International Solving Contest

In the ISC event the problems to be solved are sent out to controllers who organise a separate solving venue in each country. The controllers in each country open the envelopes and hand out the problems at the same time all over the world. A total of 288 solvers took part in this year’s event (211 in the main event and 77 in a special category for beginners), and the leading scores, out of a possible maximum of 60, were:

When points are equal, the tie-break is based on time used, which put John Nunn in first place. It was his first victory in the ISC, although he has been second on two previous occasions. Full results may be found here.

The winner, John Nunn, sent us three problems from the event for you to solve. Here now are the solutions.

Black’s rook is paralysed by the need to prevent both Qd6# and Ne7#, so Black can only move his knight and pawns. There’s a neat mate after 1...Ne6 2 Qd6+ Rxd6 3 Ne7# and it seems likely that this will be part of the solution, but the main problem is that it’s hard for White to make a threat. Indeed, it’s so difficult that one might begin to suspect that the key puts Black in zugzwang, except that it’s hard to see how the move ...b2 could weaken Black’s position. One move that makes a threat is 1 Qh6, intending 2 Nb6#, but 1...Ne6 is a good answer. However, this line provides the crucial clue: if White’s king weren’t on g2, then after 1 Qh6 Ne6 White could play 2 Qh1#. Therefore the key could well be a king move, threatening 2 Qh6 Ne6 3 Qh1#. The king has to be kept clear of annoying black checks, which suggests 1 Kg1!. A quick check reveals the following lines:

It’s interesting how the queen finds different routes to the h1-a8 diagonal after Black’s kingside pawn moves. [Click to replay]

Problems like this always look hard to solve; a lot can happen in five moves, and there are so many pieces on the board that it’s hard to know where to start. However, there’s almost always some sort of logical basis for the solution and finding this is usually the quickest way to solve the problem. One of the first things to note is that there are potential mates by Be7# and Rxe6#, but these mates are prevented by the g7-rook and g8-bishop respectively. The lines of action of these pieces cross at f7, and it looks like a typical problem idea to dump a white piece on f7 to interrupt these lines (problemists call this idea a Novotny). A quick check shows that there are in fact two white pieces which can move to f7, so it’s logical to try 1 Bf7 and 1 Rf7 and see what happens.

The problem with 1 Bf7? is that Black replies with the counter-Novonty 1...Nf6, dropping a piece onto the intersection of the lines h6-e6 and g5-e7. According to which piece White uses to take on f6, Black can then choose which piece to take on f7, thus: 2 Bxf6 Rxf7 or 2 Rhxf6 Bxf7, and in either case there is no mate within the five moves.

OK, so how about 1 Rf7?. Then Black again replies with a counter-Novotny, this time 1...Rf6 (1...Nf6 doesn’t work this time due to 2 Qxc6#). Once again White is denied a quick enough mate after 2 Rhxf6 Bxf7 or 2 Bxf6 Rxf7. This chain of logic has penetrated deeply into the problem, but we still haven’t found the solution. The Novotnys and counter-Novotnys seem likely to play an important role, but White must first do something to shift the odds in his favour.

At this point it’s worth mentioning that the knight on f1 has played no part at all in the analysis and doesn’t have an obvious function in the diagram. Could this piece be used to disrupt Black’s defences in some way? This suggests 1 Ng3, threatening immediate mate by 2 Nxe4#. Black’s only realistic defences are 1...hxg3 and 1...Qxg3 and while it’s not obvious why either of these moves should weaken Black’s position, the fact that there are two defences and two Novotny lines suggests that we might be on the right track.

First let’s look at 1...hxg3, which weakens Black’s position by interrupting the queen’s guard of e5. White must then choose the correct Novotny, which is 2 Bf7. Black uses his counter-Novotny 2...Nf6, but the big difference appears after 3 Bxf6 Rxf7 (3...Qh4 4 Bxh4 leads to mate next move), since now White has 4 Bxe5+ Kxe5 (Black can’t take with the queen) 5 Rxe6#.

It’s even less clear why 1...Qxg3 weakens Black’s position, but it turns out that it’s because the queen can’t interpose on d2. This statement may appear baffling at first sight, but let’s see how the line plays out: 2 Rf7 (Novotny) 2...Rf6 (counter-Novotny) 3 Rhxf6 Bxf7 (3...Bxb6 and 3...Nxb6 are both met by 4 Rxe6+ Kc5 5 Be7#, which explains the function of the b3-pawn) 4 Rf1. This is the key point; now that Black’s queen is on g3, there is a double threat of 5 Be7# and 5 Rd1#, and Black can’t prevent both mates.

It’s very difficult to solve a problem like this by just analysing variations; you have to understand the logic which the composers have incorporated into the problem to find the answer in a reasonable time. [Click to replay]

In this helpmate, Black moves first and both sides cooperate to help White mate Black on White’s sixth move. Note that there are two solutions to this problem (starting with different Black first moves).

Problems like this can be tough, because the usual method is to identify possible mating positions and then see if they can reached in the requisite time. A small amount of logic can cut the workload down, but there’s still a fair amount of trial and error involved. It’s easy to see that a mating position is only possible if the black pawn is on a dark square, and one’s first thought is to mate the king on g4. A mating position can arise if Black plays ...Kg4, ...f5-f4, ...Rh4, ...Qg5 and ...Bf5, with White’s king on g2 and bishop on f3. But a few moments’ thought shows this is impossible; Black can’t play ...Bf5 until the pawn has advanced to f4, but the pawn can’t advance to f4 until the king has moved to g4 and this blocks a subsequent ...Bf5. Therefore one has to try mating positions with the black king somewhere else, and it seems logical to go for f5 and e4, which are the only squares on which a mating position can be arranged in time.

Then it’s a case of working out the sequences of moves, which involves a subtlety or two as White’s pieces must reach their destinations without blocking any of the moves Black needs to play. Eventually one comes to the solutions:

(The solutions start with a black move, so the move-numbers go up to seven even though there are only six white and six black moves in the solution). [Click to replay]

John Nunn is not just a wonderful teacher of chess, he is also – as probabaly everybody who reads our pages knows – an serious amateur astronomer who is part of the GRAS project. In the above picture John, taken at the 2010 London Chess Classic, is doing a demonstration of how the remote telescopes of GRAS are operated – together with GRAS initiator Dr Christian Sasse (left).

The chess visitors were fascinated by John's astronomical lecture

Here John is showing a GRAS picture of the Crescent Nebula in the constellation
Cygnus (in the first he was explaining the relatively nearby galaxy M82 in Ursa Major)

Links to previous problem articles