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Original Problems, Julia’s Fairies – 2014 (III): September – December →Previous ; →Next ; →List 2014(III) Please send your original fairy problems to: julia@juliasfairies.com |
No.606 
No.606 by Julia Vysotska – Chameleons with Back-to-Back condition! (JV)
Definitions:
Back-To-Back: When pieces of opposite colors stand back-to-back with each other on the same file (white piece is on the top of black!), they exchange their roles. A pawn on the first rank cannot move. Any piece can make an en passant capture when it has got a role of Pawn by Back-To-Back.
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.
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No.606 Julia Vysotska |
Solutions: (click to show/hide) |
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White Kc2 Be7 Pd6 Pg3
Black Ke8 Pc4 Ph3
White Chameleon Qg2
Black Chameleon Qf7 Bd7
h#2 2 solutions (5+5) |
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What is interesting is the avoided mate in the first solution 2.cSc6? 3.cQc5! Wish there was a similar effect in the other solution!
Yes, this virtual play makes the dynamics of the problem. It would be worth to make the virtual play clear and convincing and use it as the main motivation for the real play.
The virtual content is present here but it’s not in a clear balance with the other solution.
1.cQd5/f5=cS? cQe4=cS 2.cSe3=cB cQg6=cB+ 3.cBg5=cR!
Correction of my previous post:
2…cSg6=cB+(not cQg6)
Julia has actually mentioned indirectly the virtual play:
“Position of the pairs wcSd5+bcQd4 and wcSe2+bcBe1 is determined by pawns c4/g3 preventing BTB specific defenses 3.cQd4-a4=cS / 3.cBe1-h4=cR.”
wcSe2+bcBe1 is determined because that pair as wcSe4+bcBe3 wouldn’t work due to the thematic defense 3.cBg5!
I don’t see the other potential position of the pair wcS+bcQ where the same defense would appear.
So, bPc4 does not determine the play of that pair but only the mating move 2…cSb5=cB#!(2…cSc6=cB+?).
wPg3 determines the complete play of the pair wcQg2+bcQf7.
Thus the logic behind wPg3 is much more dynamical and deeper, therefore wPg3 appears as an element which is “active” during that whole phase.
bPc4 is static/irrelevant until the very end and appears as a mere dual-stopper.
Agree with the comments above. Yes, I’ve seen the virtual play in the 2nd solution, and I wish I could have the same in the 1st solution. This problem is a bit trial problem. I was curious about a combination Chameleons + BTB, and tried to find anything specific. The complication of the white play here is that in 2-3 moves a Chameleon in any initial phase can transform into some phase to give mate without BTB. But the black play can have two interesting options: one is presented here, where the phase of the black Chameleon in BTB pair requires closing of its line for the mate; another one is black transformation into the phase where the next transformation won’t prevent the mate (like in my case it might be a pair wcSd5+bcRd4 and 3.cRa4=cQ? – still mate).
Otherwise I’ve tried to have a light problem here, with all moves made by Chameleons and a solutions “decorated” with cross-checks. I also hoped that this example would encourage some of you to try Chameleons+BTB, maybe with some other stipulation.