Cumnor Chess Club

	Serieshelpstalemate December 30, 2008
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The next puzzle involves a type of problem which I believe has never been featured in ChessBase Christmas puzzles before. Therefore I will give an example first to explain how it works.

Serieshelpstalemate in eleven

This problem (by Günter Glass, Serienzug-Rekorde 1980) is a serieshelpstalemate in 11, which means that Black plays 11 consecutive moves (without White moving at all) in order to reach a position in which White can give stalemate by playing one move. Both sides are cooperating to bring about the stalemate. There is one further rule, which is that Black is not allowed to give check except possibly on the last of his sequence of moves.

Pins are usually very important in serieshelpstalemates and he we can reasonably guess that in the final position a black piece will be pinned on the a8-h1 diagonal. Since White can make at most one capture during the solution, this means that in the final position there must be either three of four black units on the board (not counting the king).

It seems likely that Black’s king will remain on h1 (or there cannot be a pin), so something must guard or block g1. It is easy to see that a black piece on g1 will always have a move, so this means that White’s final move must be Kf2 or Kxf2. Then we can guess that a black piece must be on g2 (which must be a knight), blocking in another black piece on h2 (which must be a rook). To reach this position isn’t difficult: 1.e1R (not 1.e1N and 2.Ng2, as this blocks the rook’s route to h2) 2.Re2 3.Rh2 4.e2 5.e1N 6.Ng2 (unpinning the d5-pawn) 7.d4 8.d3 9.d2 10.d1N (to reach f2 in time) 11.Nf2 and now Kxf2 stalemate.

Black is stalemated

After this brief lesson in the art of serieshelpstalemate here is our sixth Christmas puzzle:

Serieshelpstalemate in eleven

There are a lot more pieces on the board, but this puzzle is no more difficult to solve than the one above; indeed, if you spot the idea quickly, then you may find it easier. As a hint, remember what I said about pins!

Problems selected and annotated by John Nunn.
The solutions to all puzzles will be published at the end of the series (after January 1st).
Please do not send in solutions after each problem is published.

	Forcing your opponent to mate you December 28, 2008
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Selfmate in six

The first of two selfmates this year. White is to play and force Black to mate White on Black’s sixth move. Black is trying to avoid mating White.

Long selfmates are notoriously difficult to solve, but this is an exception. Having said that, it’s not too easy. At first sight it seems unlikely that White’s king, which currently enjoys almost total freedom, might end up being mated in just six moves. Evidently the king has to approach Black’s forces so as to set up the mating position. However, a further problem for White is that Black is currently stalemated, so White can’t move his king at once. Your task is to combine two ideas: lifting the stalemate, while at the same time creating a mating position.

Problems selected and annotated by John Nunn.
The solutions to all puzzles will be published at the end of the series (after January 1st).
Please do not send in solutions after each problem is published.

	An extraordinary win in a precarious situation December 27, 2008
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White to play and win

For our third puzzle, we have an endgame study: White to play and win. Material is roughly equal, but White has an h-pawn on the verge of promotion. On the other hand, the white king is very exposed, and if Black is allowed to start checking then he will have at least perpetual check (indeed, at the moment ...Qc5+ would force mate). Therefore White must take vigorous action against Black’s counterplay before he can consider promoting the h-pawn.

To help you out, I will tell you that Q v B+N is generally won, while Q v 2N is generally drawn. Even if Black has an additional pawn, these verdicts may not change.

Problems selected and annotated by John Nunn.
The solutions to all puzzles will be published at the end of the series (after January 1st).
Please do not send in solutions after each problem is published.

	Help mate with the knight December 26, 2008
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For the second puzzle we have a helpmate in five

In a helpmate in five, Black moves first and both sides cooperate to help White mate Black on his fifth move (hence both sides play five moves in the solution).

To solve a helpmate, the best idea is to try and guess the mating position and then work out how to arrive there in the time allowed. Here White has just one piece apart from his king, so there aren’t many possibilities.

Problems selected and annotated by John Nunn.
The solutions to all puzzles will be published at the end of the series (after January 1st).
Please do not send in solutions after each problem is published.