Statistics for Fairy compositions

In chess composition there is the Fairy Chess, where we see other conditions, others pieces, other chessboards. 

For the question "which fairy elements are mostly used?", a statistic comes to our help, based on the about 785,000 problems that are contained in the database WinChloe by the French Christian Poisson. 

The statistic is published in an article of the very good composer Julia Vysotska:
https://juliasfairies.com/fairy-elements-statistics/

It is important to remember that the fairy elements were not defined at the same year, so the older elements had for more time the opportunity to be chosen for use. 

The statistic shows that there are about 2670 fairy elements (pieces and conditions) and in the first position stands the piece Grasshopper (introduced in 1913) with 14073 compositions containing it!

Another composition in CoVID19 times

Today we present a problem by Ioannis Garoufalidis, awarded Greek solver and composer.

Condition: Because of the corona virus CoVID19 … the pieces must be kept in a distance from each other!
The condition is known as Anti-ContactChess or Anti-koko.

Problem-840 FEN: 8/3P2p1/8/8/8/k7/4p3/7K h#2, (2+3) twin: bKa3 to g5	Ioannis Garoufalidis (GRE) original Black plays and helps White mate in two moves Fairy condition: Anti-koko
Condition Anti-koko: The moves, where one piece goes near another piece, are not legal and are prohibited. The capture of a piece is allowed. a) bKa3 1.e1=S d8=Q 2.Sc2 Qa5# b) bKg5 1.e1=B d8=R 2.Bg3 Rd5# An Allumwandlung (AUW) problem with model mates.

A well-timed problem, in CoVID19 times

We present today a well-timed problem by Nikos Mendrinos, awarded Greek solver, champion and new composer.

Condition: Because of the corona virus CoVID19 … the pieces must be kept in a distance from each other!
The condition is known as Anti-ContactChess or Anti-koko.

Problem-839 FEN: 4k3/b7/4S3/2q4s/K7/2p2Q2/8/1R6 #2, (4+5)

Nikos Mendrinos (GRE) original White plays and mates in two moves Fairy condition: Anti-koko

Condition Anti-koko: The moves, where one piece goes near another piece as in 1.Ka4-b3 or 1.Qf3-a8+ or 1...Qc5-a5+ or 1…Ba7-b6, are not legal and are prohibited. The capture of a piece, as 1.Qf3xh5+, is allowed. Tries: {1.Qh1? / Qg2? Sg7!}, {1.Qxh5+? Qxh5!}, {1.Rg1? Sg7!}, {1.Re1? Bb8!} Key: 1.Qe4! (waiting) 1…Sg7 2.Sc7#, 1…Sg3 2.Qg6# 1…Qc8 / Qc7 2.Sg7#, 1…Qc6+ / Qg1 / Qf2 2.Q(x)c6# 1…Bb8 2.Rxb8# The white battery, constructed with wQ and wS with the key, can be opened giving double check. The defence of the black pieces is very limited by the Anti-koko condition.

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.

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".

International Chess Composition Contest : "JT Manolas-65", C 12-VII-2015

Announcement 06-IV-2015, Last day for entries 12-VII-2015

International Chess Composition Contest : "Jubilee Tourney Manolas-65", Closing date 2015-07-12. The blogs http://chess-problems-gr.blogspot.com (in English) and http://kallitexniko-skaki.blogspot.com (in Greek) announce the International Chess Composition Contest "JT Manolas-65". Sections: A. helpmate h#2, in HotF form (Helpmate of the Future), with at least two pairs of related solutions. Judge Ioannis Kalkavouras. B. fairy #2, with accepted elements {one fairy condition} and/or {one fairy piece type}.  Judge Emmanuel Manolas. Original computer-checked problems, (no zero-positions), may be submitted by each composer to one or both sections specifying : Name & e-mail & country of the composer, diagram & FEN notation & stipulation & solution of the problem. Send documents by e-mail with subject "JT-Manolas-65" to manolas.emmanuel@gmail.com . Closing day : 12-July-2015. The participants will receive a copy of the award by e-mail. The award will be published in the above blogs.

