| Problem-833 Vito Rallo (ITA) and Emmanuel Manolas (GRE) Phenix 265, 09/2016 |
 (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 |
 (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 |
 (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. |
| Problem-823 Fadil Abdurahmanovic (Bosnia-Herzegovina) and Ioannis Kalkabouras (GRE) KobulChess.com 17.11.2015, no.269 |
 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 |
 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". |
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| 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 |
 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)). |
| Problem-2015-10 Argyrakopoulos Themis 4th Place, Marianka 2015 Source: http://www.jurajlorinc.com/chess/ma15faaw.htm |
 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 |
 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? |
| Problem-2015-02 Argirakopoulos Themis, Prentos Kostas 1st Prize (1st - 3rd place)- TT 157 Superproblem Source: https://yadi.sk/i/Iho1AFq4mNeij |
 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. |
| Πρόβλημα-2015-17 Abdurahmanovic Fadil, Kalkavouras Ioannis 1st Prize, Moskow Concours 2015 Source: http://www.selivanov.ru/download/Awards/Moscow/2015/%20mt2015-h.pdf |
 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 |
 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. |
| 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 |
 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 |
 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) |
| Problem-2015-01 Pergialis Nikos Commendation, Manolas-65 JT Source: http://chess-problems-gr.blogspot.gr/2015/08/award-for-manolas-65-jt.html |
 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 |
 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 |
 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 |
 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) |
| 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 |
 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 |
 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 |
 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 |
 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 |
 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 |
 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). |
International
Chess Composition Contest : "Jubilee Tourney Manolas-65",
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",
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.
|
| Problem-776 Manolas Emmanuel (GRE) and Prentos Kostas (GRE) Commendation, Quick Composing Ty, WCCC 2014 Bern |
 (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". |
| Problem-795 Prentos Kostas (USA) and Manolas Emmanuel (GRE) Julia's Fairies No.593, 04-09-2014 |
 (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. |
| Problem-796 Packa Ladislav and Prentos Kostas and Manolas Emmanuel Pat a Mat No.90, December 2014, page 242 Problem No.790 |
 (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. |
| Problem-797 Manolas Emmanuel (GRE) original |
 (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. |
| Problem-798 Manolas Emmanuel (GRE) original |
 (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. |
| 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 |
 (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. |
(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
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