Discrete Peaceful Encampments: Player 3 has entered the game!

Question

Here's a variation of Discrete Peaceful Encampments: 9 queens on a chessboard (which itself is a variation of Peaceful Encampments).

You have 4 white queens, 4 black queens, and 4 red queens. Place all these pieces onto a normal 8x8 chessboard in such a way that no white queen threatens a black queen, no black queen threatens a red queen, and no red queen threatens a white queen (nor vice versa).

Can you find a way to place more than 4 queens of each color "peacefully" on an 8x8 chessboard?

Answer 1

Ok, I'ma call it.

There's no way to get more than 4 of every colour. Also, there is no simple way to prove this.

This is the most efficient way to do it:

Notice how

There's room for all kinds of shenanigans: You can add
* a white queen to e1 or a2
* a red queen to g2 or h7, and
* a black queen to e7, or
* any queen to g7, if you move a1 to a2 first
Annoyingly, adding any two colours always excludes every option of adding the third colour, no matter how much you shuffle the pieces around.

This way you can get

any two colours to 5, but not all. Also, you could get white to 5 (e1) and red to 6 (g2, g7), but black still stays at 4, so you get a 4-5-6 solution.

In addition to all that:

There's so much wiggle room in the above diagram, and you can get so very very close to a 5-5-5, that any simple impossibility proof (like "there aren't enough diagonals on the chess board") is not going to work.

This all is a result of feeding this problem into a highly sophisticated self learning neural network ^*, making it start from random (and later self-selected) positions, where every improvement path always led to this position, or one of its descendants, showing that this position is at least a local optimum.

^{*_{my brain}}

Now all that is needed is

a) a brute force solution proving that this is indeed the optimum, or
b) a 5-5-5 solution, or a simple proof of its impossibility.

If anyone can provide case b, I'll happily buy that person a beer, after a solid stint of banging my head against a wall.

Answer 2

5+5+4 and 6+5+4, but no 5+5+5

The orange piece can be any color.

for the bonus:

One of the yellow square can the 9th queen of appropriate color, this then has $9+9+8$ on an $11\times11$ board.

Answer 3

Well, I can do 4+4+4:

 0 0 . . . . . .
 0 0 . . . . . .
 . . . . . . 1 1
 . . . . . . 1 1
 . . 2 . . . . .
 . . 2 . . . . .
 . . . . . . . .
 . . . . 2 2 . .

Answer 4

I believe I have obtained better results than those posted previously:

I can place 4+4+8 queens on a 8x8, sorry for the basic formatting:

..22....
..22....
......33
.......3
....3...
11......
11......
....3333

I can also do 4+5+6 on a 8x8, which beats 4+5+5 ;)

..11...1
..1....1
.......1
........
33......
33......
....222.
.....22.

And for the bonus question:

I can place 8+8+11 on 11x11:

...222.....
...222.....
...22......
.........11
..........1
...........
.33........
333........
333........
......1111.
.......1111