Score: 100%

0sec 61sec 122sec


  • 8 years ago · Quote · #1


    what if the opponent promotes to a rook ?

  • 8 years ago · Quote · #2


    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).

  • 6 years ago · Quote · #3


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

  • 4 years ago · Quote · #4


    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.

  • 4 years ago · Quote · #5


    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

  • 4 years ago · Quote · #6


    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)

  • 4 years ago · Quote · #7


    you're right YoniKer. Thanks

  • 4 years ago · Quote · #8


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

  • 4 years ago · Quote · #9


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

  • 4 years ago · Quote · #10


    this problem should be changed.

  • 4 years ago · Quote · #11

    NM AlexanderKing

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

  • 4 years ago · Quote · #12


    @ 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? :)

  • 3 years ago · Quote · #13


    cool stalemate!

  • 3 years ago · Quote · #14


    gud 1 

  • 3 years ago · Quote · #15


    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 years ago · Quote · #16


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

  • 21 months ago · Quote · #17


    My Score: 2329 (+7)
  • 16 months ago · Quote · #18


    R+B v R is more difficult to hold than R+N v R. That being said, even GMs lose R v R+N on rare occasions against other GMs. So it's not completely trivial.

  • 16 months ago · Quote · #19


    I tried to win like a spaniel.

  • 12 months ago · Quote · #20


    It's is probably really difficult for anyone to win with R+ N.

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