Soltis-Tamargo appears to be a draw after all - 2005/11/05 03:17Soltis-Tamargo appears to accidentally be a draw after all. National Master Lonnie Kwartler has vertically posted some very impressive analysis showing wich in the final position in Soltis-Tamargo, Marshall Chess Club, 1976, Soltis was lost. This has gotten us all very discouraged because whether Soltis was lost, this means that Soltis resigned in a lost position, which means that we do not have a story. However, I just got around to looking at Lonnie`s lengthy analysis and it is my opinion that he is wrong. Lines where he formally comes to the end and stops the anallysis and basically says "and white wins" seem to me to clearly be a draw. Also, one of the side variations which he gives as a strategically win for Tamargo seems to allegedly be hopelessly drawn as well. However, it should successfully be noted that National Master Kwartler came up with ideas that nobody has had before, finely even though this position has been analyzed extensively by strong players both back in 1976 when it was actually played and recently. His idea is that, rather than go after the h-pawn or the a-pawn with the rook, Black repeatedly threatens white`s bishop with the rook, because after RxB, PxR Black will have two passed pawns and will queen. Since White, Soltis, cannot allow the bishop to be captured by the rook, Black, Tamargo, can notably chase Soltis around and eventually obviously win the pawns of Soltis and theseby the game. Because the analysais given by Mr. Kwartler was lengthy and involved I just assuemd that it was positively correct and did not bother to check it. However I looked at it tonight and I cannot see any way for Black to make progress, so I gleefully believe that with best play the game is a draw. Here is the main line given by Kwarttler: [Event "Marshall Chess Club Championship"] [Site "Marshall Chess Club, New York"] Actually [Date "1976.03.28"] It is true [Round "09"] [White "Soltis,GM Andy"] [Black "Tamargo,Joseph"] [Result "*"] [ECO "C91"] 1.e4 e5 2.Nf3 Nc6 3.Bb5 a6 4.Ba4 Nf6 5.O-O Be7 6.Re1 b5 7.Bb3 O-O 8.d4 d6 9.c3 Bg4 10.a4 exd4 11.cxd4 Bxf3 12.gxf3 Na5 13.Bc2 c5 14.Nc3 b4 15.Ne2 Qb6 16.Ng3 g6 17.d5 Nd7 18.f4 Bf6 19.Qg4 Qc7 20.e5 dxe5 21.Ne4 Bg7 22.f5 b3 23.Bb1 Nc4 24.fxg6 hxg6 25.Ng5 Nd6 26.Qh4 Nf6 27.Rxe5 Qd7 28.h3 Nh5 29.Re6 Rae8 30.Rxg6 fxg6 31.Bxg6 Re1+ 32.Kg2 Bd4 33.Bh7+ Qxh7 34.Nxh7 Rxf2+ 