I was thinking about what is the maximum number of queens that can occur on the board after only legal moves. The theoretical upper bound would be 18, the original 2 queens plus all promoted pawns. However to have the pawns get past each other they would have to capture the opponents pieces but they could not capture the opponents king or queen, so thats 2 less for each side (although a capture on one side would open for the other - so gets a bit tricker than that I realize). However if so that would be 14, but with that many queens it would be easy to put the opponents king in check or checkmate I guess.

But is it possible to get 14 queens on the board in a legal game and if not what is the maximum ?

up vote 17 down vote accepted

It should be easily possible to get 18 queens. If white captures four enemy pieces, that's enough to get doubled pawns on four files (a, c, e and g, for instance). And black captures four times to get his pawns on the b, d, f and h files. Then they can all advance and promote, and it should be easy to avoid mate by storing them all in some corner.

Here, have the ugliest proof game ever :-)

[FEN ""]

1.Nc3 Nc6 2.Nb5 Ne5 3.Nd4 Nc4 4.Nf5 Na3 5.bxa3 Nf6 6.Nh6 gxh6 7.Bb2 Ng4 8.
Bf6 exf6 9.Nf3 Ne3 10.fxe3 Bb4 11.Nd4 Bc3 12.dxc3 Rg8 13.Nf5 Rg3 14.hxg3 
h5 15.Nd6+ cxd6 16.Rb1 f5 17.Rb6 axb6 18.a4 h4 19.a5 h3 20.a6 h2 21.Rg1 
h1=Q 22.a7 Rb8 23.a8=Q Qh6 24.Qa3 Qg7 25.Qb2 h5 26.a4 h4 27.a5 h3 28.a6 h2
29.a7 h1=Q 30.a8=Q Qhh8 31.Qaa1 f4 32.g4 f3 33.g3 f2+ 34.Kd2 f5 35.Bg2 
f1=Q 36.c4 Qf4 37.c5 d5 38.c6 d6 39.c7 Bd7 40.c8=Q Qfh6 41.Qcc3 f4 42.Qcb3
f3 43.c4 f2 44.c5 f1=Q 45.c6 Qf5 46.c7 Qfh7 47.c8=Q d4 48.Ke1 d3 49.Kf2 d2
50.Qf1 d1=Q 51.Qcc1 Qd5 52.g5 b5 53.g6 Qge7 54.g7 b4 55.g4 b5 56.Qd3 b3 
57.Qbc3 b2 58.Kg3 b1=Q 59.g8=Q+ Qef8 60.g5 b4 61.Qgg7 b3 62.Qgd4 b2 63.g6 
Qc2 64.g7 b1=Q 65.g8=Q Qdh5 66.Qda4 d5 67.e4 d4 68.Qda6 d3 69.Qga2 Qdf6 
70.e5 Kd8 71.e6 d2 72.Qc3a3 Kc7 73.e7 d1=Q 74.e8=Q Qd8 75.Qef7 Rb7 76.e4 
Rb8 77.e5 Rb7 78.e6 Rb8 79.e7 Rb7 80.e8=Q *

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