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shankarramNo.1170 
N.Shankar Ram (India)

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Original Problems, Julia’s Fairies – 2016 (II): July – December

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No.1170 N.Shankar Ram
India
original – 18.12.2016

Solutions: (click to show/hide)
white Se1c2a4e8g6h1 Pe2 Be6 black Ke4 Xd1b3b5d7f7h5h3f1

#4                                            (8+9)
Piece X : d1,b3,b5,d7,f7,h5,h3,f1
a) X = Knight+Antelope(3,4)
b) X = Knight+(0,5)Leaper
c) X = Wazir(0,1)+Zebra(2,3)+Giraffe(1,4)
d) X = Wazir(0,1)+Giraffe(1,4)+Corsair(2,5)
e) X = Zebra(2,3)+Giraffe(1,4)+Corsair(2,5)
f) X = Wazir(0,1)+Zebra(2,3)+Corsair(2,5)
(no wK)

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seetharaman
seetharaman
December 19, 2016 14:23

The black pieces are already in a cycle! Can the solutions be any different? 🙂 I suppose this is the record for the longest solution in Juliasfairies.

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shankar ram
shankar ram
December 20, 2016 09:39
Reply to  seetharaman

It just got longer… Departure squares added for all 2nd and 3rd moves of Black!

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Georgy Evseev
Georgy Evseev
December 20, 2016 14:02

Interesting, though pure geometry (and too much of it for my taste). It is also quite possible that by changing the positions of white knights more twins may be added. Still I do not want to check if it is really achievable.

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shankar ram
shankar ram
December 20, 2016 14:38
Reply to  Georgy Evseev

Thank you for your comments, Georgy!
As it happens, I did check all possible positions for the WSs. This was one which had an open position (mirrored BK!).

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Georgy Evseev
Georgy Evseev
December 22, 2016 09:37

Long explanation for those who do not want to go into details themselves.

There are two “knignt rectangles” (I use word “rectangle” to avoid using “square” ambiguously) around black king: d6-g5-f2-c3 and f6-g3-d2-c5.
Each black “knight” in each twin guards three of four squares of one of rectangles, so that each square is guarded three times. In twins the same piece guards another triplet of squares in same or other rectangle.
Each white “playing” knight can check black king from two squares – one from each rectangle.

Each black move after the key removes guards from one triplet, so they afterwards guarded like 3-2-2-2. White then check from the square guarded thrice, forcing to remove two guards from other squares and achieving 2-1-1 guarding (as one square cannot be reused). Then white check from square guarded twice, forcing 1-0 situation for remaining squares. One of squares is now not guarded and white can mate from it.

—
And my question in previous message (not well formulated, frankly) was the following:

If we change the position of white knights, we may make a twin when black “knights” are cyclically guarding all squares in full octagon – not in two rectangles. Still every square is guarded three times. Will this position be solvable? If yes, then how? Or will something extra be needed?

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shankar ram
shankar ram
December 22, 2016 13:08

Georgy,

we can have a full 8-cycle, with 8 B pieces each guarding 3 out of 8 squares in a cycle: ABC/BCD/CDE/DEF/EFG/FGH/GHA/HAB. But in this case, we don’t get a solvable problem. If the B piece guarding say, EFG moves, there is no way for W to continue, unless, as you put it “something extra” is added!
For a solvable “octagon”, we need 8 B pieces each guarding a different SEVEN of the 8 squares: ABCDEFG/BCDEFGH/CDEFGHA/DEFGHAB/EFGHABC/FGHABCD/GHABCDE/HABCDEF for a sound #8!
These points were discussed in my feenschach article on the Jacobs theme.
I think it’s time to write an updated article. I’m collecting (and composing!) new examples.

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Georgy Evseev
Georgy Evseev
December 22, 2016 14:59
Reply to  shankar ram

If we put black Knight+Antelope(3,4)+(0,5)Leaper+Zebra(2,3)+Giraffe(1,4)+Corsair(2,5)
units, then put four knights and some combined “horizontal/vertical leapers” (Like Horizontal Corsair+Knight b7 may play d6, d8 and g5 (giving check only in last case), but not a5, c5 and d2 – these are vertical moves),
then this silly cycle may even work.

On 9×9 board even knights are enough.

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Georgy Evseev
Georgy Evseev
December 22, 2016 22:37

And, btw, how to add user-defined piece combination?

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shankar ram
shankar ram
December 23, 2016 01:45
Reply to  Georgy Evseev

Georgy,
In winchloe, there is an option in the Solving menu called “Definition of the pieces”.
Here you can define up to 8 new pieces.

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Georgy Evseev
Georgy Evseev
December 23, 2016 08:46

Some theoretical try for #8.

