# Double chess stalemate with 60 units

Can you arrange on a chessboard 18 kings, 6 rooks and 6 bishops of each colour (i.e. 60 units, leaving 4 empty squares) so that neither side has a legal move.

Ignoring reflections, rotations and colour swaps, is the solution unique?

• Just to clarify, there is no symmetry requirement in this question, is that right? Commented Jul 25, 2022 at 11:45
• Yes no symmetry at this point. I would prefer any symmetry emerges naturally from the solution Commented Jul 25, 2022 at 12:47

## 3 Answers

Update: One ... more ...

Not at all unique:

and there are more.

My solution is

0 B K K K K R B
B R K K K K K R
K K K K K K K K
K B B B R R R 0
0 r r r b b b k
k k k k k k k k
r k k k k k r b
b r k k k k b 0


where upper case means black, lower case means white, K means king, b means bishop, r means rook and 0 means empty.

I'll admit this is a variation of the solution proposed by AxiomaticSystem, but I cannot comment yet and the problem proposed by JLee is very real, yet his fix does not take into account that changing the g3 king into a rook makes it so the g4 bishop is not pinned anymore.

Again, my solution has rotational symmetry if you swap colors simultaneously, but there are many solutions without symmetry.

More precisely, there are two possible corners you can choose on each side to be empty, for a total of 4 possibilities. Only the kings in a3,e3,f3,g3 and one of b3 and d3 have to be kings for pinning reasons, so any of the other ones can be freely swapped with the two free white rooks or the free bishop (the ones in the bottom left corner of the example). The only exception is that the bishop cannot be on b3 and the h4 king cannot be exchanged because the piece would be able to move. So choosing the b3 king to stay, 6 kings stay in place, so there are 12 possibilities for the free white bishop and then 11x10/2=55 possibilities for the white rooks. Choosing the d3 king to stay, there are only 11 possibilities for the bishop, but still 11*10/2=55 possibilities for the rooks. Since the situation is symmetric, there are also 11x55+12x55=23x55=1100+165=1265 to distribute the black pieces. So in total, there are at least 4x1265^2=2530^2=2500^2 + 2x2500x30 + 30^2 = 6250000 + 150000 + 900 = 6400900 possibilities. If you want to disregard reflection, rotation and color swaps, then this number can at most be divided by 4, so we get at least 1600225 possibilities that are similar to this.

• Yep. I realized that and deleted that comment a while back
– JLee
Commented Jul 25, 2022 at 19:07

At first I didn't see a way through, and then I realized that not every piece needs to be on the front lines:

While this particular arrangement is symmetrical, it admits not only its reflections and rotations but also arbitrary shuffles of the first two and last two ranks.

• What about either white king to h4? Or either black king to a5?
– JLee
Commented Jul 25, 2022 at 13:05