Left to right : Papastavropoulos, Anemodouras, Garoufalidis, Ilandzis, Mendrinos, Konidaris. |
The 5th Solving Contest of E.S.S.N.A. (Union of Chess Clubs in Prefecture of Attica), contest surnamed "Byron Zappas", was successfully held in Chess Club of Pagrati in Saturday, April 19.
Eighteen solvers participated, having 2h 15m available time to solve 6 problems (1 two-mover, 1 three-mover, 1 four-mover, 1 study, 1 helpmate and 1 selfmate). The contest was dedicated to the top Greek composer of chess problems Byron Zappas, who died this year. All the compositions of the contest were created by Greek problemists.
The following persons gathered the most points :
1. Nikos Mendrinos (25 grades in 30) champion of Attica,
2. Andreas Papastavropoulos (20),
3. Panagiotis Konidaris (19),
4. Leocratis Anemodouras (15),
5. John Garoufalidis (15),
6. Spyros Ilandzis (14).
Best team proved to be "Zinon Glyfadas". No prize for category "young under 20 years old" was given.
The judge of the contest, Panagis Sklavounos, selected the following problems by Greek composers.
(Problem 148) Marassoglou N., 'To Mat', 1952 Mate in 2 #2 (7+4) | |
[K3k3/P2R4/2p5/2B1P1p1/6B1/1r6/Q7/8] |
(Problem 149) Zappas Byron, Parallèle 50, 1948 Mate in 3 #3 (8+5) | |
[8/1p3R2/1pBp4/2kS1p2/RS6/1K1P4/P7/8] |
(Problem 150) Siaperas T., Problem, 1952 Mate in 4 #4 (8+9) | |
[K5S1/1p3s2/4kS1p/p2R4/3PP1Rp/5Bb1/5s1q/8] |
(Problem 151) Fragoulis K., Suomen Shakki, 1978 White plays and wins + (4+3) | |
[8/4pK2/8/pk6/SS6/8/1P6/8] |
(Problem 152) Paizis K., B.C.M., 1993 Helpmate in 3, (two solutions) h#3 (3+3) 2.1.1.1 | |
[1K3b2/4s3/5P2/4k1S1/8/8/8/8] |
(Problem 153) Moutecidis P., Gazeta Czestochowska, 1970 (Set play). Selfmate in 2 (*) s#2 (11+14) | |
[4b3/S1PkPP1R/3p4/3B4/4RPPp/2ppB1pq/3p2pB/3K1srr] |
Here are the solutions of the problems:
Problem-148, Marassoglou, #2
Tries: {1.Rd1? / Rd2? / Rd3? / Rd4? Kf7!}, {1.Rb7? Kd8!}, {1.Ba3? / Bb4? / Bd6? c5!}, {1.Bh3? / Bf5? g4!}, {1.Bh5+? Kxd7!}, {1.Qb1? Rxb1!}, {1.Qa6? Rb7!}, {1.Qa5? Rb6!}, {1.Qf2? Rf3!}
Key: 1.Rc7! (zz).
Variations: 1...Kd8 2.Rc8#, 1...Rb8+ 2.axb8=Q# / axb8=R#, 1...R~ 2.Rc8# / Qg8#, 1...Rd3 2.Qg8#, 1...Rf3 2.Rc8#.
Problem-149, Zappas, #3
Tries: {1.Rxf5? / Rxb7? / Be8? / Bd7? / Bxb7? Kd4!}, {1.d4+? Kxd4!}, {1.Sa6+? Kxc6!}, {1.Kc3? bxc6!}.
Key: 1.Kc2! [2.d4+ Kxd4 / Kc4 3.Sd3#]
Variations: 1...Kd4 2.Sa6+ Ke5 3.Re7#, 1...b5 2.Kc3 and 3.d4#, 1...bxc6 2.Rxf5 (zz) Kb5 / Kd4 / b5 / cxd5 3.Sc3# / Sa6# / Sa6# / Rxd5#.
Problem-150, Siaperas, #4
Try: {1.Rd7? Bd6!}
Key: 1.Rg7! [2.Bh5 ~ / Sf7~ (if black plays anywhere or if he plays somewhere with Sf7, then) 3.Bxf7# / Re7#]
If 1...Bb8 (Bristol line clearance)
2.Bh5 Qc7 (we understand now that the black pawns a5 and b7 were placed there to void checks from the black Queen Qc7)
3.Bg6 [4.Bf5#]
3...Qf4 / Qe5 4.Bxf7#
3...Sd6 4.Re5# (Note that in the initial position this square was guarded by three black forces!).
If 1...Bd6
2.Bh5 [3.Bxf7#]
2...Se5 3.dxe5 [4.Bf7#] Sxe4 4.Bg4#
2...Sg5 / Sh8 / Sd8 3.Be8 and 4.Bd7#
Problem-151, Fragoulis, study +
Key: 1.Sc2! Kxa4 2.Sa1!
2...Kb4 3.Kxe7 a4 4.Kd6 a3 5.Sc2+ ~ 6.bxa3 and white wins
2...e5 3.Ke6 e4 4.Kd5 e3 5.Kc4 e2 6.Sc2 e1=Q 7.b3#
Problem-152, Paizis, h#3 2.1.1.1
1.Kxf6 Se4+ 2.Kf7 Kc7 3.Ke8 Sd6#
1.Sg8 f7 2.Bc5 fxg8=Q 3.Kd6 Qe6#
Problem-153, Moutecidis, s#2 (*)
Phase of set play : (*)
1...Kxc7 2.fxe8=Q (zugzwang, zz, and black is forced to continue by giving mate) Sxe3# / c2# / Qxg4#
1...Bxf7 2.e8=S (zz)
Tries: {1.Rh5? / Rh6? / Rh8? Bxf7!}, {1.c8=B+? Kc7!}, {1.Bc6+? / f8=S+? Kxc7!}, {1.fxe8=Q+? / fxe8=S? / fxe8=R? Kxe8!}.
Phase of real play : Key: 1.Sb5! (zz)
1...Kc8 2.fxe8=B (zz)
1...Bxf7 2.e8=R (zz)
Since it has achieved the four promotions in the solution, the problem is an allumwandlung (AUW).
(This post in Greek language).
1 comment:
It is great to see the activity of Chess Solving in Greece!
I'd like to register here my regards and congratulations to Spyros Ilantzis!
Regards,
Roberto Stelling
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