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. No limit to the number of problems per author for the tourney (tournament entries are labeled "T[n]" in the table below). 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): ChessProblems.ca 2010 Award
Judge for 2011:  Paul Răican (Romania): ChessProblems.ca 2011 Award
Judge for 2012:  Arno Tüngler (Germany): ChessProblems.ca 2012 Award
Judge for 2013:  Ivan Skoba (Czech Republic): ChessProblems.ca 2013 Award
Judge for 2014:  Nicolas Dupont (France): ChessProblems.ca 2014 Award
Judge for 2015:  George P. Sphicas (USA): ChessProblems.ca 2015 Award and correction
Judge for 2016:  Hans Gruber (Germany): ChessProblems.ca 2016 Award
Judge for 2017:  Paz Einat (Israel)
Judge for 2018:  Manfred Rittirsch (Germany)

Starting 2014, all originals are published in the ChessProblems.ca Bulletin!

Participants

# Composer Country T 2010 2011 2012 2013
1 Alberto Armeni Italy 13 - 4 4 5
2 György Bakcsi Hungary 5 1 - 4 -
3 Bojan Bašić Serbia 2 1 - 1 -
4 Vlaicu Crişan Romania 0.5 0.5 - - -
5 Mihail Croitor Moldova 1 1 - -
6 Nicolas Dupont France 2 - - 1 1
7 Itamar Faybish Belgium 0.5 0.5 - - -
8 Dominique Forlot France 1.5 - - 1.5 -
9 Geoff Foster Australia 3 - 1 - 2
10 Ján Golha Slovakia 6 1.5 2.5 1 1
11 Harald Grubert Germany 1.5 - 0.5 1 -
12 Joost de Heer Netherlands 5 - - 3 2
13 Jozef Holubec Slovakia 4 - 2 1 1
14 Vladimír Janál Czech Republic 0.5 0.5 - - -
15 Branko Koludrović Croatia 3 1 - - 2
16 Vaclav Kotesovec Czech Republic 8 - - 3 5
17 Ralf Krätschmer Germany 1.5 1.5 - - -
18 François Labelle Canada 0.5 0.5 - - -
19 Alexandre Leroux Canada 0.5 0.5 - - -
20 Uwe Mehlhorn Germany 1.5 - 1.5 - -
21 Dan Meinking† USA 8 - 3.5 4 0.5
22 Miodrag Mladenović Serbia 0.5 0.5 - - -
23 Karol Mlynka Slovakia 2 - 2 - -
24 Dieter Müller Germany 5 - 1 4 -
25 Frank Müller Germany 0.5 - 0.5 - -
26 Ion Murăraşu† Romania 1 1 - - -
27 Daniel Novomesky Slovakia 5 - - 4 1
28 Cornel Pacurar Canada 10 3.5 3.5 2 1
29 Olivier Pucher France 0.5 - - 0.5 -
30 Paul Răican Romania 8.5 3 - 3.5 2
31 Mečislovas Rimkus Lithuania 8 - 3.5 3.5 1
32 Zoran Sibinović Serbia 14 4 4 3 3
33 Ivan Skoba Czech Republic 5 1.5 1.5 2 -
34 Guy Sobrecases France 0.5 - 0.5 - -
35 George P. Sphicas USA 6 1 1 - 4
36 Radovan Tomašević Serbia 5 2 3 - -
37 Arno Tüngler Germany 11 4 2.5 - 4.5
T     151 (+8 w) 30 38 47 36

Compositions:
T1-10 | T11-20 | T21-30 | T31-40 | T41-50 | T51-60 | T61-70 | T71-80
HC1-10 | HC11-20 | HC21-30 | HC31-40 | HC41-50 | HC51-60 | HC61-70


Last updated: 30.03.2014
# Composition Solutions
HC79 Vaclav Kotesovec
Czech Republic
HC79 ChessProblems.ca, 04.02.2014
dedicated to C. Poisson for his 50th birthday

ser-# 50 (7+1) C+
White Maximummer
Grasshopper d1, d6, d8, g2, g3, g7



Solution: 

1.Gd7 2.Gh2 3.Gg8 4.Gg1 5.Ga7 6.Gg1 7.Ga5 8.Ge7 9.Ga7 10.Ga3 11.Ge7 12.Ga6 13.Ga3 14.Ga2 15.Ga7 16.Ga1 17.Ga8 18.Gh1 19.Ga8 20.Ge7 21.Ga3 22.Ga2 23.Gf3 24.Gg3 25.Ge4 26.Ga4 27.Ge8 28.Gf4 29.Ga4 30.Ga5 31.Gc6 32.Gg2 33.Ge8 34.Gh1 35.Gc7 36.Gb8 37.Gf8 38.Ga3 39.Ga5 40.Ga6 41.Gc7 42.Gc6 43.Gb7 44.Ge8 45.Ga8 46.Gf8 47.Ga3 48.Ga7 49.Gc8 50.Gb8 #

T158 Zoran Sibinović
Serbia
T158 ChessProblems.ca, 31.12.2013
8/p2k4/Pp6/3PP3/6p1/6p1/8/4K3
ser-h= 24 (4+5) C+



Solution: 

1.b6-b5 2.b5-b4 3.b4-b3 4.b3-b2 5.b2-b1=B 6.Bb1-d3 7.Bd3*a6 8.Ba6-b7 9.a7-a5 10.a5-a4 11.a4-a3 12.a3-a2 13.a2-a1=B 14.Ba1*e5 15.Kd7-d6 16.Kd6*d5 17.Kd5-e4 18.Ke4-f3 19.Kf3-g2 20.Kg2-g1 21.Bb7-h1 22.g3-g2 23.Be5-h2 24.g4-g3 Ke1-e2 =

  • Kings and Pawns only. B+B promotions, ideal stalemate.
T157 Zoran Sibinović
Serbia
T157 ChessProblems.ca, 31.12.2013
8/3p4/8/3k4/8/6p1/p1P1P2p/4K3
ser-h= 14 (3+5) C+



Solution: 

1.Kd5-e4 2.d7-d5 3.d5-d4 4.d4-d3 5.d3*c2 6.c2-c1=S 7.a2-a1=Q 8.Qa1-a8 9.Sc1*e2 10.Ke4-f3 11.Kf3-g2 12.Kg2-g1 13.Qa8-h1 14.g3-g2 Ke1*e2 =

  • Kings and Pawns only. S+Q promotions, ideal stalemate.
T156 Zoran Sibinović
Serbia
T156 ChessProblems.ca, 31.12.2013
8/8/8/8/8/p3p1pp/p7/k3K3
ser-h= 15 (1+6) C+



Solution: 

1.Ka1-b1 2.a2-a1=R 3.Ra1-a2 4.Ra2-h2 5.a3-a2 6.a2-a1=Q 7.Qa1-a8 8.Kb1-c2 9.Kc2-d3 10.Kd3-e4 11.Ke4-f3 12.Kf3-g2 13.Kg2-g1 14.Qa8-h1 15.g3-g2 Ke1-e2 =

  • Kings and Pawns only. R+Q promotions.
T155 Ján Golha
Slovakia
T155 ChessProblems.ca, 28.12.2013
8/8/8/8/2bkNK2/8/8/8
pser-h# 5 (2+2) C+
Take&Make, Anti-Take&Make
2 Solutions



Solutions: 

1.Kd4*e4-f6[+wSg5] 2.Kf6*g5-f7[+wSe6] 3.Bc4*e6-g5[+wSf8]+ Kf4*g5-h6[+bBe7] 4.Be7*f8-h7[+wSg6] 5.Kf7*g6-h8[+wSf8] Sf8*h7-g6[+bBg8] #

1.Kd4*e4-d6[+wSc5] 2.Kd6*c5-d3[+wSe6] 3.Bc4*e6-g5[+wSd4]+ Kf4*g5-c1[+bBf6] 4.Bf6*d4-b3[+wSc2] 5.Kd3*c2-a1[+wSd4] Sd4*b3-c2[+bBa2] #

  • Just one parry-move in each solution but an amazing 100% (14 half-moves!) Take&Make & Anti-Take&Make thematic density! Two Wenigsteiner corner-echoes.
T154 Arno Tüngler
Germany
T154 ChessProblems.ca, 28.12.2013
4k3/1q5p/p2PpP2/3p1r2/PKp1nP2/P1P1P2P/6r1/B6b
ser-x 61 (10+11) C+



Solution: 

1.Kb4-a5 12.Bd8-b6 13.Ka5-b4 14.a4-a5 15.a3-a4 16.Kb4-a3 26.Bc1-b2 30.Kc1-d1 34.Be1-f2 36.Ke2-f3 37.Bf2-g3 39.Kg4-h4 50.Bh6-g5 52.Kh5-h6 60.Bb6-c7 61.Kh6*h7 x

  • New series-capture overall length-record. Previous record: also 61 moves but with 22 units (Cornel Pacurar & Arno Tüngler, T4 ChessProblems.ca, 2.04.2010).
T153 Branko Koludrović & Arno Tüngler
Croatia & Germany
T153 ChessProblems.ca, 28.12.2013
6b1/1q1p1p1b/p2PrP2/3k1r2/K1r2P2/P1P1P2P/P5r1/B6b
ser-x 83 (10+22) C+



Solution: 

1.Ka4-a5 14.Bc5-b4 16.Ka4-b3 18.a4-a5 19.a2-a4 20.Kb3-a3 32.Bc1-b2 36.Kc1-d1 40.Be1-f2 42.Ke2-f3 43.Bf2-g3 45.Kg4-h4 56.Bh6-g5 60.Kg7-f8 69.Bd8-e7 71.Ke8-d8 82.Bb6-c7 83.Kd8*d7 x

  • New series-capture overall length-record with promoted force. Previous record: 78 moves (Branko Koludrović & Arno Tüngler, feenschach IX 2006).
T152 Arno Tüngler
Germany
T152 ChessProblems.ca, 28.12.2013
1q6/1P5b/B3K3/1N1p1nr1/1Pk5/1RP2p1r/1PP2P2/RB1Nr1nb
ser-sF 156 (13+11) C+



Solution: 

7.Ka3-a2 9.Ra3-a5 11.Ka3-a4 13.Ra3-b3 16.Ka2-a1 17.Bb1-a2 21.Kd2*e1 25.Kb1-a1 26.Ba2-b1 29.Ka3-a4 31.Ra3-a1 33.Ka3-a2 35.Ra3-b3 44.Kf6*g5 53.Ka3-a2 55.Ra3-a5 57.Ka3-a4 59.Ra3-b3 62.Ka2-a1 63.Bb1-a2 69.Kf1*g1 75.Kb1-a1 76.Ba2-b1 79.Ka3-a4 81.Ra3-a1 83.Ka3-a2 85.Ra3-b3 96.Kg4*h3 107.Ka3-a2 109.Ra3-a5 111.Ka3-a4 113.Ra3-b3 116.Ka2-a1 117.Bb1-a2 124.Kg1*h1 131.Kb1-a1 132.Ba2-b1 135.Ka3-a4 137.Ra3-a1 139.Ka3-a2 141.Ra3-b3 152.Kg4*f3 153.Kf3-g2 154.f2-f4 155.Kg2-h2 156.Sd1-e3 + Sf5*e3 F

