Problems in July with GM Jones

This problem of mine recently won 1st Prize in a 2017 tourney of the American magazine ‘StrateGems’. (Yes, the founder of the magazine misspelt ‘stratagems’, and his mistake has been immortalized! The capital “G” is intentional, but not that vowel in front of it.) You may want to avert your gaze and read the next article in BCT straight away, because when it comes to what you can do with a set of chess men this problem is at the opposite end of the spectrum from playing a game of chess. However, I’m hoping that a few readers may read on from curiosity or morbid fascination or whatever.

The problem is ‘helpmate in 3.5 moves’. As you may know, in a helpmate we have a sequence of black and white legal moves in which Black and White conspire together to produce a position in which Black is mated. Just in order to be sound it has to be the case that it looks very unlikely that White could mate Black and that is certainly the case here! In a helpmate in 3 we’d have Black kicking off a sequence of moves leading to mate but as this is in 3.5 move we actually start with White – so we have WBWBWBW sequences. “Sequences” plural, because there is a second part: (b)Pc4>e4 – when that pawn is shifted the (a) solution no longer works but a new solution now does.

You may like to consider how on earth White could mate Black here. Or you may prefer just to read on. (Or that other option – switching to another BCT article – remains open, of course.)

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“White was pleased to get Black out of book early in the game”…A help mate in 3.5 moves

SOLUTION BELOW

The white rook is so immobilized that it looks as though it will have to be the white bishop that does the legwork. It seems that there’s a real risk that when the white bishop moves Black will have no option but to capture the white rook or even administer mate himself. The only alternative after a move by the white bishop is to interpose one of the black Knights, but these give check. There may be a possibility then, though, of playing PxN, and it might be a good thing to get White’s one other combatant, the c2P, involved in the action in this way…

Even so, the only way in which the white rook is going to be able to participate in giving mate is if the mate is by double check…

Putting this all together, you might (or may already have; if so, well done!) come to the first solution (confusingly, because in most helpmates Black begins problemists always write helpmate solutions as though Black had started the game): 1…Bxg6+ 2.Ncd3+ cxd3 3.Ne4 Bf5 4.Rhc6 dxe4. And similar ideas work in (b) 1…Bxb5+ 2.Nfd3+ Bxd3 3.Nb3 cxb3 4.Qc6 Bc4. Move by move the solutions show the same strategy (or should I say “strategems”?!), but the details are all different, which really is a prerequisite for a problem like this, and one reason why the tourney judge gratifyingly had such a favourable impression of it.

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The final position is a pleasing double check

chriscircle

Christopher Jones

Christopher holds the Grandmaster title for Chess Problem Composition and uses his skills to write a regular column for the Bristol Chess Times. He is also a longterm Horfield Chess Club player (where he is acting secretary).

Problems in April with GM Jones

Even if you don’t generally have any time for helpmates you may like to look at this remarkable position, a helpmate in 7 which appeared in the Hungarian magazine Magyar Sakkelet in 1961, by Arpad Molnar.

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A help mate on black in 7 moves.  Black to play. Try not to mate white in the process!

Solution Below

The thing that immediately strikes one is that, given that in helpmates Black is doing all he can to collaborate in being mated, it is remarkable that there is only one way in which mate can be achieved in 7 moves. Upon further inspection we realize that the difficulty is that Black is going to find it very difficult to avoid mating White! In a helpmate Black generally plays first and he has only one non-mating move – 1.f4. (In helpmates, confusingly, Black’s moves are written ‘1.f4’ and White’s moves ‘1…h8N’ – the other way round from in the score of a game.) In playing 1…h8N White avoids 1…h8Q+, forcing 2.Nc8 mate! He is preparing to give Black something else to do instead of inflicting mate; 2.f3 Ng6 3.fxg6… and as 3…g8Q+ fails as before we have 3…g8B!. Play then proceeds 4.g5 Be6 5.dxe6 d7 6.Kc7 b8R 9.e5 d8Q mate!

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Final Position!

Amazingly, White has made all 4 types of promotion (problemists refer to this by the German term ‘Allumwandlung’, or ‘AUW’ for short). It’s always quite an achievement to set up an AUW in any type of problem; in helpmates, while there are a reasonable number of examples showing black AUW, it’s virtually unheard-of to show white AUW. Bear in mind that the composer not only has to create a scenario in which each promotion has to be that specific piece, but that he has to avoid ‘cooks’ (unintended extra ways of solving the problem), the perennial bane of a helpmate composer’s life.

For some reason, Hungarians (such as Molnar) have often been the most inspired helpmate composers. So I’d like to round this column off by showing you a helpmate by another great Hungarian composer, Gabor Cseh. This problem was one of Molnar’s favourites. It also features some truly remarkable promotions. It’s a helpmate in 8, published in Thema Danicum in 1997.

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A help mate on black in 8 moves. Black to play. Try not to mate white in the process!

Solution Below

Again it looks as though White is on the verge of mating Black in very short order. And again the big obstacle is that Black will find it difficult not to move his Knight so as to give checkmate. After 1.b4 (again, the only non-mating move), White mustn’t go in for 1…e8Q+ 2.Ne3 mate! Instead we have a series of promotions to Bishop. In the case of the first such promotion it’s fairly clear why it has to be a Bishop, but in the case of the second promotion (3…f8B!!) it isn’t at all clear, as 3…f8Q wouldn’t be check. You may like examining why 3…f8Q fails. Even when you see why the intended solution doesn’t work with a Q at f8, you have to admire the great technical skill of the composer in ensuring that there are no cooks in which the f8Q would get through to mate in some other way. The full solution: 1.b4 e8B 2.b3 Bb5 3.axb5 f8B!! 4.b4 Bc5 5.f2+ Bxf2 6.Kf3 f7 7.Re8 fxe8B 8.Ke2 Bh5 mate.

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Final Position! 


chriscircle

Christopher Jones

Christopher holds the Grandmaster title for Chess Problem Composition and uses his skills to write a regular column for the Bristol Chess Times. He is also a longterm Horfield Chess Club player (where he is acting secretary).