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Susan is the tricky one, methinks.
Unless I am missing something really obvious, a simple f7 wins:
1. f7 Nf7 (Kh7/h8 2.f8Q wins)
2. Nf6 Kh8/f8
3. Ng6#
In small celular phone I was unable to see all and figure out blacks knights lost the game for black.
Kf 2 maybe one of best aproaches, closing to white knights for cornering to mate.
1. f7+! Nxf7 (1. … Kh7 2. f8Q and an easy win for white) 2. Nf6+Kf8 (or Kh8) 3. Ng6#
Requires some thinking and analizing, best move is Ke 3 as a continuing çoves follow..
This one is child’s play, but is interesting as an example of an enforced mate with what is usually called “insufficient material”.
1. f7+ Nxf7 2. Nf6+ Kf8 (or Kh8) 3. Ng6#
f7+!!
..Nxf7 Nf6+ Kh8 Nxf7#
..Nxf7 Nf6+ Kf8 Ng6#
..Kh8 f8=Q+ Kh7 Qxd8 +-
..Kh7 f8=Q +-
Kf2?? Nh5 Ke3 Nxf6=
1f7+ nxp 2Nf6+ k any 3 Ng6 mate
Lomonosov tablebases should do the trick 🙂
f7+ is the winning line.
f7+ is the winning line.
f7+ is the winning line.
I don’t get it! f7+ clearly will queen as the pawn cannot be taken.
So what is the problem…?
1.f7+ !, Nxf7 (otherwise white queens) 2. Nf6+ and black will be mated.
As Lucymarie points out, usually two knights is insufficient mating material in an ending, so here white borrows black’s knights.
In small celular phone I was unable to see all. Black knights lost the game for black.