July 17th

Zdenek Oliva (17-07-1941) Croatian composer

Zdenek Oliva composes helpmates and fairy problems, especially good series problems.

Oliva, Zdenek
feenschach, 2002

Ser-=80

1 + 16

Show Solution

1.Kh6 13.Kxa6 28.Kxc8 45.Kxa8 62.Kxc7 64.Kxa5 66.Kxb3 67.Kxa2 71.Kxe5 72.Kxd4 75.Kxf7 76.Kxg6 77.Kxh5 80.Ke2 =
World length-record with a white King minimal and this type of stipulation (serial direct stalemate).

Oliva, Zdenek & Koludrović, Branko & Pernarić, Josip
23^rd TT Problemkiste, 2002
2^nd Prize

ser-h=25

11 + 10

Show Solution

1.c1=S 2.Sd3 3.Se5 4.exf2 5.f1=R 6.Rf5 ... 9.g1=B 10.Be3 11.Bg5 12.Bf6 ... 16.gxh2 17.h1=B 18.Bc6 23.d1=Q 24.Qd7 25.Qe8 fxe8=Q=
Black AUW plus bB and wQ promotions.
Consecutive unpinnings of the black units.

Nikolajus Zujevas (17-07-1953) Lithuanian compose

Nikolajus Zujevas composes direct mates, helpmates and selfmates.

Zujevas, Nikolajus
ÚV ČSTV, 1965
3^rd Prize

#2

b) bSh4->g8

7 + 9

Show Solution

a) 1. Re2! (2. Qxd3/Se4#)
1. ... Bc5 2. Qc4#
1. ... Rc5 2. Qd4#
1. ... Rxb1+ 2. Sxb1#
1. ... Rxe2 2. Qxd3#
1. ... Bxe2 2. Se4#
(1.Bc5? Sg2!)
b) 1. Bc5! (2. Qd4/Qc4#)
1. ... Bxc5 2. Qc4#
1. ... Rxc5 2. Qd4#
1. ... Rxb1+ 2. Sxb1#
1. ... Re2 2. Qxd3#
1. ... Be2 2. S(x)e4#
(1.Re2? Se7!)

Reciprocal Grimshaw and Novotny on e2 and c5.
Rudenko.
The keys and tries are harmonious with the twinning.

Владимир Владимирович Кожакин (17-07-1957 - 16-03-2024) Russian composer (Vladimir Vladimirovich Kozhakin)

Vladimir Kozhakin was a very prolific author of more than 1900 problems. He composed direct mates (especially miniatures) and helpmates. He was International Judge for threemovers, the editor of magazine "Кудесник" ("Kudesnik"), the founder of a collection of 46,000 miniatures and trainer in chess solving and chess composing.

Кожакин, Владимир Владимирович
Probleemblad, 1989
3^rd Prize

h#2

b) bPf4->g5

4 + 14

Show Solution

a) 1.e3 Rc3 2.Bf2 Sd2#
b) 1.Kf4 Rc4 2.Sf3 Sd3#
Pin mates, line closings.

Кожакин, Владимир Владимирович
Советская Молдавия 1968
1^st Prize

#3

12 + 11

Show Solution

1.Sh8-f7 ! (2.Qd3-c4+ [A] Kd5-e4 3.Qc4xd4# [B])
1...f5-f4 2.Qd3xd4+ [B] Kd5xe6 3.Sf7xh6# [C]
1...Kd5xe6 2.Sf7xh6+ [C] Ke6-f6/Ke6-e5 3.Qd3xf5# [D]
1...Sh6xf7 2.Qd3xf5+ [D] Sf7-e5 3.c3-c4# [E]
1...Be7-f6 2.c3-c4+ [E] Kd5xe6 3.Sa8-c7# [F]
1...Sb8-c6 2.Sa8-c7+ [F] Kd5-c5 3.Qd3-c4# [A]
Six-fold cycle of White 2nd and 3rd moves AB-BC-CD-DE-EF-FA.

Tichomir Hernádi (17-07-1958) Slovak composer

Tichomir Hernádi composes helpmates and fairy problems.

Hernádi, Tichomír
174^th TT Die Schwalbe, 1987
1^st Prize

h#6

3 + 14

Show Solution

1.g1=Q bxa6 2.Qb1 a7 3.Qg6 hxg6 4.h1=Q gxf7 5.Qb7 fxe8=Q 6.Qb8 axb8=Q#
Rare achievement of 2+2 Queen promotions.

Hernádi, Tichomír
Sachova Skladba, May 1986 (470)
2^nd Prize

h#50

f8, a7, b5, d5, c4, e2: Locust

5 + 10

Show Solution

1.Kh8 Kg3 ... 11.Kh8 Ka4 12.L(xa6)a5+ Kxa5 ... 24.Kh8 Kf4 25.Kh7 Kxe3 ... 45.Kh7 Kf6 46.Kh8 Kxg6 47.e3 Kxh5 48.Kh7 g6+ 49.Kh8 g7+ 50.Kh7 g8=Q#
Logical seriesmover: White must give Black a tempo, or else the black King will be stalemated after Kxg6. Hence the capture of bPe3, which allows the tempo 47.e3.