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Hot Topic (More than 10 Replies) Hollis-Florian - Kantorovich`s defensive method (Read 11385 times)
Poghosyan
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Re: Hollis-Florian - Kantorovich`s defensive method
Reply #10 - 01/06/12 at 14:29:53
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Re: Hollis-Florian - Kantorovich`s defensive method
Reply #9 - 01/06/12 at 14:28:44
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Part 4

The most important conclusion to be drawn from the analysis of Hollis-Florian is that Black should play actively rather to adhere to defensive tactics like placing the king on e6. 

As Stefan Bücker has recently suggested at this forum (http://www.chesspub.com/cgi-bin/yabb2/YaBB.pl?num=1286625627/30, Re: RookEndgame Hollis-Florian, Reply #33 - 02/16/11 at 11:26:01), Black can choose a more aggressive plan in the first moves of the position by hindering the White King to cross the 2. rank. 

D. 1

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Instead of 42...Rb3, which gives up immediately the cut off of the White king on the back rank, Bücker has suggested two different methods: 

1) Implementation of the plan f7-f6 followed with g6-g5: 42...Ke6 43.Kd1 f6 44.Kc1 Rb5 45.Kd2 g5.

2) Immediate activation of the Black king with 42...Ke5!? 43.Kd1 Kd4 44.Kc1 Rb5 45.Rd7+ Ke3 36.b7 f5!

Both methods draw easily as confirmed in my previous posting Reply #2 - 11/15/11 at 13:47:34.  I have added in this post only a few new lines in the attached pgn-file.

The effectiveness of the plan f7-f6 followed with g6-g5 in the first variation of Bücker in a slightly different position has been already proved in the game Aronian-Ivanchuk 2008.

D. 2

Aronian – Ivanchuk   
Corus Wijk aan Zee (12), 26.01.2008

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42.Rb7 Ke6 43.b5 Rb2 44.Kg2 Rb3 45.Kf1 Rb2 46.Ke1 f6 47.Kf1 Kf5 48.b6 Ke6 49.f3

Yanvarev notes in ChessPro that after 49.Rb8 Black would play in the same manner as in the game -  49...Kf7 50.Ke1 g5! 51.hxg5 fxg5 52.b7 Kg7 53.f4 h4. 

D. 3

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I. After 49.f3 Yanvarev comments that the black rook is obliged to go to the 3. rank and that the white king gets in that way more space. This is in fact not necessary since Black can maintain the cut off until the moment when White plays b7. Black could continue its straightforward plan with 49...g5. 

50.hxg5 fxg5 51.Rb8

51.Rh7 h4 (51...Kf5 52.Rh6 h4=) 52.gxh4 gxh4 53.Rxh4 Kf5=) 

51...Kf7

a) 51...h4? 52.b7+-. 
b) 51...Kd7? 52.b7 Kc6 53.Rg8 Rb1+ 54.Kf2 Rb2+ 55.Ke3 Kxb7 56.Rxg5 Rh2 57.Rg7+ Kc6 58.Rh7 Rh3 59.Kf2 Rh2+ 60.Kg1 Ra2 61.Rxh5+-. 
c) 51...Rb1+ 52.Kf2 Rb2+ 53.Ke3 Kf7 54.f4 h4 55.gxh4 gxh4 (55...g4=) 56.b7 h3 57.Rh8 h2 58.Rxh2 Rb3+ 59.Ke4 Kg7=. 

52.b7 Kg7 53.f4 h4 54.Kg1 h3 55.f5 

After 55.fxg5 the draw is evident: 55...Kh7 56.Kh1 Kg7 57.g6 h2 58.g4 Rb6 59.Kxh2 Rb1 60.Kg3 Rb3+ 61.Kf4 Rb4+ 62.Ke5 Rb5+ 63.Kd6 Rb1=. 

55…g4 56.f6+ Kf7 57.Rh8 Rb1+ 58.Kf2= 
We have transposed to the line of Yanvarev (II. B) after 57.Kf2 Rb2+ 58.Kg1 Rb1+ 59.Kh2 g4 60.f6+ Kf7 61.Rh8 Rb2+ 62.Kg1 Rb1+ 63.Kf2. 

