In my last post, I explained the game theory mathematics of the standard game of Rock Paper Scissors (RPS), and what the Nash equilibrium is and how it works. Now I will show a similar analysis, but based on a variant of RPS where winning with rock gets you 1 point, winning with scissors gets you 2 points, and winning with paper gets you 3 points.

We can solve for the Nash equilibrium by doing a probability tree of all the different possible combinations, as before with the standard RPS. In instances where you win, add that many points. When you lose, subtract that many points (your opponent gaining points is equivalent to you losing points). As before, we will choose R, P, and S according to some random distribution, where $R$ is our probability of choosing rock, $P$ is our probability of choosing paper, and $S$ is our probability of choosing scissor. Similarly we define the variables $R'$, $P'$, and $S'$ to be the probabilities chosen by our opponent. Doing so, we get the following expected point gain each time the game is played:

$g=0RR' - 3RP' + RS' + 3PR' + 0PP' - 2PS' - 1SR' + 2SP' + 0PP'$

Remove the zero terms and then group them by $R$‘, $P'$, and $S'$:

$(-3R + 2S)P' + (R-2P)S' + (3P-S)R'$

For the equilibrium condition, we want the total to be 0, and we want it to be independent of whatever our opponent chooses for $R'$, $P'$, or $S'$. We do that by saying that each of the terms in the parenthesis must be equal to 0. This gives us 3 equations:

$\begin{matrix} -3R + 2S = 0 \\ R - 2P = 0 \\ 3P - S = 0 \end{matrix}$

Solve the 3 equations for the 3 unknowns, and you get $R = \frac{1}{3}$, $S = \frac{1}{2}$, and $P = \frac{1}{6}$. $R+P+S = 1$ as it should. You can substitute these into the original equation and prove to yourself that no matter what your opponent chooses for $R'$, $P'$, and $S'$, the expected return is always 0.

So… if you’re one of the 3 people on the entire internet that is still reading this thing, you may be asking yourself, “what about when my opponent doesn’t choose the Nash equilibrium? Is there an optimum $R, P, S$ that will maximize my expected score?”

It turns out there is, and the answer is a quite simple, though getting there takes a bit of algebra. Take the formula for the expected point gain as before (terms equal to 0 eliminated):

$-3RP' + RS' + 3PR' - 2PS' - 1SR' + 2SP'$

Use $R+P+S=1$ and $R'+P'+S'=1$ to eliminate $S$ and $S'$ so it’s only in terms of $R$, $P$, $R'$, and $P'$, and define this as a function $g(R,P)$:

$g(R,P) = -6RP' + 2P' + 6PR' - 2P + R - R'$

Now we have a surface function of two variables, $R$ & $P$, with $R'$ and $P'$ being constants. We can imagine this being in a 2D domain with R as the x-axis and P as the y-axis. So the domain in question will be a triangle with vertices at points $(1,0)$, which corresponds to $R=1$ and $P=S=0$. Similarly, $(0,1)$ corresponds to $P=1$ and $R=S=0$, and $(0,0)$ corresponds to $S=1$ and $R=P=0$.

We could then take the calculus approach and find the local maxima of $g(R,P)$, but we really don’t have to do that. Looking at $g(R,P)$, it is linear w.r.t. both $R$ and $P$, and so $g(R,P)$ is a flat plane, tilted in some direction determined by $R'$ and $P'$. Also we know from calculating the Nash equilibrium that $g\left(\frac{1}{3},\frac{1}{6}\right)=0$ (i.e. when we choose the Nash equilibrium that expected point gain $g(R,P)=0$ ), and that when $R'=\frac{1}{3}$ and $P'=\frac{1}{6}$ that the plane will be coplanar with the X-Y plane [i.e. when our opponent chooses the Nash equilibrium then $g(R,P)$ will always be zero no matter what we choose, that is only possible when the flat plane is zero everywhere].

