1st Solving Contest in Triandria

First Solving Contest 2002-12-01 Chess Club Triandria Greece

Some years ago the First Solving Contest of the Chess Club Triandria (Salonica, Greece) has been organized on the Sunday 2002-12-01.
The problems were selected by Carlo DeGrandi.
There were 3 judges : Costas Youvantsioudis, Costas Papadopoulos, Carlo deGrandi.
Committee for examination of Objections : Byron Zappas, Odysseas Vazelakis, Pantelis Martoudis.

There were two groups each having with 12 solvers, having 2 hours time to solve 4 problems (only direcmates, with 5 points per problem, 20 points maximum).
Group A had to solve a two-mover, a three-mover, a four-mover and a five-mover.
Group B had to solve (Teenagers) should solve four two-movers.

Winners in Group A were the following :
1. Costas Prentos (20 points, in 45 minutes), (Prize : Carved plate and 120€)
2. G. Papadopoulos (15, 53'), (Prize : Medal and 70€)
3. N. Papachristou (5, 67'), (Prize : Madal and 50€).
Other participants were (by order of application) : Tsikrikonis P., Cryssostomidis A., Exissoglou G., Eleftheriadis A., Kakousatze I., Fetihakis D., Kelessiadis G., Kyvelos S., Youvantsioudis C..

Participants in Group B (teenagers) (by finishing order) :
Boutsioukis N., Solakidis P., Gerochristos G., Constantinidis C., Papadopoulos P., Frangoulis L., Pahygiannakis L., Costopoulos D., Verikios G., Georgiadis B., Papavassiliou Ch., Siozos K..
All the participants were awarded with books.

Sponsors of the event : Asterias, Massoutis, Ygro Pyr, Koudigelis K., Catselis.


We now present the 8 problems of the Contest, 4 'easier' problems for the Group-B and 4 'more difficult' problems for Group-A.

---

Group B (teenagers), Time : 2 hours

	(Problem 212) Yuri Antonov, Hlas Ludu, 1981 Mate in 2 moves. #2 (8+6)
[3K4/2pR4/2p4S/2Pskp1S/4p3/4R3/6B1/3Q4]

Tries : {1.Rd6? cxd6!}, {1.Rxc7? / Sf7+? Ke6!}, {1.Re7+? Sxe7!}, {1.Sg4+? fxg4!}, {1.Rxe4+? fxe4!}, {1.Qa1+? Sc3!}, {1.Qxd5+? cxd5!}, {1.Qd4+? Kxd4!}.
Let us observe a little closer the play after the next try, in order to compare with the post-key play :
{1.Bh3?
1...S~ (a) 2.Re7# (A)
1...f4 (b) 2.Sf7# (B)
1...Ke6 (c) 2.Rxe4# (C)
1...Sf4!}

Κλειδί : 1.Qb3! (zz).
1...Kd4 2.Qc3#
1...S~ (a) 2.Sf7# (B)
1...f4 (b) 2.Rxe4# (C)
1...Ke6 (c) 2.Re7# (A)

The problem contains the theme [Lacny 2x3] (two phases and three variations with cyclic permutation of mates).

Theme Lacny pxv. It is developed over various phases (the number of phases is p). In one phase there are some (a b c ... z) defences (the number of variations is v) which are answered with the mates (A B C ... Z). In another phase the same (a b c ... z) defences are answered with a cyclic permutation of the mates (B C ... Z A). The theme is named after the composer Ludovit Lacny, who has published in 1949 the first problem with relevant content.

	(Problem 213) Alexandr Kuzovkov Second Honourable Mention, Bachernich Charpkov, 1982 Mate in 2 moves. #2 (4+4)
[8/5K2/6Bk/4B3/Q1p5/8/s7/2q5]

Tries : {1.Kg6? / Rf2? / Rg3? d5!}, {1.Qxd7? Ke5!}, {1.Qf5+? Kxf5!}, {1.Qe5+? Kxe5!}, {1.Qd5+? Kxd5!}, {1.Rg5? / Rg4+? Kf3!}, {1.Sg3+? / Sd2+? Kxe3!}.

Key : 1.Rd2! [2.Q(x)d5#]
1...Kf3 2.Qf5#

	(Problem 214) Michael Lipton, Commendation, I R T 1996-1997 Mate in 2 moves. #2 (5+2)
[4B3/8/8/3Q3R/1k6/3s2B1/8/1K6]

Tries : {1.Qb3+? Kxb3!}, {1.Qc4+? Kxc4!}, {1.Qd4+? Kb3!}, {1.Qa5+? Kc4!}, {1.Qb5+? Kc3!}, {1.Be1+? Sxe1!}.

Key : 1.Rh3! (zz)
1...Sb2 / Sc1 / Se1 2.B(x)e1# / Bd6#
1...Se5 / Sf4 / Sc5 / Kc3 2.Be1#
1...Sf2 / Ka3 2.Bd6#

	(Problem 215) Carlo deGrandi, original, 2000 Mate in 2 moves. #2 (5+4)
[8/5K2/6Bk/4B3/Q1p5/8/s4P2/2q5]

Tries : {1.Bf4+? Qxf4+!}, {1.Bg7+? Kg5!}.

Key: 1.Qd1! [2.Qh5#]
1...Qg5 2.Bg7#
1...Qf4+ 2.Bxf4#
1...Qxd1 2.Bf4#

---

Group A, Time : 2 hours

	(Problem 216) Ignaas Vandemeulebrouke, B-Merksem, 1990 Mate in 2 moves. #2 (7+7)
[8/6r1/1B3ps1/3S4/2P1kpQp/8/2K2PS1/b7]

Tries : {1.Sxf4? Se5!}, {1.Qe2+? Kf5!}, {1.Qh5? f3!}, {1.Qxf4+? Sxf4!}, {1.Sxh4? / f3+? Ke5!}.

The solution follows :
Key : 1.Sxf4! [2.Sxg6#]
1...Be5 2.f3#
1...Ke5 2.Qe6#
1...f5 2.Qe2#
1...Se5 2.Sxf6#
1...Sxf4 2.Qxf4#

	(Problem 217) Dr Hermann Weissauer, Tidskrift fuer Schack, 1987 Mate in 3 moves. #3 (9+11)
[1b3r2/1s3Sq1/2b1R3/4P1pP/1r2kpQ1/4B1ps/3R2K1/5S2]

Tries : {1.Sd6+? Bxd6!}, {1.Sh6? Qxh6!}, {1.Re8? Rxe8!}, {1.Rf6? Qxf6!}, {1.Qf3+? Kf5!}, {1.Qf5+? Kxf5+!}, {1.Bb6? Rxb6!}, {1.Bc5? Sxc5!}, {1.Kxh3? Bb5!}, {1.Rd4+? Rxd4!}, {1.Sxg3+? Kxe3+!}.

The solution follows :
Key : 1.Re7! [2.Qf3+ Kf5 3.Sxg3#]
1...Sg1 2.Bxg1 [3.Sxg3#] Rb3 / Rb4 3.R(x)d4#
1...Qxe5 2.Bg1 [3.Sxg3#]
___2...Sf2 / Sxg1 3.Sxg5#
___2...Rb3 / Rd4 3.R(x)d4#
1...Bxe5 2.Bc5 [3.Sxg3#]
___2...Sxc5 3.Sd6#
___2...Rb3 / Rd4 3.R(x)d4#

	(Problem 218) Dr Hermann Weissauer, Diagrammes, 1981 Mate in 4 moves. #4 (7+11)
[br6/3p3p/3p4/pP5p/3S1kSK/4pP2/BQ3p2/5s2]

Tries : {1.Se2+? / Se6+? Kxf3!}, {1.Bd5? Bxd5!}.

The solution follows :
Key : 1.Qa3! [2.Qxd6#]
1...Bxf3 2.Qxd6+ Ke4 3.Bb1#
1...Rxb5 / Re8 2.Qxd6+ Re5 3.Qxe5#
1...Rb6 2.Qd3 [3.Qf5#]
___2...Sg3 3.Qxe3#
___2...Be4 3.Qxe4#
___2...Rxb5 3.Bd5 [4.Se2# / Qe4# / Qf5#]
______3...Sd2 4.Se2# / Qxe3# / Qf5#
______3...Rb2 / hxg4 4.Qe4# / Qf5#
______3...Sg3 4.Qxe3#
______3...Rxd5 4.Qe4#
______3...Bxd5 4.Qf5#

	(Problem 219) A. F. Svanberg, Schachzeitung, September 1847 Mate in 5 moves. #5 (8+6)
[1S4R1/1S3s2/1p6/1k6/1p1P1s2/1P5q/K1P5/3B4]

Tries : {1.Sd6+? Sxd6!}, {1.c4+? bxc3 e.p.!}, {1.Be2+? Sxe2!}.