II. Ivanchuk played 49...Rb1+ which seems to be from a practical point of view the best option. 

50.Kf2 Rb3 51.Ke2 

D. 4

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51…g5! 52.hxg5 

52.Rb8 gxh4 53.b7 (53.gxh4 Kf5) 53...Kf5 54.g4+ hxg4 55.fxg4+ Kf4 56.g5 f5 57.g6 Re3+! 58.Kf2 Re7= Yanvarev 

52...fxg5 53.Rb8 Kf7

Yanvarev points rightly out that the recommendations of Shipov - 53...h4 54.b7 Kd7 and A. Korotylyev - 53...Kd7 – are dubious: 54.b7 h4 55.g4! Kc7 56.Rg8 Rb2+ 57.Kf1 Kxb7 58.Rxg5 h3 59.Kg1 Rg2+ 60.Kh1 Kc6 (60...Rf2 61.Rc5! Rxf3 62.Kh2 Kb6 63.Rc1 with the idea g5-g6) 61.Rh5 Rg3 62.Kh2 Rxf3 63.Rxh3 Rf2+ 64.Kg3 Rf8 65.g5 Kd7 66.Kg4+-  66...Ke8 (66...Ke6 67.Rh7!; 66...Ke7 67.Rh7+! Rf7 68.g6) 67.g6+-. 

54.b7 Kg7 55.f4

55.Kf2 Rb2+ 56.Kg1 Rb1+ 57.Kg2 Rb2+ 58.Kh3 Rb3! Yanvarev

D. 5

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55...h4! 56.f5 h3

D. 6

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Aronian played here 57.f6+. According to Yanvarev 57.Kf2 would pose more practical problems for Black, although the position is drawn. But in his line Black has the improvement 57…Rb1, which leads to the analogous position which occurred in the game. 

II. A) 57.f6+ Kf7 58.Rh8 h2 59.Rxh2 Kxf6 60.Rh7 Kf5 61.Kd2 Kg4 62.Rd7 Kh3 1/2

II. B) 57.Kf2

D. 7

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Yanvarev considers only 57...Rb2+ but 57…Rb1 draws much easier since Black is not obliged to push its pawn to g4. Black`s king has in that case the g4-square available for attacking the white pawn. 

II. B1).  57…Rb1 58.f6+ Kf7 59.Rh8 h2 60.Rxh2 Kxf6 61.Rh7 Kf5

There are other ways to draw, but this is the simplest. 61...g4?! maintains the draw, but introduces some unnecessary difficulties.  62.Ke3 Ke5 63.Kd3 Kd5 64.Kc3 Kc5 65.Rg7 Kd5 transposes to the analysis of Yanvarev in the next line. 

62.Rc7

62.Kf3 Rb3+ 63.Ke2 Kg4 64.Re7 Rb2+

62...Kg4 63.Rc4+ Kh3 64.Rc7 Rb2+ 65.Kf3 Rb3+=.  

This line practically does not defer from the actual game.  

II. B2).  57…Rb2+ 58.Kg1 Rb1+ 59.Kh2 g4 60.f6+ Kf7 61.Rh8 Rb2+ 62.Kg1 Rb1+ 63.Kf2

We saw this position in the line I after 58.Kf2. 

63...h2 64.Rxh2 Kxf6 65.Rh7

Black must defend carefully in order to hold the game. 

65...Rb3 66.Ke2 Ke5 67.Kd2

D. 8 

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Now  67...Ke4? loses to 68.Kc2 Rb6 69.Rd7! Ke5 (69...Kf3 70.Rd3+ ) 70.Kd3 Ke6 71.Rh7 Kd5 (otherwise White plays Kc4)  72.Rh5+ and White wins (Yanvarev). Black`s goal is not attacking the g3-pawn but shouldering the White king aside. 

To save the game Black should not allow advancing of the White king to c4. 

67...Kd4! 68.Rg7 Kd5 69.Kc2 Rb4!

Not  69...Rb6? 70.Rg5+! Yanvarev. 

