Originally posted by kenanOk.White's only hope is a perpetual check...I think I have one worked out.
Certainly not.
Hint: Almost all of the above posters brought up excellent points.
1. Nxg6 Nc3+ 2. Kc1 Qxa2? (black should have taken the knight with the h-pawn here)
3. Nxe5+! (discovered check)
3..Kh8 (forced)
4. Ng6+!! Kg7
5.Ne5 Kg8
(repeat moves 4 and 5 three times and it's a draw by perpetual check)
Originally posted by bosintangNx6??
Alright, looked at this position with fresher eyes today.
White plays 1. Nx6 Nc3+ 2.Kc1 and..
Black doesn't take the knight on g6:
2.. Qxa2 3. Nxe5+ Kh8 4.Qh6! Qa7 5.Ng6+ Kg8
6.Ne7+! Kh8 7. Qb7#
Black takes the knight on g6:
2..hxg6 3. Rxg6+ Kf7 4. Qh6 Nxd1 6. Rxf6+ and white will be winning.
Originally posted by !~TONY~!Correct.
I am pretty sure the solution is fg! Nc3 Qc3! bc Nf5+ and if Kg8, then Nh6 Kg7 Nf5 is a perpetual, and if Kh8 g7 Kg8 Nh6 is mate. This is a CT-ART Problem.
1.fxg6! Nc3+ 2.Qxc3!! bxc3 3.Nf5+ Kg8 (Kh8? 4.g4+ Kg8 5.Nh6# ) 4. Nh6 (or Ne7+) Kg7 5.Nf5+ 1/2, 1/2 Valbrot-Mizes, Berlin, 1894.
The next one is gonna be easier.
1. Nxh5! gxh5 2. Bxh7+ Kxh7 3. Bxg7 Kxg7 4. Qg5+ Kh7 5. Qxh5+ Kg7 6. Ng5! with a decisive attack. It's becoming a gift that I've gone through the first four levels of CT-ART what seems like 20 times.
Edit: For what it's worth, I am almost 100% sure I've seen this problem and this is the solution, but at the end of all this Black has 6...Bf5! which defends nicely. Are you sure White has an f-pawn? That would make this problem much better!