The ranking of the participants is as follows :

(1) Mendrinos Nikolaos, points 25 (minutes 25')

(2) Vlahos Elissaios, p.25 (62')

(3) Manolas Emmanuel, p.25 (86')

(4) Sklavounos Panagis, p.22 (61')

(5) Fougiaxis Harry, p.15 (77')

(6) Nikitakis J., p.09 (90')

(7) Fotopoulos G., p.07 (84')

(8) Roinos E., p.07 (90')

(9-11) Papadopoulos P., p.05 (90')

(9-11) Fotopoulos S., p.05 (90')

(9-11) Georgakis E., p.05 (90')

(12) Berk Atakak, p.00 (84')

The first five (who take part also in the Greek champioship) were considered just visitors, so the medals were given to places sixth and up.

The Judge Ioannis Garoufalidis presented (before the contest) a tutorial for new solvers, explaining the method of solution of two problems. We present these problems and a sort explanation of their solutions.

We present next the five problems of the solving contest. We will publish their solutions shortly. In the meantime, send in comments the solutions and the time you needed to solve them.

(Problem 393) Frank Healy, Canadian Illustrated News, 1876 Mate in 2 moves. #2 (7 + 3) | |

[1b6/8/4KB2/8/6S1/Q3RPk1/6p1/6S1 ] |

If the black King makes a move, we have a ready mate [1...Κf4 2.Se2#].

In problems we do not usually have a checking first move. Here Black has defenses to moves, like [1.Bh4+? Kxh4!], [1.f4+? Kxg4!], [1.Se2+? Kh3!], [1.Qd6+? Bxd6!].

In some tries we observe that the black Bishop is defending, like [1.Kf5? / Qc5? Bf4!], [1.Qa5? / Qc3? Bd6!]. Thus, we must limit these moves by the black Bishop.

**Key : 1.Qe7!**(If black Bishop plays Bc7 / Bd6, then the Queen captures the piece and mates).

1...Be5 2.Bxe5# (The white Bishop can act capturing a piece ...)

1...Bf4 2.Bh4# ( ... or exploiting the self-block of the Black).

(Problem 394) Unknown, Land and Water, 1876 Mate in 3 moves. #3 (8 + 2) | |

[7/7K/8/2S5/2p2P2/2P2BPk/6R1/2B5] |

Black has not a move and we must allow him to have one.

Using the Knight we do not achieve anything : [1.Sb3? cxb3!], [1.Sd3? cxd3!].

With the Pawn or the Rook the results are futile : [1.g4? Kh4!], [1.Rg1? Kh2!], [1.Ra2? / Rb2? / Rc2? / Rd2? / Re2? / Rf2? Kxg3!].

So, we must find the exact square of arrival for the white Bishop. There are some tries : [1.Bd1? / Be2? / Bh5? Kxg2!], [1.Ba8? / Bb7? / Bc6? / Be4? Kg4!]. Thus there is one square for the ...

...

**Key : 1.Bd5!**(zz) (After two consecutive zugzwang, Black will have X-flights).

if 1...Kg4 2.Se4 (zz)

and if 2...Kf5 / Kh5 / Kh3 / Kf3 then 3.g4# / Sf6# / Sf2# / Sf6#

And now, the problems of the contest :

(Problem 395) Maurus Ehrenstein, Oesterreichische Lesehalle, 1881 Mate in 2 moves. #2 (2 + 2) | |

[8/Q4p2/8/5K1k/8/8/8/8] |

This problem is surely for young solvers!

(Problem 396) Comins Mansfield, Good Companions, 1914 Mate in 2 moves. #2 (8 + 8) | |

[1b6/r2p1Q1K/2B3p1/1p1SB2b/4kP2/2P4R/4r3/8] |

If the Queen does leave from the seventh row, there is a Pawn-Rook battery for Black. White has got already a Knight-Bishop battery, but is it enough?

(Problem 397) Frank W. Martindale, ?, 1890 Mate in 3 moves. #3 (7 + 4) | |

[5K2/8/7p/4P1p1/1R4pk/1R6/B4P2/2B5] |

With two ambushes, two nice chameleon mates are achieved.

(Problem 398) Wilhelm Karl Heinrich Massmann 1st Prize, Die Schwalbe, 1941, Mate in 3 moves. #3 (7 + 2) | |

[3b4/8/4S1K1/2B1P3/2P1k1S1/8/4B3/8] |

If 1.Bd4? then 1...Bh4!. If 1.Be3? then 1...Be7!. It seems that thw white Bishop on c5 must move, as a key move, but the question remains : where to?

(Problem 399) C. Bent, BESN, 2008 White plays and wins. + (4 + 3) | |

[8/B7/P2b4/8/B7/p1k4K/8/8] |

Instructive study, with the black King free in the middle of the chessboard, and a very smart fourth move!

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