I was given this puzzle by Arvind Hariharan and he has a write up with a different solution approach here: https://www.starvind.com/math/hats-to-leave-you-scratching-your-head/.
The puzzle (apparently from Riddler) is as follows:
Four people are trying to escape from a room. Guards have placed a hat on each person’s head, and each hat is one of three colors: red, yellow or blue. The four people are arranged at the vertices of a square, with an obstacle in the middle. Each person can see the hats on the heads of those on adjacent vertices of the square, but they cannot see the hat of the person diagonally across from them. They also do not know the color of the hat on their own head.
Each person must guess the color of the hat on their own head. If at least one person guesses correctly, they can all escape the room together. No communication is allowed once the hats are placed on their heads, but they can coordinate on a strategy beforehand. They also know how they will be arranged in the square.
How can they be guaranteed to escape the room?
Please stop reading if you wish to solve the puzzle by yourself.
The reason for writing this article is to discuss a strange approach to solving this problem: Giving a proof of existence of a strategy using Linear Algebra!
Since there are $3$ colours, we can work mod $3$, with $0, 1, 2$ being the colours and $x + y = x + y \mod 3$ (This is the field $\mathbb{F}_3$). Note that $2x = -x$ etc.
Let us assume the four people are $A,B,C,D$ and they are given the hat colours $a, b, c, d$, with $A,C$ and $B,D$ being diagonally opposite. Thus $A$ and $C$ only have access to $b,d$ and $B, D$ only have access to $a, c$.
Let us further assume that there is strategy where each person computes some linear combination of the colours they can see and uses it as their guess.
For eg: $A$ guesses $x_a b + y_a d$ for some $x_a, y_b$ specific to $A$.
Now introduce the offset variable $\delta_a$ which is the amount $A$'s guess is off from their actual value. i.e.
$$ \delta_a = x_a b+ y_a d - a = x_a b + y_a d + 2a$$
Writing this equation for each of $A,B,C,D$ we see that the $\delta$'s can be captured as a matrix equation as follows
$$S \begin{bmatrix}a\\ b \\ c \\ d\end{bmatrix} = \begin{bmatrix}\delta_a \\ \delta_b \\ \delta_c \\ \delta_d \end{bmatrix} $$
where $S$ (short for strategy) is the 4x4 matrix
$$ \begin{bmatrix} 2 & x_a & 0 & y_a\\ x_b & 2 & y_b & 0\\ 0 & x_c & 2 & y_c\\x_d & 0 & y_d & 2 \end{bmatrix}$$
$S$ is a viable strategy if for every possible $a,b, c, d$, the atleast one of the $\delta$'s is $0$.
Now comes the key observation:
Lemma 1: For any $x, y$ at least one of $x, y, x + y, x+2y$ is $0$ (note $\mathbb{F}_3$ working mod 3). This is easily proved.
This fact can be stated in linear algebra terms
Lemma 2: Each vector in the space spanned by $u = \begin{bmatrix} 1 & 0 & 1 & 1 \end{bmatrix}$ and $v = \begin{bmatrix}0 & 1 & 1 & 2 \end{bmatrix}$ has at least one co-ordinate which is $0$, i.e. for any $x, y$, the vector $xu + yv$ has at least one co-ordinate which is $0$.
Lemma 3: Now notice that if you now have two arbitrary values $p$ and $q$ and you wanted a 1x4 vector with one of the coordinates as $p$ and other as $q$ (in some given positions), you can find (a unique) vector of the form $w = x u + y v$ which has co-ordinates $p$ and $q$ in the given positions (the other two coordinates are not "free"), as you are essentially solving $x + y = p, x + 2y = q$ or some such pair of equations from Lemma 1.
Now look at the columns of $S$
$$ \begin{bmatrix} 2 & x_a & 0 & y_a\\ x_b & 2 & y_b & 0\\ 0 & x_c & 2 & y_c\\x_d & 0 & y_d & 2 \end{bmatrix}$$
Each column has $0$ and $2$ in some positions, thus by Lemma 3, there is an $S$ where each column of $S$ can be written as a vector of the form $x u + y v$.
Thus we see that for such an $S$, the offset vector $\begin{bmatrix}\delta_a \\ \delta_b \\ \delta_c \\ \delta_d \end{bmatrix}$, which is essentially a linear combination of the columns of $S$ is thus also a linear combination of $u$ and $v$ and by Lemma 2, has to have at least one co-ordinate which is $0$.
Hence such an $S$ is a strategy which solves the puzzle. This $S$ can actually be computed easily, which we will leave to the reader.
Note that this generalizes, for example you could have a strategy for 6 people with 5 colours where each person could see all but one.
