No.9 | by Peter Wong
The problemist’s vocabulary, like the player’s, includes the occasional foreign-language term, and a prominent example is Allumwandlung. This German word, usually abbreviated as AUW, means “total promotion”. It describes a problem in which the four possible types of promotion – to queen, rook, bishop, and knight – all take place during the course of the solution. The promotions may be made by a single pawn in different lines of play, or they may be divided among a number of pawns. When both white and black pawns participate in the thematic promotions, the result is called a mixed AUW.
49. Werner Speckmann
Schach 1963, 1st Prize
Mate in 2
(b) Qh7 to a7, (c) Ke6 to c6 in (b), (d) Ke4 to c4 in (c)
Problem 49 shows an AUW produced by a white pawn. In any white AUW problem one may ask why White would promote to a rook or a bishop instead of a queen. In directmates, the motive for the underpromotions is usually the avoidance of stalemate. The two-move 49 consists of four parts that are generated by progressive twinning. In normal twinning, every change specified to create a new position is made to the initial diagram, but progressive twinning differs in that each change is applied to the starting position of the previous part. Thus to set up position (b), transfer the queen to a7 as usual, but to generate part (c), start from position (b) and make the further change of shifting the king to c6, and so on. The four phases are solved by: (a) 1.f8=B! Kf6 2.Qf5, (b) 1.f8=R! Kd6 2.Rf6, (c) 1.f8=Q! Kb5 2.Qfc5, and (d) 1.f8=S! Kd6 2.Qc5. Aggressive keys that take control of the black king’s flights are generally frowned upon, but here such unsubtle play is very much mitigated by the exceptional economy – a record four pieces – by which the AUW is accomplished.
50. Niels Hoëg
Nordiske Schackbund 1905, Hon. Mention
Mate in 3
The three-mover 50 incorporates a white AUW as second-move continuations in four separate variations. The key 1.f7! threatens 2.f8=Q and 3.Qe7. If Black self-immobilises the e-pawn, the queen-promotion threat will fail due to stalemate. So instead 1…exd4 2.f8=B Kf6 3.Ra6, and 1…exf4 2.f8=R Kd6 3.Rf6; each of these lines ends in a model mate. With 1…Kf6 the black king aims to escape to g6, forcing 2.f8=S, then 2…exd4 3.Rf7. Lastly 1…Kd6 is answered by 2.f8=Q+ Kc6 3.Qc5, or 2…Ke6 3.Qe7. The theme is precisely shown here without extraneous variations.
51. Slobodan Mladenovic
Schach-Echo 1975, Special Prize
Helpmate in 2, 4 solutions
In helpmate problems featuring white underpromotions, the selection of a rook or a bishop as the new piece is usually not aimed at preventing stalemate, but for the more general purpose of not attacking a square needed by the black king. This is illustrated in two of the solutions in Problem 51: 1.Qxc6 b8=R (avoiding 1…b8=Q+? which would give an unwanted check) 2.Qb6 Ra8, and 1.Qh8 b8=B+ (avoiding 1…b8=Q+? which would guard a8) 2.Ka8 c7. The remaining solutions see the black queen utilised further in a nice variety of ways. 1.Qe2 c7 2.Qa6 b8=Q, and 1.Qc8 bxc8=S+ 2.Ka8 c7.
52. Rolf Trautner
(after Gyula Bebesi)
Die Schwalbe 1960
Helpmate in 7
A helpmate with only one solution and no other phases of play will naturally require at least four pawns to demonstrate AUW. Problem 52 is remarkable in achieving this task with just the four pawns and two kings, without any additional material. In the initial position, White’s pawn has to be activated and the quickest way is 1.c1=S followed by the knight’s sacrifice on b3. Now the white king seems to have a choice of waiting moves, and it is not apparent until the last move why only 1…Kg8! works. 2.Sb3 axb3 3.g1=B b4 4.Bc5 bxc5 5.a2 c6 6.a1=R c7 7.Ra7 c8=Q. Had the white king moved to the 7th rank, the pawn’s mating move would have been disabled by a pin.
53. Zoran Gavrilovski
The Problemist 1990, 3rd Hon. Mention
Helpmate in 2
(b) Kd8 to c8, (c) Pd4 to b2, (d) Rf5 to h5
Problem 53 combines AUW with two other themes, star-flights and bishop-star. In the former, a king makes each of its four diagonal moves in turn. The latter is a similar star pattern formed by a bishop’s moves. The solutions of this four-part twin run: (a) 1.Ke6 e8=S 2.Bd6 Sxg7, (b) 1.Kc6 e8=R 2.Bf6 Re6, (c) 1.Kc4 e8=Q 2.Bd4 Qxb5, and (d) 1.Ke4 e8=B 2.Bf4 Bg6.
54. Zivko Janevski
Mat 1982, 6th Prize
Helpmate in 2, Set play
Set play: 1…e8=Q 2.e1=R Qxf7. Surprisingly, despite the airy position Black has no way of preserving this set line with any initial move, and the actual play is 1.Rf8 exf8=B 2.e1=S Bxc5.