## Tuesday, July 9, 2024

### A 熱中症予防 week

There were no contests that I'd like to mention last week, so I can get straight to the new format of this blog for the coming week: a travel blog! I am going to the AtCoder World Tour Finals 2024 in Tokyo. I did not manage to qualify this time, placing 14th when I needed to be in top 12, so I am going as a spectator and as a co-commentator for the stream, together with Riku, the main admin of AtCoder.

For a second year running, Tokyo welcomes the participants with an extreme weather warning in Japanese, this time for extreme heat. Please all take care, and focus on playing board games in the air conditioned hotel!

Speaking for 5 hours while 12 contestants are facing, to put it mildly, challenging problems is also not a walk in the park. Please help us by suggesting topics that we should discuss or things we should do on stream in comments!

In my previous summary, I have mentioned a Universal Cup problem: first, we draw the vertices of a regular n-gon (n<=10000) on the plane. Then we repeat the following process m times (m<=30000): take any 3 already drawn points (either the original n-gon vertices, or ones drawn during the previous steps) Ai, Bi, Ci, and draw the point Ai+Bi-Ci (the fourth vertex of a parallelogram). Finally, we need to handle multiple queries of the form: given k drawn points (the sum of k over all queries <=30000), do they form the vertices of a regular k-gon in some order?

We can get points that are really close to a regular k-gon but not exactly one in this problem, and no feasible floating point precision is enough. Therefore we need to solve it using only integer computations. Nevertheless, let us first consider how a floating-point solution might work.

We can imagine that the points lie on the complex plane, and the initial n points are ei/n. Drawing a new point corresponds to computing Ai+Bi-Ci using the complex number arithmetic. There are many ways to check if k computed points form a regular k-gon, here is one: we need to check that the k points, in some order, can be represented as x, xy, xy2, ..., xyk-1, where y is such that yk=1 and no smaller power of y is equal to 1. Note that this order does not have to be the clockwise/counter-clockwise order of the vertices: multiplying by y can also represent a jump by any coprime number of edges, and this criterion will still be valid. Also note that we can actually pick any of the vertices as x and such y will exist, moreover there will be φ(k) different values of y that work for each vertex. So one way to find y is to just try a few other vertices z of the polygon, let y=z/x, and check if the criterion is satisfied. Since φ(k) is not too small compared to k, we should find a valid y after a few attempts, let's say at most 50. Of course, we could've just said that y=ei/k, but you will see below that the y=z/x approach leads to an interesting question that I want to discuss.

If we denote the "base" point as u=e2π/n, then all other initial points are powers of u, all computed points are polynomials of u, and the checks we are making boil down to whether a certain polynomial of u with integer coefficients is equal to 0 or not (even though we also use division, checking if poly1(u)/poly2(u)=poly3(u)/poly4(u) is the same as checking if poly1(u)*poly4(u)-poly2(u)*poly3(u)=0). We could try to maintain said polynomials with integer coefficients, but since the degrees would be on the order of n, and the coefficients themselves could be huge, this is not really feasible within the time limit.

Here comes the key idea: instead of complex numbers and u=e2π/n, let us do the same computations using integers modulo a prime p and using such u that the order of u modulo p is equal to n. Such u exists if and only if p=sn+1 for some s, so we can search for a random big prime number of this form, which can be found quickly since all arithmetic progressions with coprime first element and difference have a lot of prime numbers.

This actually works very nicely and allowed us to get this problem accepted. However, why does this actually work? The order of u is the same in our two models (complex numbers and modulo p), so every polynomial of the form ut-1 is equal to 0 in both or in neither model. However, this does not guarantee the same for an arbitrary polynomial with integer coefficients, or does it?

Of course it is not true for an arbitrary polynomial. For example, the polynomial p is equal to 0 modulo p, but not equal to 0 in complex numbers. However, we can deal with this by picking the modulo p at random, and potentially also checking several moduli in case the judges actually create testcases against many potential moduli. So the real question is: are there polynomials for which being equal to 0 differs between the complex numbers and computations modulo p for all or a significant fraction of all possible values of p?

Here we need to bring in some maths. When two polynomials with rational coefficients are equal to 0 for the given u their greatest common divisor also has rational coefficients and is also equal to 0 for the given u, which means that there must exist a minimal polynomial such that a polynomial with rational coefficients is equal to 0 for the given u if an only if the minimal polynomial divides it. Such minimal polynomial for our u is called the n-th cyclotomic polynomial Φn(u).

Now, consider the equality un-1=Φn(u)*gn(u) (where gn(u) is just the result of dividing un-1 by Φn(u)). This equality is true in rational numbers, so it is also true modulo p where there is no division by p in it, so for almost all p. The left-hand side is 0 modulo p because of our choice of u, so either Φn(u) or gn(u) must be 0. However, from the structure of the cyclotomic polynomials we know that gn(u) is a product of cyclotomic polynomials of smaller order, so if it was equal to 0, it would mean that the identity ut-1=0 would hold for some t<n, which contradicts our choice of u. So we know that Φn(u)=0 modulo p, which means that every polynomial with integer coefficients that is equal to 0 for the given complex u will also be equal to 0 for the given u modulo p. So we have proven one of the two required implications.

Now let us tackle the the opposite implication. Consider a polynomial h(u) with integer coefficients that is equal to 0 for all or a significant fraction of all possible values of p (with the corresponding u). If Φn(u) divides this polynomial (as a polynomial with rational coefficients), then it is also equal to 0 for the given complex u, as needed. If Φn(u) does not divide it, then we can find the greatest common divisor of Φn(u) and h(u), again doing computations using polynomials with rational coefficients. Since Φn(u) is irreducible over polynomials with rational coefficients, this greatest common divisor will be 1, so we have 1=Φn(u)*a(u)+h(u)*b(u). The right side involves a finite number of different integers in denominators, so this equality will also hold modulo p for all p except those dividing one of the denominators, in other words for almost all p. But since both Φn(u) and h(u) are equal to 0 for all or a significant fraction of all possible values of p, this means that 1 is equal to 0 for all or a significant fraction of all possible values of p, which is a contradiction. Therefore we have also proven the opposite implication and this solution does in fact work.

There are still a few things I do not fully understand about this setup. One is the following: it turns out that when n is odd, we can actually construct a regular 2n-gon (roughly speaking using the fact that -1 helps generate the other n points; there was such example in the samples for n=3, k=6), so k does not have to divide n. In this case, the number y that we find as part of solving the problem must have order 2n modulo p. However, note that in general it is not even guaranteed that there is any number with order 2n modulo p, as we only choose p in such a way that there is a number with order n. Since we compute y=z/x, we can do this computation for any p where we can compute z and x. So it seems that the above also proves that for almost all primes p if there is a number of odd order n modulo p, there is also a number of order 2n modulo p. This observation is in fact true for a straightforward reason: almost all primes are odd, so there is an even number p-1 of nonzero remainders, therefore there is a number of order 2, and we can multiply the number of odd order n by the number of order 2 to get a number of order 2n. Still, I can't get rid of the feeling that I might be missing something here. Any comments?

The second thing I don't fully understand is whether we truly need a full understanding of the structure of cyclotomic polynomials to prove that Φn(u)=0 modulo p. It feels that maybe there is an easier way to explain this that does not require so much knowledge?

1. "Please help us by suggesting topics that we should discuss or things we should do on stream"
- I think something fun will be to discuss the differences you notice in problems from differences. Possibly from when you first competed in another country. Knowing such experiences might be fun!

2. Do you like anime?