Promotion Play – Part 2

No.16 | by Peter Wong

In the previous instalment we looked at some directmate problems that involve pawn promotion as the main feature. Here we continue our brief survey of promotion play, but turn to its occurrence in helpmates, where this special move is even more frequently used as a linking motif. Since helpmate play is not antagonistic, the most powerful moves are not necessarily the best; consequently, subtle underpromotions are more readily shown. In general, we may expect to see more intensive promotion effects in helpmates, or similar effects achieved in a more economical way.

91. Bo Lindgren
Tidskrift för Schack 1947

Helpmate in 2, Set play

Problem 91 neatly brings about a double change of promoted pieces across two phases of play. In the first part, set play, White commences the two-move sequence: 1…d8=S 2.Bd7 e8=S. The promotion mate is aided by the bishop battery, which fires again in the second part, the actual solution where Black begins: 1.Sc6 dxe8=Q 2.Sd8 exd8=Q. Besides promoting to a different piece in each phase, the two pawns also swap their promotion squares.

92. Norman Macleod
Mat 1980, 6th Prize

Helpmate in 2, 2 solutions

In contrast to the previous example, the battery formed by the thematic pawn in Problem 92 proved detrimental to White’s plan. First note it is not possible to arrange a battery mate on the seventh rank because of the flight square on d8, which cannot be easily controlled. A potential mate that covers d8 can be set up by promoting the d7-pawn to a rook or a knight, followed by Rd7 mate or Sc6 mate, but in each case White’s first move would give an unwanted discovered check, forcing the king to capture the promoted piece. Black resolves this problem by neutralising the battery: 1.h1=B d8=R+ 2.Bxb7 Rd7, and 1.b1=R d8=S+ 2.Rxb7 Sc6. White’s choice of promoted piece in each solution is determined by the initial black promotion, in that if White had picked the wrong piece, the eventual mate would be prevented by the new black piece on b7.

93. Chris Feather
British Chess Magazine 1991
4th-6th Hon. Mention

Helpmate in 3, 2 solutions

In Problem 93, the three-move limit does not allow for a mating net to be created near the middle of the board, so the black king must walk to the edge. Likely destinations for the king are a5 and a6, since these squares and the prospective flights surrounding them are already observed by the white knights. However, it’s not obvious how to stop the two black bishops from attacking the knights’ mating squares on c6 and c7. The quickest way to do this is for each bishop to sacrifice itself on the square to be unguarded, a scheme that ironically entails the immediate capture of the knight on that square. Each missing knight is then replaced with the help of a promotion. 1.Bxc6 axb7 2.Kb4 b8=S 3.Ka5 Sxc6, and 1.Bxc7 a7 2.Kb5 a8=S 3.Ka6 Sxc7. This problem illustrates the Phoenix theme, in which a player promotes to the same kind of piece as one that has been captured in the same solution.

94. Gábor Cseh
The Problemist 2001, 1st Prize

Helpmate in 5, 2 solutions

A curious situation is presented in Problem 94, where Black seems to have only a few spare moves to play before being forced by zugzwang to mate White. For example, after 1.gxf4 a8=S 2.f3 Sb6 3.f2 Sd5, Black can no longer avoid giving mate with any legal move, cutting short White’s plan to mate with 4…Sb4 rather drastically. Therefore White first has to release Black from the impending zugzwang position, before proceeding with the normal aim of mating Black. These two objectives are executed by a pair of promoted queen and bishop, which exchange their functions – including that of sacrifice – in the two solutions. 1.gxf4 f8=Q 2.f3 Qb4 3.f2 Qxb3+ 4.Kxb3 a8=B 5.Ka2 Bd5, and 1.gxh4 a8=B 2.h3 Bxg2 3.Qxg2 f8=Q 4.Qg7 Qh8 5.Qxb2+ Qxb2. Striking and difficult, this problem is called a “masterpiece” by the judge who gave it a first prize.

95. Tichomir Hernadi
Die Schwalbe 1986-87, 1st Prize

Helpmate in 6

A helpmate solution that requires Black to promote to a queen is quite paradoxical, since it seems illogical that creating the most powerful black piece would assist White in achieving mate. To show this idea, at least three black moves are necessary – the promotion itself, and two subsequent queen moves, because if the queen were to move only once, a promoted rook or bishop could well have replaced it. Problem 95 remarkably obtains two black queen promotions in one solution (using the theoretical minimum of six moves), and there is matching white play as well. 1.g1=Q bxa6 2.Qb1 a7 3.Qg6 hxg6 4.h1=Q gxf7 5.Qb7 fxe8=Q 6.Qb8 axb8=Q.

96. Fadil Abdurahmanovic
Liga Problemista 1987, 1st Place

Helpmate in 2, 2 solutions

Have a go at solving Problem 96, a straightforward but attractive helpmate-in-two.


1.c1=B Rbb6 2.g1=B Be2, and 1.g1=S Bd3 2.c1=S Rb5. A double change of underpromotions, where the solver cannot fail to notice the tries, 1.c1=S? and 1.g1=B?