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    <![CDATA[<p>你好，欢迎访问个人主页！</p><p>擅长密码学，安全分析，数字水印等技术。</p><p>你可以联系我通过:findmykexin@gmail.com或者知乎私信。</p><p>我的知乎链接：<a href="https://www.zhihu.com/people/su-chi-dan-dao" rel="noopener noreferrer" target="_blank">苏迟但到 - 知乎 (zhihu.com)</a></p><p>我的github链接：<a href="https://github.com/kexinoh" rel="noopener noreferrer" target="_blank">kexinoh</a></p>]]>
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    <title>2^k 个 均匀分布的 m位整型数据集合，如何压缩达到最大的压缩率？</title>
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    <pubDate>Mon, 06 Nov 2023 11:18:40 GMT</pubDate>
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      <![CDATA[<p data-pid="E9NnLz8k">对于任何压缩问题我们都需要计算一下信息熵。</p><p data-pid="uWw0W1Qm">对于2^32个64位的整型数据集合（无序），那么</p><p data-pid="i3YoIPTa"><img src="https://www.zhihu.com/equation?tex=H%3D%5Cfrac%7B2%5E%7B64%2A2%5E%7B32%7D%7D%7D%7B2%5E%7B32%7D%21%7D" alt="H=\frac{2^{64*2^{32}}}{2^{32}!}" eeimg="1"/> </p><p data-pid="6cEU5zu8">那么它的压缩率可以通过取对数和stirling公式进行估算</p><div class="highlight"><pre><code class="language-python3"><span class="p">(</span><span class="n">log</span><span class="p">(</span><span class="mi">2</span><span class="o">^</span><span class="mi">32</span><span class="o">*</span><span class="mi">2</span><span class="o">*</span><span class="mf">3.1415926</span><span class="p">,</span><span class="mi">2</span><span class="p">)</span><span class="o">+</span><span class="mi">2</span><span class="o">^</span><span class="mi">32</span><span class="o">*</span><span class="n">log</span><span class="p">((</span><span class="mi">2</span><span class="o">^</span><span class="mi">32</span><span class="p">)</span><span class="o">/</span><span class="mf">2.718</span><span class="p">,</span><span class="mi">2</span><span class="p">))</span><span class="o">/</span><span class="p">(</span><span class="mi">64</span><span class="o">*</span><span class="mi">2</span><span class="o">^</span><span class="mi">32</span><span class="p">)</span></code></pre></div><p data-pid="aTO0MtRl">使用sagemath运算可以得到结果是47.7%左右。</p><p data-pid="bvC3a1M3">而我认为压缩方法上面，一阶差分方法的效果可能还不够好，二阶差分法就应该接近极限了。</p><p data-pid="IqtUGC3E">因为我们知道两个数字间隔平均在2^32的间隔，那么我们应当取差分值与2^32的偏差值(像APPLE的图片一样处理)，那么就可以再节约一些，从而达到最优化。</p>]]>
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