Author: Odd Gunnar Malin
Date: 05:21:01 09/08/01
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On September 08, 2001 at 07:34:16, Peter Berger wrote: >On September 08, 2001 at 05:07:31, Odd Gunnar Malin wrote: > >>On September 07, 2001 at 23:10:53, Peter McKenzie wrote: >> >>>On September 07, 2001 at 22:08:27, Odd Gunnar Malin wrote: >>> >>>> >>>>I don't have his position but I have another 'quadrat position' like the Fine 70 >>>>with a not so obvious first move, well it is Kb1 but that must be a coincidence >>>>:) >>>> >>>>[D]2k5/2p2p2/2p1p3/2P1PpP1/1p1P1p2/1P3P2/1K3P2/8 w - - 0 1 >>>>White wins. 1.Kb1! >>> >>>Are you sure? >>>I ran it through LambChop, and the prefered move was switching between Kb1 and >>>Kc2 with a score of around 0.4 for each move. Then at depth 25 I got a fail >>>high on Kc2, returning a score of 1.6. >>> >>>As I write, the same score has popped up for Kc2 at depth 26 too... >>> >> >>The position is from Averbach's book on pawn endgames. In my german version is >>in on p.378. >>I have tried to follow other path than that Averbach's solution but they all end >>in draw. >> >>Here is his solution. >>The quadrat: >>White: e2=1, d2=2, e1=3, d1=4 >>Black: b7=1, a7=2, b8=3, a8=4 >> >>You can shifting this quadrat two sqaures so. f.ex white also has a quadrat on >>c2=1, b2=2, c1=3, b1=4. >> >>Die kürzesten Wege bestehen hier aus jeweils acht Feldern: >>c4-d3-e2-f1-g2-h3-h4-h5 bzw. a5-a6-b7-c8-d7(d8)-e7(e8)-f8-g7. >>Die Paare c4-d3 und h4-h5 können nicht als Basisfelder genommen werden, weil sie >>kein Verbindungsfeld besitzen. Zur Wahl stehen die Felderpaare d3-e2 (a6-b7 bei >>Schwarz) oder e2-f1 (b7-c8 bei Schwarz). In beiden Fällen erhälten wir das >>Quadratsystem mit übereinstimmenden Hauptzonen. >>1.Kb1! (jetzt muss Schwarz das Feld 1 oder 3 betreten) 1...Kb8 2.Kc1 Ka8 3.Kd1 >>Kb8 4.Ke1 Kb7 5.Ke2, und Weiss gewinnt. >> >>Running through www.freetranslation.com: >> >>The shortest ways consist here of respectively eight fields: >>c4-d3-e2-f1-g2-h3-h4-h5 and/or a5-a6-b7-c8-d7(d8)-e7(e8)-f8-g7. The couples >>c4-d3 and h4-h5 cannot be taken as a basis field because they possess no >>connection field. To the choice, the field couples d3-e2 (a6-b7 stand in black) >>or e2-f1 (b7-c8 in black). In both felling receiving we the square system with >>übereinstimmenden head zone. 1.Kb1! (Now must black the field 1 or 3 enter) >>1...Kb8 2.Kc1 KA8 3.Kd1 KB8 4.Ke1 KB7 5.Ke2, and white win. >> >>I'm sure V.Deepeven or A.Silver could explain this theorie, I myself must have >>some reread of it before I lay out on such a trip. >> >>Odd Gunnar > >This theorie sounds a little like dark magic to me still . Some more empirical >data : I did some tests with computer on this 1.Kc2 move and there could very well be that this win too, but much harder. One path ends in a Queen endings where the computer have White as winner (up +8). Averbach's solution is much simpler. After: 1.Kb1 Kb8 2.Kc1 Ka8 3.Kd1 Kb8 4.Ke1 Kb7 5.Ke2 [D]8/1kp2p2/2p1p3/2P1PpP1/1p1P1p2/1P3P2/4KP2/8 b - - 0 1 If Black goes to b8/c8 white grab the b-pawn. On other moves White has won a tempo on the King side and gets to h6 and grab all pawns. Odd Gunnar
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