discs are successfully moved from `A` to `C`.
Since we already know the solutions to $f(1)$ and $f(2)$, we can think from a divide-and-conquer perspective and **consider the two top discs on `A` as a unit**, performing the steps shown in Figure 12-13. This way, the three discs are successfully moved from `A` to `C`.
1. Let `B` be the target pillar and `C` the buffer pillar, and move the two discs from `A` to `B`.
2. Move the remaining disc from `A` directly to `C`.
@@ -80,7 +80,7 @@ Since we already know the solutions to $f(1)$ and $f(2)$, we can think from a di
Essentially, **we divide the problem $f(3)$ into two subproblems $f(2)$ and one subproblem $f(1)$**. By solving these three subproblems in order, the original problem is resolved. This indicates that the subproblems are independent, and their solutions can be merged.
From this, we can summarize the divide-and-conquer strategy for solving the Tower of Hanoi shown in the following image: divide the original problem $f(n)$ into two subproblems $f(n-1)$ and one subproblem $f(1)$, and solve these three subproblems in the following order.
From this, we can summarize the divide-and-conquer strategy for solving the Tower of Hanoi shown in Figure 12-14: divide the original problem $f(n)$ into two subproblems $f(n-1)$ and one subproblem $f(1)$, and solve these three subproblems in the following order.
1. Move $n-1$ discs with the help of `C` from `A` to `B`.
2. Move the remaining one disc directly from `A` to `C`.
@@ -532,7 +532,7 @@ In the code, we declare a recursive function `dfs(i, src, buf, tar)` whose role
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<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20move%28src%3A%20list%5Bint%5D,%20tar%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E7%A7%BB%E5%8A%A8%E4%B8%80%E4%B8%AA%E5%9C%86%E7%9B%98%22%22%22%0A%20%20%20%20%23%20%E4%BB%8E%20src%20%E9%A1%B6%E9%83%A8%E6%8B%BF%E5%87%BA%E4%B8%80%E4%B8%AA%E5%9C%86%E7%9B%98%0A%20%20%20%20pan%20%3D%20src.pop%28%29%0A%20%20%20%20%23%20%E5%B0%86%E5%9C%86%E7%9B%98%E6%94%BE%E5%85%A5%20tar%20%E9%A1%B6%E9%83%A8%0A%20%20%20%20tar.append%28pan%29%0A%0A%0Adef%20dfs%28i%3A%20int,%20src%3A%20list%5Bint%5D,%20buf%3A%20list%5Bint%5D,%20tar%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E6%B1%82%E8%A7%A3%E6%B1%89%E8%AF%BA%E5%A1%94%E9%97%AE%E9%A2%98%20f%28i%29%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%20src%20%E5%8F%AA%E5%89%A9%E4%B8%8B%E4%B8%80%E4%B8%AA%E5%9C%86%E7%9B%98%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E5%B0%86%E5%85%B6%E7%A7%BB%E5%88%B0%20tar%0A%20%20%20%20if%20i%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20move%28src,%20tar%29%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20%23%20%E5%AD%90%E9%97%AE%E9%A2%98%20f%28i-1%29%20%EF%BC%9A%E5%B0%86%20src%20%E9%A1%B6%E9%83%A8%20i-1%20%E4%B8%AA%E5%9C%86%E7%9B%98%E5%80%9F%E5%8A%A9%20tar%20%E7%A7%BB%E5%88%B0%20buf%0A%20%20%20%20dfs%28i%20-%201,%20src,%20tar,%20buf%29%0A%20%20%20%20%23%20%E5%AD%90%E9%97%AE%E9%A2%98%20f%281%29%20%EF%BC%9A%E5%B0%86%20src%20%E5%89%A9%E4%BD%99%E4%B8%80%E4%B8%AA%E5%9C%86%E7%9B%98%E7%A7%BB%E5%88%B0%20tar%0A%20%20%20%20move%28src,%20tar%29%0A%20%20%20%20%23%20%E5%AD%90%E9%97%AE%E9%A2%98%20f%28i-1%29%20%EF%BC%9A%E5%B0%86%20buf%20%E9%A1%B6%E9%83%A8%20i-1%20%E4%B8%AA%E5%9C%86%E7%9B%98%E5%80%9F%E5%8A%A9%20src%20%E7%A7%BB%E5%88%B0%20tar%0A%20%20%20%20dfs%28i%20-%201,%20buf,%20src,%20tar%29%0A%0A%0Adef%20solve_hanota%28A%3A%20list%5Bint%5D,%20B%3A%20list%5Bint%5D,%20C%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E6%B1%82%E8%A7%A3%E6%B1%89%E8%AF%BA%E5%A1%94%E9%97%AE%E9%A2%98%22%22%22%0A%20%20%20%20n%20%3D%20len%28A%29%0A%20%20%20%20%23%20%E5%B0%86%20A%20%E9%A1%B6%E9%83%A8%20n%20%E4%B8%AA%E5%9C%86%E7%9B%98%E5%80%9F%E5%8A%A9%20B%20%E7%A7%BB%E5%88%B0%20C%0A%20%20%20%20dfs%28n,%20A,%20B,%20C%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%97%E8%A1%A8%E5%B0%BE%E9%83%A8%E6%98%AF%E6%9F%B1%E5%AD%90%E9%A1%B6%E9%83%A8%0A%20%20%20%20A%20%3D%20%5B5,%204,%203,%202,%201%5D%0A%20%20%20%20B%20%3D%20%5B%5D%0A%20%20%20%20C%20%3D%20%5B%5D%0A%20%20%20%20print%28%22%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%E4%B8%8B%EF%BC%9A%22%29%0A%20%20%20%20print%28f%22A%20%3D%20%7BA%7D%22%29%0A%20%20%20%20print%28f%22B%20%3D%20%7BB%7D%22%29%0A%20%20%20%20print%28f%22C%20%3D%20%7BC%7D%22%29%0A%0A%20%20%20%20solve_hanota%28A,%20B,%20C%29%0A%0A%20%20%20%20print%28%22%E5%9C%86%E7%9B%98%E7%A7%BB%E5%8A%A8%E5%AE%8C%E6%88%90%E5%90%8E%EF%BC%9A%22%29%0A%20%20%20%20print%28f%22A%20%3D%20%7BA%7D%22%29%0A%20%20%20%20print%28f%22B%20%3D%20%7BB%7D%22%29%0A%20%20%20%20print%28f%22C%20%3D%20%7BC%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=12&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
As shown below, the Tower of Hanoi forms a recursive tree with a height of $n$, each node representing a subproblem, corresponding to an open `dfs()` function, **thus the time complexity is $O(2^n)$, and the space complexity is $O(n)$**.
As shown in Figure 12-15, the Tower of Hanoi forms a recursive tree with a height of $n$, each node representing a subproblem, corresponding to an open `dfs()` function, **thus the time complexity is $O(2^n)$, and the space complexity is $O(n)$**.
![Recursive tree of the Tower of Hanoi](hanota_problem.assets/hanota_recursive_tree.png){ class="animation-figure" }