A Circe Chess problem - 1: Solution
Stefanos Pantazis, editor of The US Problem Bulletin, has sent me two problems that appeared in that journal and did win a prize there.
This is solution of the first problem: You can also look at the second problem: a helpstalemate in 4 moves.
This problem was composed by George P. Sphicas, was published in the US Problem Bulletin in 1993, and won a Second Prize.
King d4; Pawn d2, e6, h4.
King g4; Queen g5; Knight f2; Pawn f7, h6, h7.
Circe. Series-selfmate in 17 moves.
(b) Move the black knight from f2 to h5, and solve the problem again.
The comments can be ignored, but may help to understand the notation.
(a) The solution of the first part is:
1. e x f7. Black pawn is not reborn because square f7 is occupied.
2. f8 =Q. Pawn promotes to Queen.
3. Q x f2 (Nb8). The black knight is reborn at b8.
4. K d3.
5. K e2.
10. d8 =Q. Pawn promotes to Queen.
11. Q d1.
12. h x g5 (Qd8). Black queen goes back to d8.
14. g x h7. Black pawn is not reborn because square h7 is occupied.
15. h8 = Q. Promotion to queen again.
16. Q x h6 (h7). Pawn is reborn at h7.
17. K e1 +.
17. ..., Q x d1 mate. Only legal move of black gives mate.
(b)The solution of the second part is:
2. e8 = R. Promotion to rook.
5. d4. 6. d5. 7. d6. 8. d7. 9. d8=R. Promotion to rook again.
10. R d4.
11. h x g5 (Q d8). Rebirth of queen on d8.
13. g x h7. Again, pawn disappears.
14. h8 = R. Promotion to rook.
15. R x h6 (h7). Pawn rebirth on h7.
16. R x h5 (Ng8). Knight rebirth on g8.
17. K e5+
17. ..., Q x d4 (Ra1) mate. While the rook is reborn on a1, this only legal move of black gives again mate.
The problem couldn't be computer tested, and the possibility of a cook (additional solution, shorter solution, ...) cannot be ruled out completely, but seems not too likely. Any reader finding such a cook is requested to contact me.
Written by Hans Bodlaender; with thanks to Stefanos Pantazis.
WWW page created: January 6, 1997. Last modified: January 14, 1997.