Διεθνής Διαγωνισμός Σκακιστικής Σύνθεσης : "Επετειακό Τουρνουά Μανωλάς-65", λήξη 2015-07-12. Τα ιστολόγια http://chess-problems-gr.blogspot.com (στα Αγγλικά) και http://kallitexniko-skaki.blogspot.com (στα Ελληνικά) ανακοινώνουν τον Διεθνή Διαγωνισμό Σκακιστικής Σύνθεσης "JT Manolas-65". Τμήματα: A. βοηθητικά h#2, σε μορφή HotF (Helpmate of the Future, Βοηθητικό του Μέλλοντος), με τουλάχιστον δύο ζεύγη συναφών λύσεων. Κριτής Ιωάννης Καλκαβούρας. B. μυθικά #2, με αποδεκτά στοιχεία {μία μυθική συνθήκη} και/ή {ένα είδος μυθικού κομματιού}. Κριτής Εμμανουήλ Μανωλάς. Αδημοσίευτα προβλήματα ελεγμένα από υπολογιστή (όχι zero-position) μπορεί να υποβάλει ένας συνθέτης σε ένα ή δύο τμήματα του διαγωνισμού, καθορίζοντας : Όνομα και e-mail και χώρα του συνθέτη, διάγραμμα και FEN συμβολισμός και εκφώνηση και λύση του προβλήματος. Στείλτε έγγραφο μέσω e-mail με θέμα "JT-Manolas-65" στο manolas.emmanuel@gmail.com . Ημερομηνία λήξης : 12-Ιουλίου-2015. Οι συμμετέχοντες θα λάβουν ένα αντίγραφο της βράβευσης μέσω e-mail. Η βράβευση θα δημοσιευθεί στα ανωτέρω ιστολόγια.

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.