  • New series self-pin overall length-record with promoted force. Previous record: 142 moves (Branko Koludrović, T134 ChessProblems.ca, 1.03.2013).
T151 Geoff Foster
Australia
T151 ChessProblems.ca, 22.10.2013

ser-h# 3 (0+1+3n) C+
Neutral Rook h2, Neutral Bishop g8, Neutral Knight d4
b) h2->d2, +c) g8->f1 +d) d2->c7
Take&Make, Anti-Take&Make (Circe-Take&Make)
2 Solutions



Solutions: 

  • a)
    1.nSd4-f3 2.Ke4*f3-g1[+nSg5] 3.Kg1*h2-h8[+nRa2] nBg8*a2-b2[+nRa8] #
    1.Ke4*d4-c6[+nSe6] 2.nBg8*e6-f4[+nSc7] 3.Kc6*c7-a8[+nSb5] nBf4*h2-g2[+nRh8] #

    b)
    1.nRd2-d1 2.Ke4*d4-c2[+nSe2] 3.Kc2*d1-h1[+nRd8] nRd8*g8-h7[+nBd5] #
    1.nRd2*d4-f3[+nSb3] 2.Ke4*f3-c3[+nRf7] 3.Kc3*b3-a1[+nSd2] nBg8*f7-f6[+nRa7] #

    c)
    1.Ke4*d4-b3[+nSf3] 2.nSf3*d2-d7[+nRa2] 3.Kb3*a2-a8[+nRe2] nBf1*e2-e4[+nRa2] #
    1.Ke4*d4-b3[+nSc2] 2.Kb3*c2-a1[+nSb4] 3.nRd2-d3 nBf1*d3-c3[+nRd1] #

    d)
    1.Ke4*d4-f3[+nSe2] 2.nBf1*e2-f4[+nSg3] 3.Kf3*g3-h1[+nSe2] nBf4*c7-c6[+nRh7] #
    1.Ke4*d4-e6[+nSb5] 2.nSb5*c7-f7[+nRc4] 3.Ke6*f7-h8[+nSg5] nBf1*c4-d4[+nRc8] #

  • ChessProblems.ca TT4 theme (Wenigsteiner Series and Parry-Series Four-Corners Echoes). See the TT4 Award here.
  • "This problem uses the same material and fairy conditions as my =1 Prize in the 4th Bulgarian Wine Tourney (section B), but that was a h#2 and only had one twin, whereas this one is a ser-h#3. Actually, with this force the only difference between a h#2 and a ser-h#3 is that in the latter the black king can move at any stage!" (Author). See the 4th Bulgarian Wine Tourney 2013 award here.
  • C+ Popeye v4.63.
T150 George P. Sphicas
USA
T150 ChessProblems.ca, 10.10.2013
8/1k6/4Qq2/2Kprp2/8/8/8/8
pser-s# 15 (2+5)



Solution: 



  • Try: 1.Qb6+? Not forced is 1... Qxb6#?? 1... Ka8? is shorter 2.Qe6 4.Kc7 5.Qc8+ etc, but try is defeated by ... Kc8! and white loses control.

    Try: 1.Qd7+? 1... Ka6?? 3.Qb6+ 1... Ka8 is shorter 2.Qe6 4.Kc7 5.Qc8+ etc, but try is defeated by 1... Kb8! , which needs one move longer: 2.Qe6 4.Kd7 6.Qc8+ Ka7 7.Kc7 8.Qb8+ etc.

    Solution: 1.Kd6 2.Kd7 3.Qd6 4.Qc7+ Ka8 (if 4... Ka6 5.Kc8 6.Qb7 etc or 5.Qb8 6.Kc7 7.Qb7) 5.Qc8+ Ka7 6.Kc7 7.Qb8+ Ka6 8.Qb7+ Ka5 9.Qc6 10.Qd6 12.Kc5 13.Qb8 14.Qb4+ Ka6 15.Qb6+ Qxb6#

    After 7... Ka6 it is premature to try to return to c5, e.g. 9.Qe6?? 11.Kc5 12.Qb6+, since Qe6 is check.

T149 George P. Sphicas
USA
T149 ChessProblems.ca, 10.10.2013
8/3np3/2pqp3/2k3p1/2p3rp/3pPp1r/3P1P2/b4Knb
ser-!xz 106 (4+16)



Solution: 



  • 17.Kxg4 35.Kxg1 54.Kxh3 74.Kxh1 94.Kxf3 95.Ke4 100.f8=R 102.Rxd7 103.Rxd6 104.Rd4 105.Ke5 106.e4 !xz

  • Minor promotion (R).
T148 George P. Sphicas
USA
T148 ChessProblems.ca, 10.10.2013
8/2pp4/2kqp3/3p2p1/4n1rp/3P1p1r/2P1PP2/b4Knb
ser-!xz 110 (5+15)



Solution: 



  • 18.Kxg4 37.Kxg1 57.Kxh4 78.Kxh1 99.Kxf3 100.Ke3 103.fxe6 105.e8=S 106.Sxc7 108.Sc3 109.Kd4 110.e3 !xz

  • Minor promotion (S).
T147 George P. Sphicas
USA
T147 ChessProblems.ca, 10.10.2013
dedicated to the Memory of Dan Mainking
8/2p1p3/2P1k3/n1P1q1b1/P1p1P1rp/2p2p1r/4PP2/5Knb
ser-!xz 117 (7+14)




Solution: 



  • 20.Kxg4 41.Kxg1 63.Kxh3 86.Kxh1 109.Kxf3 110.Kg4 111.f4 112.fxe5 113.Kxg5 116.Kd4 117.e3 !xz

  • New ser-!xz length record. Previous record: ser-!xz114, Dan Meinking & George P. Sphicas, GPSC0432 StrateGems 57, 2012.
T146 Cornel Pacurar
Canada
T146 ChessProblems.ca, 03.10.2013
8/8/8/1k5b/8/5P2/ppKp2p1/1QR1RB2
ser-h!= 4 (6+6) C+



Solution: 

1.g2*f1=Q 2.a2*b1=R 3.b2*c1=B 4.d2*e1=S + Kc2-d1 !=

  • This is probably only the second ever series-mover with full AUW by the moving side in 4 moves (with no fairy elements)! The only other known example is a series selfmate, also with 12 units, below:

Michel Caillaud
The Problemist 1994

ser-s# 4 (5+7) C+
1.a7*b8=R 2.d7*c8=B 3.b7*a8=Q 4.e7-e8=S + Ba4*e8 #

  • See a related very interesting article just published by George P. Sphicas in StrateGems 64, October-December 2013: "Seriesmovers with AUW: Shortest and Longest".
T145 Jozef Holubec
Slovakia
T145 ChessProblems.ca, 03.09.2013
bnq5/6p1/5R1P/2n1B3/p5Qp/r2R4/2P3B1/2K3k1
ser-h= 10 (8+9)
Circe, Madrasi


Solution: 

1.Sc5-e4 2.Se4-g3 3.Kg1-h2 4.Kh2-h3 5.Sg3-f1 6.Sf1-e3 7.Se3xc2 8.Sc2-b4 9.Sb4-a6 10.Sa6-c5 Rf6-a6=

  • Knight circuit.
T144 Alberto Armeni
Italy
T144 ChessProblems.ca, 16.08.2013
4k2r/q2p4/8/4P3/3r4/8/5K2/8
pser-h# 6 (2+5) C+



Solution: 

1.Rd4-h4+ Kf2-f3 2.Rh4-h7 3.0-0+ Kf3-e4 4.d7-d5+ e5*d6 ep. 5.Qa7-e7+ d6*e7 6.Kg8-h8 e7*f8=Q #

  • Valladao in miniature Parry-Series form. See also T139.
T143 Geoff Foster
Australia
T143 ChessProblems.ca, 01.08.2013
dedicated to the ChessProblems.ca TT4 participants

ser-h# 4 (1+1+2) C+
Neutral Queen b7, Neutral Bishop d7
b) b7->g5, c) d7->f6, d) d1->c1
Take&Make
PWC


Solutions: 

a) 1.nQxd7-h3[nBb7] 2.Kc7 3.Kxb7-a8[nBc7] 4.nQh2, nBxh2-h1[nQc7]#

b) 1.nBc6 2.Kxc6-h1[nBd6] 3.nQg3 4.nQxd6-b8[nBg3], nBxb8-a8[nQg3]#

c) 1.Ke6 2.Kxf6-h8[nBe6] 3.nQf7 4.nQxe6-a2[nBf7], nBxa2-a1[nQf7]#

d) 1.Kxd7-e6[nBd6] 2.nBe5 3.Kxe5-a1[nBe6] 4.nQc8+, nBxc8-h8[nQe6]#
  • ChessProblems.ca TT4 theme (Wenigsteiner Series and Parry-Series Four-Corners Echoes). See the TT4 Award here.
T142 Nicolas Dupont
France
T142 ChessProblems.ca, 28.06.2013
4B3/8/4k1P1/2P5/1B4K1/3b4/8/8
ser= 8 (5+2) C+
Back-Home
Madrasi

Solution: 

1.g7 2.Bg6 3.g8=R 4.Rf8 5.Rf5 6.Be8 7.Bb5 8.c6=

Back-Home is a new fairy condition invented by the T142's author, with the following definition:
  1. If a piece can legally move to the square it occupied in the diagram position, it must move to this back-home square.
  2. Back-home moves are prevalent to the virtual capture of the opponent King by any piece, i.e. checks are fairy.
  3. If more back-home moves are legal, the side-on-move chooses which one to play.
  4. The back-home square of a Pawn which is promoted during the solution is the initial diagram square of this Pawn.
  • See the Back-Home introductory article here.
HC78 Arno Tüngler
Germany
HC78 ChessProblems.ca, 07.06.2013
(after M. Tomašević & M. Ott)
8/2pp4/2kq2p1/5K1n/5ppr/3p4/3P2bn/4b2B
ser-s+ 110 (3+13) C+



Solution: 

18.Kd1*e1 36.Kg5*h4 57.Kg1*h2 78.Kg5*g4 98.Kf2*g2 99.Kg2-f3 102.Bf1*d3 109.Ka5-a6 110.Bd3-b5 + Kc6-d5/c5 +

  • New series selfcheck move-length record for 16 units!
HC77 Arno Tüngler
Germany
HC77 ChessProblems.ca, 07.06.2013
(after M. Tomašević & M. Ott)
8/2pp4/2kq2p1/K5rn/5ppr/3p4/3P2bn/4b2B
ser-s+ 120 (3+14) C+



