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As you know, in chess 960, in the initial position, the king lies between the two rooks. Here are also the rules for castling. The question is:

For a starting position position P, let n(P) be the minimal number of moves the Whites need to legally castle in the King wing. How many position are there such that n(P)=1? n(P)=2? And so one.

There are cases where we can castle at the first move (when the King is in f1 and the rook in g1).

There are other extreme case: King in b1, rook in c1; then we have to make space.

Maybe some engines can solve the question.

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    If you can count you can work out the answer yourself. There are only 960 positions to consider. The clue is in the name. – Brian Towers Aug 29 at 17:39
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    Yes, but 1. I am mathematician, so I can count :). 2. Since I am a mathematician, I am lazy so I would like to know whether there are general principles which would avoid to count. And how n(P) is distributed. – Davide Giraudo Aug 29 at 20:17
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    @DavideGiraudo : 1.I am mathematician, so I can count (...) 2. Since I am a mathematician, I am lazy : I could name several counterexamples for both claims ! – Evargalo Aug 30 at 10:17
  • Is n(P) the minimal number of moves before White can castle if he plays alone, or in any possible game (I.e. Black can help)? – Evargalo Aug 31 at 6:32
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    Ah I missed this aspect. I am interested in the minimum in any possible game so like the example of your answer shows, black can help to make space by taking a pawn in the second row. In this way, they do not have to spend a time. – Davide Giraudo Aug 31 at 8:17
6

The three first steps:

There are 90 positions where n(P)=1

For White to be able to castle short at once, he needs to have Kf1, Rg1 in the initial position. Then there are 5 possible spots from the second rook, 3 for one bishop and 2 for the other one, and finally 3 remaining spots for the queen. The knights go to the last two spots.

There are 64 positions where n(P)=2

36 of them with Kf1, Ng1, Rh1: White can play 1.Nf3 or 1.Nh3 and 2.0-0

28 of them with Rg1, Nf1, Ke1: White can play 1.Ng3 or 1.Ne3 and 2.0-0

There are 242 positions where n(P)=3 please double-check this part

Here are the different configurations and the number of relevant positions:

xxxxxQKR 18
xxxxxBKR 36
xxxxxRKR 18
xxxxxKQR 18
xxxxxKBR 24
xxxxKNNR 8
xxxKNNRx 6
xxxKNRNx 6
xxxKRNNx 6
xxKRxNNx 4
RKRxBNNx 2
xxxxKBRx 24
xxxxKQRx 18
xxxxKRBx 36
xxxxKRQx 18

Total:242

Once we reach n(P)=4, we cannot ignore Black's play anymore: for instance, QNRNKBBR looks like it should have n=5, but actually it has n=4 because of the sequence 1.f4 b5 2.Bxa7 Qxg2 3.Bxg2 Nc6 4.0-0. Counting by hand becomes too fastidious and hazardous: a software program is needed.


On the other hand, after clarification by the OP, I believe 7 to be the maximum value of the function P->n(P).

If Black wasn't allowed to "help" we could build only 2 positions with n(P)=9; Here is one, the other is similar with RKBBNNQR on the first rank.

enter image description here

In a series problem where White plays alone he would need 9 moves in order to castle. But since Black moves are taken into account, White actually can castle on move 7 from that diagram if the game goes something like 1.b3 d5 2.Qa3 Qe6 3.Nb2 Qxe2 4.Nd3 Qxf2 5.Bxf2 Nc6 6.Be2 0-0-0 7.0-0.

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