12. svetová súťaž – vypísanie a témy

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12. WCCT: G – Exo

Tému navrhla krajina: Izrael
Rozhodcovské krajiny: Fínsko, India, Japonsko, Švédsko, Švajčiarsko
Náhradník:

TÉMAG
Slovensky:
Úlohy HS#2-5 s expodmienkou Isardam. Žiadne iné exopodmienky alebo exofigúry nie sú povolené. Mat musí byť exomat, čo znamená, že pozícia nie je mat alebo je nelegálna v ortodoxnom šachu, alebo posledný ťah je nelegálny v ortodoxnom šachu.
Isardam: Každý ťah, vrátane brania kráľa, je v nelegálny, ak by z neho vyplynula paralýza typu Madrasi (okrem kráľov).
Anglicky:
Required are HS#2-5 problems with the fairy condition Isardam. No other fairy conditions or fairy pieces are allowed. The mate must be fairy mate, meaning that the position is not mate or is illegal in normal chess, or the final move is illegal in normal chess.
Isardam: Any move, including capture of the king, is Isardam illegal if a Madrasi-type paralysis would result from it (not including kings).

PRÍKLADYG

Ricardo Vieira
3rd Prize
6th Tzuica, Jurmala 2008
white Pa6 Sc5 Rg5 Kf4 Bc2 Re1 black Pc7 Re7 Bh7 Sd6 Kb5 Qe4 Ph4 Pc3 Rh3
1.Re1-b1 + Kb5-c6 2.Rb1-b4 Qe4-g6 3.Bc2-e4 + Qg6*e4 # 1.Bc2-a4 + Kb5-c4 2.Ba4-c6 Qe4-e6 3.Re1-e4 + Qe6*e4 # {Popeye FreeBSD-11.2-RELEASE-amd64-64Bit v4.83 (1024 MB) / 36.039 s}
[P-G01]
hs#3 (6+9)
2.1.1...
Explanations:
the wK is not in check because Qe4xf4 is illegal because Bc2 observes Bh7, and Re1 observes Re7.
The final move is mate because in a) Rb4 cannot take Qe4 because Re7 observes Re4. In b) Bc6 cannot take Qe4 because Bh7 observes Be4.

Eric Huber
Vlaicu Crisan
2nd HM
Problem Online 2007
white Kd3 Qb1 black Kb5 Qf5
1.Qb1-c2 Kb5-b4 2.Qc2-c3 + Qf5-a5 3.Kd3-c4 + Qa5-c5 # 1.Kd3-c2 Kb5-c4 2.Qb1-b3 + Qf5-d5 3.Kc2-c3 + Qd5-d3 # 1.Kd3-e4 Kb5-c4 2.Qb1-d3 + Qf5-b5 3.Ke4-d4 + Qb5-d5 # {Popeye FreeBSD-11.2-RELEASE-amd64-64Bit v4.83 (1024 MB) / 0.296 s}
[P-G02]
hs#3 (2+2)
3.1.1...
Explanations:
the wK is not in check because Qf5xd3 is illegal because Qb1 observes Qd3.
In the first solution, the move 3.Kc4+ is legal because the Madrasi doesn’t include kings. This is a check because 4.Kxb4 is a legal move.
3…Qc5# is mate because the bK threats wK and the latter cannot move because Qc5 will observe Qc3.

Mario Parrinello
2nd comm
StrateGems 2009
white Rh6 Pd4 Ke3 Rf3 Bd2 Qh2 black Ba8 Qh8 Kc7 Pg5 Pe4 Ph4 Bg3 Rc2 Rg2
a) 1.Rf3-f7 Bg3-f4 2.Qh2*f4 + Qh8-b8 3.Bd2-c3 + Rc2*c3 # b) wQh2-->c1 1.Bd2-a5 Rc2-c3 2.Qc1*c3 + Qh8-c8 3.Rf3-f4 + Bg3*f4 # {Popeye FreeBSD-11.2-RELEASE-amd64-64Bit v4.83 (1024 MB) / 2:14.148 m:s}
[P-G03]
hs#3 (6+9)
b) Dh2→c1
Explanations:
the mating position is thematic because in normal chess
the mating position is illegal (bK is in check).

Mario Parrinello
2nd Prize
Tournoi de Noël France-Echecs 2011
white Pf7 Ke6 Pe5 Pf5 Pg3 Rh3 Be2 Bf2 black Qd8 Pg7 Pg5 Pd4 Bg4 Pb3 Re3 Kf3
1.Rh3-h6 Re3*e5 2.Bf2-e3 Re5-c5 3.Rh6-h2 Rc5-c3 4.Rh2-f2 + Rc3*e3 # 1.Be2-c4 Bg4*f5 2.g3-g4 Bf5-g6 3.Bc4-f1 Bg6-h5 4.Bf1-g2 + Bh5*g4 # {Popeye FreeBSD-11.2-RELEASE-amd64-64Bit v4.83 (1024 MB) / 17:15.398 m:s}
[P-G04]
hs#4 (8+8)
2.1.1...

Luis Miguel Martin
2nd Prize special
18th Tzuica 2020
white Ke8 Pe7 Rf7 Sd6 Rd4 Be3 black Qc7 Sa6 Pd5 Kh5 Rb2 Bd1
a) 1.Rf7-f3 Qc7-b8 + 2.Ke8-d7 Bd1-a4 + 3.e7-e8=B + Qb8*e8 # b) bKh5-->e2 1.Be3-d2 Qc7-c6 + 2.Ke8-d8 Rb2-b8 + 3.e7-e8=R + Qc6*e8 # {Popeye FreeBSD-11.2-RELEASE-amd64-64Bit v4.83 (1024 MB) / 1:35.648 m:s}
[P-G05]
hs#3 (6+6)
b) Kh5→e2
Explanations:
the final move Qxe8 is illegal in normal chess because wK
is already in check.

Vlaicu Crisan
S.K. Balasubramanian
3rd Prize
Quartz 2010-15
white Rf6 Kb5 Bg5 Pd4 Ph3 Bd2 Rf2 Rh2 black Rd7 Be7 Kg3 Bb2 Rc2
1.Kb5-b4 Rc2-c3+ 2.Kb4*c3 Be7-b4 3.Rf6-g6 Rd7*d4 4.Bg5-f6+ Rd4-d3 # 1.Kb5-c6 Be7-d6+ 2.Kc6*d6 Rc2-c6 3.Bd2-e1 Bb2*d4 4.Rf2-d2+ Bd4-e5 # {Popeye FreeBSD-11.2-RELEASE-amd64-64Bit v4.83 (1024 MB) / 1:35.648 m:s}
[P-G06]
hs#4 (8+5)
2.1.1...


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