ChessProblems.ca Series-Movers Tourney


ChessProblems.ca organizes an annual Series-Movers Informal Tourney open for series-movers of any type and with any fairy conditions and pieces. Tasks and records are encouraged, but are not mandatory. Maximum four problems per author are accepted for the tourney (tournament entries are labeled "T[n]" in the table below), but compositions can also be sent for publication "hors concours" (labeled "HC[n]"). Please email your entries to Cornel Pacurar at originals (at) chessproblems.ca.

Judge for 2010:  Dan Meinking (USA)            |    2010 Award (16.05.2011)
Judge for 2011:  Paul Raican (Romania)
Judge for 2012:  Arno Tüngler (Germany)

Participants

# Composer Country 2010 2011 2012
1 Alberto Armeni Italy - 4 4
2 György Bakcsi Hungary 1 - 4
3 Bojan Bašić Serbia 1 -  
4 Vlaicu Crişan Romania 0.5 -  
5 Mihail Croitor Moldova 1 -  
6 Nicolas Dupont France - - 1
7 Itamar Faybish Belgium 0.5 -  
8 Geoff Foster Australia - 1  
9 Ján Golha Slovakia 1.5 2.5  
10 Harald Grubert Germany - 0.5 1
11 Jozef Holubec Slovakia - 2 1
12 Vladimír Janál Czech Republic 0.5 -  
13 Branko Koludrović Croatia 1 -  
14 Ralf Krätschmer Germany 1.5 -  
15 François Labelle Canada 0.5 -  
16 Alexandre Leroux Canada 0.5 -  
17 Uwe Mehlhorn Germany - 1.5  
18 Dan Meinking USA - 3.5 4
19 Miodrag Mladenović Serbia 0.5 -  
20 Karol Mlynka Slovakia - 2  
21 Dieter Müller Germany - 1 4
22 Frank Müller Germany - 0.5  
23 Ion Murăraşu Romania 1 - -
24 Cornel Pacurar Canada 3.5 3.5  
25 Olivier Pucher France - - 0.5
26 Paul Răican Romania 3 - 3.5
27 Mečislovas Rimkus Lithuania - 3.5 2
28 Zoran Sibinović Serbia 4 4  
29 Ivan Skoba Czech Republic 1.5 1.5 1
30 Guy Sobrecases France - 0.5  
31 George P. Sphicas USA 1 1  
32 Radovan Tomašević Serbia 2 3  
33 Arno Tüngler Germany 4 2.5 -

Compositions:
T1-10 | T11-20 | T21-30 | T31-40 | T41-50 | T51-60
HC1-10 | HC11-20 | HC21-30 | HC31-40 | HC41-50


Last updated: 05.04.2012
# Composition Solution
T98 Dan Meinking, USA
T98 ChessProblems.ca, 02.04.2012
to the APS Workshop!
k1K2b2/1n1Q4/5p2/8/8/3q4/8/8
aser-s# 20 (2+5)


Solution: 

1.Qh7!! 2.Kd8** Sd6 3.Kd7 4.Ke6 5.Ke5*! f5 6.Kf6 7.Kg7*! Be7 8.Kg6 9.Kxf5** Se4 10.Kg5** Sf6 11.Kh6 12.Kg7 13.Kf8*! Bd8 14.Kf7 15.Ke6 16.Kd7** Sd5 17.Kc6 18.Kb6** Sc7 19.Ka6** Sb5 20.Qa7+ Sxa7 #
  • Definitions and detailed analysis here.
T97 Paul Răican, Romania
T97 ChessProblems.ca, 02.04.2012
dedicated to Dan Meinking
7k/6nN/6PP/2q1p3/8/7b/K7/n7
aser-s# 24 (4+6)


Solution: 

1.Kb3*! Sc2 2.Kb2 3.Kc1 4.Kd2 5.Ke3* Sd4 6.Kf4* e4 7.Kg3 8.Kh4 9.Kh5* S7f5 10.Kg5 11.Kf4 (11.Kf6? 12.Ke7* Sd6 13.Ke6* S4f5 14.Sf6 15.g7+ Sxg7+ 16.Ke7!) 12.Kxe4 13.Kd3 14.Kd2 15.Ke1 16.Kf1* Bg4 17.Ke2* Sf3 18.Ke3* S5d4 19.Kf4 20.Kg5* Se5 21.Kf6 22.Ke6* Sf5 23.Sf6 24.g7+ Sxg7 #
T96 Paul Răican, Romania
T96 ChessProblems.ca, 02.04.2012
dedicated to Nicolas Dupont
2kr1b1r/4p1pp/8/5b1P/P3BnR1/N4P2/1P1PP3/2BQ1KN1
ahser-dia 43 (13+9)


Solution: 

1.a4 2.Ra3 3.Rd3 4.Rxd7 5.Rd4 6.Rh4 7.g2-g4 8.Bf1-g2 9.Be4 10.f2-f3 11.Ke1-f2 12.Kf2-e3 13.Ke3-d4* Sb8-d7 14.Kd4-c5* Sf6 15.Kc5-c6* b5 16.Kc6xc7* Qd8-d5 17.Kc7-c6* Qd5-g5 18.Kb7* Bf5 19.Kxa7* 0-0-0 20.Kb6 21.Kc5 22.Kc4* b4 23.Kb3 24.Ka3* b3 25.Ka2* bxc2 26.Sa3 27.Kb1* cxd1=S! 28.Kc2 29.Kc3* Sf2 30.Kd4* Sd5 31.Kd3* Sxh1 32.Ke3* Sf4 33.Kf2* Sg3 34.Kf1* Sh5 35.gxh5 36.Rg4 37.h4 38.hxg5 39.g6 40.gxf7 41.fxg8=Q 42.Qb3 43.Qd1 dia
T95 Alberto Armeni, Italy
T95 ChessProblems.ca, 02.04.2012
R7/2p3b1/rQ3B1P/p7/kN6/N2p4/BP1Pp3/R2nK2b
ser-h= 8 (11+9)
Take & Make Chess

