A Welcome to 2017 with cooperations

I will open the January with wishes for health and successes to all composers.

Let us hope that the new year 2017 will bring joy and happiness to all people.

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Three compositions - cooperations of mine, belonging to fairy chess because they use grasshoppers, were published last year and I have learned about it this year.

I offer my thanks for the cooperation to Vito Rallo (with colours of Italy) and to Kostas Prentos (with colours Greece/USA).

The Grasshoppers are pieces - hurdlers. They see on a row or file or diagonal another piece - hurdle, and they jump on the square exactly after the hurdle. If this square is nonempty, it could only contain an opponent piece, which is captured by the Grasshopper.

Problem-833 Vito Rallo (ITA) and Emmanuel Manolas (GRE) Phenix 265, 09/2016

8/8/8/GK6/8/3g1k2/3P4/8 (3 + 2), (Grasshoppers a5 + d3)  Helpmate h#6

1.Gg3 Ge1 2.Kf4 d4 3.Ge5 d5 4.Ke4 d6 5.Kd5 d7 6.Kd6 d8=Q# Theme Excelsior (a pawn starts from its initial position until it is promoted). Self-blocking. Miniature. A comment : A beautiful mate is presented.

Problem-834 Emmanuel Manolas (GRE) and Vito Rallo (ITA) Phenix 268, 12/2016

3g4/8/8/k7/8/7G/8/s3K3 (2 + 3), (Grasshoppers h3 + d8) Helpmate h#7, fairy condition Andernach

1.Ka4 Kd2 2.Gd1 Kc3 3.Ka3 Gb3 4.Ka2 Kb4 5.Sxb3(wSb3) Sc1+ 6.Ka1 Ka3 7.Gb1 Sb3# The fairy condition Andernach changes the colour of the capturing piece. Self-blocking. Miniature. A comment : Difficult manouver, to change the colour of the piece that will mate.

Problem-835 Emmanuel Manolas (GRE) and Kostas Prentos (Greece/USA) Strategems, 12/2016

1g6/5G1G/8/2pk4/7K/8/1pp5/5G1G (5 + 5), (Grasshoppers f1f7h1h7 + b8) Helpmate h#4

1.c4 Gb5 2.Kc6 Gd7 3.Kb7 Gc7+ 4.Ka8 Gb7# (mate at a8, northwest, NW) 1.Ke6 Gd5 2.Kf6 Gf7 3.Kg7 Kh5 4.Kh8 Kg6# (mate at h8, northeast, NE) 1.Ke4 Ge1 2.Kf3 Gf2+ 3.Kg2 Gh2 4.Kh1 Kg3# (mate at h1, southeast, SE) 1.Kc4 Gf8 2.Kb3 Gb1+ 3.Ka2 Gb3+ 4.Ka1 Gf6# (mate at a1, southwest,SW) Star of the bK, who gets mated at the four corners of the chessboard. In two variations the White activates the white royal battery with mirror symmetry.

Goodbye 2016

I will close this December wishing health and success to all composers.

Let us hope that the New Year 2017 will bring joy and happiness to everyone.

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Two compositions of mine, belonging to fairy chess, are published in the Romanian e4e5 magazine, Nr.34, December 2016, p. 559.

The first is a helpmate threemover with fairy condition Circe : Black plays first and helps White to mate. In the meantime, any captured unit is reborn on its initial square (initial as in the start of a chess game).

The second is a directmate twomover with fairy conditions Circe and Madrasi, and also two fairy pieces Nightriders. We spoke previously about Circe.
Madrasi is a condition of paralysis : When two pieces of same type and different colour (wR and bR, wQ and bQ, etc) are threatening each other, then they are paralysing each other, until the end of threat.
The Nightriders are pieces of linear way of moving, with Knight steps.
In this composition all the tries and all the white moves during solution are done by the white king.

Problem-831 Emmanuel Manolas (GRE) e4e5 Nr.34, 12/2016, p.559

4r3/1r6/K7/6p1/1p4k1/1P4PR/B1P1p3/5b2 (6 + 7) h#3, Circe

1.Bxh3(+wRh1) Rxh3(+bBc8) 2.Bf5 Kxb7(+bRa8) 3.Rxa2(+wBf1) Bxe2(+bPe7)# Black sacrifice with the key. Reciprocal captures bBf1 and wRh3. Openings and closings of lines. Selfblocking. In three moves the bBf1 must go to f5, and the wBa2 must go to e2.

Problem-832 Emmanuel Manolas (GRE) e4e5 Nr.34, 12/2016, p.559

1q4b1/P1pNPp1P/3P2P1/1Qnr3R/1pKb2S1/Prq1B3/p2Q2B1/1R4k1 (15 + 12), (Nightriders d7 + c5) #2, Circe, Madrasi

Tries : {1.Kxb3(+bRa8)+? a1=R!}, {1.Kxb4(+bPb7)? [2.Qf1#] Qxa7!}, {1.Kxc5(+bNc1)+? Ne5+!}, {1.Kxd5(+bRa8)? [2.Rh1#] Bxh7(+wPh2)!}, {1.Kxd4(+bBf8)+? Bh6!}. Key : 1.Kxc3(+bQd8)! [2.Qf2# / Qd1# / Qe1# / Qc1#] 1…a1=Q/B+ 2.Kxb3(+bRa8)# Themes Durbar (All white moves are made by the wK) and royal Option (Tries and Key are made by the wK).  This active wK stops the Madrasi paralysings, capturing black pieces in its field. Only one from six capturings is succesful, while the rest five are tries with different treatments. There are also other effects, as exposition of the wK to check, indirect Self-pin and unpins, Reciprocal pins, Crossed checks.

Playing with Composition

In this post we will start from a nice miniature by our friend Nikos Pergialis and we will enhance it with more variations.
A miniature has up to 7 pieces.
Adding pieces to it, the problem will not be a miniature anymore, which was a limiting factor for old chap Nikos, but it will get more variations, which is my goal.