Solution: 

11.Kh6*g5 28.Kd1*e1 46.Kg5*h4 67.Kg1*h2 88.Kg5*g4 108.Kf2*g2 109.Kg2-f3 112.Bf1*d3 119.Ka5-a6 120.Bd3-b5 + Kc6-d5/c5 +

  • Surprising new series selfcheck move-length record for 17 units!
T141 Daniel Novomesky
Slovakia
T141 ChessProblems.ca, 26.05.2013
8/8/2k5/8/8/7K/5P2/8
ser-!= 15 (2+1) C+
b) f2->d6
c) c6->c8
Sentinelles PionAdvers
MaximumWhite 8 MaximumBlack 3

Solutions: 

a) 1.Kh3-h2[+bPh3] 2.Kh2-g3[+bPh2] 3.Kg3-h4[+bPg3] 4.Kh4*h3[+bPh4] 5.Kh3-g2 6.Kg2-h1 7.f2*g3 8.g3*h4 9.h4-h5 10.h5-h6 11.h6-h7 12.h7-h8=R 13.Rh8-g8 14.Rg8-g3 15.Rg3-g2[+bPg3] !=

1.d6-d7 2.d7-d8=B 3.Bd8-c7 4.Bc7-h2[+bPc7] 5.Bh2-g3[+bPh2] 6.Bg3-e1[+bPg3] 7.Kh3-g2 8.Be1*g3 9.Kg2-h3[+bPg2] 10.Kh3*h2[+bPh3] 11.Kh2-g1 12.Bg3*c7[+bPg3] 13.Bc7-b8 14.Bb8*g3 15.Bg3-h2[+bPg3] !=

1.Kh3-g3[+bPh3] 2.Kg3-f3[+bPg3] 3.Kf3-e3[+bPf3] 4.Ke3*f3[+bPe3] 5.Kf3-e4 6.Ke4-d4 7.f2*e3 8.Kd4-c5[+bPd4] 9.e3*d4 10.Kc5-b6[+bPc5] 11.d4*c5 12.Kb6-a7[+bPb6] 13.c5*b6 14.Ka7-a8[+bPa7] 15.b6*a7 !=
HC76 Arno Tüngler
Germany
HC76 ChessProblems.ca, 26.05.2013
N7/P1p1p2B/2PqPr2/2rkrP2/2pr4/P4rn1/1r1PK3/b3n1r1
ser-sZ c4 202 (9+15) C+



Solution: 

1.Ke2-d1 8.Bd3-c2 9.d2-d3 11.Kd2-c3 12.Bc2-b3 14.Kb4-a4 25.Ba6-b5 29.Kb7-c8 38.Be8-d7 40.Kd8-e8 51.Bh5-f7 53.Kf8-g7 54.Bf7-g6 56.Kh6-g5 58.Bh5-g4 60.Kh4-h3 68.Ba4-c2 70.Kh2*g1 72.Kh2-h3 80.Bh5-g4 82.Kh4-g5 84.Bh5-g6 86.Kh6-g7 87.Bg6-f7 89.Kf8-e8 100.Bc8-d7 102.Kd8-c8 111.Ba4-b5 115.Ka5-a4 126.Bd1-b3 128.Kb4-c3 129.Bb3-c2 131.Kd2-c1 133.Bd1-e2 135.Kd1*e1 137.Kd1-c1 139.Bd1-c2 141.Kd2-c3 142.Bc2-b3 144.Kb4-a4 155.Ba6-b5 159.Kb7-c8 168.Be8-d7 170.Kd8-e8 181.Bh5-f7 183.Kf8-g7 184.Bf7-g6 186.Kh6-g5 188.Bh5-g4 190.Kh4-h3 198.Ba4-c2 200.Kg2*f3 201.Bc2-a4 202.d3*c4 + Rd4/Rc5/Kd5*c4 z

  • New series-selftargetsquare overall record with promoted force!
  • Previous record: 171 moves (T127 ChessProblems.ca, 23.01.2013).
HC75 Arno Tüngler
Germany
HC75 ChessProblems.ca, 26.05.2013
8/2p1p3/2PqPr2/2rB1P2/2PRBr1b/p2PrN2/1r2B3/r1nKn1bk
ser-s+ 197 (11+15) C+



Solution: 

11.Ba4-c2 13.Kd2-c3 14.Bc2-b3 16.Kb4-a4 27.Ba6-b5 31.Kb7-c8 40.Be8-d7 42.Kd8-e8 53.Bh5-f7 55.Kf8-g7 56.Bf7-g6 58.Kh6-h5 69.Bh3-g4 70.Kh5*h4 71.Kh4-g5 73.Bh5-g6 75.Kh6-g7 76.Bg6-f7 78.Kf8-e8 89.Bc8-d7 91.Kd8-c8 100.Ba4-b5 104.Ka5-a4 115.Bd1-b3 117.Kb4-c3 118.Bb3-c2 120.Kd2-d1 131.Bf1-e2 132.Kd1*e1 133.Ke1-d1 144.Ba4-c2 146.Kd2-c3 147.Bc2-b3 149.Kb4-a4 160.Ba6-b5 164.Kb7-c8 173.Be8-d7 175.Kd8-e8 186.Bh5-f7 188.Kf8-g7 189.Bf7-g6 191.Kh6-g5 197.Bh3-g2 + Rb2*g2 +

  • New series-selfcheck overall record with promoted force!
  • Previous record: 153 moves (T124 ChessProblems.ca, 10.01.2013).
HC74 Arno Tüngler
Germany
HC74 ChessProblems.ca, 26.05.2013
N7/2p1p3/2PqPr2/2rkqP2/2br3b/P2P1p1B/1r5n/q1nK2n1
ser-<> 205 (8+16) C+



Solution: 

9.Ba4-c2 11.Kd2-c3 12.Bc2-b3 14.Kb4-a4 25.Ba6-b5 29.Kb7-c8 38.Be8-d7 40.Kd8-e8 51.Bh5-f7 53.Kf8-g7 54.Bf7-g6 56.Kh6-h5 67.Bh3-g4 68.Kh5*h4 69.Kh4-g5 71.Bh5-g6 73.Kh6-g7 74.Bg6-f7 76.Kf8-e8 87.Bc8-d7 89.Kd8-c8 98.Ba4-b5 102.Ka5-a4 113.Bd1-b3 115.Kb4-c3 116.Bb3-c2 118.Kd2-d1 129.Bf1-e2 132.Kf2*g1 135.Ke1-d1 146.Ba4-c2 148.Kd2-c3 149.Bc2-b3 151.Kb4-a4 162.Ba6-b5 166.Kb7-c8 175.Be8-d7 177.Kd8-e8 188.Bh5-f7 190.Kf8-g7 191.Bf7-g6 193.Kh6-g5 195.Bh5-g4 197.Kh4-h3 205.Ba4-d1 <>

  • New ser-PW overall record with promoted force!
HC73 Arno Tüngler
Germany
HC73 ChessProblems.ca, 26.05.2013
8/2p1p3/2PqPr2/2r2P2/2Pr3b/2rPrp2/1r2n2n/kB2K1n1
ser-!=173 (7+15) C?



Solution: 

1.Ke1-d1 13.Bf1*e2 24.Ba4-c2 26.Kd2*c3 27.Bc2-b3 29.Kb4-a3 31.Ba4-b5 36.Kb7-c8 45.Be8-d7 47.Kd8-e8 58.Bh5-f7 60.Kf8-g7 61.Bf7-g6 63.Kh6-h5 74.Bh3-g4 75.Kh5*h4 76.Kh4-g5 78.Bh5-g6 80.Kh6-g7 81.Bg6-f7 83.Kf8-e8 94.Bc8-d7 96.Kd8-c8 105.Ba4-b5 110.Ka4-a3 112.Ba4-b3 114.Kb4-c3 115.Bb3-c2 117.Kd2-c1 119.Bd1-e2 123.Kf2*g1 127.Kd1-c1 129.Bd1-c2 131.Kd2-c3 132.Bc2-b3 134.Kb4-a3 136.Ba4-b5 141.Kb7-c8 150.Be8-d7 152.Kd8-e8 163.Bh5-f7 165.Kf8-g7 166.Bf7-g6 168.Kh6-g5 170.Bh5-g4 172.Kh4-h3 173.Bg4*f3 !=

  • New series-autostalemate overall record with promoted force! It extends the previous C+ record (HC72 ChessProblems.ca, below) with 8 moves, but it cannot be validated by Popeye.
HC72 Arno Tüngler
Germany
HC72 ChessProblems.ca, 24.05.2013
8/2p1p3/2PqPr2/2r1bP2/2Pr3b/3Prp2/1r2n2n/kB2K1n1
ser-!= 165 (7+15) C+



Solution: 

1.Ke1-d1 13.Bf1*e2 14.Kd1-c1 16.Bd1-c2 18.Kd2-c3 19.Bc2-b3 21.Kb4-a3 23.Ba4-b5 28.Kb7-c8 37.Be8-d7 39.Kd8-e8 50.Bh5-f7 52.Kf8-g7 53.Bf7-g6 55.Kh6-h5 66.Bh3-g4 67.Kh5*h4 68.Kh4-g5 70.Bh5-g6 72.Kh6-g7 73.Bg6-f7 75.Kf8-e8 86.Bc8-d7 88.Kd8-c8 97.Ba4-b5 102.Ka4-a3 104.Ba4-b3 106.Kb4-c3 107.Bb3-c2 109.Kd2-c1 111.Bd1-e2 115.Kf2*g1 119.Kd1-c1 121.Bd1-c2 123.Kd2-c3 124.Bc2-b3 126.Kb4-a3 128.Ba4-b5 133.Kb7-c8 142.Be8-d7 144.Kd8-e8 155.Bh5-f7 160.Kh6-g5 162.Bh5-g4 164.Kh4-h3 165.Bg4*f3 !=

T140 Arno Tüngler
Germany
T140 ChessProblems.ca, 13.05.2013
N5B1/P1p1p3/2Pqrr2/2rk1P2/2pr4/P4rn1/1r1PK3/b3n1r1
ser-Z e3 208 (8+15) C+



Solution: 

1.Ke2-d1 8.Bd3-c2 9.d2-d3 11.Kd2-c3 12.Bc2-b3 14.Kb4-a4 25.Ba6-b5 29.Kb7-c8 38.Be8-d7 40.Kd8-e8 51.Bh5-f7 53.Kf8-g7 54.Bf7-g6 56.Kh6-g5 58.Bh5-g4 60.Kh4-h3 68.Ba4-c2 70.Kh2*g1 72.Kh2-h3 80.Bh5-g4 82.Kh4-g5 84.Bh5-g6 86.Kh6-g7 87.Bg6-f7 89.Kf8-e8 100.Bc8-d7 102.Kd8-c8 111.Ba4-b5 115.Ka5-a4 126.Bd1-b3 128.Kb4-c3 129.Bb3-c2 131.Kd2-c1 133.Bd1-e2 135.Kd1*e1 137.Kd1-c1 139.Bd1-c2 141.Kd2-c3 142.Bc2-b3 144.Kb4-a4 155.Ba6-b5 159.Kb7-c8 168.Be8-d7 170.Kd8-e8 181.Bh5-f7 183.Kf8-g7 184.Bf7-g6 186.Kh6-g5 188.Bh5-g4 190.Kh4-h3 199.Bd1-e2 201.Kg2*f3 203.Kf2-e1 204.f5*e6 206.Bd1-c2 208.Kd2-e3 Z

  • An extraordinary achievement: 208 moves, with no fairy units or conditions! This is now the overall record with promoted force for all series-movers stipulations!
  • Previous record: 176 moves (T122 ChessProblems.ca, 4.01.2013)
T139 Alberto Armeni
Italy
T139 ChessProblems.ca, 13.05.2013
2R5/6N1/8/b4p1q/2pp3p/2ppp2r/bP6/1k2K2R
pser-h# 4 (5+12) C+
Take&Make


Solution: 

1.c3-c2 + b2-b4
2.c4*b4
3.c2-c1=R + Sg7*h5-d1
4.Rc1*d1-b2 0-0 #

  • Valladao.
T138 Vaclav Kotesovec
Czech Republic
T138 ChessProblems.ca, 24.03.2013

ser-# 67 (5+1) C+
WhiteMaximummer
G=Grasshopper a4, b6, d3, f8


Solution: 

1.Kc2 2.Gd1 3.Gd4 4.Ga7 5.Gc5 6.Gb4 7.Gc1 8.Gb3 9.Ge3 10.Gf4 11.Ga4 12.Gd1 13.Gd4 14.Ge4 15.Gf5 16.Gb1 17.Gc5 18.Gb5 19.Gb6 20.Gc1 21.Ge3 22.Gf4 23.Gg4 24.Gg5 25.Ga5 26.Gh4 27.Gc5 28.Gc1 29.Gc3 30.Gf4 31.Ge4 32.Gb1 33.Gb6 34.Gb7 35.Gb8 36.Gg3 37.Gh3 38.Ge5 39.Gd6 40.Gc7 41.Gc1 42.Gc3 43.Gb3 44.Gb8 45.Ge5 46.Gf6 47.Gc6 48.Gc1 49.Gf4 50.Gg5 51.Gd5 52.Ge4 53.Gb1 54.Gg3 55.Gd3 56.Gh3 57.Ge5 58.Gd6 59.Gc7 60.Gc1 61.Gd4 62.Gc3 63.Gb3 64.Ge6 65.Gc4 66.Gb3 67.Gb2 #

HC71 Vaclav Kotesovec
Czech Republic
HC71 ChessProblems.ca, 24.03.2013

ser-h# 61 (5+1) C+
RH=Rookhopper g4, h3, h4, h8
PWC


Solution: 

1.Kb2 2.Kc3 3.Kd4 4.Ke5 5.Kf6 6.Kg7 7.K*h8(RHg7) 8.Kh7 9.Kg6 10.Kf6 11.K*g7(RHf6) 12.Kg6 13.Kf5 14.K*g4(RHf5) 15.K*h4(RHg4) 16.Kg3 17.Kg2 18.K*h3(RHg2) 19.Kh2 20.Kh1 21.K*g2(RHh1) 22.Kf3 23.Ke4 24.K*f5(RHe4) 25.K*g4(RHf5) 26.Kf3 27.Ke3 28.Kd4 29.K*e4(RHd4) 30.K*f5(RHe4) 31.Ke5 32.K*f6(RHe5) 33.Ke7 34.Kd6 35.Kc5 36.K*d4(RHc5) 37.K*e4(RHd4) 38.Kd3 39.Kc3 40.K*d4(RHc3) 41.K*e5(RHd4) 42.Kd5 43.Kc4 44.Kb3 45.Kb2 46.K*c3(RHb2) 47.K*d4(RHc3) 48.Kc4 49.Kb3 50.Ka2 51.Kb1 52.K*b2(RHb1) 53.K*c3(RHb2) 54.Kb4 55.K*c5(RHb4) 56.Kb6 57.Ka5 58.K*b4(RHa5) 59.Ka3 60.K*b2(RHa3) 61.Ka1 Kc2 #

T137 Joost de Heer
Netherlands
T137 ChessProblems.ca, 19.03.2013

ser-h= 40 (3+9) C+
L=Locust d8, f8
AlphabeticChess


Solution: 

5. a1=L 10. b1=L 15. c1=L 20. d1=L 25. e1=L 30. f1=L 35. g1=L 40. h1=L Kb8 =

  • 8 Excelsiors and Locust promotions!
T136 Joost de Heer
Netherlands
T136 ChessProblems.ca, 11.03.2013
8/pppp1p2/8/4K3/4P3/6Pk/8/7N
ser-h# 33 (4+6) C+
AlphabeticChess


Solution: 

5.a1=R 7.Rh2 12.b1=B 13.B*e4 18.c1=Q 19.Qh6 24.d1=Q 26.Qdh4 27.Bg2 32.f1=S 33.S*g3 Sf2 #

  • AUW + 1, Excelsiors x 5.
T135 Paul Răican
Romania
T135 ChessProblems.ca, 04.03.2013
3n4/1pK5/7r/p1p1k3/8/p1nPb1P1/P7/8
aser-reflex= 34 (4+9)
Circe


Solution: 

1.Kb8! 2.Ka7 3.Ka6* b6 7.Kc6* Se6 8.Kb7 9.Ka6 10.Kxa5*(pa7) b5 11.Kb4* c4 12.Kb3* cxd3(pd2) (and now the black Knight c3 must be brought to g8) 13.Kxc3(Sb8) 14.Kb4 15.Kc5*! Cd4 16.Kb4 17.Ka5 18.Ka6* S8c6 19.Kb7 20.Kxc6(Sg8)* Se6 21.Kxb5(pb7) 22.Kc4 23.Kxd3(pd7) 27.Kg5* Sf4 30.Kf2* Bxd2 33.Kg5 34.g4 Be1 =

  • Anti-Parry Series Reflex Stalemate.
T134 Branko Koludrović
Croatia
T134 ChessProblems.ca, 01.03.2013
4bnr1/b7/2kp1r1n/7n/1P3p2/KPpP1P1b/2P3p1/1Bb2b2
ser-sF 142 (7+15) C+
(pin absolute and effective)


Solution: 

2.Ka2-a1 3.Bb1-a2 5.Kb1*c1 7.Kb1-a1 8.Ba2-b1 14.Ka6*a7 20.Ka2-a1 21.Bb1-a2 29.Kh2*h3 37.Kb1-a1 38.Ba2-b1 48.Kd8*e8 58.Ka2-a1 59.Bb1-a2 69.Kh4*h5 79.Kb1-a1 91.Ke7*f6 102.Ka2-a1 103.Bb1-a2 114.Kh5*h6 125.Kb1-a1 126.Ba2-b1 141.Kd4-c4 142.b4-b5 + Kc6-b7,d7,b6,c7 F

  • New overall series self-pin move length record (promoted force). Previous record: 141 moves, Arno Tüngler, HC66 ChessProblems.ca, 17.01.2013.
  • C+ WinChloe
T133 Vaclav Kotesovec
Czech Republic
T133 ChessProblems.ca, 01.03.2013

ser-h# 205 (2+7) C+
BlackMaximummer
AN=Antilope h1, G=Grasshopper e3, RH=Rookhopper g1, BH=Bishophopper e1, e2


Solution: 

1.RHg7 2.ANe5 3.ANa2 4.ANd6 5.ANg2 6.ANc5 7.ANf1 8.ANb4 9.BHa5 10.ANf1 11.ANc5 12.ANg2 13.ANd6 14.ANh3 15.ANe7 16.Ge8 17.ANa4 18.Ge1 19.ANd8 20.ANg4 21.ANc7 22.RHb7 23.ANg4 24.ANc1 25.ANf5 26.ANb8 27.ANe4 28.ANa7 29.ANd3 30.ANg7 31.RHh7 32.ANc4 33.ANf8 34.ANb5 35.BHa6 36.ANf2 37.ANc6 38.ANg3 39.ANd7 40.RHc7 41.ANa3 42.ANe6 43.Ge7 44.ANa3 45.ANd7 46.ANh4 47.ANe8 48.ANb4 49.Ga3 50.ANf7 51.ANc3 52.RHc2 53.ANf7 54.ANb4 55.ANe8 56.ANh4 57.ANd7 58.ANg3 59.Gh3 60.ANd7 61.Gc8 62.Gc1 63.Gg5 64.ANg3 65.ANc6 66.RHc7 67.ANg3 68.ANd7 69.ANh4 70.ANe8 71.ANb4 72.ANf1 73.ANc5 74.Gb5 75.ANf1 76.ANb4 77.ANe8 78.ANh4 79.ANd1 80.ANg5 81.Gh5 82.ANc2 83.RHc1 84.ANg5 85.ANd1 86.ANh4 87.ANd7 88.ANa3 89.ANe6 90.ANh2 91.ANd5 92.Gc5 93.ANg1 94.ANc4 95.ANf8 96.ANb5 97.ANe1 98.ANh5 99.ANd2 100.BHe1 101.ANh5 102.ANd8 103.ANg4 104.ANc7 105.ANf3 106.ANb6 107.ANe2 108.BHf1 109.ANh6 110.ANd3 111.ANg7 112.ANc4 113.BHb5 114.ANg1 115.ANd5 116.ANh2 117.ANe6 118.ANa3 119.ANd7 120.ANg3 121.ANc6 122.ANf2 123.RHc6 124.Ge3 125.Gg5 126.Ga5 127.BHg3 128.BHe5 129.BHd7 130.ANb5 131.ANe1 132.ANh5 133.ANd2 134.Ge1 135.ANh5 136.ANd8 137.ANg4 138.BHh3 139.ANc7 140.ANf3 141.ANb6 142.ANe2 143.ANa5 144.ANe8 145.ANb4 146.Ga5 147.ANe8 148.ANh4 149.ANd7 150.BHc8 151.ANa3 152.ANe6 153.ANb2 154.ANf5 155.BHg4 156.ANc1 157.Gf5 158.BHe6 159.ANg4 160.ANc7 161.ANf3 162.ANb6 163.ANe2 164.ANa5 165.ANd1 166.ANg5 167.ANc2 168.Gb1 169.RHc1 170.ANg5 171.Gh7 172.ANd1 173.ANh4 174.ANd7 175.ANg3 176.ANc6 177.RHc7 178.Gb7 179.ANf2 180.ANb5 181.ANf8 182.ANc4 183.ANg7 184.RHh7 185.ANd3 186.RHa7 187.ANh6 188.ANe2 189.ANb6 190.ANf3 191.Gg2 192.ANc7 193.ANg4 194.ANc1 195.ANf5 196.ANb2 197.Ga2 198.Gf7 199.RHg7 200.ANf5 201.ANc1 202.ANg4 203.ANd8 204.ANa4 205.ANe7 d*e5 #