Solution: 

1.a*b4-c2 2.S*b2-b4 3.S*a2-b3 4.S*a1-c1 5.c*b6-b3 6.K*a3-b1 7.R*f6-a1 8.B*a8-a2 h*g7-b2 =
T94 Alberto Armeni, Italy
T94 ChessProblems.ca, 02.04.2012
r2bk3/1R6/1n6/R7/2BB4/1pN5/4P3/3K4
ser-h= 6 (7+5) C+
Take & Make Chess

Solution: 

1.S*c4-d3 2.B*a5-a2 3.0-0-0 4.K*b7-b4 5.K*c3-b1 6.R*d4-a1 e*d3-b2 =
  • Take & Make Chess: Upon capturing a unit X, a unit Y (K included) must continue from the square of capturing to make a single non-capturing move with the movement of X. If such a move does not exist, X cannot be captured by Y. The square of capturing is the square occupied by Y when the capture takes place (important for e.p. captures). Pawns must neither be placed on nor played to their first rank by the take&make condition. A pawn is promoted if and only if the final square of its move is on the 8th rank. Checks are as in orthodox chess.
T93 Jozef Holubec
Slovakia
T93 ChessProblems.ca, 13.03.2012
8/6N1/6P1/2p2R1B/2r3R1/6P1/3pPQ2/K6k
ser-h= 16 (9+4)
Circe


Solution: 

1.R*g4 5.c1=B 6.d1=R 8.Rg2 10.Kh3 13.R1d4 14.Bh6 15.Rg5 16.Rh4 g4 =

T92 Mečislovas Rimkus
Lithuania
T92 ChessProblems.ca, 13.03.2012
8/8/8/p5P1/P3r1PK/1kb1n1P1/5qP1/8
ser-= 25 (6+6)
Ghost Chess


Solution: 

3.g8=R 4.Re8 5.R*e4 9.g8=R 10.Rf8 11.R*f2 16.g8=R 17.Rg3 18.R*e3 23.g8=R 24.Rg1 25.Ra1 =

T91 György Bakcsi
Hungary
T91 ChessProblems.ca, 04.03.2012
4kN2/4P3/6P1/4p3/2pPK3/2R5/Qp1P4/8
ser-h= 11 (8+4) C+
Mirror Circe



Solution: 

1. b1S 2. Sa3 3. Sb5 4. Sc7 5. Se6 6. Sxf8 (Sb8) 7. Se6 8. Sc7 9. Sb5 10. Sa3 11. Sb1 Qa3 =

T90 György Bakcsi
Hungary
T90 ChessProblems.ca, 04.03.2012
2kb2R1/1p1p4/1P1P4/2p5/1Pnp2P1/3p4/3P4/2K1B3
ser-= 8 (8+8) C+
Mirror Circe



Solution: 

1. Bh4 2. Bxd8 3. Rg5 4. Bc7 5. Rxc5 (c2) 6. Rxc4 (Sb1) 7. Rxc2 8. Kxb1 =

T89 György Bakcsi
Hungary
T89 ChessProblems.ca, 04.03.2012
Bb1bB3/2p5/2P2p2/2k5/7K/8/8/8
ser-h# 7 (4+5) C+
2 Solutions



Solutions: 

1. Kb6 2. Ka7 3. Kxa8 4. Ba7 5. Kb8 6. Kc8 7. Bb8 Bd7 #

1. Kd6 2. Ke7 3. Kxe8 4. Be7 5. Kd8 6. Kc8 7. Bd8 Bb7 #

T88 György Bakcsi
Hungary
T88 ChessProblems.ca, 04.03.2012
6rk/3q3n/3pRpBp/8/r2B1QKn/2p2R1P/8/3bb3
ser-s# 5 (7+12) C+
Madrasi


Solution: 

1. Qxd6 2. Re8 3. Bc2 4. Rf4 5. Bxf5+ Sxf6 #

T87 Dan Meinking
USA
T87 ChessProblems.ca (v), 27.02.2012
dedicated to Paul Raican
6q1/6p1/3p4/2p3p1/2k3p1/1nP1p2p/4K3/n1B2B2
aser-s*z 22 (4+11) C?


Solution: 

1.Bg2! 2.Kf2* e2 3.Bf1! 4.Kg2* h2 5.Kf3* g3 6.Kf4* g4 7.Ke5* d5 8.Bg5! 9.Kf6* g6 11.Bf8 12.Kf7* Qh8 13.Kg7* Qh5 14.Kf6 15.Bh6 16.Kg5* Qh3 19.Kxe2 21.Be1 22.Kd1+ Qxf1 *z (pin-model autocheckzug)

  • After careful preparation, the King and Bishop nudge the Queen into the target square (h3). The wK returns to its diagram square after a long journey. Interesting roles for both white Bishops as well.
  • aser-s*z22 means "anti-parry-series self-autocheckzug in 22": white plays the series and is permitted to auto-check; when anti-parrying, black will resist white's plan; white's 22nd move forces black to put white in autocheckzug. (*)
  • More details here (document revised 13-Mar-2012).
  • Version 13.03.2012.
T86 Olivier Pucher & Paul Răican
France & Romania
T86 ChessProblems.ca (v), 27.02.2012
8/2bp3p/b3k3/2pr4/3r4/p2p1K2/8/8
aser-reflex= 47 (1+10) C?



