# No.1461 (JQ)

 No.1461 James Quah (Сингапур) Оригинальные сказочные задачи JF-2019/II: июль – декабрь’2019

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 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 1.ZRa3-c6 ! threat: 2.Rc4*f4 1...Re2-e4 2.Kd6-d5 {(2.Kc7? Kf5!)} 1...Bg2-e4 2.Kd6-c7 {(2.Ke5? Kh3!)} 1...ZRg1-e4 2.Kd6-e5 {(2.Kd5? Kh4!)} 1...Bg2*c6 2.Kd6*c6 1...ZRg5-d7 2.Kd6*d7 1...Re2*e6+ 2.Kd6*e6 1...Bf8*e7+ 2.Kd6*e7 1...ZRf7-d4 2.Kd6-c5 { (C?: WinChloe has French type of Equihopper at 90º and shows some duals in addition to author's solutions (JV))} Triple Grimshaw with royal battery dual avoidance. Task: Royal battery is fired 8 times. Black has three potential flights at f5, h4, and h3, which are guarded by E(h1)-e2-f5, E(d2)-g1-h4, and E(h1)-g2-h3 / ZR(b7)-e5-h3 (double guard). The key 1.ZRc6! threatens 2.Rxf4, so black plays each of three defenders to e4. Each interference interferes with two other units, but avoids a dual by removing the guard of an E on a flight square. After 1…Re4/ZRe4, white chooses the battery opening that uses the king as a hurdle to take back the lost flight: 2.Kd5 opens E(d3)-d5-f5, and 2.Ke5 opens E(d2)-e5-h4. The case of 1…Be4 removes one but not both guards on h3, so 2.Kc7 just avoids closing ZR(b7)-e5-h3. The other five variations complete all the openings of the ZR-K battery. C+: I wrote my own program to test this, as I don’t think the equihopper-90 can be tested on Popeye. The following two problems are interesting to mention. A is the first triple Grimshaw royal battery with dual avoidance, and is now 40 years old. B is the only royal battery I have seen which is fired 8 times. (Author)

### 5 комментариев: No.1461 (JQ)

1. Julia пишет:

1) WinChloe has Equihopper-90 French type, so partial testing is possible, looks like:

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…

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. seetharaman 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).

• Julia пишет:

Sorry, thanks, the animation and the definition of Equihopper are corrected! Hope, everything correct now..

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