tiles covering a $7\times 7$ square

Question

A $7 \times 7$ board is divided into $49$ unit squares. Tiles, like the one shown below, are placed onto this board. The tiles can be rotated and each tile neatly covers two squares. Note that each single tile consists of two unit squares joined at a corner. What is the minimum number of tiles that can be placed onto the board so that every uncovered square will be adjacent to at least one covered square?

Note: Two squares are adjacent if they share a common side.

My attempt: These tiles cover a shape like $$\begin{matrix}0&0&1&0\\0&1&1&1\\1&1&1&0\\0&1&0&0\\ \end{matrix}$$where $'1'$s are the covered squares/next to covered squares. Using $3$-coloring (to the $7\times7$ board) we need at least $7$ tiles. I found a way to do it with 8 tiles. But is there a way prove $7$ isn't enough (or is it actually enough)?

Answer 1

You also cannot succeed with $8$ tiles. Each of the tiles can only cover one of the squares marked with an $\times$: $$ \begin{array}{|c|c|c|c|c|c|c|} \hline \times &\;\,&\;\,&\times&\;\,&\;\,&\times\\ \hline &&&&&&\\ \hline &&&&&&\\ \hline \times &\;\;&\;\;&\times&\;\,&\;\,&\times\\ \hline &&&&&&\\ \hline &&&&&&\\ \hline \times &\;\;&\;\;&\times&\;\,&\;\,&\times\\ \hline \end{array} $$

Here is a solution with $9$ tiles: