# How did four chessmen disappear?

I was visiting two friends yesterday and found them sitting at a chess table.

After a glance at the board, I thought I had interrupted them when they were installing the starting position and I said:

"Oh, how annoying, you have set the starting position but some pieces must be missing and you cannot play.
- What do you mean ?, asked Fabi, who was sitting behind the dark pieces. Nothing is missing, the game is already started and we have played four moves each. But Mag is thinking hard now, I must have put him in trouble already !
- No trouble at all, answered Mag, I am already two pawns up."

You will understand my mistake once you consider the position that was on the board at that time: all the pieces were on their possible starting squares, but four of them were already taken !

What was the position, and how did the game start ?

TL;DR

Find a chess position:

• that is reached after four moves by each player,
• in which four units are taken,
• where White has a two pawns material advantage,
• where all the remaining units stand on the same squares as at the beginning of a game.

Remarks

• If this proves too hard, I will post the position as a hint and only ask for how the game went.
• This puzzle is an adaptation a chess problem by Ernest Clement Mortimer & Andrei Frolkin published in Die Schwalbe 71 in october 1981.
• This puzzle is an entry in Fortnightly Topic Challenge #40: Retrograde Analysis.
• Spoiler: Mag won this game. But once this game finished, they went for a second round !
• Does "a two pawns material advantage" mean specifically "two more pawns and equal material otherwise" or are you e.g. counting minor pieces as 3 and rooks as 5? Commented Oct 17, 2018 at 15:49
• I found a way to do it ending on White's fifth move. Oof. Commented Oct 17, 2018 at 16:57
• The term "material" is used to refer to anything, while "piece" is often used to refer non-pawn material. Are you using "piece" in that sense? Commented Oct 17, 2018 at 17:26
• @Acccumulation : here "piece" means any unit, either pawn or figure. Commented Oct 18, 2018 at 7:21
• @GarethMcCaughan : We can have the less restrictive definition: a 2-points advantage. Commented Oct 18, 2018 at 7:23

This seems to work:

1. Nf3 e5
2. Nxe5 Ne7
3. Nxd7 Nec6
4. Nxb8 Nxb8

And the position looks like this:

Apart from the symmetrical solution, this might very well be unique:

For starters, white's moves 2, 3 and 4 must be captures. This greatly limits the available options and brings the problem within the limits of solvability.

White cannot move a pawn: if he does, it must get captured. This means that all the captured pieces must be pawns. This means that the white pawn must be the piece doing all white's capturing: any other piece doesn't have the time to move back to its starting position, and as noted earlier, all the captured pieces must be pawns in this case.
From there, we can deduce that if white starts with a pawn, white's fourth move must be exactly "pawn captures pawn on the seventh rank", and of course there is no way for black to capture it back with a pawn.

Because white cannot move a pawn, the knights are the only option. Since white's fourth move must be "knight captures something", there is no way to for a black piece to take it and return to its starting position. Also, since black takes the white knight, white must also take a knight. Therefore, black's fourth move must be "black knight takes the white knight at the other black knight's starting square", and there seems to be exactly two ways to achieve that if white must take pawns on the intervening two moves. (The other solution starts with the other white knight and is symmetrical.)

• The analysis sounds convincing, but this does not appear to satisfy the requirement that "all the remaining units stands on the same squares as at the beginning of a game" -- black's g8 knight is now at b8. Commented Oct 17, 2018 at 23:36
• The two Knights are indistinguishable for someone just walking into the room, though. Commented Oct 18, 2018 at 0:36
• @HenningMakholm reading the question carefully, the requirement is "all the pieces were on their possible starting squares", which is important, because otherwise there couldn't be a solution, I think.
– Bass
Commented Oct 18, 2018 at 5:31
• Excellent, this is the expected anwer ! Commented Oct 18, 2018 at 7:11
• @HenningMakholm : I've been careful to write "at the beginning of a game" and not "at the beginning of the game" ;) Commented Oct 18, 2018 at 7:19

I think I may have found another answer:

1. Nc3 d5
2. Nxd5 Be6
3. Nxe7 Qc8
4. Nxc8 Bxc8

Thoughts?

• Nice, but this doesn't meet the requirement that White has exactly a two-pawns advantage. Taking a Queen against a Knight is much more valuable (usually quantified as 6 pawns). Commented Oct 18, 2018 at 14:27
• Ah, I see. I thought two-pawns advantage just meant you took two pawns. Nevermind! Commented Oct 18, 2018 at 14:31
• maybe another puzzle for the third game? :) and in this case all the pieces really do stay in their starting position. Commented Oct 18, 2018 at 15:14

Partial:

For white to be two pawns up, and 4 pieces to have been taken in total, this implies that black has lost 2 pawns, and each player has also lost 1 of the same type of piece, meaning that black has lost 3 in total, and white just the 1.

Now to solve the rest...

By exhaustive computer search, there are known to be 10 so-called "homebase" positions reachable in a unique way after 4.0 moves. See PDB problem database. This gives 7 entries, but 3 of them have variants discussed in the text. They include the Mortimer/Frolkin & its mirror flip given by Bass, and also the try given by Markyland (which was found by Buchanan in 2004).

Of course, there is no requirement in the question for the move sequence to be unique. In principle, the question is still open whether there are other non-unique solutions, but I doubt it.