Normal Topic RR + 3 -v RB + 3 (Read 7501 times)
GMTonyKosten
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Re: RR + 3 -v RB + 3
Reply #8 - 08/23/13 at 09:18:24
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Benoniac wrote on 08/21/13 at 18:23:01:

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From this position I think the right plan should be to prepare ...f6 and then ...g5, bring the king to g6, then play ...f5, e.g. 1. Rb7 Re5 2. Bf4 Rd5 3. Re7 Rcd8 4. Kf3 (4. g4 h5 5. Kg3 hxg4 6. Kxg4 Rc8 7. Kg3 Rc4 8. Be3 f6 9. Ra7 Kh7 10. Ra3 Kg6) 4... R8d7 5. Re2 f6 6. Re4 g5 7. Be3 Kf7 8. Ra4 Kg6 9. Kg2 f5 then a well-timed ...g4, say 10. Ra2 R7d6 11. hxg5 hxg5 12. Ra7 Rc6 13. Ra8 g4 14. Ra2 Rd3 15. Bf4 Rb6 16. Be3 Rb1 17. Ra6+ Kf7 18. Ra7+ Ke6 19. Ra2 Re1 20. Bc5 Kd5 when White hasn't been able to play f3, the black king is active and can force the white rook off the 2nd rank by ...Kb3, and Black should be able to win at his leisure.
« Last Edit: 08/23/13 at 11:01:06 by GMTonyKosten »  
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GMTonyKosten
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Re: RR + 3 -v RB + 3
Reply #7 - 08/23/13 at 08:55:06
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Benoniac wrote on 08/21/13 at 18:23:01:

to get something like this: 

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In the diagrammed position black plays: ...g5. If white takes, then black has the possibillity to create a passed pawn. Also he can choose to gain space with: 1...g5  2.hxg5-fxg5 3.Be3-g4!? wich SF4 actually played ( and later got a winning position).


But what if White captures on g5 and then plays g4 himself, for instance 33... g5 34. hxg5 fxg5 35. g4 h4 36. Be3 fixing Black's g-pawn on a dark square? 
Creating a passed h-pawn doesn't appear particularly useful, bearing in mind these types of position are known draws:

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Benoniac
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Re: RR + 3 -v RB + 3
Reply #6 - 08/22/13 at 22:56:21
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P.s:
Last nights results was encouraging for the part trying to win; both H3 and SF4 won with the black pieces from the 2.  diagram. I will test more and try to understand. 

I think there is a (deep) zugzwang issue. I dont get it yet, but as soon as black get a pawn to g4, then it seem like white must either give up the control of the 1. or 2. rank, and in either case black is able to force it through. Because if the white pawn must move from f2 to f3, then its spells disaster ( the fort is not longer there).

A kind of positon that repeats itself is one like this:

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I believe this is won for black because of the pawn on g4 .  Since white dont want to trade rooks,  one scenario is that the B get pined on f2. Then if the black king can come  to c1, its an all trade on f2 and ...Kd2 will win the game. 

I also remember a theme with black having doubled rook on the 2. rank. White king on h1 felt the need to play Bg1 to cover h2. Suddenly the black Rd2 moved to d7 and threatened ...Rh7+. It forced a rook swap and the win was fairly easy after that. 

Im not sure, but it seems like its a deep winning method in these kind of positoins.  I will keep on with my research.  Smiley

Ben
« Last Edit: 08/23/13 at 09:44:30 by Benoniac »  

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Benoniac
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Re: RR + 3 -v RB + 3
Reply #5 - 08/21/13 at 18:23:01
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I might think there is some hope of winning this position. I know, its kind of hard to belive. But from a position like this:


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* * * * * * * *
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to get something like this: 

* * * * * * * *
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* * * * * * * *
* * * * * * * *
* * * * * * * *
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I dont think white can hinder black from getting here. 
The thing is that I ran a macht between H3 and Sf4 earlier. But SF4 crashed somehow in the end and the game wasnt automatically saved  Sad  But. SF4 achieved a winning position according to both engines. 