The solution follows :
Key : 1.Rg5+!
1...Qf5
___2.Rxf5+
______2...Sd5 3.Rxd5# / Be2#
______2...Se5
_________3.Bxe2+
____________3...Sxe2 4.Rxe5#
____________3...Sd3 4.Rxe5# / Bxd3#
_________3.Rxe5+ Sd5 4.Rxd5# / Be2#
___2.Be2+
______2...Sd3 3.Bxd3#
______2...Sxe2 3.Rxf5+ Se5 4.Rxe5#
1...Sd5 2.Rxd5#
1...Se5 2.Rxe5+ Sd5 3.Rxd5#
1...Sxg5 2.c4+ bxc3 e.p. 3.Sd6+
____________3...Kb4 4.Sc6#
____________3...Ka5 4.b4+ Kxb4 5.Sc6#


(This post in Greek language).

A study by Reti

The studies are orthodox problems not constrained in their stipulation for a specific number of moves.
From all the creations in the realm of chess composition, the studies are most attractive for the over-the-board chess players.
The usual stipulations are : "White plays and wins" or "White plays and draws". Since they present some unexpected or hidden properties of the pieces, the players consider studies as very instructive.

Richard Reti has composed many beautiful studies. We show one of them as Problem-210, (we have seen it in the excellent blog of Roberto Stelling).
It seems there that White can capture the black Queen in a straightforward manner. We shall see that this is not possible the first time, nor in the second time, nor in the third time...

	(Problem 210) R. Reti, Wiener Tagenblatt, 1925 White plays and wins + (3+3)
[8/8/2Q5/2K5/4S3/6p1/k5q1/8]

Do you agree that the plan is obvious [the Knight checks, then the white Queen captures the black Queen] ?

I propose to you to write down the solution you have found, and to compare it later with the solution that follows.

Solution of the Reti's study :

Key : 1.Sc3+ Ka1!
(first opportunity to capture the bQ : If now 2.Qxg2? Black is stalemated!)

2.Qa4+ Kb2
3.Qa2+ Kc1!
(second opportunity : If now 4.Qxg2? Black is stalemated!)

4.Qb1+ Kd2
5.Qb2+ Ke1!
(third opportunity : If now 6.Qxg2? Black is stalemated!)

6.Qc1+ Kf2
7.Sd1+ Kf3 (we will see with a reversal of moves what happens if bK goes to e2)
8.Qc3+ Ke2 (if 8...Kg4 9.Se3+, if 8...Ke4 9.Qd4+ Kf5 10.Se3+, if 8...Kf4 9.Qf6+ Ke4 / Kg4 10.Qc6+ / Se3+)
9.Qb2+ Kd3!
(fourth opportunity : If now 10.Qxg2? Black is stalemated!)

10.Qb3+ Ke2 (if 11...Ke4 12.Qb7+)
11.Qa2+ Kd3
(fifth opportunity : If now 12.Qxg2? Black is stalemated!)

12.Sb2+ Ke3 (if 12...Kc3 13.Sa4+, if 12...Ke4 13.Qa8+)
13.Sc4+ Kf3
14.Se5+

and White wins by capturing at last this black Queen! White needed fourteen checks and self-restraining enough to win this position.


(This post in Greek language).

Carlo de Grandi

Mr. Carlo de Grandi was born (1948) in Venice, Italy. He studied Accounting and Informatics in Athens.
He has written the following books (in Greek)
(1) "About the origin and spreading of Zatrikion (Chess) through the centuries", (Istorical report). Athens, 1985 (1st edition), 1989 (2nd edition)
(2) "Istorical report for Magic Squares and Magic Stars", Athens, 1999
and he is preparing his third book "The Mathematics of the Party".

He learned to play chess when he was 10 years old. Later he limited this activity to chess by post only.

He is a composer and a solver of chess problems.
Since he started composing, in 1989, he has created more than 120 problems.

He has organised 26 Contests of Solving Chess Problems.

He has cooperated with many chess magazines : [Gambit], [Skakistiki Kinissi (= Chess Move)], [Elliniko Skaki (= Greek Chess)], [Eleftheres Ores (= Free Hours)], [Deltio T.E.S.S.K (a Regional Bulletin)], [To Sah (= The Checking], [L’ Italia Scacchistica (= The Chess Italy)], [Black and White (an Indian magazine)], [Sah Mat (= check mate)], [Skaki gia Olous (= Chess for all)].

In problem-208, by Carlo DeGrandi, there is an interesting continuation with sacrifices of white pieces and guidance of black pieces. But the bK has always very restricted mobility, and there is only one series of moves.

	(Problem 208) Carlo DeGrandi, L`Italia Scacchistica No.1156/2002, Mate in 5 moves. #5 (8+8)
[6B1/4p3/6p1/4K1P1/3S4/1Pp1kP2/2P5/4B3]

Tries : {1.Bc4? / Bf7? / Bh7? / f4? / b4? e6!}, {1.Ke6? Kf4!}.

Key : 1.Bd5! (zz) e6
2.f4 (zz) e6xd5
3.f5 (zz) g6xf5
4.Ke5xf5 (zz) Ke3xd4
5.Bf2#

	(Problem 209) Carlo DeGrandi, Original, Mate in 3 moves. (Twin with wBb8 => h6) #3 (A) diagram, (B) -wBb8 +wBh6, (6+3)
[1B6/8/4S3/6p1/4P1k1/4K1p1/6P1/7R]

In this twin problem-209, by DeGrandi, the key changes together with the piece that guards g5, but the mechanism of mate is the same.

(A) wBb8
Tries : {1.Bxg3? (zz) Kxg3!}, {1.Sf4? (zz) gxf4+!}, {1.Sxg5? (zz) Kxg5!}, {1.e5? (zz) Kf5!}, {1.Rh5? Kxh5!}, {1.Rh2? (zz) gxh2!}.
Key : 1.Bf4! (zz) gxf4+ 2.Ke2 (zz) f3+ 3.gxf3#

(B) wBh6
Tries : {1.e5? Kf5!}, {1.Rh5? Kxh5!}, {1.Rh2? gxh2!}.
Key : 1.Sf4! (zz) gxf4+ 2.Ke2 (zz) f3+ 3.gxf3#


(This post in Greek language).

Circe (2)

In our previous post about Circe we said that it is an interesting Condition of the Fairy chess.

The world of the orthodox (directmate) problems has been described, many years ago, as a continent having very few unknown regions. Around this continent a magical sea is stretched, having many islands, some of which are uncartographed or unexplored. Many of these mythical islands are growing, just like the volcanic islands of the real world, and sometimes new islands emerge.

This is the case with “chess problem composition” (or “artistic chess” as we called it in Greece), where the fantasy of the composers endlessly creates new worlds, which the new composers are exploring, trying to map their limits.
In Homer's Odysseia, the island Aiaia (eea) of the nymphe witch Circe, doughter of Helios and Persa, was only one, but in Fairy Chess a whole archipelago of islands has been discovered with various kinds of the Circe Condition.


Let us see problem-202, by John Rice, which is composed in Zagoruiko frame, (that is, at least two defenses of the Black, always the same, are answered with different way in at least three Phases of the solution).

	(Problem 202) John Rice, Third Price, Phoenix 1994 Mate in 2 moves. Circe. #2 Circe (9+7)
[2Q1S2B/1q1BS3/3p2P1/3r1p2/3sk3/4P3/2P2Kb1/8]

There are a few tries in this problem : {1.Bxd4(+bSb8)? Rxd4(+wBc1)!}, {1.Bxf5(+bPf7)+? Rxf5(+wBf1)+!}, {1.Sxd6+? Rxd6(+wSg1)!}, {1.Sf6+? Ke5!}.
Let us make a special note for the next two tries on the square c6 :

{1.Sc6? (is guarding e5 and is threatening 2.Sf6#).
If 1...Qxc6(+Sb1) 2.Sd2#. (Defence a, Mate A).
If 1...Sxc6(+Sb1) 2.Sc3#. (Defence b, Mate B).
But there is the defence 1...Sxc2!}.