Solution: 

1.Ke3 2.Kd2 3.Kc3 4.Kb4* c4 5.Kc5* Rd6 6.Kb4 7.Kc3 8.Kd2 9.Ke3 10.Kf4* R4d5 11.Kg5* Rd4! (11...Ke5 12.Kg4 13.Kf3 14.Ke2* d2 15.Kxd2* Kd4 16.Kc1 17.Kb2 a2 18.Kb3+ c3 19.Kb2* c2 20.Kc1 Kc3=) 12.Kh6 13.Kg7 14.Kf8 15.Ke8 16.Kd8* Bb8 17.Ke8 22.Kf4* R4d5 23.Ke3 24.Kd2 27.Kc5* Rd4 28.Kb6* R6d5 29.Kc6* d6 30.Kc7* Ba7 36.Kg5* Ke5 37.Kg4* Ke4 38.Kg3 39.Kf2 40.Ke2* d2 41.Ke1* d1B/S 42.Kxd1* Kd3 43.Kc1 44.Kb2* a2 45.Kb3* c3 46.Kb2* c2 47.Kc1 Kc3 =

  • A very elaborate APS Reflex Stalemate!
  • Version 04.03.2012.
T85 Dieter Müller
Germany
T85 ChessProblems.ca, 27.02.2012
2b5/1nrN4/k1pp4/2p5/2K1pN2/4prb1/5p2/8
ser-h# 6 (3+12) C+
Strict Circe
b) bKa6->h2


Solutions: 

a) 1.Sb7-d8 2.Rc7-b7 3.Rb7-b1 4.Ka6-b7 5.Kb7-c7 6.Sd8-b7 Sf4-e6 #

b) 1.Bg3-h4 2.Rf3-g3 3.Rg3-g1 4.Kh2-g3 5.Kg3-f3 6.Bh4-g3 Sd7-e5 #

  • Umnovs, switchbacks, model mates.
T84 Dieter Müller
Germany
T84 ChessProblems.ca, 27.02.2012
8/8/1p6/k3p1b1/p1K5/pB5p/b1P1p2P/8
ser-s# 15 (4+9) C+
Reversible Promotion


Solution: 

1.Kd3 2.Be6 3.B*h3 4.Bf5! 9.h8=B! (9.h8Q?) 10.B*e5! (10.Q*e5+?) 11.Bb2=P 12.c3 13.Kc2 14.Bd3 15.b4 a*b3 e.p. #

  • The key moves are 4, 12 and 13 - an interesting "evacuation" cycle:
    Bishop (A) - Pawn (B),
    Pawn (B) - King (C,
    King (C) - Bishop (A).
  • Reversible Promotion is a recent condition invented by Almiro Zarur - a promoted (white / black) piece turns back to a (white / black) pawn when it moves to the second (if white) or seventh rank (if black).
  • The results of the related 11th Japanese Sake / 3rd Brazilian Cachaça Tourney, from the 54th World Congress of Chess Composition.
T83 Dieter Müller
Germany
T83 ChessProblems.ca, 27.02.2012
8/1k1p4/p2B1p2/P4B1p/P5p1/P3p3/4Pp2/5K2
ser!= 10 (Black) (7+8) C+
b) ser-h= 18



Solutions: 

a) 9.Kh3 10.h4 =

b) 5.K*a4 17.Kh3 18.h4 a4 =

T82 Dieter Müller
Germany
T82 ChessProblems.ca, 27.02.2012
8/1p1rb3/1P1k3p/p1NPp1N1/4p3/4p3/P2P4/3R1K2
ser-h# 7 (8+9) C+
2 Solutions



Solutions: 

1.Be7*g5 2.Bg5-h4 3.Bh4-e1 4.Be1*d2 5.Bd2-e1 6.Be1-h4 7.Bh4-e7 Sc5*e4 #

1.Rd7-c7 2.Rc7*c5 3.Rc5-c2 4.Rc2*d2 5.Rd2-c2 6.Rc2-c7 7.Rc7-d7 Sg5*e4 #

  • Line-opening, switchback and pseudo-identical mate position.
T81 Nicolas Dupont
France
T81 ChessProblems.ca, 27.02.2012
8/5P2/3P3P/8/8/1P6/8/5k2
aser-h= 23 (4+1) C?


Solution: 

3.Kc4* b4 4.Kc5* b5 5.Kb6 6.Kc7* d7 7.Kc8* d8=B 8.Kc7* Bh4 9.Kc6* b6 10.Kc7* b7 11.Kd8* Bf2 12.Ke8* f8=S 13.Kf7 14.Kg7* h7 15.Kg8* h8=S 16.Kf7* Sg6 17.Ke7* Sh4 18.Ke6* Sg6 19.Kf5* Sf3 20.Kf4* Sh4 21.Kg3* Ba7 22.Kh2* Se1 23.Kh1 b8=B =

  • Beautiful APS composed by the Anti-Parry Series' inventor!
  • The similarities between APS and Vogtländer should be noted. In fact, the composition was partially tested by using the Vogtländer condition and the pser-a=>b stipulation, excluding only the last half-move.
HC63 Ivan Skoba
Czech Republic
HC63 ChessProblems.ca, 27.02.2012
dedicated to Ondra
5k2/8/P1P5/P1pNNP2/2PBBP2/4K3/2P2P2/8
ser-h= 351 (13+2) C+
PWC




Solution: 

16.Kxa5 [+wPa4] 17.Kxa6 [+wPa5] 38.Kxa4 [+wPa3] 39.Kxa5 [+wPa4] 60.Kxa3 [+wPa2] 61.Kxa4 [+wPa3] 82.Kxa2 [+wPb1] 83.Kxa3 [+wPa2] 104.Kxb1 [+wPc1] 105.Kxa2 [+wPb1] 126.Kxc1 [+wPd1] 127.Kxb1 [+wPc1] 148.Kxd1 [+wPe1] 149.Kxc1 [+wPd1] 170.Kxe1 [+wPf1] 171.Kxd1 [+wPe1] 192.Kxf1 [+wPg1] 193.Kxe1 [+wPf1] 214.Kxg1 [+wPh2] 215.Kxf1 [+wPg1] 236.Kxh2 [+wPh3] 237.Kxg1 [+wPh2] 258.Kxh3 [+wPh4] 259.Kxh2 [+wPh3] 280.Kxh4 [+wPh5] 281.Kxh3 [+wPh4] 302.Kxh5 [+wPh6] 303.Kxh4 [+wPh5] 325.Kxh6 [+wPh7] 326.Kg7 327.Kxh7 [+wPg7] 329.Kxg7 [+wPg8=wB] 331.Kxh5 [+wPh6] 351.Kf8 Bf7 =

  • Happy Birthday Ondra!
T80 Dan Meinking
USA
T80 ChessProblems.ca (v), 21.02.2012
dedicated to Nicolas Dupont
bq6/3K4/4R3/8/8/8/p5N1/k5r1
aser-s# 25 (3+5) C?