Any deviation from those values by $R'$ and $P'$ will cause the plane to tilt in some direction. Since it is now tilting, there will be a direction that increases the value of $g(R,P)$, and since $g(R,P)$ has a constant slope everywhere, the maximum within the triangle-shaped domain must be on one of the points $(1,0)$, $(0,1)$, or $(0,0)$ [or perhaps two of the points will have the same maximum value].

So the optimal choice will either be $R=1$ (with $P=S=0$), $P=1$ (with $R=S=0$), or $S=1$ (with $R=P=0$). Simply evaluating the first equation above for these three choices and seeing which one maximizes the expected point gain will give you your choice.

So for example, here is the surface for $g(R,P)$ we get when our opponent chooses rock, paper, and scissors with equal frequency, or $(R',P',S')=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$. In our equation $S'$ was eliminated, so for calculating $g(R,P)$ we just need to say that $(R',P')=\left(\frac{1}{3},\frac{1}{3}\right)$:

In this plot the gray triangle is the original domain in the previous image, the red triangle is the actual surface of $g(R,P)$, and the black arrow is a normal vector from the plane coming from the Nash equilibrium point at $g\left(\frac{1}{3},\frac{1}{6}\right)=0$.

It appears that the red triangle is tilted along the $R=\frac{1}{3}$ line, so that both $(R,P)=(0,0)$ and $(R,P)=(0,1)$, or in other words playing scissors every time or paper every time (or some mix of the two) will give you the maximum expected point gain of $\frac{1}{3}$ every time that you play.

Now what if our opponent chooses a strategy that is close to but not quite the Nash equilibrium? How much of a point gain can we expect by using the optimal strategy?

Here is the surface for $g(R,P)$ if our opponent chooses $(R',P')=\left(\frac{1}{3}+0.01,\frac{1}{6}-0.01\right)$:

The slope of the response surface is much less, as reflected by the fact that our opponent’s choice of R,P, and S is much closer to the Nash equilibrium. In this case, it turns out that choosing either rock or paper every time will give the optimal point gain, in this case an expected point gain of only 0.03 each time we play.

Of course choosing rock or paper every time would be something that our opponent, if not a mindless robot, would be able to easily capitalize on. What if instead we only slightly perturb our choice of $(R,P)$ away from the Nash equilibrium like our opponent did, but instead perturb it in the direction that gives us the maximum expected point gain?

The distance we move from $(R,P)=\left(\frac{1}{3},\frac{1}{6}\right)$ is the same as our opponent. He moved 0.01 in the R-direction, and -0.01 in the P-direction, for a total distance of $\frac{\sqrt{2}}{100}$. We know that playing rock or paper every time will maximize our expected score, this corresponds to the line $P+R=1$. The shortest path to this line is to move perpendicular to it, which is the vector direction $(1,1)$. So this means we should choose the point $(R,P)=\left(\frac{1}{3}+0.01,\frac{1}{6}+0.01\right)$. If we do so, our expected point gain each time we play the game is $g(R,P)=0.0012$.

My next question is then, how many games do we have to play in order to have, say, a 95% confidence that our score will be greater than our opponent? Remember that what we calculated above is the expected gain, which is kind of like taking a series of many, many games and taking the average. The outcome of any 1 game is random, but after many games we expect to pull ahead slightly. However I’m really rusty on this kind of math and I don’t remember how to do it. I’ll have to review a bit.

Everyone knows the age-old game of rock, paper, and scissors, it’s been around pretty much the whole world for a century or so, and originated in China ~2000 years ago.

Generally the game is approached in 1 of 2 ways: the first is where you try and out-think your opponent, trying to exploit the inherent non-randomness of people when they play the game. This is essentially applied psychology.

The other way to approach the game is where you assume that you and your opponent are totally rational, i.e. both of you know everything about the game and there is no ‘human’ element to exploit. This is essentially game theory.

In game theory, two players are perfectly matched: neither one has any particular advantage over the other. Both players have the same chance to win each game. So the best strategy is to randomly choose rock, paper, or scissors each time (with an equal probability of each).

