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7.2 &nbsp; 二元樹走訪
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7.3 &nbsp; 二元樹陣列表示
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7.4 &nbsp; 二元搜尋樹
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第 8 章 &nbsp; 堆積
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第 9 章 &nbsp; 圖
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9.2 &nbsp; 圖基礎操作
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第 10 章 &nbsp; 搜尋
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10.1 &nbsp; 二分搜尋
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10.2 &nbsp; 二分搜尋插入點
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10.3 &nbsp; 二分搜尋邊界
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10.4 &nbsp; 雜湊最佳化策略
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10.5 &nbsp; 重識搜尋演算法
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第 11 章 &nbsp; 排序
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11.1 &nbsp; 排序演算法
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11.2 &nbsp; 選擇排序
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11.3 &nbsp; 泡沫排序
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11.4 &nbsp; 插入排序
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11.5 &nbsp; 快速排序
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11.6 &nbsp; 合併排序
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11.7 &nbsp; 堆積排序
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11.8 &nbsp; 桶排序
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11.9 &nbsp; 計數排序
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11.10 &nbsp; 基數排序
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11.11 &nbsp; 小結
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第 12 章 &nbsp; 分治
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12.1 &nbsp; 分治演算法
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12.3 &nbsp; 構建樹問題
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12.4 &nbsp; 河內塔問題
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第 13 章 &nbsp; 回溯
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13.2 &nbsp; 全排列問題
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13.3 &nbsp; 子集和問題
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13.4 &nbsp; N 皇后問題
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第 14 章 &nbsp; 動態規劃
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14.1 &nbsp; 初探動態規劃
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14.2 &nbsp; DP 問題特性
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14.3 &nbsp; DP 解題思路
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14.4 &nbsp; 0-1 背包問題
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<h1 id="33">3.3 &nbsp; 數字編碼 *<a class="headerlink" href="#33" title="Permanent link">&para;</a></h1>
<div class="admonition tip">
<p class="admonition-title">Tip</p>
<p>在本書中，標題帶有 * 符號的是選讀章節。如果你時間有限或感到理解困難，可以先跳過，等學完必讀章節後再單獨攻克。</p>
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<h2 id="331">3.3.1 &nbsp; 原碼、一補數和二補數<a class="headerlink" href="#331" title="Permanent link">&para;</a></h2>
<p>在上一節的表格中我們發現，所有整數型別能夠表示的負數都比正數多一個，例如 <code>byte</code> 的取值範圍是 <span class="arithmatex">\([-128, 127]\)</span> 。這個現象比較反直覺，它的內在原因涉及原碼、一補數、二補數的相關知識。</p>
<p>首先需要指出，<strong>數字是以“二補數”的形式儲存在計算機中的</strong>。在分析這樣做的原因之前，首先給出三者的定義。</p>
<ul>
<li><strong>原碼</strong>：我們將數字的二進位制表示的最高位視為符號位，其中 <span class="arithmatex">\(0\)</span> 表示正數，<span class="arithmatex">\(1\)</span> 表示負數，其餘位表示數字的值。</li>
<li><strong>一補數</strong>：正數的一補數與其原碼相同，負數的一補數是對其原碼除符號位外的所有位取反。</li>
<li><strong>二補數</strong>：正數的二補數與其原碼相同，負數的二補數是在其一補數的基礎上加 <span class="arithmatex">\(1\)</span> 。</li>
</ul>
<p>圖 3-4 展示了原碼、一補數和二補數之間的轉換方法。</p>
<p><a class="glightbox" href="../number_encoding.assets/1s_2s_complement.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="原碼、一補數與二補數之間的相互轉換" class="animation-figure" src="../number_encoding.assets/1s_2s_complement.png" /></a></p>
<p align="center"> 圖 3-4 &nbsp; 原碼、一補數與二補數之間的相互轉換 </p>
<p><u>原碼（sign-magnitude）</u>雖然最直觀，但存在一些侷限性。一方面，<strong>負數的原碼不能直接用於運算</strong>。例如在原碼下計算 <span class="arithmatex">\(1 + (-2)\)</span> ，得到的結果是 <span class="arithmatex">\(-3\)</span> ，這顯然是不對的。</p>
<div class="arithmatex">\[
\begin{aligned}
&amp; 1 + (-2) \newline
&amp; \rightarrow 0000 \; 0001 + 1000 \; 0010 \newline
&amp; = 1000 \; 0011 \newline
&amp; \rightarrow -3
\end{aligned}
\]</div>
<p>為了解決此問題，計算機引入了<u>一補數（1's complement）</u>。如果我們先將原碼轉換為一補數，並在一補數下計算 <span class="arithmatex">\(1 + (-2)\)</span> ，最後將結果從一補數轉換回原碼，則可得到正確結果 <span class="arithmatex">\(-1\)</span> 。</p>
<div class="arithmatex">\[
\begin{aligned}
&amp; 1 + (-2) \newline
&amp; \rightarrow 0000 \; 0001 \; \text{(原碼)} + 1000 \; 0010 \; \text{(原碼)} \newline
&amp; = 0000 \; 0001 \; \text{(一補數)} + 1111 \; 1101 \; \text{(一補數)} \newline
&amp; = 1111 \; 1110 \; \text{(一補數)} \newline
&amp; = 1000 \; 0001 \; \text{(原碼)} \newline
&amp; \rightarrow -1
\end{aligned}
\]</div>
<p>另一方面，<strong>數字零的原碼有 <span class="arithmatex">\(+0\)</span> 和 <span class="arithmatex">\(-0\)</span> 兩種表示方式</strong>。這意味著數字零對應兩個不同的二進位制編碼，這可能會帶來歧義。比如在條件判斷中，如果沒有區分正零和負零，則可能會導致判斷結果出錯。而如果我們想處理正零和負零歧義，則需要引入額外的判斷操作，這可能會降低計算機的運算效率。</p>
<div class="arithmatex">\[
\begin{aligned}
+0 &amp; \rightarrow 0000 \; 0000 \newline
-0 &amp; \rightarrow 1000 \; 0000
\end{aligned}
\]</div>
<p>與原碼一樣，一補數也存在正負零歧義問題，因此計算機進一步引入了<u>二補數（2's complement）</u>。我們先來觀察一下負零的原碼、一補數、二補數的轉換過程：</p>
<div class="arithmatex">\[
\begin{aligned}
-0 \rightarrow \; &amp; 1000 \; 0000 \; \text{(原碼)} \newline
= \; &amp; 1111 \; 1111 \; \text{(一補數)} \newline
= 1 \; &amp; 0000 \; 0000 \; \text{(二補數)} \newline
\end{aligned}
\]</div>
<p>在負零的一補數基礎上加 <span class="arithmatex">\(1\)</span> 會產生進位，但 <code>byte</code> 型別的長度只有 8 位，因此溢位到第 9 位的 <span class="arithmatex">\(1\)</span> 會被捨棄。也就是說，<strong>負零的二補數為 <span class="arithmatex">\(0000 \; 0000\)</span> ，與正零的二補數相同</strong>。這意味著在二補數表示中只存在一個零，正負零歧義從而得到解決。</p>
<p>還剩最後一個疑惑：<code>byte</code> 型別的取值範圍是 <span class="arithmatex">\([-128, 127]\)</span> ，多出來的一個負數 <span class="arithmatex">\(-128\)</span> 是如何得到的呢？我們注意到，區間 <span class="arithmatex">\([-127, +127]\)</span> 內的所有整數都有對應的原碼、一補數和二補數，並且原碼和二補數之間可以互相轉換。</p>
<p>然而，<strong>二補數 <span class="arithmatex">\(1000 \; 0000\)</span> 是一個例外，它並沒有對應的原碼</strong>。根據轉換方法，我們得到該二補數的原碼為 <span class="arithmatex">\(0000 \; 0000\)</span> 。這顯然是矛盾的，因為該原碼表示數字 <span class="arithmatex">\(0\)</span> ，它的二補數應該是自身。計算機規定這個特殊的二補數 <span class="arithmatex">\(1000 \; 0000\)</span> 代表 <span class="arithmatex">\(-128\)</span> 。實際上，<span class="arithmatex">\((-1) + (-127)\)</span> 在二補數下的計算結果就是 <span class="arithmatex">\(-128\)</span> 。</p>
<div class="arithmatex">\[
\begin{aligned}
&amp; (-127) + (-1) \newline
&amp; \rightarrow 1111 \; 1111 \; \text{(原碼)} + 1000 \; 0001 \; \text{(原碼)} \newline
&amp; = 1000 \; 0000 \; \text{(一補數)} + 1111 \; 1110 \; \text{(一補數)} \newline
&amp; = 1000 \; 0001 \; \text{(二補數)} + 1111 \; 1111 \; \text{(二補數)} \newline
&amp; = 1000 \; 0000 \; \text{(二補數)} \newline
&amp; \rightarrow -128
\end{aligned}
\]</div>
<p>你可能已經發現了，上述所有計算都是加法運算。這暗示著一個重要事實：<strong>計算機內部的硬體電路主要是基於加法運算設計的</strong>。這是因為加法運算相對於其他運算（比如乘法、除法和減法）來說，硬體實現起來更簡單，更容易進行並行化處理，運算速度更快。</p>
<p>請注意，這並不意味著計算機只能做加法。<strong>透過將加法與一些基本邏輯運算結合，計算機能夠實現各種其他的數學運算</strong>。例如，計算減法 <span class="arithmatex">\(a - b\)</span> 可以轉換為計算加法 <span class="arithmatex">\(a + (-b)\)</span> ；計算乘法和除法可以轉換為計算多次加法或減法。</p>
<p>現在我們可以總結出計算機使用二補數的原因：基於二補數表示，計算機可以用同樣的電路和操作來處理正數和負數的加法，不需要設計特殊的硬體電路來處理減法，並且無須特別處理正負零的歧義問題。這大大簡化了硬體設計，提高了運算效率。</p>
<p>二補數的設計非常精妙，因篇幅關係我們就先介紹到這裡，建議有興趣的讀者進一步深入瞭解。</p>
<h2 id="332">3.3.2 &nbsp; 浮點數編碼<a class="headerlink" href="#332" title="Permanent link">&para;</a></h2>
<p>細心的你可能會發現：<code>int</code> 和 <code>float</code> 長度相同，都是 4 位元組 ，但為什麼 <code>float</code> 的取值範圍遠大於 <code>int</code> ？這非常反直覺，因為按理說 <code>float</code> 需要表示小數，取值範圍應該變小才對。</p>
<p>實際上，<strong>這是因為浮點數 <code>float</code> 採用了不同的表示方式</strong>。記一個 32 位元長度的二進位制數為：</p>
<div class="arithmatex">\[
b_{31} b_{30} b_{29} \ldots b_2 b_1 b_0
\]</div>
<p>根據 IEEE 754 標準，32-bit 長度的 <code>float</code> 由以下三個部分構成。</p>
<ul>
<li>符號位 <span class="arithmatex">\(\mathrm{S}\)</span> ：佔 1 位 ，對應 <span class="arithmatex">\(b_{31}\)</span> 。</li>
<li>指數位 <span class="arithmatex">\(\mathrm{E}\)</span> ：佔 8 位 ，對應 <span class="arithmatex">\(b_{30} b_{29} \ldots b_{23}\)</span> 。</li>
<li>分數位 <span class="arithmatex">\(\mathrm{N}\)</span> ：佔 23 位 ，對應 <span class="arithmatex">\(b_{22} b_{21} \ldots b_0\)</span> 。</li>
</ul>
<p>二進位制數 <code>float</code> 對應值的計算方法為：</p>
<div class="arithmatex">\[
\text {val} = (-1)^{b_{31}} \times 2^{\left(b_{30} b_{29} \ldots b_{23}\right)_2-127} \times\left(1 . b_{22} b_{21} \ldots b_0\right)_2
