Thursday, July 05, 2012

Helpmate with Circe condition

During the composition process of a chess problem we may have a low target, while it may be possible to create something better by adding a few pieces. For example, we will see a weak problem, helpmate two-mover with condition Circe, and how much can be improved.
Reminder : The condition Circe says "the captured piece is reborn in its initial position in a normal OTB chess game".

Low target : wR and bK exchange places.


4k3/3p1p1p/pK6/4B3/8/R7/8/8
(3 + 5)
Problem-589
Manolas Emmanuel
original
h#2, Circe, 2111

1.Kd8 Rg3 2.Kc8 Rg8#
1.Kf8 Rc3 2.Kg8 Rc8#

The pawn bPa6 stops undesired solutions (if captured by wR, it is reborn on a7 and stops mate on a8). Similar role plays the wBe5, being active on one solution only - not good.


Higher target : In each solution we have a different castling.


4k3/3p1p1p/pK6/4B3/8/RP6/8/1r6
(4 + 6)
Problem-590
Manolas Emmanuel
original
h#2, Circe, 2111

1.Ra1 Rxa1(+bRh8) 2.0-0 Rg1#
1.Rxb3(+wPb2)+ Rxb3(+bRa8) 2.0-0-0 Rc3#

Now the problem is better, because we use actively the condition Circe, and the play has become more interesting. The white pieces are active in both solutions. The two solutions are omo-strategic and no move is repeated in the solutions.


Even higher target : In each solution we have a different castling, but a black line is closed and the square, from which the mate is given, is not observed anymore.


4k3/3p1p1p/1K6/1R2B3/1r1P4/1P6/8/2q5
(5 + 6)
Πρόβλημα-591
Manolas Emmanuel
original
h#2, Circe, 2111

1.Rxd4(+wPd2) Bxd4(+bRh8) 2.0-0 Rg5# 1.Rc4 bxc4(+bRa8) 2.0-0-0 Rc5#

A white pawn moves and closes the observation line of the bQ towards the square (g5 or c5) from which the wR gives the mate.


Other directions of composition : We keep the different castlings in each solution and we try other continuations.


4k3/3p1p1p/1K6/4B2p/4p3/3p3R/8/6r1
(3 + 8)
Problem-592
Manolas Emmanuel
original
h#2, Circe, 2111

1.Ra1 Bxa1(+bRh8) 2.0-0 Rg3#
1.Rh1 Rxh1(+bRa8) 2.0-0-0 Rc1#

The bR is captured on the squares a1 and h1 from different white pieces. Not very important.

4k3/3p1p1p/1K6/1R2B3/1r1p4/8/8/2r3b1
(3 + 8)
Problem-593
Manolas Emmanuel
original
h#2, Circe, 2111

1.Rbb1 Rxb1(+bRa8) 2.0-0-0 Rxc1(+bRh8)#
1.Rb2 Rxb2(+bRh8) 2.0-0 Rg2#

The bR is captured on the squares b1 and b2 and the theme is bicolored Bristol. It can become better.

4k3/3p1p1p/pK6/Rp2B3/rp6/3p4/RP1g4/8
(5 + 10) (Grasshopper 0 + d2)
Πρόβλημα-594
Manolas Emmanuel
original
h#2, Circe, 2111

1.Rxa2(+wRh1) Rxa2(+bRa8) 2.0-0-0 Rc1#
1.Rxa5(+wRa1) Rxa5(+bRh8) 2.0-0 Rg1#

The bR captures one wR and is captured from the other, and the bicolored Bristol is happening reciprocally in the two solutions. With repeated application of the condition Circe, white and black Rooks are reborn in the four corners of the chessboard. The black Grasshopper stops an undesirable solution.

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