This was the last round’s game between Hikaru Nakamura and Michael Rohde. White just played 44.h4 and offered a draw. Black accepted. Which side is better? How would you continue as Black?
Chess Daily News from Susan Polgar
A very interesting position
Looks drawish to me, but the only thing I’m skeptical about is that the White King is slightly safer than the Black King. I cant however see how either side can exploit the other. I’m eager to hear what the answer is to this one.
What’s with the finger?
Black could try Rxe5, then get a queen and take pawn on h4 and win.
I would guess the finger is gambit’s way to show defiance and disrespect toward the monster microsoft monopoly, eh?
As for the problem, first of all, with black’s pawn on the 7th rank, white’s queen is pinned; hence 1…RxP! has to be the key move, for if the Queen checks and wins the rook, then the black pawn queens. So that would yield:
1 (h4) Rxe5
2 Qb8+ Kh7
3 Qxe5 b1=Q
4 Qxe3 and black is up a full pawn
So instead, white can try a slightly different fork with:
1 (h4) Rxe5
2 Qc8+ Kh7
3 Qxc3 b1=Q
4 Qxe5 and again black is up a full pawn.
(I at first thought 3…Rb5!? would win, as black saves the rook and can’t be stopped from queening, but 4 Qc7 b1=Q 5 Qxf7+ with perpetual checks, or 4 Qc7 Kg7 5 Qc3+ with perpetual checks.)
Now the question here is: can black, with queens still on the board, force a win with a 3-to-2 pawn advantage, or can white force perpetual checks eventually or trade queens in such a way that the extra pawn can’t help black?
I don’t know the answer to that question, as I have poor endgame skills. Yet in any case, surely black could have declined the draw and played 1…RxP to essentially trade the rook and passed pawns for a new queen and end up a full pawn up.
Other tries by black don’t look so bad either, such as 1…Kg7. As the threat to promote the b-pawn pins the queen indefinitely (unless the king can shuffle across the board), black can move his king into a better position and threaten to play Rxe5. I just can’t figure out what to do after that, but I see nothing good for white in response. Here the position is unclear, unlike the 1…Rxe5 variation, where after simplification white is a clear full pawn down without compensation. But maybe someone can find some tricky tactics to make 1…Kg7 work. I haven’t been able to do so.
1.-Kg7 is risky due to 2.h5. If you don’t go for immediate action with 1.-Rxe5 you should try 1.-h5. Anyway, I don’t see how black can make more of it than a Q+3P against Q+2P.
Shredder 8 calls the position drawn while fritz 8 declares that black wins.
I vote draw.
Basically the rook can go most anywhere on the board. It has immunity. if the queen takes the rook then black gets a queen.
there is no practical way for white to get the 2 pawns with the queen unless the king comes to help. But there is no way for black to win the white queen and force a new queen either.
On a practical basis I think a draw is acceptable here. Black does arrive at queen and 3 pawns vs queen and 2 for white. Hard to imagine a forced win for black. I can only guess it would be a theoretical draw.
Fritz9 does not give a win. It gives -1 black for the extra pawn. This is still a draw in my view. But I am not an endgame guru.
Black is better. I’d proceed by accepting the draw offer.
I can see it’s possible to play c2 immidiately. (or after some preparation) Q takes and Rc8. Not sure if there is win for black, but white cannot do anything, thats for sure. Q is in c1 forever..
What about
1 .. Rd8
Threatening Rd1 and promoting the b pawn to queen ?
Will that work ?
Then I see that after
1 .. Rd8
White can play
2. Qe7 Rd1
3. e6 fxe6
4. Qe8+
and force perpetual check.
So I think this is yes a draw.
Forget computer analysis – there are too many pieces on the board. I think black is better. How he can win is a different story. I would think getting his king out of a check might be best then he threathens taking the pawn. He should win with those two passed pawns.
Black is a little better after Re5. Note that 2Qc8?? fails to Kh7 3Qc3 Rb5 4Qc4 Rb7! and Black wins. Hence 2 Qb8 Kh7 3Qe5 b1=Q when White should hold but only black has winning chances. Maybe Black was low on time.
Post #13 at 3:32 PM is mine.
-Justin Daniel
“Note that 2Qc8?? fails to Kh7 3Qc3 Rb5 4Qc4 Rb7! and Black wins.”
Nah.
1 …Rxe5
2 Qc8+ Kh7
3 Qxc3 Rb5?
4 Qc7!! (not the ridiculous Qc4??)
And now:
4 … b1=Q
5 Qxf7+! with perpetual checks, or
4 … Kg7 (or 4…Kh8)
5 Qc3+! with perpetual checks, or
4 … Kg8
5 Qc8+! with perpetual checks.
So 2 Qc8+ works just fine and black can’t get Rb5 in without being subjected to perpetual checks, and thus must play:
1 (h4) Rxe5
2 Qc8+ Kh7
3 Qxc3 b1=Q
4 Qxe5 and again black is up a full pawn, as I said before, but with a very difficult endgame that only a computer might be able to win against the best defense on white’s behalf.