This blog is available for the presentation of the works (preferably) of the Greek composers, but we do not exclude compositions sent to us from foreign friends.
Today we present problems by Themis Argyrakopoulos, Ioannis Garoufalidis and Joaquim Crusats.
Mr Themis Argyrakopoulos is one of the good Greek solvers (having ELO 1775, as the World Comittee WCCC has announced in 2008). From the problems that we will study here, we will see that he is a very capable composer.
His first publication was in the Finnish [Suomen Tehtavaniekat] with two helpmates, which are very interesting.
We note here that in the helpmates the Black plays first and helps White to reach his goal (in the helpmates the win, in the helpstalemates the draw).
 | (Problem 220) Themis Argyrakopoulos, Suomen Tehtavaniekat, No 4-5/2003 Helpmate in 5 moves. (Two solutions) h#5 2.1.1... (5+7) |
| [6Sq/8/3k3B/8/8/P5p1/pp4P1/rb5K] | |
Key : 1.Ke5! Kg1 2.Be4+ Bc1 3.b1=B Bb2+ 4.Kf4 Sh6 5.Qe5 Bc1#
Key : 1.Qg7! Sf6 2.Ke7 Bf4 3.Kf8 Bxg3 4.Bg6+ Kh2 5.Bf7 Bd6#
 | (Problem 221) Themis Argyrakopoulos, Suomen Tehtavaniekat, No 4-5/2003 Helpmate in 6 moves. h#6 (2+6) |
| [8/8/2k5/2b5/2P3s1/5b2/2p3s1/7K] | |
Key : 1.Ba7! c5 2.Kb7 c6+ 3.Ka8 c7 4.c1=B c8=B 5.Bf4 Bxg4 6.Bb8 Bxf3#
Mr Ioannis Garoufalidis is also one from the good Greek solvers (having ELO 1922, as the World Comittee WCCC has announced in 2008). His compositions are impressive, and he is interested in selfmate more-movers.
Here we will see a selfmate six-mover. In a future post we will see selfmates with more moves.
We note here that in selfmates the White plays first and forces Black to give mate.
 | (Problem 222) Ioannis (John) Garoufalidis, The Problemist (?) Selfmate in 6 moves. s#6 (12+4) |
| [4BB2/P5Sp/4PQsk/5Kp1/4P3/6R1/6PP/5S2] | |
Black is very tightly pressed, comes often in a zugzwang (zz) situation and it seems unable to give instantly mate to White. If the black pawn g5 is promoted, maybe the Black will have increased possibilities.
There are many tries {1.Bf8~? / Be8~? / a8=~? / e7? / e5? / Rg3~? / h3? / Sf1~? (zz) g4!}, {1.Sh5+? Kxh5!}, {1.Qxg6+? hxg6+!}.
The solution follows :
Key : 1.Rf3! (zz) g4 2.Rf4 (zz) g3 3.a8=Q (zz) gxh2 4.Qb8 (zz)
if 4...h1=Q 5.Rh4+ Qxh4 6.Qg5+ Qxg5#
if 4...h1=S 5.Sh5+ Kxh5 6.Sg3+ Sxg3#
if 4...h1=R 5.Rh4+ Rxh4 6.Qf4+ Rxf4#
if 4...h1=B 5.Rf3 (zz) Bxg2 6.Rh3+ Bxh3#
The problem is an allumwandlung (AUW), that is it has four promotions of the bP. In four variations of equal length, the promoted piece is forced to give mate. The whole impression is exceptional.
Mr Joaquim Crusats lives and creates in Catalonia, Spain. For his compositions, mainly multimover orthodox problems, he prefers cooperating with other composers.
He is a friend of this blog and, thanking him, we present next a moremover (ten-mover) problem of his.
 | (Problem 223) Joaquim Crusats, Problem Online, 15.10.2007, problem n. C0017 Mate in 10 moves. #10 (9+9) |
| [8/1p6/kPp5/p1P3p1/p3S1p1/K1p2pP1/p1p2p2/3B4] | |
Tries {1.Sxg5? / Sf6? Kb5!}, {1.Be2+? fxe2!}.
Key : 1.Bxf3! [2.Be2#] gxf3 2.Sxc3 g4 3.Sxa4 Kb5 4.Kb3 Ka6 5.c3 (zz) Kb5
6.c4+ Ka6 7.Sc3 a4+ 8.Kb4 a3 9.Sb5 cxb5 10.cxb5#
(This post in Greek language).
No comments:
Post a Comment