
大部分语言文字书写方向是 LTR(从左写到右) 方向，但是还有一些语言文字是 RTL (从右写到左) 方向，比如阿拉伯文，希伯来文、乌尔都文等。使用这些语言的 UI 界面也需要做一些细节调整。尤其是阿拉伯语，是世界上目前有 6 亿人正在使用的文字语言，相应的，如果一个软件界面使用了阿拉伯语等 RTL 语言，也需要在 UI 做一些调整。

RTL 适配的内容较为琐碎，本文就尝试做一些总结。

## UI 与功能界面

我们常用的软件界面需要把功能区按照镜像方式颠倒过来。比如：

如果导航从左向右展开，RTL 页面应从右向左展开。如果有侧边栏、侧边菜单，那么应当把它放到右边。窗口的关闭按钮同样也放在右边

比如下图，就是设置页面，两种方向的对比。

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250117122440.png)

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250117122547.png)

同样，底部菜单栏、水平菜单栏等图标排序方向，也是从右到左。

如果网站有面包屑（crumb）组件，那么排序方式也应当翻转过来，比如 `Home > Category > Subcategory` 翻转过来后就是 `Subcategory < Category < Home`

如果有时间线组件，那么记住：事件发生顺序要从右到左排列。尤其是要注意 **折线图等含有时间轴的统计图表，顺序也是从右到左。**
对于条形加载进度条，进度方向也要做出相应的调整

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250117122828.png)

+ 如果有 Toggle Switch 组件，要记住，开（on）状态的指示器应在左侧，关（off）状态在右侧。如下图，

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250117121142.png)

## 文字与符号

阿拉伯文默认对齐方式是右对齐。

对于阿拉伯文，虽然书写方向是 RTL，但是在混排情况下，还有一些内容是 LTR。比如常用的印度 - 阿拉伯数字，仍然是从左写到右。比如“一千二百三十四”这个数字，仍然写为 $1234$ 而不是 $4321$。在阿拉伯文中，与数字混写时，阿拉伯文是 RTL，但数字是 LTR。同理，如果阿拉伯文中夹杂少量英文单词，那么英文单词也是 LTR。

```plaintext
هذا نص مختلط، هذا هو الرقم 1234، هذه هي الكلمة الإنجليزية English
```

另一方面，阿拉伯人写的数学公式比较特殊，有两种情况，一是使用他们自己的阿拉伯字母作为公式符号，二是跟我们一样使用拉丁、希腊字母作为公式符号。

这两种都需要分情况处理。

**第一种，如果阿拉伯人的数学公式，里面的变量、未知数等使用的是阿拉伯字母符号，那么公式必须镜像化处理。**

不仅推导方向也要改写成从右到左，一加一等于二就不能写成 $1 + 1 = 2$ 而是 $2 = 1 + 1$. 比如下图的公式，算式是从右到左来进行推导演算的，

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250117104949.png)

还有符号的方向也要镜像化，比如下图中就是一个阿拉伯的高等数学网课截图，注意不定积分的符号方向

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250117105825.png)

还有数学根式

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250117105939.png)

**但是，数学公式里没有使用阿拉伯字母符号，而是跟我们一样，使用了国际惯例的拉丁、希腊字母符号，就不需要镜像化处理。** 比如下面的公式，里面使用了 $x,e,i$拉丁字母作为未知数和变量符号，那么在书写的时候就不需要镜像化。

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250117125222.png)

## 图标

对于图标，也有一点要注意：虽然阿拉伯文字是 RTL，**但是阿拉伯人仍然是右撇子，做事还是跟我们一样习惯使用右手。** 所以部分图标也不能简单镜像化。

比如这个放大镜图标，大部分人仍然习惯用右手去拿放大镜，所以设计的时候，手柄位置仍然在右手位置，不要镜像过来。

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250117114401.png)

还有，这个笔记图标，也是同样的道理，阿拉伯人还是用右手写字，所以设计图标的时候，笔仍然要在右手位置，不要镜像过来。

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250117114659.png)

一般来说，大部分拟物图标都是按照人类的右手习惯来设计的，在 RTL 布局的设计中，一般不需要操心图标的翻转问题。

**不过还是要注意，如果图标中涉及到文本结构，那么图标还是需要反转**，比如下图中，表达“目录树”和“文本”的图标，都需要做调整。

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250117123437.png)

## 运动与行为

在 RTL 的设计中，要注意箭头代表的运动方向，比如分页组件、幻灯片按钮中的“上一项”和“下一项”按钮的箭头，**箭头方向不是相反的，但是逻辑是相反的。**

在我们的标准中，左箭头 `⬅` 代表“上一项”，右箭头 `➡` 代表下一项。

但是在阿拉伯地区中，**左箭头 `⬅` 代表“下一项”，右箭头 `➡` 代表上一项。** 比如在[这个阿拉伯新闻网站](https://www.alarabiya.net/editor-choice/archive)的新闻列表中，底部分页组件长这样，当前页面是第一页，想跳到下一页，就需要点击左边的左箭头按钮。

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250117112944.png)

这种运动颠倒行为也可以类比，再比如轮播图组件，对于 RTL 布局，向右滑动是“上一张”，向左滑动是“下一张”。同理，导航中的“前进”和“后退”按钮也要调整过来。

还有一点需要注意，在对话框中：**取消按钮（Cancel）通常在左侧，确认按钮（Confirm）在右侧。**

比如下图中的对话框，左侧是取消按钮，右侧是确认按钮。

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250117115156.png)

同样的，在 UI 交互界面中，也要把高频使用的，或者希望用户电机的按钮放在**右侧**，把低频使用的、不希望用户按的按钮放在**左侧**。

在动画设计中也要做出相应的调整，“进入”动画要设计成“从右到左”，“退出”动画要设计成“从左到右”。

在移动端，尤其是要注意滑动、拖动方向，在拖动菜单中也要做出调整

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250117122034.png)

## 前端设计实现

如果你的前端页面支持 i18n，如果想要支持阿拉伯语、希伯来语等 RTL 文字，那么以上提到的几点需要注意。在前端设计的布局中，如何支持 RTL，有以下几点启示。

一般来说，浏览器默认是 LTR 布局的，如果你想切换到 RTL，很简单，只需要在 `<html>` 或者 `<body>` 的 DOM 元素上添加 `dir` 属性，就可以把整个页面设置为 RTL。

```html
<html dir="rtl">
  <head>
  </head>
  <body>
  </body>
</html>
```

这样就可以改变排版流方向了，比如对于本博客页面

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250117124134.png)

在 CSS 上，有两种方法可以各自适应 RTL 和 LTR。

第一种是手动法，你可以分别设置针对 RTL 和 LTR 元素的规则的选择器，比如下面的规则，分别对左右的内间距进行设置。

```css
/* 针对 LTR 的样式 */
[dir="ltr"] .container {
  padding-left: 20px;
  padding-right: 10px;
}

/* 针对 RTL 的样式 */
[dir="rtl"] .container {
  padding-left: 10px;
  padding-right: 20px;
}
```

另一种是使用 CSS 的逻辑性规则。比如 `margin-inline-start` 规则，可以自动检测当前页面的方向，然后再自动做出调整。

```css
.container {
  text-align: start; /* LTR 时等效为 left，RTL 时等效为 right */
  margin-inline-start: 20px; /* LTR 时等效为左侧间距，RTL 时等效为右侧间距 */
  margin-inline-end: 10px; /* LTR 时等效为右侧间距，RTL 时等效为左侧间距 */
}
```

在现有的 CSS 框架中，比如 Tailwind CSS，已经有开箱即用的、现成的 RTL 支持了。你可以使用 `rtl:`或者 `ltr:`前缀来自动根据当前布局应用类。

另外，**不要再使用 `ml-*,mr-*,left-*,right-*` 之类的类，也不要再使用`margin-left`之类的规则**。因为他们是针对具体的客观方向设置距离，而不是针对具体的逻辑方向设置距离。这对自动适应 RTL 非常不利。

而应当使用 `ms-*,me-*,start-*,end-*` 等类，或者 `margin-start, margin-end` 之类的规则，CSS 会自动根据当前的流方向来设置距离，以自动匹配 RTL 和 LTR。

[这是文档对此部分的说明。](https://tailwindcss.com/docs/hover-focus-and-other-states#rtl-support)

在上文中也提及到，像“前进”、“后退”按钮，还是“Toggle Switch”组件，它们的逻辑是相反的。所以，在 React 中，可以设置一个全局 State，来指示当前的布局是 RTL 和 LTR，以修改按钮的行为。并且，为每个 i18n 语言都设置对应的 RTL 和 LTR 属性，切换语言时也要自动调整布局的 State。


大部分语言文字书写方向是 LTR(从左写到右) 方向，但是还有一些语言文字是 RTL (从右写到左) 方向，比如阿拉伯文，希伯来文、乌尔都文等。使用这些语言的 UI 界面也需要做一些细节调整。

I18n: RTL 文字方向的 UI 设计细节


在 2024 年上半年，我开始了为期一年的背单词计划，每天坚持记忆、复习 200-400 个单词，坚持了将近一年，大约累计背了 16000 个单词，把 IELTS，TOFEL，专业四级和大半个 SAT 词书都背完了。一路走过来再回顾这些历程，虽然过程极度痛苦，日日夜夜，但好歹也算走过来了，想想也挺不容易的哈。

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250116194715.png)

在背单词的过程中，我也累积了一些经验。我打算把这些经验总结一下，然后再分成几个篇幅来依次说说。本次想要分享的是：挑选词典。在 2024 年上半年，我买入了四本词典，都是牛津学习词典系列。

