Wednesday, 9 July 2008
The Unmoveable King
Here is a little problem to fill in some spare minutes (hours?). Can White force checkmate?
Easy you say. However there is one stipulation. The White King cannot be moved (ie it must remain on c3). So now it becomes harder. If the answer to the question is "Yes, White can still force checkmate" then how do you do it?
Subscribe to:
Post Comments (Atom)
4 comments:
Hmm... I thought this looked very interesting. Checkmat will heppen with the black king either on the a file or first rank. Since the white king is on c3 it is symmetrical. So either way it is the same pattern whether checkmate on a file or first rank. (a file would be easiest and quickest after move 1 Qh7) In every line I could find checkmate except for 1. I force the king easily to a8 but I have found that there is no way to both prevent the black kings from running off the a8/h1 diagonal without the kings escaping out of my net. So is it impossible or is there a snaky way I am not seeing?
I am glad to see this problem surface. I was shown this many years ago at the Dubbo Tournament, the only problem was, the person showing me didn't remember the solution! They vowed that there is a solution. I have spent several hours looking at this at different times in the period since being shown and guess what, I still haven't solved it!!
I look forward to finally seeing the solution.
OK. To answer the first part, "Yes, it can be done". And indeed the solution starts with the obvious 1.Qh7 followed by pushing the King to a8 by 2.Qg7 etc.
Now to the tricky bit. To force the king up the a (or b) file (and towards the white king), you need to place the White queen on c8 when the Black king is on a7. This can be done by checking on d5 when the king is on a8. If the king goes to b8 then Qc6! is the right move, otherwise 6. ... Ka7 7.Qb5! Ka8 8.Qa6+ Kb8 9.Qc6!
Hopefully that is enough info to find the rest of the solution (which ends on move 17 with the longest resistance by Black)
Isn't it always the case, easy once you know how!
Thanks Shaun, it is a nice problem with an instructive solution.
Post a Comment