Using game theory and probability, you can actually prove this. Here’s one way it can be done. Define the following: if you win a game, you get 1 point. If you lose a game, you lose 1 point. $R$, $P$, and $S$ are the probabilities of you choosing rock, paper, and scissors respectively. Similarly $R'$, $P'$, and $S'$ are the probabilities your opponent chooses. Obviously $R+P+S=1$, $R'+P'+S'=1$, and $0\le R, P, S, R', P', S' \le 1$.

Now we calculate the expected point gain per game by taking every possible game outcome, multiplying the point gain/loss of that particular outcome times the probability of it occurring, and then summing them all together. So the points won per game is the following:

$g=0RR'-RP'+RS'+PR'+0PP'-PS'-SR'+SP'+0SS'$

Since we know that the best choice is to choose $R$, $P$, and $S$ with equal probability, then $R=P=S=\frac{1}{3}$. Since our opponent is also perfectly rational, he will also choose the same probabilities. So the points won per game is:

$g=0-\frac{1}{9}+\frac{1}{9}+\frac{1}{9}+0-\frac{1}{9}-\frac{1}{9}+\frac{1}{9}+0=0$

Everything sums up to $0$, which is what you expect when the game is evenly matched. However, this choice of $(R,P,S)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$ lends itself to a very interesting consequence. Let’s say your opponent chooses some $(R',P',S')\neq \left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$. What happens then? Let’s say he’s really stupid and chooses rock every time, or $(R',P',S')=\left(1,0,0\right)$. Of course the obvious choice for us to win would be to choose paper every time, or $(R,P,S)=\left(0,1,0\right)$. But if we don’t change our strategy, the following happens:

$g=0RR'-RP'+RS'+PR'+0PP'-PS'-SR'+SP'+0SS'$
$g=0-\frac{1}{3}(0)+\frac{1}{3}(0)+\frac{1}{3}(1)+0-\frac{1}{3}(0)-\frac{1}{3}(1)+\frac{1}{3}(0)+0$
$g=\frac{1}{3}-\frac{1}{3}=0$

Of course, this is obvious without having to calculate it. If your opponent chooses rock every time, and you are choosing each one 1/3 of the time, then 1/3 of the time you’ll tie with rock, 1/3 of the time win with paper, and 1/3 of the time lose with scissors. So let’s see what happens when we choose $(R,P,S)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$, but $R'$, $P'$, and $S'$ are still unknown:

$g=0RR'-RP'+RS'+PR'+0PP'-PS'-SR'+SP'+0SS'$

Rearrange the terms a bit:

$g=(0R+P-S)R'+(-R+0P+S)P'+(R-P+0S)S'$
$g=(\frac{1}{3}-\frac{1}{3})R'+(-\frac{1}{3}+\frac{1}{3})P'+(\frac{1}{3}-\frac{1}{3})S'=0$

So we see that for choosing $(R,P,S)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$, it doesn’t matter what kind of strategy or ratio that our opponent chooses: the expected point gain will always be zero and we will always be evenly matched with our opponent.

This is what’s called the Nash equilibrium, named after John Nash, the subject of A Beautiful Mind. As I best understand it, for a zero-sum game (of which RPS certainly is), there always exists a state where one player can force the game to remain in equilibrium, no matter what the other player does. For the standard RPS, this comes out to be $(R,P,S)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$. If you choose to play with this distribution, you and your opponent will always be in equilibrium no matter what strategy your opponent tries.

So that brings us to variants of RPS. This whole line of inquiry was inspired by a homework problem that my younger daughter had a few weeks ago. Basically it had a variant of RPS where if you win with rock you get 1 point, win with scissors you get 2 points, and win with paper you get 3 points. Since my daughter is in 1st grade the questions were very easy, i.e. “if you have 10 points and win with scissors, how man points do you have?” etc. But this got me thinking about how this game would actually work. Of course you want to win with paper, but you run into the same problem as regular RPS: you and your opponent are both trying to out-guess each other, so there’s no simple strategy that’s guaranteed to win.

So the next question is, is there a Nash equilibrium for this version of RPS, and if so what is it?