  • C+ Alybadix
T132 Vaclav Kotesovec
Czech Republic
T132 ChessProblems.ca, 01.03.2013

ser-h# 123 (2+6) C+
BlackMaximummer
G=Grasshopper a4, RH=Rookhopper e1, e3, BH=Bishophopper b7


Solution: 

1.Ge8 2.BHe4 3.BHc6 4.RHe4 5.Sf6 6.Gb5 7.RHe8 8.Ge5 9.Gb8 10.RHa8 11.RHe8 12.Gf8 13.Gf5 14.RHf8 15.Gc5 16.RHf5 17.Gf8 18.RHc5 19.Gf5 20.BHe4 21.RHc8 22.RHf8 23.Gc5 24.Gc8 25.RHb8 26.Gg8 27.Gc4 28.RHg8 29.Gc8 30.RHb8 31.Gh8 32.RHa8 33.Ge5 34.Ge8 35.RHf8 36.RHf5 37.RHf8 38.RHc5 39.RHc8 40.Gb8 41.RHg8 42.RHa8 43.Gh8 44.Ge5 45.RHh8 46.Gb8 47.RHa8 48.Gd6 49.Gg6 50.BHh7 51.BHf5 52.Ge4 53.Ge8 54.RHf8 55.Sh5 56.RHf4 57.RHd8 58.RHd4 59.RHc4 60.RHc8 61.Gb8 62.Gd6 63.Gf8 64.RHg8 65.Gf4 66.Gb8 67.Gh8 68.Gc3 69.Gc8 70.Gg4 71.RHg3 72.RHh4 73.Ge6 74.Gc4 75.RHb4 76.Gc8 77.Gg4 78.Ga4 79.Sf6 80.Se8 81.Sd6 82.Sb7 83.RHb8 84.Sc5 85.Sa6 86.Ga7 87.Gd7 88.BHc8 89.Gd4 90.Sb4 91.RHb3 92.RHh3 93.Ga4 94.Sa2 95.Ga1 96.Sb4 97.Sc2 98.Se1 99.Gf1 100.Sc2 101.Sb4 102.Sa2 103.Sc1 104.Gb1 105.Se2 106.Sf4 107.Sh5 108.RHh6 109.Sg7 110.RHg8 111.RHb8 112.Se8 113.Sd6 114.RHc6 115.Sc4 116.RHc3 117.Sb2 118.Sa4 119.Sc5 120.RHc6 121.Sa4 122.Sb6 123.Gb7 d6 #

  • C+ Alybadix
T131 Alberto Armeni
Italy
T131 ChessProblems.ca, 01.03.2013
5K2/r7/8/3k4/2N3B1/p7/R3rR2/8
ser-h# 5 (5+4) C+
b) bKd5->h4
Circe



Solutions: 

a) 1.Kd5*c4 [+wSb1] 2.Kc4-b3 3.Kb3*a2 [+wRh1] 4.Ka2-a1 5.Re2-a2 Sb1-c3 #

b) 1.Kh4*g4 [+wBf1] 2.Kg4-g3 3.Kg3*f2 [+wRa1] 4.Kf2-g1 5.Re2-h2 Bf1-d3 #

T130 Alberto Armeni
Italy
T130 ChessProblems.ca, 01.03.2013
2r2b2/rP6/8/8/8/2k2K2/1R6/8
ser-h# 4 (3+4) C+
b) bKc3->c7
AntiCirce



Solutions: 

a) 1.Ra7-a1 2.Kc3-d2 3.Kd2-e1 4.Ra1-a8 b7*a8=R[wRa8->h1] # (and not 4... b7*c8=R [wRa8->h1] + 5.Ra8-a1!)

b) 1.Bf8-b4 2.Kc7-b6 3.Kb6-a6 4.Bb4-a5 b7*c8=B[wBc8->f1] # (and not 4... b7-b8=S + 5.Rc8*b8 [bRb8->h8]!)

T129 Alberto Armeni
Italy
T129 ChessProblems.ca, 01.03.2013
8/7b/8/8/8/2Kp4/3p1ppp/2kB2N1
ser-h# 5 (3+7) C+
b) bKc1->e3



Solutions: 

a) 1.Kc1*d1 2.Kd1-c1 3.d2-d1=R 4.d3-d2 5.Bh7-b1 Sg1-e2

b) 1.f2-f1=S 2.Ke3-f2 3.Kf2*g1 4.Kg1-h1 5.g2-g1=B Bd1-f3 #

  • Zilahi, three minor promotions.
T128 Paul Răican
Romania
T128 ChessProblems.ca, 17.02.2013
Dedicated to Arno Tüngler
1qnb4/8/b1rp1k2/nKp5/2p1P1P1/4rpP1/1pP1p3/8
ser-sF 124 (5+14) C+
b) ser-!F 123




Solutions: 

a) 1.Ka4 2.c3 20.Kxd8 39.Kxa5 59.Kxc6 80.Kxa6 103. Kxc4 123.Ka3 124.g5+! K~ F (124.e5+? Rxe5!)

b) The first 123 moves of a)

  • New series self-pin and series auto-pin move length records for 19 units!
  • Previous records: ser-sF 121 moves / 20 units and 123 moves / 21 units, ser-!F 122 moves / 19 units!
  • C+ WinChloe
T127 Branko Koludrović & Arno Tüngler
Croatia & Germany
T127 ChessProblems.ca, 23.01.2013
1qnb1K1B/6RR/b1r2k2/n4P1R/2p1n1P1/b1P1r1P1/1p2n2p/2b5
ser-sZh1 171 (9+15)



Solution: 

1.Kf8-g8 3.Rh7-h3 7.Kh5-h4 9.Rh5-h7 11.Kh5-h6 13.Rh5-g5 24.Ka2*a3 35.Kh5-h6 37.Rh5-h3 39.Kh5-h4 41.Rh5-g5 48.Ke8*d8 55.Kh5-h4 57.Rh5-h7 59.Kh5-h6 61.Rh5-g5 74.Ka4*a5 87.Kh5-h6 89.Rh5-h3 91.Kh5-h4 93.Rh5-g5 101.Kd7*c6 109.Kh5-h4 111.Rh5-h7 113.Kh5-h6 115.Rh5-g5 129.Ka5*a6 143.Kh5-h6 145.Rh5-h3 147.Kh5-h4 149.Rh5-g5 159.Kd5*c4 160.Kc4-d5 165.c7*b8=Q 166.Qb8*c8 167.Qc8*c1 168.Qc1*e3 169.Qe3*e2 170.Qe2*e4 171.Qe4-b1 h2-h1~ z

  • New overall series self-target-square move length record (promoted force). The connection of two known mechanisms is seen here for the first time.
  • Previous record: 156 moves (HC67 ChessProblems.ca, 17.01.2013).
T126 Vaclav Kotesovec
Czech Republic
T126 ChessProblems.ca, 19.01.2013

ser-h# 20 (2+4) C+
Kangaroo a7, e3
Grasshopper a2, c1
PWC
3 Solutions


Solutions: 

1.Kb3 2.K:a2(Gb3) 3.Kb1 4.Ga1 5.Kc1 6.Gd1 7.Kc2 8.G:b3(Gd1) 9.Kc3 10.Gd3 11.KAb3 12.Kc4 13.Gb5 14.Kc5 15.Kb6 16.KAb7 17.Kc7 18.Kb8 19.Ka8 20.Gb8 Gf3#

1.Kc5 2.Gc6 3.Kd5 4.KAb7 5.Kc5 6.Gc4 7.Kb4 8.Kb3 9.G:a2(Gc4) 10.Ka3 11.Ga4 12.KAa2 13.Kb3 14.Gc2 15.KAf2 16.KAb2 17.KAb1 18.Ka2 19.Ka1 20.Ga2 Gf1#

1.Kc3 2.Gc4 3.Gf1 4.Gd3 5.KAb3 6.Kd4 7.Gd5 8.G:a2(Gd5) 9.K:d5(Gd4) 10.Kc5 11.KAe3 12.KAb6 13.Kb4 14.KAb7 15.Kb3 16.KAb2 17.KAb1 18.Kc2 19.Kc1 20.Gc2 Ga1#
  • C+ Alybadix
T125 Vaclav Kotesovec
Czech Republic
T125 ChessProblems.ca, 19.01.2013

ser-h# 22 (2+5) C+
Kangaroo e5, f4, g2, h3
Kangaroo-Lion c5
2 Solutions


Solutions: 

1.Ke2 2.Kd3 3.Kc4 4.Kb5 5.KAa5 6.KAd5 7.Kc6 8.KAb7 9.KAa8 10.Kd5 11.KAe4 12.KAf3 13.Kc4 14.KAb4 15.KAa4 16.Kc3 17.KAb3 18.KAa3 19.KAd6 20.KAe7 21.Kb4 22.Ka3 KALf8#

1.Kf2 2.Kg3 3.KAh2 4.Kh4 5.KAh5 6.KAh6 7.Kh3 8.KAh2 9.KAh1 10.Kg4 11.Kf5 12.Ke5 13.KAd6 14.KAc7 15.Ke4 16.KAd5 17.KAc6 18.Kd3 19.KAd2 20.KAd1 21.Kc2 22.Kc1 KALc8#
  • C+ WinChloe
T124 Dan Meinking† & Arno Tüngler,
USA & Germany
T124 ChessProblems.ca, 10.01.2013
1N2b3/B1p3n1/RRP5/2k1b1r1/1RB4n/1PRPpp1r/1PP5/N4Knb
ser-s+ 153 (14+12) C+


Solution: 

8.Ka4-a5 10.Ra4-a2 12.Ka4-a3 14.Ra4-b4 26.Kh6*g5 38.Ka4-a3 40.Ra4-a6 42.Ka4-a5 44.Ra4-b4 53.Kf1*g1 62.Ka4-a5 64.Ra4-a2 66.Ka4-a3 68.Ra4-b4 82.Kg4*h3 83.Kh3*h4 96.Ka4-a3 98.Ra4-a6 100.Ka4-a5 102.Ra4-b4 122.Ka4-a5 124.Ra4-a2 126.Ka4-a3 128.Ra4-b4 142.Kg4*f3 144.Ke4*e5 148.Kd8*c7 149.Kc7-d8 150.c6-c7 153.Sd4-e6 + Sg7*e6 +

