No.284, 284.1 
János Mikitovics (Hungary)

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Original Problems, Julia’s Fairies – 2013 (I): January – April

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red-roses2Happy Birthday to Seetharaman Kalyan! (birthday: 18.03.1949)
No.284 – János Mikitovics – A non-standard fairy condition Cage Circe here is combined with KoBul Kings
No.284.1 – János Mikitovics – An improved version of No.284! (JV)


Definitions:

KoBul Kings: When a piece (not a pawn) of his own side is captured, a King transforms into a Royal piece of the same type as the captured one. When the King is in the form of any Royal piece and there is a capture of one of the pawns of his own side, he becomes a normal King again. Сaptures are illegal if their result is self-cheсk because of the transformation of the Кings according to KoBul rules. Castling is allowed only if the KoBul King is on his initial square in the form of a normal King and if he has not already moved; however he may already have been transformed. In the case of capture by a King in AntiCirce he is reborn on his initial square and may castle. If the capture is by a King which is in the form of some Royal piece, he is reborn on the initial square of that piece.

Cage Circe (from “Le circe cage ou la genese d’une condition feerique”, Nicolas Dupont, Etienne Dupuis -Octobre 2009):

1. Captured pieces are reborn if there is one or more rebirth squares for the captured piece. When there are none, the captured piece is de nitively removed from the board. As kings can not be captured, checks and mates are orthodox.

2. Rebirth squares are called cages. A cage for a given piece is a square such that if the given piece lied on that square, its only legal moves would be captures.

3. When there are more than one cage, the side making the capture chooses the rebirth square from the available cages.

4. Cages are determined after the capture has taken place.

5. White pawns may reborn on the 1st row, from which they can move like ordinary pawns, including making a double step from the second row. Same for black pawns.

6. White pawns may also reborn as promoted pieces on the last row. In such case, the rebirth square must be a cage for the promoted piece, the type of which is choosen by the side making the capture. Same for black pawns.

7. A reborn piece is “brand new”; its move history is lost. For example, a white rook reborn on h1 retrieves the ability to castle.


No.284 János Mikitovics
Hungary
original-18.03.2013
Dedicated to Seetharaman Kalyan
in occasion of his birthday!
 
284-h=3-jm
h=3                                               (3+2)
b) Pc7→d7; c) Rd5→h8
KoBul Kings
Cage Circe
 
 
Solution: (click to show/hide)
 
No.284.1 János Mikitovics
Hungary
Improvement of No.284, 29.03.2013
Dedicated to Seetharaman Kalyan
in occasion of his birthday!
 
284.1-h=3,5-jm
h=3,5            2 solutions             (3+2)
KoBul Kings
Cage Circe
Royal Sh2
 
 
Solution: (click to show/hide)
 

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Janos Mikitovics
Janos Mikitovics
March 19, 2013 02:46

{If from b)
>Rd5→h7
h=3 KoBul Kings
1.Ke6-e7 Kh3-g4 2.Qf7-g7 + Kg4-f5 3.Ke7-f7 d7-d8=Q =
or
>Rd5→h8
h=3 KoBul Kings
1.Ke6-e7 Kh3-g4 2.Qf7-g8 + Kg4-f5 3.Ke7-f8 Kf5-f6 =}

seetharaman
seetharaman
March 19, 2013 09:15

Thanks Janos for the greeting and dedication.

Beautiful problem showing three nice pins of the black queen.

Janos Mikitovics
Janos Mikitovics
March 20, 2013 09:32
Reply to  seetharaman

From No.284 c)
Dedicated to Seetharaman Kalyan
in occasion of his birthday!
White Pd7 Pe7 Kh3
Black Qc8 Ke6
h=3 b) Pd7→h7 (3+2) C+
KoBul Kings
Cage Circe

a) 1.Ke6-f6 e7-e8=Q 2.Qc8*d7 [+wPg7] + Kh3-g3 3.Qd7*g7 [+wPh8=wQ] + Kg3-f4 =

b) 1.Qc8-f8 h7-h8=R 2.Ke6-f7 Rh8-h7 + 3.Kf7-g8 e7-e8=R =

The diagonal and horizontal pins of bQ are with the promotions Q&R.

Diyan Kostadinov
Diyan Kostadinov
March 27, 2013 03:25

When I received and published KoBul Kings + Circe Cage problem by Themis Argirakopoulos on my website I though it was the first such problem, but later I remember about your problem 284 here Janos, which probably is the first combination of these two fairy conditions.
In my opinion this type problem will be much better if the Kings transformations are more used. Your second version with R/Q promotions is also nice, probably can be even accept as an independent problem.
You can see the Argirakopoulos problem here: http://kobulchess.com/en/problems/originals2013/227-155-themis-argirakopoulos.html

Janos Mikitovics
Janos Mikitovics
March 28, 2013 22:43

Themis Argirakopoulos problem is an excellent creation! Congratulations to Author!
Gratitude for you Diyan, I found a good setting with the transformations of Kings:

Improvement of No.284
White royal Sh2
White Pd6 Rg4
Black Qd7 Kf6
h=3,5 2 solutions C+
KoBul Kings + Cage Circe

I) 1…rSh2-f3 2.Kf6-e6 rSf3-g5 + 3.Qd7*d6 [+wPe5] [wrS=rK] Rg4-e4 4.Qd6*e5 [+wPd8=wQ] + rKg5-g6 =

II) 1…Rg4-e4 2.Qd7*d6 [+wPf5] [wrS=rK] + Re4-e5 3.Qd6*e5 [+wRf4] [wrK=rR] rRh2-h5 4.Qe5*f5 [+wPe8=wQ] [wrR=rK] rKh5-h6 =

Chameleon echo-stalemates.

Diyan Kostadinov
Diyan Kostadinov
March 30, 2013 22:28

Well done Janos! The using of wK transformations is good improvement. The specific echo chameleon stalemates where the bQ can not capture the wR because of selfcheck by the transformed wK are also nice.

Janos Mikitovics
Janos Mikitovics
March 31, 2013 15:05

Thank you for the kind words of the inventor of the fairy condition(s) KoBul Kings! Thank you, Diyan!

Nicolas Dupont
Nicolas Dupont
April 1, 2013 05:01

Congratulations too, from one inventor of the other fairy condition you used! I’m pleased to see that Cage Circe is still alive and deserves such a nice stuff!

Janos Mikitovics
Janos Mikitovics
April 1, 2013 14:34

Thank you, Nicolas! Of course the Cage Circe is also a superb fairy condition which is a very good combination with the KoBul Kings and others! I will to search the new opportunities.