Long ago in 2000, Tim Krabbe presented, a now classic, problem by Harry Goldsteen in his article “The Horse Concoction” on his site. It has blossomed across the Internet, as some of us know.

The task's stipulation is complex.

Take an empty board. Put white pawns on b3, b6, h4 and h7, and a white Queen on f1. Now add ten black knights, both Kings, and the necessary material to enable White to stalemate Black in one move, and Black to win after any other move.

The retrograde analysis Tim Krabbe glosses over isn’t too hard to understand. But there is just one part that he skips over that nags at me, and I cannot figure it out.

Firstly, here is the unique position that solves the problem.

[FEN "8/7P/1P5B/2B1Q1n1/3nn2P/1PRnk1nR/3nnnK1/2B1nQBn w - - 0 1"]

Krabbe remarks that the knight jammed in the corner cannot be jammed in the corner on h8 for reasons too big for his article. But why is that so?

2 Answers 2


This is just the very beginning of what must have been Goldsteen's demonstration, but if anyone wants to carry on from here...

Let's try and build the position after White's key move.

The bNh8 cannot be pinned, so there must be "blockading" knights on f7 and g6, that are pinned themselves.

Because squares f7 and h7 are occupied, the bNg6 cannot be pinned diagonally. It cannot be pinned vertically either, otherwise we cannot pin the bNf7. Hence it is pinned horizontally, leftwards.

The bK cannot be on f6 because there wouldn't be enough space to set up a pin on the f6-h8 diagonal. It cannot be too far from the bNf7 either, so you can conclude that the bK stands on e6.

The space available (or lack thereof) allows us to define the positions of four more pinned knights and four pinning pieces, leading to this position:

[FEN "2B1R1Bn/3nnn1P/1PRnk1nR/3nn3/2B4P/1P6/8/5Q2 w - - 0 1"]

Two remarks:

  • The wQf1 (or even a wP, although it seems unlikely) might have moved with the key move, I just put it on the diagram as a reminder that it had to be there before.

  • Each wB or wR might be replaced by a wQ.

There will be two additionnal bN's, and two pinning pieces, on the e-file and on the e6-h3 diagonal.

To go further, I suppose we might discuss cases depending on the locations of these two bN's. A very important point is that Black must be winning if White doesn't stalemate, so most probably wK willl be under check by the 10th knight (as in the solution with a bNh1) to avoid allowing scores of possible first white moves. Most probably too, the bK will also have a free square to escape to, thus unpinning half of its army.

Second step:

White already has three promoted pieces on the diagram. Since adding two more of them would be illegal (due to the remaining four wP's), it means that one of the last pinning pieces must be either the wQ or the wPh7.

There are only five possible stalemating key moves to consider:

  1. Ph7xNg8:B or Ph7xNg8:Q (these ones are very unlikely because they cannot be check-parrying moves)
  2. Qf1xNh3
  3. Qf1xNe1
  4. Qf1xNe2

As a consequence, the positions of the wP's on b3, b6, and h4 are certain.

Let's consider that White's key move is 1.Qf1-takes-the-knight-that-was-checking-white-King and try now to infer where the wK might stand.

Given that the wK cannot be on f4 or g5 (these squares are controlled by other bN's), the available square that is closest to the cluster of black pieces is d4, with d3 and f3 second-order possibilities. Because it is very hard to imagine Black, who is down scores of material, to be winning against a far-away wK (say, on c1), putting the wK on d4 could be the most logical next step.

As a consequence, the pinned bN on the e6-h3 diagonal cannot be on f5, and must stand on g4. To limit wK freedom, it is tempting to put the 9th knight on e5 (controlling d3), but it might also be on e3 or e4.

Which leads to this diagram before the key move:

[FEN "2B1R1Bn/3nnn1P/1PRnk1nR/3nn3/2BK2nP/1P5B/4n3/5Q2 w - - 0 1"]

But this leads to several new difficulties:

  • The key move 1.Qxe2 is not stalemating in spite of the 8 pins, since black can play 1...Kf6 or 1...Kf5. Of course, adding more (non-promoted) units, e.g. wNg3, wBg5, might still be possible. (Notice that if we add chosen the setup wKd3/bNe4/bNe1 instead, we would also need to cover square e5; but then, the bNg4 could be on f5 blocking one flight square...)

  • We have to demonstrate a black win after the alternative capture 1.Bxe2, or make it impossible by adding a wNd3.

In both case, adding white units reduces the number of previous captures by Black, and is very likely to be impossible while keeping the legality of the position (less captures means less paths to promotions for both sides).

