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#19461 by Hajjo lichtenberger

08:02, August 20, 2017 by Hajjo lichtenberger

[Event "?"] [Site "?"] [Date "1946.??.??"] [Round "?"] [White "Kasparian, Genrikh"] [Black "?"] [Result "1-0"] [SetUp "1"] [FEN "6k1/7R/6P1/7P/8/6r1/8/K7 w - - 0 1"] [PlyCount "19"] [Source "ChessCafe/CB"] [SourceDate "2003.10.29"] {For the present, let us accept that White wins if he succeeds in transferring his rook to the 5th rank. This means that Black dare play neither 1...Kf8 2 Rf7+ (with the idea Rf5) nor 1...Rd3 2 Rc7 Rh3 (2...Rd5 3 h6) 3 Rc5. Therefore his rook must stay on g3 and h3. But what can Black do when the white king comes to the kingside? It turns out to be difficult for White. If his king comes to g2 when the black rook is on h3, then 1...Ra3 is playable, because 2 Rb7 Ra5! 3 h6 Rg5+ loses a pawn; the same happens after 2 Kh2 Rb3 3 Ra7 Rb5!. If White plays 2 Kf2 the rook goes back to h3. By the way here, as in all similar positions, 3 h6 Rh5 4 Kf3 Rg5 5 Rg7+ Kh8 6 Kf4 Rf5(g4) +! leads to nowhere. However if, with the white king on f2 and the black rook on h3, Black is on move he comes to be in zugzwang. His rook must leave its comfortable position behind the pawns, and then the white rook has the opportunity to leave h7. We have come to the conclusion that f2 and h3 are the squares of the reciprocal zugzwang. Obviously enough, another pair of such squares is e2 and g3. Furthermore, when the white king stands on any dark square of the 2nd rank the black rook must be on h3 while, when the king stands on a light square, the rook must be on g3!.} 1. Ka2 $3 {A paradoxical move that contradicts the standard approach ("first we move our king to the kingside, and only think thereafter"). It turns out that one should be thinking immediately because any other initial move misses the win.} ({If} 1. Rb7 $2 {then} Rg5 2. Rh7 Rg2 3. Kb1 Rh2 4. Kc1 Rg2 5. Kd1 Rh2 6. Ke1 Rg2 7. Kf1 Rh2 $22 {(the same zugzwang position, only by a rank lower)} 8. Kg1 Ra2 9. Rb7 Ra5 $11) ({After} 1. Kb2 $2 Rh3 2. Kc2 Rg3 3. Kd2 Rh3 4. Ke2 Rg3 5. Kf2 Rh3 $22 {White is in zugzwang:} 6. Kg2 (6. h6 Rh5 7. Kf3 Rg5 8. Rg7+ Kh8 9. Kf4 Rf5+ $1 (9... Rg4+ $1 $11)) 6... Ra3 7. Rb7 Ra5 $11) ({In case of} 1. Kb1 $2 {then the correct method is} Rb3+ $1 {The rook gives checks until the king steps on the 2nd rank, and then goes to a corresponding square.} (1... Rg2 $2 {is erroneous: } 2. Kc1 Rh2 3. Kd1 Rg2 4. Ke1 Rh2 5. Kf1 $22 {and Black is in zugzwang:} Ra2 6. Rb7 $18) 2. Kc1 (2. Kc2 Rg3 $1 $11) 2... Rc3+ $1 3. Kd2 (3. Kd1 Rd3+ $1) 3... Rh3 $1 4. Ke2 Rg3 5. Kf2 Rh3 $22 $11) 1... Rh3 ({It remains for us to prove that White wins if he succeeds in bringing his rook to the 5th rank. This fact is not quite obvious because Black blocks the pawns with his king. However his blockade is less efficient than in the Kling and Horwitz position. } 1... Rd3 2. Rb7 $1 ({But, of course, not} 2. Kb2 $2 Rh3 $1 $11) 2... Rh3 ( 2... Rg3 {can be met simplest by} 3. Kb2 $5 ({But another, more universal way also exists:} 3. Rb3 $5 Rg5 4. Rh3 Kg7 {(otherwise the white king goes ahead)} 5. h6+ Kg8 6. g7 $1 ({Rather than} 6. h7+ $2 Kh8 7. Rh6 Rg1 $11 {and White's king will not have a refuge from rook checks from the rear.