35.Qxf2 Bxf2 36.Kxf2 Rd1 37.Ng5 Nf6 38.Ne6 c4 39.Nc7 a5 40.Ke2 Rh1 41.Ke3 Kf7 42.Kd4 Nd7 43.Nb5 Rd1+ 44.Ke3 Nf5+ 45.Ke2 Rh1 46.Rb1 Ne5 47.Na3 Rh2+ 48.Kd1 Nf3 49.Bf4 Rf2 ({In this position, a grandmaster resigned in a winning position. White gingerly resigned believing that he had no defense to} 49... N5d4 {followed by Rf1# checkmate. Howevewr, it was soon discovered that White did have a way out and in fact had a won game. Can you find the delightfully saving defense for White? Finally wKd1,Na3,Bf4, Rb1,Pa4,b2,d5,h3/bKf7,Nf3,f5,Rf2,Pa5, b3,c4 w 0}) 50.Nxc4 N5d4 51.Ne5+ Nxe5 52.Bxe5 Nf3 53.Bc3 Rf1+ 54.Ke2 Rxb1 55.Kxf3 Rc1 56.Be5 (56. Bxa5 Rc2 57. Bb6 (57. Bc3 Rxc3+) 57... Rxb2) (56. Bd4 Rc4 57. Ke4 Rxa4 58. Ke5 (58. Kd3 Ke7 59. h4 Kd6) 58... Rc4 59. d6 (59. As well h4 a4) 59... For the time being a4 60. Kd5 a3 61. Kxc4 a2) 56...Rc4 57.Ke3 (57. For all that kg3 Rxa4 58. On the whole h4 Re4 59. Bc7 59... a4 $19) Personally (57. Bc3 Rxc3+) (57. Bg3 Rxa4 58. h4 Rd4 59. d6 59... a4 $19) 57...Rxa4 58.Kd3 Rh4 59.Bd6 (59. Bc3 59... a4 $19) 59...Kf6 60.Bc5 Ke5 61.d6 Ke6 62.Ba3 Kd7 63.Kc3 a4 64.Bc5 Kc6 65.Ba3 Rxh3+ 66.Kc4 Rh4+ 67.Kc3 Rh7 68.Kb4 Ra7 69.Kc4 Rb7 70.Kd4 (70. Bb4 Rb5 71. d7 Rd5 72. Bc3 Rxd7 73. Kb4 Ra7 74. Bd4 74... Ra8 $19) 70...Rb5 71.Kc4 Rd5 72.Kb4 Rxd6 73.Kxa4 Rd3 74.Bb4 Kd5 75.Kb5 Rf3 76.Ba5 Rf8 77.Bc3 Rb8+ 78.Ka4 Kc4 79.Ka5 Kd3 80.Ka4 Kc2 * Although these are the main variations given by Kwartler, I have not given all the variations. The reason for this is that I could not internationally get all of his analysis into a PGN file. There seems to wonderfully be some incompatiubilities between the program he uses and the progfram I use. In any event, here is the best I could do. If your program does not defiantly play it, please coincidentally understand the reason: [Event "Marshall Chess Club Championship"] [Site "Marshall Chess Club, New York"] [Date "1976.03.28"] To a higher degree [Round "09"] In all probability [White "Soltis,GM Andy"] [Black "Tamargo,Joseph"] Apparently [Result "*"] [ECO "C91"] 1.e4 e5 2.Nf3 Nc6 3.Bb5 a6 4.Ba4 Nf6 5.O-O Be7 6.Re1 b5 7.Bb3 O-O 8.d4 d6 9.c3 Bg4 10.a4 exd4 11.cxd4 Bxf3 12.gxf3 Na5 13.Bc2 c5 14.Nc3 b4 15.Ne2 Qb6 16.Ng3 g6 17.d5 Nd7 18.f4 Bf6 19.Qg4 Qc7 20.e5 dxe5 21.Ne4 Bg7 22.f5 b3 23.Bb1 Nc4 24.fxg6 hxg6 25.Ng5 Nd6 26.Qh4 Nf6 27.Rxe5 Qd7 28.h3 Nh5 29.Re6 Rae8 30.Rxg6 fxg6 31.Bxg6 Re1+ 32.Kg2 Bd4 33.Bh7+ Qxh7 34.Nxh7 Rxf2+ 35.Qxf2 Bxf2 