For all those who lost their lives and due to this tragic disease
CONTEXT
This question is inspired by the following question that was proposed by my math teacher Lam Nguyen, I shall cite this exactly as it's origin:
Assume that there is a $2m\times2n$ table. The governor and the Coronavirus play a game. In their respective moves, the governor and the virus quarantine a square that is formed by multiple unit squares (its size can vary). The first player who does not have a legal move loses the game. The governor moves first. Proves that no matter how the virus spreads, the governor always has a winning strategy.
Now it is very clear that the table can be divided in to 4 separate parts. If the governor picks the $2\times2$ square in the center of the table, then the table still consists of 4 identical parts. Now what the governor has to do is to pick the squares symmetrically through the table's centre.
!note that by picking at the center first, the player has avoided the fact that the center square is a reflection of itself through the center.
I doubt the applicability of this strategy in this following problem
EXPANSION 1
What is the minimal casualty that the COVID-19 virus can cause regarding a $4038\times4040$ table
EXPANSION 2
Provides a $2m\times2n$ table, and 2 players A and B starts moving from 2 opposite corners of the rectangle (2 corners that lies on the same diagonal). In each move, each player inhabit an adjacent square that:
- Is formed of unit squares
- Must share at least one unit side with one of the squares that player occupied.
The one who makes the last move wins. Player A moves first
Post script
Now I suppose that A and B cannot use the "symmetric" strategy now. But I assume that A still win because he can determine the situation by the first moves. It could also be depending on the value of m and n.
Per OP's request, I am writing a solution for the game on an $m\times n$ board under the teacher's rule (the government and the virus can occupy anywhere on the board). When $m\equiv n\pmod{2}$, the government has a winning strategy that guarantees that it occupies at least $\left(\frac{m+n}{2}\right)\min\{m,n\}$ unit squares. The virus has a strategy to infect at least $\left(\frac{|m-n|}{2}\right)\min\{m,n\}$ unit squares.
Wlog, suppose that $m>n$. The government's first move is to occupy the central $2\times 2$ square when $m$ and $n$ are even, or the central $1\times 1$ square when $m$ and $n$ are odd. Then the government copies the virus move symmetrically about the board's center. This is a winning strategy of the government.
However, the government can maximize its occupied territory with the first move being the occupation of the central $n\times n$ square. This leaves $(m-n)n$ unit squares to be dealt with, and the strategy of symmetric moves against the virus guarantees that the government can take at least have of the remaining unit squares. That is, the government can aim to occupy at least $$n^2+\frac{(m-n)n}{2}=\left(\frac{m+n}{2}\right)n$$ unit squares.
On the other hand, the virus plays the greedy algorithm. That is, in every turn, it infects a square of the largest possible size. If the first move by the government is to take over a $k\times k$ square (clearly, $k\le n$), then the virus can take at least half of the remaining area. Hence, the virus has a strategy to infect at least $$\frac{mn-k^2}{2}\geq \frac{mn-n^2}{2}=\left(\frac{m-n}{2}\right)n$$ unit squares.
P.S. The situation where $m\not\equiv n\pmod{2}$ is a bit complicated. For $(m,n)=(2k,1)$, the virus wins (and each player occupies half of the board). But for $(m,n)=(3,2)$, the government wins and is able to occupy all but one unit square. I have no idea about this problem if the OP's rule (the players start from opposite corners of the board, and each can only play a square adjacent to previously played squares) is used.