Award for 2nd TT Kobulchess Christmas 2014

(25.12.2014) Here is the Award of the 2nd KoBulChess TT - Cristmas Tourney 2014! Many thanks to all participants and to the judge IM Krassimir Gandev for his quick work. The award remains open for 1 month period. 2nd KoBulChess TT – Christmas Tourney 2014 Theme: All type problems (#/=, H#/H=, S#/S=, HS#/HS= etc.) in 2-4 moves (up to 8 moves for Series and P-Series problems) with the fairy condition Circle SneK. Other fairy pieces and conditions are not allowed. Royal pieces can be used of course. Circle SneK: When a Queen is captured - a Rook (or Royal Rook) of the same color (if exists on the board) becoming Queen; When a Rook is captured – a Bishop (or Royal Bishop) of the same color (if exists on the board) becoming Rook; When a Bishop is captured - a Knight (or Royal Knight) of the same color (if exists on the board) becoming Bishop; When a Knight is captured – a Queen (or Royal Queen) of the same color (if exist on the board) becoming Knight. Only one piece may change its type after a capture. In case of option – the capturing side choose which piece will be transformed. The capture and the change of type is a single move. If this full move result a selfcheck - the capture is forbidden.  The capture of a pawn is normal. The capture is normal also in the case when there is no piece on the board which should be transformed. Castling with Royal piece is not allowed. Entries: 22  Participants: Pierre Tritten, Manfred Rittirsch, Mario Parrinello, Kostas Prentos, Emmanuel Manolas, Rainer Kuhn, Themis Argirakopoulos, Sebastien Luce, Ralf Kraetschmer, Alain Bienabe Countries: Greece, Italy, Germany, France AWARD It was pleasure for me to be a judge of this interesting tourney. I received from the director Diyan Kostadinov 22 entries (including 2 versions) in anonymous form. I propose the following ranking: 1st Prize – Manfred Rittirsch    a) 1....Rd2? ... 3. ... Qxg3(wRh8)+ 4.hxg3(bQh5)! 1...Sf2 [~? ... 4.Sxc3(bBh3)!] 2.Rf5+ [~? ... 3.h8Q+ Rd,Rhxh8(wQf7)!] gxf5 3.h8Q+ Bxc3(wSh8)# [4.Sxc3(bBf2)??, 4.Kxf2(bSg4)??]    b) 1...Sf2? ... 3. ... Bxc3(wSh8)+ 4.Kxf2(bSh5)! 1....Rd2 [~? ... 4.hxg3(bQd8)!] 2.Sd6 [~? ... 3.h8B+ Qxh8(wBb5)!] cxd6 3.h8B+ Qxg3(wRh8)# [4.hxg3(bQd2)??, 4.Kxd2(bRb2)??] This is my favourite. Very rich Circle SneK specific logical maneuvers with reciprocal correspondence of pieces, change of promotions, cross checks and Circle SneK mates. 2nd Prize – Mario Parrinello a) 1…rQh8 2.Sa6 Sf6 3.rQa1 Sxg8(rSa1)+ 4.hxg8Q(rSh8)+ Qxa6(wSg8)# b) 1...rQc8 2.Sh6 Sc6 3.rQc1 Sxb8(rSc1)+ 4.axb8Q(rSc8)+ Qxh6(wSb8)# Wonderful Echo mates, creation of black batteries with ODT, change of functions between wSs and nice fairy play! 3rd Prize – Emmanuel Manolas 1.Bg7 fxe8Q(bQh7) 2.Bxc3(wBg1) Qxe2(bSh7)# 1.Sxc3(wBg1) fxe8S(bQh7) 2.Qd3 Sxd6(bRf8)# Model mates, selfblocks, promotions, all types of SneK conversions and SneK mates. 4th Prize – Pierre Tritten 1.Be4 Bxb3(bSf4) 2.Se2 Rxd1(bRc3)# 1.Sd2 Rxf4(bQd1) 2.Qe2 Bxg6(bBd2)# Same motivation for white captures: first one allows black transformed piece to block on e2, second one avoids black defense, change of functions, all types Circle SneK transformations. 5th Prize – Themis Argirakopoulos 1.d1B 2.b1R 3.Rb2 4.Rg2 5.Bf3 6.Bc6 Kxg2(Rc6)= [7.Rxb6(Rd6)? 7.Rxc7(Se6)? 7.Rxd6(Bc7)?] 1.d1S 2.b1Q 3.Qa2 4.Se3 5.Sg2 6.Qa8 Kxg2(Sa8)= [7.Sxb6(Rd6)?  7.Sxc7(Se6)?] A wonderful problem with AUW where the stalemate positions are possible because of SneK protections.   6th Prize – Argirakopoulos, Luce, Tritten a) 1.c2 2.c1B 3.Bf4 4.Bh2 5.g1R+ Sxg1(bRh2)# b) 1.b1S 2.Sd2 3.Sf1 4.g1Q 5.Qg4 Bxf1(bSg4)# Interchange of function between the white pieces, AUW. Special Prize – Pierre Tritten 1.Bf4+ Kxf4(bBh3) 2.Rf3+ Kxf3(bRh3) 3.Qe3+ Kxe3(bQh3) 4.Se2+ Kxe2(bSh3) 5.Sf2 Kxf2= Funny idea – the black Knight h3 plays like a Knight again after a full Circle SneK cycle of transformations 1st Honorable mention – Argirakopoulos, Tritten a) 1.Kd5 2.Kxe4(wRf8) 3.Kd5 4.Kd6 Rd8# b) 1.Sd7 2.Sxf8(wBh3) 3.Sd7 4.Sc5 Rd4# c) 1.Rc3 2.Rxh3(wRf8) 3.Rc3 4.Rc5 Rd8# Double switchbacks by three black pieces (King, Knight, Rook), Zilahi and nice white/back Forsberg suit twins.    2nd Honorable mention – Kostas Prentos 1.Qe3 Sg6 2.Bd4 Bd3# 1.Qxe5 Bc4 2.Be3 Rf4# Two Circle SneK specific mates. 3rd Honorable mention – Kostas Prentos 1.Bb1 Bg7 2.Se4 Bxe5(wBe4) 3.Bec2 Kxf5(wRc2) 4.Rf2+ Sxf2(wRb1)# 1.Rf2+ Kf7 2.Sh7+ Ke8 3.Sg5 Bxe5(wBg5) 4.Bc6+ Sxf2(wRc6)# Problem type ANI: in the 1st solution is presented a hybrid of Bristol and Indian - the first white Bishop opens the line for the second white Bishop, which on the next move will transform into Rook, creating a battery. 2nd solution reach the same mate after different active play of the white pieces. Commendations (equal rank): Com – Themis Argirakopoulos a) 1…g8R 2.e1S Rg2 3.Sc2 Rxc2(bSd3)# b) 1…h8Q 2.h1B Qh6 3.Bc6 Qxc6(bBb3)# Com – Alain Bienabe 1.Kxd5(wRe7) Ra7 2.Kc5 Ra5# 1.Kxe7(wBc4) Sc6+ 2.Ke8 Rd8# Com – Rainer Kuhn 1.Sxa1(bRd1)! Rxa1 2.a8Q+ Qxa8(wQe1)# 1.rBa8! Qa4 2.Re8+ Qxe8(rRa8)# 1.rBb7! Bxc2 2.a8Q+ Qxa8(wQe1)# Com – Sebastien Luce 1…Ba7 2.Sxa7(wBe4)+ Bxg2(bBa7) 3.Bb8 Kb6# 1…Bb6 2.Sxb6(wBe4) Bxg2(bBb6) 3.Ba7 Kc7# Com – Emmanuel Manolas 1.Qxd5(bBc3)+! (bBe6?) Kxd5(wQh5) 2.Qxe6(bSh2)+ Kxe6 3.Qxf7(bRc3)+ Kxf7 4.g8Q# Com – Rainer Kuhn 1.rSc3 Kg4 2.Bxf5+ Kxf5(rBc3) 3.rBa1 Be5# 1.rSb4 Kh3 2.rSc6 Be5 3.rSxe5(wBe8) d4# I wish to all participants and the tourney director – Merry Christmas and Happy New Year 2015! Sofia 24.12.2014        Judge: IM Krassimir Gandev

