Harvard girl

My QQ Browser home page content recommendation engine gave me some news about this “Harvard girl,” who created some sensation around 1999 in China for her admission to Harvard. Her mom ended up writing and publishing some book on that that sold well.

That news article mentioned that that girl, despite her saying that she would return to 报效祖国 (contribute to the mother country), she ended staying in America and getting US citizenship. Maybe perhaps likely she also married a white guy. There was a photo of her with a white guy in it.

As for the comments, there were maybe over a thousand. The ones with the most upvotes were mostly negative from what I remember. Of course, it’s statistically speaking quite hard for Chinese to immigrate to America. For most Chinese, it’s basically a dream. I was very fortunate in that my family actually got the green card relatively quickly and smoothly, without engaging in anything that could be truly regarded as the spineless behavior that you see in many Chinese in America, though of course, in that regard, you can always do better, but that of course, depends much on your ability. Basically, there’s no guarantee that you’ll be able to stay and if you aren’t and have to return to China, you will feel like a failure, and likely your prospects back in China won’t be all that great either. To the majority of native Chinese, the attitude towards the emigrants and the emigrants-in-trial (those without the green card) is one of a combination of envy and contempt, or actually, maybe more like apathy since those people are so far removed from the majority of the Chinese population. Some native Chinese guy when I mentioned the term “banana-man” was even like, “what is that?” I guess back in the late 90s, it was much more of envy, but now it’s much more of contempt. To be fair, my attitude towards those smart but not that smart Chinese-American kids who are foreign to the Chinese language and culture is mostly one of contempt. I don’t really care much about alienating them because they don’t have any power and pretty much never will. If they really were actually really genuinely smart, they wouldn’t be that way. Even if that person has elite academic credentials, that person’s IQ can still be somewhat questioned, or more so, that person’s (and parents’) taste and judgment.

I asked this mother under age 40 in China if she knew of that Harvard girl, and she said yes. She mostly thought that that girl wasn’t terribly exceptional in talent or ability, and that at the time, information on how to get into Harvard in China was very limited, and rumor has it that some recommendation letter from this American helped. She was certainly quite lucky, but at the same time, her level of career success in America, so it seems, pretty much matches her talents. There was not all that much she could bring to China anyway.

In contrast, that mother mentioned this girl from a small place who really was exceptional at English, winning first place again and again in some competition in English public speaking in Britain. She got into Fudan with gaokao waived but because she didn’t like the major she got into that much, she entered the English department at Nanjing University. Now she’s 40 and a prominent TV news anchor in English. I asked her to send me a video and indeed her English really was exceptional. (Mine is too, but that’s another matter, and I also grew up in America so it’s not a fair comparison. It’s also not fair to compare my Chinese to those who grew up in China.) In contrast, that social climbing slut Zhang Zetian who married JD’s founder/CEO Richard Liu, her English was really meh, despite there being “Zhang Zetian English” as a search recommendation on Baidu. I believe her father who was quite rich (not ridiculously so though) had people manufacturing her image behind the scenes. I was told that her father wanted her to marry a “third generation red,” the ones who are the true elite in China as opposed to billionaires from the grassroots like Richard Liu. Rumor has it that she dated one and got dumped.

What that mother told me that really cracked me up was

当年铺天盖地都是她的新闻,她妈妈很极端的,书里写,为了训练意志力让她手握冰块,结果好多脑残家长跟着学

Translated to English, it is

During that time, there was news of her everywhere. Her mother was really extreme. In her book, she wrote that to train her persistence, she had her grab ice blocks with her bare hands, and as a result, many braindead parents followed suit.

Like this is just absolutely fucking ridiculous. No more comment.

By the way, I also back in 2015-6 helped this Harvard girl (ethnic Chinese but not culturally Chinese at all) who won science prizes in high school prepare for coding/algorithm interviews. (I didn’t do it for free earned a little side money from that as well. I was mostly interested in meeting her at that time.) I was surprised at the gross inconsistency between her high school achievements and how little she knew about computer science as well as how slow she was to learn. Like, she kept on asking me what the difference was between a linked list and a cache. There is this interview question of simulating an LRU cache in software. I had told her that there are hardware caches as well with orders of magnitude lower latency than reading than from memory, and that really, disk -> memory -> L2 cache -> L1 cache -> register is basically a cache hierarchy. Really, caching is quite intuitively obvious. It’s like how in the home you have a small bucket for trash that you only empty to the bigger one outside when it gets full. She also couldn’t understand why quick-select was average case O(n) and worst case O(n^2). I explained to her that pivot operation (in quick-sort it’s the same) but she just couldn’t get it. I then told her that nobody in the fucking software or machine learning world would give a fuck about quick-select, but her reaction was one of

NOOOOOOO!!!!!! QUICKSELECT!!!!!!

What really brought me into disbelief was when she asked me for some homework problem how to compute \int_0^{2\pi} \cos^2 x dx and I was like “you must be trolling.” The method I instantly thought of was using the nothing esoteric trig identity 2\cos^2 x - 1 = \cos (2x). It’s an obvious 1/2 \cdot 2\pi since the \cos (2x) component is vanished by the integral by symmetry.

