## Father Ray Foundation in Pattaya Thailand

If I spell his name correctly, Elvin, a volunteer from South Africa who are working in Father Ray Foundation in Pattaya, showed us around their campus and told us many how the organization is being operated. First of all, the reason why we went there was completely random. We saw the sign for the orphanage, which I believe is the same as is mentioned in Lonely Planet, on a tuk-tuk on our way to the floating market. Then after visiting many relatively “disappointing” places, I decided to visit the place. Ironically, the tuk-tuk driver drove us to another place, which is also a charity organization pound under Father Ray Foundation, out of the difficulty in communication I guess. I should be thankful now to the driver though it cost pretty much. I can’t say how this place is different from other charities, however, I really like the way it is operated. This is the home to many disabled kids, needy kids, and homeless kids. As a general charity organization, it is not actually restricted to some special category, any kid with any problem (some with their families) may come here for help. It was founded by Father Ray, who is a western priest. http://www.fr-ray.org/en/ contains all information about it. I just want to share some of my personal thoughts about it. It actually encapsulates all the

positive impressions I have for Thailand and Thai people (really abundant). It is desirable to describe the country exists with a Buddha in heart, not just because of the number of temples across the country but the benevolence and softness deeply rooted in the personality of Thai people. Sex tourism is pretty much what we had expected here (though we are not totally coming out of that) which has been proved wrong not long after we arrived here. The interaction with the local people made me extremely comfortable and soothing for my souls since the purity has made me regrow the hope for the future of human beings.

Father Ray Foundation provides accommodation and meals for each volunteers (according to the current policies), and each one who is considering doing some charity work can consider this place. God bless Thailand!

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## Sexy

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February 4, 2012 · 8:25 am

## 思境

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## Timetable Survey

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## A note on Donald Knuth’s third research problem on stable marriage

{Problem statement: }
If the preference matrix of the men is given and if that of women is random, is the mean number of proposals maximum when all the men have the same order of preference for the women.

{When all the men have the same order of preference for the women}
It is easy to verify that for any possibility of women’s preference matrix, the number of proposals of all men is always $\frac{n(n+1)}{2}$.

{First Attempt (though the bound is not tight enough)}
In this section I try to derive an upper bound for the mean number of proposals for any arbitrary given men’s preference list. Let us consider an arbitrary man $m$, whose list is $(w_1,w_2,\dots,w_n)$. Let $A_k$ denote the event that he has made at least k proposals. Then $P(A_1)=1$. If he makes more than one proposal, then a necessary condition (not sufficient) should be $m$ should not be at the first position on $w_1$‘s list. Therefore, $P(A_2)\leq \frac{n-1}{n}$. Continue the argument like this, we get $P(A_k)\leq (\frac{n-1}{n})^{k-1}$. Therefore, E(number of proposals)=$\sum_{1\leq k \leq n} P(A_k)=n(1-(\frac{n-1}{n})^n)$. However, this bound is not enough to prove that $\frac{n(n+1)}{2}$ is the largest possible result.

{Second Attempt}
We proceed by proving one lemma first. This is done by adapting others’ result.
\subsubsection{Lemma: If women are the proposers, then for any man the expected rank of his (male-pessimal) partner is $\leq \frac{n+1}{2}$.}
\noindent Let $m$ be an arbitrary man whose list is $(w_1,w_2,\dots,w_n)$. Let $p(m)$ be the male-pessimal partner, and $r_m(p(m))$ be the rank of this woman on man $m$‘s list. We argue that $E(r_m(p(m))| r_m(p(m))>k-1)=(1-p_k) E(r_m(p(m))| r_m(p(m))>k)+ p_k \cdot k$, where $p_k$ is the probability that $w_k$ proposes to $m$ given $w_1, \dots, w_{k-1}$ do not.
Now we evaluate $p_k$. Under the assumption, $w_k$ should be rejected by $p(w_1),p(w_2),\dots,p(w_{k-1})$ since none of them is $m$. For the rest of the unproposed men, the event that $w_k$ first proposes to $m$ has probability $\frac{1}{n-k+1}$ (Note: In this case, m is guaranteed to be matched with $w_k$, but this is not the only possibility). Therefore, $p_k \geq \frac{1}{n-k+1}$.
Since $E(r_m(p(m))| r_m(p(m))>k) \geq k$, $E(r_m(p(m))| r_m(p(m))>k-1)=(1-p_k) E(r_m(p(m))| r_m(p(m))>k)+ p_k \cdot k \leq \frac{1}{n-k+1}k+\\ \frac{n-k}{n-k+1}E(r_m(p(m))| r_m(p(m))>k)$. Apply to $k=1, \dots, n$.\\
$E(r_m(p(m)))=E(r_m(p(m))|r_m(p(m))>0)\\ \leq \frac{1}{n} + \frac{n-1}{n} E(r_m(p(m))|r_m(p(m))>1)\\ \leq \frac{1}{n} + \frac{n-1}{n} (\frac{2}{n-1} + \frac{n-2}{n-1} E(r_m(p(m))|r_m(p(m))>2))\\ \leq \frac{1}{n} + \frac{2}{n} + \frac{n-2}{n} E(r_m(p(m))|r_m(p(m))>2)\\ \vdots \\ \leq \frac{1}{n} + \frac{2}{n} + \dots +\frac{n-1}{n}+ \frac{1}{n}E(r_m(p(m))|r_m(p(m))>n-1) \\ = \frac{1}{n} + \frac{2}{n} + \dots +\frac{n-1}{n}+ \frac{n}{n} = \frac{1}{n}(\frac{n+1}{2})=\frac{n(n+1)}{2}$ $E(r_m(p(m))|r_m(p(m))>n-1)=n$.

