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Twinning example for section C with 1 battery and 3 phases
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A: diagram
B: bSf6
C: bBh4
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Twinning example for section C with 2 batteries and 5 phases
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A: diagram
B: bSf6
C: bBh4
D: bSd5
E: bRd1
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F.Chlubna
Austria – Switzerland 1977 3rd Prize |
Jorge M. Kapros
2657 USPB 87-88 01-04/1993 |
Ricardo de Mattos Vieira
Best Problems 2149 - Oct 2008 |
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h2#
3+7
B: Qf2 -> g1 C: Qf2 -> h1 |
h2#
6+5
B: Sf -> c1 |
h2#
10+8
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A: 1.Sd5 Rf1 2.Rd8 Qf7#
B: 1.Bd3 Re1 2.Qd7 Qxg8# C: 1.Bd4 Rb1 2.0-0-0 Qb7# Line opening (departure effect) and line closing (arrival effect) |
A: 1.Sc3 Bc5 2.Kc4 Se3#
B: 1.Se3 Rc5 2.Kd4 c3# Block and indirect pin |
1.Se2 Sf6 2.Rb5 Sxe2#
1.Sc7 Sf5 2.Qxc3 Sxc7# Sacrifice |
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Ricardo Vieira
Mario Figueiredo UBP – Colonelli Jub.Ty – 1982 4th Honourable Mention (correction) |
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h2#
5+7
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1.Sc6 Be3! (Rf4?) 2.Qc7 Rf8#
1.Sg6 Rf4! (Be3?) 2.Re8 Ba5# |


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M. Gordian
Die Schwalbe 1931 Prize |
H.Mertes
Feenschach 1984 |
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3#
2+7
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h5#
2+2
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1.h8=G ! (2.Ga8 3.Ga3#)
1... Ge5 2.Kc3 threat: 3.Gh1# 2... Gb2 3.Kd3# |
1.Kd1 Ne3+ 2.Kc1 Kc4
3.Kb2 Kd3 4.Ka2 Kc2 5.Ka1 Kb3# |

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Milan Velimirović
Dedicated to Udo Degener Mat Plus 1995 Special Prize |
Slobodan Mladenović
Die Schwalbe 1989 2nd Place |
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3#
11+10
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3#
11+10
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1.Qe6+ !
1... Kxe6 2.Sxd3 ~ 3.Rxc6/Rxg6# 1... Kd4 2.Rc3 ~ 3.Rxd3# 2... Rd~ x 2.Qf6# A 2... dxe4 y 2.Qd6# B 2... fxe4 z 2.Qe5# C 1... Kf4 2.Rg3 ~ 3.Sxd3# 2... Rd~ x 2.Qd6# B 2... dxe4 y 2.Qe5# C 2... fxe4 z 2.Qf6# A |
1.Kd7 ! zz
1... bxc5 2.Qxc6 (3.Sg4# A) 2... Kf6 x 3.Rxf5# B 1... Bc7 2.Rxe7 (3.Rxf5# B) 2... Kf6 x 3.Bg7# C 1... b4 2.Qf1 (3.Bg7# C) 2... Kf6 x 3.Sg4# A 1... Bf7 2.Rxf7 ~ 3.Sg6# 1... Bxh7 2.Qa2 ~ 3.Qxe6/Sg4# 1... Kf6 2.Bg5+ Ke5 3.Sg6# |



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Yosi Retter
4th Place, Macedonia 2008 |
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2#
8+6
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1.Qa1? zz
1… exf5 2.Sf3# A 1… Sf~ 2.Rxe6# B 1… R~ 2.Qxd4# 1... Sd5! 1.Qh2? zz 1… exf5 2.Sg6# C 1… Sd~ 2.Rxe6# B 1… Sxf5 2.Sf3# A 1... Sc6! 1.Qg5! zz 1... Sd~ 2.Sf3# A 1... Sf~ 2.Sg6# C 1... exf5,fxg5,R~ 2.Qxf6,Bg7,Qxf4 # |


