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1. Rd5+ Kxf4
( 1…… Ke6 2. f5# )
( 1…… Kg6 2. Rg5# )
( 1…… Kg4
2. Nf6+ Kh4
3. Rh5# )
2. e3+ Kg4
4. Nf6+ Kh4
5. Rh5#
Fairly straight forward checkmate calculation.
1. Rd5+ Kxf4
(1…. Ke6 2. f5++
1…. Kg6 2. Rg5++
1… Kg4 2. Nf6+ Kxf4 3. e3++)
2. e3+ Kg4
3. Nf6+ Kh4
4. Rh5++
1. Rd5+ wins
if 1… Kg6
2. Rg5#
if 1…. Kxf4
2. e3+ Kg4
3. Nf6+ Kh4
4. Rh5#
if 1….. Kg4
2. Nf6+
now, 2… Kh4, 3. Rh5#
or 2… Kxf4 3. e3#
1.Rd5+ Kxf4 (the other choices are not better either)
(1…Ke6 2.f5#)
(1…Kg6 2.Rg5#)
(1…Kg4 2.Nf6+ Kxf4 3.e3#)
2.e3+ Kg4 (forced)
3.Nf6+ Kh4 (forced)
4.Rh5#
A true classic which is a mate in 4:
1. Rd5 Kf4
If black plays Ke6, white mates with 2.f5. If black plays Kg6, white mates with 2.Rg5. If black plays Kg4, white mates in 2 starting with 2.Nf6 and ending with either 3.e3 or 3.Rh5 depending on where the black king goes. Continuing:
2. e3! Kg4
3. Nf6 Kh4
4. Rh5#
1.Rd5+ Kxf4
1.-Ke6 2.f5++
1.-Kg6 2.Rg5++
1.-Kg4 2.Nf6+ Kxf4 (2.-Kh4 3.Rh5++) 3.e3++
2.e3+ Kg4 3.Nf6+ Kh4 4.Rh5++
i think, it´s
1. Rd5+ Kxf4 (Ke6, 2. f5#)
(Kg6, 3. Rg5#)
2. e3+ Kg4
3. Nf6+ Kh4
4. Rh5#
1. … Kg4
2. Nf6 Kxf4 (Kh4, 3. Rh5#)
3. e3#
greets, jan
1. Rd5+
1….Kxf4 2. e3+ Kg4 3. Nf6+ Kg4 4. Th5#
1….Kg4 2. Nf6+ … similar to 1. variant
1….Ke6 2. f5#
1….Kg6 2. Tg5#
Rd1-d5 QXf4
e2-e3 followed by Nf6
– Nabendu
I think the following plays itself:
1. Rd5+
a)
1. … Kxf4
2. e3+ Kg4
3. Nf6+ Kh4
4. Rh5#
b)
1. … Kg4
2. Nf6+ Kxf4 (Kh4 Rh5#)
3. e3#
c)
1. … Ke6
2. f5#
d)
1. … Kg6
2. Rg5#
The solution is not that complicated as it looks at first sight.
1Rd5+ Kxf4
1… Kg6 2Rg5#
2e3+ Kg4
3Nf6+ Kh4
4Rh5#.
Looks like it’s all been covered.
Rd3 wins the queen