# Can you mate with each of your 16 pieces on its original square?

An opinion poll conducted by problem composer Roberto Osorio over 10 master-level chess players in the Argentine Chess Club produced 100% wrong answers! :-)

When the right answer is posted here, I will post a follow-up problem.

EDIT: OK here is the follow-up:

``````[title "R.Osorio - StrateGems 2007 "]
[fen "rnb2rqb/pppp4/6p1/7p/6n1/7k/PPPPPPNP/RNBQKBNR w - - 0 1"]
``````

This is the position after White's 16th move. What was the exact sequence of moves?

• Can you mate with each of your 16 pieces on its original square? Yes. Let's say, you have all your pieces, he has just the king, then you can mate him on c2. Drive him to c8 with your pieces, put your queen to e7. Next, place your remaining pieces on their original squares, take care that you don't stalemate him. Then it is a mate in 8 with your queen on d1. A player already did this in his games and posted it to the forum on Lichess, but I didn't bookmark it. Mar 2, 2022 at 23:23

Obviously the last white move was 16. Nf3g1+.

I note that black needs at least 15 moves to place the pieces as they are:

• 4 moves to swap queen with rook as they are (various movements possible),
• 2 moves for the bishop (Bf8g7, Bg7h8),
• 2 moves for the knight (Ng8f6, Nf6g4 or Ng8h6, Nh6g4)
• 1 move each for the two pawns (g7g6 and h7h5)
• 5 moves for the king (any direct route)

Together with the information that it is black's 16th move, this means that all of black's pieces must have taken a direct route to their destination.

Note that black did not capture any white pieces, because white still has 16 pieces on the board. This means in particular that the pawns on g6 and h5 are the pawns from g7 and h7 respectively.

White has promoted the g pawn for a knight. Since black has still the original g pawn, and all pieces except for two pawns, this is only possible, if the white g pawn moved to g6, captured the black pawn on f7 and promoted on f8 for a knight.

The white pawn could only be promoted after the square on f8 was emptied, i.e. the order of moves is roughly: 1) white pawn moves to g6, 2) white pawn captures black pawn on f7, 3) black plays g7g6, and Bf8g7

Before white captured g6xf7, the black e pawn must have been gone as otherwise (with e7, Qd8, Bf8 and Ke8 present), black would have been forced to recapture the white pawn on f7 since it is check (and the black king would have no squares to escape).

Counting the minimum number of white moves now until the position after blacks's 15th move (i.e. with white knights on f3 and g2)

• white's g pawn: 5 moves (g2g4, g4g5, g5g6, g6xf7 f7f8N)
• return the promoted knight from f8 to either f3 or g2: 3 moves
• capturing the e7 pawn and returning the piece: If this was done by the rook, it would take at least 8 moves. This would make the total (5+3+8>15), so is not possible. So e7 was captured by a white knight, which returned. Both options (a) white knight from g1 capturing e7 and returning to g2 or f3 or (b) white knight from b1 capturing e7 and returning to b1 plus the move Ng1f3; would take 7 moves.

5+3+7=15, meaning that also white moved its pieces on shortest possible routes without any time for waiting moves.

As for the rest, basically it comes down to check how to achieve the main plans in order, i.e.:

1. capture the pawn on e7 with a knight and return the knight at least partially (so that the black king is able to get to e7); as mentioned this needs to be done before white captures gxf7
2. capture gxf7 and promote the pawn
3. return all white pieces to their final destination

During phase "1." black is very limited in moves since the g pawn cannot move yet (has to wait for the white pawn to play gxf7). This leaves as the only possible first moves for black 1....h5 2...Rh6 ... 3...Rf6 4....Nh6.

These four moves are just enough time for white to capture the e7 pawn with the knight on b1 and to return the knight to c3 (making space for the king on e7, i.e. giving black additional "waiting" moves (waiting for white to push the pawn to g6 and play gxf7). Note that white has to capture the e7 pawn with the knight on b1, because trying to capture it with the knight on g1 would take one move longer and black would run out of waiting moves. So the first four moves are: 1. Nc3 h5 2. Nd5 Rh6 3. Nxe7 Rf6 4. Nd5 Nh6 5. Nc3 Ke7.

