1r6/2k5/8/3np3/6N1/4B3/6K1/8 w - - 0 1
I am going to have to write up a entry discussing the issues I have with this problem.
Yes, Yancey you are right, my v-2 is wrong. Black will win the ending.I made a mistake on the third move.
It's a draw.
In the solution I found given by the composer, he has 2.....Ng6 as black's best defense, but then I have problems with the variation he gives to the alternative of 2.....Rb3- I can't resolve whether or not 2.....Rb3 is a draw for white and maybe even leads to a cooked problem in one line, though that is judgment call, in my opinion. This is one time I would like someone with access to the Lomonosov Tablebase to offer some input into this problem- it is difficult to resolve these lines since white does always best to avoid piece exchanges, and black does best to avoid such exchanges that lose the pawn, too. In other words, I can't reduce this to positions I can be sure of, even with great effort.
Alena, that is a line I never looked at, but I wonder about black's 6th move in V-2. What if black just plays 6.....Ke6? Can white avoid the exchange of the knights? Just to give you a line for exemplification, consider: 6......Ke6 7.Nh5 Nf4+ 8.Nf4 ef4 which is an easy win for black. Indeed, I can't find a way to avoid that knight exchange, and all those endings should be won black as R+P vs B endings.
It's a draw.
You should just post it, Alena, since there is an underlying point that I want to get to in the problem.
I have got the solution for this line. If DaveCubby doesn't post the solution I will post it.
That part looks good, DaveCubby, but now consider the main alternative to 1......Kd6 which is 1.....Nf4.
- Bf2! Kd6, 2. Nxe5!! Kxe5, 3. Bg3+! Nf4+, 4. Kf3 Rb3+, 5. Kg4!! Rb4, 6. Bh2 Rb2, 7. Bg3 forcing threefold repetition and a draw.