FRC: checkmates in three plies from the starting position.

[Ajedrecista]

Hello:

We all know that the shortest checkmate from the starting position in chess is the fool's mate, that is, a checkmate in four plies from the starting position.

I asked myself the following question: was a shorter checkmate possible in Fischer Random Chess, also known as Chess960? I think so, so I open this thread in a try to collect all the checkmates in three plies from FRC positions. I will use the numbering scheme from the Wikipedia, where No. 518 is the usual starting position RNBQKBNR.

I think that the only possibility to checkmate in three plies (two moves by white and one by black) is to checkmate in a way like this:

[d]rbkrbqnn/p1pppppp/Qp6/8/4P3/8/PPPP1PPP/RBKRB1NN b KQkq - 1 2

The white piece that checkmates is the queen or a bishop that goes to the sixth rank and the black king is attacked through a diagonal (a6-c8 in the example).

Focusing with the black king on c8 (other cases can be mirrored later to f8 square):

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..k.....

1.- A black rook must be on a8 or b8 for allow castling (the other rook must be on the other side of the king).

2.- The black rook can not be on b8 because it would not be checkmate (2.- ..., Rb7). Rook on a8.

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r.k.....

3.- Neither the queen (2.- ..., Qb7) or a black knight (2.- ..., Nxa6) can be on b8. Bishop on b8.

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rbk.....

4.- If the white checkmating piece must be the queen or a bishop, which starts on f1, then the same piece must be on the black side on f8. Since there is a DSB on b8, it can not be other DSB on f8, hence the queens are on f1 and f8. Queen on f8.

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rbk..q..

5.- A knight can not be on d8 (2.- ..., Nb7) and the remaining bishop can not be on d8 (other DSB again). Rook on d8.

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rbkr.q..

6.- Two knights and one bishop (a LSB) remaining for two light squares (e8, g8) and a dark square (h8). Knight on h8.

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rbkr.q.n

7.- Then, two possibilities:
RBKRBQNN (No. 920).
RBKRNQBN (No. 940).

8.- Mirror the positions:
NNQBRKBR (No. 45).
NBQNRKBR (No. 28).

There are two ways of checkmate from each starting position:

1.- e3, b6; 2.- Qa6#. (Positions No. 920 and No. 940).
1.- e4, b6; 2.- Qa6#. (Positions No. 920 and No. 940).
1.- d3, g6; 2.- Qh6#. (Positions No. 45 and No. 28).
1.- d4, g6; 2.- Qh6#. (Positions No. 45 and No. 28).

So, eight possible checkmates in three plies, the same number that the fool's mate in standard chess.

Any corrections are welcome.

Regards from Spain.

Ajedrecista.

[rjgibert]

NNQBBRKR: 1.c4 g5 2.Qxg5#

[Ajedrecista]

Hello:

NNQBBRKR: 1.c4 g5 2.Qxg5#

Great! An epaulette mate that I overlooked. There should be many more of this kind, basically all with RKR at f-g-h files with queens on c-file and without knight on e8 (2.- ..., Ng7). Then mirror the positions to RKR on a-b-c files and so on.

Thinking a bit more: RKR does not have to be only on a-b-c or f-g-h files, they could be elsewhere with the queens on the correct file to allow the checkmate and without knights next to rooks. For example, RKR could be on c-d-e files and queens on h-file, without knights on b and f files. That would lead with many possibilities, which is false, as I show right now.

Each RKR block needs queens on a specific file, leaving the four remaining files for the two knights and two bishops. Except RKR blocks in a-b-c and f-g-h files, the rest should have the two bishops next to the rooks to not allow knights to avoid the checkmate, this is, BRKRB blocks, but then the bishops would be in the same colour, which is impossible. Hence RKR blocks must be in either a-b-c or f-g-h files.

Once reached this point, the setups are KRKB*Q** and **Q*BKRK. For each setup, the remaining bishop can be in two files (one bishop in each colour), then completing the remaining files with knights.

RKRBBQNN (Position No. 921).
1.- e3, b5; 2.- Qxb5#.
1.- e4, b5; 2.- Qxb5#.

RKRBNQBN (Position No. 941).
1.- e3, b5; 2.- Qxb5#.
1.- e4, b5; 2.- Qxb5#.

NNQBBRKR (Position No. 41).
1.- d3, g5; 2.- Qxg5#.
1.- d4, g5; 2.- Qxg5#.

NBQNBRKR (Position No. 24).
1.- d3, g5; 2.- Qxg5#.
1.- d4, g5; 2.- Qxg5#.

Any corrections are welcome.

Regards from Spain.

Ajedrecista.

[Dann Corbit]

I would think that there would be many times more in FRC, because in chess you have one single opening position and in FRC there are a boatload of them.