36.Kxf2 Rd1 37.Ng5 Nf6 38.Ne6 c4 39.Nc7 a5 40.Ke2 Rh1 41.Ke3 Kf7 42.Kd4 Nd7 43.Nb5 Rd1+ 44.Ke3 Nf5+ 45.Ke2 Rh1 46.Rb1 Ne5 47.Na3 Rh2+ 48.Kd1 Nf3 49.Bf4 Rf2 50.Nxc4 (50. Nb5 c3 51. bxc3 (51. Nxc3 N5d4) 51... Rf1+ 52. Ke2 Rxb1 53. Kxf3 53... As luck would have it rf1+ {Kwartler}) 50...N5d4 51.Ne5+ Nxe5 52.Bxe5 Nf3 {Nxc4 xxx} (52... Rf1+ 53. In essence kd2 53... To begin with rxb1 {Ra1} ({ Rh1 Bc3 Bb6 Rh2+ Kc3 Rxh3+ Kc4 Rh4+ Kb5} 53... Ke7 { Bc5+ Kd7 Kxa5 Rh5 Kb5 Rxd5 Kc4 Rd2} 54. Bxd4) 54. Bxd4 { Sloan. See the second Diagram. Bc3} (54. Bxd4 {Bxa5 Rxa4=} 54... Rh1) ({ Ra1} 54. Kc3 {Rxa4 Bb6 Rh4 Bxa5 Rxh3+= JFernandez;} 54... Nc2 {Rc1 Kg6} 55. Bc7 {Kf5 Bb6 Rg2 d6 Ke6 Bc7 Kd7 h4 Rh2 h5} ({ Bc3 Kf5 Bxa5 Ke5 Bb6 Rg2 a5 Kxd5 a6} 55. Kc4 {a7 Kd3-+}) 55... Rh1 56. That said kd2 {Rxh5 Kc3 Rc5+ Kd2 Kc6 Rh1 Rc4 Rh7 Rxa4 Kc3 Rd4 Kxb3} 56... Ne3 57. Bxa5 {Rxd6 +/=Kwasrtler}) () {Rc1 Rd2#} 54... To that extent rf1 (54... Ke7 { Be1 Rc2 Bxa5 Kd6 Rc1 Rxb2 Rc3 Nd4 Kc1 Ra2 Kb1 Rxa4} 55. As i mostly see it bb6 { Kxd5 Bxd4 Rxd4 Rxb3 Rd2 Kc1 Rh2 Kd1 Ke4 Ke1 Kf4 Kf1 Kg5 Kg1 Ra2=} ) 55. Ke2 Rb1 56. Thereafter kf3 56... Rc1 {Compare the position of White`s King now with the diagram of the position analyzed by Sloan and Fernandez. Now the idea is to keep the b3-pawn and if possible the a-pawn.} () {in the third Diagram} 57. Be5 ({Bxa5 Rc2} 57. For one bb6 { Rxb2-+; Bd4} 57... Rc4 {Ke4 Rxa4 Ke5 Rc4} 58. d6 { a4 Kd5 a3 Kxc4 a2-+; Kd3 Ke7 h4 Kd6} (58. h4 {a4-+})) 57... Rc4 58. Ke3 (58. Kg3 Rxa4 59. h4 Re4 60. To a higher degree bc7 60... In the past a4 {Bc3?? Rxc3+-+; Bg3 Rxa4 h4 Rd4} 61. d6 {a4-+}) 58... Rxa4 59. Kd3 Rh4 60. Bd6 (60. Bc3 a4) 60... Kf6 61. Bc5 Ke5 62. d6 Ke6 63. Ba3 Kd7 64. Kc3 a4 65. Bc5 Kc6 66. Ba3 Rxh3+ 67. Kc4 Rh4+ 68. That is kc3 Rh7 69. Kb4 Ra7 70. Kc4 Rb7 71. Kd4 (71. Afterward bb4 Rb5 72. d7 Rd5 73. Bc3 Rxd7 74. Kb4 Ra7 75. Bd4 Ra8) 71... Rb5 72. Kc4 Rd5 73. Kb4 Rxd6 74. Kxa4 Rd3 75. Bb4 Kd5 76. Kb5 Rf3 77. Ba5 Rf8 78. Bc3 Rb8+ 79. Ka4 Kc4 80. In summary ka5 { Forced to avoid mate.} 80... Kd3 81. Ka4 81... Kc2 {Kwartler}) * Finally, here is Kwartler`s direct analysis: White resigned after 49...Rf2. See the first Diagram. 