![牛津短语动词词典，和新旧两版同义词词典](https://raw.githubusercontent.com/codemetic/pictures-host/main/imagesee97854c285b3202284930edd6fef07.webp)

![牛津习语词典](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250116170944.webp)

我选的这些词典，它与像牛津词典、朗文词典等知名的综合大词典不同，它属于“专题词典”，针对某一目的或者主题来收录词条，使用目的性较强，专门为学习者设计，使用“专题词典”来对英语进行有目的的、针对性的学习，效率更高。这也是我目前正在学习的方法。

英语的学习难点在于词汇、词法。据统计，一个成年美国人的基础词汇量是 20000-30000 左右，也是一个美国高中生的词汇水平。也就是说，如果你想毫无障碍的阅读大多数美国语料，那么你的词汇量必须达到两万以上，才能勉强不卡壳。所以学习英语的过程也是在不断征服单词的过程。

## 同义词词典

首先是介绍同义词词典。

对任何有过背单词经历的同学来讲，英语的同义词现象非常常见，每次背单词的时候经常会受到同义词的困扰，每一个“含义”几乎都有 3-4 个同义词！

因为英语在历史的发展进程中吸收了大量的外来语，所以现在几乎是世界上词汇量最大的语言，而其中的同义词、近义词就占据了大半。确实，相比其他语言如日语、法语、西班牙语等等，**英语的同义词现象时最严重的。**

举个例子，“拒绝”的表达，就有 Refuse、Reject、Deny、Rebuff、Dismiss、Turn down、Oppose、Withhold、Repudiate、Resist 等等，它们都有“拒绝”之意。如果你单独把这些词一个一个去记，那么很难把它们“内化”成自己的一部分。在从外界接收、背诵了一大堆单词之后，久而久之，大脑还是处于无组织、混乱的状态。

假如你在英语写作和英语对话中，想要表达“拒绝”时，你能在大脑中一瞬间枚举出以上所有的表达吗？而且上面的同义词，在不同语境下都会有一些很细微的差别，你能思索一番后，举一反三，选取出最合适的表达词汇吗？例如：refuse 和 decline 都表示“拒绝”，但 decline 更礼貌，而 refuse 更直接。

还有，雅思考试、托福考试对单词的考题，大都是以“同义词替换”来考察的。需要识别同义词和近义词来回答阅读和词汇题。

所以归纳总结就非常有必要。这样才能把消极词汇转化为主动词汇。在背单词时，把不同的单词按照同一个或者相似的 idea 进行总结就尤为重要。归纳同义词和近义词不仅是词汇学习的一种技巧，更是一种提升语言能力的策略。它可以让学习者从“记单词”转向“用单词”，从而在实际交流中更自信、精准地表达自己。

还有，从同义词入手，可以显著提高背单词的效率，对提升词汇量都有着不小的帮助。

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250116173639.webp)

这也是同义词词典存在的意义。我推荐牛津英语同义词词典，它按照首字母编序，为每个单词都枚举出了对应的同义词和近义词。

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250116175401.webp)

如上图，举个例子，tedious 在表达“乏味、无聊”时的同义词，就有 boring,dreary,dull,monotonous,tire 等。它把对应的意思的相关近义词、同义词全部枚举出来，对归纳总结和提升词汇量都有不小的帮助。

它由于是 2001 年的早期初版，在淘宝上可能买不到新书了，只有二手书，价格低廉，大约 15 元以内就可以包邮到家。当然，不想花钱也可以下载电子书。同义词词典的初版在 Zlibrary 上可以找到电子版：[牛津同义词词典 - Zlibrary 链接](https://z-library.sk/book/22266968/4f2cc7/%E7%89%9B%E6%B4%A5%E8%8B%B1%E8%AF%AD%E5%90%8C%E4%B9%89%E8%AF%8D%E8%AF%8D%E5%85%B8.html)。

但是这本书还有一个新版，是 2014 年出版的，它还对近义词加上了辨析内容。

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250116181118.webp)

它与初版相比，更注重于辨析，所以它只收录了常见同义词，删减了很多，而且价格也比初版更贵（160 元）

比如对“拒绝”的同义词，它的收录内容预览如下。

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250116175951.webp)

**但是，我还是建议选初版的同义词词典。** 一来，我个人推荐在大量阅读的过程中学习同义词辨析，而不是词典中。二来，新版的同义词词典只收录了一些初级的、常用的词汇，对于进阶一点的词汇都删减了。**而且价格不菲，性价比不高。** 新版的词典我在网上没有找到电子版本。

所以还是推荐使用初版的，虽然初版词典没有辨析内容，但是它相较于新版的，词量更加全面。

当然，为了学习方便，我把这本词典做成了词表，[发布在了 Github 上](https://github.com/PrinOrange/my-english-word-books/tree/main/dictionaries/%E7%89%9B%E6%B4%A5%E8%8B%B1%E8%AF%AD%E5%90%8C%E4%B9%89%E8%AF%8D%E8%AF%8D%E5%85%B8/plain)，读者可以下载，导入不背单词、墨墨单词等单词 App 上。

同义词词典适合作为在学习过程中的总结阶段中使用，总结同义词可以帮助建立单词之间的联想，形成语义网络，助于更全面地理解一个词的含义和使用场景。长期积累有助于内化语言逻辑，逐步形成地道的英语语感。

## 短语动词词典

在英语学习阶段，只背“单词”本身是不够的，因为在英语中还有大量的“短语动词”，也就是 verb + adv./prep. 的形式，这类词的含义经常让人捉摸不透。

比如 take up 这个词，就是 verb. + prep. 的组合。

按照我们朴素的理解，“take”是“带走”的意思，“up”是“上面”的意思。乍一看，我肯会以为它是“把某件物体带到上面”的意思。

但实际上并不是，它是“占据”的意思。而“up”在这里不是介词而是副词，表示“彻底”的意思。和“把某件物体带到上面”没有任何关系。也就是说，我们无法单单从“take”或者“up”上推测这个短语动词的含义。类似的还有“take off”，是“起飞”的意思。

再换一个，比如 take over, make over, pull over，你能按照朴素的理解来推测它的含义吗？

但它们的含义是“接管”、“转让”和“停车”。

这就是短语动词的恶心之处，一个含义加上另一个含义，就生成了一个完全毫无关联的、新的含义。

短语动词很多是常见的动词加上一个常见的介词或副词，给人一种“很熟悉”的感觉，但实际上完全是另一个意思。而偏偏在英语的文本中（尤其是在互联网上），短语动词的使用频率又往往会占到四成水平。短语动词广泛用于英语的口语和非正式交流中，几乎每一个母语者都会经常使用。尤其是在英语口语中，母语者往往倾向于使用短语动词而不是正式词汇，因为它们更简单、自然。

对于非母语者来说，不积累足够多的短语动词，就会造成在阅读上的南辕北辙。**对于短语动词，其实没有什么好的办法，只能死记硬背，把它们“当成一个独立的新单词”来背。这也是一个难点所在**。

也不要指望自己像做数学题那样“推测”含义，因为短语动词的组合极为灵活，在英语的发展过程中大都是“约定成俗”的用语，**甚至没有什么逻辑，并不是简单的词义相加就能推测出来的。**

所以，我推荐“牛津短语动词词典”

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250116184439.webp)

这本词典收录了大量的短语动词，并且总结出了一些常用义项。比如 run into 这个词，词典就给出了三种解释：遭遇、遇到、纠缠三重含义。

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250116184809.webp)

它也是以词为单位的，把每个动词都列举出所有对应的短语动词，并依次编排，比如 rush 这个词

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250116185052.webp)

短语动词大都是动介组合或者动副组合，其中的副词和介词又被称为小品词，像 turn out,make out，等，out 就是小品词，在短语动词中起到辅助意义的作用。

这本词典在附录中也总结了很多小品词，例举出了小品词的不同含义，总结的很全面，比如 over。

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250116191451.webp)

这本书总结了 6000+ 左右的短语动词，已经满足了九成以上的场景覆盖需求，对于学习者来说已经完全足够了。这本书的价格也很亲民，40 元左右，影印书更便宜，9.9 元包邮。当然，它在 Zlibrary 上也有电子版：[牛津短语动词词典 - ZLibrary 链接](https://z-library.sk/book/18213158/3388a4/%E7%89%9B%E6%B4%A5%E7%9F%AD%E8%AF%AD%E5%8A%A8%E8%AF%8D%E8%AF%8D%E5%85%B8-%E8%8B%B1%E6%B1%89%E5%8F%8C%E8%A7%A3%E7%89%88%E7%AC%AC%E4%BA%8C%E7%89%88-oxford-phrasal-verbs-dictionarychineseenglishenglishchinese2nd-edition.html)

## 习语词典

在进阶阶段，接触到英语电影、小说的时候，里面会经常有一些修辞性的、本土化的表达，称之为习语“idiom”，类似于中文里面的“成语”。同样也不能只从字面上理解。

比如“得陇望蜀”就是个修辞性词语，来形容一方野心勃勃、贪得无厌。但是在上个世纪中，中方对苏联的外交翻译中，因为外交工作人员的失误，直接翻译成了字面意义上的“占领了甘肃和四川”，导致双方出现战略误解。

英语习语跟成语一样，它也跟英语文化紧密结合。如果想要深入接触英语文化，尤其是阅读英语本土文学作品，理解它们的表达，那么去学习一些习语则是必要的。许多习语反映了英语国家的文化、历史、传统和价值观。例如："Spill the beans"（泄露秘密）来源于古希腊的投票习俗。通过学习习语，可以更好地了解英语文化和母语者的思维方式。

