# Permuting officers during a Chess960 game

There are... let me see... ah yes 960 different possible starting positions in Chess960.

Suppose the players never move a pawn, or make a capture, but simply move their officers so that eventually each officer is somewhere on its own back rank again. How many new but legal arrangements of the officers are possible, beyond the 960 that we are familiar with.

Assume that the final position is still symmetrical between White-Black.

So if we were talking about orthodox chess, the answer would be 3, because either rook might swap with the adjacent knight. But what’s the answer for Chess960?

Hint:

The only way that one can reach a new position involves castling. Consider each of the 16 possible bishop arrangements in turn to see what castling can happen.

• Are the players to keep the position mirror symmetrical?
– Bass
Jul 31, 2023 at 6:50
• @JaapScherphuis rot13(Lrf ohg fhpu n cbfvgvba jvyy nyernql or pbhagrq nzbat gur 960 cbfvgvbaf lbh fgnegrq jvgu!) Aug 1, 2023 at 7:51
• Oh, it was not at all clear to me that that is what your question was asking for, but on rereading it I can see that is what was meant. I thought "new" merely meant different to that particular starting position. Aug 1, 2023 at 7:55
• I thought the same as @JaapScherphuis when I read the question. Perhaps it could be reworded to make it clearer what is wanted.
– fljx
Aug 1, 2023 at 7:59
• I've added some words to hopefully make it clearer - please let me know if this isn't ok yet! Aug 1, 2023 at 8:05

If I didn't make a mistake in counting, the answer is

549.

Recall that in Chess960, the King must be in between the Rooks. So any new position we create must have both Rooks to one side of the King. Thus, any such new position must be reachable after a castling move.

Observe some constraints:

1. Bishops cannot move
2. Kings and Rooks cannot move past Bishops
3. After castling, any piece(s) between the King and its closest Rook can only be Knights
4. Any movement of the King and its closest Rook after castling must be doable by only inserting and replacing Knights (and shuffling pieces)

Consider each of the $$4\times 4 = 16$$ valid starting combinations of the Bishops.

We get some results in the following table (rows are dark-bishop locations, columns are light-bishop locations), displayed as "A-side castling combinations" + "H-side castling combinations". We let a=45, b=37, c=30, d=21, e=E=20, f=12, g=9 (proof later).

* B D F H
A c+a 0+b b+0 a+c
C 0+a 0+b 0+0 0+c
E g+d 0+d e+0 e+f
G E+0 0+0 a+0 a+0

a=45: for each of the (6 choose 2 =) 15 positions of the N, there are exactly 3 ways to rearrange the remaining KRRQ, namely (assuming A-side castling) QKRR, KRQR, and KRRQ. (Thanks Laska for the hint)

c=30: for each of the 15 positions of the N, there are exactly 2 ways remaining (i.e. deleting QKRR from the "a" list).

b=37: the configuration XYYXCXX (up to mirror symmetry), where Y are the squares occupied by the castled K and R immediately after castling, C represents any number of Bs, and X are free space. Calculation is in the below table (note: in the left of the table, an asterisk (*) denotes one of the remaining pieces (subject to constraints), r counts the number of ways for the R, and q counts the number of remaining ways for the Q).

XYYXCXX r*q
KR**C** 3*4 - 2 = 10 (-2: can't start KRRQ or KRQR)
*KR*C** 3*3 = 9
**KRC** 2*3 - 0 = 6 (-0: guaranteed N to the left of K)
KNR*C** 3*2 = 6
*KNRC** 2*2 = 4
KNNRC** 2*1 = 2


d=21: the configuration ZZZZBYYX, where one of the Zs is a B. The rightmost three squares can be RK? (9 ways), RNK (6 ways), or NRK (6 ways).

e=20: the configuration XXYYCXX.

E=20: the configuration XBYYXXBX.

f=12: the configuration XXXXBRKB. There are 4 ways for the R and 3 remaining ways for the Q.

g=9: the configuration XBKRBXXX. There are 3 ways for the R and 3 remaining ways for the Q.

• Great stuff. I think it can be a bit slicker. For each BB case, for each of qside & kside, there are always 60 candidates. Q can't be between the K&R that castle, nor can it block a castling square. So for BBxxxxxx, for the four locations of Q in RKR, there are two that work for qside & three for kside. So there are 60*(2/4 + 3/4) = 30+45 = 75. Aug 6, 2023 at 2:22
• For BxxxxxxB it's the reverse: kside 2/4 qside 3/4, so still a total of 75. For xxxxxxBB there is no kside castling, but qside is possible except for RQKR case so 3/4 = 45. And so on Aug 6, 2023 at 2:27
• Thanks for the shortcut @Laska . It saved some calculations for my "a" and "c" cases. However, it cannot be applied everywhere. I've made an updated solution. Aug 6, 2023 at 8:55
• Yea, the "a" and "c" tables aren't necessary. Aug 6, 2023 at 9:29
• @Laska corrected the "b" case :) Aug 6, 2023 at 9:40