  • Overall series self-check move length record (promoted force).
  • New "moved Zeller-trap" matrix (combined with Kemp matrix), discovered by Dan in November 2012.
  • Withdrawn from the tournament by the author, 28.12.2013.
T123 Mečislovas Rimkus
Lithuania
T123 ChessProblems.ca, 10.01.2013
r7/3p2K1/8/b7/8/BBQ5/8/1k6
ser-hs# 9 (4+4)
GhostChess


Solution: 

1.d7-d5 2.d5-d4 3.Ba5*c3 4.Ra8*a3 5.Kb1-b2 6.Kb2*b3 7.Ra3-a8[+wuBa3] 8.Ra8-h8 Kg7*h8 9.d4-d3 #

T122 Arno Tüngler
Germany
T122 ChessProblems.ca, 04.01.2013
1bn2bRB/r5q1/1r3k2/rpr4K/4bPPR/1bn2rPR/5P1B/6B1
ser-Z d6 176 (11+14) C+


Solution: 

1.f4-f5 2.Rg8*g7 3.Kh5-h6 5.Rh5-g5 7.Kh5-h4 9.Rh5-h7 11.Kh5-h6 13.Rh5-g5 23.Kb2*b3 33.Kh5-h6 35.Rh5-h3 37.Kh5-h4 39.Rh5-g5 44.Kg8*f8 49.Kh5-h4 51.Rh5-h7 53.Kh5-h6 55.Rh5-g5 67.Kb4*c5 81.Rh5-h3 83.Kh5-h4 85.Rh5-g5 93.Kd8*c8 101.Kh5-h4 103.Rh5-h7 105.Kh5-h6 107.Rh5-g5 120.Kc5*b6 133.Kh5-h6 135.Rh5-h3 137.Kh5-h4 139.Rh5-g5 148.Kc8*b8 157.Kh5-h4 159.Rh5-h7 161.Kh5-h6 163.Rh5-g5 176.Kc5-d6 z

  • New direct series target-square move length record (promoted force).
  • Withdrawn from the tournament by the author, 28.12.2013.
  • Previous record:
Branko Koludrović
K1478 Problemkiste (145) 02/2003

ser-Z c6 175 (8+15) C+
1.Kf2 2.Bh2 4.Kh1 5.Bg1 9.K*h5 13.Kh1 14.Bh2 17.Kf3 19.Bg1 28.K*b5 37.Kf3 39.Be1 42.Kh1 43.Bg1 52.K*e8 60.Kh1 61.Bh2 64.Kf3 66.Bg1 75.K*d6 84.Kf3 86.Be1 89.Kh1 90.Bg1 99.K*d8 108.Kh1 109.Bh2 112.Kf3 114.Bg1 123.K*b6 132.Kf3 134.Be1 137.Kh1 138.Bg1 149.K*b8 160.Kh1 161.Bh2 164.Kf3 166.Bg1 175.Kc6 z
T121 Arno Tüngler
Germany
T121 ChessProblems.ca (v), 04.01.2013

ser-# 148 (9+15) C+


Solution: 

1.Kc7-c8 11.Be8-d7 13.Kd8-e8 24.Bh5-f7 26.Kf8-g7 27.Bf7-g6 29.Kh6-g5 31.Bh5-g4 33.Kh4-h3 42.Bd1-e2 46.Ke1-d1 57.Ba4-c2 59.Kd2-c3 60.Bc2-b3 62.Kb4-a4 73.Ba6-b5 74.Ka4*a5 75.Ka5-a4 86.Bd1-b3 88.Kb4-c3 89.Bb3-c2 91.Kd2-d1 102.Bf1-e2 106.Kg2-h3 115.Bh5-g4 117.Kh4-g5 119.Bh5-g6 121.Kh6-g7 122.Bg6-f7 124.Kf8-e8 135.Bc8-d7 137.Kd8-c8 138.Bd7-e8 148.Ba6-b7 #

  • New direct series mate move length record (promoted force).
  • Version 26.05.2013.
  • Withdrawn from the tournament by the author, 28.12.2013.
  • Previous version:
k7/2K1p3/BPpqPr2/n1r1rP2/2Pn1r2/P2P1P2/1r5b/q1n3b1
ser-# 148 (9+15) C+
(Same solution)

  • Previous record:
Vladimír Janál
(after C.E.Kemp, Fairy Chess Review 1958)
SU 11/2006

ser-# 140 (7+15) C+
5.Kh2-h1 6.Bg1-h2 11.Kd1*c1 16.Kg1-h1 17.Bh2-g1 33.Ka5*b4 49.Kh2-h1 50.Bg1-h2 56.Kc1*b1 62.Kg1-h1 63.Bh2-g1 80.Kb4*a3 97.Kh2-h1 98.Bg1-h2 105.Kb1*a1 112.Kg1-h1 113.Bh2-g1 131.Kc5*d4 132.Kd4*c3 133.Kc3-d4 136.c5*b6 138.b7-b8=Q 140.Qf8-h6 #

T120 Zoran Sibinović
Serbia
T120 ChessProblems.ca, 31.12.2012
8/4p3/8/3p2p1/6P1/3P1k1K/p7/8
ser-= 19 (3+5) C+


Solution: 

8.Kd4*d5 10.Ke6*e7 12.Kf6*g5 13.Kg5-h4 17.g7-g8=Q 18.Qg8*a2 19.Qa2-d2 =

  • Kings and Pawns only, ideal stalemate.
T119 Zoran Sibinović
Serbia
T119 ChessProblems.ca, 31.12.2012
8/1p2p3/4P2k/8/6PK/4p3/4P3/8
ser-h# 21 (4+4) C+


Solution: 

5.b2-b1=B 7.Bd3*e2 8.Be2-d3 10.e2-e1=R 11.Re1*e6 12.Re6-g6 17.e2-e1=R 19.Re7-h7 20.Rg6-g7 21.Bd3-g6 g4-g5 # br />
  • Kings and Pawns only, Bishop and two Rook promotions, ideal mate.
T118 Zoran Sibinović
Serbia
T118 ChessProblems.ca, 31.12.2012
6k1/4p3/6PP/8/p1p5/P1P4K/8/8
ser-h# 16 (5+4) C+


Solution: 

5.e2-e1=B 6.Be1*c3 7.Bc3-h8 10.c2-c1=B 11.Bc1*a3 12.Ba3-f8 15.a2-a1=B 16.Ba1-g7 h6-h7 #

  • Kings and Pawns only, 3 Bishop promotions.
T117 T117 ChessProblems.ca, 31.12.2012 Withdrawn by the author.
T116 Bojan Bašić
Serbia
T116 ChessProblems.ca, 29.12.2012

ser-hc# 3 (10+10)
Lion g2, Bishop-Lion a1, g3, h1, b2, c6
Isardam
b) +Amnésie, ser-h# 3


Solutions: 

a) 1.Rf8-h8 2.0-0 3.c3-c2 + LIg2-a2 #

b) 1.Rd8-a8 2.0-0-0 3.d7-d5 + LIg2*g4 #

  • "The problem explores the subtle difference between consequent series-movers and ordinary series-movers with fairy condition Amnésie. Here everything revolves around the fact that pawn's double-step is illegal if the opponent can capture it en passant. Therefore, the final position in a) is mate, since Black's defence d7-d5 is illegal. Let us turn to phase b). How is 3.d7-d5 + now legal? Because of Amnésie, White cannot capture d-pawn en passant on his following move (unless he can prove that the previous move was indeed the double-step; in this case, the previous move could as well be bLa8-c6, and thus there is no proof), and therefore 3.d7-d5 + is not forbidden! This is also the reason why the solution from a) doesn't work anymore - Black now could defend with d7-d5." (Author)
  • See a related MatPlus Forum discussion.
T115 Cornel Pacurar
Canada
(after Erich Bartel)
T115 ChessProblems.ca, 26.12.2012
8/8/8/8/2pp4/1prp4/pppp1b2/nbrk1K1n
ser-r= 24 (1+15) C+



Solution: 

1.Kf1-g2 2.Kg2-f3 (2.Kg2*h1?) 8.Ka3*b2 16.Kg2*h1 18.Kg2*f2 21.Ke4*d4 22.Kd4*c3 24.Kb2*a1 c4-c3 =

  • Longest solution for a Rex Solus (w) Series Reflex stalemate.
  • The first Rex Solus (w) Series Reflex stalemate:
Erich Bartel
707 feenschach 11, 10/1972 8/8/3p4/3p4/1ppp4/1pbp4/nrrp4/nbqk3K
ser-r= 8 (1+16) C+
1.Kh1-h2 (1.Kh1-g2?) 2.Kh2-h3 6.Ke6*d5 7.Kd5*c4 8.Kc4*d3 d6-d5 =

  • Another related longer composition:
Erich Bartel, Hans Hilmar Staudte & Hansjörg Schiegl
2765 Aachener Nachrichten, 03/11/1972 8/6p1/6p1/6p1/6pp/2K1b1pb/6pr/6kr
ser-r= 13 (1+12) C+
1.Kc3-c4 5.Kf7*g6 8.Ke4*e3 11.Kf5*g5 12.Kg5*h4 13.Kh4*g3 g7-g5 =
T114 Ján Golha
Slovakia
T114 ChessProblems.ca, 24.12.2012
8/8/8/3N1R2/1nnkN3/8/8/8
ser-h= 6 (3+3) C+
Take&MakeChess ParrainCirce
2 Solutions


Solutions: 

1.Sb4*d5-e7 2.Kd4*e4-d6 [+wSd7] 3.Sc4-a5 [+wSc5] 4.Kd6*c5-a6 5.Se7*f5-f8 [+wSd6] 6.Sf8*d7-b6 [+wRb3] Rb3*b6-a4 [+wSc8] =

1.Sc4-a5 2.Sb4*d5-e3 3.Kd4-c4 [+wSc5] 4.Kc4*c5-a4 5.Se3*f5-f2 [+wSd4] 6.Sf2*e4-c5 [+wRc8] Rc8*c5-a6 [+wSc2] =

  • Two complex and laborious solutions ending with beautiful echoes!
T113 Joost de Heer
Netherlands
T113 ChessProblems.ca, 23.12.2012

exact-ser-!= 63 (1+0)
Querquisite d1
HaanerChess WhiteMaximummer



Solution: 