{1.Bc6? (is pinning Rd5 and is threatening 2.Sxd6(+d7)#).
If 1...Qxc6(+Bf1) 2.Bd3#. (Defence a, Mate C).
If 1...Sxc6(+Bf1) 2.Bxg2#. (Defence b, Mate D).
If 1...Sxc2 2.Qxf5(+f7)#. (Defence c, Mate G).
If 1...Qc7 2.Bxd5(+Ra8)#. (Defence d, Mate H).
But there is the defence 1...Qxe7(+wSg1)!}.

Key: 1.Qc6! (is pinning Rd5 and is threatening 2.Sxd6#).
If 1...Qxc6(+wQd1) 2.Qd3#. (Defence a, Mate E).
If 1...Sxc6(+wQd1) 2.Qxd5(+bRa8)#. (Defence b, Mate F).
If 1...Sxc2 2.Bxf5(+bPf7)#. (Defence c, Mate I).
If 1...Qc7 2.Qxd5(+bRa8)#. (Defence d, Mate J).

Other variations : 1...Sb5 2.Bxf5(+bPf7)#, 1...Qb7-b4 / Qb8 2.Qxd5(+bRa8)#, 1...Qxd7(+wBf1) 2.Qxd5(+bRa8)# / Bd3#.

We see that in three (3) phases of the problem there are answers A-B C-D E-F for the two (2) defences a-b, that is we have a Zagoruiko 3x2.

There are, also, changed mates G-H I-J for the defences c-d, as an additional asset of the problem.


We will see next a help-stalemate problem with condition Circe.

	(Πρόβλημα 206) Harry G. Polk, Mat 1982, Black plays and helps white to stalemate in 6 moves. Circe. h=6 Circe (15+16)
[S1s1BQ2/2RpP1rp/PB1pPpPp/1S1sPpPK/4PrP1/5p2/p1b3k1/2q3b1]

From the initial position only a white Rook is missing.
The Pawns can have this peculiar setting, because the captured pieces were not lost but regenerated, or reborn by Circe.
During the solution all black pieces are trapped and Black is stalemated.

Key : 1.Bxe4(+wPe2) exf3(+bPf7)
2.Qxc7(+wRal) Sbxc7(+bQd8)
3.Bxb6(+wBcl) Bxf4(+bRh8)
4.Rxf8(+wQdl) Qxd5(+bSg8)
5.Kxf3(+wPf2) Rxa2(+bPa7)
6.Rxe8(+wBfl) Sxb6(+bBf8) =

The following twenty captures are illegal, as the black King is exposed to check : a7xb6(+wSg1)+, Sxb6(+wSg1)+, Qxc7(+wSg1)+, Bxd5(+wQd1)+, Scxe7(+wPe2)+, Qxe7(+wPe2)+, Rxe7(+wPe2)+, Bxe7(+wPe2)+, Sgxe7(+wPe2)+, d7xe6(+wPe2)+, f7xe6(+wPe2)+, d6xe5(+wPe2)+, f6xe5(+wPe2)+, Kxf4(+wBc1)+, f7xg6(+wPg2)+, Rxg6(+wPg2)+, h7xg6(+wPg2)+, f6xg5(+wPg2)+, h6xg5(+wPg2)+, f5xg4(+wPg2)+.


Ending this post, we will see a Circe with theme Zappas.

	(Problem 207) Byron Zappas, Die Schwalbe 1987, Mate in 2 moves. Circe. #2 Circe (12+13)
[1B2R2Q/P2p1P1p/1p5r/1P1kSB2/2psS1p1/1Pp1ps1p/4b3/3R3K]

In theme Zappas, one flight is guarded by three white pieces and in two moves the flight is unguarded, and this is repeated cyclically. There must be three thematic tries.
Here we observe that the square e5 is guarded by Bb8, Re8 and Qh8.

Thematic tries :
{1.Sxc4(+bPc7)? Already Bb8 is blocked,
and the threat [2.Sxe3(+bPe7)#] will block Re8,
so there is a defense 1...Rf6! to block Qh8.
It is no good to play 1...Bxc4(+wSb1)? 2.Sxc3(+bPc7)#}.

{1.Sxg4(+bPg7)? Already Qh8 is blocked ,
and the threat [2.Sxe3(+bPe7)#] will block Re8,
so there is a defense 1...Rd6! to block Bb8}.

{1.Qf8? The Qh8 has stopped guarding e5 in order to hold c5
and now is threatening [2.Sxc3(+bPc7)#], but this blocks Bb8
so there is a defense 1...Re6! to block Re8}.

There are many more tries, making the problem difficult to solve : {1.f8=B? Rd6!}, {1.a8=Q+? / a8=B+? / a8=S? Rc6!}, {1.Sd3? / Sg6? / Sc6? cxb3(+wPb2)!}, {1.Sxf3(+bSg8)? Bxf3(+wSb1)+!}, {1.Sxc3(+bPc7)+? Kc5!}, {1.bxc4(+bPc7)+? Bxc4(+wPc2)!}.

Key : 1.f8=Q! (which is guarding c5 and is threatening [2.Sxc3(+bPc7)#] )
If 1...Rd6 2.Qxd6#
If 1...d6 2.a8=Q/B#


(This post in Greek language).

Circe (1)

One amazingly interesting condition of the Fairy chess is Circe Chess. There are many kinds of this condition. In the following we describe some of these kinds, ten kinds of Circe and one Anti-Circe, and we present an example of the first kind, the regular Circe.

1. Regular Circe : A captured piece is instantly reborn on the initial- game- square. (We name “initial- game- square” the square which is used for the initial placement of the piece when we start a classical chess game. For example, the white-squared white Bishop has f1 as initial- game- square).
If the initial- game- square is occupied, then the captured piece is out of the game (is lost).
If the captured piece is one of those which move on squares of various colors (i.e. R, S), then the piece is reborn on an initial- game- square with same colour as the capture- square. (For example a white Rook, if it is captured on the white c8 it will be reborn on the white h1, but if it is captured on the black d8 it will be reborn on the black a1).
If a Pawn is captured, it will be reborn on the initial- game- square of the same file where the capture- square is. (The initial- game- square of a white Pawn is on row-2, while the initial- game- square of a black Pawn is on row-7).
The Circe condition is not applied on the King.
Castling with a reborn Rook is allowed.
The fairy pieces (which we believe that appeared after a strange Pawn promotion) are reborn on the promotion- square of the capture- file. (We know that the promotion happens on row-8 for white pieces and on row-1 for black pieces).

2. Circe Parrain : The square, where the captured piece-A will be reborn, is specified after the move of the next same-colored piece-B. The line from the capture- square to the regeneration- square is parallel, equal in length, and has the same direction with the move of piece-B. In other words, the piece-B by making its move specifies the name of the square for regeneration of piece-A (becomes the godfather, “parrain” in French).
The Pawns can be reborn on row-1 and on row-8. It it is promotion- row for them, they are instantly promoted to any piece the owner of the Pawn desires. If the row is not promotion- row, they can make (in a later move) a step on the file.

3. Circe Rex Inclusiv : It is like the regular Circe, but the Circe condition is valid for the Kings, that is the King can be captured and reborn on its initial- game- square. (If the square is occupied, it is the end of the game).

4. Diagram Circe : It is like the regular Circe, but the captured piece is reborn on the initial- diagram- square. (We name “initial- diagram- square” the square on which the captured piece was placed in the initial position of the problem).

5. Interchange Circe (German “Platzwechsel-Circe”, PWC) : It is like the regular Circe, but the captured piece is reborn on the departure- square of the capturing piece. Capturing piece and captured piece interchange places.

6. Kamikaze Circe : It is like the regular Circe, but additionally the capturing piece is also reborn.

7. Mirror Circe : It is like the regular Circe, but the captured piece is reborn as if it were of the opposite colour.

8. Reflecting Circe : The regeneration- square is symmetrical (with axis of symmetry the line between file-d and file-f) with the one that would be valid in regular Circe.

9. Symmetrical Circe : The regeneration- square is symmetrical (with center of symmetry the center of the chess-board) with the square of the capture. (Capture on g6, regeneration on b3).

10. Volcanic Circe : The captured piece is not out of the game when the regeneration square is occupied. The piece remains inactive “in” the square. Just as the square is evacuated, the captured piece reappears there having all its power.

