The following puzzle is an old classic, due originally to the great puzzler Sam Loyd, who published the first three parts in 1859; the fourth and fifth parts are variations due to Brian Stewart and Friedrich Amelung respectively. I searched for this puzzle on PSE and couldn't find it, but such a neat chess problem really deserves a place here, so here goes!

The fact that this is an old chestnut means, of course, that you could easily find an answer on the internet in just a few seconds of Googling. But please don't! This will be much more fun to solve if you do it using your wits rather than a search engine :-) Of course I won't be able to tell whether you did or not, but you will always know ...

King Charles XII of Sweden was playing chess with one of his ministers during the Skirmish at Bender. The game reached the following position, with the king playing white and it being his turn to move.

initial position

  1. At this point in the game, the king announced a mate in three.

  2. Before he could make his move, a Turkish bullet came whizzing through the window and shattered the white knight. Unperturbed, the king announced that he could still win even from the new position, but this time with a mate in four.

  3. In an incredible stroke of bad luck, another Turkish bullet shattered the pawn at h2 before the king could make his move. This time, he announced a mate in five.

  4. The minister cleverly pointed out that if the second bullet had taken the other white pawn instead, Charles would still have had a mate in ten.

  5. Finally Charles responded by pointing out that if the first bullet had taken the rook instead of the knight, he would at that point have had a mate in six.

Can you prove each one of the five assertions above?

  • $\begingroup$ Well this is quite hard without cheating... mate in ten... $\endgroup$ Oct 11, 2016 at 0:52
  • 1
    $\begingroup$ @greenturtle3141 the moves could be forced one after another, so then 10 moves could run out very quickly. $\endgroup$
    – Matsmath
    Oct 11, 2016 at 1:40
  • $\begingroup$ Ugh, that bishop is very annoying. $\endgroup$
    – justhalf
    Oct 11, 2016 at 3:44
  • $\begingroup$ So many great puzzles... sigh I wish I actually had time to attempt some $\endgroup$
    – Areeb
    Oct 11, 2016 at 3:46

2 Answers 2


Finally got all five cases without any external help. The last case is very interesting!

The mate in 3 is already quite tough (I'm not used to writing chess notation, please help me fix any errors):

1. Rxg3 Bxe1
...(1. ... Bxg3
....2. Nf3 ...
....3. g4#;
....1. ... Kh4
....2. Rh3#)
2. Rh3+ Bh4
3. g4#

The mate in four:

1. hxg3 Be3
...(1. ... Bxg3
....2. Rxg3 Kh4
....3. Rh3#;
....1. ... Be1
....2. Rg4 Bxg3
....3. Rh4+ Bxh4
....4. g4#)
2. Rg4 Bg5
3. Rh4+ Bxh4
4. g4#

The mate in five (remove white knight and h2):

1. Rb7 Be3
...(1. ... Be1
....2. Rb1 Kh4
....3. Bxe1 Kh5
....4. Bh1#)
2. Rb1 Bg5
3. Rh1+ Bh4
4. Rh2 gxh2
5. g4#
Note that other first move for white rook will not make it in time, e.g., 1. Rd7 Bd4 2. Rd5 Be3 (2. Rxd4=) and white is delayed by one move

The mate in ten:

Start with hxg3, then similar to the "mate in four" case, in at most 3 steps the rook will capture the bishop. Then it's just a normal king+rook vs king ending. I present the worst case below:
1. hxg3 Be1
2. Rg4 Bxg3
3. Rxg3 Kh4
4. Kf4 h5
...(4. ... Kh5
....5. Rg1 Kh4
....6. Rh1#)
5. Rg2 Kh3
6. Kf3 h4
7. Rg4 Kh2
8. Rxh4+ Kg1
9. Rh3 Kf1
10. Rh1#

The mate in six:

1. Nf3 Be1 (protecting against the threat hxg3 then g4#)
...(1. ... gxh2
....2. g4#)
2. Nxe1 Kh4
...(2. ... gxh2
....3. Nf3 h1=Q
....4. g4#)
3. h3 Kh5
4. Nd3 Kh4
5. Nf4 h5
6. Ng6#


I recall seeing this, probably either in Chess Life or probably one of Soltis' books, but I do not recall the solutions.

Edited to add solutions, which I stumbled across in Irving Chernev's "The 1000 Best Short Games of Chess." The puzzle and solutions for the first three, are added, after the game 88, Alehkine v. Amateur, Vienna 1936, whose ending Chernev said were "reminiscent" of this puzzle. Converted from descriptive notation.

What I've figured out so far:

Mate in 3:

1. Rxg3, threatening if Bxg3
2. Nf3 B-any
3. g4#


1. Rxg3 Bxg1 (All others allow Rh3#)
2. Rh3+ Bh4
3. g4#

Part 2: Mate in 4, remove White's knight:

1. hxg3 Bb6 (To defend h4)
2. Rxg4 Bd8
3. Rh4+ Bxh4
4. g4#

Chernev has the following:

1. hxg3 Be3 2. Rg4 Bg5 3. Rh4+ Bxh4 4. g4#

Part 3: Mate in 5, remove Knight and h2 pawn:

1. Rb7 Be3
2. Rb1 Bg5
3. Rh1+ Bh4 4. Rh2 gxh2
5. g4#

If, instead

1. Rb7 Bb6 (leads to quicker mate)
2. Rxb6 Kh4
3. Rb1 Kh5
4. Rh1#

Alternatively, also in Chernev,

1. Rb7 Bg1 2. Rb1 Bh2 3. Re1 Kh4 4. Kg6 any 5. Re4#

As mentioned above, these two parts of the puzzle are not in the book.

Part 4: Mate in 10, remove Knight and g2 pawn:

1. hxg3 Bxg3 Anything else leads to a quicker mate.
2. Rxg3 Kh4
3. Kf4 h5
4. Rg2 Kh3
5. Kf3 h4 6. Rg1 Kh2
7. Rg4 Kh1
8. Rh4 Kg1
9. Rxh3 Kf1
10. Rh1#

And finally, for part 5, remove white's rook:

1. Nf3 Be1
2. Ne1 Kh4
3. Kf4 gxh3
4. Nf3+ Kh5
5. Kf5 h1=q
6. g4#

Or, after finally figuring out how to trap the king on the h-file, I came up with:

1. Nf3 Be1 2. Ne1 Kh4 3. h3 Kh5 (h5 4. Nf3#) 4. Kf6 Kh4 5. Kg6 h5 6. Nf3#

  • $\begingroup$ For the mate in 4, it can also be 1. ... gxh2 and you lose $\endgroup$
    – justhalf
    Oct 11, 2016 at 4:11
  • $\begingroup$ good catch. I'll have to figure that one out again. :-\ $\endgroup$
    – Herb
    Oct 11, 2016 at 4:19
  • $\begingroup$ For your part 4, 1. ... Be1 will delay you by one move $\endgroup$
    – justhalf
    Oct 11, 2016 at 6:28
  • $\begingroup$ for part 4, 1. ... Be1 2. Rg4 Bg3 3. Rxg3 Kh4 4. Ra3 Kh5 5. Rh3# $\endgroup$
    – Herb
    Oct 11, 2016 at 6:41
  • $\begingroup$ In that case 4. ... h5 is a possibility, it doesn't have to be 4. ... Kh5 $\endgroup$
    – justhalf
    Oct 11, 2016 at 6:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.