Solution: 

1.Rg6!! 5.Kh3 6.Sf4! 7.Kh2 8.Kh1** Rg2 9.Sd3! 10.Kh2** Rg3 11.Kg2** Rf3 12.Kg3** Rf4 13.Kf3** Re4 14.Kf4** Re5 15.Ke4** Rd5 16.Ke5** Rd6 17.Kd5** Rc6 18.Kd6** Rc7 19.Kc6**! Rb7 24.Kh1 25.Rg1+ Rb1 #

  • Systematic "collapsing" of eleven bR double-check batteries! White must force the Rook to climb the full "staircase", all the way to b7, from where it closes the Queen's guard of b1 (and the Bishop's guard of h1). The key-move looks a bit 'mysterious': the wR immediately unblocks e6, but doesn't move again till the very end. The wS serves to control auto-checks during the introduction, closing the g1-g4 and b8-h2 lines and opening the a8-h1 and b8-h2 lines in a precise manner.
  • aser-s#25 means "anti-parry-series self-mate in 25": white plays the series and is permitted to auto-check; when anti-parrying, black will resist white's plan; white's 25th move forces black to deliver checkmate.
  • Anti-Parry Series (APS) is a new fairy condition invented by Nicolas Dupont. The official definition and other details can be seen here (document revised 13-Mar-2012).
  • Version 04.03.2012.
T79 Dan Meinking
USA
T79 ChessProblems.ca, 13.02.2012
dedicated to George P. Sphicas 6bq/7p/6kb/8/1K6/n7/N7/8
aser-# 19 (2+6) C?


Solution: 

1.Ka4!! 4.Sd8! 5.Kb3* Sc4 6.Kc2 7.Kd2** Se3 8.Kd3 9.Kc4** Sd5 10.Kd3 11.Ke3** Sf4 12.Ke4 13.Kd5** Se6 14.Ke4! (Ke5*? Sg7!) 15.Kf4** Sg5 16.Ke5*! Kg7 17.Ke6** Sf7 18.Kf5 19.Se6 # (ideal)

  • Systematic "collapsing" of six bS double-check batteries, with lots of strategy throughout! White's first task is to force bS=>c4, but 1.Kb3*? allows 1...Kf7! and too much time is lost. The move 4.Sd8! guards f7 (to prevent ...Kf7) and eyes e6 (the mate-delivery square). White must avoid Kc3*?/Kd4*? always because black can reply ...Kf6! Even 14.Ke5*? Sg7! is too soon. Only 16.Ke5! can force ...Kg7.
  • aser-#19 means "anti-parry-series direct-mate in 19": white plays the series and is permitted to auto-check; when anti-parrying, black will resist white's plan; white forces mate on his 19th move.
  • Anti-Parry Series (APS) is a new fairy condition invented by Nicolas Dupont. The official definition and other details can be seen here (document revised 21-Feb-2012).
T78 Harald Grubert
Germany
T78 ChessProblems.ca, 11.02.2012

ser-h# 8* (9+2) C+
Nightrider a4, a5, b4, b6
2 variants


Solutions: 

1...Nb6-f4 #

1.a7*b6 2.b6-b5 3.b5*a4 4.Kh8-g7 5.Kg7-f7 6.Kf7-e6 7.Ke6-d6 8.Kd6-c5 Na5-b7 #

1.a7*b6 2.b6-b5 3.b5*c4 4.c4*d3 5.d3*e2 6.e2-e1=B 7.Be1*c3 8.Bc3-g7 Na4-e2 #
T77 Alberto Armeni
Italy
T77 ChessProblems.ca, 10.02.2012
8/5p1b/5b1n/8/4k3/8/2K3pr/8
ser-h# 6 (1+7) C+
Einstein Chess, AntiAndernach Chess



Solution: 


1.Sh6-g4=wP 2.Rh2-h6=wB 3.Bf6-h4=wS 4.f7-f5=wP 5.Bh7*f5=R 6.Rf5-f6=wB Sh4*g2=B #

  • All non-royal units change their color (black on diagram, white at the end of the solution).
HC62 Cornel Pacurar & Arno Tüngler
Canada & Germany
HC62 ChessProblems.ca, 07.02.2012
8/8/8/5k2/4n3/b2n1p2/1q3P2/3bK3
ser-+ 55 (2+7) C+



Solution: 

16.Kc4*d3 Ke1*d1 52.Ke3*f3 53.Kf3-e3 55.f3*e4 +

HC61 Arno Tüngler
Germany
HC61 ChessProblems.ca, 07.02.2012
q7/8/8/2p2k2/4n3/1n3p2/r4P2/1bK5
ser-+ 61 (2+8) C+



Solution: 

18.Kc4*b3 37.Kc1*b1 58.Ke3*f3 59.Kf3-e3 61.f3*e4 +

  • New series-check length record for 10 units. Previous record: 52 moves.
HC60 Cornel Pacurar & Arno Tüngler
Canada & Germany
HC60 ChessProblems.ca, 07.02.2012
q7/8/8/3K1k2/3pn3/3r1p2/r4P2/3bn3
ser-+ 71 (2+9) C+



Solution: 

16.Kf1*e1 32.Kc4*d3 49.Ke1*d1 66.Kc4*d4 68.Ke3*f3 69.Kf3-e3 71.f3*e4 +

  • New series-check length record for 11 units. Previous record: 64 moves.
HC59 Arno Tüngler
Germany
HC59 ChessProblems.ca, 07.02.2012
q7/8/8/2p2k2/1p1Kn3/1r3p2/r4P2/1bn5
ser-+ 79 (2+10) C+



Solution: 