11. Anti-Circe : On making a capture, the captured piece is lost and the capturing piece is reborn on its initial- game- square, but if it cannot be regenerated there the capture is forbidden.
A piece standing on its initial- game- square can make a capture remaining practically motionless!
A Pawn can capture with simultaneous promotion, if the regeneration- square of the promoted piece is unoccupied.

---

Regular Circe, with multiple promotions

Now we will study the problem-201, by N. A. Macleod, with theme AUW and condition Circe, (which problem is the enlightening example on the web-page of British Chess Problem Society after following the link “Fairy Chess”).
On writing the solution of a Circe problem, after each capture move, we note inside a parenthesis that a piece is added, (i.e. (bBc8) means that a black Bishop is added on square c8).

	(Problem 201) N. A. Macleod First Honourable Mention, Europe Echecs 1977 Mate in 2 moves. Circe. #2 Circe (10+4)
[k7/3R4/8/4B3/3p4/K2P1S2/bP1pQ3/3SR2B]

The problem has various tries :
{1.Sg1+? / Sxd2+? / Sh2+? / Sh4+? / Sg5+? / Sxd4+? Bd5!}, and we understand that it is of no use to lift the Knight Sf3 and check by discovery of the Bishop Bh1, because the black Bishop Ba2 can be interposed on d5, and if it is captured there by the wB giving check (2.Bh1xd5(+bBc8)+), the bB is reborn on c8 and can be interposed again on b7! As it seems we are out of moves since the problem is a two-mover. But suppose we insisted with (3.Bd5xb7(+bBc8)+), then the bB would be reborn on c8 and it would capture the wB (3...Bc8xb7(+wBf1)).
{1.Qe4+? Bd5!}, similar comment, as with previous tries.
{1.b3? Bxb3(+wPb2)!}.
{1.Qxd2? [2.Qa5#] Bc4!},
{1.Ra7+? Kxa7(+wRa1)!}, because during this Phase the Ra7 is unsupported.

Key : 1.Kb4! (zz). The key is not threatening something, (basically the file-a is evacuated), and is bringing Black into a zugzwang situation. If the wRe1 is captured, it will be reborn on a1 and will pin the bBa2, (and then the wSf3 will be lifted to discover wBh1, which will check-mate).

1...dxe1=Q(+wRa1)+ 2.Sxe1(+bQd8)#

1...dxe1=R(+wRa1). White cannot continue with (2.Sxe1(+bRh8)+? Rxh1(+wBf1)!), so White plays (2.Sg1#) and bRe1 is blocked.

1...dxe1=B(+wRa1)+. Here the move (2.Sxe1(+bBf8)+) is forbidden since the bB is reborn on a square where it continues to check. The correct continuation is (2.Sd2#) and bBe1 is unable to interfere.

1...dxe1=S(+wRa1), and bSe1 can interfere on the main diagonal. White is futile to continue with (2.Sxe1(+bSb8)+?), since the bS is reborn on a square from which it can again interfere on the main diagonal. The correct continuation is (2.Qe4#).

The problem has achieved the four promotions (theme Allumwandlung, AUW).
In the next variation we see a peculiarity of the Circe condition, namely that a piece can support itself!

1...B~ 2.Ra7#. Now that bB has evacuated file-a, the defense (2...Kxa7(+wRa1)?) is forbidden since wR is reborn on a1 and guards a7. That is, during the Phase of the actual play the white Rook supports itself from a great distance and from the future! We see now why the white King in the key-move left file-a!


(This post in Greek language).

Kostas Prentos

The biography you are about to read was written by Kostas Prentos in 2006, (and published in the edition "Kostas Prentos 40 Jubilee Tourney 2006" of the Greek Chess Composition Committee, December 2006)

"I was born 28/08/1966 in Thessaloniki. I graduated from the Economic University of Thessaloniki and I am currently working as an accountant. I am not married.

I learned the moves of chess at the late age of 12 but I was immediately hooked. Triantafyllos Siaperas, a chess author and journalist, influenced me in my early chess steps. His weekly newspaper columns, and the chess problems featured in them, somehow sparked my interest for chess problems long before my first club games. Before the age of 15, I composed my first problems, as puzzles for my friends to solve. I joined a chess club in 1982 and began playing tournament chess, but my strong interest for chess problems and studies remained.

During the 1984 Chess Olympiad in Thessaloniki, I had the opportunity to participate in my first solving tourney and enjoyed the experience very much. Later I successfully participated in further solving tourneys, and became a regular solver.

The friendship and frequent correspondence with the skilled composer Harry Fougiaxis taught me a great deal about helpmates and fairy problems and especially how to appreciate a chess problem aesthetically. As a result, I was involved in composing more seriously and between 1987 and 1991 I composed a few chess problems and published a dozen of them, mostly fairy helpmates and proof games. I lost interest in composing and solving after 1991, for almost 10 years. At the same time, I was playing a lot of tournament chess; I became Candidate Master in 1987, Master in 1990 and FIDE Master in 1995.

The Internet revolution changed the world in general but in particular, it gave chess a boost. I started playing chess online, and especially enjoyed playing losing chess and progressive chess. It was a small group of problemists formed in the Internet Chess Club which reinstated my lost interest in chess problems, around 2001. I started composing again and I have published more than 100 problems in the last few years, the majority of them being proof games.

In 2002, the organization of the 1st Greek Solving Championship triggered my involvement in solving again. I have participated in all World Solving Championships since then and I was awarded the title of International Solving Master in 2004."

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The continuation of the text is written by Alkinoos :

The situation now, 08/2008, is almost the same...
Kostas Prentos repeatedly takes the First Prize in every local Solving Contest he participates and he is seven times (in seven contests) Champion of Greece in Solving Chess Problems.
In the 50th WCCC, held in Rhodes Greece (October 2007), he was honoured as composer and he also (together with Nikos Kalessis) took First place in bughouse chess contest. In order for you to judge if this was a notable result, we present here the final list of the players (in parenthesis [Solver ELO if the player is a solver] and [Country]) coming from various countries :

50th World Congress of Chess Composition, Bughouse Chess Tournament, Rhodes, Wednesday 17 October 2007

01) 11 points: Nikos Kalessis (Greece) & Kostas Prentos (2492 Greece)
02-03) 10p. : Piotr Gorski (2166 Poland) & Ryszard Krolikowski (1989 Poland), Ivan Denkovski (1566 F.Y.R.O.M.) & Vladimir Podinic (2598 Serbia)
04-06) 09p. : Eric Huber (2464 Romania) & Ion Murarasu (2081 Romania), Ofer Comay (2761 Israel) & Gady Costeff (Israel), Eddy van Beers (2584 Belgium) & Andy Ooms (1904 Belgium)
07) 08p. : Noam Elkies (2605 Israel) & Yedael Stepak (~2250 Israel)
08) 06p. : David Friedgood (2358 Great Britain) & John Nunn (2860 Great Britain)
09-11) 04p. : Milan Petras (2053 Czech Republic) & Miroslav Voracek (2273 Czech Republic), Indrek Aunver (1566 Estonia) & Margus Soot (2237 Estonia), B. Piliczewski (2457 Poland) & Michal Wysocki (~1989Poland)
12) 03p. : Al. Azhusin (2450 Russia) & Andrey Selivanov (2620 Russia)
13-14) 02p. : Vlaicu Crisan (2273 Romania) & Masaki Yoshioka (2138 Japan), Mark Erenburg (2379 Israel) & Josef Retter (1968 Israel).

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Problems by Kostas Prentos

Mr Kostas Prentos, besides being an excellent solver, is an exceptional composer. Here follow some of his prized problems, which are published in the FIDE Albums 1986-88 and 2001-2003.

	(Problem 203) Kostas Prentos, Honourable Mention, feenschach, 1988 Proof Game, 18 moves. PG 18.0 (14+13)
[2krbKr1/ppp1s1pp/2pb3s/8/1B6/SP6/P2PPPPP/2R2BSR]

In a Proof Game the solver must specify the history of moves, starting from initial arrangement of the pieces for a chess game, and ending with the given position.
The solution follows :
Key : 1.c4 Sc6 2.c5 Se5 3.c6 dxc6 4.Qb3 Bh3 5.Kd1 Sg4
6.Kc2 Qd3+ 7.Kxd3 0-0-0+ 8.Ke4 f5+ 9.Kxf5 e6+ 10.Kxe6 Bc5
11.Kf7 Rd6 12.Kf8 Se7+ 13.Qg8 Sh6 14.b3 Bd7 15.Ba3 Be8
16.Bb4 Rd8 17.Sa3 Bd6 18.Rc1 Rxg8


The next Problem-204, is an example of another category of help-problems, named helpstalemate. Black plays and helps White to make a draw.

