When a composition has a defect or it happens that the judge dislikes it or it is weaker than the compositions of other composers, then it does not take a distinction. We do our best to create a very nice position but, as it happens in life, this is not always possible. Some of the compositions, when we examine them later without pressure (and with more experience), can be improved. This is something that the experienced judges see or feel, and that is why their opinion is useful.
In a previous post we saw five compositions, let us see now and the rest of them.
11th Japanese Sake / 3rd Brazilian Cachaça Tourney
Theme : Series selfmate / selfstalemate, (ss#n, ss=n), with Reversible Promotion. Neither fairy pieces nor other fairy conditions are allowed except for Maximummer.In a Series self(stale)mate of n moves, the White alone makes n-1 moves without checking and in the last move the immobility of Black is raised and the Black achieves the goal (mate or stalemate).
We know that a white pawn starts from the second line and is promoted in the eighth line (for black : seventh and first, respectively). With the condition of Reversible Promotion, a white promoted piece stepping on the second line (for black : seventh) becomes pawn again.
 k7/7P/p2pp3/K1ppp3/P1p5/8/1s2p3/1r3bb1 w (3+13) | Problem-502 Kostas Prentos - Manolas Emmanuel First Honourable Mention ss#18, Reversible Promotion, Maximummer, 2 solutions 1.h8=B 2.Bxe5 3.Bxb2=P 4.b4 5.bxc5 6.c6 7.c7 8.c8=B 9.Bd7 10.Bb5 11.Bxc4 12.Bxe2=P 13.e4 14.e5 15.exd6 16.d7 17.d8=B 18.Bb6 Bxb6# 1.h8=S 2.Sg6 3.Sf4 4.Sxe2=P 5.e4 6.exd5 7.dxe6 8.e7 9.e8=S 10.Sxd6 11.Sxc4 12.Sxb2=P 13.b4 14.bxc5 15.c6 16.c7 17.c8=S 18.Sb6+ Bxb6# |
The Japanese Judge Tadashi Wakashima commented : "Very nice presentation in two solutions of three promotions and two reversals (wB in the first solution and wS in the second one).".
The price was accompanied by a bottle of Japanese Sake.
 8/q3r2P/1b4p1/pb6/2p3p1/2p2k1p/1s1r1p1P/5K2 w (3+14) | Problem-503 Manolas Emmanuel Third Honourable Mention ss#27, Reversible Promotion 1.h8=B 2.Bxc3 3.Bxb2=P 4.b4 5.bxa5 6.axb6 7.bxa7 8.a8=R 9.Rd8 10.Rxd2=P 11.d4 12.d5 13.d6 14.d7 15.d8=S 16.Se6 17.Sf4 18.Sxh3 19.Sf4 20.h4 21.h5 22.hxg6 23.g7 24.g8=Q 25.Qxc4 26.Se6 27.Qd3+ Bxd3# |
The Japanese Judge Tadashi Wakashima commented : "Allumwandlung with two pawns".
The price was accompanied by a bottle of Brazilian Cachaça.
I worked several days to make this problem. The computer software were not of real help because the moves were too many and the time needed was too much (about 45 hours). At the end the very fast solver Kostas Prentos helped me with the validity checking, and I thank him.
The plan of White, to check with the Queen at d3 forcing bBb5 to capture it giving mate, has some obstacles. First there is no Queen, second many black pieces (Qa7, Bb6, Sb2, c4, Rd2, Re7) can intervene to a threat of mate from d3, and furthermore bK has a flight f4. And the bBb5 cannot guard d3 because bPc4 stands in the way.
The pawn bPg6 is needed for column change of wPh2 going to be promoted at g8, but if it was positioned at g5, a second unwanted solution appears with 1.h8=S.
The key should not be promotion to Queen because the second move 2.Qxc3 would check the bK.
Similarly for the move 8.a8=R, if we had a promotion to Queen it would check the bK. And then only the path Ra8-d8-xd2=P is valid, not via a2!
The Queen will be needed at the end and also a piece to guard f4. If the pawn wPh2 should be freed, we observe that bPh3 can be captured only by a Knight without giving check to bK!
In the fourteenth move the Knight cannot capture the rook (14.dxe7?) and also solve the problem in 27 moves. So bypasses it and postpones the closing of the line of bR (and also guarding f4 from there) for later. How much later?
Before e6 is occupied, the Queen must pass over it, 25.Qg8xc4, and only then 26.Sf4-e6.
Finally in this composition we have the four promotions (=allumwandlung), Bishop, Rook, Knight, Queen.
 8/1Pq5/8/8/4p1pp/4r2k/3spr1p/3bbs1K w (2+13) | Problem-504 Manolas Emmanuel Third Commendation ss#27, Reversible Promotion 1.b8=B 2.Ba7 3.Bxe3 4.Bxd2=P 5.d4 6.d5 7.d6 8.dxc7 9.c8=R 10.Rg8 11.Rxg4 12.Rxe4 13.Rxe2=P 14.e4 15.e5 16.e6 17.e7 18.e8=R 19.Rxe1 20.Rxf1 21.Rxf2=P 22.f4 23.f5 24.f6 25.f7 26.f8=Q 27.Qf3+ Bxf3# |
The Japanese Judge Tadashi Wakashima commented : "Three promotions and three reversals by a single pawn".
Here I have gathered the maximum number of obstacles (without promoted black force) to the White plan.
Thematic Tourney «The Urals Problemist» - 2011
Theme: Selfmate (s#3-5). Recovery of the initial white or black battery. 7b/7p/3S1k1P/2P4P/4R3/pp2pPQ1/p4p2/Kb3r2 w (8+10) | Problem-505 Manolas Emmanuel Version, Dedicated to Andrey Selivanov s#4 Tries : {1.f4? [2.Qg6+ hxg6 3.Re6+ Kxe6#] Bg7!}, {1.Re8? [2.Qf4+ Bf5#] e2!} Key : 1.Re5! [2.Qf4+ Bf5#] e2 2.Qh4+ Kxe5 3.Qe4+ Kf6 4.Qe7+ Kxe7# |
The Dedication is because the Judge Andrey Selivanov found that the initial position I submitted {7b/8/3S1k1P/2P3RP/Q7/pp2p3/p4p2/Kb3r2 w} had a flaw (unprovided check 1...Kxf5+ in the set play). The version here is corrected.
I hope that I have given you an idea of the creativity frenzy that exists during the world chess composition congress. Think about several other composing tourneys I did not participated and the important fact that there were many composers there better than me.
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