\]</div>
<p>轉化到十進位制下的計算公式為：</p>
<div class="arithmatex">\[
\text {val}=(-1)^{\mathrm{S}} \times 2^{\mathrm{E} -127} \times (1 + \mathrm{N})
\]</div>
<p>其中各項的取值範圍為：</p>
<div class="arithmatex">\[
\begin{aligned}
\mathrm{S} \in &amp; \{ 0, 1\}, \quad \mathrm{E} \in \{ 1, 2, \dots, 254 \} \newline
(1 + \mathrm{N}) = &amp; (1 + \sum_{i=1}^{23} b_{23-i} 2^{-i}) \subset [1, 2 - 2^{-23}]
\end{aligned}
\]</div>
<p><a class="glightbox" href="../number_encoding.assets/ieee_754_float.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="IEEE 754 標準下的 float 的計算示例" class="animation-figure" src="../number_encoding.assets/ieee_754_float.png" /></a></p>
<p align="center"> 圖 3-5 &nbsp; IEEE 754 標準下的 float 的計算示例 </p>
<p>觀察圖 3-5 ，給定一個示例資料 <span class="arithmatex">\(\mathrm{S} = 0\)</span> ， <span class="arithmatex">\(\mathrm{E} = 124\)</span> ，<span class="arithmatex">\(\mathrm{N} = 2^{-2} + 2^{-3} = 0.375\)</span> ，則有：</p>
<div class="arithmatex">\[
\text { val } = (-1)^0 \times 2^{124 - 127} \times (1 + 0.375) = 0.171875
\]</div>
<p>現在我們可以回答最初的問題：<strong><code>float</code> 的表示方式包含指數位，導致其取值範圍遠大於 <code>int</code></strong> 。根據以上計算，<code>float</code> 可表示的最大正數為 <span class="arithmatex">\(2^{254 - 127} \times (2 - 2^{-23}) \approx 3.4 \times 10^{38}\)</span> ，切換符號位便可得到最小負數。</p>
<p><strong>儘管浮點數 <code>float</code> 擴展了取值範圍，但其副作用是犧牲了精度</strong>。整數型別 <code>int</code> 將全部 32 位元用於表示數字，數字是均勻分佈的；而由於指數位的存在，浮點數 <code>float</code> 的數值越大，相鄰兩個數字之間的差值就會趨向越大。</p>
<p>如表 3-2 所示，指數位 <span class="arithmatex">\(\mathrm{E} = 0\)</span> 和 <span class="arithmatex">\(\mathrm{E} = 255\)</span> 具有特殊含義，<strong>用於表示零、無窮大、<span class="arithmatex">\(\mathrm{NaN}\)</span> 等</strong>。</p>
<p align="center"> 表 3-2 &nbsp; 指數位含義 </p>
<div class="center-table">
<table>
<thead>
<tr>
<th>指數位 E</th>
<th>分數位 <span class="arithmatex">\(\mathrm{N} = 0\)</span></th>
<th>分數位 <span class="arithmatex">\(\mathrm{N} \ne 0\)</span></th>
<th>計算公式</th>
</tr>
</thead>
<tbody>
<tr>
<td><span class="arithmatex">\(0\)</span></td>
<td><span class="arithmatex">\(\pm 0\)</span></td>
<td>次正規數</td>
<td><span class="arithmatex">\((-1)^{\mathrm{S}} \times 2^{-126} \times (0.\mathrm{N})\)</span></td>
</tr>
<tr>
<td><span class="arithmatex">\(1, 2, \dots, 254\)</span></td>
<td>正規數</td>
<td>正規數</td>
<td><span class="arithmatex">\((-1)^{\mathrm{S}} \times 2^{(\mathrm{E} -127)} \times (1.\mathrm{N})\)</span></td>
</tr>
<tr>
<td><span class="arithmatex">\(255\)</span></td>
<td><span class="arithmatex">\(\pm \infty\)</span></td>
<td><span class="arithmatex">\(\mathrm{NaN}\)</span></td>
<td></td>
</tr>
</tbody>
</table>
</div>
<p>值得說明的是，次正規數顯著提升了浮點數的精度。最小正正規數為 <span class="arithmatex">\(2^{-126}\)</span> ，最小正次正規數為 <span class="arithmatex">\(2^{-126} \times 2^{-23}\)</span> 。</p>
<p>雙精度 <code>double</code> 也採用類似於 <code>float</code> 的表示方法，在此不做贅述。</p>
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