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Позвольте, представить еще одну «crazy» 😆 .
11,2,11,2,9,11,40,21,2,10,44 enz.
Дважды 2-11, трижды попадание на поле 2, уничтожение 3-х белых дамок. Кто больше?
Masterpiece! But with flaws: 40 can be played anytime.
Nasty second solution: 40,11,2,13-8,9,21,10,3,13.Yes, that is the intended prove of legality. And no, it’s not new. Published in DP december 2003. It was meant as a task-problem, with 7 promotions during the combination. It’s a version of this one:
S. Korteling18-12 12×3 3-9 7-2 2×47 47-42 17-12 12×1 1×45 45×41 etc.
Published by the famous problem editor of «Hoofdlijn» (nr. 92), Hein Wilsens.Hello colleagues,
This afternoon I had a very crazy idea, but most probably this problem is not correct (second solution) and I don’t like the first move. Who is able to create a correct composition based on this idea ( two times 2-11)?
white: 7, 8, 12, 13, 14, 18, 19, 20, 21, 26, 27, 31, 38, 42, 43
black: 1, 3, 6, 16, 17, 25, 28, 29, 30, 33, 34, 36, 37, 40
Solution: 2, 2-11 (7) 22, 2, 21, 2-11, 23, 7, 3, 14, 48.Furthermore I am looking for the email addresses of E. Zubow and B. Morkus. Can anybody help me? Thank you!
2x 2-13.
7-2, 2-13, 11×2, 16×7, 30-25, 2-13, 48-42, 6-1, 1×3, 3×45, 45×41, 35×24, 25×5 (15-20), 5-32
In the newspaper the name of Sijbrands opponent is not given. But Sijbrands demonstrated this game a few weeks ago in the Rotterdam library before an audience of 50 people (including me). The opponent was a contender for the World Youth Championship, playing Sijbrands just before the start of the tournament. It was the extremely talented Joeri Anikeev.
(…)
Сравнил бы визуально…
Но и диаграммы не открываются… 😥 😥 😥Java Virtual Machine is needed. Download here (for Windows):
Black: 12,17,25 White 19,28,45.
(White to move)Same motive. White must be able to reach 5 in case black starts with 21. So: 14 (30), 18! , 9 (33), 4 etcetera.
I’m sorry to have added to the confusion. Now I did some research (to find the date) and I found in De Problemist december 2004, page 146 an article by Kuijper, titled «Goodbye».
Here is an excerpt:
— The second solution was found in the solving contest President Cup 2002.
-After Steef the Bruin passed away, the collection Pennings (dedicated to 46-5) came temporary in the hands of Kuijper. He was shocked to find nr 5602 in the boek 46-5 part 1. The name of the author was J. Wagter 1907-1967) publicized in Nieuwsblad van het Noorden, march 5 1960.Kuiper had registered «his» problem in the collection Pennings in december 1963. In febrari 1965 it appeared, without the author knowing, in Shashki/ Dambrete.
Kuijper gives another version:
White: 20,24,27,30,32,34,37,40,41,47,50
Black: 8,9,10,15,18,22,23,28,29,33
P. Kuiper, Het Parool april 18 1992
Solution: 19,39,38,398,5,49.Pieter Kuijper concludes with: «Time to say goodbye. Thanks, Wagter!!»
Поскольку речь идет о любимых композициях то позвольте мне предложить эту штучку с которой я обвораживаю публику более 30 лет. В голове торчит фамилия Куйкен как автор хотя это возможно не верно.
37,35,36!,24 — полный рот (6!) большинства :hearteyes:
Тут я обычно отступаю и даю возможность зрителям найти 5 последующих ударов на большинство.Eddie — could you please research who the author of this gem was?
P.S. Какие сочные краски Damon создал!
Pieter Kuiper. http://www.ericsdamsite.com/Kuiper.htm, nr 8. I read somewhere (De Problemist probably) 21, 27, 27 is a second solution.
Thank you for your kind words, Miljenko. But, after cooling down, I now think I did not add much to Viergever. All the really good stuff is his. It is an extension. A good extension, but not more.
I didn’t know C18 from the endgame tournament; the similarity is indeed striking. In my opinion C18 is also heavily indebted to Viergever.
Для меня лучшим этюдом в 100 остается следующая позиция. Не знаю кто автор. У «Мозера» есть похожая, но с отсеченными двумя первыми ходами.
Белые выигрывают.Автор этого этюда J. Viergever (Nederland)
Алексей
This was posted in the fmjd-forum today. I noticed a black can be added at 25:
Some nice variations after 13-2, and after (25-30) we have safely returned to Viergever.
This is also possible, but gives no extra variations.
Siep Korteling
