Solving in style December 30, 2002
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In 1986 I received another very important book, one I always have close at hand and which I have taken on a number of holidays with me. It is John Nunn's Solving in Style, originally published in 1985 by George Allen & Unwin. There is good news regarding this book: it has recently been reprinted as a Gambit Publications book and so, unlike the others I mentioned in this Christmas Puzzle section, it is readily available.

John's book had a profound influence on my problem chess interests. It deals with all the common problem types – studies, shorter and longer direct-mates, helpmates, self and reflex mates, series problems, retro analysis, etc. In each case many very attractive examples are given, together with some of the finest explanations I have encountered in this area. In all his books John has a great talent to anticipate exactly the questions that occur to me while reading a bit of analysis. The answers are almost always given in the next sentence.

Chapter six of Solving in Style (which that great linguistic troll Jonathan Speelman refers to as "Dissolving in Bile") deals with helpmates, the subject of our previous Christmas Puzzle page. I took one of the puzzles from John Nunn's book and quote his explanation in full.

H. Forsberg, Revista Romana de Sah, 1936

Helpmate in two moves (quintuplet)
a) Diagram b) bRa6 c) bBa6 d) bNa6 e) bPa6

This is a five-in-one problem. After solving the diagram position, replace the piece at a6 with black rook, bishop, knight and pawn in turn to make four more helpmates.

The solver‘s first aim in a helpmate is to find the mating position. Once this is done, the method of reaching the final position from the diagram can be tackled. The second part of the solving process is usually easy in a two-move helpmate, so the main task is to spot the mate. Typically the solver will visualise a series of potential mating positions based on the material in the diagram and check each one to see if it can be reached in the given number of moves.

In the above diagram the White king is too far away to take part in a mate, so White is restricted to the use of his rook and knight. These two pieces can mate on their own, but only in a corner, and White‘s knight is too far from c3 to arrange a corner mate. lt follows that all the mates involve a self-block of Black‘s king. If we imagine a series of mating positions with one self-block, we can work out afterwards which part of the problem we are actually solving!

For example, White can mate by Nc1 and Rb3, provided a4 is blocked. The blocking piece can‘t be a rook or a queen, since the White king would be in check, while a bishop or pawn could take the rook. lt follows that this is part d, with a knight at a6, and the solution runs 1 Nc5 (remember that this is a Black move) Nc1 2 Na4 Rb3. The White pieces can also mate with Rb4, Nc2 and a block at a2. The piece at a2 must be a bishop (a knight is impossible because it can‘t reach a2 while b4 is occupied by White‘s rook), so this solves part c by 1 Bc4 Ne1 2 Ba2 Nc2. A third position has Nc5 and Ra4, with a block at b2, which could be a rook or a queen. The rook can‘t reach b2 in two moves with the b-file blocked, so this gives the solution 1 Qf6 Nc5 2 Qb2 Ra4 to the diagram position.

There are just two parts left, b and e. With bRa6, the self-block can‘t be at a4 or b4 (check to wKg4) and the rook can‘t reach a2, so it must be at b2 or b3. A mate is certainly possible with bRb3 and White‘s rook on the a-file, but can this be arranged? After 1 Rb6, White's rook must clear the way from b6 to b3, so the solution runs 1 Rb6 Rb1 2 Rb3 Ra1.

Finally with bPa6, no mates are possible in two moves with bKa3 and bPa4, but Black‘s king is not obliged to stay at a3 during the solution. Helpmates involving Black king moves are generally tricky to solve, because even a shift of one square completely alters the potential mating nets and makes it hard to guess the final position. In this case, Black‘s king and pawn can meet halfway at a4 and a5, allowing White to mate by 1 a5 Rb3+ 2 Ka4 Nc5. Notice that this mate was impossible with any other Black man at a6; a rook or a queen could take at c5, while a bishop or a knight couldn‘t reach a5 in two moves.

The individual diagrams and solutions are here.

There are some important points to note when solving and especially when composing helpmates. In direct-mate problems there may be only one solution. An alternate key move immediately makes the problem worthless. Helpmates on the other hand often have more than one solution, the only condition being that each must show a distinctive mate to justify its existence. There is also a convention that each solution should begin with a different first move.

In another point the helpmate is much stricter than direct mate problems. In a helpmate the entire sequence of moves leading to mate must be completely unique. If there is even the slightest deviation at any point the problem is ruined.

Here is an example of a problem in which the move order is critical and unique. It is by my Finnish friend Mika, who is not only a world champion in solving helpmates, but also has composed a number of such problems himself.

Mika Korhonen, 1983

Helpmate in six moves

The solution is 1.e1B e4 2.b1R e5 3.Rb4 e6 4.Bh4 e7 5.Kg3 e8Q 6.Rg4 7.Qe3 mate. Here is the final position:

It's a very pretty mate, and the black moves leading to it have to be elegantly executed. Amazingly they must all be played in precisely the order given above. For instance the b-pawn cannot promote first since otherwise the rook would be checking the white king; the bishop has to move to h4 before the black king can move; and the black king must be on g3 before the black rook can move to g4 to avoid checking the white king. It is a perfect helpmate.

Here's another by the same author that should give you an impression of how in helpmates two solutions can be aesthetically pleasing (and acceptable):

Mika Korhonen, 1983

Helpmate in six moves (two solutions)

Do try to solve this problem yourself. It is not impossible, even for beginners. For instance it is clear that White is going to queen his b-pawn in five moves and then use it to deliver mate in the sixth move. So you really only need to look for black moves. Both solutions are very clever and you can feel proud if you find them. The solutions will be published after January 1st.

Finally here's a problem that will be part of our special Christmas Puzzle prize competition, where you can win a copy of Fritz autographed by a world champion.

Frederic Friedel, Computerschach & Spiele, February 2002

Helpmate in five moves

I composed this one to demonstrate some of the principles described above. After reading this page you should be able to answer a second part to the puzzle: originally I had placed the black king on b4 (instead of b7), which is in some ways more harmonious to the solution of the puzzle. Why did I have to discard this version and instead use the one shown in the diagram?

Solutions

Frederic Friedel