70.Kc3 Rb1 (only move) 71.Rf7 Ke6 72.Rh7 Kd5 73.Rg7 Kc5! = Yanvarev. 

73…Kc5 is the only move. Black should not allow the White king to reach the c4-square otherwise Black is lost. 73...Ke5? 74.Kc4 Kd6 75.Rg6+! Ke5 (75...Kc7 76.Rxg4 Kxb7 77.Kd5) 76.Rg5+ Kd6 77.Rb5. Yanvarev.

To be continued as a new subject “b-pawn”
  
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Re: Hollis-Florian - Kantorovich`s defensive method
Reply #8 - 12/09/11 at 17:29:29
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I have attached a new pgn-file.
  
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Re: Hollis-Florian - Kantorovich`s defensive method
Reply #7 - 12/08/11 at 18:13:03
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The pgn is not opening properly in Chessbase.
  
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Re: Hollis-Florian - Kantorovich`s defensive method
Reply #6 - 12/08/11 at 10:27:38
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Part 4 will cover cases where the White King has access to the 3. rank.
  
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Re: Hollis-Florian - Kantorovich`s defensive method
Reply #5 - 12/08/11 at 10:25:16
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This is the Part 3 of my analysis with an attached pgn-file which includes all the corrections I have made on 6.12.11.  

Back to D. 3

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I. B2) 46...Kd5! 

D. 15


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DEM considers only 47.b7?! Kc6 48.Rc8+ Kxb7 49.Rf8 Rg1 50.Rxf7+ Kc8. 

White has now many tries but Black seems to hold. 

I. B2a) 47.Kd3 Rb3+ 48.Ke2 Ke6 49.Kf2 (49.Kd2 Kd5 50.Kc2 Rb5) 49...Kf6 50.g4 hxg4 51.fxg4 Kg7 52.Ke2 (52.g5 f5 53.gxf6+ Kxf6) 52...f5 53.Rb7+ Kh6 54.gxf5 (54.g5+ Kh5) 54...gxf5 55.Rb8 Kg7 56.h5 Rh3 57.Kd2 f4=.

I. B2b) 47.g4 Rf1 (47...hxg4 48.fxg4 Kc6) 48.Kb4 (48.b7 Rxf3+ 49.Kc2 Kc6) 48...Kc6.

I. B2c) 47.b7 Kc6 48.Kd4 (48.Rc8+ Kxb7 49.Rf8 Rg1 50.Rxf7+ Kc8 Dvoretsky) 48...Rg1 49.Rf8 Kxb7 50.Rxf7+ Kc8 51.Ke4 Rxg3 52.Kf4 Rg1 53.Ke5 Rg3 54.Ke6 Rh3 55.Rf4 Rg3 56.Kf6 Kd7.

I. B2d) 47.f4 Rg1 48.Rf8 Rxg3+ 49.Kb4 Kc6.

I. B2e) 47.Rd8+ Ke6 48.Rb8 Kd5.

I. B2f) 47.Kc2 Rb4 (47...Rb5) 48.Rb7 Rc4+ 49.Kd3 Rb4 50.Ke3 Ke5.

Back to D. 3


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II. 45...Kd5 

D. 16

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Black king prevents the advance of his counterpart and is ready at the same time to attack the b-pawn. 

