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.
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 ...
As a starter here is a 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, or 4...Ke6 and 5...Kf5, or 3.Kf1 Kd6/Ke8 4.Bxd7 Kxd7 5.Kg2 Ke6 etc. Only five possible ...
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 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,...
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 ...
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#
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 ...
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"]
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"]
[Title "A Visual Solution Minus The Illegal Move"]
[FEN "4knq1/3p1p1p/4pPpK/6P1/8/2p5/1b6/8 b - - 0 1"]
1... Qg7+ 2. fxg7 c2 3. g8=Q c1=Q ...
1...R8xe3 wins a piece (the knight on e3), because either recapture leads to worse things for White, since the d2 bishop is pinned to the white queen, and the f-pawn recapture opens the possibility of mate on g2:
[fen "4rbk1/pp3pp1/1nq4p/8/1PP5/P3N1P1/Q2BrP1P/3R1RK1 b - - 0 1"]
1...R8xe3 2.Bxe3 (2.fxe3 Qg2#) Rxa2
Anthony Stewart Mackay Dickens found another solution, also with 105 mating moves, but with only 17 units in the diagram (16 white and the black king):
[title "Anthony Stewart Mackay Dickens, The Problemist, Jan 1970. 105 mates"]
[fen "2Q1Q3/2Q4Q/Q4Q2/3k4/Q5Q1/1R6/B1NBQ3/K2R1N2 w - - 0 1"]
This may be found here on PDB.
Black's last move must have been ......
I'm going to start from scratch even though the OP posted a partial answer in the question, so I will be covering some familiar ground.
I began to break the problem down by assigning black to the upper king, then making every piece which attacked him black. Since the g4 knight attacks a king, no other piece can be giving check to either king.
[title "White ...
Disclaimer: This solution is not reachable from the starting position, and is not reachable in a game of Chess960 (thanks Rewan!).
[FEN "3bBNRN/2pPpPKQ/2P1P1PR/7P/p7/rp1p1p2/qkpPpP2/nrnbB3 w - - 0 1"]
Why does the solution here not work?
This is clearly not reachable from the starting position (because of the bishops stuck on the first rank), but the ...
After 1.Ra6, Black is in Zugzwang - the purpose of Ra6 is to prevent Black's a-pawn from moving. Now Black must move his rook. If he moves it off the h file, then White plays Rh6#. However, after 1... Rh8, White mates anyway with Ng7.
Black can hold (after a fashion) if his Rook can control g7 and h6. After 1. Ra6, he will have to leave one of those ...
This is a familiar tactic I often get to use myself in another opening line. If black plays 1... Rf8, then comes 2. Bc7! Qe8 3. Nd6 Bxd6 4. Bxd6 and black loses the exchange anyway.
[Title "White to move"]
[FEN "r1bqr1k1/pp1nbpp1/4pn1p/2pp4/2PP1B2/P1NBPN2/1P3PPP/R2Q1RK1 w - - 0 1"]
1. Nb5 Rf8 2. Bc7! Qe8 3. Nd6 Bxd6 4. Bxd6
Chess is about patterns. Since ...
Just to prove that this is possible, here's an opening bid of 9 moves. It's ugly because black makes several time wasting moves, but white needs to get his king to somewhere where it can be mated and also remove black's pawn (I don't see how to get a mating position with 7 black pawns on their original squares), so it takes time.
1.e4 e5 2.Ke2 Ke7 ...
If black plays QxR then white plays Qc3+ with mate to follow. Black's Qg7 stops the mate. With the black queen on b2 white cannot play Qc3. The Re2 move is to divert the black queen
[FEN "5r1k/1pQ4p/3nB1p1/3P4/5p2/1P5P/rq5P/4R1RK w - - 0 1"]
1. Re2 Qxe2 2. Qc3+ Qe5 3. Qxe5+ Rf6 4. Qxf6#
I think I found a double-self-smothered-stalemate that could occur via sequence of legal moves from the normal starting position.
[FEN "5bnk/4ppbp/4P2p/5P1P/p1p5/P2p4/PBPP4/KNB5 w - - 0 1"]
1. c3 f6
Each player has seven pawns and two dark-square bishops, so one pawn from each player must have been promoted. Once White's b pawn has captured to a3 and ...
Very nice puzzle! Looks like the solution is as follows:
[FEN "8/8/6p1/5pP1/5P1K/5PpP/4p2p/2Q2Bkr w - - 0 1"]
1. Qe1 g2 (1... exf1=Q 2. Kxg3 Qxe1#) (1... exf1=R 2. Qxg3#) (1... exf1=N 2. Qf2+ Kxf2 (2... gxf2)) (1... exf1=B 2. Kxg3) 2. Bxe2# 1-0
Interestingly, in the different lines, following results are achieved:
White gives a checkmate
Black gives a ...
There is only one black piece left, so the last move was with the king. Black's last move was obviously Kg1-h2.
The king could not have come from h3, because there is no way the white pawn could have given check from g2 (its initial position). All other squares around h2 are occupied by white pieces.
Since the black king came from g1, it was in check from ...