Selections 7


First prizes are often really the best problems in the relevant tourneys, but not all tourneys are equal in terms of overall quality. Moreover sometimes it is easy not to agree with the judge's views. Still, usually it makes sense to study the 1st Prizes as we do in today's selection.
Herbert Ahues
1st Prize Neue Osnabrücker Zeitung 1993-94

1.Bh2? th. 2.Qe6#
1...Re4!

1.Bf4? th. 2.Qe6#
1...Se3!

1.Bd4? th. 2.Qe6#
1...Re4 2.Qxa2#
1...Sf4!

1.Bd6! th. 2.Qe6#
1...Se3 2.Rf5#
1...Sf4 2.Rd3#
1...Re4 2.Qb5#
1...Re5 2.Qxe5#

If Be5 departs, White threats 2.Qe6# and there are dual avoidance double-check mates prepared for two defences by Sg2. However, Black has strong defence 1...Re4 that has to be either disabled or provided for. 1.Bf4? interferes with bR, but also with wR, allowing refutation. 1.Bd4? interferes with bR, allowing a new checkmate, but again with wR too, with a new refutation. Finally, 1.Bd6! intereferes with bB, leaving all important lines open.









#2 (7+9)

Jaroslav Hronec
1st Prize UV CSTV 1958

1.Bg8! th. 2.Re6+ Kf5 3.Qd3#
1...Re5 2.Qc7 th. 3.Qf7#
    2...Re7 3.Qxf4#
1...Be5 2.Qc5 th. 3.Qf8#
    2...Kf5+/Bd6 3.Qxe5/Re6,Qxg5#
(1...Kf5/Be3 2.Qd3+,Qc5+,Kg7,d5/Qxe3,Qc7)

This old threemover uses quite limited white material. The unprovided flight e7 is replaced by f5 by the key, with threat that allows interferences on e5. Not only that of wR, but also Grimshaw exploited by wQ quiet moves. The rest is as expected, with a few dualistic variations.

The first prize received reflects perhaps the low level of strategic threemovers back in 50s in the former Czechoslovakia.









#3 (7+7)

Stefan Sovik
1st Prize Pravda 1985-86

1...cxb3 2.Rb2 c4 3.Kb1 cxd3 4.Rc1#
1...cxd3 2.Rd2 c4 3.Kd1 cxb3 4.Rc1#

1.Ra1?
1...cxb3 2.Kb1 c4 3.Ra5 b2/cxd3 4.Ra3/Rc5#
1...cxd3 2.Ra3 d2+ 3.Rxd2 c4 4.b4#
1...Kxb3!

1.Re1!
1...cxb3 2.Re3 b2+ 3.Rxb2 c4 4.d4#
1...cxd3 2.Kd1 c4 3.b4 d2 4.Re3#

Rather rare sight in fourmover, three-phase change of two variation in classic mutate form: set play, try and solution, even with flight-giving key. Obviously, there is no big strategy involved and there is almost complete vertical symmetry, but the result is noteworthy anyway.

(The full award is available on the net too.)









#4 (8+5)

Mechislovas Rimkus
1st Prize 137th TT SuperProblem C 25.4.2015

1.Rc6+ Kb8 2.Qd8 Ka7 3.Kd6 Se4+ 4.Kc7 Sc3 5.Rd7 Sb5#

1.Be6 Sg8 2.Rd7+ Kc6 3.Qd5+ Kb6 4.Kd6 Se7 5.Be5 Sc8#

1.Qe4 Sd5 2.Rf6 Kd8 3.Rf4 Ke7 4.Bf5 Sb4 5.Rd5 Sc6#

1.Ke6 Sd5 2.Kf7 Kxd6 3.Kg8 Ke7 4.Qh8 Ke8 5.Bg7 Se7#

1.Bf5 Sd7+ 2.Ke6 Kc8 3.Sd5 Kd8 4.Qf6+ Ke8 5.Be5 Sc5#


Stunning helpmate in five has 5 solutions. Not all of them are equal, while the first three of them show nice echo of mid-board model mates, the fourth is also a model mate, however on the edge, and the last one is just a mate. But still, the form of five solutions is quite unusual, moreso in five moves. I wonder, how this would fare in the running Klemanic 50 JT...









h#5 (2+9)
5.1.1...