This post is already too long, so I’ll show how to find the Nash equilibrium for this version in my next post.

Back in our younger days, my younger brother and I were quite the aficionados for low-budget martial arts movies. We’re not talking Jackie Chan movies here, those movies are grade-A top quality compared to some of the stuff we regularly watched.

For example, I’ve seen pretty much everything by JKVD made before 2000, almost everything that Cynthia Rothrock has been in, all 3 of the Best of the Best movies (the first is the best of the Best of the Best, though #2 is more entertaining imho), all of the Master Ninja series before I had even heard of MST3K (same goes for Quest of the Delta Knights!), tons of Godfrey Ho Hong Kong drek that was made by editing together unfinished parts of movies (Ninja Death Squad being my favorite example. I could only find that one scene online, but it’s a very representative example), and everything that Billy Blanks did long before he did Tae Bo (which seems to be pretty much dead now).

And speaking of Billy Blanks, probably my favorite fight scene from a martial arts movie is from one of his movies, Showdown from 1993. It’s pretty much a remake of Karate Kid, with Billy in the Mr. Miyagi role and more-or-less unknown Kenn Scott in the Daniel role (seriously, the only mainstream thing he’s done was the suit actor for Raphael in the old TMNT movies).
(more…)

Take a look at all these photos of city streets in the rain. I found them all on google image search and I am shamelessly hotlinking them:

I was driving home from work today in the rain, and saw many similar scenes. However I noticed something interesting that I have never noticed before. The street lights, car lights, traffic lights, etc. all reflect off of the wet surface of the road. However since the road is not a mirror-flat surface, the reflection is smeared out or diffused due to the rough texture of the road. So my question is this: why is the smearing or diffusion of the reflection almost entirely in the vertical direction? Certainly the surface of the asphalt or concrete does not have a sufficiently heterogeneous asperity (i.e. the roughness is uniform from all directions) to account for this. Shouldn’t the light spread out as far horizontally as it does vertically? Any insights here?

With my wife and kids in Japan, I have been spending a lot more time reading, watching movies and TV shows, etc. Although reading books is usually a personal experience, watching TV or movies is often a social experience. But when you’re a lone, it becomes personal again and I find that I have been exhibiting some strange behavior in that regard. Here are some examples:

1. Reading the book The Name of the Wind, recommended by my brother. So far seems to be an interesting and well-written fantasy novel. However about 1/3 of the way through the book, where the main character is telling his life story (making the entire novel into a story-within-a-story, it seems), when he starts telling about his first attempt at romance, I find myself putting the book down and being unable to continue. I’m not sure why, but since I know it will not end well (since we know from the beginning of the novel that he is not with this girl that obviously it didn’t end well, even though I haven’t read the actual account yet), I find it difficult to continue reading the book. I’ll get back to it eventually, but in the meantime I’ve been reading some ‘lighter’ fare, such as David Edding’s fantasy series.

2. Friday night I watched the brainless action movie ‘Dead or Alive’, based on the titular (that was a joke, get it?) video game of the same name. It’s utterly vapid in such a degree that I couldn’t even conceive of a better villain than Eric Roberts for the movie, but if you’re expecting a brainless and nearly plotless fighting movie that’s heavy on the fanservice, it’s pretty entertaining. However there is a scene where the hacker/nerd character (fitting every Hollywood trope and stereotype of the hacker/nerd) tries to woo one of the supermodel beauties that he has a crush on. I found myself pausing the movie and taking a break, because I knew that he was going to get shot down hard, and I didn’t want to watch that.

Later I continued watching the movie, and it turned out that the writer subverted the trope in this case, and the nerd did get the girl in the end. Probably a wise choice considering the target audience of this movie.

