chess peices

Chess Peices

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bunta 16 ( +1 | -1 )
help mate puzzle -> www.chessbase.com

I havent read it, its chessbases's christmas present =D
ionadowman 68 ( +1 | -1 )
Here's the puzzle: Black to play; helpmate in 2 (White delivering mate).

b


If it's White to play, the thing's pretty quick:
1.Re4 Rg7 2.Rh4#
Interestingly, if the WN were absent from the board, the position would still be checkmate. A clue here, maybe?
My usual approach to this kind of thing is to look for a mating configuration, but I'm finding it difficult (except for the pseudo-solution shown) to discover one... The problem is that it's hard to arrange for the g6-rook to be unpinned whilst retaining the bishop's presence on that diagonal. I think I'll go away and think about it for a while... Trust John Nunn to come up with something challenging...
Cheers,
Ion

ionadowman 15 ( +1 | -1 )
Help in 3 is easy: ...but there are several ways of achieving it. The tastiest is:
1...Bh4+ 2.Nxg6 Bf6 3.Rc6 Bg7 4.Nf8#

Stick a pawn at h6 and the 2-mover is simple:
1...Bxe5 2.Rg4 Bg7 3.Bxg6#

Still no cigar, though... :-/
sf115 3 ( +1 | -1 )
Answer 1... Bh4 2. Bd1 Rg7 3. Rxh4++ is the answer to the helpmate in 2
sf115 19 ( +1 | -1 )
ionadowman said, "If it's White to play, the thing's pretty quick:
1.Re4 Rg7 2.Rh4#" that's illegal as the rook on g7 is pinned by the bisiop.
vanleeuwenhoek 8 ( +1 | -1 )
No, that was the point of the Re4 move, to block the bishop's diagonal and unpin the rook.
chrisp 8 ( +1 | -1 )
Not the answer, sf115 After 1.....Bh4,

white cannot play Bd1 as he is in check, so that solution isn't correct!!
juve_leo 13 ( +1 | -1 )
...but there are several ways of achieving it. The tastiest is:
1...Bh4+ 2.Nxg6 Bf6 3.Rc6 Bg7 4.Nf8#
your numbers are wrong Nxg6 should be Nxf6

1...Bh4+ 2.Nxf6 Bf6 3.Rc6 Bg7 4.Nf8#
ionadowman 121 ( +1 | -1 )
No... 2.Nxg6 is correct. But in general, we agree the line ends in a 3-move helpmate.
vanleeuwenhoek is correct: 1.Re4 blocks the bishop's action and hence releases the pin on the rook, allowing it to retire (1...Rg7). But this obviously isn't the solution: the White knight doesn't have a role.
It is very easy to get things wrong in this kind of thing. I had another "3-mover" I was about to post, but I saw, just in time, it wasn't a mate.
Something to point out, too, in puzzles of this nature. Every element is there for a reason, and every move is essential. Suppose 2.Bd1 in sf115 his solution were legal. Would it have been the only move that led to mate? No. 2.Bb3 would have done just as well. Not a solution then.
The solution begins with a "key move" by Black. So far, I still haven't found it. Anything other than a Bishop move seems to free up Black's King, but a bishop move is a check, which limits White's response. I almost wish the position were incorrect and that there was supposed to be a Black pawn on h6!
Cheers,
Ion
ionadowman 76 ( +1 | -1 )
It might help... ... to try out this "quintet" puzzle - got from the same site...

Black to play, helpmate in 2:

b

The difficulty here is that although R+N can mate unaided in the corner, such a setup would take more than 2 moves here. So the Black Q has to cooperate.

It gets better, though. having solved it, replace the queen with a black rook, and solve. Then solve for a bishop, then a knight, then a pawn. For the last, remember thet the pawn is 5 squares away from queening, not two!
I found these 5 a lot easier than than the one given to begin this thread (though the bishop one is a bit tricky). But its ideas may be a help in solving the latter.
Cheers,
Ion

vanleeuwenhoek 4 ( +1 | -1 )
Aha! The solution! So don't scroll down if you don't want to see it..













1...Qa8 2. Bb3 Rh6 3. Rc7#
honololou 12 ( +1 | -1 )
Well done… vanleeuwenhoek. Now I can get a good night's sleep tonight.
ionadowman 27 ( +1 | -1 )
Very neat... ... I hardly looked at 1...Qe8 at all (though I did wonder - very briefly - if there might be something in ...Qa8 and ...Rb8!). I did look at Bb3 and Rc7 as possibilities within the sequence but not in conjunction with Black's Qa8. Thanks, vanleeuwenhoek. The beauty of this solution is that every piece participates in the final mate.
;-)
Ion