# No.235 (JM)

 No.235 János Mikitovics (Hungary) Original Problems, Julia’s Fairies – 2013 (I): January – April →Previous ; →Next ;  →List 2013(I) Please send your original fairy problems to: julia@juliasfairies.com

No.235 – s#7 by János Mikitovics – A nice fairy 5-men! (JV)

Definitions:

Maximummer – Black must play the geometrically longest move or may choose from among longest moves of equal length, distances being measured from the center of each square. Diagonal and oblique distances are measured from the orthogonal coordinates by using Pythagora’s theorem (take the square root of the sum of the squares of the orthogonal distances). All other orthodox chess rules apply.

KoKo – A move is possible only if the piece moved arrives on a square next to another unit.

 No.235 János Mikitovics Hungary original-21.01.2013 Dedicated to Petko A. Petkov   s#7                                             (3+2)Maximummerb) Maximummer; KoKoNightrider g8     Solutions: (click to show/hide)   a) 1.Ng8-f6 zz. Rf2-a2 2.Bh5-f3 + Ra2-g2 3.Bf3-e2 zz. Rg2-g8 4.Ke3-f2 zz. Rg8-g1 5.Nf6-b4 zz. Rg1-g8 6.Kf2-f1 + Kh1-h2 7.Nb4-f2 Rg8-g1 # b) 1.Ke3-e2 zz. Rf2-f8 2.Ng8-e7 zz. Rf8-f1 3.Ke2-f2 zz. Rf1-e1 4.Kf2-f1 zz. Re1-e6 5.Kf1-g1 + Kh1-g2 6.Ne7-g6 zz. Re6*g6 7.Bh5-f3 + Kg2-f1 # (Py 4.61) Mutual batteries and zugzwangs, self-blockades, selfpin-unpin, Baby-aristocrat. (Author)

### 10 Responses to No.235 (JM)

1. Nikola Predrag says:

Repeated moves (g8-g1-g8-g1) of bR in a) are not so interesting as the route (f2-f8-f1-e1-e6-g6) in b).

wN nicely helps wK to reach f1 with check (disabling the premature switcback of bR to g1) in a) and disables 3…Rg1 in b), 7.Bf3+ is the climax – forces a battery check 7…Kf1+ and prevents the flight of wK to f2. 8.Bg2?? would parry the check by bR but then bK would become the checking piece.

Altogether, very good and interesting play with only 5 pieces.

2. Paul Raican says:

With Chameleon Bishop h5 we have a third solution:
1. cBh5-d1=cR + ! Kh1-g2 2.Ng8-c6 Rf2-f8 3.Ke3-e2 Rf8-f1 4.cRd1-d2=cQ Rf1-f8 5.Ke2-d1 + Rf8-f2 6.Nc6-e2 Rf2-f8 7.Kd1-e1 Rf8-f1 #

3. Janos Mikitovics says:

The main theme is: creations of the mutual batteries (by black & white) with the same pieces. Unfortunately, this condition is not satisfied with Chameleon Bishop h5.

• Nikola Predrag says:

What do I not see? I see one white royal battery in a), one white antibattery and black royal battery+antibattery in b). Paul’s comment presents one white royal battery with another rear piece.

4. Janos Mikitovics says:

Dear Nikola,
As I wrote, the main theme is: creations of the mutual batteries (by black & white!!) with the same pieces.
Paul does not presents a black battery with Chameleon Bishop h5. This is a partial solution only, which I do not like…
Best,
Janos

• Nikola Predrag says:

Dear Janos,
I agree that Paul did not present BLACK battery with a WHITE piece, actually I can not imagine how such a thing could be possible at all, except with some additional condition like Andernach.
Perhaps I do not understand what you want to say or I simply fail to see it. If there is indeed a black battery with a white piece in your problem, I will be grateful if you can explain it. Is it so obviously clear to everyone except me?

5. Janos Mikitovics says:

Dear Nikola,
– The original position (with wB on h5) have solutions with black and white batteries.
– The second position (with wcB on h5) have one solution with a white battery only.
“Paul does not presents a black battery with Chameleon Bishop h5.” Here: the Chameleon Bishop h5 = the second (twin) position! The wcBh5 does not presuppose a black battery with white pieces. How may you have thought of this?
I hope that you are rested already…
Best,
Janos

• Georgy Evseev says:

I would agree with Nikola that there is no black battery in a). I would also say, there is no white battery in b).

Do we have a definition discrepancy here?

• Nikola Predrag says:

Dear Janos,
there is no white battery and also no black battery in twin b) of the original. So, I can not understand at all what you are talking about. There is also no black battery in a).

• Janos Mikitovics says:

Dear Friends,

I repeat my answer in more detail:

I. The original position (with Bishop on h5) has two solutions with conditions a) Maximummer b) Maximummer + Koeko:

Maximummer
a) 1.Ng8-f6 zz. Rf2-a2 2.Bh5-f3 + Ra2-g2 3.Bf3-e2 zz. Rg2-g8 4.Ke3-f2 zz. Rg8-g1 5.Nf6-b4 zz. Rg1-g8 6.Kf2-f1 + Kh1-h2 7.Nb4-f2 Rg8-g1 # [with a white battery]

Maximummer + Koeko
b) 1.Ke3-e2 zz. Rf2-f8 2.Ng8-e7 zz. Rf8-f1 3.Ke2-f2 zz. Rf1-e1 4.Kf2-f1 zz. Re1-e6 5.Kf1-g1 + Kh1-g2 6.Ne7-g6 zz. Re6*g6 7.Bh5-f3 + Kg2-f1 # [with a black battery (the white b. is a gift)]

II. The second position (with Chameleon Bishop on h5) has one solution with condition Maximummer:

Maximummer
a) 1.cBh5-d1=cR + ! Kh1-g2 2.Ng8-c6 zz. Rf2-f8 3.Ke3-e2 zz. Rf8-f1 4.cRd1-d2=cQ zz. Rf1-f8 5.Ke2-d1 + 5…Rf8-f2 6.Nc6-e2 zz. Rf2-f8 7.Kd1-e1 Rf8-f1 # (with a white battery)

Maximummer + Koeko
b) I do not see another solution with a black battery.

The main theme is: creations of the mutual batteries (by white & black) with the same pieces. Unfortunately, this condition is not satisfied in the second position only with white.

I’m sorry, but I can not present my idea more clearly with my little English knowledge.

Best wishes,
Janos