No.543, 543.1 Congratulations to Emil with the first publication on JF! |
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No.543 by Michal Dragoun, Ladislav Salai jr. & Emil Klemanič – A very interesting and difficult cyclic h#2 by three famous authors! Welcome to Emil and thanks to Michal and Ladislav for one more excellent publication! (JV)
No.543.1 by Michal Dragoun, Ladislav Salai jr. & Emil Klemanič – Improved version of No.543, dedicated to Nikola Predrag for his notes! (JV)
Definitions:
Lion(Li): Moves along Queen lines over another unit of either color to any square beyond that unit. A capture may be made on arrival, but the hurdle is not affected.
Rook-Lion(RL): (0,1) Lion. Moves along Rook lines over another unit of either color to any square beyond that unit. A capture may be made on arrival, but the hurdle is not affected.
Bishop-Lion(BL): (1,1) Lion. Moves along Bishop lines over another unit of either color to any square beyond that unit. A capture may be made on arrival, but the hurdle is not affected.
No.543 Michal Dragoun, Ladislav Salai jr. & Emil Klemanič
Czechia / Slovakia / Slovakia
original – 10.05.2014
h#2 3 solutions (7+8)
Lion a2 Rook-Lions: d7, h6 Bishop-Lions: b1, a4, b4 Solutions: (click to show/hide)
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No.543.1 Michal Dragoun, Ladislav Salai jr. & Emil Klemanič
Czechia / Slovakia / Slovakia
Version of No.543 – 13.05.2014
Dedicated to Nikola Predrag for his notes!
h#2 3 solutions (7+7)
Lion a3 Rook-Lions: c7, d1 Bishop-Lions: h8, b5 Solutions: (click to show/hide)
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The white bishop-lion and rook-lions each need a hurdle. The B1 moves provide hurdles for 2 of the wBL/RLs, but one of the wBL/RLs must jump over the hurdle in order to guard its squares. The B2 moves provide a hurdle for the remaining wBL/RL. In addition, each of the black moves guards a square, preventing one of the mates: Rd3 guards d5 (2…LIa5?), Bg6 guards e4 (2…LIe2?) and BLd6 guards f4 (2…LIh2?).
The bBLa4 acts as a hurdle for 2…LIa5, and also prevents 2.Rc2 in the third solution. Unfortunately the wBL can’t be on c2 because it would block the wLI. By the way, the position is (7+8), not (8+9)!
Very nice cycle, very bad bBLa4.
bPd2 might serve as a hurdle on a4.
You are right, problem has then three solutions as well. But:
– cycle of black moves is lost
– RLd2 allows move of LIa2, which has no analogy in other solutions
– mate by LIa5 is repeated.
Sorry, my mistake, number of pieces is corrected! (7+8), not (8+9) as it was shown!
But I looked once more closely and problem is sound with Kd1 to b6, BLb1 to c2, bpa4 instead of bBLa4 and without pc6 and pd2 (and without captures then). I will ask co-authors about their opinion about new position.
Thank you, Nikola.
Hi Michal, wKb6 would be nice to avoid the need of capture of bPc6. But after 1.Bg6 BLh7 there would be no hurdle for wLI to show the wrong final checks.
I thought simply about bPd2 to a4, wKd1 to b2.
This position is o.k. Very nice cycle. Excellent
I am very honoured by the dedication although my role in this was rather insignificant.
For several recent years I did not manage to come to Slovakian solving championships and I’m unhappy about that. Wonderful impressions from several earlier championships in which I did participate, include the rare pleasure of meeting and talking with Michal, Ladislav and Emil in the same place.
In the solving rankings, my name used to come after theirs but without a dedication. Now, my status has been improved 🙂
Thanks!