In the previous column (Part One) we looked at some directmate problems that involve pawn promotion as a main feature.

Here we will continue our brief survey of promotion play, but turn to its appearance in helpmates, where this special move is even more commonly found as a recurring motif.

Because 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 more economically.

In contrast to the previous example, the battery formed by the thematic pawn in Problem 92 proved detrimental to White’s plan.

First, it is not possible to arrange a battery mate on the seventh rank because of the flight-square on d8, which cannot be guarded.

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 Nc6 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(N)+ 2.Rxb7 Nc6

92. Norman Macleod
Mat 1980
6th Prize

Helpmate in 2
2 solutions

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 controlled 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 only way to do so in one move is for each bishop to sacrifice itself on the square to be un-guarded, and this ironically entails capturing a knight – which White then replaces by promotion.

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(N) 2.f3 Nb6 3.f2 Nd5, Black no longer can avoid giving mate with any legal move, cutting short White’s intention to mate with 4…Nb4 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, who exchange their functions – including that of sacrifice – in the two solutions.

94. Gábor Cseh
The Problemist 2001
1st Prize

Helpmate in 5
2 solutions

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 unlikely that creating such a 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.Qa1 a7 3.Qg6 hxg6 4.h1(Q) gxf7 5.Qb7 fxe8(Q) 6.Qb8 axb8(Q).

Have a go at solving Problem 96, a straightforward but attractive helpmate-in-two.  The solution will appear next month.

96. Fadil Abdurahmanovic
Liga Problemista 1987
1st Place

Helpmate in 2
2 solutions

90. Allan Werle
Tidskrift för Schack 1945

Mate in 4

Solution to Problem 90 in the previous column:

The obvious 1.e8(Q)? is refuted by 1…d1(N)+! 2.Kg3 Ne3 3.Qxe3 stalemate (or 2.Kf1 Ne3+ 3.Qxe3 Kh2 4.?).

Only 1.e8(R)! solves in four moves, 1…d1(N)+ 2.Kg3 Ne3 3.Rxe3 Kg1 4.Re1.

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