Problem-825 Nikos Pergialis (GRE)

8/8/8/rS6/3b4/1Q6/p1K5/k7 (3 + 4) #2

Try : {1.Qb4? [2.Qe1# / Qxd4#], 1…Bb2 2.Qxb2#, 1…Bf2 / Be3 2.Qb2# / Qc3#, 1…Bc3 2.Qxc3#, 1…Rxb5!} Key : 1.Sc3! [2.Qb2#], 1…Bxc3 2.Qxc3#, 1…Rb5 2.Qxa2# The key is quasi-sacrificial. It closes one line allowing one mate, and also supports the wQ for the second mate. There are three mates in total.

The pieces are relocated two columns to the right.
A wRb6 was added, to keep bK from escaping to the second column.
A wBb1 was added, to allow promotion of bPc2 to knight, stoping the threat. A bRa1 was need to stop some duals from occuring. When bPc2 captures, the wQ mates from square d1.
A bPe4 can capture the wQ with check, but then a wSe5 mates.
The bQ can deter the threat, but then wRh7 mates. The presence of this rook offers a try, which allows us to see the Pseudo-Le Grand theme.

Problem-826 Emmanuel Manolas (GRE) (after Nikos Pergialis) original

7q/7R/1Rr5/3SS1b1/4p3/3Q4/2p1K3/rBk5 (7 + 7) #2

Tries : {1.Rxh8? [2.Rh1#], 1…Bd2 2.Qxd2#, 1…Be3 2.Qxe3#, 1…Bh4 2.Qd2# / Qe3#, 1…Rxb6 / Rf6 / Rh6 2.Qxc2#, 1…exd3+ 2.Sxd3#, 1…Bh6!}, {1.Qd4? [2.Qg1# / Qb2#], 1…cxb1=Q 2.Qd1#, 1…cxb1=R 2.Rxc6# / Qd1#, 1…Rxb6!} Key : 1.Se3! [2.Qd2#], 1…Bxe3 2.Qxe3#, 1…Rxb6 / Rd6 2.Qxc2#, 1…Qd8 2.Rh1#, 1…cxb1=S 2.Qd1#, 1…exd3+ 2.Sxd3# The problem works with similar mechanism, as the previous one, only it has now 14 pieces and there are six different mates.  There is also the Theme Pseudo-Le Grand : "Two threats (A, B) in two phases, reappear crosswise as mates after different black responses (a, b) in these phases". In the present problem the relevant moves are as follows 1.Rxh8? [2.Rh1# (A)] Bh4 (a) 2.Qd2# (B), 1…Bh6! 1.Se3! [2.Qd2# (B)] Qd8 (b) 2.Rh1# (A)

Composer Cooperations (6)

It is very probable that there exist more compositions with recent cooperations, and if any Greek composers want to inform me, I will update this post. I select two of the cooperating teams for today.

Problem-823 Fadil Abdurahmanovic (Bosnia-Herzegovina) and Ioannis Kalkabouras (GRE) KobulChess.com 17.11.2015, no.269

8/8/2b5/8/8/2Ppk2p/pp5P/3sK2R (4 + 7) h#4, 2 solutions

1.Sf2 0-0 (Rf1??) 2.Kd2 Kxf2 3.Kc2 Ke1 4.Kb1 Kd2# 1.Kf3 Rf1+ (0-0??) 2.Kg2 Rf3 3.Kxh2 Kf1 4.Kh1 Rxh3# In one variation the castling is a good move, in the other it is not effective. The "wrong" moves are noted with double question mark.

Problem-824 Emmanuel Manolas (GRE) and Ioannis Kalkavouras (GRE) 3rd Hon. Mention, Moskovski Concurs 2016

8/3p4/k2r2PK/1r6/3ss3/p4bb1/2S1p3/6q1 (3 + 11) h=10, Helpstalemate Fairy condition : Circe

1.Ka7 Sxa3 2.Ka8 Sxb5 3.Kb8 Sxd4 4.Kc8 Sxf3 5.Kd8 Sxg1 6.Ke7 Sxe2 7.Kf8 Sxg3 8.Kg8 Sxe4 9.Kh8 Sxd6 10.Kg8 g7= It is a help-stalemate. Initially the Black helps a lot for his pieces to be captured, and at the end the White helps a lot tryin to avoid winning!  The comment of the judge : "Circe, with nine pieces captured without rebirth! The essence of the problem : marching towards a stalemate trap, the bK passes from the rebirth squares of the black pieces, allowing the wS to capture all these pieces which would inhibit the stalemate, without any of them being reborn".

Cooperations of composers (5)

We will see here the result of a cooperation of Ioannis Kalkavouras and Emmanuel Manolas, aiming to the composition of a helpmate in three and a half moves (that is, White plays first) with the following restrictions :
(a) The White has got only a Bishop and a Knight.
(b) The white pieces, B and S, will form with pericritical moves a battery which will fire with double check.
(c) The black in two variations makes different castling and is mated in the eighth row.
(d) No replacements of pieces (for twinning) is allowed.

Problem-815 Frank Fiedler Gaudium, 2013 Source: WinChloe, no. 487666

r3k2r/8/8/4b3/8/4S1q1/8/1K1B4 (3 + 5) h#3, 2 solutions

1.0-0 Sd5 2.Bh8 Bb3 3.Qg7 Sf6# 1.0-0-0 Sf5 2.Bb8 Bg4 3.Qc7 Sd6# Here is shown the idea in simple form. The wB could be positioned on e2.  The wK, to avoid checks by bR and bQ, could be only on columns b and e. But, with wKb5 the solutions are 1507, with wKe6 the solutions are 1283 and with wKe4 the solutions are 32. The wK is very well placed on b1!  The white movements are not pericritical, they simply form the battery. If we force the restriction h#3,5, then many unwanted solutions appear, 68 in total.