对英语学习者来说是非常重要的，因为习语不仅是语言的重要组成部分，还能够丰富表达方式，帮助更好地理解英语文化和母语者的思维方式。

另一方面，英语习语有比正是表达更为灵活微妙的表达能力，更有感情色彩，这是常规表达中难以传递的。例如："It's raining cats and dogs."（下着倾盆大雨）、"Hit the nail on the head."（一针见血），"a piece of cake"（小菜一碟）。

学习习语时，你会接触到许多表达方式，增强对英语的敏感度和语感。例如："Break the ice" 不仅可以用作“打破僵局”，还能帮助你理解类似的表达如 "icebreaker"（破冰者/活动）。

除此之外，使用习语可以让你的英语表达更加生动自然，贴近母语者的语言风格。例如：与其说 "I'm very tired," 不如说 "I'm worn out."（我累坏了）。熟练使用习语能让你的语言表达从“标准化”提升到“地道化”。

这本书收录的也相对较为全面，选取一页来看

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20250116191057.webp)

它收录了大约一万句常用习语，不仅可以用来查阅，它也给出了很多用法例句，也非常适合在茶余饭后，没事的时候，来消遣时随意翻上一两页（不用刻意去背），坚持一年，对自己的表达能力的提升都有极大的帮助。

不过这本书在网上我也没找到电子版，但是价格也比较便宜，30-40 左右，也可以去拼夕夕拿 9.9 包邮的影印书，非常实惠。

以上推荐的三本词典大约如此。这些主题词典都会对英语进阶有很大的帮助。当然，更重要的一条：

**注重积累**。


根据我的一段学习英语的历程，本期推荐三本好用针对非母语者的英语学习词典：牛津短语动词词典、牛津习语词典和同义词学习词典。适合进阶阶段使用。

英语学习词典推荐


The time I start writing blog is when I was enrolled in University after the College Entrance Matriculation in 2020. And I had my some thoughts over technology and programming, which I would like to put them down as fruition for studying progress.

Now the 2025 is coming, also it's my fifth year since I starting my blog.

During my teenhood when I'm a senior high school student, my parent restricted my time spending on computer so that I have less chances with my computer. At that time, I didn’t have the ability to build my own blog website. So I back then prefer using QZone, or writing my thoughts on specified notebook.

And for a long time I accrued more valuable notes about programming and thinking, they entails with C++, Java and some software theories until 2021. But unfortunately, my QZone was banned for some censorship and most of my blog was deleted, years of hard work went down the drain, its a great pity, and that's when I made a decision of building my own blog site. I can not tolerate the deteriorating cyber environment and living in censorship with trepidation.

So I learned React.js, next.js and frontend UI design, finalize this blog website as my own my blog, I have my own free writing space now.

Throughout the years topics for my blog have always been "Thoughts on Programming and Life" but in fact the point is, my blog touches more programming than life experience.

That's because, the contents on life experience would be better on social media platforms of microblog, like twitter, weibo and QZone. I expect more fallbacks from my cyber friends when I publish my content to get more emotional values, because most of my friends are online in social media platforms rather than distributed blog site. I would hard to earn this effect if I publish on blog site.

Secondly, the most of contents over life experience are fragmentation that I would write them down anytime and anywhere. For example, when you cooked culinary delicacies, got out of kitchen and want to post them in public on mood. It's convenience for everyone to open App in several seconds and type them in mobile, and publish, all in one go. Rather than write them in blog site, I must pay more focus on the text formatting after I power up computer and launch blog platform. And naturally the long blog text is not suitable for fragment writing for no convenience.

Thus I write my life experience on social media(more on [my twitter](https://x.com/codemica)) and technologies on this blog site. But the cons are also remarkable, long time fragmentation writing conduces less deep thinkings and expression, I also noticed that my expression becoming more superficial than years before.

The next steps I will take efforts to do is pick my ability of deep writing up again, especially the content over life experience, which I would put more thoughts in.


This is my fifth year since I started writing blog. After these years I noticed that I have lost more about writing.

The Fifth Anniversary for My Blog


This issue can be seen as a classic case of historical inertia combined with outdated reforms in the education system. This encapsulates the core of the problem.

Frankly, I find it difficult to tolerate the myriad justifications people offer to defend this situation. It is merely a residual consequence of the glacial pace of educational reform, yet some have framed it as "developing logical thinking" or claimed that "university isn't a training camp." Such arguments are utterly untenable.

In reality, even if computer science majors studied more mathematics instead of physics, it would be far more beneficial.

Computer science fundamentally diverges from traditional engineering disciplines such as civil engineering, mechanical engineering, electronic engineering, and chemical engineering. These traditional fields are rooted in physics, whereas computer science is deeply grounded in mathematics.

Traditional engineering disciplines primarily engage with the physical world. Their foundations lie in natural sciences, particularly physics, and they boast histories spanning decades, if not centuries. The primary focus of these disciplines is the transformation of the physical realm.

Consider this: as a mechanical engineering major, it is inconceivable to bypass the study of mechanics—how else would you design gears? If you are in precision instrumentation, knowledge of optics and electronics is indispensable for designing sensors. For chemical engineering students, quantum mechanics and thermodynamics are fundamental to understanding molecular interactions. Similarly, hydraulic engineering requires mastery of fluid mechanics.

Each traditional engineering discipline is built upon natural sciences like physics, dedicated to leveraging these principles to reshape the material world.

In stark contrast, computer science follows an entirely distinct trajectory. It is a "formal science," entirely man-made, with most of its concepts abstracted directly from mathematics.

Modern computer science theory originated with Alan Turing, whose work on computability theory employed pure mathematical constructs to define the concept of a "computer." These theoretical foundations were later implemented to create various types of computing systems.

Computer science, at its core, studies theoretical frameworks. While engineers apply these frameworks to design chips and systems, the discipline itself focuses on abstract structures and models rather than the tangible, physical world.

Formal sciences—such as mathematics, logic, and computer science—emphasize abstract, logical constructs instead of physical phenomena. They rely on deductive reasoning and logical proofs rather than empirical observation, which aligns computer science closely with mathematics.

For instance, many foundational concepts in computer science, such as Turing machines, operating system models, and the Von Neumann architecture, are mathematical abstractions. These constructs do not exist in nature; they are products of human ingenuity.

Even programming concepts such as functions, pointers, parameters, and operators—or areas like complexity analysis—are rooted in mathematical models. Algorithm design exemplifies this reliance on mathematical abstraction.

So, what justification exists for requiring computer science majors to study physics? Will they be calculating the forces involved in moving servers? Or perhaps determining the melting point of silicon to fabricate CPUs?

Let us be candid: computer science is fundamentally a repository of mathematical abstractions. This alignment with mathematics is entirely unsurprising, given the discipline's origins.

And for those who argue, "We study electronics to design circuits for CPUs," it is crucial to clarify: digital circuits and logic gates are mathematical abstractions as well. The physical implementation of chips falls under the domain of electronic engineering, not computer science. The role of computer science is to provide the models, theories, and blueprints, while other fields handle physical realization.

Furthermore, most computer science graduates today pursue careers in software development. It is unrealistic to assume they are designing CPUs.

In China, computer science was only formalized as a university discipline in the 1980s, following the reinstatement of the national college entrance exams. However, the theoretical foundations of computer science had already been established during World War II. By the time China integrated it into its educational system, we were already lagging behind Western nations by no less than half a century.

During that era, computers were a rarity in China, and expertise in teaching the subject was even scarcer. Lacking domestic models to emulate, we resorted to copying. Since physics was a staple of other STEM fields, it was incorporated into computer science curricula as well.

And yet, we remain entrenched in these antiquated practices. Despite nearly four decades, our approach to computer science education continues to stagnate. For example, certain universities still teach C++ using VC6, a tool that is unequivocally obsolete by contemporary standards.

In conclusion, the shortcomings of computer science education in China can be attributed to historical inertia and outdated reforms. Unlike traditional engineering disciplines rooted in physics, computer science is grounded in mathematics and focuses on abstract models. The inclusion of subjects like physics in the curriculum reflects a failure to adapt to the unique nature and practical demands of this field.


China's computer science education is outdated, often including irrelevant subjects like physics, due to historical and systemic inertia.

Should Computer Science Majors Learn College Physics


在美国，它的日期格式是世界上独具一格、特立独行的存在。它不同于欧洲的「由小及大」的 DMY 和东亚国家的「由大及小」的 YMD，它居然是 MDY。而且也不像其他欧洲人一样喜欢使用纯数字日期格式，而是半数字半英文格式：英文月份 + 日期数字 + 年份数字。比如今日是 2024 年 11 月 27 日，其他欧洲人如英国人可能会写成 27/11/2024，但是美国人的写法是 November 27, 2024!

一下是世界各地的时间格式地图，

![](https://raw.githubusercontent.com/codemetic/pictures-host/main/images20241128101343.webp)

其中蓝色地区为 DMY，黄色地区为 YMD，红色地区为 DMY 和 MDY 混用，绿色地区为 DMY 和 YMD 混用，粉色地区只用 MDY，且使用这种日期格式的，只有美国。

曾经就有英国作家吐槽过美国人的这种习惯：
> “大多数英国人先写日，后写月。商界人士和老女人仍然坚持把月份放在前面。” \
> —— _Have His Carcase_, by Dorothy Leigh Sayers

我其实非常疑惑，美国日期格式的逻辑到底是什么？很多曾经的英国殖民地，或者英联邦国家，大都沿用了英国的 DMY 格式。但为什么美国曾经作为英国的殖民地，饱受英国文化影响，就没有继承英国的 DMY 格式呢？

其实美国对日期给“固执己见”，我还听过另一个笑话：一个英国女人和美国男人的情侣，准备挑选一个日期去约会，他们给各自的信选的日期是“10/12/1997”。然而到最后谁也没有见到。为什么？因为美国男人理解的是 10 月 12 日，而英国女人理解的是 12 月 10 日！

假如，世界上只存在 YMD 和 DMY 两种格式的日期，哪怕是亚洲人和西方人阅读各自的文本时，也不会造成歧义。

因为类似“2024-11-11”日和“11-11-2024”两种日期，只需要看到年份在哪个位置，很快就能判断出日期格式，也就能找到对应的月份、日号数字，这个世界仍然会美好的很多。

但是，坏就坏在，美国在计算机、科学技术上话语权的强势，强行把他们的文化，包括这反人类的 MDY 日期格式强行推广到了世界。也推广到了欧洲。

如果有一份英语文本，里面有一个日期：11/11/2024，但是你无法判断它是美国人写的还是英国人写的，那么这个日期你会怎么理解呢？

所以很多英语国家包括很多欧洲国家也不得不被逼得妥协：对于日期造成歧义时，月份应当使用单词简写。比如上面的日期，英国人就要写成 11 November, 2024.或 11 Nov. 2024.

