Fairy chess composition

# Original Problems (27)

## Original Problems (page 27)

Original fairy problems published during 2012 will participate in the informal tourney JF-2012

The site is mostly about fairies, but h# and s# are also welcomed for publication! Please send your problems to my e-mail: julia@juliasfairies.com

### Go to →List of Problems; →Page 26 ;  →Page 28

No.68 – h#3.5 by Peter Harris – Аn interesting miniature with three fairy conditions and surprising content which demonstrates a lot of specific promotions. Probably – complicated for solving? Pay attention that in initial position the b.K is not under check, because the moves 1. fxg8=Q??, 1.fxg8=R??, 1.fxg8=B?? and 1.fхg8=S?? are illegal! (JV)

Definitions:

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.

Anti-Andernach: A piece (excluding King) changes its color after any non-capturing move. After capture, the piece retains its color. Rooks on a1, h1, a8 and h8 can be used for castling, provided the usual other rules for that move are satisfied. After castling, Rooks do not change color, If White makes a non-capturing move with neutral or halfneutral piece, that piece becomes black and vice versa.

AntiCirce: antiCirce 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. antiCirce Cheylan: As antiCirce Calvet except that captures on the rebirth square are not allowed.

You can сlick on “Solutions” to show or hide the solutions!

 – No.68 Peter Harris South Africa original-06.08.2012 – h#3.5        b) Sb1↔Qd1       (6+1) Maximummer Anti-Circe Anti-Andernach (no w.K)   – Solutions: (click to show/hide) a) 1…Qd1-c1=bQ + 2.Kg8-h7 Bf1-h3=bB 3.Qc1*h6[bQh6->d8] Sb1-d2=bS 4.Qd8-g5=wQ Rh1*h3[wRh3->h1] # b) 1…Rh1-h2=bR + 2.Kg8-h7 f7-f8=bR 3.Rh2-a2=wR Ra2-f2=bR 4.Rf8-a8=wR Sd1*f2[wSf2->g1] # –

The diagrams are made on WinChloe and its Echecs font is used for Logo design

Subscribe
Notify of