Claude Beaubestre

Original Fairy problems
July – December’2018

Definitions: (click to show/hide)

No.1363 Claude Beaubestre

original – 26.12.2018

Solution: (click to show/hide)

White Qa8 Ba7 Ka6 Black Ke8 Bf2 Qh1

ser-Z(b8)13                                (3+3)

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Jacques Rotenberg
Jacques Rotenberg
December 28, 2018 03:08

what does it mean ser-Z(b8) ?

December 28, 2018 12:48

Z stands for ‘Zielfeld’, ‘field to be reached’. You have to get a piece to b8 in 13 consecutive moves, doesn’t matter which piece.

Jacques Rotenberg
Jacques Rotenberg
December 28, 2018 15:01

so, why not 5.Kxf2 and 6.Bb8 ?

Arno Tüngler
Arno Tüngler
December 28, 2018 16:18

Seems that was not correctly checked as Popeye also confirms the cooks.

Dmitri Turevski
Dmitri Turevski
December 28, 2018 21:53

So basically this?

Stipulation ser-za213
black Bh8 Pf7f5e4d3c2g6
white Kc1
1.Kc1-d2 2.Kd2-e3 3.Ke3-f4 4.Kf4-g5 5.Kg5-h6 6.Kh6-h7 7.Kh7-g8 8.Kg8*f7 9.Kf7-e6 10.Ke6-d5 11.Kd5-c4 12.Kc4-b3 13.Kb3-a2 z

Sergey Shumeiko
Sergey Shumeiko
December 29, 2018 07:37

Without a pawn.
1.Kb5 2.Kc4 3.Kd3 4.Ke2 5.Kf1 6.Kхg1 7.Kf2 8.Ke3 9.Kd4 10.Kc5 11.Kb6 12.Ka7 13.Kb8 z

Arno Tüngler
Arno Tüngler
December 29, 2018 04:14

The “orthodox” length record for this stipulation with 6 units is 49 moves:'P1113601‘. If you want it to be without captures, you can remove the pieces on f7 and h7, move the wK to h8 and get a ser-Zf7 in 17 moves.

Georgy Evseev
Georgy Evseev
December 29, 2018 10:57

I think that original idea was to have a clear geometry and attractive use of madrasi effects.

So the following version seems nearer to author’s thoughts, though I was not able to resolve all issues).

White : Kd1 Qb2 Rf4
Black : Kf6 Qb4 Rc4 Pa2

Sergey Shumeiko
Sergey Shumeiko
December 29, 2018 20:45
Reply to  Georgy Evseev

The author’s concept can be save by little shifting of pieces:
ser-zb813 Madrasi (3+3)
1.Kb5 2.Kc4 3.Kd3 4.Ke2 5.Kf1 6.Kg1 7.Kh2 8.Kg3 9.Kf4 10.Ke5 11.Kd6 12.Kc7 13.Kb8 z
If captures will be added then the solution length can be increased, for example:
ser-za255 Madrasi (3+10)
1.Kh5 2.Kхh6 3.Kh7 4.Kg8 5.Kf8 6.Ke8 7.Kd8 8.Kхc8 9.Kd7 10.Kd6 11.Kхd5 12.Kd6 13.Ke7 14.Kf8 15.Kg8 16.Kh7 17.Kh6 18.Kg5 19.Kхf4 20.Kg5 21.Kh6 22.Kh7 23.Kg8 24.Kхf7 25.Ke6 26.Kd5 27.Kc4 28.Kb4 29.Kхa3 30.Kb4 31.Kc4 32.Kd5 33.Ke6 34.Kf7 35.Kg8 36.Kh7 37.Kh6 38.Kg5 39.Kf4 40.Ke3 41.Kd2 42.Kхc1 43.Kd2 (43.Kxb1??) 44.Ke3 45.Kf4 46.Kg5 47.Kh6 48.Kh7 49.Kg8 50.Kf7 51.Ke6 52.Kd5 53.Kc4 54.Kb3 55.Ka2 z
wK, like a “shuttle”, moves 5 times through right upper corner of the board.

Nikola Predrag
Nikola Predrag
December 29, 2018 22:04

Why so many captures? Just a simpler scheme (after Georgy):
White Qa7 Kf5 Rg3
Black Be8 Qg7 Pa6 Sc6 Rg6 Se4 Ke3
Stipulation ser-za853
Condition Madrasi

1.Kf5-e6 2.Ke6-d5 3.Kd5-c4 4.Kc4-b3 5.Kb3-c2 6.Kc2-d1 7.Kd1-e1 8.Ke1-f1 9.Kf1-g2 10.Kg2-h3 11.Kh3-h4 12.Kh4-h5 13.Kh5-h6 14.Kh6-h7 15.Kh7-g8 16.Kg8-f8 17.Kf8*e8 18.Ke8-f8 19.Kf8-g8 20.Kg8-h7 21.Kh7-h6 22.Kh6-h5 23.Kh5-h4 24.Kh4-h3 25.Kh3-g2 26.Kg2-f1 27.Kf1-e1 28.Ke1-d1 29.Kd1-c2 30.Kc2-b3 31.Kb3-c4 32.Kc4-d5 33.Kd5*c6 34.Kc6-d5 35.Kd5-c4 36.Kc4-b3 37.Kb3-c2 38.Kc2-d1 39.Kd1-e1 40.Ke1-f1 41.Kf1-g2 42.Kg2-h3 43.Kh3-h4 44.Kh4-h5 45.Kh5-h6 46.Kh6-h7 47.Kh7-g8 48.Kg8-f8 49.Kf8-e8 50.Ke8-d8 51.Kd8-c8 52.Kc8-b8 53.Kb8-a8 z

Claude Beaubestre
Claude Beaubestre
January 4, 2019 18:09
Reply to  Julia

Sorry, Julia, it’s my fault, I checked the prepublication badly.

Nikola Predrag
Nikola Predrag
December 30, 2018 04:29

5 captures/124 moves, but Madrasi is here a rather mechanical tool:
White Ka7 Rf7 Qb1
Black Qb7 Rc7 Ba6 Rc6 Pd6 Kf5 Sc4 Pf4 Ba3 Pe3 Pf3 Sc1
Stipulation ser-za1124
Condition Madrasi

20.Kxc6-41.Kxa6-60.Kxc4-82.Kxa3-102.Kxc1 etc.

Claude Beaubestre
Claude Beaubestre
January 6, 2019 11:32

Thank you for your feedback ; they interested me a lot and learned. I propose the following problem with the same idea. But 2 pieces less and 1 more move.
White: Ka8, Ba7
Black: Kd7, Bg1


1.Kb7 2.Ka6 3.Kb5 4.Kç4 5.Kd3 6.Ké2 7.Kf1 8.K×g1(×a7) 9.Kf2 10.Ké3 11.Kd4 12.Kç5 13.Kb6 14.Ka7

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