A game with checkers Alice puts checkers in some cells of a $8 \times 8$ board such that :


*

*There is at least one checker in any $1\times 2$ or $2\times 1$ rectangle.

*There are at least two adjacent checkers in any $7\times 1$ or $1\times 7$ rectangle.


Find the least amount of checkers satisfying these conditions.
 A: The optimal is $37$ checkers.
Here is a solution with $37$ checkers:
$\begin{array}{ccccccccc}
-&O&O&-&O&-&O&-\\
O&-&O&O&-&O&-&O\\
-&O&-&O&O&-&O&-\\
O&-&O&-&O&O&-&O\\
O&O&-&O&-&O&O&-\\
-&O&O&-&O&-&O&O\\
O&-&O&O&-&O&-&O\\
-&O&-&O&O&-&O&-\\
\end{array}$
Now for the proof of optimality: it is easy to see that every line or column must have at least $4$ checkers. Moreover, there are only three patterns that achieve this, they are $-O-OO-O-$, $-OO-O-O-$, and $-O-O-OO-$. Two of these patterns cannot be two consecutive columns, because of the first line which would violate condition $1$. In particular, we have at most four $4$-checkers columns.
If we have at most three $4$-checkers columns, then we have at least $3\cdot 4+5\cdot 5=37$ checkers, so we are good.
We now look at the possible patterns for four $4$-checkers columns.
If they alternate, the first line has every other space blank, which violates condition $2$. So it means that they leave two columns in between two of them. 
We look at the pattern on these two columns. The first and last line must be $OO$ to fulfill condition $2$. Then we have a $2\times 6$ grid to fill. We need at least $6$ checkers to do it in a checkerboard way, but this is not enough, because a checkerboard will violate condition $2$ on two $1\times 7$ columns, so we need at least $7$ checkers in this $2\times 6$ grid, bringin the total to at least $37=4\cdot 4+2\cdot 5+ 4+7$.
This concludes the proof that $37$ is optimal.
A: to meet condition 1. any column must have 4 checkers, but to meet condition 2, there must be five checkers if column has a checker at the end.
Since the repeat of this pattern is odd, i.e if say 00-0-0-0 is selected for column 1 then the and cycling round, i.e c2 = -00-0-0- then colum 8 will be a be the same as column 1, ( since shifting the pattern will at some point split the double around pos 1 and pos 8, which is invalid).
thus, there are 5 columns of 5 and 3 columns of 4, giving 37 checkers.
Not much proof, involved, but not sure where to go from there.
