No.592 (IK)

No.592 
Igor Kochulov
(Russia)

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Original Problems, Julia’s Fairies – 2014 (III): September – December

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Please send your original fairy problems to: julia@juliasfairies.com


No.592 by Igor Kochulov – Bristol by two white Rooks! A nice play of Chameleons with Anti-Circe. I’m grateful for the dedication! (JV)


Definitions:

Chameleon: On completing a move, a Chameleon (from classical standard type) changes into another piece, in the sequence Q-S-B-R-Q… Promotion may be to a chameleon at any stage in the cycle.

Anti-Circe (Anti-Circe Calvet (the default type)): After a capture the capturing piece (Ks included) must immediately be removed to its game array square (necessarily vacant, else the capture is illegal). Captures on the rebirth square are allowed. Game array squares are determined as in Circe.

Circe: Captured units (not Ks) reappear on their game-array squares, of the same colour in the case of pieces, on the file of capture in the case of pawns, and on the promotion square of the file of capture in the case of fairy pieces. If the rebirth square is occupied the capture is normal.


No.592 Igor Kochulov
Russia

original – 03.09.2014
Dedicated to Julia Vysotska

Solution: (click to show/hide)

white kc8 rg1h1 pc5c7d6e2f6g2h2 white Chameleon bb3 qf8 re3 sf3h4 black ke6 re4f7 sg7 bh8 pc6h5

#2                                           (15+7)   
Chameleons: Qf8,Re3,Bb3,Sf3,Sh4
Anti-Circe


3 Responses to No.592 (IK)

  1. Georgy EvseevGeorgy Evseev says:

    Unfortunately, the key seems to take flight e7.

  2. Nikola Predrag says:

    Hmm, one square-vacation + 5-fold Bristol clearance by wRg1.
    Quintuple-avoidance – 5 combinations of 4 squares which have to be avoided + one square particularly required (d1), for the arrival of wRh1.
    Exemplary use of Chameleons!

  3. Geoff Foster says:

    Initially 1…Ke7 is possible, because after 2.dxe7? or 2.fxe7? the rebirth square e2 is blocked, while after 2.cQxe7=cS? the rebirth square g1 is blocked. In addition, initially the bK is not in check from the cBb3, because after 1.cBxe6=cR?, the cR’s rebirth square (h1) is blocked.

    After 1.Ra1 there is no threat, even though it seems that any move by the wRh1 would mate the bK (cBxe6=cR[cRe6>h1] would be possible). However each time the wR blocks a rebirth square, as follows:

    2.Rh1-b1? Kxd6[ bKd6>e8]! (3.cQxe8=cS[cSe8>b1] impossible)
    2.Rh1-c1? Ke5! (3.cSxe5=cB[cBe5>c1] impossible)
    2.Rh1-d1? Rc4! (3.cRxe6=cQ[cQe6>d1] impossible)
    2.Rh1-e1? Kd7+! (3.Kxd7[wKd7>e1] impossible)
    2.Rh1-f1? Kf5! (3.cSxf5=cB[cBf5>f1] impossible)
    2.Rh1-g1? Ke7! (3.cQxe7=cS[cSe7>g1] impossible)

    However Black is in zugzwang, and any move by the bRf7, bSg7 or bRe4 blocks one of the bK’s escape squares. Also 1…Kd5 allows the double-check 2.Rd1.

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