# No.347,348 (PH)

 No.347, 348 Peter Harris (South Africa) Original Problems, Julia’s Fairies – 2013 (II): May – August  →Previous ; →Next ; →List 2013(II) Please send your original fairy problems to: julia@juliasfairies.com

No.347, 348 – Two problems by Peter Harris demonstrate Anti-Andernach Chess condition in interesting combinations with other Fairy conditions and unusual fairy pieces. (JV)

Definitions:

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.

Ultra-Patrol Chess – A piece can move, capture or give check only if it is observed by a piece of its own side.

Anti-Circe: After a capture the capturing piece (Ks included) must immediately be removed to its game array square (necessarily vacant, else the capture is illegal). R, B & S go to the square of the same colour as the capture; Ps stay on the file of capture.

Orphan (O): An orphan can move only when observed by an enemy piece; when so observed it can move like the piece(s).

Non-Stop Equihopper (NE): Moves along any line over another unit of either colour to a square situated such that the hurdle stands at the mid-point between the Equihopper’s departure and arrival squares. The English (standard) Equihopper cannot pass over an obstruction other than the hurdle when playing along Queen-lines. The non-stop/French Equihopper does not have this restriction.

Equistopper (QE): moves to any square that is exactly halfway towards any other unit of either colour, which must (naturally) be an even number of rows (ranks and files) away orthogonally.

 No.347 Peter Harris South Africa original-14.07.2013   h#5                                               (1+3) Anti-Andernach-ChessUltra-PatrolOrphan a2     Solution: (click to show/hide)   1.Rh8-h2=wR Rh2-b2=bR 2.Rb2-b3=wR Rb3-b8=bR 3.Ra8-a3=wR Ra3-b3=bR 4.Rb3-b2=wR Rb2-b1=bR 5.Rb8-b3=wR Rb3xb1 # No.348 Peter Harris South Africa original-14.07.2013   h#4        b) wLIb2→bLIb2       (5+1)Anti-CirceAnti-Andernach ChessOrphan b8Non-Stop Equihopper b1Equistopper b7Lion b2Locust b3     Solution: (click to show/hide)   a) 1.Kb5-c5 LIb2-b6=bLI 2.LIb6-d4=wLI NEb1-f7=bNE 3.Kc5xd4[bKd4->e8] QEb7-d7=bQE 4.Ke8-d8 Lb3xf7-g8[wLg8->f8] # b) 1.Kb5-c6 NEb1-b5=bNE 2.NEb5-d7=wNE QEb7-c7=bQE 3.Kc6xd7[bKd7->e8] Ob8-g3=bO 4.LIb2-b7=wLI Lb3xg3-h3[wLh3->g8] #

### 3 Responses to No.347,348 (PH)

1. Dmitri Turevski says:

Could someone explain No.347 for me, please?

After 1.Rh8-h2=wR the wRh2 is able to move (because it will make the next move)
Therefore it is observed by some white unit.
This unit can only be wOa2 (there are no other white units).
Therefore wOa2 is able to observe as a rook.
Therefore it observes bKa1
Therefore the move 1.Rh8-h2=wR must have been an illegal self-check.

2. peter harris says:

Thank you Dmitri; you are quite right. Last night after having sent the problem, something at the back of my mind was worrying me and then I remembered. Some years ago I experienced this ultrapatrol / orphan bug in Popeye and decided to avoid the combination in future but as you see forgot about it. To preempt comments I was going to write this morning – but Dmitri was too quick for me! It is a phenomenon that a computer program can have the affect of suspending one’s own senses – a sort of hypnosis whereby one is induced into accepting nonsense.

• Thank you Mr. Harris for your explanation. I have submitted a bug-report to the popeye development team.