In the diagrammed position black plays: ...g5. If white takes, then black has the possibillity to create a passed pawn. Also he can choose to gain space with: 1...g5  2.hxg5-fxg5 3.Be3-g4!? wich SF4 actually played ( and later got a winning position).

Maybe the fact that the white pawn is on h4 is a telling factor. I dont know.  I will run this again tonight and hopefully have some answers in the near future...

Ben  
« Last Edit: 08/21/13 at 19:24:42 by Benoniac »  

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Keano
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Re: RR + 3 -v RB + 3
Reply #4 - 04/18/13 at 08:45:44
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I think it should be defendable:

check: Van Wely - Iturrizaga  Istanbul ol 2012

In that case Black set up a fortess with his pawn on f7 and a White squared Bishop. That seems like it should be OK to me, another question is if Black had a Black squared Bishop, in that case it is possibly lost, but who knows, could also be defendable - for example if Black could play ...f6, and ...g5 and stick the Bishop on e5 he has good chances for a fortress I think. But that is a big if of course.

Heinemann,Thies (2469) - Grandelius,Nils (2536) 
Bundesliga 2011 Black also defended the endgame when he had the "right" coloured Bishop.









  
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Re: RR + 3 -v RB + 3
Reply #3 - 04/18/13 at 07:46:43
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IMJohnCox wrote on 04/17/13 at 23:57:08:
Does the panel think that in general R + R + fgh against R + B + fgh should be a win?

I had this in a recent game (something like W: Be3 Ra7 P f2 g3 h4 Kg2: B R c8 e8 Kg8 P f7 g7 h6).

As it happened the dreaded adjournments were in operation (don't ask) and it happened that a draw won the match for my team, so my correct course seemed clear (I had the two rooks).

Against my will, though, I've become interested. It seemed to me at first that this must be winning, but the more I've tried, the more I haven't been able to come up with any remotely sensible plan.

Does anyone have any ideas? Or know of any discussion of this ending anywhere - my books don't seem to have anything.

I am pretty sure that you should be able to win this endgame in the assumption you have sufficient time on the clock despite the adjournement. It is impossible to analyse all possibilities of this endgame in the adjournement so the correct strategy is to put continuous pressure on the opponent without really caring about the quickest winning plan. Let him run out of time and then finish him with some small tactics of which he will certainly miss one. I've adopted this method several times myself with success, see my blogarticle: http://schaken-brabo.blogspot.be/2012/05/eindspelen-met-een-kwaliteit-meer.html
  
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Re: RR + 3 -v RB + 3
Reply #2 - 04/18/13 at 07:35:36
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The main thing it alters is that it makes it much harder to activate Black's king. The two rooks don't seem to be able both to stop checks and to protect Black's own pawns while you try and do this.

Swopping rooks would be a win, of course, but how to achieve that?
  
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Re: RR + 3 -v RB + 3
Reply #1 - 04/18/13 at 07:17:35
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There are some examples of R+R vs R+B with pawns in "Practical Endgame Play - beyond the basics" by Glenn Flear, although not fgh-pawns on each side.

According to Awerbach rook vs bishop with fgh on board gives very good winning chances for the side with the rook. I am not sure what an additional rook on both sides alters - there should be a way for the stronger side to exchange it. 
  
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RR + 3 -v RB + 3
04/17/13 at 23:57:08
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Does the panel think that in general R + R + fgh against R + B + fgh should be a win?

I had this in a recent game (something like W: Be3 Ra7 P f2 g3 h4 Kg2: B R c8 e8 Kg8 P f7 g7 h6).

As it happened the dreaded adjournments were in operation (don't ask) and it happened that a draw won the match for my team, so my correct course seemed clear (I had the two rooks).

Against my will, though, I've become interested. It seemed to me at first that this must be winning, but the more I've tried, the more I haven't been able to come up with any remotely sensible plan.

Does anyone have any ideas? Or know of any discussion of this ending anywhere - my books don't seem to have anything.
  
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