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Evgenij M. Bogdanov
МИНИАТЮРЫ - 2002 |
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3#
4+2
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1.Kc8 ? Kc6!
1... a5 2.Kd7 Kb7 (a4) 3.Rb5# 1.Se8, Se6? Kb7! 1... a5 2.Sc7 Kb7 (a4) 3.Rb5# 1.Sa6? Kxa6! 1... Kb7 2.Sc7 Kb6 3.Rb5# 1.Rd4 ! 1... Kc5 2.Se6+ Kb5 (Kb6) 3.Tb4# 1... a5 2.Rd5 Kb7 (a4) 3.Rb5# |

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Evgenij M. Bogdanov
РОБІТНИЧА ТРИБУНА 1985 |
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3#
5+2
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1.Rc3 !
1... Kd6 2.Sb3 2... e3 3.Rc5 e2 4.Qf6# 2... Ke5 3.Rc5 Kd6 4.Qf6# 3... d5 4.Rd5# |

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Miroslav Kasár
JT Metlickij 60 2nd Prize |
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3#
8+4
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1.Sc6! (2.Se7+ Ke6 3.Bf5#)
1... Sf5 2.Sb4+ Ke4 3.Rc4# 1... Sc4 2.e4+ Kc5 3.Be3# |

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Robert A. Lincoln
Melnichuk-JT, 2007 |
Robert A. Lincoln
Original |
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2#
4+3
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2#
5+2
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1.Qd6 ! (2.Rgxf4#)
1... Kxf3 2.Qxf4# 1... Kf5 2.Rfxf4# 1... Be5 2.Qd3# |
1.Se4 ! (2.Sf2#)
1... Bxg3 2.Qxg3# 1... Bg5 2.Rxg5# 1... Bf6+ 2.Sxf6# |
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Evgenij M. Bogdanov
РОБІТНИЧА ТРИБУНА 1985 |
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3#
5+2
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1. h4 !
1... Kg1 2.Qh3 e2 3.Bf2#
1... Kh1 2.Qf1+ Kh2 3.Bg3# 1... Kh2 2.Qf1 e2 3.Bg3# 1... e2 2.Qf2+ Kh1 3.Sg3# 2... Kh3 3.Qg3# |
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Evgenij M. Bogdanov
Pod Wieza 1999 Special Prize |
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4#
4+2
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1.Kc4 !
1... b2 2.Kc3 Ka3 3.Ra1+ ba1~ 4.Rxa1# 2... bc1~ 3.Rxc1 Ka3 4.Ra1# 2... b1~ 3.Rxb1 Ka3 4.Ra1# 1... Ka3 2.Ra1+ Kb2 3.Re1 Kc2 4.Re2# 1... Kb2 2.Rd5 Kxc1 3.Kb3 Kb1 4.Rd1# 2... Ka2 3.Kc3 (b2) 4.Ra5# 2... Ka3 3.Kc3 Ka4 4.Ra1# 3... Ka2 (b2) 4.Ra5# |

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Chris J. Feather
Suomen Tehtäväniekat 1994 |
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h3#
5+7
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1.Kc4 Bxc3 2.Rd6 Rd5
3.Bd3 Rc5# 1.Ke3 Rxd2 2.Bf6 Be5 3.Be4 Bf4# |


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Aaron Hirschenson
Macedonian Problemist 2003 1st Prize |
Paz Einat
Israel-France 1983 1st Place |
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2#
12+7
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2#
9+7
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1.Re7 ? zz 1... Rd5 !
1 … Be4~ a 2.Qe3# A 1 … Bd5 b 2.Sxf5# B 1.c5 ! zz 1 ... Be4~ a 2.Sxf5# B 1 ... Bd5 b 2.Qe3# A |
1... Se2 2.Qe5# A
1... Sfe6 ! 2.Qc6# B 1.Sf3 ! (2.Rd4#) 1... Se2 2.Se7# 1... Sfe6 2.Sb6# 1... Sb3 2.Qe5# A 1... Sce6 ! 2.Qc6# B |
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Movement
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Phases of problem
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1 (Key)
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A
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B
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C
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D
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2
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B
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C D A
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D A
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A C
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3
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C
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D A D
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A B D
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B A B
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4#
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D
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A C C
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B A B
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C B A
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Evgenij M. Bogdanov
Echo 2009 |
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4#
10+4
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1.Bf1 ? A
1... Kh6 2.g5 B Kh5 3.g4+ C Kh4 4.g3# D 3... Kg6 4.Bd3# 2... Kg6 ! 1.g5 ! B 1... Kh5 2.g4+ C Kh4 3.g3+ D Kh3 4.Bf1# A |

