# Оригинальные задачи (32)

## Оригинальные задачи (страница 32)

Оригинальные сказочные задачи, опубликованные в течение 2012-ого года, участвуют в неформальном конкурсе JF-2012!

Сайт в основном посвящен сказочному жанру, но h# и s# тоже будут опубликованы! Пожалуйста, присылайте свои произведения на адрес: julia@juliasfairies.com

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No.75 – ser-h=2 by Peter Harris – A short, but difficult play, which gives Anti-Super-Circe condition. (JV)

No.76 – ser-h=4 by Peter Harris – This problem is of course more difficult – but the play is far more interesting. (Author)

Определения:

Circe: Captured units (not Ks) reappear on their game-array squares, of the same colour in the case of pieces, on the file of capture in the case of pawns, and on the promotion square of the file of capture in the case of fairy pieces. If the rebirth square is occupied the capture is normal.

Anti-Super-Circe: After a capture, the side making the capture can place the capturing unit on any empty square. [Placement is compulsory]. A Pawn placed on its first rank is immovable.

Sentinels Pion advers:   When a piece (not a Pawn) moves, a Pawn of the colour of the opposite side appears on the vacated square if it is not on the first or the last rank, and if there are less than 8 Pawns of that colour on the board.

Вы можете нажать на “Solutions”(решения) чтобы увидеть или спрятать решения!

 – No.75 Peter Harris South Africa original-08.08.2012 – ser-h=2          b) Sc1→h2         (3+5) Circe Anti-Super-Circe   – Solutions: (click to show/hide) a) 1.d4*c3 [+wQd1][bPc3->c2] 2.e2*d1=R [+wQd1][bRd1->a1] Qd1-g4 = b) 1.d4*c3 [+wQd1][bPc3->b2] 2.Kh1*h2 [+wSg1][bKh2->a1] Sg1*e2 [+bPe7][wSe2->e6] = – I have deliberately made it very short – Ser-H=2. Black makes 2 moves, then White stalemates. I hope this shortness will encourage visitors to solve. (Author) – No.76 Peter Harris South Africa original-08.08.2012 – ser-h=4         2 solutions         (4+4) Circe Sentinels Pion advers Anti-Super-Circe   – Solutions: (click to show/hide) I. 1.Ka1*b1 [+wQd1][bKb1->f2] 2.b2-b1=B 3.Bb1*h7 [+wBf1][bBh7->a1] 4.Kf2*f1=K [+wBf1][bKf1->b2][+wPf2] Bf1-a6 = II. 1.Ka1*b1 [+wQd1][bKb1->b8] 2.a2-a1=R 3.Ra1*d1=R [+wQd1][bRd1->c7] 4.b2-b1=R Qd1*b1 [+bRa8][wQb1->a6] = – This problem is of course more difficult – but the play is far more interesting.(Author)

Все диаграммы сделаны на  WinChloe и ее фонт Echecs использован для дизайна Лого.