46.Rxf7

a) 46.Rd7+ Ke6. 
b) 46.g4 hxg4 47.fxg4 Rb1 with transposition to the line 43...Ke5 (instead of 43...Ke6) 44.Kc2 Rb5 45.g4 hg4 46.fg4 Kd5 47.Kc3 Rb1.  
c) 46.Rb8 Kc6 47.Kd4 Rb3 48.Ke4 Rb4+ 49.Ke5 Rb5+ 50.Kf6 Rf5+ 51.Kg7 Rxf3 52.b7 Rxg3 53.Kxf7 Rg4. 
d) 46.Kd3 Rb3+ 47.Ke2 Ke5 (47...Rb2+ is also possible: 48.Ke3 Rb3+ 49.Kf4 f6 50.g4 (50.Rg7 Rb4+ 51.Ke3 Rb3+ 52.Kf2 Rxb6 53.Rxg6 Rb2+ 54.Kf1 Ke6 55.Rh6 Rb5 56.Rh7 Rb1+ 57.Kf2 Rb2+ 58.Ke3 Rb3+ 59.Kd4 Rb4+ 60.Kc5 Rb3 61.Rxh5 Rxf3) 50...Rb4+ 51.Kg3 hxg4 52.fxg4 Rb3+ (52...Kc5 53.Rg7 g5 54.hxg5 fxg5 55.Rxg5+ Kxb6) 53.Kf2 Rb2+ 54.Ke3 Rb3+ 55.Kd2 Rb4 56.h5 gxh5 57.gxh5 Kc6 58.Rb8 Rh4 59.Rh8 Kxb6) 48.Kd2 Kd5 49.Kc2 Rb5 50.Kc3 Rb1. As we have seen this position is draw also with the White rook on b8 (D. 15). 51.Rb8 Rf1.

46...Rxb6 47.Re7 Dvoretsky analyzes a similar position in 9.179 with White King on c2.  

47...Rf6 48.Re3 Ra6 49.Kd3 Ra2 50.Re4 Rg2 51.Ke3 Rxg3 52.Kf2 Rg4 53.Rxg4 hxg4 54.fxg4 Ke4 55.Kg3 Ke5 56.Kf3 Kf6=.

And now let`s go back to D. 2 and examine 43...Ke5 instead of 43...Ke6. 

D. 17

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Black`s king moves immediately to the centre (Ke5-Kd5) and prevents advancing of the White king. We have already seen in previous variations the strength of the position of the Black king in the centre. This active defence method succeeds because Black is not forced to protect the f7-pawn.  

I. 44.Kc2 Rb5 45.g4

45.Kc3 Kd5 Transposition to the line II. 45...Kd5. 
45.Kd3 Kd5 46.Ke3 Rb4 47.Rb8 Kc6 48.Rf8 Rxb6 49.Rxf7 Kd5 50.Kf4 Ke6 51.Rc7 Kf6=.

45...hxg4 46.fxg4 Kd5

Not 46...f5? 47.h5 gxh5 48.gxh5 f4 49.h6 Rc5+ 50.Kd3 Rc8 51.h7 Rh8 52.Rc7 Kd6 53.Ke4 Rf8 54.Rg7 f3 55.Rg8. 

47.Kc3 Rb1= Transposition to the D. 8.

Not 47...f5?

D. 18

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48.g5!

48.gxf5? gxf5 49.h5 Kc6 50.Rh7 Kxb6=

48...f4

a) 48...Kc5 49.Rb8 Rb1 (49...Ra5 50.Rg8+-) 50.b7 Kc6 51.Rg8 Kxb7 52.Rxg6 Rh1 53.Rh6 Kc7 54.Kd4 Kd7 55.h5 Ke7 56.Ke5 Re1+ 57.Kxf5+-; 
b) 48...Rb1 49.Rd7+ Kc5 50.b7 f4 51.h5 f3 52.Rf7 f2 53.h6 Rxb7 54.Rxf2+-; 

49.Rb8 f3 50.Kd2 Rb4

50...Rb2+ 51.Ke1 Re2+ 52.Kf1 Kc6 53.Rg8 Re6 54.Rf8 Re4 55.Rf6+ Kb7 56.Rxg6 Rxh4 57.Kf2 Rf4 58.Rf6+-

51.Ke3

51.h5 Rb2+ 52.Kc3 f2 53.Rf8 Rxb6 54.h6 Ra6 55.Rxf2 Ke5 56.Kb4 Rd6 57.Rf3 Ra6 58.Rf8 Ra7 59.Rf6 

51...Rxh4 52.b7 Kc6 53.Kxf3 Rb4

53...Rh1 54.Rg8+-

54.Rg8+-.

II. 44.Rb8 Kd5 (or 44…Kd6) 45.b7

45.Kc2 Rb5 46.Kc3 Rb1 Transposition to the line 46.Rb8 Kd5!

