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Comments


  • 6 years ago · Quote · #1

    gimce

    what if the opponent promotes to a rook ?

  • 6 years ago · Quote · #2

    camembert

    Then it's just a draw: it's impossible to force a win with rook and knight versus rook unless the weaker side is already in a very bad position (which isn't the case here).

  • 3 years ago · Quote · #3

    WindowsEnthusiast

    I was greedy with 1.Rg5, which intends Ra5 and Ra1#. Black can however play d2 and White loses.

  • 2 years ago · Quote · #4

    YoniKer

    interesting is that even after 1..Nf5?? (allowing Re1#) 2.Rxe4?? d2 it is still a draw: 3.Re2! d2-d1=Q 4,Rh2+ Kg1 Rh1+! =

    I bother to mention this variation with two redicilous moves because there is sa similar study where both variations are possible for black,making this one...boring/irrelevant.

  • 22 months ago · Quote · #5

    seidel

    what if white plays 1.Rg2. I thought it was mate after Ra2 and Ra1# I couldn't find a way for black to stop it

  • 22 months ago · Quote · #6

    YoniKer

    after 1..e4 2.Rg2?? it is quite easy to see that 2..e3 3. Ra2 Kg1 (only move but also winning) cannot be a mate or a win for white. What that was more interesting for me was to see how black wins from that position (otherwise Re3 would not be the only solution) : 

     

    4. Kg3 Nf5+ 5.Kf3 (other moves,for examples Kf4 would lose simply to e2 followed by d2 [Ra1+ Kf2]) and now.... (to be continued)

  • 22 months ago · Quote · #7

    seidel

    you're right YoniKer. Thanks

  • 21 months ago · Quote · #8

    gsi07924

    2. e1+ does not work because Black can capture and promote to a Rook leaving White's King free to move.

  • 21 months ago · Quote · #9

    YoniKer

    in case of 2.Re1+?,promoting to a knight will work as well...

  • 19 months ago · Quote · #10

    Fighter4Christ

    this problem should be changed.

  • 18 months ago · Quote · #11

    NM RelaxingNaked

    this problem should be extended! instead of queening, Black has the try ...Nf5, after which White has to find Rg4!

  • 18 months ago · Quote · #12

    YoniKer

    @ RelaxingNaked 

    You are outright wrong-after Nf5 white has both Rg4?! (with the idea of Rg1+! and no matter how black captures it is a stalement) AND Re2 with the idea d2-d1=Q(Only try otherwise RxP) Rh2+ and Rh1+.

    So after Nf5 there is a double solution-impossible to extend this problem. Do you agree now? :)

  • 13 months ago · Quote · #13

    _36darshan--

    cool stalemate!

  • 9 months ago · Quote · #14

    alakazam1999

    gud 1 

  • 8 months ago · Quote · #15

    mattisks

    Kasparov has beaten Judit in R+N vs R. I think there might be some way to try and play that for win even here.. Problem is it would perhaps take more than 50 moves?! 

    It was thought before that R+B vs R is nearly always a draw and nowadays? You can see plenty of GM games with R+B vs R where "stronger" side succeeds!

  • 3 months ago · Quote · #16

    pakitine

    Q has to take R to get out of chk: therefore stalemate. Other pathways result in new Q winning.

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