1.QQd1-d8 2.QQd8-d2 3.QQd2-h6 4.QQh6-a6 5.QQa6-g6 6.QQg6-e5 7.QQe5-f6 8.QQf6-h4 9.QQh4-a4 10.QQa4-g4 11.QQg4-e3 12.QQe3-d4 13.QQd4-a1 14.QQa1-a3 15.QQa3-d3 16.QQd3-b1 17.QQb1-c3 18.QQc3-a5 19.QQa5-d5 20.QQd5-h1 21.QQh1-e1 22.QQe1-f2 23.QQf2-g1 24.QQg1-f3 25.QQf3-e2 26.QQe2-f1 27.QQf1-h3 28.QQh3-h2 29.QQh2-g2 30.QQg2-f4 31.QQf4-g3 32.QQg3-e4 33.QQe4-f5 34.QQf5-c8 35.QQc8-e6 36.QQe6-f7 37.QQf7-g8 38.QQg8-e7 39.QQe7-d6 40.QQd6-b4 41.QQb4-a2 42.QQa2-c2 43.QQc2-b3 44.QQb3-c1 45.QQc1-b2 46.QQb2-c4 47.QQc4-b5 48.QQb5-c7 49.QQc7-b8 50.QQb8-c6 51.QQc6-a8 52.QQa8-a7 53.QQa7-b7 54.QQb7-c5 55.QQc5-b6 56.QQb6-d7 57.QQd7-e8 58.QQe8-f8 59.QQf8-g7 60.QQg7-h5 61.QQh5-g5 62.QQg5-h7 63.QQh7-h8 !=
T112 Joost de Heer
Netherlands
T113 ChessProblems.ca, 23.12.2012

ser-Zd4 44 (1+0) C+
Elephant c1
HaanerChess WhiteMaximummer



Solution: 

1.ETc1-h6 2.ETh6-a6 3.ETa6-f1 4.ETf1-f8 5.ETf8-a3 6.ETa3-h3 7.ETh3-c8 8.ETc8-f2 9.ETf2-a7 10.ETa7-h7 11.ETh7-b1 12.ETb1-g6 13.ETg6-c2 14.ETc2-c7 15.ETc7-h2 16.ETh2-b5 17.ETb5-h8 18.ETh8-a1 19.ETa1-g7 20.ETg7-b2 21.ETb2-h5 22.ETh5-b8 23.ETb8-e2 24.ETe2-e8 25.ETe8-e3 26.ETe3-a5 27.ETa5-g8 28.ETg8-a2 29.ETa2-f7 30.ETf7-b3 31.ETb3-d7 32.ETd7-g1 33.ETg1-e5 34.ETe5-g3 35.ETg3-c5 36.ETc5-e1 37.ETe1-g5 38.ETg5-c3 39.ETc3-e7 40.ETe7-d5 41.ETd5-h1 42.ETh1-e4 43.ETe4-a4 44.ETa4-d4 z

T111 Joost de Heer
Netherlands
T111 ChessProblems.ca, 23.12.2012
8/8/8/8/8/8/8/Q7
ser-Zg2 49 (1+0) C+
HaanerChess WhiteMaximummer



Solution: 

1.Qa1-h8 2.Qh8-b2 3.Qb2-g7 4.Qg7-a7 5.Qa7-g1 6.Qg1-b6 7.Qb6-h6 8.Qh6-c1 9.Qc1-c8 10.Qc8-h3 11.Qh3-a3 12.Qa3-f8 13.Qf8-f1 14.Qf1-a6 15.Qa6-e2 16.Qe2-e8 17.Qe8-a4 18.Qa4-h4 19.Qh4-b4 20.Qb4-g4 21.Qg4-d7 22.Qd7-d1 23.Qd1-d6 24.Qd6-h2 25.Qh2-e5 26.Qe5-a5 27.Qa5-d5 28.Qd5-h1 29.Qh1-e4 30.Qe4-b1 31.Qb1-d3 32.Qd3-g3 33.Qg3-e1 34.Qe1-c3 35.Qc3-c7 36.Qc7-c4 37.Qc4-a2 38.Qa2-b3 39.Qb3-c2 40.Qc2-d2 41.Qd2-g5 42.Qg5-d8 43.Qd8-f6 44.Qf6-f2 45.Qf2-c5 46.Qc5-e3 47.Qe3-f4 48.Qf4-f3 49.Qf3-g2 z

T110 Dominique Forlot
France
T110 ChessProblems.ca, 12.2012
Dedicated to the memory of Dan Meinking
8/1KpP4/4P3/3k4/1Q6/8/4R3/8
aser-h= 5 (5+2)
2 Solutions



Solutions: 

1.Kd5-c5+ Qb4-f4 2.Kc5-d6+ Qf4-f7 3.Kd6*d7+ e6-e7 4.Kd7-d8+ e7-e8=B 5.c7-c6 Be8*c6 =

1.Kd5*e6+ Re2-f2 2.Ke6-e7+ Qb4-d4 3.Ke7-e8+ d7-d8=B 4.Ke8-d7+ Qd4-f6 5.Kd7-e8 Kb7*c7 =
T109 Mečislovas Rimkus & Dominique Forlot
Lithuania & France
T109 ChessProblems.ca (v), 04.12.2012
Dedicated to the memory of Dan Meinking
8/6p1/3k4/8/4PK2/8/1N6/8
Pser-h# 11 (3+2) C+
b) g7->h7
Ghost Chess



Solutions: 

a) 1.Kd6-c5 2.g7-g5 + Kf4-e5 3.g5-g4 4.g4-g3 5.g3-g2 6.g2-g1=S 7.Sg1-f3 + Ke5-e6 8.Kc5-d4 9.Kd4xe4 10.Sf3-d4 + Ke6-e7 11.Ke4-e5[+wuPe4] Sb2-d3 #

b) 1.h7-h5 2.h5-h4 3.h4-h3 4.h3-h2 5.h2-h1=Q 6.Qh1*e4 + Kf4-g5 7.Qe4-g2[+wuPe4] + Kg5-f6 8.Qg2*b2 + Kf6-f7 9.Kd6-e5 10.Qb2-f2[+wuSb2] + Kf7-e7 11.Qf2-d4 uSb2-d3 #
  • Version 22.12.2012 (twin by Dominique Forlot).
T108 Daniel Novomesky Slovakia
T108 ChessProblems.ca, 04.12.2012

ser-h# 8 (4+1) C+
b) Shift a1->a2
c) Shift a1->e2
d) Shift a1->f1
PWC, Take&MakeChess
Grasshopper a5, b3, c3, c4

Solutions: 

a) 1.Ka4*a5-d2 [+wGa4] 2.Kd2-c1 3.Kc1-b2 4.Kb2*b3-d3 [+wGb2] 5.Kd3*c4-c2 [+wGd3] 6.Kc2*d3-b3 [+wGc2] 7.Kb3*c2-c4 [+wGb3] 8.Kc4*c3-a1 [+wGc4] Ga4-d4 #

b) 1.Ka5-b6 2.Kb6*a6-d3 [+wGb6] 3.Kd3-e4 4.Ke4-d5 5.Kd5*c4-c6 [+wGd5] 6.Kc6-c7 7.Kc7-b8 8.Kb8-a8 Gb4-b7 #

c) 1.Ke5-f5 2.Kf5*g4-g6 [+wGf5] 3.Kg6*g5-e5 [+wGg6] 4.Ke5*e6-g4 [+wGe5] 5.Kg4*f4-f6 [+wGg4] 6.Kf6-f7 7.Kf7-g8 8.Kg8-h8 Gg4-g7 #

d) 1.Kf4-g4 2.Kg4*h3-f3 [+wGg4] 3.Kf3-e4 4.Ke4*f5-h3 [+wGe4] 5.Kh3*g4-g2 [+wGh3] 6.Kg2*h3-f3 [+wGg2] 7.Kf3-f4 8.Kf4*e4-h1 [+wGf4] Gh4-e4 #
T107 Daniel Novomesky Slovakia
T107 ChessProblems.ca, 04.12.2012

ser-h# 9 (4+1) C+
b) SWd5->b1
c) Md3->f5
d) EAd4->b5
PWC
Grasshopper d6, Sparrow d5, Eagle d4, Moose d3

Solutions: 

a) 1.Kf3-e2 2.Ke2*d3 [+wMe2] 3.Kd3-c4 4.Kc4*d5 [+wSWc4] 5.Kd5*d6 [+wGd5] 6.Kd6-e6 7.Ke6*d5 [+wGe6] 8.Kd5-e4 9.Ke4*d4 [+wEAe4] Me2-c5 #

b) 1.Kf3-e4 2.Ke4*d4 [+wEAe4] 3.Kd4-e3 4.Ke3*e4 [+wEAe3] 5.Ke4-e5 6.Ke5*d6 [+wGe5] 7.Kd6-c5 8.Kc5-c4 9.Kc4*d3 [+wMc4] SWb1-c3 #

c) 1.Kf3-g4 2.Kg4*f5 [+wMg4] 3.Kf5-e6 4.Ke6-d7 5.Kd7-c6 6.Kc6*d5 [+wSWc6] 7.Kd5-e5 8.Ke5*d4 [+wEAe5] 9.Kd4-e4 SWc6-d4 #

d) 1.Kf3-e3 2.Ke3*d3 [+wMe3] 3.Kd3-e4 4.Ke4-f5 5.Kf5-e6 6.Ke6*d5 [+wSWe6] 7.Kd5-c6 8.Kc6*d6 [+wGc6] 9.Kd6-e5 EAb5-d5 #
T106(v) Daniel Novomesky Slovakia
T106(v) ChessProblems.ca, 29.09.2012

ser-h# 18 (5+1) C+
b) wCYb4->a7
c) wCYb4->f3
d) wCYa4->e4
PWC
Charybdis a4, b4, e3, c1

Solutions: 

a) 1.Kc4*b4 [+wCYc4] 3.Kc3*c4 [+wCYc3] 7.Ka5*a4 [+wCYa5] 9.Kb3*c3 [+wCYb3] 11.Kd3*e3 [+wCYd3] 16.Ka4*a5 [+wCYa4] 18.Kb4-a3 CYd3-b4 #

b) 2.Kd3*e3 [+wCYd3] 4.Kd4*d3 [+wCYd4] 6.Kc2*c1 [+wCYc2] 10.Kb4*a4 [+wCYb4] 13.Ka6*a7 [+wCYa6] 16.Ka5*a6 [+wCYa5] 18.Kb5-a4 CYd4-b5 #

c) 2.Kb4*a4 [+wCYb4] 5.Kc2*c1 [+wCYc2] 7.Kb2*c2 [+wCYb2] 9.Kd3*e3 [+wCYd3] 11.Kd2*d3 [+wCYd2] 14.Kb1*b2 [+wCYb1] 16.Kc1*b1 [+wCYc1] 18.Kc2-d1 CYb4-c2 #

d) 2.Kb3*b4 [+wCYb3] 4.Ka3*b3 [+wCYa3] 8.Ke2*e3 [+wCYe2] 11.Ke1*e2 [+wCYe1] 14.Kc2*c1 [+wCYc2] 16.Kd1*e1 [+wCYd1] 18.Kd2-c1 CYe4-d2 #
  • 2 x Chameleon Echoes.
  • Version 10.10.2012.
T105 Daniel Novomesky Slovakia
T105 ChessProblems.ca, 29.09.2012

ser-h# 28 (4+1) C+
2 Solutions
PWC
Skylla d3, d5, e4

Solutions: 