There is of course also noticing that by symmetry \int_0^{2\pi} \cos^2 x dx = \int_0^{2\pi} \sin^2 x dx. Sum the two to get 2\pi \cdot 1 and divide by two.

I think I also told her that one can substitute \cos x with \frac{e^{ix}+e^{-ix}}{2}, square that and integrate. Anyone who’s studied Fourier series should instantly tell that we only care about the constant coefficient, which is an obvious 2 \cdot \frac{1}{4}.

I spoke of this case to a former Harvard PhD student from China and he was like,

Maybe she’ll eventually become director of research at Google. And write a book like Lean In.

Only then, did I learn that Lean In was some book written by Jewish Facebook exec Sheryl Sandberg who also went to Harvard. Her husband was also an exec, and my reaction to that case was basically one of

How the fuck do you die running on a treadmill. That guy must have been unusually clumsy or had some serious health problems to begin with.

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A special case of Vandermonde’s identity

Waking up this morning, I was somehow reminded of this combinatorial identity that appeared on an exam in a “math problem solving” class I took, which I didn’t actually solve during the test because back then I was an idiot. It was

\displaystyle\binom{2n}{n} = \displaystyle\sum_{k=0}^n \binom{n}{k}^2.

Basically, it’s observing that \binom{n}{k}^2 = \binom{n}{k}\binom{n}{n-k} and then seeing that we have an instance of Vandermonde’s identity. The square is basically a form of obfuscation.

This stuff feels so obvious to me now yet it wasn’t back then. To make this entirely self-contained, I will prove Vandermonde’s identity as well for this specific case.

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RSA公开密钥加密算法

工作中用某library写了些RSA公钥密钥对生成,则有空时就重新阅读并思考了下这个已被广泛运用的加密技巧。我是高中时得知这个算法的,当然完全是似懂非懂,主要是那时候对其所基于的数论还未形成最基本的直觉,所以必然无法真正搞懂,不久就忘掉如没学一样。好的是,用现在更成熟的眼光看,这个算法真挺直接的,要点就那么几个。

先简单讲讲它的背景。他的被公布的发现发明是七十年代末,由美国的Rivest, Shamir, Adleman。维基百科的英文页也说这算法也是七十年代初就被英国情报局的一位数学家发现,但它直到九十年代末才被公布。那时候的计算力量还不足以让这个算法可被实现在我们日常生活中,但是好像八九十年代Rivest, Shamir, Adleman为此创办了一家公司,把它非常成功的商业化了,也赚了不少钱。毫无疑问,这算是比较重要的新技术和发展,美国也占了它的牛耳。这些如何进入别的国家(或是被别人也独立发现而至今为公布)我就不太清楚了,我读到过美国对密码学技术出口法律上有严格限制的,九十年代在美国好像有个做这个的人由于公开某加密技术而被美国政府调查,这个人的名字是Zimmerman

RSA算法

现在,我把RSA的技巧简单解释一下,细节也不必讲的太透彻,主要讲讲他的思路和启发。

基本数论有个欧拉定理(欧拉18世纪就发现了),它的断言为

a^{\phi(n)} \equiv 1 \mod n

\phi为totient函数,\phi(n) = |\{1 \leq k \leq n : \gcd(k, n) = 1\}|

通过这个可以观察到在a^m,若m \equiv 1 \mod \phi(n),就可恢复a。那如果d \cdot e \equiv 1 \mod \phi(n),可以试试以a^d加密编码为a信息,收信方以e为自己的钥,以其解密。在\mod nd次方是计算复杂度很高的计算问题,则即使第三方知道被加密的信息是某数的d次方,也难以进行反过来的操作,而取d次方确是个复杂度为\log d的问题。 Continue reading “RSA公开密钥加密算法”

不知为何,突然想起测度论里的不可测度的维塔利集合

复制以我写的知乎文章

我在知乎上写的目前竟是一些有关于美国华人和ABC和犹太人的政治话题,自己快成了民族活动家了,其实对于学理工科的人而言,民族活动家比较贬义。民族活动家似的言论与活动,尤其在美国,其实是自然被有能力的人所藐视的,这原则很简单,它根本就是不“专业”的表现,甚至可以说是一种流氓耍诬赖的作为。在美国,中国人政治上都是特别老实的,从来不闹事儿,不抗议,就服从性的低调的埋头苦干。相反,我看到过一位根本不黑但有黑人血统的数学研究生,他的数学水平其实很差的,与其他人相比,可是他却公开的支持Black Lives Matter,然后学校媒体却非常支持他,以他宣传自己的diversity,公布的视频里还有他说I didn’t have to think about race。有意思的是他根本不黑,要他不说,其实都看不出来他是黑人。所预料,这些在学校没人敢说的,说了都怕给自己惹麻烦,其实好多人都为此感到不满,但不得不不了了之,最终政治赢者是谁就毫无疑问了。

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Big Picard theorem

I’ve been asked to prove the Big Picard theorem, assuming the fundamental normality test. Assuming the latter, it is a very short proof, and I could half-ass with that. I don’t like writing up stuff that I don’t actually understand for the sake of doing so. There’s little point, and if I’m going to actually write up a proof of it, I’ll do so for real, which means that I go over the fundamental normality test in its entirety.

