Monday, January 05, 2009

The twelve puzzles of Xmas

problem by John Beasley

The White knight is on its home square (g1) but its objective is to play to Black's king's home square (e8). However, after each move Black may place a mine on any vacant square other than e8 itself - and if the knight subsequently sets foot on that square it is blown up. Can the knight succeed, or can Black eventually surround it with a cordon of mined squares?


Anonymous said...

The answer is yes, the White knight can get to e8.

I'll try to explain why.

First, there are four squares providing direct access to e8. These are c7, d6, f6 and g7. Once the White knight gets to one of these it will be able to achieve its goal. White will win as Black won't be able to cover all four of these squares.

White plays 1 Ne2.

Black then effectively has a choice of two moves: f6 and g7. The moves d6 and c7 can be discounted as they are mirrors of f6 and g7; it only makes sense to cover a square which gives direct access to e8.

Taking the 1... f6 bomb first. After this White plays 2 Nd4. If Black then covers c7 White can play Nf5 to give access to both d6 and g7. If, instead, Black covers d6 then White can play Ne6 to give access to c7 and g7. And, if Black covers g7 then White can play Nb5 giving access to c7 or d6.

Black's second attempt after 1 Ne2 is to bomb g7. But this fails to 2 Nc3 followed by 3 Nb5, 3 Nd5 or 3 Ne4 as appropriate.

Make sense?... I'm sure Justin can explain more succinctly.


ejh said...

Not before coffee I can't.

Neill Cooper said...

I agree with Angus.

Imagine black has pawns on c7, d6, f6 and g7. The knight has to take one, so a fork is required. After Ne2 which ever pawn black removes, white can still move to a square where he is threatening to fork all 3 remaining pairs of pawns. Any pawn is removed and white can fork the remaining two and so win one, and thus get to e8. [Each pawn black removes is replaced by a bomb.]

As Angus says, remove g7 and Nc3 threatens forks on b5, d5 and e4. Remove f6 and Nd4 threatens forks on b5, e6 and f5. And symmetry for c7 and d6.

[I now ralise that all I've done is say what Angus already had said. But I enjoyed working it out myself first!]