World Solving Championship – solutions (2)

39th World Chess Solving Championship (2)

By John Nunn

Christopher Jones, Original for WCSC, 2015

Helpmate in three moves: a) Diagram
b) Move the black queen to d6
c ) Move the black queen to d5

In a helpmate, both sides are cooperating to help White mate Black, although the moves still have to be legal. Black moves first, with White mating Black on his third move (so the sequence of moves goes B W B W B W mates). There are three parts to this problem. The first is the diagram position, and for the second and third you must move the black queen to d6 and d5 respectively (all three parts are helpmates in three).

Helpmates generally have either multiple solutions or twins (positions which are a modified version of the diagram position). Generally speaking, there is some connection between the solutions. Here the fact that there are three parts and three pawns on the fourth rank is already suggestive. Tackling the diagram position first, the obvious question is: what can White do when the queen is on d7 that he cannot do when the queen is on d6 or d5? There are several possible answers, but one of the most obvious is that White can play Bxd7. This leads to the idea that Bxd7 might actually be the mating move, and then the mating pattern with the black king on f5 and white rook on g4 arises naturally. Black’s knight has to be used to allow the king to cross the fourth rank and the solution runs 1...Neg4 2.exf5 Ke4 3.Rxg4+ Kxf5 4.Bxd7#.

Now we can look for the possible connection between the three parts. In the first solution, a white pawn makes a capture on the fifth rank, and the black king follows behind, first occupying the square vacated by the pawn and then taking the pawn. We can guess that this pattern will hold in the remaining two parts.

It looks likely that the second part will involve dxe5, with the black king being mated on e5, while the third part will be based on cxd5 with the king being mated on d5. With all this extra information, it’s not hard to find the other two solutions. With the queen on d6, the solution is 1...fxe4 2.dxe5 Kd4 3.Bg6 Kxe5 4.Rxe4# and with the queen on d5 we have 1...Nc6 2.cxd5 Kc4 3.e5 Kxd5 4.Bf7#.

Replay the solutions of all three helpmates

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Ralf Krätschmer, Original for Schach in Schleswig-Holstein, 2015

White to play and mate in six

It’s hard to know what to expect with longer problems – sometimes there’s complex strategy, but other problems are based on a forcing line leading to an attractive mate. This problem falls into the latter category, and there’s not really a better way to solve it than to try a few forcing continuations and hope that something turns up. Black has some pretty strong defensive moves available, such as ...c3, giving the king access to c4, or ...Nd3 shutting out the white queen, so White’s moves are unlikely to be all that subtle.

The first point is that White can transfer his knight to e2 with gain of tempo by checking on f4 and e2, and this looks like a good start for two reasons: firstly, if White later advances his d-pawn (and why else would it be there?) Black won’t have a check on a2, and secondly it clears e6 for a possible mate threat.

So let’s start with 1.Nf4+ (the immediate 1.Qf5 is well met by 1...c3) 1...Kd4 2.Ne2+ Kd5. Now it’s time to bring the queen into play before it gets shut out by ...Nd3. White plays 3.Qf5 with the brutal threat of 4.Be6#. There’s only one defence, namely 3...Rb5, and now you just have to see the finish: 4.d4 (threatening both 5.Be6# and 5.Qxe5#) 4...cxd3 (4...c5 5.Qxe5+ Kc6 6.Qe6#) 5.Qe4+! Kxe4 (5...Kc5 6.Qd4#) 6.Nf6# with an elegant mid-board mate.

Richard Dunn, The Problemist 2002

Selfmate in three moves

In a selfmate, White is trying to commit suicide by forcing Black to mate him. Black is doing his best to avoid mating White. White plays first, and must force Black to give mate on his third move.

This problem proved challenging for a variety of reasons. It’s fairly clear that there are only two real possibilities of forcing Black to deliver mate. One is by Qg2+, with the aim of compelling either ...Nxg2# or ...hxg2#. This doesn’t work at the moment because of the e4-bishop, which can take on g2 after ...hxg2. The other possibility is to somehow get rid of the bishop and queen and then play Nf3+ to force ...Nxf3#. At the moment, this seems like a long shot because the bishop on b7 can also take on f3. However, if the e4-bishop vanished and Black could be forced to play ...Bxf3 (possibly by the threat of Qg2+), then Nxf3+ would almost work, the only flaw being that ...Nxf3+ isn’t mate because the b8-rook can take on b1.

All this preliminary work suggests a plan for White of first removing the rook from b8, then playing Bd3 (the only way to get the bishop off the long diagonal), by which time White is genuinely threatening Qg2+. This threat can only be met by ...Bxf3, whereupon Nxf3+ forces ...Nxf3#. This plan looks horribly slow, but Black isn’t in a position to do much about it. The most effective counter would be to dismantle the battery along the first rank, but he cannot achieve this in two moves. Selfmates which have a quiet (that is, non-checking) threat are often tough to solve, even in only three moves, and this is no exception.

It only remains to guess the destination for the b8-rook. There’s no obvious reason to prefer one square rather than another, but experienced solvers will almost certainly go for 1.Rh8, simply because the line-up with the black king looks as if it might be relevant.

The other tricky thing about this problem is that it’s not easy to see what Black’s defences are, and some care is needed to make a complete list. There are two rather uninteresting lines: 1...Qxf5 2.Bxf5 (threatening 3.Qg2+) 2...Bxf3 3.Nxf3+ Nxf3# and 1...Nf2 (this meets the threat because 2.Bd3 Nd1! 3.Qg2 Nxg2 isn’t mate) 2.Qh1+ Nxh1 3.Nf3+ Nxf3#.

The first of two key lines is 1...Qd5 (now 2 Bd3 fails to 2...Qxf3+ 3 Nxf3 Bxf3 – no mate) 2 Rh5! (the queen move has deprived Black of the pinning ...Qxf4, which previously refuted this move) and now there is no defence to 3.Qg2+ because Black has to take with the knight (or with the queen after 2...Qxe4).

The second main line runs 1...Nxe5 (to meet 2.Bd3 by 2...Nxf3 3 Nxf3+ Bxf3) 2 Rxh6 (this is why the rook had to move to h8; Black no longer has the reply ...Nxh6 because he moved the knight away) 2...N5xf3+ (or else Qg2+) 3 Nxf3+ Nxf3#.

The top individual scores at the 39th World Solving Championship in Ostroda were
Kacper Piorun 80.5 (out of a possible 90), John Nunn 80 and Jorma Paavilainen 73.

The senior prize-winners, John Nunn (80 points), Tadashi
Wakashima (49.5 points) and Aleksandr Feoktistov (48 points)

Photos by Franziska Iseli