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Thierry Le Gleuher
Phénix 1995-96 Commendation |
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We must add the wK. Both Kings cannot be in check. On the d2-f4 diagonal, the wK would be in an impossible check. On d5, he forbids the discovered check by the wR. On d7 he allows this discovered check, but there is a trap: The black pawns b7, c7, f7 and g7 make an impassable barrier for the wK! The wK shall be added on d3 and the last moves were: n-2.Nd4b3+(dec) n-1.d2d4 e4xd3(ep) n.Ke3xd3. |
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Add 1 piece
8+14 |
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Thierry Le Gleuher
Phénix 1995-96 3rd Honiurable Mention |
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The last move is O-O-O. The bB on d3 was promoted at b1 and the only white capture was e2xRf3 to allow the Rook going on f2. Black can have made only 4 captures since the missing pawn is white. If the wPa is still on his file, the bPa avoided it and has made 3 captures, but this is impossible. So the missing pawn is on the h-file, and the wPa couldn't promote on his file, so it was captured on his file so that the bPa could promote. The wBa2 proves that the bPa was promoted with a2xb1=B. The wQB was captured on f6 (dark square). The bKR must have got out before freeing the wKB then the wQB. We note that the wQ is not free at the beginning of the game and that we must start with h7xNg6. After the opening g7xBf6 we see that the bK cannot pass by g5 and g4 due to the pawns f4 and f3 already in position. So the bK must have gone through h6-h5-h4. The wPh must have gone at least as far as h4 to let the wKR pass, and must not hinder the bK on his way. The wP is on h7! |
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Where is the missing pawn?
11+15 |
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Raymond Smullyan
The chess mysteries of Sherlock Holmes, 1979 |
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The last move is c7xXd8=R+. We cannot have captured a Rook or a black Queeen on d8 with the last move because of the illegal check to the wK! The piece captured on d8 can thus be only a bN (3 black Knights then) or a bB on dark square resulting from promotion (bBf8 taken at home). bPh was thus promoted (and h4 is not a bP) by capturing once at least (hxg-g1) and the bPs thus captured 5 times. The piece h4 cannot be black because a bQ or bR would give also check to wK (there is already check to the bK) and a bB or bN needs a black promotion moreover (impossible). The piece on h4 is thus white and the bPs took the other white available units. wBc1 (running on black squares) could not be taken by the bPs (all captures on white squares), it is thus on h4! |
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Add a piece on h4
10+10 |
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Vasil Markovcij
Troll 1997 |
Vasil Markovcij
Smena 1999 Prize |
Vasil Markovcij
The Problemist 2009 |
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2#
5+2
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2#
4+2
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2#
5+2
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1... Kc5 2.Qc3# A
1... ed 2.Qb4# 1.Rc6 ! 1... Ke5 2.d4# 1... ed 2.Qf4# 1... e3 2.Qc3# A |
1.Bc4 ? 1... d5 !
1... Kd4 2.Qf4# 1... Kf3 2.Bd5# 1.Bf1 ? 1... d5 ! 1... Kd4 2.Rc4# 1... Kf3 2.Bg2# 1.Qf6 ! 1... Ke3 2.Re2# 1... Kd5 2.Bb7# |
1.Bf3 ? B 1... Kd2 !
1... Kf1 a 2.Qa1# A 1.Qa1 ? A 1... c3 ! 1... Kf1 a 2.Bf3# B 1... Kd2 2.Sd3# 1.Sh3 ! 1... Kf1 a 2.Rg1# 1... Kxd1 2.Qa1# A |
