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free_fall you said it.
Black will play then +Nf1 !!!
That took couple of goes
The continuation would be...
4... Nf1+ 5.) Kf2 or f3, NxQ 6.) Ng6+, Kh7 7.) BxR, and now White is up an outside passed pawn a bishop (black can run with the knight and white runs with his knight, or black takes and white takes as well). White wins.
Any problem with 2.Kh2 ?
Seems possible as well. Might only be a little bit more complicated.
(one nice line: 1.Rxf6 Qc1+ 2.Kh2 Kh7 3.Qxh6+! gxh6 4.Rf7#)
Hardest puzzle in awhile!!!
That was different...
But the idea is simple: the 1st move threatens checkmate next. Black then finds a tricky way to avoid it, by sacrificing his Queen.
PS: for those interested in the continuation, it would be:
4. ... Nf1+ 5. Kf3 Nxe3 6. Ng6+ Kg8 7. Bxd8,
and White is winning in the endgame.
" cheap trick "
I showed this one to neighbours kids... I ´ll wait them .
Certainly 3. Qg2 is a better try. It breaks the pin, and if 3...Nf4, forking queen and rook, then 4. Rd8+ wins the queen. What would be the continuation after 3. Qg2?
Interesting that my comment before seeing the title was going to be that it was difficult to solve because it required alot of calculation. Before checking the solution I solved it through 3. Kg3 but missed the fact that after 3...Qxf6, black is threatening 4...Nf1+.
This shows two things. First that it required more calculation then I had even thought and second it reminds me that calculation is a tremendous weakness in my game.
Hmm, my first thought too. But after Qg2, Qb1+, forces Qg1, followed by Qe4+ capturesthe rook, if the queen goes to g2 to stop, you can just take the queen followed by Nf4+ forking rook and king
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