[Eddie]

[/quote]

Back in business again.

My solution for this problem is: 7, 14, 4 (9) 2 A (339) 30 (48) 18, 34, 24. But A: 1 looks winning too, even 30 is an option.

[Eddie]

volk wrote:

I do not understand the discussion, but (if I may add this) in this position:
* 41, 372 (35) 1 (14) is =
* 24, 25 wins a piece

[Eddie]

Very old position.
In De Humorist, 5-12-1940) published on name of C.K. Kaan [22 (28) 41, 21 (40) 1 (35) 394 (19) 30, 40, 40], but a recent discovery of mine indicates that in 1935 A.A. Polman made it earlier.

[Eddie]

(PR?)

The exact problem was sent to me in may 2006 by S.M. van Eijk. It was also published in his collected 7×7’s, part 12 [position 6384], which was edited summer 2007.

[Eddie]

(ПР?)[/quote]

Probably this motiv is from J. Scheijen. I found it in more 7×7’s. Two examples:

01/06/09/13/19/25/32 — 22/24/34/39/40/42/43
D. van den Berg (De Problemist, january 1966)

01/06/09/13/19/25/33 — 22/24/28/39/43/44/45
G. van der Linde (Damnieuws, 11-6-1994)

[Eddie]

Problem is already made by B. Sjkitkin, Alex. See the book Mir Miniatur !

[Eddie]

A lot of interesting and difficult endgames in theses files !
For humans it’s hard to check these problems, but for computer programs it’s much easier.
I tried position 73. From mr. Stanowski, I understand.

black: D 2 and pieces 13 + 22
white: D 16, D 25, D 49.

Winning is: 25-3 etc.
The computer gives a second solution: 25-20 and for instance:
1] (27) 49×21 (18) 33 (23) 22 (35) 2 (28) 50 (40) 21-17 +
2] (2- 3 (2) 26 (18) 11, 26-21, 21 or 16 +

Also nice.

[Eddie]

Just curiosity. In the info about the concours there are several problems given without the name of the authors. Are they known ?

[Eddie]

Second solution: 43, 34, 2, 46

[Eddie]

It’s not in my files.

[Eddie]

If I understand well, mr. Evgraf told that the miniature 05/25/29/32/33/34/36 — 10/14/18/19/20/24/42 looked very familiar to him.
Indeed he himself posted it here on 6-3-2005.

[Eddie]

Fenix wrote:

Tsvetov wrote:

Пётр, а почему из №141 не сделать мини? Например:

40…х

Mini is very interesting and offers a lot to analyze. Question: is there a dual after 5 ? I did not find it till now.

* Of course black can not answer (50) 449, 32, 9, 44.
* So: (21) is a possible move. Now after 439, 49 (9) 1 (50) and =.
* Let’s try something else: (21) 140 (40) [still not (50) 449, 4, 9] 35 (50) 4, and now again black saves himself: (9) 33 and =

Maybe anyone has other ideas ?

[Eddie]

dual.

Yes, from three positions only one survived 😆

Of course, Alex, one of the others has the same problem too. I was too hasty when I was planning to go to bed.
After the alternative 5×16 (28) 11 (32) 12-7 ( 1 wins, but also 2 (12) and now simply 6 or 16 (37) 2-11.

[Eddie]

№2. Белые начинают и выигрывают

Always creating, Alexander ! That’s good. But the last one has a dual.

[Eddie]

The position 10/19/21/27/28/29/33 — 37/39/41/43/44/45/50 was already made in 1975 by mr. B. Sjkitkin and was in his notebook (1986), that was used for the book Mir Miniatjur. Although it was not published in this book (or anywhere else, but of course I do not see many East-European sources, so this is not certain) the «manuscript» of mr. Sjkitkin is in the meantime included in the Groeneveld archive.

[Автор]

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