White knights (instead of present ones):a2, a6, c8, g6, g8

White zebras (add): a1, b1, a4, e8, f5

Pawn (add): white c7, black a7

Black “knights” are NWCCL: non-wazir color-changing leaper – can leap to any square of another color except nearest ones.

Everything else is “unchanged” (WPe2 and WBe6).

Solution is 1.Zd3!

Thematic squares are c1, c3, c5, d6, f6, g1, h2, h6.

“Bad thematic squares” are b6 and c7. I have disabled them in easiest possible way, though WPc7 is suspicious.

The position is not computer-tested, as I am still not sure how to define additional pieces. For test, NWCCL may be defined as combination of 9 or so leapers (enough for this problem)

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Georgy Evseev
Georgy Evseev
December 24, 2016 09:58

I should stop writing to this topic, but still…

The following problem looks correct (WinChloe), though I have only tested the key (no cooks, no variations displayed) and defence 1…Oh7-h8 (correct variation tree).

White : Sg8a6c6a2g2 DBf5f3 We6c4g4e2
Black : Ke4 Oh8d7f7b5b3c2b1

DB=Dabbaba
W=Wazir
O=(0,1)+(1,2)+(1,4)+(2,3)+(2,5)+(3,4)

#7

1.We2!

There is place for white Sc8 and black Og6 (#8 with same key), but I think my PC is unable to test it.

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shankar ram
shankar ram
December 24, 2016 13:32
Reply to  Georgy Evseev

Very nice non-symmetrical setting, Georgy!
Oh8 position seems wrong though?

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shankar ram
shankar ram
December 24, 2016 13:44
Reply to  shankar ram

Oh8 should be on h7. On second look, the position is actually symmetrical too!
Think should try to avoid white Dabbabas.

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shankar ram
shankar ram
December 24, 2016 16:53

Took a deeper look. Realised you had shown position after B1 move h7-h8. Seems no way to avoid the W Dabbabas.
I “feel” the #8 setting should be sound, with key 1.We2.
So, congratulations, Georgy! I didn’t think an 8x Jacobs would be possible on the normal board.
Next challenge: 9x or 10x!

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Georgy Evseev
Georgy Evseev
December 25, 2016 11:37

Yes, I had posted position after the key 1.Wf2-e2 and black move Oh7-h8 by mistake (that was the last position checked).

In principle, we may try putting white Oh4 and Oe7 instead of dabbabas, though the duration of computer check greatly increases in this case. I have put to test the same 6 move position, but it is expected that it will require around 2 days of testing on my pc.

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shankar ram
shankar ram
December 25, 2016 12:00

Oh4 and Oe7 won’t work. They’ll be captured by the BOs. In fact, the BOs will capture ANY W piece on a black square.

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Georgy Evseev
Georgy Evseev
December 25, 2016 13:43
Reply to  shankar ram

That is the reason there are two of them: each controls all four star-shaped white squares. So after the capture the squares are still guarded, while all remaining moves are checks and black do not have time to capture second unit.

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shankar ram
shankar ram
December 25, 2016 16:45
Reply to  Georgy Evseev

You’re right. Sorry! Didn’t notice.
In that case, Os are preferable to Dabbabas.
The question of solving remains. Need to find a high powered PC running winchloe. Too bad winchloe doesn’t have something like Popeye’s startmovenumber option.

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Georgy Evseev
Georgy Evseev
December 25, 2016 19:59
Reply to  shankar ram

Unfortunately, white Os create cooks. (wOxbO and the threat from some far square like h2 which not only checks bK, but again covers all star-flights.)

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shankar ram
shankar ram
December 26, 2016 04:44
Reply to  Georgy Evseev

Maybe 4 WSs on e1, b4, e7, h4?

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Georgy Evseev
Georgy Evseev
December 26, 2016 08:18
Reply to  shankar ram

This may work. Still, even more symmetry and even longer computer check. I have put 8-mover with dabbabas on WinChloe (no variations – only key) and by rough estimate it will take around 10 days for the check (and afterwards I’ll have to check 8 7-movers to be sure). I am very doubtful I’ll have enough patience.

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Georgy Evseev
Georgy Evseev
January 5, 2017 16:20

The below #7 is C+ with reasonable key, distorted symmetry and no dabbabas. The key was checked and 7 defences of type Oxd4.

White : Sd8b7h7b3f3h3b1f1 Pd3 Wd7b5f5
Black : Kd5 Og8f7e6g6a2c2e2

1.Sbd4! with 5040 branched variations

I am more or less sure of #8 with dabbabas, but as I have written before, it is extremely difficult for me to fully test it. I am not so sure of position with knights.

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