Phase two starts and white has to push the g pawn forward, because playing for instance 6. Nb1 black would run out of "waiting" moves and would not manage to regroup queen/rook/bishop in time. So, the next three and a half moves (6. g4 Ke6 7. g5 Kf5 8. g6 Kg4 9. gxf7 ) are basically black pushing the king on the shortest available route towards h3 and white pushing the g pawn forward. As mentioned black has to wait with playing g6 until white captures gxf7, so black can only push the king during this phase.

The final phase (9....g6 10. Nf3 Bg7 11. f8=N Bh8 12. Ne6 Qg8 13. Nf4 Rf8 14. Ng2 Kh3 15. Nb1 Ng4 16. Ng1#) is basically just moving the pieces to their final squares. There are no alternative move orders possible. For instance white cannot play Nb1 earlier, because with 14. Ng2 he is just in time to block the bishop from attacking h3, allowing 14. ... Kh3.

• White has enough moves left to capture the e7 pawn with the queen's knight and return. However, I can find only three waiting moves (h5, Nh6, Rh7) which Black can play before the King must move out. Apr 18, 2018 at 18:05
• @Glorfindel: Black could move the rook via h6-f6 to f8, giving him an extra waiting move. Apr 18, 2018 at 18:11

The thoughts mentioned in user1583209's answer more or less summed up mine, but I couldn't find a way to get the king out in time without requiring extra moves. The final trick is

to have the b1 knight capture the e7 pawn; this opens the way for the black king. Black has just enough moves (thanks again @user1583209) he can make with his kingside:

``````[FEN ""]
1. Nc3 h5 2. Nd5 Rh6 3. Nxe7 Rf6 4. Nd5 Nh6 5. Nc3 Ke7 6. g4 Ke6 7. g5 Kf5 8. g6 Kg4 9. gxf7 g6 10. Nf3 Bg7 11. f8=N Bh8 12. Ne6 Qg8 13. Nf4 Rf8 14. Ng2 Kh3 15. Nb1 Ng4 16. Ng1#
``````
• Why is it not possible to capture e7 with the g1 knight? Apr 18, 2018 at 18:15
• Takes one more move, so one more 'waiting' move from Black. Apr 18, 2018 at 18:16
• Ah, I see, the e7 pawn needs to be captured quickly and only the b1 knight can do that. Amazing problem and all of it forced as far as I can see. Apr 18, 2018 at 18:17
• @user1583209 Great team work, glor & user! Roberto's theme is 3 white knight circuits. I will give the tick to user because the detail of his response is educational for anyone trying to learn about proof games. (This is a retro area of chess problem composition where we are definitely in the "golden age" right now, thanks to composers like Roberto, Nicolas Dupont, Silvio Baier, Andrei Frolkin, Michel Caillaud etc.) Apr 19, 2018 at 5:57

I'd say

yes, but you need to allow at least one pawn to promote (and return to its original square). E.g. promote the white a pawn to a queen, move it back to a2 and put a knight on a3. Then, move the black king to a4, black pawns to a5 and b4 and a black rook to b5. In the meantime, White can just play Ng1-h3-g1 etc. The move Na3-b1 will then be checkmate, and the white pieces will all be on their original square.

• The question should be "Can you mate with 15 out of your 16 pieces on their original squares?" Apr 17, 2018 at 19:39
• @Glorfindel yes that's the solution. Clearly one might ask "but is a promoted pawn the same piece?" This is a question of semantics, and we don't want to spoil Roberto's excellent joke. Now I will post one of his problems. If you solve it, you definitely get the "tick mark" for correct answer. If someone else solves it, I'll have to decide if you get the tick mark or they do :-) Apr 18, 2018 at 2:10
• Great! It's better to post the follow-up as a new question. With the diagram, you're already spoiling the solution. Apr 18, 2018 at 5:40
• @Glorfindel: thanks for your comment. If I had posted as 2 questions then someone might come along later and say that Q1 is just a subset of Q2 and vote to close it. I don't think your solution is "spoiled" at all. It remains good and you have got a decent amount of reputation from it. I did point out in my initial post that there would be a second part. Bottom line: from a long term stackexchange perspective, it is better to have one post about Roberto's idea rather than two. Apr 18, 2018 at 17:46

After some thinking, as I did not look at the other answers, I concluded that it is indeed possible to checkmate with all 16 pieces on their original squared. Doing this requires a promoted piece to travel back to its “original square.”