18.Kd1*c1 36.Kc4*b3 55.Kc1*b1 76.Ke3*f3 77.Kf3-e3 79.f3*e4 +

  • New series-check length record for 12 units. Previous record: 68 moves.
HC58 Cornel Pacurar & Arno Tüngler
Canada & Germany
HC58 ChessProblems.ca, 07.02.2012
q7/8/3p4/5k2/1b1Pn3/3r1p2/r4P2/3bnK2
ser-+ 84 (3+10) C+



Solution: 

14.Kb5*b4 29.Kf1*e1 45.Kc4*d3 62.Ke1*d1 81.Ke3*f3 82.Kf3-e3 84.f3*e4 +

  • New series-check length record for 13 units. Previous record: 80 moves.
HC57 Arno Tüngler
Germany
HC57 ChessProblems.ca, 07.02.2012
q7/8/3p4/2p2k2/4nP2/1n1r1p2/r4P2/2bbK3
ser-+ 91 (3+11) C+



Solution: 

16.Kc4*d3 33.Ke1*d1 50.Kc4*b3 68.Kd1*c1 88.Ke3*f3 89.Kf3-e3 91.f3*e4 +

  • New series-check length record for 14 units. Previous record: 81 moves.
HC56 Cornel Pacurar & Arno Tüngler
Canada & Germany
HC56 ChessProblems.ca, 07.02.2012
q7/8/3p4/4pk2/1b2n3/3r1p2/r4P2/3bnK2
ser-F 88 (2+11) C+



Solution: 

14.Kb5*b4 29.Kf1*e1 45.Kc4*d3 62.Ke1*d1 81.Ke3*f3 82.Kf3-e3 84.f4*e5 85.e5*d6 87.d7-d8=Q 88.Qd8-d3 F

  • New series-pin length record for 13 units. Previous record: 85 moves.
T76 Ivan Skoba
Czech Republic
T76 ChessProblems.ca, 26.01.2012
2r5/4r3/3K4/7R/4NkNb/7P/8/1bB4R
ser-h# 215 (7+5) C+
Holes



Solution: 


7.Kd8 9.Rg8 11.Kf8 13.Re7 23.Ka1 24.Ba2 26.K*c1 28.Ka1 29.Bb1 39.Kf8 41.Rc8 43.Kd8 45.Re7 52.Kf4 55.Bg1 62.Kd8 64.Rg8 66.Kf8 68.Re7 78.Ka1 79.Ba2 86.K*h1 93.Ka1 94.Bb1 104.Kf8 106.Rc8 108.Kd8 110.Re7 117.Kf4 120.Bh4 127.Kd8 129.Rg8 131.Kf8 133.Re7 143.Ka1 144.Ba2 151.K*h3 158.Ka1 159.Bb1 169.Kf8 171.Rc8 173.Kd8 175.Re7 182.Kf4 185.Bg1 192.Kd8 194.Rg8 196.Kf8 198.Re7 208.Ka1 209.Ba2 215.Kg2 Se3 #

  • Pendulum of black King passing Zeller and Bishop traps. The entry tries to introduce basic series-mover manoeuvres that allow composing long problems. By using holes, the construction becomes very simple and light. We should also notice that the solution has three times 60 or more moves without capture!
T75 Alberto Armeni
Italy
T75 ChessProblems.ca, 26.01.2012
r5b1/8/4rP2/5P2/2P1N3/bpnP1p2/1np1pP2/1k2K3
ser-h== 12 (7+11) C?
Einstein Chess


Solution: 


1.R*f6=Q 2.B*c4=R 3.Rc-c8=B 4.B*f5=R 5.Rb5=B 6.B*d3=R 7.Rd1=B 8.Sd3=P 9.Bc1=S 10.Ra1=B 11.S3-a2=P 12.Qb2=R Sd2=P ==

  • Nice double-stalemate finale!
T74 Paul Răican
Romania
T74 ChessProblems.ca, 26.01.2012
8/1pp2r2/3b4/1pnk2rp/2pp4/Pp3nPp/1P3K2/5N2
ser-sxz 41 (5+14) C+
Isardam


Solution: 


5.Kb4*b5 7.a4-a5 14.Ke8*f7 26.Ke2*f3 29.Sg4-e5 31.Kf4*g5 32.Kg5*h5 37.g7-g8=R 39.Rg3*b3 40.Rb3-c3 41.b2-b4 c4*b3 ep. x

  • The first Series Self-Cap-Zug Isardam!
  • Isardam ("Madrasi" spelled backwards) is a variant of Madrasi in which only moves that do not lead to a Madrasi paralysis are legal.
  • C+ Popeye v4.55
HC55 Cornel Pacurar
Canada
HC55 ChessProblems.ca, 07.01.2012
8/8/1P6/8/8/1k1p4/3P4/K7
ser-r+ 20 (3+2) C+



Solution: 

9.Kc5-b5 11.b7-b8=S 12.Kb5-c5 14.Sa6-b4 19.Ke1-d1 20.Sb4-c2 d3*c2 +

  • New series reflex-check length record for 5 units.

Also, a new series reflex-check length record for 6 units:

Cornel Pacurar
HC56 ChessProblems.ca, 07.01.2012 8/8/1P6/8/8/1k1p4/3P4/rK6
ser-r+ 21 (3+3) C+
1.Kb1*a1 10.Kc5-b5 12.b7-b8=S 13.Kb5-c5 15.Sa6-b4 20.Ke1-d1 21.Sb4-c2 d3*c2 +

Previous record:

M. Tomašević
K177 Problemkiste (122) 04/1999 8/8/8/5p2/8/2p2k2/2P5/1BK5
ser-r+ 20 (3+3) C+
12.Kd4*c3 13.Kc3-d3 18.c7-c8=S 20.Sd6-e4 f5*e4 +
T73 Mečislovas Rimkus
Lithuania
T73 ChessProblems.ca, 07.01.2012
k5K1/1r6/P7/8/8/1p6/7P/r7
ser-s# 14 (3+4) C+
GhostChess



Solution: 