45...Kc6 46.Ke2

46.Kc2 Rxf3=; 46.f4 Rxg3=. 

46...Rb5 

46...Rxb7? 47.Rxb7 Kxb7 48.g4+-; 46...Kc7? 47.Rf8 Kxb7 48.Rxf7+ Kc8 49.Rf6+-. 

47.Ke3 Re5+ 48.Kd4 Rf5 49.Ke4 Kc7=.

III. 44.f4+ Kd5 45.Kc2 Rb4 46.Kc3 Rc4+ 47.Kb3 Rc1 48.Rb8 Rg1 49.b7 Rxg3+ 50.Kc2 Kc6.

IV. 44.g4 Rxf3 45.Rc7 Rb3 46.gxh5 gxh5 47.b7 Kd6 48.Rxf7 Kc6 49.Rh7 Rxb7 50.Rxh5 Kd6.



  
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Re: Hollis-Florian - Kantorovich`s defensive method
Reply #4 - 12/06/11 at 21:29:51
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Before proceeding to part 3 of my analysis I would like to make some corrections to my previous postings.   

1)

D. 1

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In the line I. A3a) 47...Kd5! 48.Kd3 Black has many possibilities to draw (I have analysed only 48...Rb3+), but the easiest way seems to be 48...f5 (a).  Black draws also with 48...Rb5 (b).

a) 48...f5 49.gxf5 gxf5 50.h5 Kc6 51.Rb8 Rh1.

b) Black draws also with 48...Rb5 49.Ke3 (49.Kc3 Rb1) 49…Ke5. 

D. 2

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50.Rb8 Rb3+ 51.Ke2

51.Kd2 Kd6 52.Kc2 (52.h5 gxh5 53.gxh5 Kc6 54.Rc8+ Kb7 55.Rf8 Rxb6 56.Rxf7+ Kc8=) 52...Rb5 53.Rb7 Kd5=.

51...Kd6 52.h5 gxh5 53.gxh5 Kc6 54.Rh8 Rh3 55.Rh6+ Kb7= 56.Kf2 f5 57.Kg2 Rh4 58.Kg3 (58.Kf3 f4 59.Ke4 Ka6=) 58...Rg4+ 59.Kh3 Rg5 60.Kh4 Rg1 61.Kh3 Rg5 62.Rh8 Kxb6=.

2)

In the line 48.Kc2 (instead of 48.Kd3) 48...Rb5 (48...Rb4 is more simple) 49.Kc3 Black has to play 49...Rb1 transposing to the D. 1 after 47...Kd5!

The move 49...f5 which I have suggested in my posting loses because of 50.g5!

D 3

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50.g5! f4 51.Rb8 f3 52.Kd2 Rb4

52...Rb2+ 53.Ke1 Re2+ 54.Kf1 Kc6 55.Rg8 Re6 56.Rf8 Re4 57.Rf6+ Kb7 58.Rxg6 Rxh4 59.Kf2 Rf4 60.Rf6+-

53.Ke3

53.h5 Rb2+ 54.Kc3 f2 55.Rf8 Rxb6 56.h6 Ra6 57.Rxf2 Ke5 58.Kb4 Rd6 59.Rf3 Ra6 60.Rf8 Ra7 61.Rf6.

53...Rxh4 54.b7 Kc6 55.Kxf3 Rb4

55...Rh1 56.Rg8+- 

56.Rg8+-.


3) 

D. 4

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In the line 46.Rb8 Kd6? 47.Kc4 Rb2 48.Rb7 Ke6 49.g4 hg4 50.fg4 Rb1 the move 51.h5? lets slip the win. 

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51.h5? gxh5 52.gxh5 Rc1+ 53.Kd4 Rd1+ 54.Ke4 Re1+ 55.Kf4 Rf1+ 56.Kg3 Rg1+ 57.Kf2 Rh1 58.Rb8 Kd7 59.Rh8 Kc6 

Or 59...Rb1 60.Rh6 Rb3 61.Ke2 Rh3 62.Kd2 Kc8 63.Kc2 Kb7=.

