Julia's Fairies

No.955
N.Shankar Ram
(India)

Original Problems, Julia’s Fairies – 2015 (II): July – December

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No.955 by N.Shankar Ram – “6×6 cycle of W & B moves” (NSR). Kamikaze Chess is implemented in WinChloe only. Applying of Kamikaze behavior to all pieces for Popeye doesn’t work as well, as Kamikaze seems to not work with fairy pieces. (JV)

Definitions:

Kamikaze Chess: Excepting Ks, pieces disappear after making a capture.

Diagram Circe: A captured piece is reborn on the square it occupied in the diagram position.

Ibis(I): (1,5) Leaper. (notation “15” for Popeye).

Triton(TR): Marine piece operating along Rook lines: without capture moves as Rook, with capture – as Locust (capturing an enemy unit, arrives on the square immediately beyond that unit, which must be vacant).

No.955 N.Shankar Ram
India
original – 22.11.2015

Solution: (click to show/hide)

White Kd1 Rh2 Pf2f7 TRe1g1 Black Kf1 Sb1 Be8 Rb2c3a5b6c7c8d8 Pd3d4d5d6d7 IBd2

s#7 (6+16)
Kamikaze Chess
Diagram Circe
Ibis d2
Triton e1, g1

11 Comments

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Kjell Widlert

November 23, 2015 00:45

It seems so simple when you have the mechanism – but it takes a stroke of genius to invent that!
White wants to force all pieces between Rd8 and Kd1 to capture out of the line. But if White (after the key) plays 2.TRe2+, for example, he can never force dxe3 as the Rc3 will capture instead. So White must open the only line that has no bR behind the d-file piece, and the threat is 2.TRe3+ dxe3. At this point, White doesn’t have to fear the opening of the third row as dxe3 has already occurred, so he can continue 3.TRe2+ dxe3, after which the second row may be opened by 3.TRe7+ Ixe7; etc.
In the variations, Black puts a R on the fourth row but in turn leaves another row empty – so White can play in the same manner as before, but starting at a different point of the circle of e-file squares.

A wonderful idea, that doesn’t need a bigger board and a larger cycle to impress.

seetharaman

November 23, 2015 11:56

Strikingly simple when you understand it! Shankar had shown a 3×3 cycle in orthodox S#3 earlie (1st Pr. The Problemist,1983) which is considered a pioneer for the theme. It has taken more than 30 years and his genius to extend that scheme into a 6×6 cycle. Bravo Shankar and welcome back to your old form. I hope priming in time for the 10-WCCT !

Ganapathi

November 23, 2015 12:42

Reply to seetharaman

A great problem indeed, Shankar! Would not have understood the beauty without Kjell Widlert’s comments.
Can Seetha show Shankar’s Problemist S#3 please?

seetharaman

November 23, 2015 13:57

Reply to Ganapathi

It is a S#4…
http://i.imgur.com/yEPFoMt.png

Georgy Evseev

November 23, 2015 13:17

It is interesting that exactly fourth row should be “rook free” in diagram position – otherwise leaper move will become refutation.

shankar ram

November 23, 2015 17:43

Thanks! Kjell, Seetharaman and Ganapathi for your comments!
And Julia, for publishing so quickly!
Georgy: the 5th row/rank can also be “rook free”, by moving Ra5 to a4. Here, too 1…Ie7 doesn’t defeat the threat. In fact it allows additional duals.

Luce Sebastien

November 23, 2015 18:47

I am very happy that Julia publishes a new problem after some difficult days…
And it is a beaufiful one, celebrating the thought,
which is the greatest richness of humanity.
Congratulations M.Shankar Ram
(hope my english is good)

shankar ram

November 24, 2015 05:30

Reply to Luce Sebastien

Thank you, Luce!
I think you’re referring to Paris?
We here faced similar thing in Mumbai, 2008. Our hearts are with all of you in Paris and France.
Your English is not just good, it’s better than my French!

Luce Sebastien

November 24, 2015 13:47

Yes I am from Paris, very close from St Denis (north periphery of Paris). Thank you for you mail, M. Shankar Ram,
best to you
Sebastien

seetharaman

November 24, 2015 19:43

Reply to Luce Sebastien

We still cannot get over the horror of Mumbai 2008 (26/11). Repetitive barbaric attacks on Paris is heart-rending. Naturally all of us are with the people of France.

shankar ram

November 24, 2015 15:35

The board size for larger cycles is actually “n div 2 + 5 x n+2”, n>3 (div = integer division), and not “n+1 x n+2”, n>6 (only works for n=7,8!).
Generalised or “infinite” problems were pioneered by T.R.Dawson. Some famous examples are his “Lunar Q” and “Infinite Nowotny” problems.