5.h7-h8=R 7.Rh7*b7 12.Kc4*b3 13.Kb3-b2[+buPb3] 14.Kb2*a1 uPb3-b2 #


T72 Arno Tüngler
Germany
T72 ChessProblems.ca, 26.12.2011
8/pp5p/P7/kB6/P7/2KQ4/PNP5/bR5B
ser-h!= 30 (10+5) C+
VerticalMirrorCirce



Solution: 


4.h3-h2 5.b7*a6 6.a6*b5 [+wBc1] 7.b5*a4 9.a3*b2 12.Ka3*a2 16.Ka5-b6 21.a2*b1=B 22.Bb1*c2 [+wPf2] 23.Bc2*d3 [+wQe1] 25.Be4*h1 26.Bh1-f3 27.h2-h1=R 28.Rh1*e1 29.Re1-f1 30.b2*c1=Q + Kc3-d3 !=

  • Excelsior x 3 (BQR)!
T71 Ján Golha & Cornel Pacurar
Slovakia & Canada
T71 ChessProblems.ca, 19.12.2011
8/8/8/8/8/2Nk4/8/3B4
ser-h= 8 (2+1) C+
EquipollentsCirce
b) bKc4-->h1
c) bKc4-->d7
d) wBd6-->g3


Solutions: 

a)
1.Kc4-d3 2.Kd3*e4 [+wSf5] 3.Ke4*f5 [+wSg6] 4.Kf5-g5 5.Kg5-h6 6.Kh6*g6 [+wSf6] 7.Kg6-g7 8.Kg7-h8 Bd6-f8 =

b) bKc4->h1
1.Kh1-g2 2.Kg2-f3 3.Kf3*e4 [+wSd5] 4.Ke4*d5 [+wSc6] 5.Kd5*d6 [+wBd7] 6.Kd6-c7 7.Kc7-b7 8.Kb7-a8 Bd7-c8 =

c) bKc4->d7
1.Kd7-e6 2.Ke6-f5 3.Kf5*e4 [+wSd3] 4.Ke4-e3 5.Ke3*d3 [+wSc3] 6.Kd3-c2 7.Kc2-b2 8.Kb2-a1 Bd6-a3 =

d) wBd6->g3
1.Kc4-d4 2.Kd4*e4 [+wSf4] 3.Ke4-f3 4.Kf3*g3 [+wBh3] 5.Kg3*f4 [+wSe5] 6.Kf4-g3 7.Kg3-h2 8.Kh2-h1 Se5-f3 =

  • Wenigsteiner, 4-corner echoes.
  • No white King.
  • Version 20.12.2011.
T70 Cornel Pacurar
Canada
T70 ChessProblems.ca, 19.12.2011
8/3b4/8/3K4/k7/8/6P1/8
pser-h= 8 (2+2) C+
ParrainCirce, BrunnerChess
2 Solutions



Solutions: 


1.Ka4-b5 2.Kb5-c6 + Kd5-e4 3.Bd7-f5 + Ke4*f5 4.Kc6-d5 [+bBg4] + Kf5*g4 5.Kd5-e4 [+bBh3] + Kg4-f3 + 6.Bh3*g2 + Kf3*e4 [+wPf3] 7.Bg2-h1 8.Bh1*f3 + Ke4*f3 [+wPg2] =


1.Bd7-f5 2.Bf5-e4 + Kd5*e4 3.Ka4-b3 [+bBf3] + g2*f3 4.Kb3-c4 [+bBg4] 5.Bg4*f3 + Ke4-d3 [+wPe2] + 6.Bf3*e2 + Kd3*c4 [+wPd3] 7.Be2-f1 8.Bf1*d3 + Kc4*d3 [+wPe2] =

  • Wenigsteiner Parry-Series echoes.
HC54 Arno Tüngler
Germany
HC54 ChessProblems.ca, 12.12.2011
8/8/5K2/8/8/2B5/N1kN4/8
ser-h= 37 (4+1) C+
Vertical Mirror Circe


Solution: 

1.Kc2-d3 2.Kd3-e3 3.Ke3-f4 4.Kf4-g4 5.Kg4-h5 6.Kh5-h6 7.Kh6-h7 8.Kh7-g8 9.Kg8-f8 10.Kf8-e8 11.Ke8-d7 12.Kd7-c6 13.Kc6-b5 14.Kb5-a4 15.Ka4-a3 16.Ka3*a2 [+wSg1] 17.Ka2-a3 18.Ka3-a4 19.Ka4-b5 20.Kb5-c6 21.Kc6-d7 22.Kd7-e8 23.Ke8-f8 24.Kf8-g8 25.Kg8-h7 26.Kh7-h6 27.Kh6-h5 28.Kh5-g4 29.Kg4-f4 30.Kf4-e3 31.Ke3-d3 32.Kd3*c3 [+wBf1] 33.Kc3-d4 34.Kd4-e3 35.Ke3-f2 36.Kf2*g1 [+wSb1] 37.Kg1-h1 Sd2-f3 =

  • Serieshelpstalemate Vertical Mirror Circe move-length record for 5 units.
HC53 Arno Tüngler
Germany
HC53 ChessProblems.ca, 12.12.2011
7B/8/8/8/7k/1N3K2/8/8
ser-h= 27 (3+1) C+
Vertical Mirror Circe


Solution: 

1.Kh4-g5 2.Kg5-f5 3.Kf5-e6 4.Ke6-d5 5.Kd5-c4 6.Kc4*b3 [+wSg1] 7.Kb3-c4 8.Kc4-d5 9.Kd5-e6 10.Ke6-f7 11.Kf7-g8 12.Kg8*h8 [+wBf1] 13.Kh8-g7 14.Kg7-f6 15.Kf6-e5 16.Ke5-d4 17.Kd4-c3 18.Kc3-d2 19.Kd2-e1 20.Ke1*f1 [+wBc1] 21.Kf1*g1 [+wSb1] 22.Kg1-f1 23.Kf1-e1 24.Ke1-d1 25.Kd1*c1 [+wBf1] 26.Kc1-d1 27.Kd1-e1 Bf1-e2 =