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On grad school, science, academia, and also a problem on Riemann surfaces

I like mathematics a ton and I am not bad at it. In fact, I am probably better than many math graduate students at math, though surely, they will have more knowledge than I do in some respects, or maybe even not that, because frankly, the American undergrad math major curriculum is often rather pathetic, well maybe largely because the students kind of suck. In some sense, you have to be pretty clueless to be majoring in just pure math if you’re not a real outlier at it, enough to have a chance at a serious academic career. Of course, math professors won’t say this. So we have now an excess of people who really shouldn’t be in science (because they much lack the technical power or an at least reasonable scientific taste/discernment, or more often both) adding noise to the job market. On this, Katz in his infamous Don’t Become a Scientist piece writes:

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Vector fields, flows, and the Lie derivative

Let M be a smooth real manifold. A smooth vector field V on M can be considered as a function from C^{\infty}(M) to C^{\infty}(M). Every function f : M \to \mathbb{R} at every point p \in M is by a vector field (which implicitly associates a tangent vector at every point) taken to some real value, which one can think of as the directional derivative of f along the tangent vector. Moreover, this varies smoothly with p.

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Sheaves of holomorphic functions

I can sense vaguely that the sheaf is a central definition in the (superficially) horrendously abstract language of modern mathematics. There really does seem to be quite a distance, between crudely speaking, pre-1950 math and post-1950 math in the mainstream in terms of the level of abstraction typically employed. It is my hope that I will eventually accustom myself to the latter instead of viewing it as a very much alien language. It is difficult though, and  there are in fact definitions which take quite me a while to grasp (by this, I mean be able to visualize it so clearly that feel like I won’t ever forget it), which is expected given how long it has taken historically to condense to certain definitions golden in hindsight. In the hope of a step forward in my goal to understand sheaves, I’ll write up the associated definitions in this post.

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Construction of Riemann surfaces as quotients

There is a theorem in Chapter 4 Section 5 of Schlag’s complex analysis text. I went through it a month ago, but only half understood it, and it is my hope that passing through it again, this time with writeup, will finally shed light, after having studied in detail some typical examples of such Riemann surfaces, especially tori, the conformal equivalence classes of which can be represented by the fundamental region of the modular group, which arise from quotienting out by lattices on the complex plane, as well as Fuchsian groups.

In the text, the theorem is stated as follows.

Theorem 4.12.  Let \Omega \subset \mathbb{C}_{\infty} and G < \mathrm{Aut}(\mathbb{C}_{\infty}) with the property that

  • g(\Omega) \subset \Omega for all g \in G,
  • for all g \in G, g \neq \mathrm{id}, all fixed points of g in \mathbb{C}_{\infty} lie outside of \Omega,
  • for all K \subset \Omega compact, the cardinality of \{g \in G | g(K) \cap K \neq \phi\} is finite.

Under these assumptions, the natural projection \pi : \Omega \to \Omega / G is a covering map which turns \Omega/G canonically onto a Riemann surface.

The properties essentially say that the we have a Fuchsian group G acting on \Omega \subset \mathbb{C}_{\infty} without fixed points, excepting the identity. To show that quotient space is a Riemann surface, we need to construct charts. For this, notice that without fixed points, there is for all z \in \Omega, a small pre-compact open neighborhood of z denoted by K_z \subset \Omega, so that

g(\overline{K_z} \cap \overline{K_z}) = \emptyset \qquad \forall g \in G, g \neq \mathrm{id}.

So, in K_z no two elements are twice represented, which mean the projection \pi : K_z \to K_z is the identity, and therefore we can use the K_zs as charts. The gs as Mobius transformations are open maps which take the K_zs to open sets. In other words, \pi^{-1}(K_z) = \bigcup_{g \in G} g^{-1}(K_z) with pairwise disjoint open sets g^{-1}(K_z). From this, the K_zs are open sets in the quotient topology. In this scheme, the gs are the transition maps.

Finally, we verify that this topology is Hausdorff. Suppose \pi(z_1) \neq \pi(z_2) and define for all n \geq 1,

A_n = \left\{z \in \Omega | |z-z_1| < \frac{r}{n}\right\} \subset \Omega

B_n = \left\{z \in \Omega | |z-z_2| < \frac{r}{n}\right\} \subset \Omega

where r > 0 is sufficiently small. Define K = \overline{A_1} \cup \overline{B_1} and suppose that \pi(A_n) \cap \pi(B_n) \neq \emptyset for all n \geq 1. Then for some a_n \in A_n and g_n \in G we have

g_n(a_n) \in B_n \qquad \forall n \geq 1.

Since g_n(K) \cap K has finite cardinality, there are only finitely many possibilities for g_n and one of them therefore occurs infinitely often. Pass to the limit n \to \infty and we have g(z_1) = z_2 or \pi(z_1) = \pi(z_2), a contradiction.