1.Kc4*d5 [+wSKc4] 4.Kb5*c4 [+wSKb5] 5.Kc4*d3 [+wSKc4] 6.Kd3*e4 [+wSKd3] 9.Kc6*b5 [+wSKc6] 10.Kb5*c4 [+wSKb5] 11.Kc4*d3 [+wSKc4] 16.Ka5*b5 [+wSKa5] 17.Kb5*c4 [+wSKb5] 23.Kb7*c6 [+wSKb7] 28.Ka7-a6 SKb5-a7 #

3.Kc2*d3 [+wSKc2] 4.Kd3*e4 [+wSKd3] 5.Ke4*d5 [+wSKe4] 8.Kb3*c2 [+wSKb3] 9.Kc2*d3 [+wSKc2] 10.Kd3*e4 [+wSKd3] 15.Kc1*c2 [+wSKc1] 16.Kc2*b3 [+wSKc2] 22.Ke2*d3 [+wSKe2] 28.Ke1-d1 SKc2-e1 #
  • Echoes.
T104 Mečislovas Rimkus Lithuania
T104 ChessProblems.ca, 29.09.2012
8/pR6/8/8/8/7p/7K/kN6
ser-hs# 14 (3+3)
GhostChess


Solution: 

5.a2*b1=Q 11.Ka6*b7 12.Kb7-a8[+wuRb7] 13.Qb1-g1[+wuSb1] + Kh2*g1 14.h3-h2 #
T103 Ivan Skoba, Czech Republic
T103 ChessProblems.ca, 22.08.2012

ser-h# 252 (11+6) C+
Holes


Solution: 

1.Kf8 2.Kg8 3.Kh7 4.Qd8 5.Re8 6.Rh8 7.Qg8 8.Re8 9.Re7 10.Qc8 11.Qxb7 12.Qc8 13.Qg8 14.Re8 15.Rc8 16.Qd8 17.Re8 18.Re7 19.Qh8 20.Kg8 21.Kf8 22.Ke8 23.Kd8 24.Re8 25.Rg8 26.Ke8 27.Kf8 28.Re8 29.Re7 30.Ke8 31.Kd8 32.Kc8 33.Kb7 34.Kxb6 35.Kb7 36.Kc8 37.Kd8 38.Ke8 39.Kf8 40.Re8 41.Rc8 42.Ke8 43.Kd8 44.Re8 45.Re7 46.Ke8 47.Kf8 48.Kg8 49.Kh7 50.Qd8 51.Re8 52.Rh8 53.Qg8 54.Re8 55.Re7 56.Qc8 57.Qb7 58.Qb5 59.Qxa4 60.Qb5 61.Qb7 62.Qc8 63.Qg8 64.Re8 65.Rc8 66.Qd8 67.Re8 68.Re7 69.Qh8 70.Kg8 71.Kf8 72.Ke8 73.Kd8 74.Re8 75.Rg8 76.Ke8 77.Kf8 78.Re8 79.Re7 80.Ke8 81.Kd8 82.Kc8 83.Kb7 84.Kb6 85.Kb5 86.Ka4 87.Kxa3 88.Ka4 89.Kb5 90.Kb6 91.Kb7 92.Kc8 93.Kd8 94.Ke8 95.Kf8 96.Re8 97.Rc8 98.Ke8 99.Kd8 100.Re8 101.Re7 102.Ke8 103.Kf8 104.Kg8 105.Kh7 106.Qd8 107.Re8 108.Rh8 109.Qg8 110.Re8 111.Re7 112.Qc8 113.Qb7 114.Qb5 115.Qa4 116.Qa2 117.Qb1 118.Qc2 119.Qxd1 120.Qc2 121.Qb1 122.Qa2 123.Qa4 124.Qb5 125.Qb7 126.Qc8 127.Qg8 128.Re8 129.Rc8 130.Qd8 131.Re8 132.Re7 133.Qh8 134.Kg8 135.Kf8 136.Ke8 137.Kd8 138.Re8 139.Rg8 140.Ke8 141.Kf8 142.Re8 143.Re7 144.Ke8 145.Kd8 146.Kc8 147.Kb7 148.Kb6 149.Kb5 150.Ka4 151.Ka3 152.Ka2 153.Kb1 154.Kc2 155.Kd1 156.Ke2 157.Kf1 158.Kg2 159.Kxg3 160.Kg2 161.Kf1 162.Ke2 163.Kd1 164.Kc2 165.Kb1 166.Ka2 167.Ka3 168.Ka4 169.Kb5 170.Kb6 171.Kb7 172.Kc8 173.Kd8 174.Ke8 175.Kf8 176.Re8 177.Rc8 178.Ke8 179.Kd8 180.Re8 181.Re7 182.Ke8 183.Kf8 184.Kg8 185.Kh7 186.Qd8 187.Re8 188.Rh8 189.Qg8 190.Re8 191.Re7 192.Qc8 193.Qb7 194.Qb5 195.Qa4 196.Qa2 197.Qb1 198.Qc2 199.Qd1 200.Qe2 201.Qf1 202.Qg2 203.Qxg4 204.Qxh5 205.Qg4 206.Qg2 207.Qf1 208.Qe2 209.Qd1 210.Qc2 211.Qb1 212.Qa2 213.Qa4 214.Qb5 215.Qb7 216.Qc8 217.Qg8 218.Re8 219.Rc8 220.Qd8 221.Re8 222.Re7 223.Qh8 224.Kg8 225.Kf8 226.Ke8 (initial position without Sb7-Pb6-Pa4-Pa3-Qd1-Pg3-Pg4-Bh5) 227.Kd8 228.Re8 229.Rg8 230.Ke8 231.Kf8 232.Re8 233.Re7 234.Ke8 235.Kd8 236.Kc8 237.Kb7 238.Kb6 239.Kb5 240.Ka4 241.Ka3 242.Ka2 243.Kb1 244.Kc2 245.Kd1 246.Ke2 247.Kf1 248.Kg2 249.Kg3 250.Kg4 251.Kh5 252.Bf4 + Sxf4 #
  • Pendulum of black King and Queen passing Zeller trap using Holes, model mate. Without white pieces on route (Sb7-Pb6-Pa4-Pa3-Qd1-Pg3-Pg4-Bh5) the solution is trivial, but the black King has to pass through the Zeller trap and continue marching to h5: 1.Kd8 3.Rg8 5.Kf8 7.Re7 25.Kh5 26.Bf4 + Sxf4 #.
T102 Cornel Pacurar Canada
T102 ChessProblems.ca, 13.08.2012

pser-h== 7 (2+2) C+
Royal Lion e1, e2, Lion a7, d5
2 Solutions

Solutions: 

1.rLIe1-e4 2.rLIe4-b7 3.LIa7-e7 4.rLIb7-e4 + rLIe2-e5 5.rLIe4-e6 + rLIe5-a5 6.LIe7-e5 + LId5-f5 7.LIe5-g5 + LIf5-h5 ==

1.rLIe1-e3 2.LIa7-f2 3.LIf2-a2 4.LIa2-e6 + rLIe2-e4 5.rLIe3-e5 + rLIe4-a8 6.LIe6-e4 + LId5-f3 7.LIe4-g2 + LIf3-h1 ==
  • Orthogonal - diagonal echoes in two well-balanced solutions. Lions-only Wenigsteiner Parry-Series.
  • Marianka 2012 Fairy TT was dedicated to compositions with Royal Lion - here is the preliminary award.
T101 Vaclav Kotesovec Czech Republic
T101 ChessProblems.ca, 13.08.2012

ser-h# 26 (2+5) C+
Kangaroos
2 Solutions

Solutions: 

1.Ke1 2.KAb1 3.KAf1 4.Ke2 5.KAc4 6.Kd2 7.Kc3 8.KAc5 9.KAc2 10.Kb4 11.Kb5 12.KAa6 13.Kc5 14.Kd4 15.KAb4 16.Kd3 17.KAe2 18.KAf1 19.Kc4 20.KAb5 21.Kb3 22.KAb2 23.KAb1 24.Ka2 25.Ka1 26.KAa2 Kg1 #

1.Ke3 2.Kf4 3.Kg5 4.KAg6 5.KAg3 6.Kf4 7.Kf3 8.KAh3 9.KAe3 10.Ke2 11.Kd2 12.KAf4 13.Kc3 14.KAb3 15.Kc4 16.KAb4 17.KAa4 18.KAd4 19.Kc5 20.KAb6 21.KAa7 22.Kc6 23.Kb7 24.Ka8 25.KAb7 26.KAb8 Kg2 #
  • Similar to the length-record T100, but two solutions this time around!
T100 Vaclav Kotesovec Czech Republic
T100 ChessProblems.ca, 01.07.2012

ser-h# 41 (2+5) C+
Kangaroos

Solution: 

1.Kg6 2.Kf5 3.Ke4 4.Kd3 5.Kc2 6.KAd1 7.Kd2 8.Ke1 9.KAf1 10.Kd2 11.Kc2 12.Kb1 13.KAa1 14.Kc2 15.KAa4 16.Kc3 17.Kb4 18.Ka5 19.KAa6 20.KAa3 21.Kb4 22.Kc3 23.KAd3 24.Kc2 25.KAd1 26.Kd2 27.KAd4 28.Kc3 29.KAa3 30.Kd3 31.KAe3 32.Kc4 33.KAb4 34.Kc5 35.KAb6 36.KAa7 37.Kc6 38.Kb7 39.Ka8 40.KAb7 41.KAb8 Kg2 #
T99 Vaclav Kotesovec Czech Republic
T99 ChessProblems.ca, 24.06.2012

ser-h# 16 (1+5) C+
2 Solutions
Grasshoppers, no white king

Solutions: 

1.Ka3-b4 2.Kb4-c5 3.Kc5-c6 4.Gf3-b7 5.Gb5-b8 6.Gh6-b6 7.Kc6-d5 8.Kd5-e4 9.Ke4-f3 10.Gb7-g2 11.Kf3-f2 12.Gb6-g1 13.Kf2-g3 14.Kg3-h2 15.Kh2-h1 16.Gc2-h2 Gc8-a8 #

1.Ka3-b3 2.Gc2-a4 3.Ga4-c6 4.Gh6-b6 5.Gb6-d6 6.Gc6-e6 7.Gd6-f6 8.Gf3-f7 9.Ge6-g8 10.Gf7-a2 11.Kb3-c3 12.Gf6-b2 13.Kc3-c2 14.Kc2-b1 15.Kb1-a1 16.Gb5-b1 Gc8-h8 #
  • Beautiful play and mirrored corner-echoes!
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.

Compositions:
T1-10 | T11-20 | T21-30 | T31-40 | T41-50 | T51-60 | T61-70 | T71-80
HC1-10 | HC11-20 | HC21-30 | HC31-40 | HC41-50 | HC51-60 | HC61-70