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<li>Let the index of the current tree's root node in <code>inorder</code> be denoted as <span class="arithmatex">\(m\)</span>.</li>
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<li>Let the index interval of the current tree in <code>inorder</code> be denoted as <span class="arithmatex">\([l, r]\)</span>.</li>
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</ul>
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<p>As shown in the Table 12-1 , the above variables can represent the index of the root node in <code>preorder</code> as well as the index intervals of the subtrees in <code>inorder</code>.</p>
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<p>As shown in Table 12-1, the above variables can represent the index of the root node in <code>preorder</code> as well as the index intervals of the subtrees in <code>inorder</code>.</p>
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<p align="center"> Table 12-1 Indexes of the root node and subtrees in preorder and inorder traversals </p>
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<div class="center-table">
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@@ -3662,7 +3662,7 @@
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<li><strong>Divide (partition phase)</strong>: Recursively decompose the original problem into two or more sub-problems until the smallest sub-problem is reached and the process terminates.</li>
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<li><strong>Conquer (merge phase)</strong>: Starting from the smallest sub-problem with a known solution, merge the solutions of the sub-problems from bottom to top to construct the solution to the original problem.</li>
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</ol>
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<p>As shown in the Figure 12-1 , "merge sort" is one of the typical applications of the divide and conquer strategy.</p>
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<p>As shown in Figure 12-1, "merge sort" is one of the typical applications of the divide and conquer strategy.</p>
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<ol>
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<li><strong>Divide</strong>: Recursively divide the original array (original problem) into two sub-arrays (sub-problems), until the sub-array has only one element (smallest sub-problem).</li>
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<li><strong>Conquer</strong>: Merge the ordered sub-arrays (solutions to the sub-problems) from bottom to top to obtain an ordered original array (solution to the original problem).</li>
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<p><strong>Divide and conquer can not only effectively solve algorithm problems but often also improve algorithm efficiency</strong>. In sorting algorithms, quicksort, merge sort, and heap sort are faster than selection, bubble, and insertion sorts because they apply the divide and conquer strategy.</p>
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<p>Then, we may ask: <strong>Why can divide and conquer improve algorithm efficiency, and what is the underlying logic?</strong> In other words, why are the steps of decomposing a large problem into multiple sub-problems, solving the sub-problems, and merging the solutions of the sub-problems into the solution of the original problem more efficient than directly solving the original problem? This question can be discussed from the aspects of the number of operations and parallel computation.</p>
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<h3 id="1-optimization-of-operation-count">1. Optimization of operation count<a class="headerlink" href="#1-optimization-of-operation-count" title="Permanent link">¶</a></h3>
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<p>Taking "bubble sort" as an example, it requires <span class="arithmatex">\(O(n^2)\)</span> time to process an array of length <span class="arithmatex">\(n\)</span>. Suppose we divide the array from the midpoint into two sub-arrays as shown in the Figure 12-2 , then the division requires <span class="arithmatex">\(O(n)\)</span> time, sorting each sub-array requires <span class="arithmatex">\(O((n / 2)^2)\)</span> time, and merging the two sub-arrays requires <span class="arithmatex">\(O(n)\)</span> time, with the total time complexity being:</p>
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<p>Taking "bubble sort" as an example, it requires <span class="arithmatex">\(O(n^2)\)</span> time to process an array of length <span class="arithmatex">\(n\)</span>. Suppose we divide the array from the midpoint into two sub-arrays as shown in Figure 12-2, then the division requires <span class="arithmatex">\(O(n)\)</span> time, sorting each sub-array requires <span class="arithmatex">\(O((n / 2)^2)\)</span> time, and merging the two sub-arrays requires <span class="arithmatex">\(O(n)\)</span> time, with the total time complexity being:</p>
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<div class="arithmatex">\[
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O(n + (\frac{n}{2})^2 \times 2 + n) = O(\frac{n^2}{2} + 2n)
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\]</div>
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@@ -3708,7 +3708,7 @@ n(n - 4) & > 0
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<h3 id="2-optimization-through-parallel-computation">2. Optimization through parallel computation<a class="headerlink" href="#2-optimization-through-parallel-computation" title="Permanent link">¶</a></h3>
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<p>We know that the sub-problems generated by divide and conquer are independent of each other, <strong>thus they can usually be solved in parallel</strong>. This means that divide and conquer can not only reduce the algorithm's time complexity, <strong>but also facilitate parallel optimization by the operating system</strong>.</p>
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<p>Parallel optimization is especially effective in environments with multiple cores or processors, as the system can process multiple sub-problems simultaneously, making fuller use of computing resources and significantly reducing the overall runtime.</p>
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<p>For example, in the "bucket sort" shown in the Figure 12-3 , we distribute massive data evenly across various buckets, then the sorting tasks of all buckets can be distributed to different computing units, and the results are merged after completion.</p>
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<p>For example, in the "bucket sort" shown in Figure 12-3, we distribute massive data evenly across various buckets, then the sorting tasks of all buckets can be distributed to different computing units, and the results are merged after completion.</p>
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<p><a class="glightbox" href="../divide_and_conquer.assets/divide_and_conquer_parallel_computing.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="Bucket sort's parallel computation" class="animation-figure" src="../divide_and_conquer.assets/divide_and_conquer_parallel_computing.png" /></a></p>
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<p align="center"> Figure 12-3 Bucket sort's parallel computation </p>
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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>
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<div class="admonition question">
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<p class="admonition-title">Question</p>
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<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>
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<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>
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<ol>
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<li>A disc can only be picked up from the top of a pillar and placed on top of another pillar.</li>
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<li>Only one disc can be moved at a time.</li>
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<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>
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<h3 id="1-consider-the-base-case">1. Consider the base case<a class="headerlink" href="#1-consider-the-base-case" title="Permanent link">¶</a></h3>
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<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>
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<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>
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<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"><1></label><label for="__tabbed_1_2"><2></label></div>
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<div class="tabbed-content">
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<div class="tabbed-block">
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</div>
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<p align="center"> Figure 12-11 Solution for a problem of size 1 </p>
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<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>
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<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>
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<ol>
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<li>First, move the smaller disc from <code>A</code> to <code>B</code>.</li>
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<li>Then move the larger disc from <code>A</code> to <code>C</code>.</li>
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<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>
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<h3 id="2-decomposition-of-subproblems">2. Decomposition of subproblems<a class="headerlink" href="#2-decomposition-of-subproblems" title="Permanent link">¶</a></h3>
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<p>For the problem <span class="arithmatex">\(f(3)\)</span>, i.e., when there are three discs, the situation becomes slightly more complicated.</p>
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<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>
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<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>
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<ol>
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<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>
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<li>Move the remaining disc from <code>A</code> directly to <code>C</code>.</li>
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<p align="center"> Figure 12-13 Solution for a problem of size 3 </p>
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<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>
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<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>
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<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>
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<ol>
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<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>
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<li>Move the remaining one disc directly from <code>A</code> to <code>C</code>.</li>
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@@ -4113,7 +4113,7 @@
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<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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<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></p>
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</details>
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<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>
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<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>
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<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>
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<p align="center"> Figure 12-15 Recursive tree of the Tower of Hanoi </p>
|
||||
|
||||
|
||||
Reference in New Issue
Block a user