No.1461 (JQ)

No.1461
James Quah
(Сингапур)

Оригинальные сказочные
задачи JF-2019/II:
июль – декабрь’2019


Определения: (показать/спрятать)


No.1461 James Quah
Singapore

original – 11.12.2019

Solutions: (click to show/hide)

White Kd6 Rc4 Be7 Pe6 ZRa3a8b7h6 EQd2d3h1 Black Kg4 Re2 Bf8g2 Qh5 Pa4f4g3 ZRf7g1g5

#2                                           (11+11)
Zebra-Rider a3,a8,b7,h6,f7,g1,g5
Equihopper at 90º (English) d2,d3,h1


5 комментариев: No.1461 (JQ)

  1. JuliaJulia пишет:

    Separately – author’s reply to the following my comments:

    1) WinChloe has Equihopper-90 French type, so partial testing is possible, looks like:
    1461-comm-WinChloe.png
    I guess, the reason of the duals is EQ behaviour, jumping over obstacles.

    2) If Zebra-Rider is necessary? If Zebra would work here too, or any other white piece to guard f4? Equihoppers used as technical pieces only…

    Reply by author:
    Some zebra-riders could be zebras, of course, but it’s better to have just one type. Yes, only a8, b7 and g1 need to be zebra-riders. The others just guard squares, but it’s nice that there are no knights. The key ZR needs to go to c6 to prevent 1…Bxb7! as well as guard f4. So 1.ZRd1? fails to 1…Bxb7!

    Equihopper-90 must be English (not non-stop = French) because some moves like E(d3)-d1 are too strong. The equihopper-90s are needed to guard squares and enable the dual avoidance, otherwise white would have two mates after each interference on e4. Why are they equihopper-90s and not normal equihoppers? The most important reason is that ZR(g1) needs to be a hurdle, and no other hopper seems possible to use. They are very difficult to handle, and a lot of care must be taken to keep hurdles out of their way. I kept them down to three.

    The solution found by WinChloe is correct other than the details of the duals. Not sure why it said ‘d’ when it should have been ‘h’. But these assume the French type, and they don’t happen, because I’ve specified the English type. It’s interesting that it found the cyclic dual avoidance.
    1…Fxc6 2.Rxc6 dual E90hxc6 not counted
    1…ZNd4 2.Rc5 dual E90hg8/E90dd1 not counted

    My program was able to check it, so I was sure it was sound. But we can also consider it C+ by WinChloe, because the intended mates do not depend on the non-stop property. Notice that we have B(e7) rather than pawn, because it needs to guard g5 after 1…ZRd7. E90(d2)-g2-g5 is blocked.

    So the full solution (as given in the file) is:
    1.ZRc6! (2.Rxf4)

    Triple Grimshaw with dual avoidance:
    1…Re4 2.Kd5 (2.Kc7? Kf5!)
    1…Be4 2.Kc7 (2.Ke5? Kh3!)
    1…ZRe4 2.Ke5 (2.Kd5? Kh4!)

    Other king moves to complete all 8 king moves to fire the battery:
    1…Bxc6 2.Kxc6
    1…ZRd7 2.Kxd7
    1…Rxe6+ 2.Kxe6
    1…Bxe7+ 2.Kxe7
    1…ZRd4 2.Kc5

    This theme was shown only in the example by Hans Peter Rehm (1979), which might be nice to show. I was inspired by this problem.

  2. Seetharamanseetharaman пишет:

    I thought non-stop Equihopper is the French type. The English EQ is stoppable!

  3. Kjell Widlert пишет:

    The animation is wrong in one place:
    1…ZRe4 is really ZRg1-e4 but is shown as ZRf7-e4 (a zebra-rider can’t make that move).

  4. shankar ram пишет:

    Hello, James!
    I’ve done an 8-fold Royal Battery(feenschach, 1986). See below.
    p.s: it also shows an 8-fold Lacny 😉

    https://imgur.com/rmcJ5EZ

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