{"version":"https://jsonfeed.org/version/1.1","title":"苏迟但到的主页","home_page_url":"https://kexohproject.pages.dev","feed_url":"https://kexohproject.pages.dev/json/","description":"<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>","icon":"https://kexohcdn.gptapi.cyou/kexohproject/production/images/channel-2e54d141ee195646ca12a9d16507a908.jpg","favicon":"https://kexohcdn.gptapi.cyou/kexohproject/production/images/favicon-340a2925d02a0386f3b954a032834917.jpg","authors":[{"name":"苏迟但到"}],"language":"zh-cn","items":[{"id":"OE6tflFLR-S","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？","content_html":"<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>","content_text":"对于任何压缩问题我们都需要计算一下信息熵。\n\n对于2^32个64位的整型数据集合（无序），那么\n\nH=\\frac{2^{64*2^{32}}}{2^{32}!}\n[https://www.zhihu.com/equation?tex=H%3D%5Cfrac%7B2%5E%7B64%2A2%5E%7B32%7D%7D%7D%7B2%5E%7B32%7D%21%7D]\n\n那么它的压缩率可以通过取对数和stirling公式进行估算\n\n(log(2^32*2*3.1415926,2)+2^32*log((2^32)/2.718,2))/(64*2^32)\n\n使用sagemath运算可以得到结果是47.7%左右。\n\n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。\n\n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。","date_published":"2023-11-06T11:18:40.000Z","_microfeed":{"web_url":"https://kexohproject.pages.dev/i/2k-m-OE6tflFLR-S/","json_url":"https://kexohproject.pages.dev/i/OE6tflFLR-S/json/","rss_url":"https://kexohproject.pages.dev/i/OE6tflFLR-S/rss/","guid":"OE6tflFLR-S","status":"published","itunes:title":"New Article Title for iTunes","date_published_short":"Mon Nov 06 2023","date_published_ms":1699269520000}}],"_microfeed":{"microfeed_version":"0.1.2","base_url":"https://kexohproject.pages.dev","categories":[{"name":"Education","categories":[{"name":"Language Learning"}]},{"name":"Technology"}],"subscribe_methods":[{"name":"RSS","type":"rss","url":"https://kexohproject.pages.dev/rss/","image":"https://kexohproject.pages.dev/assets/brands/subscribe/rss.png","enabled":true,"editable":false,"id":"4KlfbtkEfzy"},{"name":"JSON","type":"json","url":"https://kexohproject.pages.dev/json/","image":"https://kexohproject.pages.dev/assets/brands/subscribe/json.png","enabled":true,"editable":false,"id":"DVFm7TYiNSq"}],"description_text":"你好，欢迎访问个人主页！\n\n擅长密码学，安全分析，数字水印等技术。\n\n你可以联系我通过:findmykexin@gmail.com或者知乎私信。\n\n我的知乎链接：苏迟但到 - 知乎 (zhihu.com)\n\n我的github链接：kexinoh","copyright":"©2024","itunes:type":"episodic","items_sort_order":"newest_first"}}