Matching problem of slightly different chessboard

Question

I try to prove that it is impossible to cover figure 1 with tiles of size $1\times 2$ and $2 \times 1$. When I abstract figure 1 as a graph I realize that it is a bipartite graph. I think this equation might help me. $\sum_{a \in X} \deg(a) = \sum_{a \in Y} \deg(a)$ where $X$ and $Y$ are the bipartition of the graph.( I already proved this formula) Unfortunately, I don't see how I could argue any further now.

modified chessboard

Answer 1

This is called the "Mutilated Chessboard Problem" https://en.wikipedia.org/wiki/Mutilated_chessboard_problem

Assume you had a colored chessboard. The removed squares in opposite corners would be the same color. So instead of 64 squares: 32 black and 32 white, you would have 62 squares: 32 black and 30 white.

Each $1 \times 2$ or $2 \times 1$ tile would need to cover one black square and one white square. Since there are not equal numbers of black and white squares, you can't cover the board using only the $2 \times 1$ and $1 \times 2$ tiles. One tile would have to cover two black squares, which can't happen.