所以这个世界就出现了 MDY，YMD，DMY 三方混战的景象。

YMD 和 DMY 两种日期格式的来源体现了东西两种文明长期生产实践的需要，为什么有这两种差异，具体原因众说纷纭没有定论，不过目前有两种令人信服的说法，分别从文化需要和 [认知语言学](https://zh.wikipedia.org/zh-hans/%E8%AA%8D%E7%9F%A5%E8%AA%9E%E8%A8%80%E5%AD%B8) 的角度上进行了解释。

第一种是文化需要的解释。

很长一段时间以来，中国人没有采用西方的公元数字纪年法，而采用年号纪年法。年号在中国古代政治中有很重要的意义，体现着国家和帝王的意志。汉武帝之后朝廷公文文书中，常常会把当年的年号放在开头首位，以体现重视。而中国长期以来是农业立国的传统，中国人发明了农历历法，农业生产往往以月份作为播种、培植、收割的阶段性单位。所以中国人往往会把年份和月份放在前面。其他东亚国家如朝鲜、日本、蒙古等都受到了中国农历的影响。

而欧洲人则不一样。欧洲在中世纪时宗教氛围浓厚，欧洲人的生活中常常会参与宗教活动。如礼拜日、讲经日等，教会活动通常注重节日、宗教活动的具体日期，月份被看作日期的附属信息，而年份的变化频率较低，因此放在最后。例如：“Die 31 mensis Decembris anno Domini 2024” （2024 年主历 12 月 31 日）。
这种注重“日”优先的传统在许多欧洲国家延续了下来，影响了日期格式。

第二种是认知语言学的角度。

从语言学角度看，西方日期格式采用 DMY 是因为欧洲语言普遍遵循从具体到抽象的逻辑顺序，日期作为修饰信息优先表达，贴近月和年的核心词。在许多印欧语言中，修饰语（如描述时间的具体日期或特征）通常放在主要信息之前。在西方国家的地址格式也有类似的特点，都是先写房屋、街道，再写具体的城市、州、省等。这种格式符合欧洲语言中“修饰语靠近被修饰对象”的语法习惯，同时也与其文化中注重逻辑性和线性时间流动的特点相契合。中文采用从大到小的逻辑（年→月→日），这种顺序强调时间的递进关系，与汉语注重“背景到具体”的叙事结构相符。

在我眼里，最优秀的日期格式应当是对于人类和计算机都友好的表示方式，只有 YMD，也是东亚国家流行的格式。因为 YMD 对于人类来讲更符合直觉，在整理文书文件时最方便。

YMD 和 DMY 格式的差异尚且可以从多个角度找到合理的解释，但是美国人的 MDY 格式逻辑是什么？使用 MDY 的理由又是什么？

这个问题我在网上搜索了大量资料，也没有找到具体原因。

我严重怀疑这是某个美国人随手一拍脑袋就定下的日期格式。

因为大部分语言（如中文和英文）的阅读顺序，都是从左到右。想象一下，假如你需要整理一堆文书，想把它们按照日期排序，当你看到“20241111”这个日期时，只需要顺序扫描 2-0-2-4-1-1-1-1 这六个数字，很容易就能判断出它的年份、月份前缀，并归到一个“大范围”内，更容易细分整理。但假如日期格式是英国的 DMY 格式，“12072024”，那么你需要打断阅读流，从日期尾部开始，扫描 4-2-0-2，再把顺序纠正过来成“2-0-2-4”，判断出是年份，然后再后扫描“7-0”，纠正出是“0-7”，判断出是七月份。所以当你在扫描 DMY 格式的日期时，提取出年份和月份就要连续换 4 次方向。短期还好，但是如果你的工作量非常大的时候，对人脑和心智都是一笔不小的负担，会极大提高因为马虎而出错的概率。

YMD 格式对计算机也非常友好，因为 YMD 格式的日期前缀在计算机中可以直接排序，字符串形式的“2024-07-12”，因为编码是顺序的，非常容易实现词典排序。

想象一下你有两份文件

```plaintext
2024-11-12-document1.docx
2024-12-07-document2.docx
```

因为有日期格式前缀，所以就能保证前者的文件名词典序大于后者文件名。也就利于文件资源管理器直接借助文件名就能实现按照时间日期排序。

数字形式的“20240712”，也非常容易实现两个日期的大小比较。直接使用计算机内部的 `>`和 `<` 比较符即可，也不需要再实现其他的处理过程。

以上优势，是 DMY 不具备的。

美国人的顽固不化不仅体现在日期格式上，美国的度量标准也是一坨。比如华氏度、英制单位（英里、英寸、英尺、加仑、盎司等）。英制单位的雏形出现在在欧洲罗马帝国时期，长期以来就是一笔糊涂账，没有固定的标准，随着不同历史时期的变化而变化。到了欧洲中世纪才被英国定位标准单位。

但是英国在定制英制单位时并没有标准的、严格的尺寸根据。比如一英寸就被定义为三颗大麦排成一起的长度，一英尺就是英国国王爱德华鼻端到手端的直线距离。问题是，用麦粒定义英寸，一千种麦粒有一千种长度，用身体部位做长度定义，一千种人也有一千种长度。仍然是一笔糊涂账。

长期以来的度量衡混乱直到近代的法国自然科学取得一定发展后才取得严格定义。所以才有了现代的严格公制单位，公制单位采用可复现的物理实验方法定义，比英制单位严谨得多。

所以，当今全球，除了美国，都在采用公制单位。但是美国就是一个异类，仍然在固守英制的光荣传统。并冠以“传承文化”、“维护自由多元”为名。拒绝公制单位此事上，美国人仍表现得过量情绪化，就像《格列夫游记》里争执鸡蛋大小端的打破位置一样不可理喻。属是奇葩。


在美国，它的日期格式是世界上独具一格、特立独行的存在。它不同于欧洲的「由小及大」的 DMY 和东亚国家的「由大及小」的 YMD，它居然是 MDY。

吐槽一下美制时间格式

实现在 React.js 中渲染你的 Markdown 文本，并从 Markdown 编译过程中的一些概念入手，说明 unified.js 和一些插件的用法和用例。

如何在 React 中渲染 Markdown 文档


British economist John Maynard Keynes and Austrian economist Friedrich Hayek are two of the most renowned figures in the history of 20th-century economic thought. Their ideas not only had a profound impact on the past economic and social landscape, but they will continue to influence current and future economic and social development until a more convincing economic ideology takes hold. The debate between Keynes and Hayek is known as “the most classic and famous duel in economic history.” American journalist Nicholas Wapshott, who was a senior editor at *The Times* of London and *The New York Sun*, systematically examined this debate in his book *Keynes Hayek: The Clash that Defined Modern Economics*. As we revisit this classic academic feast, we also reflect on the serious economic question: “To what extent should the government intervene in the market?”

Today, people are already quite familiar with the starkly opposing economic views of Keynes and Hayek. Keynes believed that the root cause of an economic crisis is the lack of confidence among capitalists in investment prospects, leading to reduced investment, and thus advocated for effective government intervention and control over social economic activities. Hayek, on the other hand, believed that the root cause of economic crises was over-investment by capitalists and argued that government intervention would only damage the mechanisms of market development and functioning. Therefore, he firmly defended the idea of a free market order. Keynes argued that the government could expand public spending through fiscal deficits and national debt, increase state investment and consumption, and indirectly advocated for the "nationalization of currency." Hayek, by contrast, proposed that currency issuance should be handed over to private banks, which would provide sound and stable money based on their own interests, advocating for the "denationalization of currency."

Keynes supported inflationary policies to increase corporate profits, stimulate investment, expand employment, and achieve stable economic development. Hayek argued that inflation would widen the income gap, cause asset prices to soar, and eventually lead to greater unemployment and economic vicious cycles. He believed that business cycles would naturally return to full employment. Keynes argued for increased investment in public works, improving social welfare, and providing relief spending for those living below the poverty line. Hayek, however, believed that public goods do not directly satisfy any specific demand but provide certain conditions under which individuals and smaller groups can enjoy better opportunities to meet their respective needs.

The debate between Keynes and Hayek is characterized by a vast ideological chasm, and throughout their lives, they could never reconcile with each other’s viewpoints. Their debates were later carried on by their passionate disciples, such as John Kenneth Galbraith and Milton Friedman. In terms of personality, the two were also very different. Keynes was highly articulate and charismatic, often painting an optimistic vision of the world, which was embraced by an entire generation of politicians and economists on both sides of the Atlantic. In contrast, Hayek was meticulous and persistent, swimming against the tide, but found support among market advocates and libertarians. To this day, the two distinct schools of thought in economics continue their “battle.”