Problem-816 Anatoly Styopochkin JT Feoktistov-50, Shakhmatnaya Kompozitsiya 1998-2000, 1st Honourable Mention Source: WinChloe, no. 161976

r3k2r/s7/1p2B3/p2S2q1/4p3/3b4/8/b2K4 (3 + 10) h#3.5, 2 solutions

1…Bg8 2.0-0-0 Bh7 3.Kb7 Bxe4 4.Ka6 Sc7# 1…Bc8 2.0-0 Ba6 3.Bh8 Bc4 4.Qg7 Sf6# The nice movements of the wB are pericritical and the bishop forms the battery in both variations. But the two mates do not happen on the eighth row.

Problem-817 Ioannis Kalkavouras and Emmanuel Manolas original

r3k2r/8/8/3Sbp2/4b1B1/4Ksq1/8/8 (3 + 8) h#3.5, 2 solutions

1…Bh3 2.0-0 Bf1 3.Bh8 Bc4 4.Qg7 Sf6# 1…Ke2 2.0-0-0 Se3 3.Bb8 Sxf5 4.Qc7 Sd6# In one variation the white battery is formed by the wB moving clockwise, and in the other by the wS moving anti-clockwise. The two mates appear symmetrical in the eighth row (vertical mirror, echo mates(0,4)).

Theme Elmgren

(Initial date of this post was 21/12/2015)

In this post we will see the Theme Elmgren under a different light. This theme is named after the Swedish composer Bertil Elmgren, born 25/05/1912.

In this theme, all the black pieces (3 or more), with a possible exception of the bK, are actively refuting white tries!
A try is an attempt by the White to win, but the Black has a unique defense.
The White has at least three such tries, and all the black pieces take their turn to refute.
With more black pieces on board, the probability all of them refuting a try diminishes, making it more difficult for the composer to achieve an Elmgren.
Likewise, the more the black pieces, the more are the tries, making the problem relatively more difficult for a solver.

Having all the black pieces struggling in defense, symbolizes a common battle where no one stays behind the trenches, waiting to see what will happen!
There are no idle pieces, simply blocking a flight or stopping an unwanted second solution.
Everything is useful and active!

I have published in this blog some problems containing the Elmgren theme.

15/10/2011, Manolas Emmanuel, problem_497
http://chess-problems-gr.blogspot.gr/2011/10/manolas-emmanuel-4-compositions-from.html

16/11/2012, Pergialis Nikolaos, problem_625
http://chess-problems-gr.blogspot.gr/2012/11/nikos-pergialis-5-compositions.html

27/01/2013, Retter Yosi, problem_687
http://chess-problems-gr.blogspot.gr/2013/01/isc-9-compositions-solved-and-some.html

07/06/2014, Manolas Emmanuel, problem_771
http://chess-problems-gr.blogspot.gr/2014/06/some-two-movers-for-you-to-solve.html

27/12/2014, Manolas Emmanuel, problem_792,
http://chess-problems-gr.blogspot.gr/2014/12/farewell-to-2014.html


We shall see now some rare compositions with theme Elmgren, with at least five black pieces refuting white tries.

5 black pieces

Problem-808 Termaat Nicolaas I. J. 2nd Prize, Probleemblad 1951

1s6/SQ2pb2/1P6/P1S4R/2kp2B1/K1P2R2/1P1P4/2s5 (12 + 6) #2

Tries: {1.Sd3? [2.Rc5#] Sb3!}, {1.Se6? [2.Rc5/Qd5#] Bxh5!}, {1.Sa6? [2.Rc5#] Sd7!}, {1.Se4? [2.Rc5#] dxc3!}, 1.Sb3? [2.Rc5#] Sd3!}, {1.Sd7? [2.Rc5#] e5!}. Key: 1.Sa4! [2.Rc5#] 1...Bxh5 2.Be6#, 1...Bd5 2.Qxd5#, 1...e5 2.Qxf7#, 1...dxc3 2.Qe4#, 1...Sb3 2.d3#, 1...Sd3 2.b3#, 1...Sd7/Sa6 2.Q(x)a6# Themes Option and Elmgren.

Problem-809 Burbach Johannes J. Shakend Nederland, 1981

4Q3/8/3p4/1pp1Ppp1/4P3/RPP1kPPR/4SSP1/3KB3 (14 + 6) #2

Tries: {1.c4? [2.b4#] b4!}, {1.b4? [2.c4#] c4!}, {1.exf5? [2.exd6#] d5!}, {1.g4? [2.f4#] f4!}, {1.f4? [2.g4#] g4!}. Key: 1.Qxb5! [2.Qd3#] 1...c4 2.Qb6#, 1...fxe4 2.Sg4# Theme Elmgren.

6 black pieces

Problem-810 Manolas Emmanuel (GRE) original dedicated to Kostas Prentos

rB6/1sK1S2P/2P2p2/1R1Pk1P1/6R1/1B4P1/6bQ/6r1 (12 + 6) #2

Tries: {1.Sg6+? Kf5!}, {1.Bc2? [2.Sg6#] Sc5! (The square d5 is guarded by three white pieces Rb5 Bb3 Se7. One leaves making the try-move, one will leave to make mate, so the Black shuts off the third guardian : Theme Zappas simple)}, {1.d6+? Sc5!}, {1.Re4+? Kxe4!}, {1.Kxb7+?/Kd7+? Rxb8+!}, {1.Qxg1? [2.Qd4#] Be4!}, {1.Qxg2? [2.Re4#/Qb2#/Qe2#/Qe4#] Rxg2!}, {1.Qh6?/h8=Q?/h8=B? Rf1!}, {1.h8=S? [2.Shg6#/Sf7#] fxg5!}. Key: 1.Rf4! [2.Sg6#] 1...Be4 2.Qb2#, 1...Bxd5 2.Qe2# (horizontal-diagonal transformation) Theme Elmgren (here the bK is also refuting tries). In the phases {1.h8=S?} and {1.Rf4!} there is Anti-Dual with moves Be4 and Bxd5. (There are some more technical merits to examine: changed mates in two phases, a pin-mate, Urania, Bartolovic, Dombrovskis paradox, Anti-reversal, Anti reversal-menace).