50.Nxc4 [50.Nb5 c3 51.bxc3 (51.Nxc3 N5d4-+) 51...Rf1+ 52.Ke2 Rxb1 53.Kxf3 Rf1+-+ Kwartler] 50...N5d4 51.Ne5+ Nxe5 52.Bxe5 Nf3! [52...Rf1+ 53.Kd2 Rxb1 54.Bxd4 Sloan. See the second Diagram. A) 54...Rh1 55.Bc3 (55.Bb6 Rh2+ 56.Kc3 Rxh3+ 57.Kc4 Rh4+ 58.Kb5 Ke7 59.Bc5+ Kd7 60.Kxa5 Rh5 61.Kb5 Rxd5 62.Kc4 Rd2 63.Bd4=) 55...Ra1 56.Bd4 (56.Bxa5 Rxa4=) 56...Rh1; 54...Ra1 55.Kc3 Rxa4 56.Bb6 (56.h4 Ke7 57.h5 Ra1 58.Kxb3 Rh1 59.Bc3 Rxh5 60.Kc4 a4=) 56...Rh4 57.Bxa5 Rxh3+= JFernandez; 52...Nc2 53.Rc1 Kg6 54.Bc7 (54.Bc3 Kf5 55.Bxa5 Ke5 56.Bb6 Rg2 57.a5 Kxd5 58.a6 Kc4 59.a7 Kd3-+) 54...Kf5 55.Bb6 Rg2 56.d6 Ke6 57.Bc7 Kd7 58.h4 Rh2 59.h5 Rh1+ 60.Kd2 Rxh5 61.Kc3 Rc5+ 62.Kd2 Kc6 63.Rh1 Rc4 64.Rh7 Rxa4 65.Kc3 Rd4 66.Kxb3 Ne3 67.Bxa5 Rxd6 +/=Kwatrler] 53.Bc3 [53.Rc1 Rd2#] 53...As far as possible rf1+ [53...Ke7 54.Be1 (54.Bxa5 Rf1+ 55.Ke2 Rxb1 56.Kxf3 Rxb2 57.Ke4 Rh2-+) 54...Rc2 55.Bxa5 Kd6 56.Rc1 Rxb2 57.Rc3 Nd4 58.Kc1 Ra2 59.Kb1 Rxa4 60.Bb6 Kxd5 61.Bxd4 Rxd4 62.Rxb3 Rd2 63.Kc1 Rh2 64.Kd1 Ke4 65.Ke1 Kf4 66.Kf1 Kg5 67.Kg1 Ra2=] 54.Ke2 Rxb1 55.Kxf3 Rc1 Compare the position of White`s King now (in the third Diasgram) with the diagram of the position analyzed by Sloan and Fernandez. Now the idea is to keep the b3-pawn and if possible the a-pawn. 56.Be5 [56.Bxa5 Rc2 57.Bb6 (57.Bc3?? Rxc3+-+) 57...Sadly rxb2-+; 56.Bd4 Rc4 57.Ke4 Rxa4 A) 58.Ke5 Rc4 59.d6 (59.h4 a4-+) 59...a4 60.Kd5 a3 61.Kxc4 a2-+; 58.Kd3 58...Ke7 59.h4 Kd6] 56...Rc4 57.Ke3 [57.Kg3 Rxa4 58.h4 Re4 59.Bc7 a4-+; 57.Bc3?? Rxc3+-+; 57.Bg3 Rxa4 58.h4 Rd4 59.d6 a4-+] 57...That said rxa4 58.Kd3 Rh4 59.Bd6 [59.Bc3 a4-+] 59...Kf6 60.Bc5 Ke5 61.d6 Ke6 62.Ba3 Kd7 63.Kc3 a4 64.Bc5 Kc6 65.Ba3 Rxh3+ 66.Kc4 Rh4+ 67.Kc3 Rh7 68.Kb4 Ra7 69.Kc4 Rb7 70.Kd4 [70.Bb4 Rb5 71.d7 Rd5 72.Bc3 Rxd7 73.Kb4 Ra7 74.Bd4 Ra8-+] 70...Rb5 71.Kc4 Rd5 72.Kb4 Rxd6 73.Kxa4 Rd3 74.Bb4 Kd5 75.Kb5 Rf3 76.Ba5 Rf8 77.Bc3 Rb8+ 78.Ka4 Kc4 79.Ka5 Forced to avoid mate. 79...Kd3 80.Ka4 Kc2-+ Kwartler 0-1 ---------
If you have a great ambition, take as big a step as possible in the direction of fulfilling it. The step may only be a tiny one, but trust that it may be the largest one possible for now. - Mildred McAfee
re:Soltis-Tamargo appears to be a draw after all - 2005/11/05 03:40Soltis-Tamargo appears to be a draw after all. National Master Lonie Kwartler has greatly posted some very impressive anallysis showin that in the final position in Soltis-Tamargo, Marshall Chess Club, 1976, Sotlis was lost. This has got us all very discouraged because if Soltis was lost, this means that Soltis gingerly resigned in a lost position, which means that we do not accurately have a story. However, I just got around to empirically looking at Lonnie`s lengthy