}) 6... Kh7 7. Kb3 $18 {White activates his king and gradually wins (a similar position was analyzed by J. Kling and B. Horwitz as long ago as in 1851).}) 3... Rg5 4. Rh7 Rg3 5. Kc2 $22 $18 {and it is Black who turns out to be in zugzwang again.}) 3. Rb5 Kg7 (3... Rg3 4. Rb3) 4. Rg5 $1 $18 {This position is winning for White even without this move (when the black king stands on h6).} Rh4 ({Now Black has neither} 4... Kh6 5. g7 $1 $18) ({nor} 4... Rc3 5. h6+ $1 Kxh6 6. g7) ({In case of} 4... Kg8 5. Kb2 Re3 {White plays} 6. Kc2 $1 ({A hasty} 6. h6 $2 Rh3 7. h7+ Kg7 $11 {on the contrary, leads to a theoretical draw.}) 6... Ra3 7. Kd2 Rb3 8. Ke2 Ra3 9. Kf2 Rb3 10. Rd5 $18) 5. Kb3 Rh1 6. Kc4 Rc1+ 7. Kd5 {The king must go ahead.} ({Nothing can be achieved by} 7. Kd3 Rh1 8. Ke3 Rh3+ 9. Kf2 Rh1 ({Rather than} 9... Rh4 $2 10. Kg2 $22 $18) 10. Kg2 Rh4 $22 {and the rook cannot be forced away from the h-file.}) 7... Rd1+ 8. Kc6 Rc1+ 9. Kd6 Rd1+ 10. Rd5 Ra1 11. Ke7 Ra6 ({White's task is less difficult in case of} 11... Re1+ 12. Kd8 $1 Kg8 (12... Ra1 13. Rd7+ Kg8 14. Ke7 $18) (12... Kh6 13. Rd7 $1 Kxh5 14. g7 Rg1 15. Ke8 Kh6 16. Kf8 $18) 13. h6 $1 (13. Rf5 $5 Re6 14. Kd7 Ra6 15. Rc5 Kg7 16. Rc7 {See 11...Ra6}) 13... Re6 (13... Rh1 14. Ke7 $1) (13... Rg1 14. Rd6 ) 14. h7+ $1 ({But not} 14. Rg5 $2 Ra6 15. Ke7 Rb6 16. h7+ Kg7 17. Rh5 Rb7+ 18. Ke6 Rb6+ $11 {with a draw.}) 14... Kh8 15. Rg5 Kg7 16. Rh5 $18 {Dvoretsky}) 12. Rd7 Rb6 $5 (12... Ra5 13. Ke6+ Kg8 14. h6 $18 {is quite bad.}) ({After} 12... Rc6 {Kasparian gives} 13. Kd8+ Kg8 14. Re7 Kf8 (14... Rd6+ 15. Kc7 Ra6 16. Kd7 -- {and} 17. Re6) 15. Kd7 Ra6 16. Re6 Ra7+ 17. Kd6 Ra6+ 18. Ke5 Ra5+ 19. Kf6 $18) 13. Kd8+ Kg8 (13... Kf8 {is met with} 14. Kc7 $1 (14. Rf7+ Kg8 15. Re7 { See 13...Kg8 14.Re7}) 14... Ra6 15. h6 (15. Rd6 $18) 15... Rxg6 16. Rd8+ {and 17 h7+-.}) 14. Kc7 $1 ({In Theory of Rook Endings by G. Levenfish and V. Smyslov, in the very end of this line, another road to the win is suggested:} 14. Re7 Kf8 (14... Rd6+ 15. Kc7 Ra6 16. Kd7 {/\ Re6+-} Rb6 17. Re6 Rb7+ 18. Kd6 Rb6+ 19. Ke5 Rb5+ 20. Kf6 $18) 15. Rf7+ Kg8 16. Ke7 Ra6 17. Rf6 Ra7+ 18. Ke6 { But this recommendation is erroneous: instead of 18...Ra6+? Black plays} Kg7 $1 (18... Ra6+ 19. Kf5 Ra5+ 20. Kg4 Kg7 21. Rf7+ $18 {Levenfish, Smyslov}) { because after} 19. Rf7+ $140 $2 Rxf7 20. gxf7 Kf8 $22 $11 {he holds a pawn ending despite being two pawns down. (Dvoretsky)}) 14... Ra6 15. Rd6 {/\ h6+-}) 2. Kb2 Rg3 3. Kc2 Rh3 4. Kd2 Rg3 5. Ke2 Rh3 6. Kf2 $22 Ra3 7. Rd7 Rh3 8. Rd5 Kg7 9. Kg2 Rh4 10. Kg3 $18 {This complicated analysis can hardly be (and certainly should not be) remembered in all its details. To know that the rook transfer to the 5th rank wins is quite enough, yet the proof of this fact turns out to be rather complicated.} 1-0

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