Problem-811 Witt Andreas Commendation, Die Shwalbe 2005

4s1R1/4p2R/4pQ2/4Bsb1/B1P4r/K2SP3/2S5/1k6 (10 + 7) #2

Tries: {1.Bh2? [2.Qa1#/Qb2#] Rd4!}, {1.Bg3? Bxf6!}, {1.Bf4? Sd4!}, {1.Bd6? e5!}, {1.Bc7? exf6!}, {1.Bb8? Sxf6!}, {1.Rxe8? [2.Rb8#] Bxe3!}, {1.Kb3? [2.Sa3#] Sd4+!}. Κλειδί : 1.Ba1! [2.Qb2#] (line opening Bristol) 1...Rd4 2.Rh1#, 1...Bxf6 2.Rg1#, 1...Sd4 2.Qf1#, 1...e5 2.Qb6#, 1...exf6 2.Rb7#, 1...Sxf6 2.Rb8# Theme Elmgren. Multiple captures of the bQf6.

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Upgrade on 12-01-2016

Motivated by this post, the Italian composer Alberto Armeni decided to compose a problem with more than 6 black pieces, all refuting tries. He has initially set the record to 8 pieces, which we were glad to present here, but then he created a task with 11 pieces (see Problem-814 below).

8 black pieces

Problem-812 Armeni Alberto International Composing Tourney "e4-e5" 2015-2016, section 2# Source: http://www.chessplayer.ro/tourneys_2015.html (with click on the blue 24)

1s5b/3B1PP1/p4Rr1/2Bk4/3P4/1pK1R1rb/pP2PS2/2Q2S2 (13 + 9) #2

Set play : {1...Sc6 2.Bxc6#, 1...Rxg7/g5 2.Rd6#, 1...R6g4 2.Rd6/f5#, 1...Rxe3+ 2.Sxe3#, 1...R3g4 2.Rf5# / Re5#, 1...Rg2/g1 2.Re5#, 1...Bg2/xf1/f5 2.R(x)f5#} Tries : {1.Sxg3? [2.Re5#] Sb8xd7!}, {1.g8=S? [2.Se7#] Bh8xf6!}, {1.Kb4? [2.Re5# / Qc4#] a6-a5+!}, {1.f8=Q? [2.Qd6#] Rg6xf6!}, {1.Qc2? [2.Qxb3# / Qe4#] b3xc2!}, {1.Rd3? [2.e4#] Rg3xd3+!}, {1.Rxg6? [2.Rd6#] Bh3-e6!}, {1.Kxb3? [2.Qc4#] a1=S+!}, {1.Ref3? [2.e4# / Se3#] Rxf3+!}, {1.Rxg3? [2.e4# / Se3#] Rxg3+!}, {1.Sd3? [2.Sb4# / Sf4# / Re5#] Rxe3!}, {1.Qd1? [2.Qxb3#] a1=S!} Key : 1.Kd3! [2.Qc4#] 1...Rxe3+ 2.Sxe3#  1...Bf5+ 2. Rxf5#. Theme Elmgren.

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Update on 11-04-2016

The composer Alberto Armeni has created a tak for theme Elmgren, with 11 black pieces!

Task with 11 black pieces

Problem-814 Armeni Alberto “Problemas”, April 2016, problem n. 174, page 350 Source: https://drive.google.com/file/d/0B5lHTACLJK1fZnZjUjZzWjZ1RUk/view

2b5/p2p3R/1p1Qp3/5p2/5Pp1/3p2Kp/r2S3R/b2S2kB (8 + 12) #2

Tries: {1.Bb7? / Ba8? Bc8(x)b7!}, {1.Qxb6+? a7xb6!}, {1.Bc6? d7xc6!}, {1.Qc5+? b6xc5!}, {1.Bd5? e6xd5!}, {1.Be4? f5xe4!}, {1.Sf3+? / Bf3? g4xf3!}, {1.Re2? d3xe2!}, {1.Rg2+? h3xg2!}, {1.Rf2? Ra2xd2!}, {1.Qd4+? / Qe5? Ba1xd4! / Ba1xe5!}. Key: 1.Bg2! [2.Rh1#] hxg2 2.Rxg2# Theme Elmgren. Task with 11 black pieces defending tries.

A selfmate, with strip-tease.

Selfmate are the compositions where White plays first and forces Black to mate, while Black tries to avoid it.

Theme Strip-tease : One by one some pieces are removed from the chessboard, leaving a new problem to be solved.

This theme, Stip-tease, is contained in Problem-777 (see here) which has got a distinction in the World Congress of Bern.

Today we will see an easy selfmate. Black has a light-squared Bishop who can mate the cornered in h1 white king, and the matter is how can we empty the diagonal from the pieces that block (direct obstacles wQb7, bPc6, bKe4) or can inhibit the mate (possible obstacles wPe2, wRh5, wRg4, wBa4) the mate.

Problem-813 Manolas Emmanuel (GRE) Original

b3S3/1Qp5/2p5/2B4R/B1P1kPR1/8/3PPp1p/5SbK (12 + 7) s#5, Twins a) diagram, b) = a) -wPc4, c) = b) -wRg4

a) 1.Be3! (the obstacle wPe2 is gone) Bxb7 (the obstacle wQb7 is gone) 2.Bxc6+ (the obstacle bPc6 is gone) Bxc6 (the obstacle wBa4 is gone) 3.f5+ (the obstacle wRh5 is gone) Ke5+ 4.Re4+ Kxe4 (the obstacle wRg4 is gone) 5.d3+ Ke5# (the last obstacle bKd4 is gone) b) -wPc4 1.Qxc7! Bb7 2.Bxc6+ Bxc6 3.f5+ Kd5 4.f6+ Ke6+ 5.Re4+ Bxe4# c) -wPc4, -wRg4 1.Qb1+! Kxf4 2.Qe4+ Kxe4 3.Be3 c5 / Bb7 4.B(x)c6+ Bxc6 5.Re5+ Kxe5#

In these three twin compositions, White loses a pawn and a rook, one after the other. In each occasion, sacrificing pieces and threatening the bK, White forces Black to mate in 5 moves.

