An earlier version was published at The Chess Cafe, December 2000

In its first issue of the year 2000, the Dutch Probleemblad nominated 32 compositions in eight categories for 'problem (or study) of the millennium'. Among the nominees, there was a guy I've been coming across in the streets and the café's of Amsterdam all my life: Harry Goldsteen, now 62 years old.
   Alongside names like Saavedra, Kasparyan, Mansfield, Yarosh, Visserman, Loshinsky and Loyd, this was of course a great honor, but then: in his field, Goldsteen is a living legend. That field is retrograde analysis, although Goldsteen has covered other grounds as well: practical and composed endgames, sometimes even coarse subjects like middle games. But whatever arouses his passion, he is always an incredibly deep analyst, and he never uses a computer. When for instance I thought I understood the phenomenon of the Rambling Rook (my term for the rook that keeps giving checks and cannot be taken on account of stalemate), Goldsteen sent me a 30-page manuscript about one particular Rambling Rook that proved to me I understood next to nothing about the subject.
    All of his enthusiasms yield similar manuscripts.
    Around the time of his nomination, I had once again bumped into him in a busy shopping street. One of his more agreeable sides is that he will not waste time over How are you's and I am fine's, but that he will immediately talk business. In this case, he answered my 'Hi, Harry!' with: 'Have you ever wondered whether it would be possible to stalemate ten knights?' He had just finished a new problem he called The Horse Concoction, and while people with their arms full of packages were bumping into us, he phrased the task: '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.'
    I had no clue, but I was lucky that Goldsteen agreed to spend an afternoon with me, in which he explained the history of the Horse Concoction, and the Horse Concoction itself. (But he had forgotten to bring the satchel of black knights that he had prepared.) At this point I must warn the reader that in my account of that afternoon, however brief, we will be removed as far as is humanly possible from what an ordinary chessplayer might innocently think that chess is.
N. Höeg, 1935
Add ten white Rooks; Black mates in 1
    The Horse Concoction is a development of an idea that began in 1935 with a problem by the Danish composer Niels Höeg. (See Diagram) Retrograde-analysis will prove that these Rooks can only stand on a1, a2, a7, b4, c1, c2, c5, c7, d7 and f7, allowing Black to mate with Qxb4.
    In the last few years this old idea, where all the pawns of one or even both sides promote to one kind of piece, had been taken up again by the Dane Henrik Juel and the Dutchman Guus Rol - first with Bishops, then with Knights. And when Goldsteen saw Rol's latest version of an earlier composition by Juel, something unusual happened to him: he was jealous. He was just pissed off that he hadn't thought of Juel's and Rol's ten-knight problem (add ten black Knights to a given position so that White can mate in 1), and so he formulated the audacious idea that he told me in the street: stalemate ten black Knights! And let Black win if White doesn't.

We return to the initial position of Goldsteen's problem. (See Diagram)
    First the stalemate. Clearly, as many knights as possible must be pinned: eight, along all the diagonal and orthogonal lines the black King stands on. This means there must be eight pinning pieces; at least three promoted ones. That leaves two other knights to be bridled. One of them will be captured with the stalemating move, but the tenth knight can only be stalemated by being jammed. That can only happen on a corner square (the geometry of the chessboard forbids other squares; they would necessitate jamming knights on squares where they could not all be pinned) and this in turn means that both squares a knight's jump away from the corner square, must also be occupied by knights. Therefore, a8 and a1 are impossible. 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. But in the bottom right corner square, it can be done. And proudly, and with many thanks to Harry Goldsteen, I hereby announce the international premiere of his Horse Concoction. (See Diagram)

The Horse Concoction
White stalemates or Black wins
Harry Goldsteen, 2000
White is in check; With 1.Qxe1 he can stalemate Black. A sight to be savoured.
    The first question that arises is whether the diagrammed position is legally possible. That is where the retrograde analysis comes in. Blacks has ten knights, so all of his pawns must have promoted. Therefore, his b- and h-pawns must each have captured at least once. As there are thirteen White pieces remaining, this leaves one other capture for Black. Black's remaining eleven pieces leave five captures in White's past. With two of them, his b- and h-pawns became doubled - therefore he has three past captures which, together with the remaining capture by Black, should explain that the white c2-f2 pawns and the black c7-f7 pawn could have passed each other on the way to their promotion squares. That is easily possible, but as we have four white Bishops of the dark squares, we still have to explain that at least three white pawns have promoted on black squares. Is that possible? Just. Black has captured dxe at one point, and White cxd, exd and fxg; precisely enough for three white Bishop promotions on d8, and one Queen promotion on g8.
    But why so much trouble, one might ask, to promote to Bishops; wouldn't Queens have helped just as well in the stalemating? They could have promoted on white squares just as well. Sure, but we mustn't forget the second part of the stipulation: if White does not stalemate, Black must win. And with a Queen instead of even one of White's Bishops, Black could not do that.
    So that is that. The position is possible, White can stalemate, and we now turn to Black's task of winning after any other move than 1.Qxe1. In fact, White only has one: 1.Kh2 The way in which the nine remaining horses, all of whom were pinned only a moment ago, now break loose massively, is almost as impressive as the whole Horse Concoction itself. Black plays 1...Kf3!, a move we would be tempted to call silent if we didn't hear the deafening whinnying of the four Knights that are now unpinned. Nxf1 mate is threatened, and after extensive analysis, Goldsteen has concluded that White will be mated no later than at move 12. The main variation is: 2.Rxd3+ Kg4!! (That beautiful move would also have followed 2.Qxf2+) 3.Rhxg3+ Nhxg3 4.Rxg3+ N2xg3 5.Qxg3+ Nxg3 6.Qxe1 N2f3+ 7.Kg2 Nxe1+ 8.Kxf2 Ndc2 9.Bxg5 Seven Knights have now perished, but the remaining three have concocted a mate: 9...Nd3+ 10.Kg2 Nce1+ 11.Kh2 Ne5 and mate next move by 12...Nf3.
    There are numerous side variations.

© Tim Krabbé, 2001

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