But instead of 5. Qh8+ white has 5. Bg6 and Black will half to jetison his queen, and at least a rook, to hold off checkmate.
The immediate 3. — Qc7 may be better for Black, but he is still completely lost. For example:
3. —- Qc7 4. Qh7+ Kf8 5. Bg6 Qf7 6. Bxf7 Kxf7 7. Bf6 Rg8 8. Qh5+ Kf8 9. Bxd8 And white has a queen and pawn for a rook and knight, and black is still going to lose the other rook.
So it appears that there are two winning moves (Rxg7 & Qxh6). But the mate in 4 is obviously what Susan wanted us to find.
1.rg7 kg7 if kh8 then qh6 mate
2.bf6 kg8
3.qh6 any
4.qg7#
1. Rxg7+ (1…Kh8 2. Qxh6+ Nh7 3. Qxh7++) Kxg7
2. Bf6+ Kg8
3. Qxh6 (and ++ via 4. Qh8 or 3…Ng6/Nh7 4.Qg7++)
Qxh6 looks good after a few different continuations
1.QxRP N-N3
2.RxN PxR
3.QxP(onN3)
Rg7+! is the move. Qxh6 doesn’t do the job because of Ng6 with nothing better than a draw.
1. Rxg7+ mates in 4 moves.
But it appears that 1. Qxh6 also wins, though it is long and obtuse. For example:
1. Qxh6 Ng6
2. Rxg6 fxg6
3. Qxg6
So far all this is obvious. White is now down 2 rooks for 2 pawns, but the Black king is exposed.
Mark M is correct in claiming that White has a draw by repetition after
3. —- Kf8
4. Qh5 Qc7
5. Qh8+ Kf7
6. Qh5+ Kf8
7. Qh8+ Kf7
etcetera
But instead of 5. Qh8+ white has 5. Bg6 and Black will half to jetison his queen, and at least a rook, to hold off checkmate.
The immediate 3. — Qc7 may be better for Black, but he is still completely lost. For example:
3. —- Qc7
4. Qh7+ Kf8
5. Bg6 Qf7
6. Bxf7 Kxf7
7. Bf6 Rg8
8. Qh5+ Kf8
9. Bxd8
And white has a queen and pawn for a rook and knight, and black is still going to lose the other rook.
So it appears that there are two winning moves (Rxg7 & Qxh6). But the mate in 4 is obviously what Susan wanted us to find.
Rg7+ Kh8
Bf6 Ng6
Rxg6+ Kh7
Rg3
wolverine