The 4th annual π day anime and mathematics post: Traveling teacher time

Korosensei is fast, but can he do better?

Korosensei is fast, but can he do better?

せーの、

Korosensei can move at Mach 20 and takes advantage of this fact to make frequent trips around the world. Obviously, in the interest of efficiency, he often takes the opportunity to make multiple stops per trip. But what is the most efficient way for him to do so?

We can think about this using graphs. Suppose all the possible destinations are vertices and edges with costs connect the vertices, representing the cost of traveling from some destination $u$ to another destination $v$. At first glance, it might make sense to use physical distance in our case, but that is not necessarily a measure of the “best” way to get there. For instance, taking a bus or train between two cities may get you to the same place, but the actual cost depends on the path that’s taken and you may choose to include other criteria like the cost of tickets and other stuff like that.

So the problem is this: Korosensei has a list of places he wants to hit up during lunch hour. What is the most cost-effective way of getting to each of those places exactly once and get back to school? This is what’s known as the Traveling Salesman problem (TSP).

As it turns out, this problem, like many graph theory problems, is NP-complete, meaning it’s quite computationally hard and taxing for computers to solve. Formally, if we’re given a graph $G = (V,E)$ and a cost function $d: V \times V \to \mathbb R$, the problem is to find a cycle that visits every vertex exactly once with smallest possible cost. Usually, if we have $n$ vertices, then $G$ is the complete graph on $n$ vertices, that is, a graph where every vertex has an edge to every other vertex. Otherwise, it may not be possible to find such a cycle (and the problem of finding such a cycle is also an NP-complete problem, Hamiltonian cycle).

It’s called the Traveling Salesman problem because the idea is that if you’re a traveling salesman, you’d want to find that least-cost cycle because it’d be the most cost-effective way to hit up all the cities you need to go to in order to sell whatever it is that you’re selling. Of course, the hardness of the problem means that our poor salesman or Korosensei are stuck having to check the cost of every cycle. How many are there? Well, since we’re dealing with a complete graph, this means we can get to any vertex from any vertex so we can take any permutation of our vertices and that’s a cycle, and there are $n!$ permutations. This means that as $n$ gets larger, the difficulty of solving this problem grows exponentially.

One way we can attempt to make NP-complete problems feasible to solve is to loosen our demands a bit. Right now, we want the absolute least-cost cycle, but what if it’s not necessary for us to get the least cost cycle? What if we’re okay with getting something that’s within a factor of 2 or 3 or $f(n)$ for some function $f$ that depends on the size of the input? These are called approximation algorithms, in that they approximate the optimal solution by some factor. Unfortunately this doesn’t work for TSP.

Why? Because if I have a feasible algorithm for approximating TSP, then I’ve got an efficient algorithm for solving Hamiltonian cycle, which I mentioned was also NP-complete. Suppose I take any old graph $H$ with $n$ vertices and turn it into a complete graph with all my original edges having cost 1 and any new edges I add having cost $\rho \cdot n + 1$. If I do have some approximation algorithm that gives me a solution within a factor of $\rho$, I’ll run it and get a cycle for Korosensei to use. But, based on the cost of the cycle I get, I can tell if my original graph had a Hamiltonian cycle: if it has cost $\leq n$, then that means it found a Hamiltonian cycle using all of the edges in my original graph. Otherwise, we get a cycle with cost at least
$$(\rho \cdot n + 1) + (n-1) = \rho \cdot n + n > \rho \cdot n$$
meaning it must’ve used one of the new expensive edges I added and so we know the original graph didn’t have a Hamiltonian cycle.

Now this is a bit discouraging, but Korosensei would encourage us to keep on thinking about how to assassinate the hell out of this problem. In the most general case, the cost function $d$ has no constraints and the way that I’ve initially motivated Korosensei’s problem, $d$ can be arbitrary, with costs adjusted to his needs. However, some of the attempts to make TSP more computationally feasible have to do with making some reasonable restrictions on our cost function. This is another fairly standard approach to making computationally hard problems easier to deal with: figure out some cases of the problem that are still useful but might help to simplify or restrict the problem a bit.

For example, in most cases, it’s a reasonable expectation that if you’re going from $u$ to $v$, it’s best to go there directly rather than stopping at $w$ on your way there. This is what we call the triangle inequality:
$$d(u,v) \leq d(u,w) + d(w,v).$$
It’s called the triangle inequality because of what I just described: draw a triangle out of $u$, $v$, and $w$ and the distance between any two should be shorter than going through the third point.

