Related to Open Problem #1: Knights and Knights

The right-hand diagram shows that it is possible to do this with a $7 \times 7$ square.

If we tried to place the knights on a $6\times6$ board, then (every knight has to attack four squares) no knight can go on the corner squares or the eight squares adjacent to the corners. Although the other edge squares attack four squares, they cannot be used either, since they attack the squares adjacent to the corners, and so a knight there would have to attack another knight that could only attack at most three other knights. Thus no knight can go on the edge, and so we are in fact forced to try to achieve the arrangement of knights on a $4\times4$ board.

If we tried to place the knights on a $5\times5$ board, then (for exactly the same reasons as above) we could not place a knight on any edge square, and so we would be attempting a knight arrangement on a $3\times3$ board.

If we tried to place the knights on a $4\times4$ board, then (again) we could not use any of the edge squares, so would be restricted to a $2\times2$ board.

No knight on a $3\times3$ board or smaller can attack four other knights.

Thus the size of the smallest possible square board for which this arrangement of knights is possible is $\boxed{7\times7}$ .

Any other possible solutions?

1. Is this the only configuration?

2. Is this the only configuration if we're trying to minimize the board size?

3. How many ways are there of doing this?

So I think it's well-proven that the minimal board size is $\boxed{7\times7}.$

However, some questions arise to our brilliant minds. Then how about we go deeper?

---

1. Is this the only configuration?

No. There are other ways.

One way is to just simply copy the configuration and then paste it on a pretty distant place.

---

2. Is this the only configuration if we're trying to minimize the board size?

Yes, you plug in the colors in the below diagram.

It is found by trying to minimize the whole size with one type of knights. (since one knight must threaten exactly 4 other knights)

and thus this is the only-smallest configuration.

---

3. How many ways are there of doing this?

If we consider rotating/flipping as another way of making the diagram, then there are $\boxed{384}$ ways of doing this.

Lemme explain.

First, pick any point from the diagram. [there are $\color{#E81990}16$ ways of doing this.]

Next, pick any point that is connected by a line and eliminate all other points that were the option. [there are $\color{#E81990}4$ ways of doing this.]

Then do it again. [there are $\color{#E81990}3$ ways of doing this.]

Again! [there are $\color{#E81990}2$ ways of doing this.]

Again... [there are $\color{#E81990}2$ ways of doing this.]

Again...? [there are $\color{#E81990}2$ ways of doing this.]

...again. [there are $\color{#E81990}2$ ways of doing this.]

After that, revive all the red points, and then select from the two options that you've got. [there are $\color{#E81990}2$ ways of doing this.]

Then you'll realize you've made a $\color{#D61F06}c\color{#EC7300}l\color{#CEBB00}o\color{#20A900}s\color{#3D99F6}e\color{#69047E}d~\color{#E81990}p\color{#624F41}a\color{#BBBBBB}t\color{#333333}h!$

The rest of the places are, as you may have assumed, for white knights to occupy. :P

Therefore the number of cases is $16\times4\times3\times2\times2\times2\times2\times2=6144.$

Wait, no. This is a $\color{#D61F06}c\color{#EC7300}l\color{#CEBB00}o\color{#20A900}s\color{#3D99F6}e\color{#69047E}d~\color{#E81990}p\color{#624F41}a\color{#BBBBBB}t\color{#333333}h$ with length $8.$

So we must divide that number by $16,$ to avoid duplicates.

Hence the number of cases is $6144\div 16=\boxed{384}.$

I guest 7 and it was right the first try.

It wasn't really a solution. I just looked at it then guessed 7, although I mentally did extrapolate the coverage of a board with two knights 'at full reach' which clearly requires a minimum of seven squares.