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Original Problems, Julia’s Fairies – 2015 (I): January – June →Previous ; →Next ; →List 2015(I) Please send your original fairy problems to: julia@juliasfairies.com |
No.764 by Vladislav Nefyodov – Cycle of white and black moves with a rich additional strategy! (JV)
Definitions:
NAO (NA): (1,2) Chinese. Chinese piece operating along the lines of Nightrider: moves as Nnightrider, but captures only by hopping over a hurdle to any square beyond.
Nightrider(N): (1,2) Rider. Operates along straight lines with squares lying a Knight’s move away from each other.
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No.764 Vladislav Nefyodov |
Solutions: (click to show/hide) |
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White Kb4 Rh3 Rf1 Bh5 Sf3
Black NAd1 Ke4 Rf8 Bh6 Bd5 Sa4 Sb2 Pc3 Pd3 Pg2
h#2 3 solutions (5+10) |
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It is C+ Winchloe too !
SL
Very nice! There is also the cycle of black interferences and cycle of change of roles of the thematic black pieces!
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-“There is also the cycle of black interferences and cycle of change of roles of the thematic black pieces!”-
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There is one cycle of 3 black pieces+2 black functions (selfblock “s” and interference “i”).
The CHAIN of phases I-II-III automatically includes the s-CHAIN of selfblocks and the i-CHAIN of interferences.
It is the combined chain, made of s-chain and i-chain, which is closed into the presented CYCLE.
Each phase connects one “piece+function” with the other “piece+function”.
I=(B+s)-(R+i) or Bs-iR
II=(R+s)-(N+i) or Rs-iN
III=(N+s)-(B+i) or Ns-iB
The combined chain is connected through a certain piece which has “i” function in one phase and “s” function in the next phase making a 3×2 cycle:
Bs-iRs-iNs-iBs-iRs-iNs-…etc., (in a circle)
White pieces present a complete 3×3 cycle of pieces+functions:
Rf1,Rh3,Bh5 +
guard (f3), pin, mate
The following version saves a few pieces:
White Bb6 Rd6 Kf6 Sd4 Rh4
Black Ba6 Ra4 Kc3 Bb2 Pd2 NAh1
Curiously, two selfpinning black 2nd moves remove second guard from mating lines (Ra4-a5, NAh1-g3), while bishop does not. To rectify this slight defect, one more piece might be necessary, in the version (with additional unguard Bb5-c6, having full analogy again):
White Bb6 Rd6 Kf6 Sd4 Rh4
Black Rd8 Bb5 Ra4 Kc3 Bb2 Pd2 NAh1
Bravo Juraj! Nao instead of nightrider seems to make a world of difference!
Bad news this time. Is the following too much of anticipation?
Gerard Smits
3rd Prize Themes-64 1984-1985
White : Kg2 Re3b2 Be8
Black : Kc4 Rd6 Bh8b5 Sa7 Pa6c5d3d2 Na5b1c1
h#2
3+0 nightrider
b) d3->b4
c) b5->b3
I see the only important difference in the form (3 solutions better than twinning), but the basic scheme is obviously the same.