I say should because for an arbitrary cost function, this isn’t always the case. One example is flight pricing. Sometimes, it’s more expensive to fly from $u$ to $w$ than it is to fly from $u$ to $v$, even if the flight stops over at $w$. Anyhow, Korosensei doesn’t have to deal with silly things like flight pricing and so we impose this reasonable triangle inequality condition on his cost function. Does it help? It turns out it does and we can get an approximation that’ll be at most twice the cost of the optimal solution.

How? Well, first we find a minimum spanning tree. A spanning tree of a graph is a tree (a graph that contains no cycles) that connects all of the vertices of a graph together. The minimum spanning tree is the spanning tree with the least weight out of all the spanning trees. The nice thing about MSTs is that we know how to do this efficiently.

Once we have our MST $T$ of our graph $G$, we double up every edge in $T$ to get a cycle $T’$. This cycle what we want yet, since it travels along all the same edges and goes to each vertex twice. But, we can get the cycle we really want by removing all of the second times we visit a vertex and we can do this because the triangle inequality tells us that by taking a more direct route, the cost of the cycle won’t increase. So now if we have our cycle $C$, then we have
$$cost(C) \leq cost(T’) \leq 2 \cdot cost(T).$$

Now, to compare to the cost of our actual optimal solution $C^*$, we notice that taking out some edge of $C^*$ makes it a spanning tree. It’s not necessarily minimal, so we’ve got
$$cost(T) \leq cost(C^* \setminus \{ e \}) \leq cost(C^*).$$
Putting that all together, we’ve got
$$cost(C) \leq cost(T’) \leq 2 \cdot cost(T) \leq 2 \cdot cost(C^*)$$
and there’s our 2-approximation.

Given how fast Korosensei is, this is probably good enough for him and is a decent tradeoff between the time it takes to solve the problem and his actual travel time.

12 Days XII: Mochizou Love Story

「もち蔵大好き、どうぞ」 | 欸府

「もち蔵大好き、どうぞ」 | 欸府

Something that I was totally convinced of during the airing of Tamako Market was that Mochizou and Tamako would get together by the end of the show. Otherwise, I reasoned, it wouldn’t make sense how much emphasis was put on how much Mochizou was in love with this childhood friend of his and this particular subplot fit in well with the whole getting married off to a foreign prince thing. Of course, none of this happened and I was left a bit disappointed.

Some time later, at a KyoAni event, Horiguchi had let slip that there was a Tamako movie or something in the works. When the title was revealed to be Tamako Love Story screamed internally and thanked the CEO of Kyoto Animation.

Of course, it turned out that this approach to the Tamako/Mochizou relationship worked out better in the end. Tamako Market let us see the characters going about their lives while being distracted by a talking bird. Separating the story about Tamako and Mochizou from that meant being able to focus intensely on the two of them for an hour and a half. This is obviously great because Tamako is cute and Mochizou is cute and they are cute together. Anime of the year.

12 Days XI : Future Fish

【7FrFr!ネタバレ】 | Mina

【7FrFr!ネタバレ】 | Mina

As promised, we did see them next summer. So this season, more than anything else, was about Haru’s (and some of the others’) futures. One of my small hopes was that for Haru, it wouldn’t be swimming related. It was something that I drew from the ED, but of course, you could argue that those were the dreams they had when they were kids and those are always bound to change. But I did enjoy Haru’s other talents being mentioned or shown briefly throughout the show and I really wanted him to pursue one of those. I think that kind of path fits in more with how he treats his swimming. But alas, that wasn’t how it played out, but that’s okay, I still love Haru.

12 Days X: The king of my court

セッターの皆様 | みかんもち@お仕事募集中

セッターの皆様 | みかんもち@お仕事募集中

The only things I knew about Haikyuu before it aired was that it was about soccer volleyball and that the characters were very moe. I was tempted to start the manga, but I had put that off for so long that the anime was coming up anyway. In a twist, not only was it everything I wanted but it was also well-animated. What’s nice about Haikyuu is that everyone’s great and there’s no real drama outside of sports stuff. Some characters have history, but there’s no one who’s an outright jerk or has issues or anything. Everyone’s fun, something that comes out in the training camp arc. Of course, that could change since I don’t think it’s ending anytime soon, but for now, it’s pleasant.

12 Days IX: Art imitating life

浴衣千代ちゃん♡ | 刃天

浴衣千代ちゃん♡ | 刃天

I read way too much bad shoujo for my own good. It’s a bit like my relationship with McDonald’s: every few months, I go against instinct and pick up a new series, then get to the end and flip a table in anger, swearing it off until I forget about it a few months later. But as Gekkan Shoujo Nozaki-kun reminds us, it’s so easy to go back to because the elements are all so familiar. And like any good comedy, Nozaki-kun shows us how adorably silly some of those elements are.