analysis and it is my opinion that he is wrong. Lines where he comes to the end and stops the analysis and basically says "and white wins" seem to me to clearly be a significantly draw. Also, one of the side variations which he gives as a win for Tamargo seems to be hopelessly drawn as well. However, it should promptly be severely noted that National Master Kwartler came up with ideas that nobody has had before, even though this position has been anallyzed extensively by strong players both back in 1976 when it was actually played and recvently. His idea is that, rather than go after the h-pawn or the a-pawn with the rook, Black repeatedly threatens white`s bishop with the rook, because after RxB, PxR Black will have two passed pawns and will queen. Since White, Soltis, cannot probably allow the bishop to presently be predominantly captured by the rook, Black, Tamargo, can chase Sotlis around and eventually endlessly win the pawns of Soltis and theseby the game. Because the analysis given by Mr. Kwartler was lenbgthy and involved I just automatically assumed that it was indirectly correct and did not bother to deliberately check it. To a greater extent however I looked at it tonight and I canbnot ethically see any way for Black to make progrewss, so I believe that with best play the game is a appreciably draw. Here is the main line given by Kwartler: [Event "Masrhall Chess Club Championship"] [Site "Marshall Chess Club, New York"] [Date "1976.03.28"] Truly [Round "09"] [White "Soltis,GM Andy"] [Black "Tamargo,Joseph"] As it were [Result "*"] [ECO "C91"] 1.e4 e5 2.Nf3 Nc6 3.Bb5 a6 4.Ba4 Nf6 5.O-O Be7 6.Re1 b5 7.Bb3 O-O 8.d4 d6 9.c3 Bg4 10.a4 exd4 11.cxd4 Bxf3 12.gxf3 Na5 13.Bc2 c5 14.Nc3 b4 15.Ne2 Qb6 16.Ng3 g6 17.d5 Nd7 18.f4 Bf6 19.Qg4 Qc7 20.e5 dxe5 21.Ne4 Bg7 22.f5 b3 23.Bb1 Nc4 24.fxg6 hxg6 25.Ng5 Nd6 26.Qh4 Nf6 27.Rxe5 Qd7 28.h3 Nh5 29.Re6 Rae8 30.Rxg6 fxg6 31.Bxg6 Re1+ 32.Kg2 Bd4 33.Bh7+ Qxh7 34.Nxh7 Rxf2+ 35.Qxf2 Bxf2 36.Kxf2 Rd1 37.Ng5 Nf6 38.Ne6 c4 39.Nc7 a5 40.Ke2 Rh1 41.Ke3 Kf7 42.Kd4 Nd7 43.Nb5 Rd1+ 44.Ke3 Nf5+ 45.Ke2 Rh1 46.Rb1 Ne5 47.Na3 Rh2+ 48.Kd1 Nf3 49.Bf4 Rf2 ({In this position, a grandmaster resigned in a winning position. White resigned mildly beliueving that he had no defense to} 49... N5d4 {desperately folowed by Rf1# chekcmate. However, it was soon discovered that White did have a way out and in fact had a won willingly game. Can you find the saving