The alternative, in case adding a wB would still allow all the promotions, is:

[FEN "2B1R1Bn/3nnn1P/1PRnk1nR/3n1n2/2B1n2P/1P1K3B/8/B3nQ2 w - - 0 1"]

[Ba1 could be on b2,c3,d4,g7] [If this is legal - edit: it is not]

Then we really have a stalemate after 1.Qf1xe1, but demonstrating a Black win after the alternative 1.Ke2 is a long shot, since only two black units can move! The best try would be to put wBa1 on c3 and play 1...Nxc3+ followed by 2...Ke5 or 2...Kf6, but that would allow yet another variation starting with 1.Bxe1! And still... (funnily enough, 1...Ng3+ 2.Kxe1 Nxf1 3.Kxf1 is another stalemate, and if Black cannot win the wQ at once he should not dream of a long-term win and has to checkmate quite forcefully)

The third step should be retrograde analysis to prove that we cannot add a white knight nor a dark-squared bishop.

Edit: this step is actually pretty easy: Black needs three captures to promote all his pawns: one by the b-pawn, one by the h-pawn, and one by the e-pawn or the c-pawn to allow the opposing white pawn to promote on a light square.

As a consequence, the third diagram above is illegal. With a key move by the wQ, even without the stipulation that it should be the only move to prevent a defeat, I am thus unable to even reach a legal stalemate. [but I will not pretend I am 100% confident about all this reasonning; dear reader, please double check!]

The possible key moves that remain, because they allow more pieces to be present, are the non-check-parrying 1.h7xNg8:B and 1.h7xNg8:Q. I am pretty convinced that they cannot be mandatory to prevent defeat (and thus, that Harry Goldsteen is right - not a surprise!), but step four will be to...

Step four

build a legal 9-knights stalemate starting with 1.h7xNg8:B

I believe this is achieved by the following position:

[FEN "2B1R1nn/3nnn1P/1PRnk1nR/3nn3/2BK2nP/1P5B/8/4RQ2 w - - 0 1"] 

1.h7xNg8=B is stalemate

However, Black is clearly not winning after any other White move. Indeed, White has a direct mate in 1 with

1.Bxg4# or 1.Bxd5#

Convincing oneself that it is really mate is a little mind-boggling with all these pinned defenders who cannot defend.

I may spend a bit more time and sweat to prove properly that this last position is legal, but I will stop thereafter. After reaching that point, I am totally convinced that Goldsteen's assertion that "there is no solution to the Horse Concoction with a bNh8" is absolutely correct.


Here is a game that proves the position is legal. Due to the page freezing, it can be played out on apronus.com.

Move List: 1. a4 a5 2. h4 h5 3. g4 e5 4. g5 Rh6 5. gxh6 d5 6. f4 Bd6 7. d4 e4 8. Na3 g5 9. Nc4 e3 10. Nd2 exd2+ 11. Kf2 Be5 12. dxe5 g4 13. e4 Qg5 14. fxg5 f5 15. b3 Ra6 16. Ba3 b5 17. h7 Rb6 18. g6 b4 19. g7 Rb5 20. axb5 bxa3 21. c4 c5 22. Nh3 Bd7 23. b6 Bb5 24. cxb5 g3+ 25. Ke2 g2 26. Nf2 f4 27. Ng4 hxg4 28. b7 f3+ 29. Ke3 d4+ 30. Kf4 g3 31. e6 f2 32. e5 g1=N 33. Bh3 g2 34. e7 Kf7 35. e6+ Kf6 36. e8=B a2 37. e7 a4 38. Bf7 a3 39. e8=B c4 40. Bed7 c3 41. Bc8 Nd7 42. b8=R d3 43. Bb7 c2 44. Re8 Ne7 45. Bc4 f1=N 46. g8=R c1=N 47. b6 Nf3 48. Bb5 Nfe5 49. Qe1 Nf7 50. Rh2 g1=N 51. Rb1 d1=N 52. Rg5 Nf2 53. Rh5 Ng4 54. Rc2 d2 55. Ke4 d1=N 56. Rcc5 Nc3+ 57. Kd4 Ncd5 58. Bc4 Nfe3 59. Rb2 N3f5+ 60. Ke4 N5h6 61. Kd4 Nhg8 62. Ke4 a1=N 63. Rd2 Nf3 64. Qe2 N3e5 65. Qe1 N5g6 66. Qe2 Ngh8 67. Qe1 Nc2 68. Qe2 Ne1 69. Rd1 a2 70. Qg2 a1=N 71. Qg1 Nac2 72. Qf1+ Ke6 73. Bc8 Nce3 74. Kd4 N3f5+ 75. Ke4 N5d6+ 76. Kd4 Ncd3 77. Rc6 N3e5 78. Rc1 N1d3 79. Re1 N3f4 80. Rh6+ Nfg6


The article mentions:

Square h8 also being impossible is the result of incredibly complex considerations (Goldsteen spent a month and a half on this problem; 'Not much, for me') that would be beyond the scope of even this article.

There's no obvious reason why h8 doesn't work; a complete proof could even require more characters than fit in a Stack Exchange answer.


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