Tuesday, January 04, 2011

The twelve puzzles of Xmas

Buzandian, 1969

Mate in eleven

Martin S. said...

Something wrong here as 1. Ra8 is mate in 3, no?

Richard James said...

If you go Ra8 I'm going Bc8. Then I suppose you're going to take on a5.

Martin S. said...

Richard,
You mean we have to mate in ten, even if it's possible to do it quicker?!

Jack Rudd said...

I've seen this one before. I can't remember the exact solution (except that it starts 1.Ra8 Bc8), but the idea is that the we get the following sequence (with the white king on b6 and both black bishops on the back rank)

...Bc7+
Kb5 Bb8/d8
Kxa4 Bd7+
Kxa3

...leaving black with an extra pawn move to ruin his stalemate defence.

The key thing is how to manouevre the king within these ideas - it can never go to c5 with the white-squared bishop on the back rank, because then black just plays ...Bd6+ and ...Bf8.

Martin S. said...

Sorry - the penny has dropped.
Back to square one.

Martin S. said...

Got it. 1. Ra8 was right after all.

I'd say it has Geometry and Flow (though most problems with long solutions are likely to have that, I suspect). And then some slightly Paradoxical footwork around the a pawns, so as to win a couple to give black a move.

ejh said...

OK. Jack has it, basically.

The puzzle was originally published in Shakhmaty v SSSR, 1969, and my source was Nunn, Solving in Style, Gambit 2002, p.71.

The main line is 1.Ra8 Bc8 2.Kd1! Bf4+ 3.Kd3! Bb8 4.Kc4! Be6+ 5.Kc5+ Bc8 6.Kb6 Bc7+ 7.Kb5 Bb8/d8 8.Kxa4 Bd7+ 9.Kxa3 and mate in eleven.