i looked at Rd8 to start things off but found this line to be a problem Rd8 Bxd8 Rxd8 Bd7.ive been playing alot of poker latly so my solving might not be as strong as a few weeks ago.
I am neither convinced by Nxg7 because black mustn’t (and should’t of course) take it and after Bd7 it seems to me at if the knight is lost because it is pinned (white has to play with two figures down), nor I am convinced by the posted lines to 1. Rd8 which is the move I “found”, too.
I also think it is Rd8 but there is still some work to do which nobody did until now (probably there are posts which haven’t been published yet, but while writing there aren’t). 1. Rd8 and there are two variations. 1.-, Bd8 (take everything what we can get!) 2. Rxd8, g6 (what else?) 3. Qc3+ (or 3.Qg5 e5 seems to lead to the same variation – but I do more like Qc3+ because it’s with check) e5 (what else?) 4. Qxe5+, f6 5. Nxf6! and now black is without defens (5.-, Rxd8 6. Nh5/e8+ and mate soon).
But there is also 1.-, g6 (without taking the rook – the bishop is needed to defend e5 and I do not see the winning variation now. Another defense trick is the following: 2. Qg5 e5 (forced) 3. Qh6 (threating Nf6 as long as the rook is pinned that is deadly) Qe4! and again the bishop is needed to attack f4.
So I think if Rd8 is correct there is still a lot of work to do if black takes not the rook with his bishop.
2…Bxd8, 3 Rxd8 e5 (….Rxd8, 4 Qf6+ followed by Qg7#), 4 Rxg8+ Kxg8, Qd8+ mates next)
2… e5, 3 Qh6! (simplest, 3 Qxe5+ looks flashy but leads to many complications) gxh5 (forced to stop Qg7#), 4 Rxg8+ Kxg8, 5 Qxc6 wins easily, as Black must lose at least another Bishop.
Rd8 g6
Qg5
yep rd8 looks pretty crushing.
Mike M
1. Rd8 Bb7 2. Qxg7#
1. Rd8 Bxd8
2. Rxd8 g6
now what?
rd8 bd8
rd8 g6
qc3
This comment has been removed by the author.
Nxg7 Rxg7
Rd8+ Bxd8
Rxd8+ Rg8
Qxg8++
i looked at Rd8 to start things off but found this line to be a problem Rd8 Bxd8 Rxd8 Bd7.ive been playing alot of poker latly so my solving might not be as strong as a few weeks ago.
i looked at Rd8 to start things off but found this line to be a problem Rd8 Bxd8 Rxd8 Bd7.
um, Rd8 Bxd8, Rd8 Bd7, Qg7#.
The rook may be en pris but it can still pin the Black rook!
ive been playing alot of poker latly so my solving might not be as strong as a few weeks ago.
I hope you’ve been winning 🙂
I am neither convinced by Nxg7 because black mustn’t (and should’t of course) take it and after Bd7 it seems to me at if the knight is lost because it is pinned (white has to play with two figures down), nor I am convinced by the posted lines to 1. Rd8 which is the move I “found”, too.
I also think it is Rd8 but there is still some work to do which nobody did until now (probably there are posts which haven’t been published yet, but while writing there aren’t).
1. Rd8 and there are two variations.
1.-, Bd8 (take everything what we can get!)
2. Rxd8, g6 (what else?)
3. Qc3+ (or 3.Qg5 e5 seems to lead to the same variation – but I do more like Qc3+ because it’s with check)
e5 (what else?)
4. Qxe5+, f6
5. Nxf6! and now black is without defens (5.-, Rxd8 6. Nh5/e8+ and mate soon).
But there is also 1.-, g6 (without taking the rook – the bishop is needed to defend e5 and I do not see the winning variation now.
Another defense trick is the following:
2. Qg5 e5 (forced)
3. Qh6 (threating Nf6 as long as the rook is pinned that is deadly) Qe4! and again the bishop is needed to attack f4.
So I think if Rd8 is correct there is still a lot of work to do if black takes not the rook with his bishop.
Greetings
Jochen
I think Rd8 and Nxg7 need black to comply for an easy mate. i did look at Rd8 g6 Qc3 e5 this seems to throw everything into lots of variations.
jochen, in the line 1. Rd8 g6 2. Qg5 e5
3. Qh6, isn’t the last move threating Qg7# and hence 3. … Qe4 doesn’t help?
1 Rd8 g6
2 Qg5 Bxd8
3 Rxd8 e5
4 Qxe5+ f6
5 Nxf6 Rxd8
6 Nh5+ followed by the mate
I think it’s the longest variation for black
Does the back rank mate work?
1. Rd8 g6
2. Qg5! e5
3. Rxg8+ Kxg8
4. Rd8+ Bxd8
5. Qxd8 and mates
Or
2. – Bxd8
3. Rxd8 e5
4. Rxg8 etc.
Or
3. – Qe8
4. Qf6 Rg7
5. Rxf8 mate
fpt1. Rd8
a)1..Rd8 2. Qg7#
b)1..Bd8 2. Rd8
i)2..R,Q,B,a/c/d/e/h-pawn-any, 3.Qg7#
ii)2..g6 3. Qc3+ A)3..f6 4.Qf6#
B)3..e5 4.Qe5+ f6
5.Nf6 (6.Rg8#)
if 5..Rd86.Ne8+Kg8
7.Qg7#
if5..Kg7 6.Ng4+Kf7
7.Nh6#
c)1..Bf4+ 2. Kb1
i)2..B-any except Bc7, 3.Rg8+Kg8
4.Rd8+
ii)2..Bc7 3.Qg7#
iii)2..g6 3.Qf4 with Qf6+ etc
iv)2..Qf4?! Bb7 2. Qg3/g4, Qg2 etc
1.Rd8! g6?
2.Qc3+ …
imho :))
Hey anonymous, of course Qg7 is a mate – too simple to see.
Thanks. 🙂
Jochen
I don’t know if somebody had already said this, but:
1. Rd8, g6?;
2. Qc3+, e5;
3. Qxe5!, Bxe5;
4. Rxg8, Kxg8;
5. Rd8
After 1.Rd8 Bb7 I cannot find effective attack for white
> Anonymous said…
>
> After 1.Rd8 Bb7 I cannot find
> effective attack for white
As yuly (3rd post) already pointed out:
1. Rd8 Bb7 2. Qxg7#
1. Rd8 leaves Black two reasonable lines of defense against the threat of mate at g8: Rxd8 and g6.
1…Rxd8, 2 Rxd8 g6 (forced), 3: Qg5 (3 Qc3+ also wins) e5 (forced), 4 Qxe5+ f6 (forced), 5 Nxf6 mates soon.
1…g6, 2 Qg5 e5, and now:
2…Bxd8, 3 Rxd8 e5 (….Rxd8, 4 Qf6+ followed by Qg7#), 4 Rxg8+ Kxg8, Qd8+ mates next)
2… e5, 3 Qh6! (simplest, 3 Qxe5+ looks flashy but leads to many complications) gxh5 (forced to stop Qg7#), 4 Rxg8+ Kxg8, 5 Qxc6 wins easily, as Black must lose at least another Bishop.