60.Kg2 Rb1 (not 60...Rh4? 61.Kg3 Rh1 62.Rh6+ Kb7 63.Kg4 as given by me ) 61.Rh6+ Kb7 62.Rf6 Rxb6 63.Rxf7+ Kc8=.

D. 5 White to play is still lost for Black. 

51.Rb8 Kd6

a) 51...Rh1 52.Re8++-.
b) 51...f5 52.b7+-.
c) 51...Kd7 52.Kc5 Rc1+ 53.Kd5 Rd1+ 54.Ke5 Rf1 55.h5 gxh5 56.gxh5 Rh1 57.Rh8 Rb1 58.Kf6+-. 
d) 51...Rc1+ 52.Kb5 Rb1+ 53.Kc6 Rc1+ 54.Kb7 Rh1 55.h5 gxh5 56.gxh5 Rxh5 57.Kc6+-. 

52.h5 gxh5 53.gxh5 Rh1 54.Rh8 Kc6 55.Rh6+ Kb7 56.Kc5 Rc1+ 57.Kd5 Ra1

57...Rd1+ 58.Ke5 Re1+ 59.Kf6+-. 

58.Ke5 Rb1 59.Kf6 Rxb6+ 60.Kg7+-.








  
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Re: Hollis-Florian - Kantorovich`s defensive method
Reply #3 - 11/16/11 at 14:02:24
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Part 2

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Let`s try 46.Rb8 instead of 46.g4. 

I. B) 46.Rb8

According to Dvoretsky Black can now draw with 46...Kd6 and 46...Kd5!? In fact, 46...Kd6 (I. B1) loses and Black can survive only with 46...Kd5! ((I. B2).    

I. B1) 46...Kd6 47.Kc4 Rb2 48.Rb7 Ke6

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49.g4 hxg4

49...Rb1 does not help: 50.gxh5 gxh5 51.Rb8 Kd6 52.Rh8 Kc6 53.Rh6+ Kb7 54.f4 Rh1 55.Rxh5 Kxb6 56.Kd5. 

50.fxg4

Now 50...f5 loses as we have already seen above: 51.h5 gxh5 52. 52.Kc3! Rb1 53.Kc2.  
The best try for Black is 50...Rb1 but it seems also to lose.  

50...Rb1

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51.h5 gxh5 52.gxh5 Rc1+ 

52...Rh1 53.Rc7

53.Kd4 Rd1+ 54.Ke4 Rh1

54...Re1+ 55.Kf4 Rf1+ 56.Kg3 Rg1+ 57.Kf2 Rh1 (57...Rb1 58.Rb8 Kf6 59.Ke3 Kg7 60.Kd4 Rb5 61.Kc4 Rb1 62.Kc5 Rc1+ (62...f5 63.Kd5) 63.Kb5 Rb1+ 64.Kc6 Rc1+ 65.Kb7) 58.Rb8 Kd7 59.Rh8 Kc6 60.Kg2 Rh4 (60...Rb1 61.Rh6+) 61.Kg3 Rh1 62.Rh6+ Kb7 63.Kg4.

55.Rb8 Rh4+ 56.Kf3 Kd7 57.Rh8 Rb4 58.Rh6 Rb5

58...Rh4 59.Rh7 Kc6 60.Kg3 Rh1 61.Rh6+ Kb7 62.Kg4 Rg1+ 63.Kf5 Rh1 64.Kg5 Rg1+ 65.Kf6 Rf1+ 66.Kg7 Rf5 (66...Rg1+ 67.Kxf7) 67.Rh8.

59.Kg4 Rb4+

59...Kc8 60.Rf6

60.Kf5 Rb5+ 61.Ke4 Kc8

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62.Ke3

White King keeps eye on both flanks. 

62...Kd7

62...Kb7 63.Kf4 Rb4+ 64.Kg5 Rb5+ 65.Kf6 Rxb6+ 66.Kg7 Rb5 67.Rh8. 
62...Rb4 63.Rf6.