This economic battle has left government decision-makers oscillating between the two scholars' ideas, influencing the ups and downs of the world economy, affecting the lives and livelihoods of billions. The global economic crisis of 1929–1933 led to the widespread adoption of Keynesianism, represented by the "New Deal" of Franklin D. Roosevelt, and it seemed that Keynes had triumphed over Hayek. However, the oil crisis of the 1970s plunged the global economy into the dilemma of "stagflation," and the neoliberalism represented by the "Washington Consensus" became the dominant policy-making ideology, giving Hayek a comeback. By the end of the 20th century, the failure of neoliberal economic policies in Latin America, Russia, Southeast Asia, and Eastern Europe, coupled with the impact of the 2008 global financial crisis, suggested that Keynesianism might once again be on the rise.

In terms of policy practice, the struggle between Keynesianism and neoliberalism may continue, and it is still too early to conclude who will ultimately prevail. Interestingly, the Nobel Prize in Economics, which is seen as a "weather vane" for academic trends, has also been swayed by economic crises and shifts in government decision-making. Since the Nobel Prize in Economics was first awarded in 1968, Keynes, who had passed away by then, was naturally ineligible. Hayek, however, won the prize in 1974. During the 1970s and 1980s, when neoliberalism was on the rise, figures like Milton Friedman, Robert Lucas, Finn Kydland, and Edward Prescott also won the prize. Later, however, new Keynesians like Joseph Stiglitz and Peter Diamond, who carried on Keynes' legacy, were awarded the prize as well. In 2012, with the resurgence of neoliberalism, Alvin Roth and Lloyd Shapley took home the Nobel Prize in Economics.

In fact, despite their dramatic disagreements, Keynes and Hayek had similar aims. They both studied the same subject: the cycles of boom and bust in commercial activity that emerged from the ruins of World War I. Both sought to establish a theoretical framework that combined real and monetary phenomena to explain the dynamics of the economy, especially the business cycle. Their economic theories both originated from Knut Wicksell's theory of monetary equilibrium and the concept of "natural interest rates." Furthermore, they shared a deep commitment to the basic principles of a market economy governed by freedom, democracy, and the rule of law. If we look purely at logical reasoning, both of their theoretical conclusions and policy recommendations could explain the causes of economic recessions and offer solutions. The historical choices made by policymakers ultimately explain the root cause of their disagreements—it was a matter of experience and practical judgment.

It’s worth mentioning that although their debate was often heated, with sharp criticisms exchanged, the mutual respect they maintained beyond academia, and their personal friendship, is even more commendable. Although Hayek had differing views on credit from Keynes, he praised Keynes' *A Treatise on Money* as "an outstanding description and classification of different forms of money." Similarly, Keynes was deeply moved by Hayek's great work *The Road to Serfdom*. Keynes not only provided housing for Hayek when he first arrived in London, but in their debate letters, they openly discussed their friendship. Hayek once said, “Although I still disagreed with Keynes' views and had fierce debates with him, we maintained the best of personal friendships. And as a man, I held him in the highest esteem in many ways.”

Their ideas not only had a profound impact on the past economic and social landscape, but they will continue to influence current and future economic and social development until a more convincing economic ideology takes hold.

The Historical Debate Between Hayek and Keynes


## The Largest Ear of Wheat

It is said that Socrates once led his students for a walk in a wheat field. His students asked him: how can one find the most suitable life partner?

Socrates looked at the wheat field and gave them a task: "Go into the wheat field and find the largest ear of wheat, but as you walk, you can only move forward and cannot turn back."

The students began walking while choosing, and the field was full of large ears of wheat. But which one was the largest? They walked with their heads down, checking one ear of wheat after another. They shook their heads at this one; they shook their heads at that one. They always believed that the largest ear was still ahead. Although they picked a few, they were not satisfied and threw them away. They thought there were plenty of chances and no need to decide too early.

The students continued walking with their heads down, carefully selecting as they went, for quite a long time. Suddenly, they heard Socrates' deep, resounding voice: "You have reached the end." At that moment, the students, empty-handed, realized what had happened and felt disheartened.

Socrates said to them: "There is certainly a largest ear of wheat in this field, but you may not encounter it; and even if you do, you may not make the correct judgment. Therefore, the largest ear is the one you just picked."

This story tells us that life is like walking in a wheat field, also searching for the largest ear of wheat. Some people, upon seeing a full and ripe ear, seize the opportunity and pick it without hesitation; others keep looking around and miss their chances again and again. Of course, the goal should be the largest ear, but holding the one in hand is the most tangible reality.

**So, the question arises: Is there a method that can help us find the largest ear of wheat as efficiently as possible?**

## Treasure Box Game

Let’s consider another example. Imagine there is a gambling game with the following rules: There are $m$ treasure boxes, each containing a random prize amount between $0$ and $x$ dollars, uniformly distributed. **However, you don’t know what the maximum amount $x$ in the boxes is.** Once you open a treasure box and see the prize, you have two choices: either take the prize and stop, or discard the box and open another one. However, once discarded, a box cannot be chosen again.

Now the question arises: how should you strategize to get the most money possible? Naturally, the more money, the better.

What would happen if we didn’t use any strategy and randomly picked a treasure box?

Let’s assume the amounts in the treasure boxes range from $1000, 999, 998, ..., 0$, with the top prize being $1000$. What is the probability of drawing the top prize?

Since each specific amount (including the top prize of $1000) has an equal chance of appearing, and the prize is uniformly distributed, the probability of drawing the top prize can be calculated as follows:

$$
P_{Jackpot} = \frac{1}{(\text{MaxMoney} - \text{MinMoney} + 1)}
$$

In this case, the maximum prize is $1000$ and the minimum is $0$, so:

$$
P_{Jackpot} = \frac{1}{(1000 - 0 + 1)} = \frac{1}{1001}
$$

Therefore, the probability of randomly picking a treasure box containing the top prize of $1000$ is $0.0999\% ≈ 0.1\%$. The odds are very low indeed.

Since chasing the jackpot is unrealistic, let’s lower our expectations a bit. What if we aim for a prize of at least $800$? What is the probability of winning that?

We can use the formula for uniform distribution. For a prize uniformly distributed between $0$ and $1000$, the probability density function is constant. The formula for probability is:

$$
P(a \leq X \leq b) = \frac{b - a}{\text{range}}
$$

In this case, the range is from $0$ to $1000$, so the total range is $1000$. We want to calculate the probability for the interval from $800$ to $1000$, so:

$$
P(800 \leq X \leq 1000) = \frac{1000 - 800}{1000} = \frac{200}{1000} = 0.2
$$

Thus, the probability of randomly selecting a treasure box with a prize between $800$ and $1000$ is $20\%$.

While the probability has increased significantly, $20\%$ is still relatively low.

So, if you randomly pick a treasure box and decide to keep it, the chances of getting a high prize are still slim, which clearly isn’t the optimal strategy.

## Optimization

**In most cases, we don’t know the maximum prize inside the treasure boxes.** Just like in the story of Socrates' students picking the largest ear of wheat, they didn’t know how big the largest ear was.

Let’s change our approach. Even though we don’t know the maximum prize, we do know that if we sample a portion of the treasure boxes, the probability of finding the jackpot in the sample is equal to its probability in the entire set. By sampling some treasure boxes, we can estimate the value of the jackpot based on the maximum prize in the sample.

Let’s denote the jackpot in the entire set as $M$. Since the jackpot is uniformly distributed among the treasure boxes, we can take a sample from the boxes and record the maximum prize in that sample, denoted as $M'$. Clearly, this sample maximum $M'$ will be close to the overall jackpot $M$.

Sampling a portion of the boxes to estimate the jackpot $M$ is feasible, and the more samples we take, the closer $M'$ will be to $M$.

However, the sample size cannot be too small, or the difference between $M'$ and $M$ will be large, making the estimate ineffective. On the other hand, the sample size cannot be too large either, because there will be fewer treasure boxes left to choose from, reducing the chance of finding the jackpot. Consider an extreme example: if we sample $999$ treasure boxes out of $1000$, $M'$ will likely be almost equal to $M$, but with only one treasure box remaining, the chance of hitting the jackpot becomes negligible.

Thus, it is important to balance the sample size with the total number of treasure boxes. Is there an optimal sample size where $M'$ is close to $M$ and the probability of winning a high prize is maximized afterward?

Next, we will use mathematical methods to derive this “balance point.” Suppose the total number of treasure boxes is $m$, and the sample size is $r$. We will find the expression for the optimal $r$.

## Mathematical Derivation

Assume we have a total of $N$ treasure boxes, and we select the first $r$ boxes as a sample. In this scenario, we aim to find the optimal sample size $r$ that maximizes the probability of selecting the highest prize after discarding the sample.

### Probability Derivation

For a fixed $r$, suppose the highest prize appears in the $k$-th treasure box. To win the highest prize after discarding the sample, it is necessary that the second-highest prize appears in the first $k-1$ boxes, and specifically, this second-highest prize must also appear within the sample (the first $r$ boxes). The probability of this happening is $\frac{r}{k-1}$.

This can be visualized with a diagram, where the red arrow indicates the positions where the second-highest prize occurs:

![](https://raw.githubusercontent.com/PrinOrange/pic-host/master/myblog/20240922115506.webp)

Thus, we derive the following equation:

$$
P = \frac{1}{N} \sum_{k=r+1}^{N} \frac{r}{k-1} = \frac{r}{N} \sum_{k=r+1}^{N} \frac{1}{k-1}
$$

In this equation, $\frac{1}{N}$ represents the probability of the highest prize appearing in the $k$-th treasure box, and $\frac{r}{k-1}$ represents the probability that the second-highest prize is within the sample of the first $r$ boxes. $P$ is the probability of winning the highest prize after discarding the sample.