Circle SneK condition - Christmas 2014

In fairy chess, various conditions are introduced each year. Some of them attract the interest of the composers, some stay as curiosities.
The Bulgarian composer Diyan Kostadinov likes to invent new conditions (KoBul Kings, SneK Chess) and recently has announced a tourney for Christmas 2014 with the new condition Circle Snek.

Circle Snek : When a Queen is captured, a Rook of the same side becomes a Queen. When a Rook is captured, a Bishop of the same side becomes a Rook. When a Bishop is captured, a Knight of the same side becomes a Bishop. When a Knight is captured, a Queen of the same side becomes a Knight. Only one piece may change its type after a capture. In case of option – the capturing side chooses which piece will be transformed. The capture and the change of type is a single move. If this full move results in a selfcheck - the capture is forbidden. The capture of a pawn is normal. The capture is normal also in the case when there is no piece on the board which might be transformed. Castling with Royal piece is not allowed.

You will find the announcement of the 2nd KoBulChess Thematic Tourney (Christmas Tourney 2014) here : http://kobulchess.com/en/tournaments/announcements/676-kobulchess-2nd-tt-christmas-tourney-2014.html

Theme: All type problems (#/=, H#/H=, S#/S=, HS#/HS= etc.) in 2-4 moves with the fairy condition Circle SneK. Other fairy pieces and conditions are not allowed. Royal pieces can be used of course.

The deadline is 20-Dec-2014.

Wishing you happy holidays, I present an original miniature with the Circle SneK condition, designed in a way to be easily solved. My computer (using WinChloe v3.31) could not see the solution instantly and spent 4 hours and 12 minutes to finish.

Problem_289 Manolas Emmanuel (GRE) original

Ks3r2/8/8/k5b1/8/7b/7P/8 (2 + 5) h#5, Circle SneK

Select inside the brackets to see the solution [1.Bd7 h4 2.Kb6 hxg5(Bb8) 3.Kc7 g6 4.Kd8 g7 5.Bc7 gxf8=Q(Rd7)#].

