Some h#n selected from the WCCI

by Chris J. Feather

(JL: This is a second part of Chris' selection, the first part dedicated to h#2 was published about a month ago.)

The longer helpmates in this competition varied considerably in quality. A great many suffered from an excess of colourless moves, making them seem like the elongated ghosts of shorter problems! Here is a small selection, not of those gaining the most points, but simply of some of the ones which amused me.
Christer Jonsson
1nd Prize Magadan 60 JT 1999

1.Bxd6 Rh3 2.Bg3 Bd3+ 3.Kf3 Rxg3#

1.Rfxd6 Rf3 2.Rd3 Bc4 3.Re3 Rf4#

1.Rbxd6 Bb5 2.Rd4 Re3+ 3.Kd5 Re5#

Attractive echoes with a little added unity provided by the first moves, which are all captures on d6. "Such problems usually delight (almost) all solvers... but they should not overestimate them, as unfortunately even some judges do" - those (in translation) are the words of Friedrich Chlubna in his excellent book 'Schach für Nußknacker’, describing a different problem in the same style. He goes on to point out how much easier it is to create such echoes in the helpmate than in the directmate. The roles of solver and judge are hard to balance!

h#3 (5+10)

Tichomir Hernadi
4th Prize Moment 1999

a) 1.Qh4+ Rh6 2.Kg3 Rh5 3.Kh2 Rh6 4.g3 Rxh4#

b) 1.Be4+ Bf5 2.Kf3 Bg6 3.Be3 Bf5 4.Rf2 Bxe4#

Entertaining switchback shuffles on the pinlines. Because of the limited white moves the problem almost solves itself, but that does not detract from the amusement value.

h#4 (3+12)
b) e6 = wB

Dieter Müller
1st Prize Magadan 1999-2000

a) 1.Rg7 Rxf6 2.Kf3 Rf8 3.Rg2 Bf7 4.Sg3 Bh5#

b) 1.Rb7 Bxe6 2.Ka2 Bg8 3.Rb1 Rf7 4.Sb2 Ra7#

The position may not be pretty, but the strategy is as intensive as that found in many shorter problems, not an easy thing to achieve in view of the technical difficulties. Yet again we may laugh at the folly of Dawson’s remark about h#4s with WR and WB always being cooked!

h#4 (3+14)
b) g2 -» b2

Valerij Kopyl
Comm StrateGems 2000

I enjoyed solving this more than any other helpmate in the competition. For those who want to solve it now I will not spoil the experience by providing a description. In any case it seems that I must be mistaken in my enthusiasm for this problem, since I gave it as many points as the other two WCCI judges put together; and the original tourney judge was lukewarm about it!

Solution see below.

h#5 (3+7)

Zlatko Mihajloski
2nd HM Makedonski Problemist 2000

a) 1.Qd7 Kf4 2.Qb5 Kxg4 3.Sd5 Kf3 4.Sc3 Ke3 5.d5 cxd3#

b) 1.Be4 Ke2 2.Sd5 Kd1 3.Sc3+ Kd2 4.Bd5 Ke3 5.Qd3+ cxd3#

Two round trips by the WK have been shown before, but this is a neat and economical setting, with the changed capture on d3 substantially compensating for that inherent drawback of this kind of problem, the repeated mate. It is a pity that one Rundlauf involves a capture, though.

h#5 (2+11)
b) d6 -» b5

Alexandr Semenenko
1st Place Kievskoye Mnogoborye 2000

1.Bc7 Kg4 2.Ra6 Kf5 3.Ra8 Bg4 4.Be6+ Kxe6 5.Rb8 Ke7#

1.b6 Bf1 2.Rh6 Ke4 3.Bd5+ Kxd5 4.Rh8 Kc6 5.Rd8 Ba6#

As a judge one has to say that this is a rather untidy piece of work, and the 1.b6 solution is superior to the other one. Nevertheless it makes good use of the board and material, and the unifying BR manoeuvres are both unusual and visually attractive.

h#5 (2+10)

Dan Meinking
Prize Ideal Mate Review 2000

1.Sc6+ Kc8 2.Sd5 Kd7 3.Se5+ Ke6 4.Sf3 Kf5 5.Se3+ Kxf4 6.Sg2+ Kg3 7.Sg1 Bxg2#

Not difficult, but a fascinating sequence. The BK in check is hardly a drawback in this kind of problem… but wouldn’t it have been wonderful if there had been a unique last white move instead of the seven possibilities here?

h#7 (2+4)

Solution of h#5 by Valerij Kopyl:

1.b1R Rb2 2.Raa1 Ra2 3.Sc7 Rxa6 4.Rb7 Rxd6 5.Raa7 Rd8#

Comments to Juraj Lörinc.
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