	(Problem 204) Kostas Prentos & Dan Meinking, Third Prize, StrateGems, 2001 Helpstalemate in 6.5 moves, h=6.5 (3+11)
[4b3/7r/8/8/4p1p1/1K3prp/5Rqk/5Bss]

This is a help-problem, where normally Black plays first, but the moves are 6.5 so White plays first. Furthermore, the goal is not the win, but the draw.
The stipulation could be "White plays and with the help of Black draws in 6.5 moves".

During the solution we see, twice, opening of a line by a white piece, reverse move of a black piece on this line, and switchback of the white piece only to be annihilated. The solution is :

Key : 1...Ba6!
2.Bb5 Kc2
3.Bf1 Be2
4.fxe2 Rf8!
5.Rf7 Kd2
6.Rf2 Rf3
7.exf3 Ke1 = (black is stalemated).

	(Problem 205) Kostas Prentos & Kevin Begley, First Prize, StrateGems 5th Anniversary Tourney, 2003 Helpmate in 5 κινήσεις. (2 solutions). h#5 2111... (4+14)
[8/3K1b2/2p3r1/2p1p3/1Ppbrp2/1ppPkp2/2Pq4/8]

The way that bK makes his triangle, f2-e2-e3 or e2-f2-e3, specifies the file of promotion of the Pawn b4. Behind the King are hiding two linear pieces with mutual roles : the one must not support the Pawn to be captured, the other must block a flight. The solution follow :
Key : 1.Kf2 b5 2.Re1! b6 3.Ke2 b7 4.Bf2 b8=Q 5.Ke3 Qxe5#
Key : 1.Ke2 b5 2.Bg1! bxc6 3.Kf2 c7 4.Re2 c8=Q 5.Ke3 Qxc5#

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The 36th issue of the Greek magazine "Skaki gia Olous" (=Chess for everybody) is in circulation, since 19-06-2008, from the editions Kedros. The front page of the issue contains a photo of Kostas Prentos, who was nominated Champion of Greece for seventh consecutive year in the Seventh Contest of Solving Chess Problems! The column of artistic chess of this issue is dedicated to the contest of this year and to the success of the multi-champion from Thessaloniki.

(This post in Greek language).

7th Solving Contest in Greece, 2008-06-01, G.C.F., Aegaleo

The seventh Chess Problems Solving Contest in Greece, organized by the Greek Chess Federation (E.S.O.), was successfully held at the hospitable Chess Club of Aegaleo (Sunday, June 01, 2008).
Twenty one solvers have participated, having 4 hours time to solve 12 problems (the Group-A) or 8 problems (the Group-B). The problems were selected by the IM Harry Fougiaxis.

The most points in Group-A were gathered by :
1. Kostas Prentos (56 points from 60 maximum, in 227 minutes), Champion of Greece for the seventh time in a row,
2. Andreas Papastavropoulos (43.2, 240),
3. Nikos Mendrinos (35, 238),
4. Ioannis Garoufalidis (28.2, 240),
5. Dimitris Skyrianoglou (27.2, 239),
6. Panagiotis Konidaris (26.5, 240).
With less points follow Panagis Sklavounos, Alexandros Kostouros, Kostas Mitsakis, Leokratis Anemodouras, Lefteris Markessinis, Spyros Ilantzis, Giorgos Tsolakos, Emmanuel Manolas, Aggelos Sandalakis, Theodora Koutsogiannopoulou, Vassilis Blazos, Theopistos Nikitidis, Apostolos Tsiropoulas, Ilias Georgakis.

In Group-B, Panagiotis Papathanassiou took first place.

In Women, Miss Theodora Koutsogiannopoulou had the best result.

The winners were awarded with Cups, Medals, Books. The Champion received also an artistic chessboard.

We present the 20 problems of this contest, 8 'easier' problems for Group-B and 12 'harder' problems for Group-A. The Contest was divided into two Rounds of two hours each.

Observe in the solutions the way points are distributed in variations.
The points are written in boldface inside brackets.
When we write the solution, we must not omit important variations.

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Group B, 1st Round, Time : 2 hours

	(Problem 181) Franz Pachl, Third Prize, Schach-Echo, 1979 Mate in 2 moves. #2 (8+7)
[5q2/8/8/QB6/3r1R1B/s2p1p2/2sRP3/3S1k1K]

Tries : {1.eхd3? [2.Rf2#] Rb4!}, {1.eхf3? [2.Rf2#] Qb4!}, {1.e3? [2.Rf2#] Sb4!}, {1.Rxc2? Sxc2!}. The solution follows :

Key : 1.e4! [5.0] [2.Rf2#]
1...Qb4 2.Rxf3#, 1...Rb4 2.Bxd3#, 1...Sb4 2.Se3#

	(Problem 182) Norman A. Macleod & David Friedgood Prize, Quick Composition Contest, Ramsgate, 1984 Mate in 2 moves. #2 (8+7)
[8/8/3S1Kp1/2p3P1/2Bkb2Q/2ps4/3s2P1/B4S2]

Set play : 1...S3~ 2.Qf2#, 1...S2~ 2.Qxe4#

Tries : {1.Ke6? Sf4+!}, {1.Sb5+? Kxc4!}, {1.Qe1? Sxe1!}, {1.Qf2+? Sxf2!}, {1.Qxe4+? Sxe4+!}, {1.Qf4? Sxf4!}, {1.Sh2? Ke3!}, {1.Sxd2? Ke3!}, {1.Bxc3+? Kxc3!}, {1.Bb2? cxb2!}.

Key : 1.Ke7! [5.0] (zz)
1...Ke5 2.Bxc3#, 1...S3~ / Se5 2.Qh8# / Qf2#, 1...S2~ 2.Qxe4#

	(Problem 183) Theodor Begheijn, Third Prize, Tijdschrift, 1960 White plays and wins. + (5+3)
[S4k1K/5r1P/1P6/8/1p6/8/1P6/8]

The solution follows :
Key : 1.Sc7! Rf6 2.b7 Rb6 3.Se6+ [1.0] Kf7 4.Sd8+ Kf8 5.b3 [1.0] Rb5
6.Se6+ Kf7 7.Sc5 [1.0] Rb6 8.Se4 [1.0] Kf8 9.Sd6 [1.0] +-

	(Problem 184) Valery Karpov, Problem, 1973 Selfmate in 2 moves. s#2 (11+6)
[4RS2/B2p2Q/2P1P3/5k2/3K1s2/4P2P/1pr5/bB3R2]

There are tries : {1.Ra8? / Rd8? / e7? d5!}, {1.Qf6+? Kxf6!}, {1.Qg4+? Kf6!}, {1.Qg5+? Kxg5!}, {1.Qf7+? Kg5!}. The solution follows :

Key: 1.Rb8! [1.0] (zz)
1...dxc6 2.Rb6 [1.0] c5#
1...dxe6 2.Qh6 [1.0] e5#
1...d5 2.Rxb2 [1.0] Bxb2#
1...d6 2.Qe5+ [1.0] dxe5#

The solution shows the four variations with moves of the bPd7.

