Original Problems, Julia’s Fairies – 2013 (III): September- December
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No.378 by Paul Rãican – A complicated for solving problem with interesting addition of material on the board applying Sentinels condition! (JV)
Chameleon Chess: All pieces on the board which are displayed as orthodox Q, R, B, S, are Chameleons. A Pawn can promote only in Chameleon-pieces.
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.
Sentinels: when a piece (Ks included but not pawns) moves, a pawn of the same colour appears on the vacated square unless that square is on the first or eighth ranks or there are 8 pawns of that colour on the board already.
Maximummer – Black must play the geometrically longst move or may choose from among longest moves of equal length, distances being measured from the center of each square. Diagonal and oblique distances are measured from the orthogonal coordinates by using Pythagora’s theorem (take the square root of the sum of the squares of the orthogonal distances). All other orthodox chess rules apply.
No.378 Paul Rãican
1w -> ser-hs=15 (1+1)
Solution: (click to show/hide)
Try: 1.Ke5-d5[+wPe5]? then 1.b7-b5 2.b5-b4 3.b4-b3 4.b3-b2 5.b2-b1=Q 6.Qb1-h7=S 7.Sh7-f8=B[+bPh7] 8.Bf8-a3=R 9.Ra3-h3=Q[+bPa3] 10.Qh3-c8=S[+bPh3] 11.Sc8-d6=B 12.Bd6-b8=R[+bPd6] 13.Rb8-b1=Q 14.Qb1-g6=S 15.Sg6-e7=B[+bPg6] 16.Be7-h4=R[+bPe7] & 1.e5-e6 Rh4-a4=Q[+bPh4] = too long.
Solution: 1.Ke5-f6[+wPe5]! then 1.b7-b5 2.b5-b4 3.b4-b3 4.b3-b2 5.b2-b1=S 6.Sb1-c3=B 7.Bc3*e5=R[+bPc3] 8.Re5-a5=Q[+bPe5] 9.Qa5-d8=S[+bPa5] 10.Sd8-f7=B 11.Bf7-a2=R[+bPf7] 12.Ra2-h2=Q[+bPa2] 13.Qh2-b2=S[+bPh2] 14.Sb2-d3=B[+bPb2] 15.Bd3-h7=R[+bPd3] & 1.Kf6-g5[+wPf6] Rh7-h3=Q =
(С+ by Popeye 4.55)
Two model stalemates. (Author)