Category: Blarghing

My god.; this weekend was utterly wasted for the most part. I spent a good 8 hours in a training session on Saturday, then afterwards started a game of Civ5 as Poland on emperor difficulty.

My position started out pretty precarious, with two warmongering civilizations Songhai AND Mongols neighboring me. But luckily, it seems that they started going to war against each other with the Mongols capturing a city-state to the east of me, and the Songhai’s capital actually.

I got my first three cities running, then just started rolling over the Mongols once I researched and built crossbowmen. Not sure where the Khan’s troops were, but a combination of longswords men and crossbowmen really just steamed rolled the Mongols.

By that time, I obtained the technology for producing winged Hussars and cannons. This duo was literally impossible to stop by the AIs. I conquered the capital of Byzantine, then rolled over to the Shoshones. And then somehow everyone in the world decided to declare war on me.

And then I conquered all of them by the 1950s in-game-time.

I guess the most fun part of civ 5 is reaching late game; by then, the juggernaut becomes impossible to stop. The AI is so bad at naval combat that my battleships were shelling every city with no counterplay from them.

It’s been a rough few days for my body. My left arm has some ulnar nerve impingement issues, and have been extra sensitive.

This past Sunday I fell from my bike and scraped my right knee, and injured my right arm’s wrist.

I have one remaining non-impaired limb left.

(Work has been going slow lately… hence all these math posts.)

Let $n$ be a positive integer. Starting with $1, \frac{1}{2}, \cdots, \frac{1}{n}$, form a new sequence of $n-1$ entries $3/4, 5/12, \cdots, (2n-1)/(2n(n-1))$ by taking the averages of two consecutive entries in the first sequence. Repeat the averaging process on the second sequence to obtain a third sequence of $n-1$ entries, and continue until the final sequence of only a single number $x_n$. Show that $x_n < 2/n$. Solution: Given a sequence $a_1, a_2, \ldots, a_n$, the averaging process will give us $(a_1 + a_2)/2, \ldots, (a_{n-1} + a_n)/2$. If we proceed with the averaging procedure, it is not hard to see that the binomial coefficients show up. In fact, we can see that \begin{align*} x_n &:= \frac{\sum_{k=1}^n \binom{n-1}{k-1} \frac{1}{k}}{2^{n-1}} \\ &= \frac{\sum_{k=1}^n \frac{(n-1)!}{(k-1)!(n-k)!} \frac{1}{k}}{2^{n-1}} \\ &= \frac{\sum_{k=1}^n \frac{1}{n}\binom{n}{k}}{2^{n-1}} = \frac{2^n - 1}{n 2^{n-1}} < 2/n. \end{align*} (Definitely one of the easier problems.)

Good news; my advisor and I submitted our first paper this past Monday. I’ve uploaded a draft here for those curious.

 

NYTimes Article

I think I always forget how privileged I am. I benefited from many points laid out in the article by Mr. Levy: my parents explicitly moved to one of the best school zones in Tallahassee just for me to get a better shot at college applications, I also had the benefit of them expediting the citizenship process so I could apply as an American citizen, and they provided enough money for all the extracurricular I needed. All this after living in graduate student housing for 2 years.

I like to think that I can do what I am doing without my parents’ support, but the truth is I can’t. So much of success is dependent on having a good family. I don’t think any of the people currently in my graduate program came from an under-privileged home… and to be honest, all of my friends are quite affluent.

Do I agree with the article? Yes. Do I think it’s enough? No. College is the end of the road which starts from pre-K. Certainly universities can offer more aide, or allocate more slots for under-represented students, but will this really pull them up? I don’t think so; the system right now is too fundamentally broken for a top-down approach.

1. Went to RPI applied math days this weekend, and gave a talk about preconditioning. It went pretty darn well, but I really do have to feel bad about the international grad students who don’t have a mastery over the English language… because they can’t bullshit out of anything like I can.
2. On the way there, I saw for the first time a cop pulling over someone going too slow in the left lane. I rejoiced.
3. On the way back from the gym, I saw a cancer patient in a car by the traffic lights. Chemo really does take the life out of people’s eyes.
4. Being ghosted sucks ass, especially when you thought you clicked well with the other person.
5. Taxes suck ass.

I really liked this problem that my friend Shamile gave during dinner:

Consider 50 people lined up in a random order. Show that there exists a subsequence of length 8 such that the height is non-increasing or non-decreasing.

The solution is quite elegant. Assign an ordered pair to person in the line such that the first coordinate is the length of the longest subsequence which is increasing from person , and the second coordinate is the length of the longest subsequence which is decreasing from the same person.

Let’s assume by contradiction that the longest such length is less than 8; this implies that the set of acceptable coordinates is from (0, 0) to (7,7). There is only 49 total combinations, so by pigeonhole principle, there exists a pair of the same number. But this cannot be! The rest is left to the reader.

I always forget this fact, so hopefully typing this out will help.

Let be a triangle with vertices , then there exists unique linear functions such that . This is the so called barycentric coordinates.

The fact here is that where is the length of the edge opposite vertex , and is the area of the triangle. The vector is the unit outward normal.

The derivation is simple. The direction of the gradient is towards the highest growth, and as the barycentric coordinates are linear functions, it’s clear that is the correct direction. The scaling is to simply note the area as where is the shortest distance from the edge to the vertex. This gives us the inverse of the slop as the barycentric coordinate has a value of 1 at the vertex.