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en/chapter_divide_and_conquer/hanota_problem/index.html
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<p>In both merge sorting and building binary trees, we decompose the original problem into two subproblems, each half the size of the original problem. However, for the Tower of Hanoi, we adopt a different decomposition strategy.</p>
<div class="admonition question">
<p class="admonition-title">Question</p>
<p>Given three pillars, denoted as <code>A</code>, <code>B</code>, and <code>C</code>. Initially, pillar <code>A</code> is stacked with <span class="arithmatex">\(n\)</span> discs, arranged in order from top to bottom from smallest to largest. Our task is to move these <span class="arithmatex">\(n\)</span> discs to pillar <code>C</code>, maintaining their original order (as shown below). The following rules must be followed during the disc movement process:</p>
<p>Given three pillars, denoted as <code>A</code>, <code>B</code>, and <code>C</code>. Initially, pillar <code>A</code> is stacked with <span class="arithmatex">\(n\)</span> discs, arranged in order from top to bottom from smallest to largest. Our task is to move these <span class="arithmatex">\(n\)</span> discs to pillar <code>C</code>, maintaining their original order (as shown in Figure 12-10). The following rules must be followed during the disc movement process:</p>
<ol>
<li>A disc can only be picked up from the top of a pillar and placed on top of another pillar.</li>
<li>Only one disc can be moved at a time.</li>
@@ -3624,7 +3624,7 @@
<p><strong>We denote the Tower of Hanoi of size <span class="arithmatex">\(i\)</span> as <span class="arithmatex">\(f(i)\)</span></strong>. For example, <span class="arithmatex">\(f(3)\)</span> represents the Tower of Hanoi of moving <span class="arithmatex">\(3\)</span> discs from <code>A</code> to <code>C</code>.</p>
<h3 id="1-consider-the-base-case">1. &nbsp; Consider the base case<a class="headerlink" href="#1-consider-the-base-case" title="Permanent link">&para;</a></h3>
<p>As shown below, for the problem <span class="arithmatex">\(f(1)\)</span>, i.e., when there is only one disc, we can directly move it from <code>A</code> to <code>C</code>.</p>
<p>As shown in Figure 12-11, for the problem <span class="arithmatex">\(f(1)\)</span>, i.e., when there is only one disc, we can directly move it from <code>A</code> to <code>C</code>.</p>
<div class="tabbed-set tabbed-alternate" data-tabs="1:2"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">&lt;1&gt;</label><label for="__tabbed_1_2">&lt;2&gt;</label></div>
<div class="tabbed-content">
<div class="tabbed-block">
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</div>
<p align="center"> Figure 12-11 &nbsp; Solution for a problem of size 1 </p>
<p>As shown below, for the problem <span class="arithmatex">\(f(2)\)</span>, i.e., when there are two discs, <strong>since the smaller disc must always be above the larger disc, <code>B</code> is needed to assist in the movement</strong>.</p>
<p>As shown in Figure 12-12, for the problem <span class="arithmatex">\(f(2)\)</span>, i.e., when there are two discs, <strong>since the smaller disc must always be above the larger disc, <code>B</code> is needed to assist in the movement</strong>.</p>
<ol>
<li>First, move the smaller disc from <code>A</code> to <code>B</code>.</li>
<li>Then move the larger disc from <code>A</code> to <code>C</code>.</li>
@@ -3664,7 +3664,7 @@
<p>The process of solving the problem <span class="arithmatex">\(f(2)\)</span> can be summarized as: <strong>moving two discs from <code>A</code> to <code>C</code> with the help of <code>B</code></strong>. Here, <code>C</code> is called the target pillar, and <code>B</code> is called the buffer pillar.</p>
<h3 id="2-decomposition-of-subproblems">2. &nbsp; Decomposition of subproblems<a class="headerlink" href="#2-decomposition-of-subproblems" title="Permanent link">&para;</a></h3>
<p>For the problem <span class="arithmatex">\(f(3)\)</span>, i.e., when there are three discs, the situation becomes slightly more complicated.</p>
<p>Since we already know the solutions to <span class="arithmatex">\(f(1)\)</span> and <span class="arithmatex">\(f(2)\)</span>, we can think from a divide-and-conquer perspective and <strong>consider the two top discs on <code>A</code> as a unit</strong>, performing the steps shown below. This way, the three discs are successfully moved from <code>A</code> to <code>C</code>.</p>
<p>Since we already know the solutions to <span class="arithmatex">\(f(1)\)</span> and <span class="arithmatex">\(f(2)\)</span>, we can think from a divide-and-conquer perspective and <strong>consider the two top discs on <code>A</code> as a unit</strong>, performing the steps shown in Figure 12-13. This way, the three discs are successfully moved from <code>A</code> to <code>C</code>.</p>
<ol>
<li>Let <code>B</code> be the target pillar and <code>C</code> the buffer pillar, and move the two discs from <code>A</code> to <code>B</code>.</li>
<li>Move the remaining disc from <code>A</code> directly to <code>C</code>.</li>
@@ -3689,7 +3689,7 @@
<p align="center"> Figure 12-13 &nbsp; Solution for a problem of size 3 </p>
<p>Essentially, <strong>we divide the problem <span class="arithmatex">\(f(3)\)</span> into two subproblems <span class="arithmatex">\(f(2)\)</span> and one subproblem <span class="arithmatex">\(f(1)\)</span></strong>. 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.</p>
<p>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 <span class="arithmatex">\(f(n)\)</span> into two subproblems <span class="arithmatex">\(f(n-1)\)</span> and one subproblem <span class="arithmatex">\(f(1)\)</span>, and solve these three subproblems in the following order.</p>
<p>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 <span class="arithmatex">\(f(n)\)</span> into two subproblems <span class="arithmatex">\(f(n-1)\)</span> and one subproblem <span class="arithmatex">\(f(1)\)</span>, and solve these three subproblems in the following order.</p>
<ol>
<li>Move <span class="arithmatex">\(n-1\)</span> discs with the help of <code>C</code> from <code>A</code> to <code>B</code>.</li>
<li>Move the remaining one disc directly from <code>A</code> to <code>C</code>.</li>
@@ -4113,7 +4113,7 @@
<p><div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="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=472&codeDivWidth=350&cumulative=false&curInstr=12&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
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</details>
<p>As shown below, the Tower of Hanoi forms a recursive tree with a height of <span class="arithmatex">\(n\)</span>, each node representing a subproblem, corresponding to an open <code>dfs()</code> function, <strong>thus the time complexity is <span class="arithmatex">\(O(2^n)\)</span>, and the space complexity is <span class="arithmatex">\(O(n)\)</span></strong>.</p>
<p>As shown in Figure 12-15, the Tower of Hanoi forms a recursive tree with a height of <span class="arithmatex">\(n\)</span>, each node representing a subproblem, corresponding to an open <code>dfs()</code> function, <strong>thus the time complexity is <span class="arithmatex">\(O(2^n)\)</span>, and the space complexity is <span class="arithmatex">\(O(n)\)</span></strong>.</p>
<p><a class="glightbox" href="../hanota_problem.assets/hanota_recursive_tree.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="Recursive tree of the Tower of Hanoi" class="animation-figure" src="../hanota_problem.assets/hanota_recursive_tree.png" /></a></p>
<p align="center"> Figure 12-15 &nbsp; Recursive tree of the Tower of Hanoi </p>