Since the moves are only 5, the wPh2 must start with a double step and give mate when it reaches the 8th row and be promoted. (Theme Excelsior). So, the bBh3 must make the first half-move (move B1), (and surely prepare the blocking of bK). The wP will possibly capture one or two black pieces while marching to promotion and it will mate the bK when he is blocked in the eighth row. The bK needs 3 moves to reach the eighth row. The remaining black half-move is needed to complete the block. I believe that some solvers do not really need to see the hidden solution!

Greek compositions in WCCC 2014, Bern

This blog has special interest in Greek composers.
In this post we will see compositions that received distinctions in the composing tourneys of WCCC 2014, in Bern. In a previous post we saw that Argirakopoulos Themis, Manolas Emmanuel and Prentos Kostas had 1, 3 and 11 distinctions respectively. It so happens, that we had published a picture of these Greek composers together, in WCCC 2010.

Problem-775 Argirakopoulos Themis (GRE) 2nd Prize, Juica Ty Fairy Section, WCCC 2014 Bern

White : Kh6 Qf3, Black : Kg8 Rf2 Be1 Pd5c3, Neutral : Ra6 Bf6, (2 + 5 + 2) hs#2.5, Circe Kamikaze a) diagram, b) f6 = fairy bishop

A: 1...Rh2+ 2.Qh5 Beh4 3.Qe8+ Bxf6(Bf8;nBc1)# B: 1...Bd2+ 2.Qe3 Rf4 3.Qe8+ Rxf6(Rh8;nfBf8)#

Neutral pieces take the color of the playing side. Circe Kamikaze : When a capture occurs, the capturing piece and the captured piece in this order (King excluded, unless otherwise stated) must be replaced on their rebirth square if it is empty, otherwise, the piece vanishes. The Judge said : "The solutions of this problem culminate in a fabulous quadruple check, which is already a highly noticeable record with only 9 units on the board. Besides, the diagonal-orthogonal correspondence is perfectly realized and we find, as in many problems, the traditional reciprocal battery creation with Rook and Bishop. The wQ arrives on the same square e8 at W3, but since the routes the wQ takes are different, it is not a defect. One defect however would be the passive nRa6".

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

2Sk4/3p4/8/3K4/8/8/1Y2r3/8 (3 + 3) h#2, two solutions b2 = hurdle-color changing Lion (hL)

1.Rf2 hLg2(wRf2) 2.Ke8 hLa8# 1.Re5+ hLg7(wRe5) 2.Kxc8 Re8#

Lion : Moving in Queen lines, it jumps over a hurdle and lands / captures in any free square immediately after the hurdle. The hurdle-color changing Lion, changes the color of the hurdle (except King or neutral piece) when jumping over it.  The Judge said : "Nice miniature. The white rook and Lion exchange functions in the mate".

Problem-777 Manolas Emmanuel (GRE) Commendation, Japanese Sake Ty, WCCC 2014 Bern

8/8/8/7r/3k4/8/1PKb2p1/5srq (2 + 7) h#2, Back-To-Back a) diagram, b) = a) -Rh5 c) = b) -Fd2 d) = c) -Cf1 e) = d) -Tg1

a) 1.Kc4 b4 2.Rc5 bxc5# b) 1.Bc3 b3 2.Bb2 d5# c) 1.Sd2 b4 2.Sb3+ d5# d) 1.Rb1 b5 2.Rb4 d5# e) 1.Ke4 Kd2 2.Qb1 e5#

Back-to-Back : When a white piece is just one rank above a black piece on the same file, they exchange their way of moving / capturing. White loses pieces one after another. We see Echo mates, Chameleon mates, Model mates. Some readers might remember a similar problem by Sam Loyd, accompanied by a tale, where bullets strike off the board the pieces one by one : http://en.chessbase.com/post/chebase-puzzles-a-dangerous-game-171013

