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第 8 章 &nbsp; 堆積
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第 9 章 &nbsp;
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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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10.2 &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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11.2 &nbsp; 選擇排序
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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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11.10 &nbsp; 基數排序
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第 12 章 &nbsp; 分治
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第 13 章 &nbsp; 回溯
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14.1 &nbsp; 初探動態規劃
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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 note">
<p class="admonition-title">Note</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>
</div>
<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><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><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><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></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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