### Summation Approximation

The above equation contains a harmonic series, which does not have a simple closed form but can be approximated using the following formula:

$$
\sum_{k=1}^{N} \frac{1}{k} = \ln(N) + \gamma
$$

where $\ln(N)$ is the natural logarithm and $\gamma$ is Euler's constant. Here, we want to approximate the summation from $r+1$ to $N$. To do this, we rewrite the sum as:

$$
\sum_{k=r+1}^{N} \frac{1}{k-1} = \sum_{k=1}^{N-(r+1)+1} \frac{1}{k} = \sum_{k=1}^{N-r} \frac{1}{k}
$$

Thus, the sum starts from $k=1$ and follows the same harmonic series approximation:

$$
\sum_{k=r+1}^{N} \frac{1}{k-1} = \left( \ln(N) + \gamma \right) - \left( \ln(r) + \gamma \right) = \ln(N) - \ln(r)
$$

Finally, using logarithmic properties, we simplify this to:

$$
\ln(N) - \ln(r) = \ln\left(\frac{N}{r}\right)
$$

### Substituting into the Original Equation

Now, we substitute the simplified sum into the original probability equation:

$$
P = \frac{r}{N} \ln\left(\frac{N}{r}\right)
$$

Next, we introduce a new variable $x = \frac{r}{N}$ to represent the proportion of the total boxes chosen as the sample, yielding:

$$
P = x \ln\left(\frac{1}{x}\right)
$$

### Maximizing the Probability

To maximize $P$, we take the derivative of $P = x \ln\left(\frac{1}{x}\right)$ with respect to $x$:

$$
\frac{dP}{dx} = \frac{d}{dx} \left(x \ln\left(\frac{1}{x}\right)\right)
$$

Using the product rule:

$$
\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}
$$

where $u = x$ and $v = \ln\left(\frac{1}{x}\right)$, we calculate the derivatives:

$$
\frac{du}{dx} = 1
$$

and

$$
\frac{dv}{dx} = \frac{d}{dx} \ln\left(\frac{1}{x}\right) = \frac{d}{dx} \left(-\ln(x)\right) = -\frac{1}{x}
$$

Substituting these into the product rule:

$$
\frac{dP}{dx} = x \left(-\frac{1}{x}\right) + \ln\left(\frac{1}{x}\right) \cdot 1 = -1 + \ln\left(\frac{1}{x}\right)
$$

To find the maximum probability $P$, we set the derivative equal to zero:

$$
-1 + \ln\left(\frac{1}{x}\right) = 0
$$

Thus, we have:

$$
\ln\left(\frac{1}{x}\right) = 1
$$

Exponentiating both sides, we get:

$$
\frac{1}{x} = e
$$

Therefore, $x = \frac{1}{e}$, which means the probability is maximized when $\frac{r}{N} = \frac{1}{e}$.

### Maximum Probability

At this optimal point, the maximum probability is:

$$
P = P_{max} = \frac{1}{e} \approx 0.368
$$

This shows that the optimal strategy is to sample approximately $\frac{1}{e}$ (about 36.8%) of the total treasure boxes and then select the next box that is greater than all the ones in the sample. This strategy maximizes the chance of selecting the highest prize.

## Program Simulation

We can use TypeScript to simulate the process described above.

To verify the probability of winning the highest prize using this strategy, we can calculate the ratio of successful trials where the highest prize is obtained to the total number of trials. Below is a TypeScript program that simulates this process:

```typescript
function simulateGamblingGame(m: number, x: number, trials: number): number {
  let successCount = 0;

  for (let trial = 0; trial < trials; trial++) {
    const boxes = Array.from({ length: m }, () => Math.random() * x);
    const firstPart = boxes.slice(0, Math.floor(m * 0.37));

    // Remember the highest prize in the first 37% of boxes
    const p = Math.max(...firstPart);

    // Find the maximum prize in all boxes
    const maxPrize = Math.max(...boxes);

    // Check from the 37% onward if a prize higher than p is found
    for (const prize of boxes.slice(Math.floor(m * 0.37))) {
      if (prize > p) {
        if (prize === maxPrize) {
          successCount++;
        }
        break; // Stop once a higher prize is found
      }
    }
  }

  return successCount / trials; // Return the success probability
}

// Set parameters
const m = 100; // Total number of boxes
const x = 1000; // Maximum prize in each box
const trials = 100000; // Number of trials

const successProbability = simulateGamblingGame(m, x, trials);
console.log(`Probability of winning the highest prize: ${successProbability}`);
```

### Explanation

1. **Parameters**:
   - `m`: Total number of treasure boxes.
   - `x`: Maximum prize in each box.
   - `trials`: Number of simulation trials. More trials increase accuracy.
2. **Simulation Process**:

   - Randomly generate the prize for each box.
   - Calculate the maximum prize in the first 37% of the boxes (`p`).
   - Find the highest prize (`maxPrize`) in all boxes.
   - Starting from the 38% box, check if there is a prize higher than `p` and if it equals `maxPrize`.

3. **Result**:
   - Return the probability of winning the highest prize after many trials.

Running this simulation produces a result like this:

```plaintext
Probability of winning the highest prize: 0.367999
```

As shown, the probability of winning the highest prize increases to around 37%, which is a significant improvement from 0.1%.

---

To further verify, let's simulate the probability of winning the top $t\%$ prizes by dividing the prize ranges into $10\%$ increments (e.g., top 10%, top 20%, etc.).

### Modified TypeScript Code for Different Prize Ranges

```typescript
function simulateGamblingGameWithRanges(
  m: number,
  x: number,
  trials: number
): number[] {
  const successCounts = Array(10).fill(0);
  // Array to store the number of successes in each range

  for (let trial = 0; trial < trials; trial++) {
    const boxes = Array.from({ length: m }, () => Math.random() * x);

    const k = Math.floor(m * 0.37);
    const firstPart = boxes.slice(0, k);

    const p = Math.max(...firstPart);

    let chosenPrize = 0;

    // Find the first box greater than p after the first 37% boxes
    for (const prize of boxes.slice(k)) {
      if (prize > p) {
        chosenPrize = prize;
        break;
      }
    }

    // If no box greater than p is found, select the last box
    if (chosenPrize === 0) {
      chosenPrize = boxes[boxes.length - 1];
    }

    const sortedBoxes = [...boxes].sort((a, b) => b - a);

    // Determine the prize range that the chosen prize falls into
    for (let i = 1; i <= 9; i++) {
      const thresholdPrize = sortedBoxes[Math.floor((m * i * 10) / 100)];
      if (chosenPrize >= thresholdPrize) {
        successCounts[i - 1]++;
      }
    }

    // The top 100% prize is always chosen, so increment this count
    successCounts[9]++;
  }

  // Convert counts to probabilities
  const successProbabilities = successCounts.map((count) => count / trials);

  return successProbabilities;
}

// Set parameters
const m = 100; // Number of boxes
const x = 1000; // Maximum prize
const trials = 100000; // Number of trials

const successProbabilities = simulateGamblingGameWithRanges(m, x, trials);

// Output the probabilities for each prize range
for (let i = 1; i <= 10; i++) {
  console.log(
    `Probability of winning a prize in the top ${i * 10}%: ${(
      successProbabilities[i - 1] * 100
    ).toFixed(2)}%`
  );
}
```

### Results from the Program

```plaintext
Probability of winning a prize in the top 10%: 66.96%
Probability of winning a prize in the top 20%: 70.67%
Probability of winning a prize in the top 30%: 74.42%
Probability of winning a prize in the top 40%: 78.06%
Probability of winning a prize in the top 50%: 81.70%
Probability of winning a prize in the top 60%: 85.45%
Probability of winning a prize in the top 70%: 89.14%
Probability of winning a prize in the top 80%: 92.84%
Probability of winning a prize in the top 90%: 96.60%
Probability of winning a prize in the top 100%: 100.00%
```

We see that, using this strategy, we maximize the chances of getting a high-value prize.

## Conclusion

The simulation verifies that selecting 37% of the boxes first and then choosing the next box with a prize greater than the maximum from the first 37% is the optimal strategy. This maximizes the probability of winning the highest prize, known as the **37% rule**.

This strategy can also be applied in real-life scenarios, such as making optimal choices in uncertain situations.


This article proposes the classic 37% rule in game theory and its mathematical derivation process.