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Evgenij Bogdanov
ECHO 2000 3rd–4th HM |
Evgenij Bogdanov
Reklama 2001 (v) |
Evgenij Bogdanov
Miniatures 2002 |
Evgenij Bogdanov
Robotnichna tribuna 1992 |
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3#
4+2
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3#
5+2
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3#
3+2
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3#
3+3
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1.Qd4 ? 1... Kf5 !
1… Kf3 2.Se1+ A Ke2 3.Qxd3# 1.Se1 ? A) 1... d2 ! 1… Ke4 2.Bg3 d2 3.Qe5# B 1… Kg4 2.Qg5+ Kh3 3.Qg3# 1.Sd4 ! 1… Ke3 2.Qe5+ B Kd2 3.Qe1# 1… d2 2.Qf5+ D Ke3 3.Bf2# C 1… Kg4 2.Qg5+ Kh3 3.Qg3# 1… Ke4 2.Bf2 C Kf4 (d2) 3.Qf5# D |
1.Qa7 !
1… Kg4 2.Qh7 Kg3 3.Rg1# 1… g4 2.Qg7 g3 3.Rf1# 1… Kg6 2.Qf7+ Kh6 3.Rh1# 1… Ke5 2.Qe7+ Kf5 3.Qe6# 1… Kf6 2.Qf7+ Ke5 3.Qe6# |
1… Kh5 2.Qg5#
1.Qf7 ? 1... Kh3 ! 1… Kh4 2.Kf3 - 3.Qh7# 1… g2 2.Qf4+ Kh3 3.Sg5# 2… Kh5 3.Qg5# 1.Sg5 ! 1… g2 2.Sf3 Kg3,Kh3 3.Qh4# 2… Kh5 3.Qg5# 1… Kh4 2.Kf4 Kh5 3.Qh7# 1… Kh5 2.Kf5 Kh6,Kh4 3.Qh7# |
1.Qb6 !
1… Kg5 2.Qf6+ Kh5 3.Sg3# 1… Kh5 2.Qg6+ Kh4 3.Qh6# 1… g3 2.Qh6+ Kg4 3.e3# 1… Kh3 2.Qf2 g3 3.Qxg3# |
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Evgenij Bogdanov
Pod Wieza 1995-1997 Special Prize |
Evgenij Bogdanov
Schachmatna misal 2002 |
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5#
4+3
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4#
5+2
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1.Bf2 !
1… Kb6 2.Be2! Ka7 3.Rc7+ Kb8 4.Bb6 1… b6 2.Bc8+ Ka7 3.Rc3 Ka8 4.Bg3 1… b5 2.Rc7 Ka5 3.Bc5 Ka4 4.Bd1+ |
1.Rc3 ! (2.Rgc6 Ke5 3.R3c4)
1… Ke4 2.Rg5 Kd4 3.Rb3 1… Ke5 2.Rc4 Kf5 3.Rh6 1… c5 2.Re3 Kc4 3.Rb6 |
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Evgenij M. Bogdanov
3964 Chess Leopolis 2009 |
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3#
12+5
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1.Kc1? c6!
1… Kc4 (Ke4) 2.Rfd1 e6 3.Rd4# 1… Ke2 2. Kc2 e6 3.Rbe1# 1.Ke1? c6! 1… Kc4 (Ke4) 2.Rbd1 e6 3.Rd4# 1… Kc2 2. Ke2 e6 3.Rfc1# |
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Example 1
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Example 2
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2#
7+5
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2#
7+4
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1.B~? 2.Sxh5# 1... Sg3!
1.Bg4! ~ 2.Sxh5# 1... Sg3 2.e3# P1 = Bf5 x = 1...Sg3 L1 = g6-g3 F1 = g3 |
1.Rf ~? 2.Qc8# 1... Se6!
1.Rd5! ~ 2.Qc8# 1... Se6 2.Bf8# (1... f5 2.Qh4# 1... Sd7 2.Qxd7# 1... Be6 2.Rxf8#) P1 = Tf5 x = 1...Se6 L1 = h8-d8 U1 = Lg8 |
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Evgeny Fomichev
Valery Kirillov Wola Gulowska 2002 1st Prize |
Mario Parrinello
Idee & Form 2005 5th HM |
Francesco Simoni
O. Bonivento 90 JT 2005 Commendation |
Chris Feather
Scrapings 1999 |
Vitally Medintsev
Diagrammes 2007 Prize |
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h2#
6+9
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h2#
6+9
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h2#
7+11
B: Pb5 ->c5 |
h2#
6+10
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h2#
7+9
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1.fxe2 Rxe5+ 2.Sc5 Rxc5#
1.gxf5 Bxd3+ 2.Sc4 Bxc4# Indirect self-unpinning by capturing the pinner 1.f2 Sc4 2.Sb2 Scxb2# 1.g5 Sc5 2.Sd7 Scxd7# Indirect unpinning |
1.Ka5 Re6 (Qxe3?)
2.Rd4 Kxd4# 1.Kb5 Qxe3 (Re6?) 2.Bd6 Kxd6# Direct self-unpinning |
a) 1.Kd5 Bxb6 2.Sed4 Rxd4#
b) 1.Kc5 Rg4 2.Sbd4 Bxd4# Masked direct self-unpinning |
1.Bb6 Sd3 2.Rc6+ dxc6#
1.Bf4 Kf6 2.Qe6+ dxe6# Indirect self-unpinning |
1.Sf2 fxg5 2.Bd5+ cxd5#
1.Be2 c5 2.Se5 fxe5# Masked direct self-unpinning |