Awarded chess problems by Greek composers, 2015

Awarded Compositions GR, 2015

(Last Update : 03/01/2016)

In this post we have gathered the artistic chess compositions of the Greek composers, which have earned a distinction in composing tourneys of the year 2015.
There were more publications, but here we are limited only to awarded ones.
(A note by Alkinoos: If some award is omitted, which is possible since I do not read everything, the interested composer is kindly requested to send to me the necessary information to append it in this anniversary post. Thanks!).

In twelve (12) international composing tourneys there were seventeen (17) awarded compositions, atistic creations of five (5) Greek composers (sometimes with co-authors).

We stress here a fact, seldom happening in Greece : This year we had organized here in our country one (1) international composing tourney : JT Manolas-65.

The entries to the composing tourneys are judged by a Judge and they get distinctions (by descending order): Prize or Place, Honourable Mention, Commendation.
In some interesting cases the distinction might be preceded by the word 'Special'.
The distinctions may be numbered or not.

Argirakopoulos Themis

Problem-2015-10 Argyrakopoulos Themis 4th Place, Marianka 2015 Source: http://www.jurajlorinc.com/chess/ma15faaw.htm

2B5/3Pp3/2p4p/3R4/7k/2K4p/6p1/8  (4 + 6) hs+2,5 twins: a) diagram, b) wKc3 to d4, c) wKc3 to e1, d) wKc3 to h2 Messigny

a) wKc3 1…g1=Q 2.d8=Q Qd8↔Qg1 3.Qe1+ Qe1↔Qd8+ b) wKd4 1…g1=R 2.d8=R Rd8↔Rg1 3.Rg4+ Rg4↔Rd8+ c) wKe1 1…g1=B 2.d8=B Bd8↔Bg1 3.Bf2+ Bf2↔Bd8+ d) wKh2 1…g1=S 2.d8=S Sd8↔Sg1 3.Sf3+ Sf3↔Sd8+ Messigny = A piece (King included) can also swap places with an opposite piece of the same nature. Neither of the two pieces must have swap its place the previous move. Theme Babson, (AUW, promotions of the same type by White and Black).

Problem-2015-13 Argirakopoulos Themis 2nd Honourable Mention, 13° Tzuika, Ostroda, 2015 Source: http://www.wccc2015.com/docs/congress_bulletin.pdf

5(SL)Bk/2pp2pP/5p(EL)1/5p2/1p(LE)p2P1/PP6/K2P4/2q5 (8 + 11) hs#3 twins: a) diagram, b) bPb4 to a4 Sentinelles : (g6 Elan EL), (c4 Leo LE), (f8 Super-Leo SL)

a) bPb4 1.LEe2(+c4) SLd6 2.LEg2(+e2) SLh2(+d6) 3.LEh3(+g2)+ ELxh3(+g6)# b) bPa4 1.LEd5(+c4) SLe8 2.LEf7(+d5) SLe6 3.LEf8(+f7)+ ELxf8(+g6)# Sentinelles : When a piece (Pawn excluded) leaves a square outside the first and last rows, it leaves a Pawn of the color of the side that played unless 8 Pawns in this color are already on the board. The triple pin mate is achieved with the help of a fairy unit, the SuperLeo, which can capture over 2 hurdles. You certainly need an eagle eye and a sharp mind to anticipate the mate with three pinned units, of which two will be sentinels that will appear on the board during the solution. On the downside, the price paid by this ambitious achievement is the heavy position and the lack of interplay. Question: can anyone obtain a five-fold sentinel presentation of this splendid idea?

Argirakopoulos Themis, Prentos Kostas

  

Problem-2015-02 Argirakopoulos Themis, Prentos Kostas 1st Prize (1st - 3rd place)- TT 157 Superproblem Source: https://yadi.sk/i/Iho1AFq4mNeij

6K1/6b1/1sp1rp2/1b1k4/p1p1q3/2r1pp2/1p1P3p/Q1s5  (3 + 16) h#2, 1 solution per twin twins: a) diagram, b) wQa1 to a8, c) wQa1 to h8, d) wQa1 to h1

a) wQa1 1.Kc5 dxc3 2.Qd5 Qa3# b) wQa8 1.Kd6 d4 2.Sd5 Qd8# c) wQh8 1.Ke5 dxe3 2.Kf5 Qh5# d) wQh1 1.Kd4 d3 2.Kxd3 Qd1# Model mates. WP4 from wPd2, with bK cross and the wQ visiting the four corners.

Abdurahmanovic Fadil, Kalkavouras Ioannis

  

Πρόβλημα-2015-17 Abdurahmanovic Fadil, Kalkavouras Ioannis 1st Prize, Moskow Concours 2015 Source: http://www.selivanov.ru/download/Awards/Moscow/2015/%20mt2015-h.pdf

4s3/4s3/8/8/4k2p/1p1p2PP/bp1K1p2/qB6  (4 + 10) h#5, 2 solutions

1.Sf5 Bc2 2.b1=Q Bd1 3.Qe5 Be2 4.Qba1 Bf1 5.Qad4 Bg2# 1.Kf3 Kc3 2.Ke2 Bxd3+ 3.Kd1 Be4 4.Kc1 Kd3 5.Kb1 Kd2# Wb and Bk Platzweschel, wK triangular Rundlauf and wB zig zag from b1 to g2.