Socrates: The Story of Searching for the Largest Ear of Wheat


## Wallis's Formula

Let's start with a definite integral:

$$
\int_{0}^{\frac{\pi}{2}} \sin^n x \, dx
$$

We can derive the following result:

$$
\int_{0}^{\frac{\pi}{2}} \sin^n x \, dx = -\int_{0}^{\frac{\pi}{2}} \sin^{n-1} x \, d(\cos x)
$$

Using integration by parts, where $\int u \, dv = uv - \int v \, du$, we get:

$$
-\int_{0}^{\frac{\pi}{2}} \sin^{n-1} x \, d(\cos x) \\ = -\left( \sin^{n-1} x \cos x \right) \Big|_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} \cos x \, d\left(\sin^{n-1} x \right)
$$

$$
= \int_{0}^{\frac{\pi}{2}} \cos^2 x \cdot (n-1) \sin^{n-2} x \, dx \\ = (n-1) \int_{0}^{\frac{\pi}{2}} \left( \sin^{n-2} x - \sin^n x \right) \, dx
$$

Thus:

$$
\int_{0}^{\frac{\pi}{2}} \sin^n x \, dx = (n-1) \int_{0}^{\frac{\pi}{2}} \left( \sin^{n-2} x - \sin^n x \right) \, dx
$$

We get:

$$
n \int_{0}^{\frac{\pi}{2}} \sin^n x \, dx = (n-1) \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x \, dx \\ \implies \int_{0}^{\frac{\pi}{2}} \sin^n x \, dx = \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x \, dx
$$

Therefore:

**For even $n$, i.e., $n = 2k$:**

$$
\int_{0}^{\frac{\pi}{2}} \sin^{2k} x \, dx = \frac{2k-1}{2k} \int_{0}^{\frac{\pi}{2}} \sin^{2k-2} x \, dx \\ = \frac{2k-1}{2k} \cdot \frac{2k-3}{2k-2} \cdots \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin^0 x \, dx \\= \frac{2k-1}{2k} \cdot \frac{2k-3}{2k-2} \cdots \frac{1}{2} \cdot \frac{\pi}{2}
$$

**For odd $n$, i.e., $n = 2k + 1$:**

$$
\int_{0}^{\frac{\pi}{2}} \sin^{2k+1} x \, dx \\ = \frac{2k}{2k+1} \int_{0}^{\frac{\pi}{2}} \sin^{2k-1} x \, dx \\ = \frac{2k}{2k+1} \cdot \frac{2k-2}{2k-1} \cdots \frac{2}{3} \int_{0}^{\frac{\pi}{2}} \sin^1 x \, dx \\ = \frac{2k}{2k+1} \cdot \frac{2k-2}{2k-1} \cdots \frac{2}{3}
$$

Here, we need the double factorial:

$$
m!! = m (m-2) (m-4) \cdots
$$

Thus:

$$
\int_{0}^{\frac{\pi}{2}} \sin^{2k} x \, dx = \frac{(2k-1)!!}{(2k)!!} \cdot \frac{\pi}{2}
$$

$$
\int_{0}^{\frac{\pi}{2}} \sin^{2k+1} x \, dx = \frac{(2k)!!}{(2k+1)!!}
$$

Since $\sin^{2k+1} x < \sin^{2k} x < \sin^{2k-1} x$ for $x \in (0, \frac{\pi}{2})$, we have:

$$
\int_{0}^{\frac{\pi}{2}} \sin^{2k+1} x \, dx < \int_{0}^{\frac{\pi}{2}} \sin^{2k} x \, dx < \int_{0}^{\frac{\pi}{2}} \sin^{2k-1} x \, dx
$$

Substituting the above formulas:

$$
\frac{(2k)!!}{(2k+1)!!} < \frac{(2k-1)!!}{(2k)!!} \cdot \frac{\pi}{2} < \frac{(2k-2)!!}{(2k-1)!!}
$$

We can examine the difference between the bounds for $\int_{0}^{\frac{\pi}{2}} \sin^{2k} x \, dx$:

$$
\lim_{k \rightarrow \infty} \left( \frac{(2k-2)!!}{(2k-1)!!} - \frac{(2k)!!}{(2k+1)!!} \right) \\ = \lim_{k \rightarrow \infty} \left( \frac{(2k-2)!!}{(2k-1)!!} - \frac{2k}{2k+1} \cdot \frac{(2k-2)!!}{(2k-1)!!} \right) \\ = \lim_{k \rightarrow \infty} \frac{1}{2k+1} \cdot \frac{(2k-2)!!}{(2k-1)!!} = 0
$$

Note: Here $\frac{2k-2}{2k-1} < 1$

For the inequality:

$$
\frac{(2k)!!}{(2k+1)!!} < \frac{(2k-1)!!}{(2k)!!} \cdot \frac{\pi}{2} < \frac{(2k-2)!!}{(2k-1)!!}
$$

Multiply all terms by $\frac{(2k)!!}{(2k-1)!!}$:

$$
\frac{(2k)!!^2}{(2k+1)!! (2k-1)!!} < \frac{\pi}{2} < \frac{(2k-2)!! (2k)!!}{(2k-1)!!^2}
$$

We can further examine the difference between the bounds for $\frac{\pi}{2}$:

$$
\lim_{k \rightarrow \infty} \left( \frac{(2k-2)!! (2k)!!}{(2k-1)!!^2} - \frac{(2k)!!^2}{(2k+1)!! (2k-1)!!} \right) \\= \lim_{k \rightarrow \infty} \left( \frac{1}{2k} - \frac{1}{2k+1} \right) \cdot \frac{(2k)!!^2}{(2k+1)!! (2k-1)!!} \\= \lim_{k \rightarrow \infty} \frac{1}{2k} \cdot \frac{(2k)!!^2}{(2k+1)!!^2} = 0
$$

Thus, we obtain **Wallis's Formula**:

$$
\lim_{k \rightarrow \infty} \frac{(2k)!!^2}{(2k+1)!! (2k-1)!!} = \frac{\pi}{2}
$$

**This formula introduces Wallis's result in preparation for deriving Stirling's formula.**

## Stirling's Formula

Stirling's formula reveals a relationship between $n!$, $n^n$, and $e^n$.

> As $n$ approaches infinity, $n!$ is approximately $\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$.

First, introduce a limit:

$$
\lim_{n \rightarrow \infty} \left( \frac{n!}{n^n} \right)^{\frac{1}{n}} = \frac{1}{e}
$$

Proof:

$$
\lim_{n \rightarrow \infty} \left( \frac{n!}{n^n} \right)^{\frac{1}{n}} = \lim_{n \rightarrow \infty} e^{\frac{1}{n} \ln \left( \frac{n!}{n^n} \right)}
$$

which is equivalent to proving:

$$
\lim_{n \rightarrow \infty} \frac{1}{n} \ln \left( \frac{n!}{n^n} \right) = -1
$$

We have:

$$
\lim_{n \rightarrow \infty} \frac{1}{n} \ln \left( \frac{n!}{n^n} \right) \\ = \lim_{n \rightarrow \infty} \frac{1}{n} \left( \ln \frac{1}{n} + \ln \frac{2}{n} + \cdots + \ln \frac{n}{n} \right) \\ = \int_{0}^{1} \ln x \, dx = (x \ln x - x) \Big|_{0}^{1} = -1
$$

QED.

Thus,

$$
\frac{n!}{n^n} \sim \frac{1}{e^n}
$$

We would like to verify:

$$
\lim_{n \rightarrow \infty} \frac{n!}{n^n} e^n = 1
$$

Let’s test this.

**Using the ratio test:**

$$
a_n = \frac{n!}{n^n} e^n, \quad a_{n+1} = \frac{(n+1)!}{(n+1)^{n+1}} e^{n+1}
$$

$$
\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = \frac{e}{\left(1 + \frac{1}{n}\right)^n} > 1
$$

Although the limit of the expression is $e$, it is always approaching $e$ from below, so it is still greater than 1, indicating that $a_n$ is divergent.

Mathematicians have attempted various approaches and found:

$$
a_n = \frac{n!}{n^{n + \frac{1}{2}}} e^n
$$

To prove that $a_n$ converges, use the **ratio test** again:

$$
a_n = \frac{n!}{n^{n + \frac{1}{2}}} e^n, \quad a_{n+1} = \frac{(n+1)!}{(n+1)^{n + \frac{3}{2}}} e^{n+1}
$$

$$
\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = \frac{e}{\left(1 + \frac{1}{n}\right)^{n + \frac{1}{2}}}
$$

Compare this with:

$$
\left(1 + \frac{1}{n}\right)^{n + \frac{1}{2}} = e^{(n + \frac{1}{2}) \ln \left(1 + \frac{1}{n}\right)}
$$

and $e$.

Thus, compare:

$$
(n + \frac{1}{2}) \ln \left(1 + \frac{1}{n}\right)
$$

with 1.

Specifically, compare $\ln \frac{n+1}{n}$ with $\frac{1}{n + \frac{1}{2}}$:

Since:

$$
\frac{1}{n + 1} < \ln \left(1 + \frac{1}{n}\right) < \frac{1}{n}
$$

**Note:** The right side follows from the inequality $\ln(1 + x) < x$ (for $x \neq 0$).

**The left side follows from:** $1 - \frac{n}{n+1} < \ln \left(\frac{n+1}{n}\right)$, where $\frac{n}{n+1}$ is considered as a whole. This is high school knowledge.

Because $\frac{1}{x}$ is a concave function:

$$
\int_{n}^{n+1} \frac{1}{x} \, dx = \ln \frac{n+1}{n} > \frac{1}{n + \frac{1}{2}}
$$

Therefore:

$$
\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = \frac{e}{\left(1 + \frac{1}{n}\right)^{n + \frac{1}{2}}} < 1
$$

Thus, $a_n$ is monotonically decreasing and $a_n > 0$, so it must converge, i.e., it has a limit.

Let:

$$
\lim_{n \rightarrow \infty} \frac{n!}{n^{n + \frac{1}{2}}} e^n = a
$$

Then, we need to find this limit.

From the earlier Wallis formula:

$$
\lim_{k \rightarrow \infty} \frac{(2k)!!^2}{(2k+1)!! (2k-1)!!} = \frac{\pi}{2}
$$

$$
(2n)!! = 2n (2n-2) \cdots 2 = 2^n n!
$$

$$
(2n-1)!! = \frac{(2n)!}{(2n)!!} = \frac{(2n)!}{2^n n!}
$$

$$
(2n+1)!! = \frac{(2n+2)!}{(2n+2)!!} = \frac{(2n+2)!}{2^{n+1} (n+1)!}
$$

Thus, Wallis' formula becomes:

$$
\lim_{n \rightarrow \infty} \frac{16^n (n!)^4 (2n+2)}{(2n)!(2n+2)!} = \frac{\pi}{2}
$$

Using:

$$
n! \sim a e^{-n} n^{n + \frac{1}{2}}
$$

Substitute into:

$$
\lim_{n \rightarrow \infty} \frac{a^2 e^2}{4 \left(\frac{n+1}{n}\right)^{2n + \frac{3}{2}}} = \frac{\pi}{2}
$$

Solve for:

$$
a = \sqrt{2 \pi}
$$

Thus, as $n$ approaches infinity:

$$
n! \sim \sqrt{2 \pi} e^{-n} n^{n + \frac{1}{2}}
$$

or equivalently:

$$
n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n
$$


Stirling's formula finds a relationship of n!,nⁿ,eⁿ while n trends to infinity.

