Wednesday, April 02, 2008

Version of an Unsolvable

Dear readers, I propose to you to solve a problem with no-solution.

The composer Dr. Kurt Dittrich has published it in 1918 with stipulation “White plays and mates in 4 moves” but on square b3 he had put a black soldier and the problem had no solution. This problem has reappeared in chess books as an example of problem without solution [See Jean Bertin, ”Initiation au Problėme d’ échecs” (Introduction to chess problems), Insolubilité (lack of solution), p. 30].

(Problem 22)
Dr. Kurt Dittrich,
”Deutsches Wochenschach“, 1918,
version by Iatridis Stavros
White plays and mates in 4 moves
#4 (7+10)

The Olympic winner Iatridis Stavros (1887-1976) has corrected the problem and the stipulation is valid.

The solution of problem-22 contains line clearances, line closings, and Theme Turton.

Key: 1.Qb2! [2 Qd2#]
2.Rg8 (Two pieces (Q and B) have left g-file allowing the move Rg8. The threat now is [3.Sg4#]).
3.Qg7 (The linear piece R going up the g-file has passed over the critical square g7, then on g7 has arrived  the linear piece Q, which will move now to the opposite direction going down the g-file with threat [4.Qg1#], thus we have theme Turton. Two pieces (B and R) have moved clearing g-file and allowing the move Qg1).
3...Bg5 (The queen has come back on its previous place (switchback of wQ), but the bishop that has moved can not come back because the pawn has closed the diagonal. Now black is forced to find a different defense, giving new possibilities to white).
4.Qa7# (Two pieces (pawn and B) have left row-7 allowing the move Qa7).

The correction is the following:
On b3 must stand an obstacle, because the queen must not go from b2 to b6 to give check. The composer Dittrich had put on b3 a black pawn, and the problem had no solution:
(1.Qb2? Bd3! 2.Rg8 f5 3.Qg7 Bxe2!) and no mate exists on the fourth move.

The Olympic winner Iatridis Stavros corrected the defect of the problem with a very simple manner, changing the colour of the pawn b3 to white, thus creating the continuation:
(1.Qb2! Bd3 2.exd3 ~ 3.Qf2+ Kxd3 4.Qd2#) and this nice problem is saved.

[This post in Greek language].

No comments: