Chessville - Chess Problems and Compositions in Peter's Problem World with FIDE Master of Chess Composition Peter Wong

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Longer Helpmates
Peter's Problem World with FIDE Master of Chess Composition Peter Wong

New terms introduced here are also added to Peter's Glossary of Chess Problem Terms.

This month we continue with our introductory survey of helpmates, the problem type in which Black assists White in giving mate. Longer compositions of three moves or more are presented this time.

In solving helpmates, regardless of their length, you should first consider whether the black king is likely to be mated on its initial square or elsewhere. Look for possible mating configurations with the available pieces. Ask what are the functions of the pieces and which will take part in the mate directly.

Lastly, try out the moves that would lead to the mating positions, keeping in mind the limited move number, of course.

In Problem 31, below, the black king is already confined completely, but a knight mate on c1 or c3 would be thwarted by the black queen. If the black rook were to replace the queen as the blocking piece on a1, the knight could mate on c3.

Gábor Cseh
Ideal-Mate Review 1997
Hon. Mention

Helpmate in 3
2 solutions

31. Toma Garai
Thema Danicum 1976
Prize

Helpmate in 3
(b) BBf8

Black must, however, relocate the queen to just the right square, since after 1.Rf1 Ng3, for example, any queen move will either hinder Ra1 or keep c3 guarded.

The solution is to hide the queen on g1: 1.Qg1 Ng3 2.Rf1 Ne4 3.Ra1 Nc3. The queen makes a clearance move on the rank to allow the rook to travel on the same line and in the opposite direction – this is the Turton manoeuvre.

In the twin position, where a black bishop starts on f8 instead, we may expect the same manoeuvre to occur on the long diagonal: 1.Qh8 Nf2 2.Bg7 Nd3 3.Ba1 Nc1.

Black commencing play in helpmates is merely a convention, and there’s no reason why White cannot make the first move in this genre.

Problem 32 illustrates this less common type; the stipulation of ‘Helpmate in 3½’ indicates that White begins and mates on the fourth move.

In this position, the White pieces cannot cope with the flight on b6 (using four moves), so Black has to block it with a promoted piece. Only three moves are available to Black, and if the pawn promotes on c1, the new piece will not reach b6 in time.

With White’s help, though, Black can promote on another square to prepare for this self-block: 1…Ba2 2.c2 Bb1 3.cxb1(R) Ne4 4.Rb6 Nc5.

32. Christer Jonsson & Rolf Wiehagen
Ideal-Mate Review 1997
Hon. Mention

Helpmate in 3½
(b) Pc3 to f3

In part (b), the pawn on f3 similarly requires a white sacrifice to promote on the right square, enabling the new piece to reach b6 in one move: 1…Ne2 2.f2 Ng1 3.fxg1(B) Be6 4.Bb6 Bc8.

33. Torsten Linß
Die Schwalbe 1995
1st Hon. Mention

Helpmate in 3
2 solutions

Longer helpmates in which White’s main force is a rook and a bishop often culminate with a double-checkmate given by these pieces. Besides taking care of many flight squares, such mates carry another advantage, that the black pieces obstructing other flights won’t be able to stop the mate by interposing.

Problem 33 is quite difficult to solve, even with this double-checking scheme in mind. White’s rook and bishop are initially pinned, and both need to be unpinned before they could arrange themselves to give a discovered attack. Black seems to require too many moves to free them, eg. 1.Rd1 2.Rc1 3.Bg5 4.Bh6.

The solutions show two elegant ways of unpinning both pieces in just three moves. 1.Rd3 Rd7 2.Re3 Rf7 3.Re7 Bd4, and 1.Bg5 Be7 2.Bf4 Bc5 3.Bd6 Rf4.

In Problem 34, let us simply assume that the black king will remain on a2, to be mated by Nc3. It doesn’t take long to work out that Black cannot block both flights on a3 and b3 within the five-move limit, but the white king is capable of reaching a4 in four steps to guard these squares. Black’s main objective, then, is to create a safe passage for the white king.

1.Rg8+ Kd7 2.Rb7+ Kxc6 3.Rbg7+ (to interfere with the bishop’s guard of c3) Kb5 4.Rb8+ Ka4 5.Rb2 Nc3.

Black checks four times consecutively – an ironic idea in helpmates, though the problem’s main theme is a splendid exchange of positions by the two black rooks.

34. Jean-Michel Trillon
feenschach 1971
2nd Prize

Helpmate in 5

35. Markus Ott
feenschach 1993
1st Hon. Mention

Helpmate in 4
2 solutions

The diagram of 35 provides few clues as to where the black king will be mated. Testing the squares in the king’s vicinity, we find a good prospect in e5: the king placed there would give a near-complete mating position, requiring only an additional block on f4.

Black’s obvious choice for obstructing f4 is the f8-bishop, and coupled with the king’s need to access e5, Black’s play forces the white bishop on g7 make an interesting trip.

1.Kf4 Bf6 2.Bh6 Be7 3.Ke5 Bf8 4.Bf4 Bg7.

The white bishop executes a Rundlauf, i.e. a roundabout trip by a piece that finishes on its original square.

In the second solution, the king is mated on e4, with blocks necessary on f4 and f3. And this set-up means that the other white bishop has to go on a round-trip.

1.Qf4 Bg6 2.Bd1 Be8 3.Ke4 Ba4 4.Bf3 Bc2.

Such rundlaufs are especially attractive when, as in this example, they are not motivated by the need to capture another piece.

36. Gábor Cseh
Ideal-Mate Review 1997
Hon. Mention

Helpmate in 3
2 solutions

Have a go at solving Problem 36, which is fairly easy – especially if you take into account the comments given with Problem 33! The solution will appear next month.

Solution to Problem 30 in the previous column:

White has to sacrifice a rook each time to create a “hideaway” square for the black bishop.

1.Be5 Rxf4 2.Bxf4 Rb8, and

1.Bg7 Rxh6 2.Bxh6 Rb8.

30. Nikolaj Dolginowitsch
Die Schwalbe 1993