  • Serieshelpstalemate Vertical Mirror Circe move-length record for 4 units.
T69 Mečislovas Rimkus
Lithuania
T69 ChessProblems.ca, 11.12.2011
4k3/8/3Kp3/8/7P/8/8/8
ser-h# 7 (2+2) C+
Take & Make Chess, Ghost Chess
b) wKd6->g6



Solutions: 

a)

5.e2-e1Q 6.Qe1xh4=Qh5 (gPh4) 7.Qh5-g5 gPh4xg5=g8Q (gQg5) #

b) wKd6->g6

5.e2-e1B 6.Be1-d2 7.Bd2-g5 h4xg5=Qd8 (gBg5) #
T68 Ivan Skoba & Arno Tüngler
Czech Republic & Germany
T68 ChessProblems.ca, 28.10.2011
4k3/1p4K1/8/8/8/8/2P5/8
pser-h= 12 (2+2) C+
Vertical Mirror Circe
b) wKg7->a5



Solutions: 

a)

1.b7-b5 2.b5-b4 3.b4-b3 4.b3*c2 [+wPf2] 5.c2-c1=Q 6.Qc1-g1 + Kg7-f6 7.Qg1-g6 + Kf6-e5 8.Qg6-g3 + f2-f4 9.Qg3-g5 + f4-f5 10.Qg5-g7 + f5-f6 11.Qg7-c7 + Ke5-e6 12.Qc7-e7 + f6*e7 =

b) wKg7->a5

1.b7-b6 + Ka5*b6 [+bPg7] 2.g7-g5 3.g5-g4 4.g4-g3 5.g3-g2 6.g2-g1=Q + Kb6-c6 7.Qg1-b6 + Kc6-d5 8.Qb6-b3 + c2-c4 9.Qb3-b5 + c4-c5 10.Qb5-b7 + c5-c6 11.Qb7-d7 + c6*d7 + 12.Ke8-d8 Kd5-d6 =

  • Parry-Series Wenigsteiner and chameleon-echoes.
  • For a few more Vertical Mirror Circe Parry-Series please see ifaybish.com TT7 tourney results.
T67 Alberto Armeni, Italy
T67 ChessProblems.ca, 11.10.2011
4N2k/5p2/8/p7/2pNP1pr/6p1/6P1/3Kb2n
ser-= 10 (5+9) C+
Einstein Chess

Solution: 

1.Sd4-b3=P 2.b3*c4=S 3.Sc4*a5=B 4.Ba5*e1=R 5.Re1*h1=Q 6.Qh1-e1=R 7.Re1-e2=B 8.Be2-f3=S 9.Sf3*h4=B 10.Bh4-f6=S =
  • Again the full spectrum of Einstein transformations!
HC52 Ivan Skoba
Czech Republic
HC52 ChessProblems.ca, 11.10.2011

Position a:
6r1/3p4/rn3P2/1pbp1p2/1P1p4/np1k4/3P2Bq/3KB2r

Position b:
6r1/3p4/1n3P2/2bp1p2/1P1p4/1p1k4/3P2Bq/3KB2r
ser-a=>b 112 (6+14)/(6+11) C+





Solution: 

2.Kc1-b2 4.Bf2-g1 9.Kf2-f3 11.Bf2-g3 12.Kf3-f4 15.Bh5-g6 16.Kf4-g5 20.Bg6-e8 22.Kg6-f7 24.Bf8-d6 29.Kb7*a6 34.Ke7-f7 36.Bf8-g7 38.Kg6-g5 42.Bd6-g3 43.Kg5-f4 46.Bf3-g2 47.Kf4-f3 49.Bf2-g1 55.Kb2*a3 61.Kf2-f3 63.Bf2-g3 64.Kf3-f4 67.Bh5-g6 68.Kf4-g5 72.Bg6-e8 74.Kg6-f7 76.Bf8-d6 82.Ka6*b5 88.Ke7-f7 90.Bf8-g7 92.Kg6-g5 96.Bd6-g3 97.Kg5-f4 100.Bf3-g2 101.Kf4-f3 103.Bf2-g1 108.Kc1-b2 110.Bf2-e1 112.Kc1-d1 a=>b

T66 Dan Meinking, USA
T66 ChessProblems.ca, 12.09.2011

pser-h# 5 (2+2) C+
Orphan h1
b) bKg4->b4
c) wKb2->g6

Solutions: 

a)

1.d2-d1=S + Kb2-c2 2.Sd1-e3 + Kc2-d2 3.Se3-f1 + Kd2-e2 4.Sf1-g3 + Ke2-f2 5.Kg4-h3 Oh1*g3 #

b) bKg4->b4

1.d2-d1=R 2.Rd1-b1 + Kb2-c2 3.Kb4-a3 4.Ka3-a2 5.Ka2-a1 Oh1*b1 #

c) wKb2->g6

1.d2-d1=Q 2.Qd1-b1 + Oh1-e4 + 3.Qb1-g1 4.Kg4-h4 + Kg6-f5 5.Qg1-g4 + Oe4*g4 #

  • Wenigsteiner, three promotions and nice Orphan chameleon-echoes.
  • The second Parry-Series employing an Orphan (the first being: Cornel Pacurar, F471 Problem Paradise, 2010).
T65 Dan Meinking, USA
T65 ChessProblems.ca, 12.09.2011

pser-h= 11 (2+2) C+
Take & Make Chess, PWC
b) wKd1->f2

Solutions: 

a)