{On finishing the proof}
By the lemma, in male-pessimal instance, the mean sum of the ranks $\leq \frac{n(n+1)}{2}$. Since in male-optimal instance, the number of proposals equals the sum of the ranks, and the sum of the ranks is always $\leq$ that in the male-pessimal instance. Therefore we derive our desired result E(number of proposals)$\leq \frac{n(n+1)}{2}$, it can be achieved when all men have the same order of preference for the women.

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## Aku cinta Indonesia

It is a shame to use Google Translate for my title. But my ability to get over shame is deemed as one of my biggest prides throughout my life.

I went with Islam El-Ashi, my four-month brought-to-heart acquaintance and Nick Yang, a fellow that is really interesting for this pre-exam holiday.

The wonders of the nature, primarily from Mount Bromo and the serene temple yard (which might not be marked out in the tourist map) have the overwhelming effects on me, kindling my humility as a human being and my reverence for the nature almighty. I used to listen to a song called “heal the world”, but now I realize human beings can never heal the world,  instead, we could only be healed by the world (or nature). A Chinese proverb, Ren Ding Sheng Tian, 人定勝天 , humans are bound to defeat nature, is the most ridiculous nonsense I have ever heard of. It shows too much of arrogance, complacency and ignorance. No wonder there is a famous Jewish quote that goes “Human plans and thinks, God laughs”.

How much am I related to Indonesia? This is a country that may break my grandma’s heart. She was forced to leave the country at the age of 8 along with her dad because of the astonishing Chinese Massacres in Indonesia. If you know a bit about the history of the sufferings of Chinese people in Indonesia, you may be surprised at the “apartheid” policies which once made Chinese people there the Jews in the Holocaust.  I don’t go into the details of the Massacre since the web sources are rich. I am talking about my observations in Jakarta. Of course, I may be wrong. We went to Chinatown at night only to find there is no single Chinese store or restaurant in operation. From my observations in Singapore the night Chinatown(or night market in China) is the most prosperous spot where there are almost as many stores or restaurants or services as you can expect, some even beyond your expectation. While the Chinatown in Jakarta offered us the Chinese stores with the doors shut tight. It was even hard to peek through the slot of the doors. I know Chinese people are all over the world and losing the prosperity of one particular spot should not be that disappointing. However, the fact is if the history was rewritten, I might be now an Indonesian Chinese instead of a Chinese Chinese. I was experiencing what I might have been experiencing throughout my life and the contrast was sharp compared to everything I have now.  But anyway, that is my taste on history. After all, nowadays Chinese people don’t have to go through as tedious visa application procedures as hell if they want to go to Indonesia.

My observations are just as superficial as you might expect. The trip not only exposed me to the mysteries of nature but invoked my deep thinking about the human-nature relationship and inter-racial relationship.

Go back to my multivariable calculus.

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## For the test of LaTeX typesetting

(It is an easy problem for the sake of testing. )
How to prove $f(x)=x\cdot sin(x)$ is not uniformly continuous?
By sequential criterion all we need to do is to find two sequences $(x_n)$ and $(y_n)$ such that given $\epsilon >0$, $lim|x_n-y_n|=0$, $|f(x_n)-f(y_n)|\geq \epsilon$.
Now we take $x_n=2\pi n+\frac{1}{n}$ and $y_n=2\pi n$, it can be easily verified that the two sequences will generate the desired results (simply making use of the fact that $\lim_{x\to 0} \frac{sin(x)}{x}=1$).
Good day;)