Then black plays Kf7 and white resigns because after Ng5+ black plays Kg8 and black is two rooks ahead. I think post number one has the correct solution with:
Without looking at any posted comments, I came up with the same analysis as Post #2, and then I read this:
“I don’t think Post#2 above is correct.”
D’oh! I think you’re correct! Post #2 and I got it wrong. Continuing with the correction:
“1. Rxe7 2. Qxe7 2. Nh7+ Then black plays Kf7 and white resigns because after Ng5+ black plays Kg8 and black is two rooks ahead.”
Uh, no. for after: 2 (Nh7+) Kf7? 3 Ng5+ Kg8?, black may be two rooks up, but he gets mated because he’s also down a queen!!! 4 QxQ! with a mate on h7 in a couple of moves.
However, you’re still right that my (and poster #2’s) 1 RxB is mistaken, I’ve now concluded, but black has to play 2…QxN! not Kf7?? So we have: 1 (RxB) (QxR) 2 (Nh7+) QxN! 3 QxQ Nb1+ and black has two rooks and a knight for the queen, with a strong advantage and won endgame for black.
So it’s back to the drawing board, and unfortunately, now that I have read the other posts, I have to agree with you that:
“I think post number one has the correct solution with:
1. Ne4+ fxe4 2. f4″
Unfortunately I didn’t find this also before looking, thinking that RxB was the answer, but yep, the knight check sure looks convincing:
1 Ne4+ PxN 2 f4 e3+ 3 Kc1 Rae8 (the queen is lost, as if it leaves sight of the bish, then white plays QxB++) 4 PxQ+ dxe5 5 Qe6+ Kg5 and now white can either play RxB right away, or perhaps even more sharply play Rf7 threatening to win the bishop with a queen fork if black trades rooks, or play RxR or RxB if black does something else.
Either way, black’s king is trapped with no squares to move to, and black has much less compensation for the queen than in the other line (a whole rook less after white wins the bish); plus with a rook and not just a queen, white has lots of mate threats.
the check is not that meaningless. 2 … e3+ 3. Kd1 (Ke1 Re8 4. fe+ Kg5 5. Rxe7 Nc2+ 6. Kd1 Nb4 7. Kc1 Rf1 8. Kb2 Rf2 9. Q(R)ee8 Rxe2=) Re8 4. fe+ de 5. Qe6+ Kg5 looks like -+ (5. Qg4 looks promisisng and maybe winning) also there is 3. Kc1 avoiding some tacktical resources that black has
so I would go for the draw 1. Re7 ; if I have 15 min to calculate Ke4+ perhaps I would go for it…
Assuming 1. Ne4+ fxe4 2. f4 e3+ 3. Kc1 (even without …Rae8 we could have …Qxd5) 4. Qxe7 is not mate (…Kf5)
Maybe 1. Ne4+ … is right, but I would do like anonymous 9:52:00 PM and proceed attempting this draw: 1. Rxe7 Qxe7 2. Nh7+ Kf7 3. Ng5+ Kf6 4 Nh7+ Kf6 5. Ng5+ Kf7
“Anton said… Hi, Assuming 1. Ne4+ fxe4 2. f4 e3+ 3. Kc1 (even without …Rae8 we could have …Qxd5) 4. Qxe7 is not mate (…Kf5)
Maybe 1. Ne4+ … is right, but I would do like anonymous 9:52:00 PM and proceed attempting this draw: 1. Rxe7 Qxe7 2. Nh7+ Kf7 3. Ng5+ Kf6 4 Nh7+ Kf6 5. Ng5+ Kf7″
Why take a draw by perpetual checks when you can win? As for
1 Ne4+ fxe4 2 f4 e3+ 3 Kc1 you write, “(even without …Rae8 we could have …Qxd5)” No, black is mated with ..Qxd5: 3 …Qxd5?? 4 QxB+ Kf5 5 Qg5+ Ke6 6 Re7++
So black must play Rae8 and give up the queen. So why again would you want a draw when you could have a queen for a rook and a knight (the knight being stuck on the edge of the board away from the action)?
If white doesn’t do something forcing or blocking black has at least a perpetual check with Qe3 check, so
1. Ne4+ pxe4
2. f4
The Queen can not leave the E file or else QxB is mate so the black queen is lost followed soone by mate.
Black is way ahead in material so
1…. Qxe4
is also possible.
2. dxe4 then also lead to mate with Qe6+ followed by f4. Black can throw in a meaningless check with the night but that’s about it.