1.Ke4-e3 2.Ke3-f2 3.Kf2*e2-e4 [+wPf2] 4.Ke4-f3 5.e5-e4 6.e4-e3 7.e3-e2 + Kd1-d2 8.e2-e1=Q + Kd2-d3 9.Qe1-e3 + f2*e3-e8=Q [+bQf2] 10.Qf2-f1 + Qe8-e2 + 11.Kf3*e2-e1 [+wQf3] + Qf3*f1-f3 =

b) wKd1->f2

1.Ke4-d4 2.Kd4-c3 3.Kc3-d2 4.Kd2*e2-e4 [+wPd2] 5.Ke4-f4 6.e5-e4 7.e4-e3 + Kf2*e3-e2 [+bPf2] 8.f2-f1=Q + Ke2*f1-g1 [+bQe2] 9.Qe2-e3 + d2*e3-e8=Q [+bQd2] 10.Qd2-d4 + Qe8-e3 + 11.Kf4*e3-h3 [+wQf4] + Qf4*d4-f4 =

  • Chris Feather and Geoff Foster have recently published nine Take&Make + PWC series-helpmates in a special issue of Fairings (15b) - this is a first Parry-Series Take&Make + PWC!
  • Wenigsteiner.
T64 Dan Meinking & Cornel Pacurar, USA & Canada
T64 ChessProblems.ca, 25.08.2011

pser-hs# 7 (2+2) C+
Princess c6
Andernach Grasshopper h3
b) wPRc6->c5

Solutions: 

a)

1.Kg5-h5 2.AGh3-h6 3.AGh6-b6 [bPRc6] 4.PRc6-d4 + Kf2-g2 5.PRd4-e3 + Kg2-h3 6.AGb6-f2 [wPRe3] PRe3-f4 + 7.AGf2-f5 [bPRf4] #

b) wPRc6->c5

1.Kg5-h4 2.AGh3-h5 3.AGh5-b5 [bPRc5] + Kf2-f1 4.PRc5-d3 + Kf1-g1 5.PRd3-e2 + Kg1-h2 6.AGb5-f1 [wPRe2] PRe2-f3 + 7.AGf1-f4 [bPRf3] #

  • Wenigsteiner, Chameleon Echoes, Parry-Series Help-Selfmate.
T63 Cornel Pacurar, Canada
T63 ChessProblems.ca, 25.08.2011

pser-hs# 5 (2+2) C+
Isardam
2 Solutions
b) wKd3->e2

Solutions: 

a)

1.Kf5-f6 2.Qg5-d2 + Kd3-e4 3.Qd2-g2 + Qb8-e5 + 4.Qg2-g7 Ke4-f5 + 5.Qg7-g5 #

1.Kf5-f4 2.Qg5-g3 + Kd3-d4 3.Qg3-f2 + Qb8-e5 4.Qf2-g3 Kd4-e4 + 5.Qg3-e3 #

b) wKd3->e2

1.Qg5-d2 + Ke2-f3 2.Qd2-d5 + Qb8-f4 3.Kf5-e5 4.Qd5-d6 Kf3-e4 + 5.Qd6-d4 #

1.Qg5-g4 + Ke2-f2 2.Qg4-d4 + Qb8-g3 3.Kf5-f4 4.Qd4-e5 Kf2-f3 + 5.Qe5-e3 #

  • Wenigsteiner, Echoes, Parry-Series Help-Selfmate.
T62 Geoff Foster, Australia
T62 ChessProblems.ca, 22.08.2011

ser-h# 8 (2+1) C+
PWC, Take & Make Chess
Lion b6
2 Solutions
b) wLIb6->g6

Solutions: 

a)

1.Ke8*f8-d7 [+wSe8] 2.Kd7*e8-d6 [+wSd7] 3.Kd6*d7-c5 [+wSd6] 4.Kc5*b6-e6 [+wLIc5] 5.Ke6-d5 6.Kd5*c5-e7 [+wLId5] 7.Ke7*d6-b7 [+wSe7] 8.Kb7-a8 Se7-c6 #

1.Ke8-e7 2.Ke7*f8-e6 [+wSe7] 3.Ke6*e7-d5 [+wSe6] 4.Kd5*e6-d4 [+wSd5] 5.Kd4*d5-c7 [+wSd4] 6.Kc7-b7 7.Kb7*b6-g1 [+wLIb7] 8.Kg1-h1 Sd4-f3 #

b) wLIb6->g6

1.Ke8*f8-d7 [+wSe8] 2.Kd7*e8-d6 [+wSd7] 3.Kd6*d7-e5 [+wSd6] 4.Ke5*d6-e4 [+wSe5] 5.Ke4*e5-f7 [+wSe4] 6.Kf7-g7 7.Kg7*g6-b1 [+wLIg7] 8.Kb1-a1 Se4-c3 #

1.Ke8-e7 2.Ke7*f8-e6 [+wSe7] 3.Ke6*e7-f5 [+wSe6] 4.Kf5*g6-d6 [+wLIf5] 5.Kd6-e5 6.Ke5*f5-d7 [+wLIe5] 7.Kd7*e6-g7 [+wSd7] 8.Kg7-h8 Sd7-f6 #

  • Beautiful 4-corner echoes!
T61 Ivan Skoba, Czech Republic
T61 ChessProblems.ca, 22.08.2011

serc-!= 75 (8+10)
Madrasi
Consequent


Solution: 

See the solution and detailed explanation, as received from the author, here.
  • An interesting interpretation of the Series Consequent genre! More details will be posted soon.
HC51 Zoran Sibinović
Serbia
HC51 ChessProblems.ca, 10.04.2011
1NB5/2p5/2P2K1k/BR1P3Q/1P6/1R1N2P1/3PP1P1/8
ser-h# 51 (15+2) C+



Solution: 

1.Kh6*h5 8.Kd8*c8 17.Kg4*g3 18.Kg3*g2 20.Kf1*e2 21.Ke2*d2 25.Kc4*b5 27.Kc4*d5 39.Kc8*b8 43.Kb5*c6 44.Kc6-d6 47.c4*d3 49.d2-d1=R 51.Rc1-c6 Rb3-d3 #

  • New ideal serieshelpmate length record - last move by Rook.
  • For previous record see T52.
  • 01.06.2011: New record by Vladimír Janál - 54 moves

Compositions:
T1-10 | T11-20 | T21-30 | T31-40 | T41-50 | T51-60
HC1-10 | HC11-20 | HC21-30 | HC31-40 | HC41-50