TXBaseballFan wrote:
Rocky64 wrote:
Arisktotle wrote:
An improved version of post # 109. It's a triplet of problems with a switch of 2 units between them. All problems require a checkmate in 1 move. Solutions follow later...............................................
The retro-analysis for these is similar to the earlier version's....................
Pf6? correct notation: fxe6#
Not sure I understand you. I don't see a move Pf6 anywhere, only a unit. "Pf6" stands for "the pawn placed on f6 in the diagram". "Rf4" and "Bf3" also refer to units rather than moves. These annotations describe not solutions, only differences between the 3 diagrams.
An improved version of post # 109. It's a triplet of problems with a switch of 2 units between them. All problems require a checkmate in 1 move. Solutions follow later.
1. Diagram 1.
2. Diagram 2 = Diagram 1 plus a switch of Rf4 and Pf6
3. Diagram 3 = Diagram 1 plus a switch of Bf3 and Pf6
The retro-analysis for these is similar to the earlier version's. Again Black still has all 16 units and must have promoted 7 Ps. The main difference is that White has 4 promoted pieces rather than 5. To get out of the way of 4 white promoting Ps, 4 black Ps had to make a capture each. Because of the white P still on the f-file, Black's original f-P also had to make a capture in order to promote. These 4 + 1 black P captures account for all 5 missing white units, so there are no spare white units to be captured by either black N as the last move in the diagram.
1. Black has no possible last move, so it's Black to play and the mate is 1...Qa7.
2. Black's last move was ...Ne6 to answer the discovered check, Re6-f6+, so White is to play: 1.Rxe5.
3. Black's last move was ...Ne5 to answer the discovered check, Be5-f6+, so White is to play: 1.fxe4.
Pf6? correct notation: fxe6#