Problem-2015-03 Abdurahmanovic Fadil, Kalkavouras Ioannis 3rd Prize, ЮК «С.Билык-50» Bilyk-50 Source: http://sachmatija.puslapiai.lt/sites/default/files/Bilyk2015.pdf

RS6/Br6/8/6K1/8/8/P7/3k4  (5 + 2) h#2, 2 solutions

1.Rxa7 Sc6 2.Rxa2 Sd4 3.Rd2 Ra1# 1.Rxb8 Be3 2.Rb2 Rc8 3.Re2 Rc1# Bicolour Bristol, Annihilation of white pieces, Self-blockings, Model mates. Judge (Bilyk) :Аннигиляция белых фигур, правильные маты на краю доски, чередование функций белых коня и слона. Белая пешка вводит диссонанс: в одном решении она уничтожается чёрной ладьёй – двойная аннигиляция на вертикали ”a”, а во втором матовом финале участия не принимает, выполняя роль технической фигуры.

Manolas Emmanuel

Problem-2015-07 Manolas Emmanuel 6th Honourable Mention, Gennady Kozura-60 JT Source: http://www.chess-kopyl.com.ua/images/2015/12_2015/1_Козюра-60_2_оконч.pdf

R7/8/8/1kB5/8/8/1p4P1/1K1B4  (5 + 2) h#2, 2 solutions per twin twins: a) diagram, b) bKb5 to h1 (after Youness Benjelloun, 8/8/2pP4/2Bk4/5PR1/8/2p5/2KB4, (6+3), h#2, 2sols, Problem Paradise vol18 January-March 2015, problem H706)

a) bKb5 1.Kc4 Be2+ 2.Kb3 Ra3# 1.Kc6 Rc8+ 2.Kd7 Bg4# b) bKh1 1.Kh2 Bf3 2.Kh1 Rh8# 1.Kxg2 Rg8+ 2.Kf1 Rg1# Section h#2 HotF (Helpmate of the Future)

Problem-2015-08 Manolas Emmanuel Special Honourable Mention, Gennady Kozura-60 JT Source: http://www.chess-kopyl.com.ua/images/2015/12_2015/1_Козюра-60_2_оконч.pdf

3S4/8/3p2p1/2pkr2b/3P4/3K4/Pr4P1/8  (5 + 7) h#2, 4 solutions

1.Bf3 gxf3 2.Re4 fxe4# 1.Rb3+ axb3 2.c4+ bxc4# 1.Bg4 Sc6 2.Be6 Se7# 1.Rb6 Se6 2.Rc6 Sf4# Section h#2 HotF (Helpmate of the Future)

Pergialis Nikos

Problem-2015-01 Pergialis Nikos Commendation, Manolas-65 JT Source: http://chess-problems-gr.blogspot.gr/2015/08/award-for-manolas-65-jt.html

8/8/4P3/8/2qkp3/8/8/2Q1S2K  (4 + 3) h#2, 2 solutions per twin twins: a) diagram, b) wQc2 to g8

a) wQc1 1.Qd3 Qc6 2.e3 Sf3# 1.Qxe6 Sd3 2.Kd5 Qc5# b) wQg8 1.Ke3 Sc2+ 2.Kf2 Qg2# 1.Ke5 Sf3+ 2.Kf6 Qf7# Section Α, HotF (Helpmate of the Future). Judge (Kalkavouras Ioannisς) : "Preventive selfblocks and nice model mates; what else one might expect from a little precious stone?"

Problem-2015-04 Pergialis Nikos Special Honourable Mention, Gennady Kozura-60 JT Source: http://www.chess-kopyl.com.ua/images/2015/12_2015/1_Козюра-60_2_оконч.pdf

8/4q3/1ps2B2/3PP1q1/p1BkP3/1r/3Ss3/5K2  (7 + 8) h#2, 4 solutions

1.Qgxe5 Bh4 2.Sc3 Bf2# 1.Qexe5 Bd8 2.Rc3 Bxb6# 1.Ke3 Bxe7 2.Sed4 Bxg5# 1.Kc5 Bxg5 2.Scd4 Bxe7# Section h#2 HotF (Helpmate of the Future). Judge (Valery Kopyl ) : Чёткий HOTF, замечательная игра полей, но… практически полная симметрия

Problem-2015-05 Pergialis Nikos 11th Honourable Mention, Gennady Kozura-60 JT Source: http://www.chess-kopyl.com.ua/images/2015/12_2015/1_Козюра-60_2_оконч.pdf

8/5P2/K2p2p1/3rS1p1/3k4/2p5/b7/R7  (4 + 7) h#2, 4 solutions

1.Bb1 Ra4+ 2.Kc5 Rc4# 1.Kc5 Rb1 2.Rd4 Rb5# 1.Ke3 f8=Q 2.Kd2 Qf2# 1.Kxe5 Re1+ 2.Kf6 f8=Q# Section h#2 HotF (Helpmate of the Future)

Problem-2015-06 Pergialis Nikos Special Honourable Mention, Gennady Kozura-60 JT Source: http://www.chess-kopyl.com.ua/images/2015/12_2015/1_Козюра-60_2_оконч.pdf

R2K4/5k2/8/8/5B2/4S3/8/8  (4 + 1) h#2, 2 solutions per twin twins: a) diagram, b) bKf7 to d2

a) bKf7 1.Kf6 Ke8 2.Ke6 Ra6# 1.Kg6 Ra6+ 2.Kh5 Rh6# b) bKd2 1.Kc1 Be5 2.Kb1 Ra1# 1.Ke2 Ra2+ 2.Ke1 Bg3# Section h#2 HotF (Helpmate of the Future)

Prentos Kostas

Problem-2015-09 Prentos Kostas 1st Place, 36° R.I.F.A.C.E. (St-Germain au Mont d'Or, 22-25 mai 2015) Source: http://phenix-echecs.fr/Messigny/RIFACE_2015_jugement_retros.pdf