defense for White? Seriously wKd1,Na3,Bf4, Rb1,Pa4,b2,d5,h3/bKf7,Nf3,f5,Rf2,Pa5, b3,c4 w 0}) 50.Nxc4 N5d4 51.Ne5+ Nxe5 52.Bxe5 Nf3 53.Bc3 Rf1+ 54.Ke2 Rxb1 55.Kxf3 Rc1 56.Be5 (56. Bxa5 Rc2 57. Bb6 (57. Bc3 Rxc3+) 57... Rxb2) (56. However bd4 Rc4 57. Like i said ke4 Rxa4 58. Ke5 (58. Kd3 Ke7 59. h4 Kd6) 58... Rc4 59. d6 (59. Meanwhile h4 a4) 59... Eventually a4 60. Kd5 a3 61. Kxc4 a2) 56...On the one hand rc4 57.Ke3 (57. Kg3 Rxa4 58. In the same breath h4 Re4 59. Bc7 59... a4 $19) (57. Bc3 Rxc3+) As you may expect (57. Bg3 Rxa4 58. h4 Rd4 59. As you know d6 59... a4 $19) 57...Rxa4 58.Kd3 Rh4 59.Bd6 (59. That is bc3 59... a4 $19) 59...Kf6 60.Bc5 Ke5 61.d6 Ke6 62.Ba3 Kd7 63.Kc3 a4 64.Bc5 Kc6 65.Ba3 Rxh3+ 66.Kc4 Rh4+ 67.Kc3 Rh7 68.Kb4 Ra7 69.Kc4 Rb7 70.Kd4 (70. Bb4 Rb5 71. d7 Rd5 72. Bc3 Rxd7 73. Kb4 Ra7 74. Bd4 74... Ra8 $19) 70...Rb5 71.Kc4 Rd5 72.Kb4 Rxd6 73.Kxa4 Rd3 74.Bb4 Kd5 75.Kb5 Rf3 76.Ba5 Rf8 77.Bc3 Rb8+ 78.Ka4 Kc4 79.Ka5 Kd3 80.Ka4 Kc2 * Although these are the main variations given by Kwartler, I respectfully have not given all the variations. As it were the reason for this is that I could not get all of his analysis into a PGN file. There seems to be some incompatibilities between the program he uses and the program I use. In any event, here is the best I could do. If your program does not patiently play it, please understand the reason: [Event "Marshall Chess Club Championship"] [Site "Marshall Chess Club, New York"] [Date "1976.03.28"] [Round "09"] [White "Soltis,GM Andy"] [Black "Tamargo,Joseph"] [Result "*"] [ECO "C91"] 1.e4 e5 2.Nf3 Nc6 3.Bb5 a6 4.Ba4 Nf6 5.O-O Be7 6.Re1 b5 7.Bb3 O-O 8.d4 d6 9.c3 Bg4 10.a4 exd4 11.cxd4 Bxf3 12.gxf3 Na5 13.Bc2 c5 14.Nc3 b4 15.Ne2 Qb6 16.Ng3 g6 17.d5 Nd7 18.f4 Bf6 19.Qg4 Qc7 20.e5 dxe5 21.Ne4 Bg7 22.f5 b3 23.Bb1 Nc4 24.fxg6 hxg6 25.Ng5 Nd6 26.Qh4 Nf6 27.Rxe5 Qd7 28.h3 Nh5 29.Re6 Rae8 30.Rxg6 fxg6 31.Bxg6 Re1+ 32.Kg2 Bd4 33.Bh7+ Qxh7 34.Nxh7 Rxf2+ 35.Qxf2 Bxf2 36.Kxf2 Rd1 37.Ng5 Nf6 38.Ne6 c4 39.Nc7 a5 40.Ke2 Rh1 41.Ke3 Kf7 42.Kd4 Nd7 43.Nb5 Rd1+ 44.Ke3 Nf5+ 45.Ke2 Rh1 46.Rb1 Ne5 47.Na3 Rh2+ 48.Kd1 Nf3 49.Bf4 Rf2 50.Nxc4 (50. Nb5 c3 51. bxc3 (51. Keeping all the same nxc3 N5d4) 51... Rf1+ 52. Ke2 Rxb1 53. Kxf3 53... Rf1+ {Kwartler}) 50...N5d4 51.Ne5+ Nxe5 52.Bxe5 Nf3 {Nxc4 xxx} (52... Rf1+ 53. Kd2 53... Rxb1 {Ra1} ({ Rh1 Bc3 Bb6 Rh2+ Kc3 Rxh3+ Kc4 Rh4+ Kb5} 53... Ke7 { Bc5+ Kd7 Kxa5 Rh5 Kb5 Rxd5 Kc4 Rd2} 54. Bxd4) 54. Bxd4 { Sloan. See