Anonymous 8:34PM, Black doesn’t need to take the rook on d5, so it’s not a forced mate line. I’d suggest: 1.Re8+ Rxe8 2.Rxe8+ Rxe8 3.f7+, and now if 3…Re5, 4.f8=Q#; if 3…Ne5, then 4.Bxe5+ Rxe5 5.f8=Q#.
1. Re8+ Rxe8 (only) 2. Rxe8+ Rxe8 (only) 3. f7+ Ne5 (Re5? f8=Q#) 4. fxe8=Q+ Kg7 (only) 5. Qxe5+ Black deprived of most of his material, and if it wasn’t for a soon mate, white should also have an easy win with 6.Bxb2. This looked almost too simple…
Not sure if I’m missing anything, but I’m sure Rd5 RxR would not work since f7 would not be a checkmate, Rxd4 would solve the problem. Then white would be queen vs R+2N which is quite bad.
Rd5 RxR f7# Easy peasy.
Mate in 5!
1 Re8
2 Re8
3 f7
4 Be5
5 f8mate
1.Re8+ Rxe8
2.Rxe8+ Rxe8
3.f7+ Ne5 (if 3… Re5 4f8=Q#)
4.Bxe5+ Rxe5
5.f8=Q#
Anonymous 8:34PM,
Black doesn’t need to take the rook on d5, so it’s not a forced mate line. I’d suggest: 1.Re8+ Rxe8 2.Rxe8+ Rxe8 3.f7+, and now if 3…Re5, 4.f8=Q#; if 3…Ne5, then 4.Bxe5+ Rxe5 5.f8=Q#.
1. Re8+ Rxe8
2. Rxe8+ Rxe8
3. f7+ (A) Ne5
4. Bxe5+ Rxe5
5. f8=Q#
OR (B)
3… Re5
4. f8=Q#
1. Re8+ Re8
2. f7#
Re8+ Rxe8
Rxe8+ Rxe8
f7+ Ne5
Bxe5+ Rxe5
f8=Q++
Very Wonderfull
This one is cool.
1. Re8+ Rxe8
2. Rxe8+ Rxe8
3. f7+ Ne5
4. Bxe5+ Rxe5
5. f8=Q#
Rd5 RxR f7# Easy peasy Doesn’t work because RxR isn’t forced.
1)Re8+ Rxe8 2)Rxe8+ Rxe8 3)f7+ (if …Ne5 then 4)Bxe5+ Rxe5 5)f8=Q# or if 3)…Re5 then 4)f8=Q# Forced all the way.
1.Re8+ Rxe8
2.Rxe8+ Rxe8
3.f7+ Ne5
4.Bxe5+ Rxe5
5.f8Q#
Rd5 RxR f7# Easy peasy doesn’t work because RxR isn’t forced.
1)Re8+ Rxe8 2)Rxe8+ Rxe8 3)f7+ (if Ne5 then 4)Bxe5+ Rxe5 5)f8=Q#) if 3)f7+ Re5 then f8=Q# Forced all the way.
1)Re8+ Rxe8 2)Rxe8+ Rxe8 3)f7+ Ne5 4)Bxe5+ Rxe5 5)f8=Q mate
if 3)…Re5 4)f8=Q mate
1. Re8+ Rxe8
2. Rxe8+ Rxe8
3. f7+ Ne5
4. Bxe5+ Rxe5
5. f8=Q++
Easy peasy should take a look at the pawn on b2.
Busy cheese!
1.Re8+ Rxe8 2.Rxe8+ Rxe8 3.f7+ Ne5 4.Bxe5+ Rxe5 5.f8=Q#
1Re8 rxr 2Rxr rxr 3 f7+
either knight or rook interpose
cannot save the mate Qf8
1. Re8+ Rxe8
2. Rxe8+ Rxe8
3. f7+ Ne5
4. Bxe5 Rxe5
5. f8=Q
If 3 .. Re5 then 4. f8=Q
–S
Re8 ch ; Rxe8
Rxe8 ch ; Rxe8
f7 ch ; Ne5
Bxe5 ch ; Rxe5
f8(Q) mate
Not that hard at all!!
Re8+ RxR Re8+ RxR f7+
1. Re8+ Rxe8 (only)
2. Rxe8+ Rxe8 (only)
3. f7+ Ne5 (Re5? f8=Q#)
4. fxe8=Q+ Kg7 (only)
5. Qxe5+
Black deprived of most of his material, and if it wasn’t for a soon mate, white should also have an easy win with 6.Bxb2.
This looked almost too simple…
I found
1. Re8+ Rxe8
2. Rxe8+ Rxe8
3. f7+ Ne5 (or 3 … Re5 4. f8>Q#)
4. Bxe5+ Rxe5
5. f8>Q#
Not sure if I’m missing anything, but I’m sure Rd5 RxR would not work since f7 would not be a checkmate, Rxd4 would solve the problem. Then white would be queen vs R+2N which is quite bad.
San
Re8 better
Hi Susan Polgar,
Well,Very simple puzzle.
White wins the game in ease.
Example
=======
1.Re8+ R*Re8
2.R*Re8 R*Re8
3.f7+ Re5
4.f8(Q)++ Mate
White wins the game in ease.
By
Venky [ India – Chennai ]
1. Re8+ Rxe8 2.f7#
As usual, I had overlooked something.
After
1. Re8+ Rxe8
2. Rxe8+ Rxe8
3. f7+ Ne5
of course the most efficient continuation is:
4. Bxe5+ Rxe5
5. f8=Q#
Still, my solution was at least better than the pitfall:
1. Re8+ Re8
2. f7#
f7 played too soon!
It’s necessary to trade off both rooks here first!