Chessboard domination problem for a new piece (called Wazir) I am trying to solve a chessboard domination problem, but for a new piece, which is called Wazir (thanks to @bof for pointing that out) which only threatens the neighbouring squares meaning the ones that share an edge with each other. Also the size of the chessboard is not $8\times 8$ but is arbitrary; $M\times N$. What I have found so far is the calculation for small cases by try and error for $4\times 2,4\times 3,4\times 4, 4\times 5, 4\times 6, 4\times 7$ and for $5\times 2,5\times 3,5\times 4, 5\times 5, 5\times 6$.
Updated chart:
    1   2   3   4   5   6   7
4   2   3   4   4   6   7   7
5   2   3   4   6   7   8   9(thanks to @bof)

Can I get any hint for this?
This problem is supposed to be a math project for a college student, and the idea I have in my mind is complex for college level math:
One can see that we define a grid graph corresponding to the chessboard of size $(M-1)\times (N-1)$ by connecting the associated vertices to the squares of checkerboard if they are neighbour on checkerboard.
UPDATE:
The set that we are after is the dominating set for this grid graph. It seems the problems is open as there exist only bounds for it:
http://www.emis.de/journals/EJC/Volume_18/PDF/v18i1p141.pdf
I would be very happy if somebody could acknowledge what I am saying here.
 A: The area covered by this piece (the "wazir") is the shape of a plus sign.  It's straightforward to tile the plane with five-square plus signs, so your "piece fraction" approaches $0.2$ as the board gets big.  Each open square is attacked by exactly one of these pieces well away from one of the edges.  These wazirs would be arranged in a tilted square lattice that could be traversed by a chess knight.
The periodicity of this tiling pattern is $5 \times 5$.  In other words, if you displace the entire pattern by $5a \hat{x} + 5b \hat{y}$, it will map onto itself.
The edges of the board are where the deviations happen.  Extra wazirs have to be placed to cover the squares where the infinite plane symmetry is broken.  Specifically, every fifth square on the edge will need to be "plugged up" by a wazir because it's left unattacked.
The problem now is to minimize the sum of the number of edge squares that need to be plugged up, plus the number of non-edge squares that need to be occupied by a wazir.
Since the pattern repeats by displacing it a multiple of $5$ squares in either direction, we only need to consider the grid size modulo $5$ to determine the deviations at the edge.
The answer will be of the form
$$K(M,N) = \lfloor\frac{MN + 2(M+N)}{5}\rfloor + E(M (\text{mod } 5), N (\text{mod } 5)),$$
where $E(M (\text{mod } 5), N (\text{mod } 5)) = E(N (\text{mod } 5), M (\text{mod } 5))$ is an adjustment function.
So for a grid size of $M \times N$ let's look at the cases.  By symmetry, we can consider only those for which $M \leq N$ (mod $5$).
By explicit calculation for $M = 10 ... 14, N = 10 ... 14,$ I found:
$$E(0,0) = E(1,0) = E(2,0) = E(3,0) = E(4,0) = 0;$$
$$E(1,1) = E(1,2) = E(1,3) = E(1,4) = -1;$$
$$E(2,2) = -2; E(2,3) = E(2,4)= -1;$$
$$E(3,3) = E(3,4) = -1;$$
$$E(4,4) = -2.$$
This likely breaks down for small $M, N$.  I was looking more to shed light on the general case above.
For small boards, there are other behaviors.  $M=1$ is interesting in that the wazirs can have two blanks between them in the optimal configurations for $N > 3$.  (They must for $N > 5$.)  This is related to the fact that the board size limits their attack field to two squares.  Hence the piece fraction $\geq 1/3.$  So $K(M, 1) = \lceil M/3 \rceil.$
For $M=2$, the board size limits the attack field to three squares, and the piece fraction hence $\geq 1/4$.  From my observations, for $M \geq 2$, $K(M,2) = \lceil (M+1)/2 \rceil.$