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Kjell Widlert
Probleemblad 1969 1st Honourable Mention |
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3#
7+9+1
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1.Qa4 ! (2.nPe8=nQ+ Kf5 3.Qf4#) 1... cxd5 2.nPe8=nR+ Kf6,f5 3.Qf4# 1... Kd7 2.nPe8=nB+ 2... Kxc8/Kxe8/Ke6 3.Qa8/Sf6/Qe4# 1... Sd7 2.nPxd8=nS+ Ke5,f5 3.Qf4# |

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A. C. Reeves
British Chess Magazine 1965 2nd Honourable Mention |
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The arrival of a white piece on e5 unpins wRd6, introducing a threat of 2.Rd4. Try 1.Be5? Bf2(x) 2.Sd2 but 1...Sc2!(y) Try 1.Sce5? Sc2(y) 2.Bxd3 but 1...Bf2!(x) Key 1.Sge5! Bf2(x) 2.Qg4; 1…Sc2(y) 2.Qxf3 The two tries and key must be made by different White units to the same square. The logic of tertiary correction is found when the consequences of the thematic defence x (1...Bf2) are analysed. A mate follows x in first try but this is refuted by y (1...Sc2!) The second try corrects by arranging a mate for y but now 1...x! refutes. The key again provides for y but now x is met by a changed mate. This change is crucial to the tertiary logic. The changed mate following y is pleasing but not a thematic necessity. |
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2#
8+10
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Colin Sydenham
Die Schwalbe 1978/II 4th Prize |
Colin Sydenham
The Problemist, 1977/II 1st Prize |
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2#
11+9
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2#
5+2
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1.d4 ? (2.Qe4,Rxe5#)
1... Rf5 2.Se7# 1... Sf5 ! 1.S2d4 ? (2.Qe4,Rxe5#) 1... Sf5 2.Be6# 1... Rf5 ! 1.S6d4 ! (2.Qe4,Rxe5#) 1... Rf5 2.Bb7# 1... Sf5 2.Be6# |
1.Sxd7 ? (2.Se6#)
1... cxd3 x 2.Rxb4# 1... Sxd3 ! y 1.cxd7 ? (2.Se6#) 1... Sxd3 y 2.Sc6# 1... cxd3 ! x 1.Rbxb7 ! (2.Se6#) 1... cxd3 x 2.Qxb4# 1... Sxd3 y 2.Qxd5# |
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A. C. Reeves
The Problemist 2009 |
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A “neutral” capture of bPd4 eliminates the possibility of an en passant defence. Try 1.Bxd4? (>2.e4-A) but 1...Sc5! Try 1.Sfxd4? (>2.Bxc6-B and not 2.e4?-A) but 1...Rxc5! Key 1.Sexd4! (>2.e6-C and not 2.e4?-A nor 2.Bxc6?-B) 1...Rxc5 2.e4-A; 1…Sxc5 2.Bxc6-B (1...Kxc5 2.Rb5) The two tries and key must be made by different White units to the same square. The logic of tertiary threat correction is to be found in the sequence of threat avoidance; Try? (>2.A) Try? (>2.B-2.A?) and Key! (>2.C-2.A?/2.B?) The transfer of A and B to variation mates post-key is highly desirable but not a thematic necessity. Another feature here is that these transfers occur following the defences that refuted the tries. This is pleasing but once again not a thematic necessity. |
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2#
10+10
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A. C. Reeves
The Problemist 2009 |
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B: After the first black move in [a], a1=>a8 C: After the first black move in [b], a1=>a8 D: After the first black move in [c], a1=>a8 A: 1.c4 Qb7+ 2.Kc5 Rd5# B: 1.d5 Qe6+ 2.Kg5 Rxd5# C: 1.e4 Qg2+ 2.Ke3 Rxe4# D: 1.d3 Rb4 2.d2 Qc4# |
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h2#
4+4
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Jean-Marc Loustau
Jacques Rotenberg diagrammes 1985-87 1st Prize |
Uri Avner
TT Derby 2005 2nd HM |
Petko A. Petkov
dedicated to D. Brown StrateGems 2000 1st Comm |
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2#
11+6
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h2#
6+9
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h2#
4+4
B: Pd5 -> d7 |
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1... Be2 a 2.LEaxe2# A
1... Bc2 b 2.LE6e2# B 1... Be4 c 2.LEb2e2# C 1... d6 d 2.Rfxd3# D 1.LEee5 ? (2.Rfxd3# D) 1... d6 ! d 1.LEb2e5 ! (2.Rfxd3#) 1... Be2 a 2.LE6xe2# B 1... Bc2 b 2.LE5e2# C 1... Be4 c 2.LEae2# A 1... d6 2.f5# |
1.Bd6 LEf2 2.Bb4+ Bf6#
1.Sd6 LEf7 2.Sb5+ d6# |
A: 1.Sc8 Ke6 2.Sd6 nLEf6
3.nLEhh6+ Ke7# B: 1.Sd5 Ke4 2.b6 nLEg2 3.nLEhh1+ Ke5# |
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2#
5+11
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1.Sxf6 ! (2.Rc4#) 1... d5 2.Sd7# 1... Bd5 2.Se4# (1... Kxc3 2.Sd5#) |
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Roman Zalokotskiy
Суворовский натиск 1965 |
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h2#
4+10
B: Pc2 -> h7 C: Pa6 -> h7 D: Pe5 -> h7 |
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A: 1.Qa4 Bf5 2.Qd7 Bg6#
B: 1.Qf2 Bd1 2.Qf7 Ba4# C: 1.Rf4 Be2 2.Rf7 Bb5# D: 1.Rf5 Bf3 2.Rf7 Bc6# |
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Dan Meinking
George P. Sphicas StrateGems 2009 |
Dan Meinking
StrateGems 2009 |
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pser-h#14
3+2
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pser-s#11
6+8
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5.g1Q 6.Qg2+ e4 7.Qb2+ Kc7
8.Qh2+ e5 9.Qh7+ Kc8 10.Qf5+ e6 11.Qc5+ Kd8 12.Qg5+ e7 13.Qg8+ e8Q 14.Qg1 Qe2# |
1.Bb1 2.Be3+ Kh5 3.Bc1
5.Bxe2+ Kh4 6.Bd1 8.Bf2+ Kh3 9.Bg1 10.Bh5 11.Rg4 (zz) Rxg1# |
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Gianni Donati
Original |
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PG in11,5
15+15
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1.e4 d5 2.Bb5+ Bd7 3.c4 d4
4.Sc3 dxc3 5.d4 Bc6 6.Bf4 Sd7 7.Qd2 Bd5 8.OOO Be6 9.Kb1 Bg4 10.Rc1 Bd1 11.Ka1 Ba4 12.Bxa4 |

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Toma Garai
Die Schwalbe 1990 |
Viktor Abrosimov
Aleksandr Pankratiev 6th HM Diagrammes 2001 |
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h3#
7+9
B: Kc2 -> d2 |
h3#
3+7
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A:
1.Bd2+ Bb5 2.Qd4 Sc5 3.Qb2 Bd3# B: 1.Be3+ Sg5 2.Rf2 Bf5 3.Re2 Sf3# |
1.Kd3+ Sc4 2.Ke2 Bg4+
3.Kf1 Se3# 1.Kc5+ Bg4 2.Kb6 Sc4+ 3.Ka6 Bc8# |
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H.Bernleitner
H.Zajic The Problemist 2004 2nd Honourable Mention |
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h3#
7+8
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1.Be4 Gd5 2.Gd4 Sd6+
3.Kd8 Se6# 1.Bd5 Gc6 2.Gb6 Sd6+ 3.Kf6 Sd7# how unblocks and end in echo-style model mates |