3. Last night I started watching the 1982 movie The Challenge, fish out of water tale where the hapless American protagonist in Japan gets stuck in the middle of the warring factions of a powerful family over the possession a priceless sword. It’s directed by John Frankenheimer (The Manchurian Candidate, Ronin), and the supporting actor is none other than bad-ass legend Toshiro Mifune (who so far hasn’t spoken a single line in English!) Mifune plays Toru Yoshida, the head of the ‘good’ branch of the family who runs a martial arts school. There is a scene where the protagonist Rick (played by Scott Glen), is at a formal dinner with Yoshida’s family and students. Rick gets drunk on sake and does a classic Ugly American, shouting out inane and racist comments in English, completely ignoring the alienating and hostile atmosphere he himself has created, etc. Just watching the scene, I became very uncomfortable and embarrassed. Again I found myself stopping the movie and taking a break, reading a book for a little while.

I actually didn’t finish the movie last night, I ended up going to bed early. But I do intend to watch the rest of the movie this evening.

So what is the deal with me getting uncomfortable enough while watching a movie or reading a book that I actually put it down? It isn’t anything that I find distasteful, sickening, overly erotic, or that I’m simply not enjoying it, but because I find a scene to be embarrassing for the character, and somehow I feel embarrassed by proxy and find it difficult to continue.

Perhaps I’m even projecting myself onto the characters, and recalling experiences in my past where I have been embarrassed or humiliated. Well, looking at the three examples above:

1. First love ending badly? Check.
2. Rejection when asking out a girl who is ‘out of your league’? Check.
3. Being a culturally-insensitive American in Japan? Check.

Yes, all three are embarrassing things I have experienced int he past, and even now I don’t like to recall them. Although my experience with #3 was nowhere near as bad as the scene in the film, when I look back on some of the things I said and did back during my first time in Japan when I home-stayed with a family, I feel a lot of shame and embarrassment.

Is embarrassment and shame by proxy due to self-projection into characters and scenes from books and movies a common occurrence for other people?

Since I’m all by myself for the next few weeks with the wife and the girls in Japan, I’ve been finding projects and such to do that I normally never find the time to get around to.

Today I tackled the task of trying to retrieve data from my old old laptop computer, an Acer Travelmate 512T. Given as a gift to me by my parents when I returned from my LDS mission in 1999, it has some impressive specs: 400×800 12.1″ LCD monitor, blazing speed of 366 MHz with an awesome 32MB of RAM and a 4Gb HD. Truly an impressive machine. I used it daily from 1999-2004 or so, about the time I graduated with my BS. Since I came to Austin we had a better home computer for the family and I used it progressively less and less. The last time I remembered using it was the summer and fall of 2007, when my wife and the girls were in Japan for an especially long summer. That summer I used it primarily to dink around with some linux distributions, and to write some journal entries. It was those journal entries I was interested in retrieving, since that was during the time when my wife’s mother suddenly passed away, and so I had a lot to think and write about as I dealt with the emotional difficulties of that time.

So I dug out the laptop from the back of the closet upstairs. Battery was shot, but I still had the AC adapter. That was OK. The hinge in the screen/lid is broken, but the display still works and I’m able to get it open. I turn it on – and it gives me a BIOS error. The BIOS battery is dead! Fortunately though, it gives me the option to continue with the boot defaults. So I boot it with the default BIOS, and it takes me to a GRUB screen for an Ubuntu installation – that must have been the last linux distro that I installed on it. However, at the GRUB screen it locks up. It doesn’t recognize any keyboard input for some reason, and after a few taps it starts beeping: the familiar sound of when the keyboard buffer is full. I wait a few minutes, but nothing changes. It looks like perhaps GRUB or the boot sector is corrupted.

Trying to fix this is beyond my expertise, so I decide try a different approach: a live CD. I have a bunch of old Ubuntu, Knoppix, and other linux live CDs that I can try. Since it was an Ubuntu installation, I try one of my old Ubuntu CDs. I didn’t want to try a newer one, because with a whopping 366 MHz of CPU speed and 32 Mb of RAM, there’s no way it could handle the more recent versions of Ubuntu. I had an Ubuntu v3 CD so I tried that. Well, it worked – sort of – but it literally took 30 minutes for it to boot up. When it finally finished, I went to the file browser to see what was there. It had 3 hda’s: hda1 is generated by the Ubuntu live CD to house the OS files and actually resides in memory. hda2 was a few hundred megabytes, and hda3 was the rest of the hard drive. Maybe hda2 housed the critical OS files, or maybe it was originally a swap drive, but it didn’t recognize it as such. However, Ubuntu was unable to mount either of them: it recognized they were there, but couldn’t recognize any kind of file system.