Problem-778 Manolas Emmanuel (GRE) Commendation, Bulgarian Wine Ty, WCCC 2014 Bern

8/4K3/8/1r4S1/1k6/bP6/PB6/8 (5 + 3) h#2, two solutions SneK chess

1.Rxg5(SKe7) Bd4 2.Rb5 SKc6# 1.Bxb2(wBg5) Kd6 2.Ba3 Bd2#

SneK chess : When a Queen is captured, a Rook of the same side becomes a Queen. When a Rook is captured, a Bishop of the same side becomes a Rook. When a Bishop is captured, a Knight of the same side becomes a Bishop. When a Knight is captured, the King (but not another royal piece) of the same side becomes a royal Knight. When a Pawn is captured, the royal piece of the same side becomes a King. The Judge said : "Switchbacks of the bR and bB for re-blocking, Ideal mates. A little but lovely problem".

Problem-779 Prentos Kostas (GRE) 1st Prize, Champagne Ty Section A, WCCC 2014 Bern

q5sr/2pppp1p/1psr3b/p1k4R/P5p1/1P1BRP1P/1BPKP2S/1SQ4b (14 + 16) SPG 18.5

1.a4 a5 2.Ra3 Ra6 3.Rc3 Rd6 4.b3 b6 5.Bb2 Bb7 6.Qc1 Bxg2 7.Sf3 Sc6 8.Rg1 Bh1 9.Rg5 Qa8 10.Rh5 g5 11.Bh3 Bh6 12.Bf5 Kf8 13.h3 Kg7 14.Sh2 Kf6 15.f3 Ke5 16.d4+ Kxd4 17.Kd2 g4+ 18.Re3+ Kc5+ 19.Bd3+

SPG : Given a game position, find all the moves since the start of the game. There is an upper limit on the number of the moves. The Judge said : "« Only » 4 thematical checks but of the same nature : they are all battery checks without capture, the most sophisticated nature of thematical moves. Very « professional » realization.".

Problem-780 Prentos Kostas (GRE) 3rd Honourable Mention, Champagne Ty Section A, WCCC 2014 Bern

r2k1bsr/pp1s1ppp/5P2/P7/1K1q3p/p3Qp2/bPP1PPPP/RSB2BSR (16 + 16) SPG 9, Circe Perrain

1.d4 c5 2.dxc5 Qb6(a3) 3.Kd2 d5 4.cxd6 e.p. Be6(f3) 5.Kc3 Sd7 6.dxe7 Qe3(h4)+ 7.Kb4 Bxa2 8.Qd4(a5) Kxe7 9.Qxe3(f6)+ Kd8(Qd4)+

SPG : Given a game position, find all the moves since the start of the game. There is an upper limit on the number of the moves. Circe Parrain : In the next move following a capture, the captured unit (except a King) accomplish (from its capture square) an exact copy of that next move. If the arrival square is occupied or if the journey brings it out of the board, the captured unit vanishes. The Judge said : "Cross-double check is clearly impossible in orthodox chess. Possibly other fairy conditions than Circe Parrain allow to do it, but this problem will be a pioneer".

Problem-781 Prentos Kostas (GRE) Commendation, 17th Sabra Ty, WCCC 2014 Bern

8/3r4/1pp2p2/1s3K2/1p2Bp2/1Pk2b2/2Pqp1P1/bR3rBQ (8 + 13) h#2, two solutions

1.Bxg2 Bxc6 2.Bxc6 Qxc6# 1.Rxg1 Rxa1 2.Rxa1 Qxa1#

Orthogonal-Diagonal transformation. Bicolored Bristol. Pseudo white-Sacrifices. Quasi black-sacrifices.