1s1qkbsr/P1P1P3/2rPbPPP/8/6pp/1Qpp1pp1/1p2pB1P/RS2KBSR  (16 + 16) Proof game in 17,5 moves Anti-(Take and Make)

1.a4 b5 2.axb5(b4) a5 3.bxa6 e.p.(a4) Rxa6(a7) 4.c4 bxc3 e.p.(c5) 5.b4 axb3 e.p.(b5) 6.Qxb3(b2) d5 7.cxd6 e.p.(d4) c5 8.bxc6 e.p.(c4) Rxc6(c7) 9.e4 dxe3 e.p.(e5) 10.d4 cxd3 e.p.(d5) 11.Bxe3(e2) f5 12.exf6 e.p.(f4) e5 13.dxe6 e.p.(e4) Bxe6(e7) 14.g4 fxg3 e.p.(g5) 15.f4 exf3 e.p.(f5) 16.Bf2 h5 17.gxh6 e.p.(h4) g5 18.fxg6 e.p.(g4) Proof game : Find the unique moves, from the initial position of the pieces when starting a game, to the position of the given diagram.  Anti Take & Make : When a piece is "captured" (King excluded), it must move without capturing from its vanishing square. The capture is impossible if the captured piece can't be reborn. 13 en-passant captures.

Problem-2015-11 Prentos Kostas 7th Honourable Mention, 18° Sabra, Ostroda, 2015 Source: http://www.wccc2015.com/docs/congress_bulletin.pdf

6Q1/5pr1/4p1bp/3p1rBs/3pR2K/5k2/8/8  (4 + 10) h#2 twins: a) diagram, b) wQg8 to h7

a) wQg8 1.Bh7 Be3 2.Rg4+ Qxg4# b) wQh7 1.Rf6 Re2 2.Bd3 Qxd3# Bicolour Bristol, Orthogonal-Diagonal Transformation, Black line opening by White and by Black.

Problem-2015-12 Prentos Kostas 2nd Prize, 13° Tzuika, Ostroda, 2015 Source: http://www.wccc2015.com/docs/congress_bulletin.pdf

b3rrQ1/3p4/8/7B/s2K4/p7/S1k5/8  (4 + 7) hs#3,5 twins: a) diagram, b) -wSa2

a) with wSa2 1…Bh1 2.Qg2+ Kb3 3.Kd5 Rf1 4.Bd1+ Rxd1# b) without wSa2 1…Re1 2.Be2 Sb2 3.Ke3 Kc3 4.Qc4+ Sxc4#    The most economic achievement of the tourney. In this problem too all three different white and black pieces reach the pin line during the solutions. The epitome of elegance and refinement, in an unbelievable Meredith setting and long moves played by both sides. The slight mismatch in the motivation of black moves doesn’t detract at all the artistic impression.

Problem-2015-14 Prentos Kostas 3rd Recommendation, 3° Azemmour, Ostroda, 2015 Source: http://www.wccc2015.com/docs/congress_bulletin.pdf

q7/4s3/1p2B1bS/4k1pp/2KR2p1/4pPr1/5s2/b7  (5 + 12) h#2,5, two solutions

1…Rxg4 (A) 2.Sc6 f4+ 3.Ke4 Bd5# (B) 1…Bg8 (B) 2.Bf5 Sf7+ 3.Ke6 Rd6# (A)  The White is closing white lines. Critical squares f4 and f7. Pieces wB and wR are exchange their roles in the two solutions.

Problem-2015-15 Prentos Kostas 2nd Honourable Mention, 15° Sake, Ostroda, 2015 Source: http://www.wccc2015.com/docs/congress_bulletin.pdf

White : Kh1, Black : Kh3, Neutral : Qb4 Re4 Bf8  (1 + 1 + 3) h#2 twins: a) diagram, b) nQb4 to f4 Face to Face

a) nQb4 1.nRe6 nQb3+ 2.nBb4 nBe7# b) nQf4 1.nBa3 nQf5+ 2.nRg4 nRa4# Face-to-Face: When a white piece is just one rank below a black piece on the same file, they exchange their walk. Miniature. Reciprocal Anti-batteries. Orthogonal-Diagonal Transformation. There are several entries that tried to show ODT in a few pieces, but we think this one is the best and the most elegant in spite of the slight discrepancies between two solutions (FTF-specific battery in a) and ordinary battery in b)).

Problem-2015-16 Prentos Kostas 4th Prize, 4° FIDE Cup in Composing, 2015 Source: http://www.wfcc.ch/wp-content/uploads/E-4FIDECUP-fin.pdf

6b1/6sp/prP1B1p1/r1R5/pS2kPP1/2P1P3/1K6/5s2  (9 + 10) h#2.5 2 solutions

1…Bb3 2.Bc4 Bd1 3.Bb5 Re5# 1…Rh5 2.Rf5 Rh3 3.Rf7 Bd5# Reciprocal interception of the pair Ra5/Bg8 on two different squares. That is the novelty for this matrix. See pdb/P0579541 and yacpdb/383043. “Bicolor Bristol line opening and ODT. Although the white piece that opens the line will move again on the second move, the motivation for the Bristol is quite clear, as becomes evident by the "tries": {1...Bc4? (2…Be2) 3.??} and {1...Rd5? (2…Rd3) 3.??}” (author).

Adventure in composition with Andernach Lions

In this post we will observe the attempt for composing problems having specific fairy pieces, namely Andernach Lions.

Lion : Is a hurdle jumper, moves at Queen lines, jumps over a hurdle and steps on one of the empty squares (or captures a piece there) which are exactly behind the hurdle.
The "hurdle-color changing" Lion (or Andernach Lion) when passes over a hurdle changes the color of the hurdle (except it is a King or a neutral piece).

Last year we had seen, among the awarded problems in the World Congress on Chess Composition (WCCC) in Bern, the following problem with one Andernach Lion.
The composition tourney asked from the composers a helpmate two-mover with hurdle-color changing Lions. We had two ideas but the deadline was only three hours, so we barely had the time for checking one composition, the problem-776.