Ivan Soroka
1st Prize N. Ovechkin 100 MT 2011

1.Rf3! zz
1...d5 2.Sf2 Kxg2 3.Sg4+ Kg1 4.Re3 dxe3 5.Kc2 d4 6.Kd1 d3 7.Qg2+ Kxg2#
1...d6 2.Qd2 d5 3.Kc2 d3+ 4.Kc1 d4 5.Sc3 dxc3 6.Rf1+ Kxf1 7.Qf2+ Kxf2#

Incarcerated Rh1 hints that the checkmate will be given to wK standing on the first rank. Moreover, black pawns on d-file look like they could be used for guarding potential flights on the second rank. It is just necessary to move wK there, wait for pawns to enter the 3rd rank and then give check forcing royal battery firing. But it turns out things are not totally easy and it is necessary to build two echoed different mating nets. Nice work.









s#7 (5+5)

Vasil Dyachuk
1st-2nd Prize e.a.
Problemist Ukrainy 2010

1...f4 2.Rxd4#
1...Bxd3 2.Rxd4#

1.Rxc3? th. 2.Rxd4#
1...f4 2.Gxd4#
1...Bd3 2.Rxc5#
1...Ga4!

1.Rh5? th. 2.Rxd4#
1...f4 2.Rxe5#
1...Bxd3 2.Gxd4#
1...Ga4!

1.Gxd4! th. 2.Gf6#
1...f4 2.Rxc3#
1...Bxd3 2.Rh5#

The main fight here is about square d4 that is in the diagram position blocked, guarded by two white rooks and can be also entered by wG. It is quite unusual that piece allowed to capture on the square does not guard it when flight there is considered, but that's the property of grasshopper jumping over the enemy king.

Two rooks checkmate by capture on d4 in set play. There are two additional potential flights, coming into action in two tries after lateral moves of white rook. Finally, in solution, White unblocks d4 by grasshopper capture.

On formal side the problem shows change of 2 mates in four-phases (also known as Zagorujko 4x2), spiced by 2 threat paradoxes between set play and tries as well as key-mate reversals between tries and solution. Very SAT-like mechanism, with maybe just one weak point - the same refutation of both tries.









#2 (12+12)
SAT
1+1 grasshopper

Klaus Wenda
1st Prize Shakhmatna Misl 2004

a) 1...g8B 2.bxa1Q(Qd8) Rxc7(Ra1) 3.Qb6 Bxa2(Bf1)#

b) 1...g8Q 2.bxa1B(Bf8) Rxe7(Ra1) 3.Bb4 Qxa2(Qd1)#

Position of bK determines white piece to be promoted at g8, so that after capture on a2 he would be under double check. It also determines the square to be blocked by black promoted piece, reciprocal to white one. This in turn decides about square where Rd7 captures, to transport itself to a1 as well as open black line. Nice analogy at the cost of twinning by moving black king (see the very 1st TT CCM) that is not very popular with everybody.









h#2,5 (5+10)
Anticirce
b) a6 -» a4

Borislav Gadjanski
1st Prize 9th Tzuica
Jesi 2011

1...Rxa4 2.Rb4 Rb1 3.Rh4 Rab4 4.Se6 Qa5 5.Sg4+ Rxg4#

1...Bxb1 2.Bc2 Qa2 3.Bg6 Bc2 4.Sxh3 Rc1 5.Sf5+ Bxf5#


Absolutely stunning orthogonal-diagonal transformation with excellent analogy of complicated line manoeuvres. White line pieces R/B guard h5 from squares initially occupied by their black counterparts. In the meantime Black guards some squares and starts creation of battery, finished in B4 shortly before action. Note exchange of functions of 5 pairs of pieces and mirror mates.









hs#4,5 (10+10)
2.1.1...

Gaspar J. Perrone
1sy Prize Problem Observer 2005

1.c1B e8S 2.d1R Sf6 3.Rd4 h8Q 4.Bf4 Qxh3=

Swift creation of final position requires 4 promotions and their moving to positions. Maybe a bit too simple for the 1st Prize after all previous full-weight fairies?









h=4 (3+4)
Enemy Sentinelles

Comments to Juraj Lörinc.
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