Problem-782 Prentos Kostas (GRE) 4th Prize, 14th Sake Ty, WCCC 2014 Bern

3K4/8/7R/3k3b/8/1S4B1/6r1/8 (4 + 3) h#2, two solutions, Back-to-Back

1.Kc6 Bg6 2.Bb5 Be4# 1.Kd6 Rf4 2.Rd5 Rf6#

Back-to-Back : When a white piece is just one rank above a black piece on the same file, they exchange their way of moving / capturing. Orthogonal-Diagonal transformation. Reciprocal white batteries with Anderssen moves (A white piece wA intercepts another white piece, Black moves, then wA moves again giving an indirect check). Auto-blocking. Indirect Pinning and Unpinning. The Judge said : "A highly polished ODT with reversal of roles between wR/B and also bR/B. It’s a pity the final positions are orthodox doublechecks".

Problem-783 Prentos Kostas (GRE) 2nd Honourable Mention, 14th Sake Ty, WCCC 2014 Bern

7s/2R2P1P/P2k2BK/1p1P4/2p2pb1/5p1b/6p1/r2r4 (7 + 11) h#2, four solutions, Back-to-Back

1.Bd7 f8=R 2.Ke7 Re8# 1.Re1 f8=S 2.Re6 Sd7# 1.Rxd5 f8=B+ 2.Ke6 Be7# 1.Kxd5 f8=Q 2.Ra5 Qd6#

Back-to-Back : When a white piece is just one rank above a black piece on the same file, they exchange their way of moving / capturing. Orthogonal-Diagonal transformation. Auto-Pinning. Pin mate. Allumwandlung (AUW). The Judge said : "A task: AUW with specific BTB mates in all solutions. The setting is rather heavy".

Problem-784 Prentos Kostas (GRE) 3rd Honourable Mention, 14th Sake Ty, WCCC 2014 Bern

6s1/5K2/4p1S1/5k2/2r3p1/Pbs5/7B/2b5 (4 + 8) h#2, two solutions, Back-to-Back

1.Sa2 c2+ 2.Re4 Se5# 1.Ba2 c5+ 2.Sd5 Bd6#

Back-to-Back : When a white piece is just one rank above a black piece on the same file, they exchange their way of moving / capturing. Direct Self-Pinning. Pin mate. The Judge said : "Pinning of BTB black piece by another BTB white piece. Nicely done".

Problem-785 Prentos Kostas (GRE) 4rd Honourable Mention, 14th Sake Ty, WCCC 2014 Bern

6b1/1p6/1R4s1/P1k1qS2/2b2P2/5s2/8/3K4 (5 + 7) h#2, two solutions, Back-to-Back

1.Qe4 Se7 2.Se5 e6# 1.Bd3 Sd6 2.Qd4 d5#

Back-to-Back : When a white piece is just one rank above a black piece on the same file, they exchange their way of moving / capturing. The Judge said : "Anticipatory selfblock of the BTB black piece. Please note that in both solutions, black cannot capture the mating wP by B because it turns wS into B.".

Problem-786 Prentos Kostas (GRE) Commendation, 14th Sake Ty, WCCC 2014 Bern

s3r3/5P2/P3P3/3k4/K1p2p2/5pS1/4p3/8 (5 + 7) h#2, Back-to-Back a) diagram, b) bKd5 to d6

a) 1.e1=R f8=Q 2.Re5 Qd6# b) 1.e1=B f8=S 2.Ba5 Sd7#

Back-to-Back : When a white piece is just one rank above a black piece on the same file, they exchange their way of moving / capturing. Allumwandlung (AUW).

Problem-787 Prentos Kostas (GRE) 6th Prize, 5th Bulgarian Wine Ty, WCCC 2014 Bern

8/1R6/8/1pp4p/2PB1K2/8/p1b2s2/1S3k1s (5 + 8) h#2, two solutions, SneK chess

1.cxd4(Bb1) Bxc2(Bh1) 2.Bxb7(Rc2) Rxf2(SKf1)# 1.bxc4 Bxf2(SKf1) 2.Bxb1(SKf4) Rxb1(BKf1)#

SneK chess : When a Queen is captured, a Rook of the same side becomes a Queen. When a Rook is captured, a Bishop of the same side becomes a Rook. When a Bishop is captured, a Knight of the same side becomes a Bishop. When a Knight is captured, the King (but not another royal piece) of the same side becomes a royal Knight. When a Pawn is captured, the royal piece of the same side becomes a King. The Judge said : "Complicated combination of transformations to reach the final mating positions, each time by a different white Rook forming a Zilahi. Mutual captures by the wSb1/bBc2.".