Then I looked at Roberto’s excellent problem and thought “there must be a faster way to do that.” Of course, his problem is special in its own right because there is only one way to reach the position. But the effect that it shows is beatable without the uniqueness

There are two variants to consider. Type #1 focuses on the least possible disturbance, meaning that Black has 13 pieces (think about it) at home when mated. Type #2 is without regard to that. @Rebecca J. Stones did much-appreciated work on this in the comments. Here are the results.

``````[Title "Rebecca J. Stones, chess.stackexchange.com 5/10/2019, Type #1, Non-Unique Proof Game In 11.5 Moves"]
[FEN ""]

1. g4 g5 2. Nf3 f5 3. gxf5 Bh6 4. f6 Kf8 5. f7 Kg7 6. f8=N Kf6 7. Ne6 Kf5 8. Nf4 Kg4 9. Ng2 Kh3 10. Ne5 Bf8 11. Nf3 g4 12. Ng1#
``````

``````[Title "Rebecca J. Stones, chess.stackexchange.co 5/11/2019, Type #2, Non-Unique Proof Game In 7.5 Moves"]
[FEN ""]

1. a4 f6 2. a5 Kf7 3. a6 e5 4. axb7 Ne75. bxc8=Q Kg8 6. Qxd7 Ng6 7. Qa4 Nd7 8. Qa2#
``````

Secondly, it is also of interest what can be done with discovered checks.

``````[Title "me, chess.stackexchange.com 5/10/2019, Type #1, Non-Unique Proof Game In 15.5 Moves"]
[FEN ""]

1. h4 g5 2. hxg5 f6 3. g6 Nh6 4. g7 Kf7 5. g8=R Ke6 6. Rg3 Kf5 7. Rgh3 Kg4 8. R3h2 Ng8 9. Nh3 Kh5 10. g4+ Kh4 11. g5 Nh6 12. g6 Ng8 13. g7 Nh6 14. g8=R Kh5 15. Rgg2 Ng8 16. Ng1#
``````

``````[Title "Rebecca J. Stones, chess.stackexchange.co 5/11/2019, Type #2, Non-Unique Proof Game In 10.5 Moves"]
[FEN ""]

1. Nf3 h5 2. e4 h4 3. e5 f6 4. exf6 Kf7 5. fxe7 Kg6 6. exd8=Q Kh5 7. Qe7 g6 8. Qee2 Rh6 9. Nc3 Be7 10. Nb1 Bg5 11. Ng1#
``````
• Here's a 12-mover: `1. g4 g5 2. Nf3 f5 3. gxf5 Bh6 4. f6 Kf8 5. f7 Kg7 6. f8=N Kf6 7. Ne6 Kf5 8. Nf4 Kg4 9. Ng2 Kh3 10. Ne5 Bf8 11. Nf3 g4 12. Ng1#` It also achieves the minimum number of black pieces not on their original squares. May 10, 2019 at 22:30
• I found this 8-mover if we don't care about black's pieces, where the a pawn promotes to a queen: `1. a4 f6 2. a5 Kf7 3. a6 e5 4. axb7 Ne7 5. bxc8=Q Kg8 6. Qxd7 Ng6 7. Qa4 Nd7 8. Qa2#`. May 11, 2019 at 1:06
• Good job on getting it faster. Do note that the problem by Osorio is a proof game. What it lacks in speed, it makes up for with uniqueness. There is no other way to reach the diagrammed position after white's 16th move with any other move sequence than the solution. May 11, 2019 at 1:19
• I found one in 11 moves: `1. Nf3 h5 2. e4 h4 3. e5 f6 4. exf6 Kf7 5. fxe7 Kg6 6. exd8=Q Kh5 7. Qe7 g6 8. Qee2 Rh6 9. Nc3 Be7 10. Nb1 Bg5 11. Ng1#`. I'm not sure how good this is. May 11, 2019 at 14:55
• The Osorio problem is not only a proof game (perhaps 1 in 1000 positions admit a unique proof game?) but also includes the artistic feature that three white knights leave their starting squares and come back (of course one was not a knight when it set out on that journey) Jun 8, 2020 at 6:01