Longer Helpmates

No.6 | by Peter Wong

In this instalment, 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. When solving a helpmate (of any length), your first aim generally is to find the mating configuration with the available pieces. Consider if the black king is likely to be mated on its initial square or elsewhere. Ask how each white piece will contribute to the mate – as a guard, the mating piece, or can it be sacrificed? When a potential mating position is found, try out the moves that would lead to it. In this step, the requirement for a precise move order in a helpmate solution is itself often a hint. Finally, in a multi-phase helpmate, uncovering one solution often means that the other parts can be worked out straightforwardly, because of the analogous strategy involved.

31. Toma Garai
Thema Danicum 1976, Prize

Helpmate in 3
(b) BBf8

In Problem 31, Black’s king is already confined completely, but the queen prevents the white knight from mating on c1 or c3. If the black rook were to replace the queen as the blocking piece on a1, the knight mate on c3 would work. However, Black must take care in relocating the queen, since after 1.Rf1 Sg3, for example, any queen move will either hinder Ra1 or keep c3 guarded. The solution is to hide the queen on g1: 1.Qg1 Sg3 2.Rf1 Se4 3.Ra1 Sc3. 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 Sf2 2.Bg7 Sd3 3.Ba1 Sc1.

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

Helpmate in 3½
(b) Pc3 to f3

The convention that Black commences play in a helpmate is sometimes broken, as there is no particular reason why White should not make the first move in this genre. Problem 32 illustrates this less common form – its stipulation of ‘Helpmate in 3½’ indicates that White begins and mates on the fourth move. Here the four moves available are not sufficient for White to both control the b6-flight and position a mating piece, so Black has to arrange a self-block on that square. Black only has three moves to do this with a promoted piece, and if the promotion were to occur on c1, the new piece would not reach b6 in time. But with White’s help, Black can promote on another square so as to quicken the task: 1…Ba2 2.c2 Bb1 3.cxb1=R Se4 4.Rb6 Sc5. In part (b), the black pawn similarly makes use of a white sacrifice to promote on the right square, enabling the piece to reach b6 in one step: 1…Se2 2.f2 Sg1 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 them giving a double-checkmate. Such a mate carries the advantages of covering many flight squares, and being impervious to interposing by the black pieces blocking other flights. Problem 33 is difficult to solve, even with this double-checking scheme in mind. White’s rook and bishop are initially pinned, and both need to be freed before they could set up a discovered attack. Black seems to require too many moves to unpin them, e.g. 1.Rd1 2.Rc1 3.Bg5 4.Bh6. The solutions show two elegant ways of untying them in just three moves, and effect an orthogonal-diagonal transformation. 1.Rd3 Rd7 2.Re3 Rf7 3.Re7 Bd4, and 1.Bg5 Be7 2.Bf4 Bc5 3.Bd6 Rf4.

34. Jean-Michel Trillon
feenschach 1971, 2nd Prize

Helpmate in 5

In Problem 34, let us simply assume that the black king will remain on a2, to be mated by …Sc3. It doesn’t take long to establish that Black cannot block both flights on a3 and b3 within the five-move limit, but White’s king can reach 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 control of c3) Kb5 4.Rb8+ Ka4 5.Rb2 Sc3. Black checks four times consecutively – an ironic idea in cooperative play, but the problem’s main theme is a splendid exchange of positions by the two black rooks.

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 end up. 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 choice for obstructing f4 is the f8-bishop. The play of this piece, coupled with the king’s need to access e5, force the white bishop on g7 to take 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 entails another round-trip, now made by the second white bishop. 1.Qf4 Bg6 2.Bd1 Be8 3.Ke4 Ba4 4.Bf3 Bc2. Such rundlaufs are especially attractive when they are not motivated by the need to capture another piece, as in this example.

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!


1.Qc4 Rb8 2.Kd3 Rb5 3.Qe4 Rd5, and 1.Qd5 Bc8 2.Kd3 Bd7 3.Qe4 Bb5. In each solution, the black king and queen swap places, while White arranges a double-checkmate by moving only one piece, which returns to its initial line of attack on d3.