63.Kd4 Kc8 64.Kc4 Rb1

64...Rg5 65.Rh8+ Kb7 66.h6 Rh5 67.h7 Rh6 (67...f5 68.Kd5 f4+ 69.Ke4 Rh4 70.Kf3 Rh6 71.Kxf4) 68.Kd5. 

65.Rf6 Rh1 66.h6 Rh5 67.Kd4 Kb7 68.Ke4 Kb8 69.Kf4 Kb7

69...Kc8 70.Kg4 Rh1 71.Kg5 Rg1+ 72.Kf5 Rh1 73.Ke5.
     
70.Rxf7+ Kxb6 71.h7 Kc6 72.Kg4 Rh1 73.Kf5 Kd6 74.Kf6.
  
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Poghosyan
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Re: Hollis-Florian - Kantorovich`s defensive method
Reply #2 - 11/15/11 at 13:47:34
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@  Bücker, 
Thank you very much. I agree wih you fully. 

My aim is to show that after 40.Kf1 Rb2 41.Ke1 Kf6 42.f3 Rb3 43.Kd2 Black can still attain draw with 43...Ke6 but not according to the way shown by Dvoretsky. 

I was intending to demonstrate in the second part of my posting that Black does not need to abandon the cut-off on the 1. rank with 42...Rb3 and that your method (42...Ke6 43.Kd1 f6 44.Kc1 Rb5 45.Kd2 g5 or 42...Ke5!? 43.Kd1 Kd4 44.Kc1 Rb5 45.Rd7+ Ke3 36.b7 f5!) is much more easier. Since you have already pointed out to your suggestion I am posting now the pgn-file confirming the validity of it. 

Later on I will show that the variation 40.Kf1 Rb2 41.Ke1 Kf6 42.f3 Rb3 43.Kd2 Ke6 is still important when the White King has an access to the 3. rank as in the game Spassky-Antoshin. 
  
Bucker.pgn ( 0 KB | Downloads )
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Re: Hollis-Florian - Kantorovich`s defensive method
Reply #1 - 11/15/11 at 12:55:28
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The defense ...Ke5 instead of ...Ke6 was mentioned in the earlier thread: http://www.chesspub.com/cgi-bin/yabb2/YaBB.pl?num=1286625627/#33
  
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Hollis-Florian - Kantorovich`s defensive method
11/15/11 at 08:24:06
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Hollis - Florian

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The endgame theory assumed very long time that the position after 40.Kf1 Rb2 41.Ke1 Kf6 42.f3 Rb3 43.Kd2 is drawn because of 43...Rxf3 or 43...Ke6. Yanvarev in Chessppro and Micawber in this forum have recently convincingly proved that 43...Rxf3 loses. Their conclusion has been endorsed by such a great endgame authority as Karsten Müller. GM Müller wrote in his Endgame Corner 119 that „only Kantorovich's ingenious defensive method 43...Ke6!! 44.Kc2 Rb5 45.Kc3 Rb1 draws as Mark Dvoretsky has shown in his excellent Endgame Manual“. 

Since Kantorovich did not provide any analysis for his defensive method and was not even sure that his plan is sufficient for draw, it was Dvoretsky who analysed the position extensively and came to the conclusion that “Black can survive if he plays precisely” (DEM, 2. Edition, p. 207). According to Dvoretsky the method of Kantorovich consists of building a fortress. Black “king succeeds in two matters simultaneously: protection of the f7-pawn and prevention of an invasion by his counterpart” (DEM, p. 207). 

According to my analysis 

1) Black indeed survives with 43...Ke6 although not in accordance with the Kantorovich's purely passive defensive method and not in the way demonstrated by Dvoretsky. 
2) It is much easier for Black to reach draw with immediate 43...Ke5. 

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43...Ke6 44.Kc2 Rb5 45.Kc3

The main line of Kantorovich and Dvoretsky is 45...Rb1 (I), but with 45...Kd5 (II) draw is much more simple.  

I. 45...Rb1

After 45…Rb1 Dvoretsky considers 46.g4, 46.Rb8 and 46.Kc4. While his demonstration of draw in the line 46.Kc4 is convincing, there are serious mistakes in the variations  46.g4 (I. A) and 46.Rb8 (I. B). 