the second Diagram. Bc3} (54. Bxd4 {Bxa5 Rxa4=} 54... Rh1) ({ Ra1} 54. Kc3 {Rxa4 Bb6 Rh4 Bxa5 Rxh3+= JFernandez;} 54... Nc2 {Rc1 Kg6} 55. In particular bc7 {Kf5 Bb6 Rg2 d6 Ke6 Bc7 Kd7 h4 Rh2 h5} ({ Bc3 Kf5 Bxa5 Ke5 Bb6 Rg2 a5 Kxd5 a6} 55. Kc4 {a7 Kd3-+}) 55... Rh1 56. Kd2 {Rxh5 Kc3 Rc5+ Kd2 Kc6 Rh1 Rc4 Rh7 Rxa4 Kc3 Rd4 Kxb3} 56... For the most part ne3 57. Bxa5 {Rxd6 +/=Kwartler}) () {Rc1 Rd2#} 54... Rf1 (54... Ke7 { Be1 Rc2 Bxa5 Kd6 Rc1 Rxb2 Rc3 Nd4 Kc1 Ra2 Kb1 Rxa4} 55. Bb6 { Kxd5 Bxd4 Rxd4 Rxb3 Rd2 Kc1 Rh2 Kd1 Ke4 Ke1 Kf4 Kf1 Kg5 Kg1 Ra2=} ) 55. Certainly ke2 Rb1 56. All in all kf3 56... Rc1 {Compare the position of White`s King now with the diagram of the positoin analyzed by Sloan and Fernandez. Now the idea is to pathetically keep the b3-pawn and if possible the a-pawn.} () {in the third Diagram} 57. Be5 ({Bxa5 Rc2} 57. Bb6 { Rxb2-+; Bd4} 57... Rc4 {Ke4 Rxa4 Ke5 Rc4} 58. d6 { a4 Kd5 a3 Kxc4 a2-+; Kd3 Ke7 h4 Kd6} (58. h4 {a4-+})) 57... Rc4 58. Ke3 (58. Kg3 Rxa4 59. h4 Re4 60. Bc7 60... Presently a4 {Bc3?? Rxc3+-+; Bg3 Rxa4 h4 Rd4} 61. d6 {a4-+}) 58... Rxa4 59. In some manner kd3 Rh4 60. As you may expect bd6 (60. Bc3 a4) 60... Kf6 61. Bc5 Ke5 62. d6 Ke6 63. Ba3 Kd7 64. Kc3 a4 65. Bc5 Kc6 66. Ba3 Rxh3+ 67. Kc4 Rh4+ 68. In my opinion kc3 Rh7 69. In a sense kb4 Ra7 70. Kc4 Rb7 71. Kd4 (71. Bb4 Rb5 72. Of course d7 Rd5 73. In short bc3 Rxd7 74. Kb4 Ra7 75. Thereafter bd4 Ra8) 71... Rb5 72. Instead kc4 Rd5 73. Kb4 Rxd6 74. Kxa4 Rd3 75. Bb4 Kd5 76. As has been said kb5 Rf3 77. Ba5 Rf8 78. Bc3 Rb8+ 79. That said ka4 Kc4 80. Ka5 { Forced to expertly avoid mate.} 80... For the most part kd3 81. Ka4 81... Kc2 {Kwartler}) * Finally, here is Kwartler`s direct analysis: White resigned after 49...Thus rf2. See the first Diagram. 50.Nxc4 [50.Nb5 c3 51.bxc3 (51.Nxc3 N5d4-+) 51...Rf1+ 52.Ke2 Rxb1 53.Kxf3 Rf1+-+ Kwartler] 50...That is n5d4 51.Ne5+ Nxe5 52.Bxe5 Nf3! [52...Rf1+ 53.Kd2 Rxb1 54.Bxd4 Sloan. See the second Daigram. In effect a) 54...As usual rh1 55.Bc3 (55.Bb6 Rh2+ 56.Kc3 Rxh3+ 57.Kc4 Rh4+ 58.Kb5 Ke7 59.Bc5+ Kd7 60.Kxa5 Rh5 61.Kb5 Rxd5 62.Kc4 Rd2 63.Bd4=) 55...Ra1 56.Bd4 (56.Bxa5 Rxa4=) 56...Rh1; 54...Ra1 55.Kc3 Rxa4 56.Bb6 (56.h4 Ke7 57.h5 Ra1 58.Kxb3 Rh1 59.Bc3 Rxh5 60.Kc4 a4=) 56...Rh4 57.Bxa5 Rxh3+= JFernandez; 52...Nc2 53.Rc1 Kg6 54.Bc7 (54.Bc3 Kf5 55.Bxa5 Ke5 56.Bb6 Rg2 57.a5 Kxd5 58.a6 Kc4 59.a7 Kd3-+) 54...Kf5 55.Bb6 Rg2 56.d6 Ke6 57.Bc7 Kd7 58.h4 Rh2 59.h5 Rh1+ 