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Group B, 2nd Round, Time : 2 hours

	(Problem 185) Juan C. Morra, Práca, 1955 Mate in 2 moves. #2 (9+7)
[BK6/4p3/b7/2R5/S2k2S1/1Q6/srsP4/Br1R4]

Tries : {1.Kc7? / Ka7? / Bh1? / Bg2? / Bb7? e6!}, {1.Rc4+? Bxc4!}, {1.Rc8? Bxc8!}, {1.Rd5+? Ke4!}, {1.Bxb2+? Rxb2!}. The solution follows :

Key : 1.Rc7! [5.0] [2.Rd7#]
1...Scb4 2.Qe3#, 1...Sab4 2.Qc3#, 1...Se3 2.dxe3#, 1...Sc3 2.dxc3#, 1...Bc8 / Bc4 2.R(x)c4#, 1...Bb5 2.Qd5#, 1...Bb7 2.Qc4#

	(Problem 186) Chithathur Gopalan Sathya Narayanan, Telescacco 2000 Mate in 2 moves. #2 (7+10)
[5S2/1ss2b2/2Qp2pK/b5Rp/4Sk2/1r3p2/5B2/1B6]

Set play : 1...Rb4 2.Qc1#

Tries (virtual play) : {1.Sxg6+? Bxg6!}, {1.Qc1+? Bd2!}, {1.Qxd6+? Sxd6!}, {1.Sc3? Rb4!}, {1.Sg3? Re3!}, {1.Sf6? d5!}, {1.Sxd6 ? Sc5!}, {1.Sc5? Sd5!}, {1.Bg3+? Ke3!}. The solution follows :

Key : 1.Sd2! [5.0] [2.Qe4#]
1...Rb4 2.Qxf3#, 1...Bd5 2.Sxg6#, 1...d5 2.Qf6#, 1...Sd5 2.Qc4#, 1...Re3 2.Bg3#, 1...Sc5 2.Qxd6#

	(Problem 187) Genrikh M. Kasparyan, Zarya Vostoka, 1931 Mate in 3 moves. #3 (7+7)
[8/4p3/6QP/sPp1k3/8/2K1Spp1/4P3/3Sr3]

Tries : {1.Qf5+? / Qh5+? / Qg5+? Kd6!}, {1.Qe6+? Kxe6!}, {1.Sg4+? Kf4!}. The solution follows :

Key : 1.b6! [1.0] [2.Qf5+ [1.0] Kd6 3.Qd5#]
1...Rxd1 2.Sg4+ [1.5] Kf4 / Kd5 3.e3# / e4#
1...e6 2.Qxg3+ [1.5] Kf6 / Ke4 3.Qg7# / exf3#

	(Problem 188) Christer Jonsson, feenschach, 2006 Helpmate in 2 moves. (2 solutions). h#2 2111 (5+6)
[5B2/8/K1s5/1p5p/4kb1R/3r4/R7/1B6]

Key : 1.Kd5! Rd2 2.Be5 Ba2# [2.5]
Key : 1.Ke3! Bh6 2.Rd4 Rh3# [2.5]

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Group A, 1st Round, Time : 2 hours

	(Problem 189) Jacques Savournin, Third Commendation, Phénix, 1996 Mate in 2 moves. #2 (9+8)
[4B3/p1r1P2q/Q7/1p1kSsR1/3bSKP1/8/7p/3R4]

Tries : {1.Bc6+? Ke6!}, {1.Bf7+? Qxf7!}, {1.Qa2+? Rc4!}, {1.Qe6+? Kxe6!}, {1.Qd6+? Sxd6!}, {1.Qc6+? Rxc6!}, {1.Sf3? Qg7!}, {1.Sc6? Rd7!}, {1.Sf6+? Kc5!}, {1.Rxd4+? Kxd4!}. The solution follows :

Key : 1.Sc4! [5.0] [2.Se3#]
1...bxc4 2.Qd6#, 1...Kxc4 2.Qxb5#, 1...Rxc4 2.Qd6#, 1...Rc6 2.Qxc6#, 1...Qh3 2.Bf7#

	(Problem 190) Friedrich Chlubna, Arbeiter Zeitung, 1964 Mate in 3 moves. #3 (7+10)
[sK3B2/1B6/4k1P1/3R1s1S/2p1rpS1/2pp4/8/2b1r3]

Tries : {1.Bc8+? Kxd5!}, {1.Sg7+? Sxg7!}, {1.Rd6+? Sxd6!}. The solution follows :

Key : 1.Bc6! [1.0] [2.Rd6+ [1.0] Sxd6 3.Sg7#]
1...Ba3 2.Re5+ [1.5] Rxe5 3.Sxf4#
1...Rd4 2.Bd7+ [1.5] Kxd5 3.Sf6#

	(Problem 191) Yakov Vladimirov, First Prize, Uralsky Problemist, 2002 Mate in 4 moves. #4 (11+10)
[5R2/2psp3/1p2B2p/3PP1b1/1P1kS1r1/1PS1p1p1/4KR2/B7]

Tries : {1.Rf5? g2!}, {1.Sd2? exd2!}, {1.Sxg3? Rxg3!}, {1.Sxg5? hxg5!}, {1.Sc~+? Kxe4!}, {1.Rf1? Sxf8!}, {1.Rf4? Bxf4!}. The solution follows :

Key : 1.R8f6! [0.5] [2.Sb5+ Kxe4 3.Bf5+ [0.5] Kxd5 4.Sxc7#]
1...Bxf6 2.Sd2 [3.Sc~#] exd2 3.Sb1+ [0.8] Ke4 4.Sxd2#
1...Sxf6 2.Sc5 [3.Sc~#] bxc5 3.Sa4+ [0.8] Ke4 4.Sxc5#
1...exf6 2.Sd6 [3.Sc~#] cxd6 3.Sb5+ [0.8] Ke4 4.Sxd6#
1...gxf2 / exf2 2.Sxf2 [3.Sc~#] exf2 / gxf2 3.Sd1+ [0.8] Ke4 4.Sxf2#
1...Rf4 2.Sxg3 [3.Sc~#] Rxf2+ 3.Rxf2 [0.8] [4.Sc~#]

	(Problem 192) Jan Rusinek, First Prize, New Statesman, 1971 White plays and draws. = (5+4)
[2K5/2Psk1P1/PP6/8/2b5/2s5/8/8]

The solution follows :
Key : 1.a7! (a) Ba6+ 2.b7 Se4 3.g8=S+! [1.0] Ke8 4.Sf6+! [1.0] Sexf6 5.a8=B! [1.5] (b) Se5
6.Kb8 Sc6+ 7.Kc8 Bf1 8.b8=R! [1.5] (c) Ba6+ 9.Rb7 Se4 = (stalemate)

Notes :
(a) 1.Kb7? Bd5+ 2.Ka7 Sb5#
(b) 5.a8=Q? Sd5 6.Qxa6 Se7#
(c) 8.b8=Q? Ba6+ 9.Qb7 Se4 10.Qxa6 Sd6#, or 8.b8=S? Se7+ 9.Kb7 Bg2+ 10.Ka7 Sc8+

	(Problem 193) Valery Karpov, Third Honourable Mention, Shakhmatny, 1967 Selfmate in 3 moves. s#3 (10+8)
[s7/2p2p2/1pSp1P2/1P1K2BQ/B2pR3/3k4/1P2p1R1/8]

Tries : {1.Se5+? dxe5!}, {1.Re8? e1=B!}, {1.Bc2+? Kxc2!}. The solution follows :

Key : 1.Re7! [1.0] (zz)
1...e1=Q 2.Qf3+ Qe3 3.Rd7 [1.0] Qxf3#
1...e1=R 2.Qh7+ Re4 3.Re6 [1.0] fxe6#
1...e1=B 2.Qh3+ Bg3 3.Rxc7 [1.0] Sxc7#
1...e1=S 2.Bc2+ Sxc2 3.Sb4+ [1.0] Sxb4#

The problem is an allumwandlung (AUW), achieving the four promotions.

	(Problem 194) Nikolay Popkov, Shakhmatnaya Poeziya, 2008 Helpmate in 4 moves. (3 solutions). h#4 3111... (2+11)
[8/3s4/6p1/1sS1p3/pqkrp1K1/p3p3/8/8]

The solution follows :
Key : 1.Qa5 Sxa4 2.Kb3 Sb6 3.Rb4 Sxd7 4.Ka4 Sc5#
Key : 1.Rd3 Sxe4 2.Sd4 Sc5 3.Kd5 Sxd7 4.Ke4 Sf6#
Key : 1.Kd5 Sa6 2.Ke6 Kg5 3.Qe7+ Kxg6 4.Sd6 Sc7#

One solution = [2.0], two solutions = [4.0], three solutions = [5.0] points.

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Group A, 2nd Round, Time : 2 hours

	(Problem 195) Arthur J. Mosely, First Prize, Australian Columns, 1921 Mate in 2 moves. #2 (8+9)
[8/2Sp1Sp1/1p3sB1/pR4R1/K2Bpk1s/6p1/6P1/8]

Many tries : {1.Sd5+? Sxd5!}, {1.Se6+? dxe6!}, {1.Bh7? g6!}, {1.Rg4+? Kxg4!}, {1.Rf5+? Sxf5!}, {1.Rb3? b5+!}, {1.Rf5+? Sxf5!}, {1.Re5? b5+!}, {1.Bg1? e3!}, {1.Bf2? gxf2!}, {1.Be3+? Kxe3!}, {1.Be5+? Ke3!}, {1.Bxb6? e3!}, {1.Ka3? a4!}, {1.Kb3? a4+!}. The solution follows :

Key : 1.Se5! [5.0] (zz)
Variations : 1...e3 2.Sd3#, 1...Kxg5 2.Be3#, 1...Sf3 2.Rf5#, 1...Sxg2 2.Rf5#, 1...Sxg6 2.Sxg6#, 1...Sf5 2.Rxf5#, 1...Sf~ 2.R(x)g4#, 1...d5 / d6 2.Se6#.