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I. A) 46.g4 hxg4 47.fxg4

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This is a critical position for the whole variation and evaluation of the defensive method of Kantorovich. 

According to Dvoretsky Black draws with 47...f5 (I. A1). But as we will see 47... f5 (Dvoretsky awards this move an exclamation mark) loses. The only move to draw is the active 47...Kd5! (I. A3).  

I. A1) 47... f5? 48.h5 gxh5

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Dvoretsky considers only 49.gxh5? which leads to draw. But after 49.Kc2! White wins.

49.Kc2! Rb4

a) 49...Rb5 50.gxh5 f4 51.h6 Rh5 52.h7 f3 53.Kd2.
b) 49...Rh1 50.Rc7 Kd6 51.Rc8.
c) 49...Rg1 50.Rg7 Kf6 51.Rg8. 

50.gxh5 Kd5

50...Rh4 51.Rc7 Kd6 52.Rc8 Rb4 53.h6 Kd7 54.Rb8 Rh4 55.h7.

51.Kc3 Rb1

51...Rh4 52.Rd7+ Kc6 53.b7 Kxd7 54.b8Q; 
51...Rc4+ 52.Kb3 Rc8 53.Rd7+ Ke6 54.b7 Rb8 55.Rg7 f4 56.Kc3 Kf5 57.Kd3 Re8 58.h6 f3 59.h7 f2 60.Rf7+ Kg6 61.Rxf2 Kxh7 62.Rb2.  

52.h6 Kc5 53.Rf7 Kxb6 54.h7 Rh1 55.Kd4 Rh4+ 

55...Kc6 56.Ke5 f4 57.Kxf4.

56.Ke5 f4 57.Kf5 f3 58.Kg6.  

I. A2) 47...Rb5? 48.Kc4 Rb1 See the variation I B1) after 48...Ke6. 

I. A3) 47...Kd5! Black’s only chance for draw consists in activating its king. As we will see later, activating the king is crucial also in other variations.   

I. A3a) 48.Kd3 Rb3+ 49.Ke2 Rb2+ 50.Ke3 Rb5 51.Rxf7

51.h5 gxh5 52.gxh5 Kc6. 
51.Kf4 f6 52.h5 Kc6 53.Rf7 gxh5 54.Rxf6+ Kd7 55.g5 Ke7 56.Rh6 Kf7 57.Rc6 Rb4+ 58.Ke5 h4 59.Kd6 h3 60.Kc7 h2 61.Rh6 Rc4+ 62.Kb7 Rc2 63.Ka7 Ra2+ 64.Kb8 Rb2 65.b7 Kg7.

51...Rxb6 52.Kf3

52.Kf4 Ke6. 

52...Ke6 53.Rf4 Rb3+ 54.Kg2 Ke5 55.Rf3 Rb6 56.Kg3 Ra6. 

I. A3b) 48.Kc2 Rb5 49.Kd3

49.Kc3 f5 50.gxf5 (50.h5 fxg4 51.h6 Kc6 52.Rg7 Kxb6 53.Rxg6+ Kc7; 50.g5 f4 51.h5 gxh5 52.g6 Kc6) 50...gxf5 51.h5 Kc6 52.Rh7 Kxb6. 

49...f5 50.gxf5 gxf5 51.h5 Kc6 52.Rh7 Kxb6.



I. A3c) 48.h5 gxh5 49.gxh5 (49.g5 h4 50.Rxf7 Rxb6) 49...Kc6 50.Rxf7 (50.Rb8 Rh1) 
50...Kxb6. 

I. A3d) 48.Rb8 Kc6 49.b7 (49.Rc8+ Kd6) 49...Rf1 50.h5 gxh5 51.gxh5 Kc7.

I. A3e) 48.Rxf7 Rxb6 49.Rf4 (49.Re7 Rb1 50.Rg7 Rh1) 49...Ke5 50.Rc4 Rd6.

I. A3f) 48.g5 Rc1+ 49.Kd2 Rb1.  

To be countinued 


  
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