## Saturday, July 21, 2012

### A helpmate composition and an Exercise

During the World Congress of Chess Composition (WCCC) of 2010 in Crete, there were various composition contests.
For the 13th Sabra Composing Ty, we should compose an orthodox helpmate two-mover, where
B1 (first black move) : a black piece captures a white which closes a line
W1 (first white move) : a white piece, of line movement, arrives on this line
B2 (second black move) : the black piece, which made the capture, moves.

I made the following problem, with no distinction :

 3bk3/3p2p1/3P2P1/4PP2/PrP5/1P5b/2K5/5R1B (10 + 6) Problem-601 Manolas Emmanuel original h#2 a) Diagram, b) bKe8→c8 a) 1.Bxf5+ Be4 2.Bxg6 Bxg6# b) 1.Rxa4 Ra1 2.Ra8 Rxa8# Bicolour Bristol, Orthogonal - Diagonal transformation, Black sacrifices, Duels wB with bB, wR with bR.

The Exercise is the following :
For the following position, (heterodoxe, because on d8 there is a Grasshopper), there are solutions when the bK is placed on squares e8 and g8.
Try to find a very economical modification (translocation or addition or removal of a piece), with which the problem has solution when the bK is placed on square c8.

 3gk3/3p2p1/3P2P1/r4P2/2P5/7b/4KRB1/8 (7 + 6) (Grasshopper 0 + d8) Problem-602a Manolas Emmanuel original h#2 a) Diagram, b) bKe8→c8, c) bKe8→g8 a) 1.Bxf5 Be4 2.Bxg6 Bxg6# b) 1.?? ?? 2.?? ?? c) 1.Rxf5 Rxf5 2.Gh8 Bd5#

You may send the modification and the solution, the latest Tuesday 24-July-2012, by email to manolas.emmanuel@gmail.com. The correct solutions will be published.

### Solution of the exercise

The correct answer was send by the Slovakian composer Jaroslav Stun (visit his page here) :
a) bRa5b5, which is the most economical, and
b) +bPb6
Thank you chess-friend Jaroslav!

 3gk3/3p2p1/3P2P1/1r3P2/2P5/7b/4KRB1/8 (7 + 6) (Grasshopper 0 + d8) Problem-602 Manolas Emmanuel original h#2 a) Diagram, b) bKe8→c8, c) bKe8→g8 a) 1.Bxf5 Be4 2.Bxg6 Bxg6# b) 1.Bxf5 Rxf5 2.Rb8 Rc5# c) 1.Rxf5 Rxf5 2.Gh8 Bd5# Bicolour Bristol, ODT, Black Quasi-sacrifices. (Not very good as a problem, due to repetition of moves).

 3gk3/3p2p1/1p1P2P1/r4P2/2P5/7b/4KRB1/8 (7 + 7) (Grasshopper 0 + d8) Problem-602b Manolas Emmanuel original h#2 a) Diagram, b) bKe8→c8, c) bKe8→g8 a) 1.Bxf5 Be4 2.Bxg6 Bxg6# b) 1.Rxf5 Rxf5 2.Ga5 Rf8# c) 1.Rxf5 Rxf5 2.Gh8 Bd5#

## Friday, July 20, 2012

### International Chess Day

 Today is International Chess Day. I wish the best for chess lovers : gamers, composers and solvers.

## Monday, July 16, 2012

### Solving contest in tourney "Ikaros 2012"

The solving contest tends to be a regular feature in the international tourney "Ikaros 2012" in Greece, (more for the tourney here : http://ikariachess.blogspot.gr/).

This year's contest was prepared by Dimitris Skyrianoglou and contained direct mate problems in 2, 3, 4 moves.
Judge was the international arbiter Odysseas Vazelakis. In the first place finished with ease Nikos Kampanis and followed the solvers Nikitas Georgiadis, Kyriakos Lemonis, Stergios Sgouletas, Stergios Gemelas, Xanthoula Andrianatou, Argyris Andrianatos, Giannis Sgouletas, Christos Kazalas.

### The problems of the solving contest

 1k6/1P6/4K3/8/1Q6/8/8/8 (3 + 1) Problem-595 Kovalenko, Vitaly Karpati Igaz 1968 #2 1.Qb5! Ka7 / Kc7 2.b8=Q# (There is a Forsberg twin with wRb4 and key 1.Rb6!) 6k1/r7/1b3QK1/3S4/8/8/8/8 (3 + 3) Problem-596 Hoeg, Niels Eskilstuna Kuriren 1937 #2 1.Sc7! [2.Qd8# / Qg7#] Novotny intersection. 8/6B1/8/3BSk2/8/2K5/8/2Q5 (5 + 1) Problem-597 Loyd, Sam New York Commercial Advertiser 1895 #2 1.Bf8! Kxe5 / Kf6 2.Qg5# / Qf4# The key allows 2 flights for the besieged bK. k7/P6Q/8/8/8/8/p6p/K6R (4 + 3) Problem-598 Williams, Philip H. Carte de voeux 1904 #3 1.Kb2! a1=Q+ 2.Rxa1 h1=Q 3.Qxh1# 8/8/4B3/4p3/2p1k3/8/2P2QK1/8 (4 + 3) Problem-599 Jensen, Wilhelm E. Nationaltidente 1885 #3 Try : {1.Qa7? [2.Kf2 A [3.Qe3#] Kf4 3.Qe3#] Kf4! a} Key : 1.Qg1! 1...Kf4 a 2.Kf2 A [3.Qg4‡] e4 / Ke4 3.Qg3# / Qg4# 1...c3 2.Kf2 [3.Qg4#] Kf4 / Ke4 3.Qg4# / Kf3# Royal battery, Dombrovskis (paradox) 3B4/4Q1p1/pr3sK1/2R1R3/4P2k/5bpP/6P1/3q4 (8 + 8) Problem-600 Manolas, Emmanuel original #4 Tries : {1.Bxb6? Bg4!}, {1.Qd7? Re6!}, {1.Qf8? / Qe8? / Qxg7? Qxd8!} Key : 1.Rf5! [threat : 2.Rf4+ Bg4 3.Rh5#] 1...Bg4 2.Rf4 [threat : 3.Rh5#] Rb5 3.Qxf6+ gxf6 4.Bxf6# Mutual pins, Bristol line opening, mate with double pin.

## 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. 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 (not good). (This composition was made for a tourney, which allowed one white piece to be idle).