Problem 495: Anton Bidlen - Fairy (Ser-HS#) |
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1.Kc8-d8 2.Kd8-e8 3.Ke8-f8 4.Kf8-g8 5.Kg8-h8 6.Bh7-g8 7.Kh8-h7 8.Kh7-h6 9.Kh6-h5 10.Kh5-h4 11.Kh4-h3 12.Kh3-h2 13.Kh2-g1 14.Kg1-f1 15.Kf1-e1 16.Ke1-d1 17.Kd1-c2 18.Kc2-b3 19.Kb3-a4 20.Ka4xb5 21.Kb5-a4 22.Ka4-b3 23.Kb3-c2 24.Kc2-d1 25.Kd1-e1 26.Ke1-f1 27.Kf1-g1 28.Kg1-h2 29.Kh2-h3 30.Kh3-h4 31.Kh4-h5 32.Kh5-h6 33.Kh6-h7 34.Kh7-h8 35.Bg8-h7 36.Kh8-g8 37.Kg8-f8 38.Kf8-e8 39.Ke8-d8 40.Kd8-c8 41.Kc8xb8 42.Kb8-c8 43.Kc8-d8 44.Kd8-e8 45.Ke8-f8 46.Kf8-g8 47.Kg8-h8 48.Bh7-g8 49.Kh8-h7 50.Kh7-h6 51.Kh6-h5 52.Kh5-h4 53.Kh4-h3 54.Kh3-h2 55.Kh2-g1 56.Kg1-f1 57.Kf1-e1 58.Ke1-d1 59.Kd1-c2 60.Kc2-b3 61.Kb3-a4 62.Ka4-b5 63.Kb5xa6 64.Ka6-b5 65.Kb5-a4 66.Ka4-b3 67.Kb3-c2 68.Kc2-d1 69.Kd1-e1 70.Ke1-f1 71.Kf1-g1 72.Kg1-h2 73.Kh2-h3 74.Kh3-h4 75.Kh4-h5 76.Kh5-h6 77.Kh6-h7 78.Kh7-h8 79.Bg8-h7 80.Kh8-g8 81.Kg8-f8 82.Kf8-e8 83.Ke8-d8 84.Kd8-c8 85.Kc8-b8 86.Kb8xa8 87.Ka8-b8 88.Kb8-c8 89.Kc8-d8 90.Kd8-e8 91.Ke8-f8 92.Kf8-g8 93.Kg8-h8 94.Bh7-g8 95.Kh8-h7 96.Kh7-h6 97.Kh6-h5 98.Kh5-h4 99.Kh4-h3 100.Kh3-h2 101.Kh2-g1 102.Kg1-f1 103.Kf1-e1 104.Ke1-d1 105.Kd1-c2 106.Kc2-b3 107.Kb3-a4 108.Ka4-b5 109.Kb5xc6 110.Kc6xd5 111.Kd5-e6 112.Ke6xf6 113.Kf6-e6 114.f7-f5 115.f5-f4 116.Ke6-f5 117.Bg8-c4 118.Bc4xd3 119.e5-e4 + Qe3xe4+ 120.Bd3xe4# |
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Comments
Should not the stipulation be Ser-HS#119? After the last black help-move 110.e4+ white forces HS#1 by Qe3xe4+. Perhaps Popeye & Winchloe have programmed it differently.
The solution can also be written shortly as 1.Kd8 5.Kh8 6.Bg8 20.Ka4xb5 41.Kc8xb8 63.Kb5xa6 86.Kb8xa8 109.Kb5xc6 110.Kxd5 112.Kxf6 113.Ke6 115.f4 116.Kf5 117.Bc4 118.Bxd3 119.e4+ Qxe4+ 120.Bd3xe4#.
But the most important here is that this scheme was used in other Series problems with different stipulations before, so looks anticipated. Here is one example (many thanks to Sebastien Luce for that):
1) This is a ser-hs#119 - a series of 119 black moves followed by s#1, that is. Popeye's solution notation is just incorrect and misleading. For instance, Popeye correctly solves White Qc2 Kh1, Black Kg5 Ph4 Qb2 as ser-hs#2, but writes the solution as 1.Kg5-g4 2.Kg4-h3 Qc2-g2 + 3.Qb2*g2 #. For more explanations see the article "Wie lang is?..." by Arno Tungler in feenschach 201.
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