Problem-788 Prentos Kostas (GRE) 2nd Honourable Mention, 5th Bulgarian Wine Ty, WCCC 2014 Bern

2Kbk3/4S3/2sP4/1r2p3/8/8/8/8 (3 + 5) h#2, two solutions, SneK chess

1.Bxe7(SKc8) dxe7(Bc6) 2.Bd7+ SKd6# 1.Rb7 Sxc6(SKe8) 2.Rf7 Kxd8(BKe8)#

SneK chess : When a Queen is captured, a Rook of the same side becomes a Queen. When a Rook is captured, a Bishop of the same side becomes a Rook. When a Bishop is captured, a Knight of the same side becomes a Bishop. When a Knight is captured, the King (but not another royal piece) of the same side becomes a royal Knight. When a Pawn is captured, the royal piece of the same side becomes a King. The Judge said : "Specific fairy mates and selfblocks".

Fairy condition "BackHome"

The BackHome fairy condition is invented by Nicolas Dupont.

Definition : "If a piece can move to the square it occupied in the diagram position, it must move to this back-home square. BackHome moves have priority over the virtual capture of the opponent king by any piece, i.e. checks are fairy. If more BackHome moves are possible, the side-on-move chooses which move to play. The BackHome square of a pawn which is promoted during the solution is the-initial-diagram-square of this pawn."

The fifth thematic tourney (TT5) of ChessProblems.ca, ending 31-XII-2013, required original chess compositions employing the BackHome fairy condition. For the Announcement and the Award please visit this link. You will find excellent material if you follow the link to the bulletin.

The condition is included in programs WinChloe 3.24 and Popeye 4.65 .

I will present here two helpmate problems of mine, one with normal pieces and one using a Grasshopper, where the BackHome condition is applied.

Problem_761 Emmanuel Manolas original

3r4/8/2r5/b3S3/3k1K2/7b/5P2/8, (3 + 5) h#7, BackHome

1.Rc3 Sc4 (the wS blocks the return path for the bRc3) 2.Re3 Se5 (the wS returns BackHome) 3.Rd5 Sd7 (the wS blocks the return path for the bRd5) 4.Rc5 Se5 (the wS returns BackHome) 5.Be6 Sg4 (the wS blocks the return path for the bBe6) 6.Bd5 Se5 (the wS returns BackHome) 7.Bc3 fxe3# The wS, visiting 3 different squares and then returning BackHome, blocks the BackHome path of three black units which block their King. The four black pieces go to their destinations one after another in strict order.

Problem_762 Emmanuel Manolas original

8/8/7k/8/g6p/2p4P/2K4R/7R, (4 + 4) (Grasshopper 0 + a4) h#6, BackHome a) diagram, b) bGa4 to h8

a) diagram 1.Gd1 (the wG cannot return to a4) Rf1 2.Gg1 (blocks the return path of bRf1) Rf4 3.Kg5 Rxh4 (inhibits the BackHome move of the bK) 4.Gg6 Rh8 5.Gg4 Rg8+ 6.Kh6 (the bK makes a switchback BackHome) hxg4# (the wRh1 is in BackHome position, the check is valid, it is mate). b) bGa4 to h8 1.Gh5 (the wG cannot return to h8) Rd2 2.cxd2 Kd3 3.d1=B (inhibits the BackHome move of the bK to c2) Ke4 4.Gxh3 Kf5 5.Ge6 Rxh4 (this is not a check) 6.Kh5 (the bK is not in threat) Rh1# (wR moves away from bK and makes a switchback to BackHome position to give check. Three pieces are cleared from the h-file to make this mate possible, and the promoted bBd1 is unable to interfere). There are sacrifices, of bGg4 in (a), of wRd2 in (b). There are reciprocal captures, wPxbG in (a), bGxwP in (b).