It's checkmate in 20 moves. White's queens circle around the board giving checks, and Black interposes horizontally/vertically moving pieces. Black only has one choice because the other piece is pinned from the previous check. That goes well, until the pawn needs to move sideways:
[FEN "3Q4/7Q/3rp3/2rkr3/2rrr3/7K/8/8 w - - 0 1"]
1. Qb7+ Rc6 2. Qa5+...
Here is an example of a 12-move game after which White (to move) is stalemated. All 32 units (pieces and pawns) are still on the board. The original version of this concept game was created by Charles Henry Wheeler, and published in Sunny South in 1887, according to Edward Winter's C.N. 3679. Samuel Loyd is often, and wrongly, given credit.
Here is a starter solution in 7 moves.
Can we do better?
1.g3 e5 2.Bh3 Ke7 3.Bxd7 Kxd7 4.Kf1 Ke7 5. Kg2 Ke6 6.Kh3 Kf6+ 7.Kh4 Kg6#
If this is best, interestingly, the play is almost forced. The only variations are 4... Kd6 or 4... Ke7 in the line above. There is also 4... Ke6 and 5... Kf5, or 3. Kf1 Kd6/Ke8 4. Bxd7 Kxd7 5.Kg2 Ke6, etc. ...
White is clearly in a dire situation, since Black is threatening mate in one on either h7 or h8 if White doesn't do anything. But the White king can't move move nor can White get rid of any of the pawns that surround the king. White's rook, the only White piece, is too far away to do anything. The move 1. Ra8? to try and stop 1... Qh8# fails ...
Double check is only possible by using discovered check. So either the rook check or the bishop check was discovered by moving something in between on the previous move.
I don't see how that's possible with the rook check, but it is possible with the bishop -- if the board is shown with black at the bottom, contrary to what is usually done. Then there could ...
If you promote to a queen with 1. b8Q, black has:
1...Re2+ 2. Kd1 Rb2
attacking the queen and hinting at mate with Rb1++. If white takes the rook 3. Qxb2 it is stalemate.
Because of the mate threat white does not have any other good square for the queen either (no good check and no square that would defend b1).
If you promote to a rook, black does ...
It's impossible to checkmate faster than 7 turns (handicap on black) or 8 turns (handicap on white).
Proof by elimination
I'm going to argue from the perspective that white is helping and black is handicapped, since otherwise it would take 1 more move.
Only Queen, Bishops and Rooks can be used by the black player to threaten the white king, as only they ...
It's quite a fun problem to think about, before getting to the
calculation of long variations, try to first spot the key idea needed to crack the problem.
Here are the first observations that come to my mind which eventually
led me to spot the solution, let's break them down step by step:
a) With our bishop eyeing g7 and our doubled pawn formation on g6-g7,...
White is one tempo short of catching the pawn - if White could make two moves immediately it would be a draw as white would just take the black pawn. But they can't, so white has to find a threat which black has to respond to which gains them that move. The only threat they can make is to queen their pawn, and apparently black can stop that with their bishop ...
In my thread on the English Chess Forum, which seemed to make the world go crazy on the subject, I gave all the major and minor events in the history of the “legal” triple check that my extensive research has uncovered. This loophole was even talked about recently in Episode 20 of "The Chess Pit" at the 12:13 mark. Here is all information in ...
Use the fact that the rook is pinned, and that the king has few squares left;
I'm thinking that 1.Rd8 Kd3 (only possible move) 2.Nc5# should be the solution.
[FEN "7R/1B1N4/8/3r4/1K2k3/8/5Q2/8 w - - 0 1"]
1.Rd8 Kd3 2.Nc5#
Assuming you allow promoted material (since you didn't say anything :-),
this (on Page 13 of the PDF) is the (unfortunately, extremely unknown) finite record since ages. It shows the record for the longest sequence of only 1 legal move for each side, with use of promoted material.
[Title "Karl Scherer, Feenshach 1980, Page 13"]
r2q1rk1/p3b1p1/2Q1p2p/2npp3/4P3/2PPBN1P/PP3PP1/R3R1K1 b - - 0 1
1... Rc8 2. Qb5 Rb8 3. Qc6 Rb6!
Now white's queen is trapped.
As @Joffan has pointed out in the comments and as many have probably seen, immediately trapping the queen is not the end to this game:
r2q1rk1/p3b1p1/2Q1p2p/2npp3/4P3/2PPBN1P/PP3PP1/R3R1K1 b - - 0 1
1... Rc8 2. Qb5 Rb8 3. Qc6 Rb6 4....
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.
The first half of white's move is forced - d8. Then white has a choice of knight, bishop, rook or queen. Black has two threats which white must parry to get a draw. The first is to play Bf4 and then move the king out of the way to deliver checkmate. The second is to take the white h pawn and then queen the black h pawn.
Let's look at what happens if white ...
Depending on whether occupied squares need to be covered as well, the number is:
[Title " 12 knights, Without Covering Occupied Squares"]
[FEN "8/5N2/1NN1NN2/2N5/5N2/2NN1NN1/2N5/8 w - - 0 1"]
[Title " 14 knights, With Covering Occupied Squares"]
[FEN "8/2N1NN2/2N1N3/2N3N1/2N1N3/1NN1NNN1/8/8 w - - 0 1"]
Problems like this are called domination problems and ...
The check must come from the knight (D'uh!)
The black king must be on c5 for the mate
The white king must be used to cover any empty squares to the right of the black king - thus the white king must move, thus there must be at least one non-checking move
On a non-checking move Black can try to release the prison by Ra4
The only way white can cover this ...
If White can get the Black king to the first rank, then it will not be fast enough to catch the g-pawn from promoting. White starts by playing 1. Qg5, and after 1...Kh7 2. Qf6 Kg8 3. Qh6, the White queen can simply imitate the Black king's movements until she can start forcing the king down towards the first rank.
[FEN "7k/8/8/8/6p1/4QpPb/5PpP/6K1 w - -...
It's probably a trick problem with a promotion to a black knight.
Such promotions to the wrong colour are not allowed, and never were. In the official rules it is now specifically pointed out that the new piece has to have the same colour as the promoted pawn.
FIDE's laws of chess, Article 3.7 e:
When a pawn reaches the rank furthest from its starting ...
The Matt Bengtson problem Prof. Elkies mentions is:
[Title "Matt Bengtson, Chess Braintwisters (Burt Hochberg), no. 103. White to move & draw."]
[FEN "4kn2/3p1pPp/4pPpK/6P1/8/2p5/1b6/8 w - - 0 1"]
However, the problem is actually cooked with a win for Black starting with “3... f6!”, and there is no stalemate for White. The solution and the cooking line ...
This is actually a rather typical retrograde problem, just start with the most basic observations:
We see that black is missing both rooks and the f8 bishop. Given black's pawn structure it's easy to see that neither the f8 bishop nor the a8 rook could have escaped the structure, thus they must have been captured on the 8th row (by a white knight for ...
[FEN "1B1Q1Q2/2R5/pQ4QN/RB2k3/1Q5Q/N4Q2/K2Q4/6Q1 w - - 0 1 "]
105 mates — Nenad Petrovic, Sahovski Vjesnika 1947 (Chess Problem Database)
In this position any check is mate. There are 3 knight mates (c4, g4, f7), 23 discovered mates (14 moves for the rook on c7, 9 for the bishop on b5), and 79 queen mates: 1 on a1, 2 on b2, 3 on c3, 4 on c5, 6 on d4, 3 on ...