Just to make it more interesting — is there another way that white could win? Um, almost, but not quite. Black almost loses, but can force a draw by threefold repetition. 1. Rd8 Ng3+ 2. Kg1 Ne2+ 3. Kh2 Qf6 4. Rxf8+ Kxf8 5. c6 Nd4 6. Qc8+ (6. Qxf7+ Qxf7 7. Bxf7 Nxc6 =) 6… Ke7 7. Qd7+ Kf8 8. c7 Qf4+ 9. Kg1 Qc1+ 10. Kh2 Qf4+ 11. Kg1 Qc1+ 12. Kh2 Qf4+ {Draw. Threefold repetition.}
1. Qf7+ Rf7 (Kh8 2. Qf8#)
2. Rd8#
1) Qxf7+, rxq
2) d8#
1. Qxf7 Rxf7 2. Rd8#
Qxf7+ Rxf7
Rd8#
or
Qxr7+ Kh8
Qxf8#
1. ♕xf7+
1. … ♚h8
2. ♕xf8#
1. … ♜xf7
2. ♖d8#
1. Qxf7+
If 1… Kh8 then 2. Qxf7#
If 1… Rxf7 then 2. Rd8# because black’s rook is pinned.
This seems too easy. What am I missing?
1. Qxf7+
If 1. … Kh8 2. Qxf8#.
If 1. … Rxf7 2. Rd8# (the black rook is pinned by the white Bishop).
Yep. It is easy.
1. Qxf7+ Rxf7
2. Rd8#
Just to make it more interesting — is there another way that white could win? Um, almost, but not quite. Black almost loses, but can force a draw by threefold repetition.
1. Rd8 Ng3+
2. Kg1 Ne2+
3. Kh2 Qf6
4. Rxf8+ Kxf8
5. c6 Nd4
6. Qc8+
(6. Qxf7+ Qxf7 7. Bxf7 Nxc6 =)
6… Ke7
7. Qd7+ Kf8
8. c7 Qf4+
9. Kg1 Qc1+
10. Kh2 Qf4+
11. Kg1 Qc1+
12. Kh2 Qf4+ {Draw. Threefold repetition.}