Award of 21st TT Chess Composition Microweb C 5.5.2007

1.Kb7 2.Kc6 3.Kd5 4.Ke4 5.Kf3 6.Kg2 7.Kxh1(Ra8) 8.Kg2 9.Kf3 10.Ke4 11.Kd5 12.Kc6 13.Kb7 14.Kxa8 15.Rah1(b) 16.BHa1(b) 17.Kb7 18.Kc6 19.Kd5 20.Ke4 21.Kf3 22.Kg2 23.Kxh1(Ra8) 24.Kg2 25.Kf3 26.Ke4 27.Kd5 28.Kc6 29.Kb7 30.Kxa8 31.Rh1(b) 32.Kb7 33.Kc6 34.Kd5 35.Ke4 36.Kf3 37.Kg2 38.Kxh1(Ra8) 39.Kg2 40.Kf3 41.Ke4 42.Kd5 43.Kc6 44.Kb7 45.Kxa8 46.Kb7 47.Kc6 48.Kd5 49.Ke4 50.Kf3 51.Kg4 52.Kh5 53.d8Q(b) Qxg5(pg2)#

The author managed to more than double Syzonenko's almost 1.5 diagonal. The means used are quite strong, but it is worth noting that Double Maximummer use is not so mechanical as is often usual, rather it is in the core of problem.

a) 1.CHxd6(R) 2.CHd2(Q) 3.CHf4(S) 4.Kf5 5.Ke4 6.CHe2(B) 7.Kd3 8.Kxc3 9.CHb5(R)+ CHxb5(S)#

b) 1.CHe7(R) 2.CHxe5(Q) 3.CHxd6(S) 4.CHf5(B) 5.Kf6 6.Ke5 7.Kd4 8.Kc3 9.CHb1(R)+ CHxb1(S)#

Two different ways of wK to c3 made possible by different shields and two different (but similar) mates.

1...a3#

1.d5 a4 2.d4 a5 3.dxc3(Ke1) Kd2 4.cxd2(Ke1)+ Kxd2(d7) 5.Kxa5(a2) Kc3 6.Kb4 a3#

Certain humour in this problem. Thanks to Circe Rex Inclusiv kings might be unparalyzed and thus they take role in the repetition of initial position - with White to move after Return of Kings (by chance there is just Lord of Rings in Czech TV).

1.LIg6 Gg7 2.Kd5 Gg5#

1.LIe6 Kd6 2.Gf5 Gxe6#

1.LIg2 Gg1 2.Kd6 Gg3#

1.LIe4 Kd5 2.Gf4 Gxe4#

As the author says in his commentary: small study in geometry. 4 different mates in HOTF form.

1.Rxc4 2.Kd5(b4) 3.Rxb4 4.Kc4(a3) 5.Kb3 6.Kxa4 7.Rb2(Ra2) 8.Ka5 axb2#

1.Kd4 2.Kc3 3.Rxc4 4.Kb3(b4) 5.Rxb4 6.Kxa4(a5) 7.Kxa5 8.Rb6(a7)+ axb6#

Again two different mates with wK on the same square. Once checkmating pawn mates directly, once it opens battery.

No. 30 a) 1.PAxa7 LIe7 2.VAxe7 LIh5 3.Kc5+ VAg5#

b) 1.VAc1 LIf4 2.PAxf4 LIg8 3.Kc4+ PAf7#

The first of good strategic problems in the tourney. Echo diagonal-orthogonal with point in the final move. Black lions already stand on the future anti-battery lines, but as they stand exactly on a square where the hurdle will land, they have to be replaced. The only slight imbalance is in the motivation of the first move: 1.PAxa7 allows move of wK to c5, while 1.VAc1 prevents parrying the final white check by VAe3.

1...Gg1 2.rRa1 rRb1#
1...rRg2 2.rRh1 rRg1#

1.d2 Gc3 2.rRc1 rRc2#
1.rRh1 rRf2 2.rRg1 rRf1#
1.rRg1 Gh1 2.rRa1 rRb1#
1.rRf8 rRb7 2.rRe8 rRe7#
1.rRf7 rRb6 2.rRe7 rRe6#
1.rRf6 rRb5 2.rRe6 rRe5#
1.rRf5 rRb4 2.rRe5 rRe4#

Echos, echos, echos. The author managed to put some life into potentially very boring idea.

