In chess, what position has the highest number of checkmates in 1 move for a given side? You can have as many of each piece as you want, except for the king (you get exactly one). Bishops can be same color. Other color can have only the king, or other pieces if you want to. Rules same as chess rules. Pieces can be:

1 King (a must)

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    $\begingroup$ what do you mean by the highest number of mate in 1s for 1 color? $\endgroup$
    – Novarg
    Commented Apr 6, 2016 at 12:30
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    $\begingroup$ makes sense to me. Put pieces on the board such that the number of moves which constitute a mate in 1 is maximised, for either black or white $\endgroup$ Commented Apr 6, 2016 at 12:41
  • $\begingroup$ Edited question to be clearer and voted to reopen. Please tell me if this isn't what you meant. $\endgroup$
    – Paul Evans
    Commented Apr 6, 2016 at 13:15
  • $\begingroup$ define 'legitimate position'? $\endgroup$ Commented Apr 6, 2016 at 13:26
  • $\begingroup$ @PaulEvans You added the phrase 'legitimate position'. What did you mean, or what do you think OP meant? $\endgroup$
    – Lacklub
    Commented Apr 6, 2016 at 13:39

3 Answers 3


The highest number of one-move mates that I have found is currently 91. It is based off of the following idea:


Here, each dot or letter represents a square on the board. There are 7 positions that the bishop can move to, so I would count this as 7 different moves that end in check mate, ie. 7 different 1 move mates. When counting the total number of moves that mate from each piece in the board, I will represent it like this:


I found the following board based off of this idea that has a total of 91 different moves which result in checkmate:


Where many pieces have several possible mates (a=10, b=11):


There might be a small permutation of this board which increases in incrementally, but I am interested in what other ideas people have from scratch.

  • $\begingroup$ How have you found this board ? manually or computer ? $\endgroup$
    – Fabich
    Commented Apr 6, 2016 at 15:08
  • $\begingroup$ @Lordofdark Manually. I started off with the exterior rooks and bishops that will mate the king, then added in their blockers. I then added queens anywhere I could to get a few extra moves, and turned some of the exterior pieces to queens to get the last 2 mates. $\endgroup$
    – Lacklub
    Commented Apr 6, 2016 at 15:12
  • $\begingroup$ Replacing the d8 rook with a queen and moving it to d7 gives one extra checkmate Q7-e5++ $\endgroup$ Commented Jan 9, 2023 at 8:52

A bit late to the party, but here's 142 checkmates:

enter image description here

The K in the middle is the black king. Every number is a white queen, with the number indicating the number of ways to mate with that queen. I got this through the following reasoning:

  1. arbitrarily choose black to be mated
  2. The question asks for the highest number of mates for 1 side, not both sides, so clearly the mated side should only have a king and no other pieces, as any other pieces would just get in the way.
  3. the black king should be in one of the center squares (doesn't matter which), as that allows the greatest scope for mating
  4. the black king should not have anywhere to move, to allow maximum mates. note this would be stalemate in a real game, but there's no stipulation it has to be a realistic position.
  5. Queens and knights are the only pieces worthy of being considered for white, as they can emulate the moves of every other piece, whereas the reverse is not true.

From the above, I devised a kind of modified greedy algorithm (manual as i cant be bothered to code it), as follows:

  1. assign a number B to each square which is not yet taken, where B represents the number of mating moves which either end on that square or pass through that square. Thus B represents the number of mates which would be subtracted if you put a piece on that square. Initially all squares are 0.

  2. For each square which is not on a checking line, work out A-B=X, where A is the maximum number of mates you could add by adding a piece. Thus X is the 'net profit' from adding a piece. (queen or knight, although I soon worked out that only queens are really effective - knights cost more than they contribute in the long run).

  3. add a piece to whatever square has the most profit. in the event of a tie, favour corner squares first, side squares second and other squares third, since corner/side squares will interfere less with the greedy algo going forward.
  4. repeat from 1 until there's no more profit to be had.

The 142 solution above is one of the local maxima using this algo. I can't prove it's the global maximum.

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    $\begingroup$ You get the same number of mates from just putting a queen in every non-checking square, as all that really matters is that every checking square has a queen in each direction that isn't in line with the king. (So yes, this is the global maximum, at least unless using the blocking strategy in the other answer can beat it.) $\endgroup$
    – Zandar
    Commented Apr 7, 2016 at 18:01
  • $\begingroup$ Ah yes. Now I feel very stupid! $\endgroup$ Commented Apr 7, 2016 at 18:17
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    $\begingroup$ Please add a king for the mating side and I will present da green check to you. $\endgroup$
    – Anonymous
    Commented Apr 21, 2016 at 12:45
  • $\begingroup$ Needs more cowbe... ehr, White King. $\endgroup$
    – SQB
    Commented Jun 6, 2016 at 15:52

Without a king the maximum is 143 mates in one, which was achieved in 1947.

Nenad Petrovic, Sahovski Vjesnik 07/1947

enter image description here

Thanks to @JSI in the comments, with a slight modification to the 1947 position, the highest possible is 142 mates with a king.

enter image description here

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    $\begingroup$ If you took your 143 board and replaced the upper left queen with a white king, wouldn't you get 142? $\endgroup$
    – JS1
    Commented Sep 24, 2019 at 22:16
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    $\begingroup$ Similarly, if you modify your 141 board in the following way: move queen on h7 to g7, add white pawn on h7, add black knight on h8, then the following mates would change: 2 mates added h7xg8/q and h7xg8/b, one mate removed g6-g8, making 142 mates. $\endgroup$
    – JS1
    Commented Sep 24, 2019 at 22:24
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    $\begingroup$ Correction: Above comment should say black knight on g8 $\endgroup$
    – JS1
    Commented Sep 25, 2019 at 2:14

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