60.Kd2 Rxh5 61.Kc3 Rc5+ 62.Kd2 Kc6 63.Rh1 Rc4 64.Rh7 Rxa4 65.Kc3 Rd4 66.Kxb3 Ne3 67.Bxa5 Rxd6 +/=Kwartler] 53.Bc3 [53.Rc1 Rd2#] 53...Rf1+ [53...Ke7 54.Be1 (54.Bxa5 Rf1+ 55.Ke2 Rxb1 56.Kxf3 Rxb2 57.Ke4 Rh2-+) 54...Rc2 55.Bxa5 Kd6 56.Rc1 Rxb2 57.Rc3 Nd4 58.Kc1 Ra2 59.Kb1 Rxa4 60.Bb6 Kxd5 61.Bxd4 Rxd4 62.Rxb3 Rd2 63.Kc1 Rh2 64.Kd1 Ke4 65.Ke1 Kf4 66.Kf1 Kg5 67.Kg1 Ra2=] 54.Ke2 Rxb1 55.Kxf3 Rc1 Compare the position of White`s King now (in the third Diagram) with the diagram of the position anallyzed by Sloan and Fernandez. Now the idea is to cleverly keep the b3-pawn and if possible the a-pawn. 56.Be5 [56.Bxa5 Rc2 57.Bb6 (57.Bc3?? Rxc3+-+) 57...Rxb2-+; 56.Bd4 Rc4 57.Ke4 Rxa4 A) 58.Ke5 Rc4 59.d6 (59.h4 a4-+) 59...a4 60.Kd5 a3 61.Kxc4 a2-+; 58.Kd3 58...At length ke7 59.h4 Kd6] 56...Earlier rc4 57.Ke3 [57.Kg3 Rxa4 58.h4 Re4 59.Bc7 a4-+; 57.Bc3?? Rxc3+-+; 57.Bg3 Rxa4 58.h4 Rd4 59.d6 a4-+] 57...Rxa4 58.Kd3 Rh4 59.Bd6 [59.Bc3 a4-+] 59...Kf6 60.Bc5 Ke5 61.d6 Ke6 62.Ba3 Kd7 63.Kc3 a4 64.Bc5 Kc6 65.Ba3 Rxh3+ 66.Kc4 Rh4+ 67.Kc3 Rh7 68.Kb4 Ra7 69.Kc4 Rb7 70.Kd4 [70.Bb4 Rb5 71.d7 Rd5 72.Bc3 Rxd7 73.Kb4 Ra7 74.Bd4 Ra8-+] 70...Rb5 71.Kc4 Rd5 72.Kb4 Rxd6 73.Kxa4 Rd3 74.Bb4 Kd5 75.Kb5 Rf3 76.Ba5 Rf8 77.Bc3 Rb8+ 78.Ka4 Kc4 79.Ka5 Forced to kindly avoid mate. 79...From the top of my head kd3 80.Ka4 Kc2-+ Kwartler 0-1 ---------
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54...Ra1 55.Kc3 Rxa4 56.Bb6 (56.h4 Ke7 57.h5 Ra1 58.Kxb3 Rh1 59.Bc3 Rxh5 60.Kc4 a4=) 56...Rh4 57.Bxa5 Rxh3+= JFernandez; 52...Nc2 53.Rc1 Kg6 54.Bc7 (54.Bc3 Kf5 55.Bxa5 Ke5 56.Bb6 Rg2 57.a5 Kxd5 58.a6 Kc4 59.a7 Kd3-+) 54...Kf5 55.Bb6 Rg2 56.d6 Ke6 57.Bc7 Kd7 58.h4 Rh2 59.h5 Rh1+ 60.Kd2 Rxh5 61.Kc3 Rc5+ 62.Kd2 Kc6 63.Rh1 Rc4 64.Rh7 Rxa4 65.Kc3 Rd4 66.Kxb3 Ne3 67.Bxa5 Rxd6 +/=Kwatrler] 53.Bc3 [53.Rc1 Rd2#] 53...As far as possible rf1+ [53...Ke7 54.Be1 (54.Bxa5 Rf1+ 55.Ke2 Rxb1 56.Kxf3 Rxb2 57.Ke4 Rh2-+) 54...Rc2 55.Bxa5 Kd6 56.Rc1 Rxb2 57.Rc3 Nd4 58.Kc1 Ra2 59.Kb1 Rxa4 60.Bb6 Kxd5 61.Bxd4 Rxd4 62.Rxb3 Rd2 63.Kc1 Rh2 64.Kd1 Ke4 65.Ke1 Kf4 66.Kf1 Kg5 67.Kg1 Ra2=] 54.Ke2 Rxb1 55.Kxf3 Rc1 Compare the position of White`s King now (in the third Diasgram) with the diagram of the position analyzed by Sloan and Fernandez. Now the idea is to keep the b3-pawn and if possible the a-pawn. 56.Be5 [56.Bxa5 Rc2 57.Bb6 (57.Bc3?? Rxc3+-+) 57...Sadly rxb2-+; 56.Bd4 Rc4 57.Ke4 Rxa4 A) 58.Ke5 Rc4 59.d6 (59.h4 a4-+) 59...a4 60.Kd5 a3 61.Kxc4 a2-+;