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.

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.

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.

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).