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Problem 495: Anton Bidlen - Fairy (Ser-HS#)
anton.bidlen(07.01.2015) Here is the longest original published in the website. Long maneuvers by black leads to Help-Selfmate. The author Anton Bidlen claims that the problem is computer checked.









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#


+1 #1 Seetharaman Kalyan 2015-01-07 08:36
Nice problem in the typical style of series helpmates!

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#.
+1 #2 Diyan Kostadinov 2015-01-07 12:56
The author really sent me the problem with indicated 119 moves, but the mate is done on the 120th move, so I published it in 120. Other opinions which is correct?

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):

+2 #3 Seetharaman Kalyan 2015-01-07 17:42
Good find by Sebastian Luce. Though the final moves are different (and there is a capture of another bishop) the scheme is almost the same.
+1 #4 Cornel Pacurar 2015-01-08 04:12
A few comments (split into 3 messages because of message size limitations):

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.
+1 #5 Cornel Pacurar 2015-01-08 04:14
2) It is likely that author's intention was a series move-length record. Generally speaking, when it comes to move-length records, the therm "anticipation" does not really have the normal weight it has for regular non-task compositions, and most of the time it is out of place. In the end, the number of matrices that could be successfully used is limited, and some have been heavily used over the last 40 or so years, so anticipation is not necessarily, or by default, the real issue here. Sure, it is nice to acknowledge close predecessors, when they exist, especially for the same (series) stipulation (e.g. after John Doe).
+1 #6 Cornel Pacurar 2015-01-08 04:17
3) In this presumed context, the value of the problem is dramatically reduced by the fact that this is not an actual length record (see the table of records at lengthrecords.chessproblems.ca/)! There are longer problems even with 18 units, and the ser-hs# length record for 22 units (which is also the overall ser-hs# record) is 133 moves. I am the author of that record (a feenschach 72 TT entry published in feenschach 206 - HS24, page 491) which, in fact, better uses this very matrix, based on the well-known Kemp mechanism. feenschach 72 TT was a thematic tourney dedicated to ser-hs# move-length records (to which Mr. Anton Bidlen had unsuccessfully participated), an "Arno Tungler vs. the rest of the world" match won by Arno Tungler with the score of 11 to 8. The award was published in feenschach 206 and a review of the tournament was also included in ChessProblems.ca Bulletin Issue 3 - bulletin.chessproblems.ca/. The ser-hs#133 with 22 units is SC21, page 90.
+1 #7 Vlaicu Crisan 2015-01-10 15:04
C+ Popeye 4.69 using the [much better!] stipulation 119->s#1

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