It didn’t look good, perhaps the entire file system or disk was corrupt. Well, Ubuntu was no good, so I decided to try one of my other distributions: Knoppix, Minty, Puppy, Mandriva, an old Mandrake (before it became Mandriva), Red Hat, and DSL. Mandriva, Mandrake, and Red Hat were not live CDs. Of these, I figured DSL was the best choice for an old small machine.

So DSL only took about 5 minutes to boot up (a significant improvement!), and though it is a very simplified GUI, it seemed to work flawlessly. So the real question was: would it mount the drive? It was unable to mount hda2, but hda3 successfully mounted! So what was on it? I couldn’t find a graphical file browser, so I had to go to the good ol’ command line. I had trouble remembering the linux filesystem structure, but after a few minutes of scratching my head I remembered that the various mounted volumes are in /mnt/, so /mnt/hda3/ took me to the drive in question. Inside that there was a /home/derek/journal/. Jackpot! I went there and cated a few of the files there, I had written them in plain text so they were easily readable: except where I had used Japanese characters. DSL didn’t seem to be able to do unicode, but as long as I could copy the files as-is I figured there wouldn’t be a problem.

So the next question was: how to copy them? This laptop had a PCMCIA ethernet card, a 3.5″ floppy drive, and one single USB port. I didn’t want to mess with trying to get the ethernet working and configured, not to mention even if I did, how would I copy it to my main computer? SAMBA or something? I didn’t have the slightest clue of how to do that. I figured the USB port was the easiest. I plugged in a thumb drive, mounted it without a problem, and copied the whole directory over. I also did a quick look at what else was there. The most interesting was a bunch of configuration files for DOSBox, and a file called ‘BattleTech – The Crescent Hawks Revenge.zip’. I copied it over for good measure. That game was way fun, though fiendishly hard.

We all know about the Pirates vs. Ninjas meme, but what happens when the ninjas are pirates? That’s what happened with last year’s incarnation of the Super Sentai, Kaizoku Sentai Goukaiger (Pirate Warriors Goukaiger). So the Super Sentai are the original form of what becomes the Power Rangers in the US, before they strip out all the scenes with Japanese actors and re-do it all with American (or more recently, New Zealand) actors just keeping the fighting scenes. So ever since Kyōryū Sentai Zyuranger was re-broadcast as the original Power Rangers in 1993, the Power Rangers have always been Ninja-based. They had ninja zords, ninja powers, ninja this and ninja that. In the original Japanese the yearly incarnation wasn’t always so ninja-centric, only Kakuranger in 1994 and Hurricanger in 2002 were specifically ninja-based. Others have been police/detective based, car based, Kung Fu based, and even Samurai based.

However last year’s incarnation was absolutely based around pirates. Here’s the intro if you’re interested. Although Power Rangers have been brought to the US since 1993, in Japan it’s been going since 1975-76, and for the 35th anniversary of the Super Sentai Toei Studios decided to do something special: the gimmick is that the ‘Pirate Rangers’ have stolen the powers of all the previous Rangers, and can call upon their powers during battles. This basically means lots of nostalgia for adults that grew up watching the show, as well as numerous cameos from actors and actresses from former years of the show.

