WCSC 2018 - Open r1

1

Almiro Zarur
Boletim da União Brasileira de
Problemistas 3rd Pr. 1990–91

white Bb7h2 Kb3 Qa6 Pe3 Sb5f3 Rd8d1 black Bf1c1 Kc5 Qe8 Pe7e6g6f4b6 Sd7d3 Rg4

#2(9+12)

Solution

{
Essai
} 1.Sb5-d6 ? Sd3-f2 ! {

} 1.Sb5-c3 ! {(} 2.Sc3-e4 # {)} 1...f4*e3 2.Qa6-b5 # 1...Sd3-f2 2.Sc3-a4 # 1...Sd7-f6 2.Qa6-c4 # 1...Kc5-d6 2.Qa6*b6 # {

= 5p
}

2

Friedrich Fricke,
Deutsche Schachzeitung
1977

white Qe7 Pg2c3c5b6a6 Sd8a4 Bf2a2 Ka5 black Pf7f6c2c4b4b7a7 Se2 Rh6h1 Kd5

#3(11+11)

Solution

{
Essai
} 1.b6*a7 {(} 2.Sa4-b6 # {)} f7-f5! {

} 1.g2-g4 ! {(} 2.Qe7-d7 + {)} 2...Kd5-e5 {ou} 2...Kd5-e4 3.Qd7-f5 # 1...Rh6-h5 2.b6*a7 { (} 3.Sa4-b6 # {)} 2...Rh5-e5 3.Qe7*b7 # 1...Rh1-h5 2.Ka5*b4 { (} 3.Ba2*c4 # {)} 2...Rh5-e5 3.Qe7*b7 # 1...Se2-d4 2.c3*d4 { (} 3.Qe7*b7 # {)} 2...f6-f5 3.Qe7-e5 # 2...a7*b6 + 3.Sa4*b6 # 1...Se2-g3 2.Qe7*b7 + {)} 2...Kd5-e5 3.Bf2*g3 # 1...Se2-f4 2.Qe7*b7 + {)} 2...Kd5-e5 3.Bf2-d4 # {

1 = 0.75p 2 = 1.5p 3 = 2.5p 4 = 3.25p 5 = 4p 6 = 5p}

3

Karl-Heinz Ahlheim,
Die Schwalbe
1987

white Pf6g2d2b2c5 Se8d3 Bf1 Ke5 Rb7 black Pf5a6 Kc4 Bf8a2 Sc2

#4(10+9)

Solution

{

} 1.g2-g3 ! {zz} 1...f5-f4 2.Ke5-e4 {(} 3.Bf1-h3 {(} 4.Bh3-e6 # {)} 3...Sc2-e3 {ou} 3...Sc2-d4 {ou} 3...Sc2-b4 4.Rb7*b4 # 2...a6-a5 3.Sd3-e5 + {} 3...Kc4*c5 4.Rb7-b5 # 1...Ba2-b3 2.Sd3-c1 + 2...Kc4*c5 3.Sc1*b3 + {} 3...Kc5-c6 4.Bf1-g2 # 1...a6-a5 2.b2-b3 + 2...Ba2*b3 3.Sd3-b4 + {} 3...Kc4*c5 4.Rb7-b5 # {
autre variante hors barême
} 1...Bf8-e7 2.f6*e7 {menace} 3.Se8-d6 # 2...Sc2-b4 3.Rb7*b4 # {

1 = 0.75p 2 = 1.5p 3 = 2.5p 4 = 3.25p 5 = 4p 6 = 5p}

4

Yuri V. Bazlov,
Tidskrift för Schack
2nd H.M. 1974

white Sa2 Bh5a1 Kd1 black Pf3h2 Ke4 Bg1

=(4+4)

Solution

{

} 1.Sa2-c3+ {
(et ici } 1...Ke4-f4 2.Sc3-d5+ Kf4-e4 3.Sd5-f6+ Ke4-f4 4.Ba1-e5+ Kf4*e5 5.Bh5*f3 {fait nulle plus vite)
} 1...Ke4-e3 2.Sc3-d5+ {} 2...Ke3-f2 3.Ba1-d4+ {(sur } 3...Kf2-g3 {il y a} 4.Bd4-e5+ {)} 3...Kf2-g2 4.Sd5-f4+ {} 4...Kg2-g3 5.Sf4-e2+! {} 5...f3*e2+ 6.Kd1*e2 {(menace } 7.Bh5-f3 {)} 6...h2-h1=Q 7.Bd4-e5+ Kg3-h4 8.Bh5-f3 {} 8...Qh1-h3 9.Be5-f6+ {} 9...Kh4-g3 10.Bf6-e5+ Kg3-h4 11.Be5-f6+ {=

= 1p }

5

Juraj Lörinc,
Original
2018

white Pe4h4h2g2 Kh3 Rg4 Bg3 black Pg7e5b7a5a4a2 Kc6 Bc7c8 Sg8 Ra3

h#4 3sol.(7+11)

Solution

{

} 1.Ra3-f3 g2*f3 2.b7-b6 Bg3*e5 3.Bc8-a6 Rg4*g7 4.Ba6-b5 Rg7*c7 # {

} 1.Bc8-f5 e4*f5 2.Ra3-b3 Rg4-d4 3.Rb3-b5 Bg3-f2 4.Kc6-b6 Rd4-d6 # {

} 1.Ra3-d3 h4-h5 2.Rd3-d7 Bg3-f2 3.Bc7-b8 Rg4-g6 + 4.Kc6-c7 Bf2-b6 # {

1 = 1,5p 2 = 3,25p 3 = 5p}