	(Problem 196) Jan Kotrc, Shahmat, 1884 Mate in 3 moves. #3 (5+11)
[5S1Q/4pk2/p1B4p/2p2b2/r1pP3K/2bs2p1/8/8]

Tries (without interest) : {1.Qg7+? Kxg7!}, {1.Qg8+? Kxg8!}, {1.Be8+? Kxe8!}. The solution follows :

Key : 1.Se6! [0.5] [2.Sd8+ [0.5] Kg6 3.Be8#]
1...Bh3 / Bg4 2.Qg7+ [1.0] Kxe6 3.Qg6#
1...Se5 2.Be8+ [1.0] Kxe6 3.Qxe5#
1...Ba5 2.Bd7 [1.0] [3.Qg7#. Not {2.d4? c3+!}] Bxe6 3.Be8#
1...Kxe6 2.d5+ [1.0] Kf7 / Kd6 3.Be8# / Qb8#

	(Problem 197) Viktor Lukashov, Special Honourable Mention, Shakhmatnaya Poeziya, 2007 Mate in 4 moves. #4 (7+7)
[8/K1p2B2/3p2r1/8/p2kSP2/4s1P1/5B2/1Q4s1]

Tries : {1.Bc4? Kxc4!}, {1.Bd5? Kxd5!}, {1.Bxe3+? Kxe3!}, {1.Qd3+? Kxd3!}, {1.Qc2? d5!}, {1.Qb4+? Kd3!}, {1.Qd1+? / Qc1? Kxe4!}. The solution follows :

Key : 1.Ba2! [1.0] [2.Qd1+ Kxe4 3.Bb1+ [1.0] Sc2 4.Bxc2#]
1...Sf3 2.Qb5 [3.Qd5# / Qc4#] Kxe4 3.Qd5+ [1.0] Sxd5 4.Bb1#
1...Se2 2.Qb4+ Kd3 3.Qc4+ Sxc4 [1.0] 4.Bb1#
1...Rxg3 2.Qc2 ( > 3 [3.Qc4#] d5 3.Sxg3 [1.0] [4.Sf5#]

	(Problem 198) Nikolay Ryabinin, 64, 1990 (version, after Valery Khortov) White plays and draws. = (3+4)
[4R3/r6k/2K5/8/6Ps/8/8/7s]

If Black has two Knights, with a white Pawn on g4 he can win. If the white Pawn steps forward, only then should White try to exchange Rooks.
The solution follows :
Key : 1.Re5 [2.Rh5+ [3.RxS]] Sg3
2.Re3 [1.0] Sf1
3.Rh3 [1.0] Ra6+
4.Kb7 Rh6
5.Rc3 [1.0] (is threatening continuous check, or [wRxh6 bKxh6 wPg5+] and the position theoretically is a draw) Sg6
6.Rc5 Sg3
7.Rh5 [1.0] Sxh5
8.g5 [1.0] = (this is a nice picture of Rook trapping).

	(Problem 199) Aleksandr Azhusin, First Honourable Mention, Shakhmatnaya Poeziya, 2000 Selfmate in 3 moves. s#3 (7+11)
[4Q3/5R2/1pP2p2/pS1k1p2/p7/K2S4/bpP5/rrb5]

Tries : {1.Qe6+? Kxe6!}, {1.Rd7+? Kc4!}, {1.c4+? Kxc4!}. The solution follows :

Key : 1.Qg8! [0.5] [2.Re7+ Kxc6 3.Qc4+ [0.5] Bxc4#]
1...Kxc6 2.Rc7+ Kxb5 3.Qc4+ [0.8] Bxc4#
1...Ke4 2.Qg2+ Ke3 3.Re7+ [0.8] Be6#
1...Kc4 2.Ra7+ Kxb5 3.Qd5+ [0.8] Bxd5#
1...Ke6 2.Rf8+ Ke7 3.Qf7+ [0.8] Bxf7#
1...Be3 2.Rd7+ Kxc6 / Ke4 3.Qd5+ [0.8] Bxd5#

	(Problem 200) Lars Larsen, Skakbladet, 2002 Helpmate in 3 moves. (3 solutions) h#3 311111 (3+12)
[4sb2/3p2pb/B1p1r3/3Sp3/4k1p1/4rpK1/8/8]

The solution follows :
Key : 1.f2+ Kh2 2.Kf3 Bf1 3.Re2 Bg2#
Key : 1.Re1 Sf6+ 2.Ke3 Kh2 3.Kf2 Sxg4#
Key : 1.Rg6 Se7 2.d5 Bc8 3.d4 Bf5#

One solution = [2.0], two solutions = [4.0], three solutions = [5.0] points.


(This post in Greek language).

6th Solving Contest in Greece, 03-06-2007, G.C.F., Aegaleo

Training is useful for solvers.
Try to solve these problems, given in an actual Solving Contest in Greece.

Participating is also very important.
The practice in a real situation improves the abilities of a solver.

Observe in the solutions the way grading points are distributed in variations.
The points are written with boldface numbers inside brackets.
When we write down the solution, we must not omit important variations.

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Sixth Solving Contest of Chess Problems in Greece
Organizers : Greek Chess Federation, Chess Club of Aegaleo, Municipality of Aegaleo
Date : June 03, 2007
Controller : Fougiaxis Harry

Instructions : The correct and complete solution of each problem is graded with 5 points. The incomplete solution takes less points.
#2 : Write only the first move (the "key").
#3 : Write the key, the threat (if exists) and all the variations up to the second white move.
More-mover : Write the key, the threat (if exists) and all the variations up to the last-but-one white move.
Study : Write all the moves up to the position of the obvious winning of White.
Selfmate : Write the key, the threat (if exists) and all the variations up to the last white move.
Helpmate : Write all the moves (for all the solutions). Black plays first.

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First Round, Time : 2 hours

	(Problem 169) Touw Hian Bwee, Fourth Prize, 113th TT British Chess Federation, 1966-67 Mate in 2 moves. #2 (8+11)
[b3R2r/p4sB1/3kSR2/1Q1p1B2/7q/3sP2p/5p2/b4K2]

There are a few tries : {1.Sd4+? Qxf6!}, {1.Sg5+? Bxf6!}, {1.Qd7+? Kxd7!}, {1.Qb8+? Kc6!}, {1.Qb6+? axb6!}, {1.Qc5+? Sxc5!}. The solution follows :

Key : 1.Qa5! [5.0] [2.Qc7#]
1...Kd7 2.Sc7#, 1...Kc6 2.Sd8#, 1...Ke5 2.Sf8#, 1...Qc4 2.Sd4#, 1...d4 2.Sg5#

	(Problem 170) M. Marandyuk & V. Melnichenko, Fourth Prize, S. Brehmer MT, Die Schwalbe 1996-98 Mate in 3 moves. #3 (9+8)
[8/3BBbp1/p4b2/1P1kP3/1P2R1s1/1K1Pp3/2S3s1/8]

If the bK is moved, then the wK will be checked by the bBf7.
There are a few tries : {1.Bd6? Bxe5!}, {1.Kb2? Bxe5+!}, {1.Bc6+? Ke6!}, {1.exf6? axb5!}, {1.Rd4+? Kxe5+!}, {1.Sd4? Sxe5!}. The solution follows :

Key : 1.Ka3! [1.0] [2.Rd4+ [1.0] Kxe5 3.Bd6#]
1...Bxe5 2.Sd4 [1.0] [3.Bc6#] Be8 3.Be6#
1...Bg5 / Bh4 2.Bd6 [1.0] [3.Rd4#]
1...Bxe7 2.Bc6+ [1.0] Ke6 3.Sd4#

	(Problem 171) Michael Herzberg, Second Prize, Problem-Forum 2000 Mate in 5 moves. #5 (7+7)
[8/1B3p2/4r1p1/2pS4/P1k2p2/5S1K/1pPB4/8]

Tries : {1.Sc7? Rd6!}, {1.Sb6+? Rxb6!}, {1.Ba5? [2.Sd2+ Kd4 3.Bc3#] b1=S!}. The solution follows :

Key : 1.Sxf4! [1.0] [2.Bd5#]
1...Rd6 2.Sd5! [1.0] [3.Se3#]
2...Re6 3.Ba5! [1.0] [4.Sd2+ [1.0] Kd4 5.Bc3#]
3...b1=S 4.Bc6 [1.0] [5.Sb6#]
4...Rxc6 5.Se3# (Mate is given from a doubly guarded square).