Problem-776 Manolas Emmanuel (GRE) and Prentos Kostas (GRE) Commendation, Quick Composing Ty, WCCC 2014 Bern

2Sk4/3p4/8/3K4/8/8/1(whL)2r3/8 (3 + 3) (hL hurdle-color changing Lion b2 + 0) h#2, two solutions

1.Re2-f2 hLb2-g2(wRf2) 2.Kd8-e8 hLg2-a8# 1.Re2-e5+ hLb2-g7(wRe5) 2.Kd8xc8 Re5-e8#

The Judge wrote : "Nice miniature. The white rook and Lion exchange functions in the mate".

The second idea lead us to the Problem-795. The idea here was to put the black Lions as epaulets from both sides of the mating piece, so they cannot capture it. This is achieved with Orthogonal - Diagonal Transformation, as you can see in the solution. 

The prominent composer Kostas Prentos, international maitre and multi-champion of Greece in solving contests, has moved to Uinted States of America, that is why his name is appended with USA.

Problem-795 Prentos Kostas (USA) and Manolas Emmanuel (GRE) Julia's Fairies No.593, 04-09-2014

8/8/(whL)5b1/2(bhL)p4/3pk3/1K6/1(bhL)P1r3/8 (3 + 7) (hL, Andernach Lion a6 + b2 c5) h#2, a) Diagram, b) wPc2 to g2

a) 1.hLc5-f2(wPd4) Kb3-c3 2.hLb2-d2(bPc2) hLa6-f1(wRe2)# b) 1.hLc5-g5(wPd5) Rb3-c4 2.hLb2-g7(wPd4) hLa6-h6(wBg6)#

For this problem we had many comments, mainly because the black pieces bRe2 and bBg6 stayed inert, waiting to change color. It is not forbidden for black pieces to stay inactive in some of the solutions (for white pieces this is forbidden), nevertheless the composition could be altered.

The composer Ladislav Packa had proposed to us the idea to convert the inactive pieces to Andernach Lions. In the following Problem-796, besides Lions (moving in Queen lines) we use a Rook-Lion (moving in Rook lines). To achieve what we want, we start with a white move, so the problem is now h#2,5 .

Problem-796 Packa Ladislav and Prentos Kostas and Manolas Emmanuel Pat a Mat No.90, December 2014, page 242 Problem No.790

7b/3(bhRL)4/2p1p3/4(bhL)(bhL)2/P1kpP(whL)2/8/2p1p3/4K3 (4 + 10) (hL, Andernach Lion f4 + e5 f5) (hRL, Andernach Rook-Lion 0 + d7) h#2.5, two solutions

1…hLf4-c7(whLe5)+ 2.hRLd7-b7(bhLc7) Ke1xe2 3.hLf5-d7(wPe6) hLe5-b8(whLc7)# 1…hLf4-f7(whLf5)+ 2.hRLd7-g7(bhLf7) Ke1-d2 3.hLe5-e7(wPe6) hLf5-f8(whLf7)#

Here there is Orthogonal - Diagonal Transformation and the idea of Lions as epaulets is well presented.

Since the moves have become 2.5, the question arises if we can make something different for helpmate twomover. I have tried the position which is shown as Problem-797.

Problem-797 Manolas Emmanuel (GRE) original

3K1(whL)2/8/8/2SP2(bhL)1/3k4/1P6/5br1/1s4B1 (6 + 5) (Andernach Lion f8 + g5) h#2, a) Diagram, b) bRg2 to b4

a) 1.hLg5xg1(wRg2) Rg2-g5 2.hLg1-a1(wSb1) hLf8-f1(wBf2)# b) 1.hLg5xc5(bPd5) Bg1-h2 2.hLc5-e3 hLf8-a3(wRb4)#

Here we see the Orthogonal - Diagonal Transformation, but not the epaulets, which may be abandoned. Black pieces are used that stay inactive in some phase (not forbidden, but not economical enough), and the solutions have not the same number of color changing.

Maybe I can do better. Let us try a new position.

Problem-798 Manolas Emmanuel (GRE) original

8/8/8/1(bhL)pk4/p7/4Kp2/1P2(bhL)1(bhL)1/8 (2 + 7) (hL Andernach Lion 0 + b5 e2 g2) h#2, a) diagram, b) bPa4 to e6

a) 1.hLg2-c2(whLe2) b2-b4 2.hLc2-c6(wPc5) hLe2-a6(whLb5)# b) 1.hLb5-f1(whLe2) b2-b3 2.hLf1-f4(wPf3) hLe2-h2(whLg2)#

Here White has minimal force, just one Pawn, but he can find the required forces to make two mates, one in an orthogonal way and one in a diagonal way. In each solution there are three changes of color.

Can this become better? The answer is yes. 

Problem-799 Manolas Emmanuel (GRE) and Prentos Kostas (GRE) dedicated to Ladislav Packa The Problemist vol.25 No.1, January 2015, problem F3184 Commendation, vol.27 No.6, November 2019, p.242

4K3/8/8/1(bhL)pk1p2/p2p4/5p1(whL)/1p2(bhL)1(bhL)1/8 (2 + 10) (hL Andernach Lion h3 + b5 e2 g2) h#2, two solutions

1.hLg2-c2(whLe2) hLh3-e3(wPf3) 2.hLc2-c6(wPc5)+ hLe2-a6(whLb5)# 1.hLb5-f1(whLe2) hLh3-c8(wPf5) 2.hLf1-f4(wPf3) hLe2-h2(whLg2)#

The composition is not in a twin form, it has two solutions. There is orthogonal - diagonal transformation. White has only an Andernach Lion. In each solution there are four color changes and they are all necessary! Comment by Judge Michal Dragoun : Full analogy in recolouring of the pieces.