1Rxe7 if Qxe7 then
2.Nh7+ Qxh7
3.Qxh7
I don’t think Post #2 above is correct. If
1. Rxe7 2. Qxe7
2. Nh7+
Then black plays Kf7 and white resigns because after Ng5+ black plays Kg8 and black is two rooks ahead. I think post number one has the correct solution with:
1. Ne4+ fxe4
2. f4
Without looking at any posted comments, I came up with the same analysis as Post #2, and then I read this:
“I don’t think Post#2 above is correct.”
D’oh! I think you’re correct! Post #2 and I got it wrong. Continuing with the correction:
“1. Rxe7 2. Qxe7
2. Nh7+
Then black plays Kf7 and white resigns because after Ng5+ black plays Kg8 and black is two rooks ahead.”
Uh, no. for after:
2 (Nh7+) Kf7?
3 Ng5+ Kg8?, black may be two rooks up, but he gets mated because he’s also down a queen!!!
4 QxQ! with a mate on h7 in a couple of moves.
However, you’re still right that my (and poster #2’s) 1 RxB is mistaken, I’ve now concluded, but black has to play 2…QxN! not Kf7?? So we have:
1 (RxB) (QxR)
2 (Nh7+) QxN!
3 QxQ Nb1+ and black has two rooks and a knight for the queen, with a strong advantage and won endgame for black.
So it’s back to the drawing board, and unfortunately, now that I have read the other posts, I have to agree with you that:
“I think post number one has the correct solution with:
1. Ne4+ fxe4
2. f4″
Unfortunately I didn’t find this also before looking, thinking that RxB was the answer, but yep, the knight check sure looks convincing:
1 Ne4+ PxN
2 f4 e3+
3 Kc1 Rae8 (the queen is lost, as if it leaves sight of the bish, then white plays QxB++)
4 PxQ+ dxe5
5 Qe6+ Kg5 and now white can either play RxB right away, or perhaps even more sharply play Rf7 threatening to win the bishop with a queen fork if black trades rooks, or play RxR or RxB if black does something else.
Either way, black’s king is trapped with no squares to move to, and black has much less compensation for the queen than in the other line (a whole rook less after white wins the bish); plus with a rook and not just a queen, white has lots of mate threats.
proceed with caution!
speaker #1:
the check is not that meaningless. 2 … e3+ 3. Kd1 (Ke1 Re8 4. fe+ Kg5 5. Rxe7 Nc2+ 6. Kd1 Nb4 7. Kc1 Rf1 8. Kb2 Rf2 9. Q(R)ee8 Rxe2=)
Re8 4. fe+ de 5. Qe6+ Kg5 looks like -+ (5. Qg4 looks promisisng and maybe winning) also there is 3. Kc1 avoiding some tacktical resources that black has
so I would go for the draw 1. Re7 ; if I have 15 min to calculate Ke4+ perhaps I would go for it…
Hi,
Assuming
1. Ne4+ fxe4
2. f4 e3+
3. Kc1
(even without …Rae8
we could have …Qxd5)
4. Qxe7 is not mate (…Kf5)
Maybe
1. Ne4+ … is right, but I
would do like anonymous 9:52:00 PM
and proceed attempting this draw:
1. Rxe7 Qxe7
2. Nh7+ Kf7
3. Ng5+ Kf6
4 Nh7+ Kf6
5. Ng5+ Kf7
“Anton said…
Hi,
Assuming
1. Ne4+ fxe4
2. f4 e3+
3. Kc1
(even without …Rae8
we could have …Qxd5)
4. Qxe7 is not mate (…Kf5)
Maybe
1. Ne4+ … is right, but I
would do like anonymous 9:52:00 PM
and proceed attempting this draw:
1. Rxe7 Qxe7
2. Nh7+ Kf7
3. Ng5+ Kf6
4 Nh7+ Kf6
5. Ng5+ Kf7″
Why take a draw by perpetual checks when you can win? As for
1 Ne4+ fxe4
2 f4 e3+
3 Kc1 you write, “(even without …Rae8 we could have …Qxd5)”
No, black is mated with ..Qxd5:
3 …Qxd5??
4 QxB+ Kf5
5 Qg5+ Ke6
6 Re7++
So black must play Rae8 and give up the queen. So why again would you want a draw when you could have a queen for a rook and a knight (the knight being stuck on the edge of the board away from the action)?