The Stirling Formula


## Content

The Box-Muller transform is a method used to generate normally distributed random variables from uniformly distributed random variables.

Specifically, it involves selecting two random variables, $U_1$ and $U_2$, which are uniformly distributed over the interval $[0,1]$. The random variables $X$ and $Y$ are constructed as follows:

$$
X = \cos(2\pi U_1) \sqrt{-2\ln U_2}
$$

$$
Y = \sin(2\pi U_1) \sqrt{-2\ln U_2}
$$

Then, $X$ and $Y$ follow a normal distribution with a mean of 0 and a variance of 1.

## Derivation

Assume that $X$ and $Y$ are independent, normally distributed random variables with a mean of 0 and a variance of 1. Let $p(X)$ and $p(Y)$ be their probability density functions. Thus, we have:

$$
p(X) = \frac{1}{\sqrt{2\pi}} e^{-\frac{X^2}{2}}, \\ p(Y) = \frac{1}{\sqrt{2\pi}} e^{-\frac{Y^2}{2}}
$$

Since $X$ and $Y$ are independent, their joint probability density function is:

$$
p(X,Y) = \frac{1}{2\pi} e^{-\frac{X^2 + Y^2}{2}}
$$

We now perform a coordinate transformation by defining:

$$
X = R \cos(\theta), \\ Y = R \sin(\theta)
$$

Thus, the joint distribution in polar coordinates becomes:

$$
\int_{0}^{\infty} \int_{0}^{2\pi} \frac{1}{2\pi} e^{-\frac{R^2}{2}} R \, d\theta \, dR = 1
$$

From this, we can derive the distribution functions for $R$ and $\theta$, denoted as $P_R$ and $P_\theta$:

$$
P_R(R \leq r) = \\ \int_{0}^{r} \int_{0}^{2\pi} \frac{1}{2\pi} e^{-\frac{R^2}{2}} R \, d\theta \, dR = 1 - e^{-\frac{r^2}{2}}
$$

$$
P_\theta(\theta \leq \phi) = \\ \int_{0}^{\phi} \int_{0}^{\infty} \frac{1}{2\pi} e^{-\frac{R^2}{2}} R \, dR \, d\theta = \frac{\phi}{2\pi}
$$

It is evident that $\theta$ follows a uniform distribution over $[0, 2\pi]$. Let

$$
F_R = 1 - e^{-\frac{r^2}{2}}
$$

The inverse function is given by:

$$
R = F_R^{-1}(z) \\ = \sqrt{-2 \ln(1 - z)}
$$

When $z$ is uniformly distributed over $[0,1]$, the distribution function of $R$ is $F_R(r)$. Therefore, we can select two random variables $U_1$ and $U_2$, uniformly distributed over $[0,1]$, such that:

$$
\theta = 2\pi U_1, \\ 1 - z = U_2, \\ R = \sqrt{-2 \ln U_2}
$$

Substituting these into the earlier expressions, we obtain:

$$
X = R \cos(\theta), \\ Y = R \sin(\theta)
$$

Thus, the original expressions for $X$ and $Y$ are recovered, where they follow a normal distribution with mean 0 and variance 1.

## Verification

To generate a random variable $X \sim N(0,1)$ following a standard normal distribution, we note that the inverse of the error function $\mathrm{erf}^{-1}$ is not elementary. Using a series expansion to approximate this inverse can introduce significant computational error.

Elementary functions, on the other hand, have well-established high-precision algorithms. Thus, we aim to construct a normally distributed random variable from a uniformly distributed one using elementary functions.

Let $U, V \in (0,1)$ be uniformly distributed random variables. Consider the following transformations:

$$
X = \sqrt{-2\ln U} \cos(2\pi V), \\ Y = \sqrt{-2\ln U} \sin(2\pi V)
$$

Next, we verify that $X$ and $Y$ are independent and follow a standard normal distribution.

Define:

$$
R = \sqrt{X^2 + Y^2}, \\ \Theta = \arctan\left(\frac{Y}{X}\right)
$$

Note that $\Theta = 2\pi V$ and $R = \sqrt{-2\ln U}$. We now show that $R$ and $\Theta$, under the polar coordinate transformation, are independent and that they yield standard normal random variables.

Consider the probability density function of one variable when the other

is fixed. The probability density function of $\Theta$ is:

$$
P_\Theta(\Theta) = \frac{1}{2\pi}
$$

which is straightforward. For $R$, the cumulative distribution function is:

$$
P(R \leq R_0) = \\ P\left(U \leq \exp\left(\frac{-R_0^2}{2}\right)\right)
$$

Thus, the probability density function of $R$ is:

$$
P_R(R) = R e^{-\frac{R^2}{2}}
$$

Hence, the joint probability density of $R$ and $\Theta$ is:

$$
P(R, \Theta) = \frac{1}{2\pi} R e^{-\frac{R^2}{2}}
$$

Since the polar coordinates yield:

$$
R \, dR \, d\Theta = dX \, dY
$$

we obtain the joint probability density of $X$ and $Y$:

$$
P(X,Y) = \frac{1}{2\pi} e^{-\frac{X^2 + Y^2}{2}} \\ = \left(\frac{1}{\sqrt{2\pi}} e^{-\frac{X^2}{2}}\right) \left(\frac{1}{\sqrt{2\pi}} e^{-\frac{Y^2}{2}}\right)
$$

This is precisely the probability density function of two independent, normally distributed random variables $X$ and $Y$, each following a standard normal distribution.

## Code Implementation

```typescript
function randInterval(): number {
  return Math.random();
}

function boxMuller(mu: number, sigma: number): [number, number] {
  const u = randInterval();
  const v = randInterval();
  const x = Math.cos(2 * Math.PI * u) * Math.sqrt(-2 * Math.log(v));
  const y = Math.sin(2 * Math.PI * u) * Math.sqrt(-2 * Math.log(v));
  return [x * sigma + mu, y * sigma + mu];
}

// Usage example
const [value1, value2] = boxMuller(0, 1);
console.log(value1, value2);
// Outputs two normally distributed random numbers
```

To verify that the `boxMuller` function generates numbers that follow a normal distribution, one can generate a large set of random numbers and plot a histogram to compare with a standard normal distribution. You can use libraries such as `plotly` to visualize the histogram and evaluate the output.

Here is a test program that generates a dataset using the `boxMuller` function and performs basic statistical tests, along with plotting the histogram. Assuming you are running in a Node.js environment, you can use `plotly-nodejs` or directly use `plotly.js` in the browser to generate the graph.

First, install the required dependencies:

```bash
npm i plotly.js-dist
```

Then, execute the following code:

```typescript
import * as fs from "fs";
import * as plotly from "plotly.js-dist";

// Generate random numbers following a normal distribution using Box-Muller
function randInterval(): number {
  return Math.random();
}

function boxMuller(mu: number, sigma: number): [number, number] {
  const u = randInterval();
  const v = randInterval();
  const x = Math.cos(2 * Math.PI * u) * Math.sqrt(-2 * Math.log(v));
  const y = Math.sin(2 * Math.PI * u) * Math.sqrt(-2 * Math.log(v));
  return [x * sigma + mu, y * sigma + mu];
}

// Generate n normally distributed random numbers
function generateNormalDistribution(
  mu: number,
  sigma: number,
  n: number
): number[] {
  const values: number[] = [];
  for (let i = 0; i < n / 2; i++) {
    const [value1, value2] = boxMuller(mu, sigma);
    values.push(value1, value2);
  }
  return values;
}

// Generate the data
const mu = 0;
const sigma = 1;
const sampleSize = 10000; // Sample size
const values = generateNormalDistribution(mu, sigma, sampleSize);

// Calculate the mean and variance of the generated random numbers
const mean = values.reduce((acc, val) => acc + val, 0) / values.length;
const variance =
  values.reduce((acc, val) => acc + (val - mean) ** 2, 0) / values.length;

console.log(`Mean: ${mean}`);
console.log(`Variance: ${variance}`);

// Plot the histogram
const trace = {
  x: values,
  type: "histogram",
  xbins: {
    size: 0.1, // Bin size for the histogram
  },
  marker: {
    color: "blue",
  },
  opacity: 0.7,
  name: "Box-Muller Distribution",
};

const data = [trace];

const layout = {
  title: "Generated Normal Distribution",
  xaxis: { title: "Value" },
  yaxis: { title: "Frequency" },
  bargap: 0.05,
};

const graphOptions = {
  filename: "box-muller-distribution",
  fileopt: "overwrite",
};
fs.writeFileSync("box-muller.html", plotly.plot(data, layout, graphOptions));

console.log("Box-Muller distribution data saved as box-muller.html.");
```

This program uses the `boxMuller` function to generate 10,000 random numbers that follow a normal distribution. It calculates the mean and variance of the generated dataset. The random numbers are visualized using a histogram, allowing for a direct comparison with a standard normal distribution. The histogram is saved as a file named `box-muller.html`, which can be opened in a browser for viewing.


The Box-Muller transformation is a method to construct a random variable that follows a Gaussian distribution by using a random variable that follows a uniform distribution.

NOTE & BLOG

✨但为君故

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