	(Problem 172) Nikolay Kralin & Leonid Sokolenko, First Prize, Themes-64, 1983 White plays and wins. + (4+5)
[3k4/1pS5/8/2p5/K3R3/1p6/P7/1r6]

Key : 1.Se6+ (threatening the Pawn of the file-c) Kc8
2.axb3 b5+
3.Ka3 c4 (protecting the Pawn of file-c)
4.b4 [1.0] Rb3+ (takes the opportunity to capture the wP)
5.Ka2 Rxb4 (but the bR has restricted mobility)
6.Sc5! [1.5] Kd8 (cannot go through c7, since Sa6+ and bR is lost)
7.Ka1! [1.5] Kc8 (now bK will be retricted)
8.Rd4 [1.0] Kc7 / Kb8
9.Sa6+ +-

	(Problem 173) Frank Richter, First Honourable Mention, Schach 1984 Selfmate in 3 moves. s#3 (9+7)
[5S2/1p5b/2P5/1Qbk1P2/P2s4/7p/6pB/1B1R2K1]

It seems that, if the Queen goes away and the black Knight is forced to move, a mate by the black Bishop Bc5 is possible.
Tries : {1.Qb3+? Kxc6!}, {1.Qxc5+? Kxc5!}, {1.Rxd4+? Kxd4!}, {1.Be4+? Kxe4!}, {1.Ba2+? Ke4!}.

Key : 1.c7! [1.0] [2.Qxb7+ Kc4 3.Rc1+ [1.0] Sc2#]
1...Bxf5 2.Ba2+ Ke4 3.Qe2+ [1.5] Sxe2#
1...b6 2.Qb3+ Kc6 3.Qf3+ [1.5] Sxf3#

	(Problem 174) Evgeny Fomichev, Anatoly Skripnik & Christopher J.A. Jones Special Honourable Mention, Suomen Tehtavaniekat 2002-03 Helpmate in 3 moves. (3 solutions). h#3 3.1.1.1.1.1 (5+13)
[3K4/2p3p1/2P2p2/1s3sp1/1S1k4/2b2Rp1/1r3pq1/3rB3]

1.Kc5+ Rd3 2.Qxc6 Rd6 3.Bd4 Rxc6#
1.Kc4+ Sd5 2.Sbd4 Sxc7 3.Rb4 Rxc3#
1.Ke4+ Bd2 2.Sfd4 Bxg5 3.f5 Re3#

One solution = [2.0], two solutions = [4.0], three solutions = [5.0] points.

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Second Round, Time : 2 hours

	(Problem 175) Erich Brunner, First Prize, Leipziger Tageblatt 1924 Mate in 2 moves. #2 (10+9)
[K1B4b/p6q/P3R2r/r3k4/6p1/B1p1Q1Ps/1R1SS3/8]

The problem is a complete block. Any black move, from the 34 available for Black, is met with a mate. For example, 1...Qb1 2 Bb7#.
From the 21 tries, we show one : {1.Rb8? (zz) Qb1!}.

Key : 1.Rb7! [5.0] (zz)
Variations : 1...Q~b1 (anyplace towards b1) / Bg7 2.Rd7#, 1...Q~7 (anyplace on row-7) / Qe4 / Rg6 2.Q(x)e4#, 1...Qxb7+ 2.Bxb7#, 1...S~ 2.Sf4#, 1...Rxe6 2.Qxe6#, 1...Rh4 / Bf6 2.Rd6#, 1...Bd4 2.Qxd4#, 1...Be5 2.Qxe5#, 1...c~ 2.Qb3#, 1...Rc5 2.Qxc5#, 1...Rb5 / Rxa3 2.R(x)b5#.

	(Problem 176) Mikhail Kuznetsov, First Prize, ‘64’, 1985 Mate in 3 moves. #3 (9+8)
[6K1/3S2S1/4R1p1/s2kp3/BR3pP1/7Q/q1P1b3/3s4]

Tries : {1.Se8? Qc4!}, {1.Sb6+? / Sf6+? Kc5!}, {1.Rxe5+? / Rd6+? K(x)d6+!}, {1.Bc6+? Sxc6!}, {1.Qg2+? / Qh1+? f3!}, {1.Qf3+? Bxf3!}, {1.Qd3+? Bxd3!}, {1.Qc3? Sxc3!}, {1.Qb3+ Qxb3!}.

The white Queen forms an indirect battery with the Rook Rb4. This is not a direct battery, because it does not aim towards the black King but towards its flights. The Black will try to place something on c4 to create a flight to d4. If the black Queen leaves the diagonal a2-g8, the white King is not threatened with a possible discovered check and he can take advantage of that.

Key : 1.Qa3! [0.5] [2.Sb6+ [0.5] Kc5 3.Rc4#]
1...Qc4 2.Rxe5+ [1.0] Kd6+ / Kd4+ 3.Rxc4# / Se6#
1...Bc4 2.Qf3+ [1.0] Kd4 / e4 3.Qd3# / Qxe4#
1...Sc4 2.Bc6+ [1.0] Kd4 3.Rd6#
1...Qxa3 2.Rxe5+ [1.0] Kd6 3.Se8#

	(Problem 177) Angel Zlatanov, Fourth Honourable Mention, Zadachi i Etyudy, 2000 Mate in 4 moves. #4 (4+9)
[6r1/p3Q1p1/K1k3S1/2S3p1/82p5/4pp2/3s4/]

Tries : {1.Qd6+? Kxd6!}, {1.Qe4+? / Qe6+? / Qd7+? Kxc5!}, {1.Qe5? Re8!}, {1.Se5+? / Se4? / Sb7? Kd5!}.

Key : 1.Se6! [0.5] [2.Qc5+ Kd7 3.Qd5+ [1.5] Kc8 / Ke8 4.Qb7# / Sc7#]
1...Se3 2.Qc7+ Kd5 3.Qd7+ [1.5] Kc4 / Ke4 4.Qb5# / Sxg5#
1...Kd5 2.Qd7+ Ke4 3.Sc5+ [1.5] Ke3 / Kf3 4.Qd3# / Qh3#

	(Problem 178) Dmitry Petrov, Vecherny Novosibirsk, 1979 White plays and wins. + (6+7)
[5b1q/7p/4P1Rp/1B4pk/3P4/4p1K1/8/1R6]

Key : 1.e7! Bxe7 (It does not allow the Pawn promotion. Not [1...Qxd4 2.Rxh6+ and exf8=Q+ / Be8+])
2.Bd3 [1.5] Qxd4
3.Rh1+ Qh4+ (gets rid of the bQ...)
4.Rxh4+ gxh4+ (...and presses the bK)
5.Kf4 Bg5+
6.Kf3 [2.0] hxg6
7.Be2 [1.5] +-

	(Problem 179) Vukota Nikoletic, Prize, Mat Plus, 1995 Selfmate in 3 moves. S#3 (9+9)
[8/1r2SR1p/3B3P/4rBSK/5k2/2p2pq1/2p1R2Q/7b]

Key : 1.Re1! [1.0] [2.Qd2+ cxd2 3.Sh3+ [1.0] Qxh3#]
1...Qxh2+ 2.Bh3+ Kg3 3.Rxf3+ [1.0] Bxf3#
1...Bg2 2.Sh3+ Bxh3 3.Sg6+ [1.0] hxg6#
1...f2 2.Sd5+ Bxd5 3.Bd7+ [1.0] Bxf7#

	(Problem 180) Chris Feather, Shakhmatna Misl, 2005 Helpmate in 4 moves. (2 solutions). h#4 2111... (3+9)
[3k4/1pS5/8/2p5/K3R3/1p6/P7/1r6]

Key : 1.Rg8! Re6 2.Rg7 fxg7 3.Bc6 g8=Q 4.Bb5 Qa8# [2.5]
Key : 1.Re4! Re7 2.Rxe7 fxe7 3.Ra8 e8=S 4.Ra7 Sc7# [2.5]


(This post in Greek language).