1.a1G 2.Ge5 3.Gg3 4.Ghf2 5.Gd3 6.Gd1 7.Gd5 8.Gd7 9.Gf5 10.Ke4 Kd5==

As the author says: there is a cycle of paralyses Ga4-Gd7-Gg4-Gd1-Ga4. This paralysis frame employs 2+2 units as hurdles and thanks to the position of kings hurdles cannot move as this would cause selfcheck by unparalysis. Fresh idea.

1.LOxf4-e3 Kg5 2.RLg4 RLf1 3.Rd3+ LIf4#

1.LOxf6-e6 Kg6 2.RLg5 RLf8 3.Rd5+ LIf6#

Very good D-O echo with reciprocal exchange of roles RLf4-RLf6. Lb3 paralyses checking white locust, the check is given by Madrasi specific antibattery cutting the paralysis line. Also note anti-Indian manoeuvres 2...RLf6-f1 3...LIh4-f4# and 2...RLf4-f8 3...LIh4-f6#. I just regret the crowded position in the rectangle, where the main actions take place.

See at the end of the award the new version by Paul Raican.

1.Se5 a7 2.Kc5 Sa6#

1.Sc6 h4 2.Kd5 Sf4#

1.Bxd3+ Kd4 2.Ke5 Sxd3#

1.Bxb4 Kb5 2.Kc6 Sxb4#

Madrasi specific HOTF with enough differences between non-analogous phases and very good between analogous ones. Not very economical though, as predicted in h#s by one of competitors.

1.LIh4 LOxg4-h5 2.Kxd2 LOxc5-b5 3.LId3+ LOxd3-e2#

1.LIh6 LOxb4-a5 2.Kxe2 LOxc5-d5 3.LId3+ LOxd3-d2#

The main idea of this helpselfmate lies in triangular round trips of black locusts. W1 paralyses LIh8, W2 paralyses kings and W3 is antibattery check. Almost without blemishes - just RLb3 is one in my view. It was even possible to turn it into (normal) lion and thus reduce number of fairy elements in the problem.

a) 1.Ke4 RLd4 2.LIxf6 RLh4 3.LIc7+ LIg4#

b) 1.Ke5 BLd4 2.LIxf4 BLh8 3.LIc7+ LIg7#

Another excellent helpselfmate has rich strategy. W1 paralyzes kings and provides hurdle for B1, that is the first part of "pendulum" manoeuvre, providing hurdle for W2. W2 paralyzes LId6 needed as hurdle in check. B2 is critical move preparing final antibattery. W3 checks by closing line (Madrasi typical antibattery) and in B3 black must paralyze LId7, thus creating mating antibattery.

1.Kxd4? th. 2.Kd5#
1...Kd6 2.Kxc5#
1...Ke5 2.Sc4#
1...Kd5 2.Nb4#
1...Be5+ 2.Kxe5#
1...e2!

1.Kxc5! th. 2.Kd5#
1...Kd6 2.Sc4#
1...Ke5 2.Kxd4#
1...Kd5 2.Nc3#
1...Bd6+ 2.Kxd6#

Something for Cyclone lovers. The keys and variations 1...Kd6 and 1...Ke5 form Kiss cycle using a lot of batteries created by keys. There are a few more thematic mates, both kings play very actively. Unfortunately I cannot oversee two faults: complete symmetry of the play (this kind of Cyclone themes is called "optical form" in Slovakia) and lesser economy of some white pieces.

1...hxg5 2.Be7 Bc5 3.Kxd4 Sd2 4.Re6+ Rxe7#

1...axb3 2.Rb6 Re6 3.Kxe4 Sf3 4.Bc5+ Bxb6#

Worthy winner with perfect and rich Madrasi analogy. B1 unparalyzes bS and economizes wS. In W2 white line mover crosses the square where it is immediately paralyzed in B2, but also closes the line of final check. W3 paralyzes kings and economizes bS, B3 paralyzes white knight, that could capture mating black piece, but also opens final checking line and prevents it reclosing by bS. W4 is a check forcing capture of paralyzing white line mover... simply you have to admire author's imagination. Note also formal reciprocal relationship between B2 and W4.

1.Kb7 ... 7.Kxh1(Ra8) 8.Kg2 ... 14.Kxa8 15.Kb7 ... 20.Kg2 21.Kf1 g2#

Diagonal switchback of wK with minimal material.

1.Lxe4-d3 Kf5 2.RLf4 RLe1 3.c3+ LIe4#

1.Lxe6-d6 Kf6 2.RLf5 RLe8 3.c5+ LIe6#

Paul Raican has provided this version of the 2nd HM addressing the economy issue (2 wp instead of wR, one piece less) as well as judgement reservation about crowded rectangle.