Even though I only knew incarnations from 1992 and onward, it was interesting – and at times appalling – to see the real old-school characters. Probably the most bizarre was Battle Fever J from 1979. Partly inspired by Marvel character Captain America, and partly inspired by the disco craze of the late 1970′s, Battle Fever J had 5 heroes: Battle Japan, Battle France, Battle Cossack (Soviet Union), Battle Kenya and Miss America. It pretty much has to be seen to be believed. Also keep in mind that this came right after the Japanese Spiderman (which btw, can be watched in its entirety on the Marvel website), of which the most normal reaction for Americans seeing it for the first time is this:

Also in Battle Fever J, it’s painfully obvious that during the fighting scenes that Miss America is being played by a guy. Well, I guess it’s just following a long tradition of Japanese theater. (On the other hand, in Kyōryū Sentai Zyuranger, which became the original Power Rangers, the Yellow Ranger was actually a guy. I wonder what Thuy Trang felt about that?)

So that was a little strange, but I also got so see a bit of some of the really good Super Sentai that have never been translated into English. Specifically, Jetman was the year before Kyōryū Sentai Zyuranger, but is regarded by many as the best of all of them in terms of story. Ryoko and I in fact watched all of it and it was a lot of fun to watch, though much of it hasn’t aged well to tell the truth. Also the original Time Rangers from 2000 is a favorite of many, which in a way is unfortunate because that became Power Rangers: Time Force which was frankly pretty crappy.

If you’re interested in watching Gokaiger, the entire series (with English subtitles), can be torrented here.

I just got back from 3 months working in Japan on an assignment. It was a very interesting experience, but I’m glad to be back with my family.

One thing I noticed that I had never seen before: looking to buy a souvenir for a coworker back in Austin, my boss and I were in a high-scale department store in Fukuoka. Since we were looking for something for my bosses admin, we were were in the beauty/bath section. It was a little strange, since were 1) the only males I could see anywhere, and the only foreigners I could see anywhere.

One other thing that I thought was strange though: I saw a girl wearing HUGE plastic horn-rimmed glasses, that quite obviously had absolutely no lenses at all. I had no idea what to make of that, but evidently I’m not the first person to notice this trend.

Most people have heard of the Bacon number, the number of degrees of separation an actor has from Kevin Bacon, based of the Six Degrees of Separation idea. Well, this concept of looking up and assigning numbers based on the separation distance was actually borrowed from the Erdős number, which gives the degrees of collaboration between the person and the mathematician Paul Erdős.

I was thinking the other day that there is a good chance I have an Erdős number. I have published a peer-reviewed paper while in grad school, and my co-author has published dozens of papers in several fields, and his PhD advisor and then his PhD advisor are/were very famous in fluid dynamics, which is closely tied to mathematics, so there is a good probability of me having a fairly low Erdős number. You can look up the Erdős number for a person here, but since it only indexes from mathematics journals it’s not nearly as robust and complete as it could be. So while myself and my PhD advisor aren’t listed as having Erdős numbers, his advisor is, so I can just add 2 to that number to get my Erdős number. So my Erdős number trace is the following:
(more…)

My flight from Seoul back to Fukuoka was at 6:30 pm, so that meant I had until about 2:30 to 3:00 to do something on Saturday morning. At one of my friend’s suggestion, I decided to go on a guided tour of the DMZ, the border with North Korea.

During my short trip there I was constantly seeing things that reminded me of Japan, but there is one way that South Korea is totally unique: having a border with North Korea. It’s strange to think that after 50 years the war is still technically ongoing, i.e. there has never been a formal peace accord, just a cease-fire. As the tour bus slowly approached the DMZ, there were obvious changes that the tour guide pointed out to us: fences topped with razor wire along the Han river (the river starts in North Korea and becomes the border as it flows south and then west, so it has been an popular method for saboteurs, commandos, etc. to try and slip into South Korea), guard posts and pillboxes every 500m with ROK soldiers standing guard. Another method that the North Koreans set up was tunnels, and one of the discovered tunnels was the first stop on the tour. Before we got there though, we had to pass a checkpoint at the entrance to the army-controlled zone. Though still outside of the actual DMZ, no one goes in or out without the soldiers checking your ID (passports for all of us on the tour bus) and that you have a reason to be there. Although farms are maintained in the area, there is a curfew so that the farmers there can only work during the day and have to leave before nightfall.
(more…)

Next Page »