For a permutation \(\pi: [n]\rightarrow [n]\), we define the displacement of \(\pi\) to be \(\sum_{i\in [n]} |i-\pi(i)|\).

For given \(k\), prove that the number of even permutations of \([n]\) with displacement \(2k\) minus the number of odd permutations of \([n]\) with displacement \(2k\) is \((-1)^{k}\binom{n-1}{k}\).

The best solution was submitted by 홍의천 (수리과학과 2017학번). Congratulations!

Here is his solution of problem 2020-09.

Another solution was submitted by 고성훈 (수리과학과 2018학번, +3).

In the problem 2019-08 (https://mathsci.kaist.ac.kr/pow/2019/2019-08-group-action/), we considered a group G acting by isometries on a proper geodesic metric space X properly discontinuously and cocompactly. Such an action is called a geometric action. The conclusion was that a geometric action leads to that G is finitely generated.

Would this conclusion still hold in the case the space X is not necessarily proper?

The best solution was submitted by 홍의천 (수리과학과 2017학번). Congratulations!

Here is his solution of problem 2020-08.

A binary maze consists of \(n\) separate rooms. Each room has a teleportation machine but no doors. The numbers \( a_{i,j}\in [n] \) are given for all \( (i,j)\in [n]\times \{0,1\} \). If you shout a number \( j\in \{0,1\} \) while you are in the room \( i \), then the teleportation machine will teleport you to the room \(a_{i,j}\).

You don’t know the numbers \(a_{i,j}\), but it is given that for any \(i\neq i’ \), there exists a way to reach room \( i’ \) from room \( i \) by shouting numbers \( 0 \) and \( 1 \) in some order.

At the beginning, your enemy will teleport you into one of the rooms while your eyes are closed. Your goal is to visit all rooms at least once with your eyes closed. As your eyes are closed, you don’t know which rooms you have visited before and you don’t know which room you are currently at.

So, you decide to pick a sequence \( b=(b_1,\dots, b_s) \in \{0,1\}^s \) before entering the binary maze and decide to shout the numbers \( b_1,\dots, b_s \) in order. Find a lower bound \( \ell(n) \) and an upper bound \( u(n) \) on the minimum length of a sequence which guarantees that you can visit all \( n \) rooms. If your \( \frac{u(n)}{\ell(n)} \) is smaller than some polynomial of \( n \) for all \( n\in\mathbb{N} \) , then you will get full points.

The best solution was submitted by 홍의천 (수리과학과 2018학번). Congratulations!

Here is his solution of problem 2020-06.

Another solution was submitted by 강한필 (전산학부 2016학번, +3).

Let \( f_n : [-1, 1] \to \mathbb{R} \) be a continuous function for \( n = 1, 2, 3, \dots \). Define

\[

g_n(y) := \log \int_{-1}^1 e^{y f_n(x)} dx.

\]

Suppose there exists a continuous function \( g: \mathbb{R} \to \mathbb{R} \) and \( y_0 \in \mathbb{R} \) such that \( \lim_{n \to \infty} g_n(y) = g(y) \) for all \( y \neq y_0 \). Prove or disprove that \( \lim_{n \to \infty} g_n(y_0) = g(y_0) \).

The best solution was submitted by 홍의천 (수리과학과 2017학번). Congratulations!

Here is his solution of problem 2020-04.

Other solutions were submitted by 길현준 (수리과학과 2018학번, +3), 김기택 (수리과학과 2015학번, +3), 이준호 (2016학번, +3).

For a given positive integer \( n \), find all non-negative integers \( r \) such that the following statement holds:

For any real \( n \times n \) matrix \( A \) with rank \( r \), there exists a real \( n \times n \) matrix \( B \) such that \( \det (AB+BA) \neq 0 \).

The best solution was submitted by 홍의천 (수리과학과 2017학번). Congratulations!

Below is his solution (in text) of problem 2020-01.

An incomplete solutions was submitted by 고성훈(수리과학과 2018학번, +2).

Let’s prove the statement in the problem holds if and only if 2r>=n.
Let V=R^n and think of n*n real matrices as linear operators on V.

AB(V) and A(V), so also BA(V) has dimension at most r.
(AB+BA)(V) has dimension at most 2r so if the statement holds then 2r>=n.

Let’s assume that 2r>=n.
Let the null space and range of A be U and W, respectively, and X the intersection of the two.
Let {a_1, a_2, … , a_k}, {a_1, a_2, … , a_k, b_1, b_2, … , b_(n-r-k)},
{a_1, a_2, … , a_k, c_1, c_2, … , c_(r-k)} be bases for X, U, W, respectively.
The subspace spanned by {c_1, c_2, … , c_(n-r-k)} (here we use 2r>=n) does not contain
a non-zero element of U so {d_1, d_2, … , d_(n-r-k)}
(d_i=A(c_i), 1<=i<=n-r-k) is linearly independent.
Let {d_1, d_2, … , d_(n-r-k), e_1, e_2, … , e_(2r-n+k)} be another basis for W.
The subspace spanned by {b_1, b_2, … , b_(n-r-k)} does not contain a non-zero element of W
so {b_1, b_2, … , b_(n-r-k), d_1, d_2, … , d_(n-r-k), e_1, e_2, … , e_(2r-n+k)}
is linearly independent and we may obtain a basis for V by adding f_1, f_2, … , f_k.

Let’s define B by determining it’s values on the basis elements.
Set B(b_i) (1<=i<=n-r-k) so that AB(b_i)=c_i (which is possible since c_i is in W),
let B(d_i)=c_i+b_i (1<=i<=n-r-k) (so that AB(d_i)=d_i),
set B(e_i) (1<=i<=2r-n+k) so that AB(e_i)=e_i,
let B(f_i)=0 (1<=i<=k).
Then for any w in W, AB(w)=w.

Let’s show the range of AB+BA is the whole of V.
For 1<=i<=n-r-k, (AB+BA)(b_i)=c_i+B(0)=c_i, (AB+BA)(c_i)=c_i+B(d_i)=2c_i+b_i
so b_i is in the range of AB+BA.
For any x in X, (AB+BA)(x)=x+B(0)=x and so U is in the range of AB+BA.
For any v in V, ABA(v)=A(v) so BA(v)-v is in U.
For any w in W, (AB+BA)(w)-2w=BA(w)-w is in U so W is in the range of AB+BA.
For any v in V, (AB+BA)(v)-AB(v)-v=BA(v)-v is in U so v is in the range of AB+BA.
The whole of V is in the range of AB+BA so det(AB+BA) is not zero.