Problem 250: Nikola Predrag - Fairy (Patrol, PAO, VAO, Hoppers) |
1.VAf6? (zz) 1...BHg4 x(a) 2.Kg5# X(B)+Y(B) 1...RHe5 y(e) 2.Kg5# X(B)+Y(B)+Z(D) 1...BHe4!
1.PAd4! (zz) 1...RHf6 y(a) 2.Kg5# X(B)+Z(D) 1...RHg5 x(b) 2.Ke4# Y(C)+Z(D) 1...BHe4 y(c) 2.Kg4# X(D)+Z(B) 1...BHg4 x(d) 2.Kf6# Y(A)+Z(B)
Cyclic transfer of 2 reciprocally exchanged functions (x/X,y/Y) by Black/White on 4 squares (a/A,b/B,c/C,d/D): x/X=block of bK's flight; y/Y=line-interference as a hurdle (flight-blocks are complemented by flight-guards=Z)
...->hurdle f6(Black)+block g5(White)-> ->block g5(Black)+hurdle e4(White)-> ->hurdle e4(Black)+block g4(White)-> ->block g4(Black)+hurdle f6(White)->...
Initially, wK attacks g5+g4, due to the 'observer' wSg3. Respectively, after black selfblocks on g5/g4, wK abandons the guard of g4/g5 but plays to observe (as a hurdle) PAd4/VAe7 which will attack g4+h4/g5+h4. Respectively, after black hurdles on f6/e4, the mates must be changed-> VAe7/PAd4 do not attack but only observe g5/g4, so wK must block these squares but now (due to a new 'observer') wK can attack h4 and continue to attack g4/g5. The specific features of Patrol chess are presented by Black/White cyclic exchanges of hurdle/block functions and the reciprocal exchanges of attacking/observing functions of white pieces. (Author) Patrol Chess: A piece can capture or give check only if it is observed by a piece of its own side. PAO: Moves as Rook, but captures only by hopping over a hurdle to any square beyond. VAO: Moves as Bishop, but captures only by hopping over a hurdle to any square beyond. Rookhopper: Moves along Rook-lines over another unit of either color to the square immediately beyond that unit. A capture may be made on arrival, but the hurdle is not affected. Bishophopper: Moves along Bishop-lines over another unit of either color to the square immediately beyond that unit. A capture may be made on arrival, but the hurdle is not affected. |
KOBULCHESS.COM
site for chess composition
General editor:
Diyan Kostadinov
Co-editor:
Seetharaman Kalyan
Comments
Question:
Why in variation 1...RHf6 y(a) 2. Kg4#? is not mate
It's not a "full-blooded" 2-mover but I hope it might be amusing to the others as it was to me. The difference between the attacked and only observed squares is a simple but basic idea of the problem, together with specific blocks and hurdle effects. I tried to arrange these simple elements into one cohesive idea as intensively as I was able.
After 1...RHf6 2.Kg4?, flight g5 is observed but not attacked, wK is not observed by wSg3 any more and wPAd4 is not observed (so it only observes but does not attack h4). 2.Kg4? is not even a check to bK.
RHf1-f6>RHf6-f4,RHa5-g5>RHg5-e5,BHh7-e4>BHe4-g6,BHc8-g4>BHg4-e6.
Hurdle-removal from f5 looks accidental in the mates but some imprecise white attempts in 2nd move reveal the hidden defences: 1...BHg4 2.PAe4(f4)+? BHe6!; 1...RHg5 2.VAf6+? RHe5!(RHg2!)
These attempts fit well in the concept-> wK remains observed on f5 and PA/VA approach him to become observed but hurdle on f5 serves Black.
Very specific problem with combination of some of my favourite elements: immobilization of hurdle, antibattery, patrol chess, ...
Many thanks for dedication, Nikola!
The cycle of squares is used to link the BW exchange of functions into a cycle. Such exchange is my favourite element :)
Cheers!
RSS feed for comments to this post