Hrvoje Bartolovic (1932-2005)

Recently I have received a message from Zvonimir Hernitz. Sad news was that composing grandmaster Croatian Vojko Bartolovic is no longer among us. While I never met him, I knew well he was mainly orthodox composer. I will probably forever remember his twomover with mighty move 1...d4! In the selection below I tried to show more varying set of genres in which he (succesfully) composed.
Hrvoje Bartolovic
Sveto Stambuk

1st Prize Problem 1953

1.g4? th. 2.Qf5#
1...d4! (defends by pinning of wQ)

1.Rb8? th. 2.Re8#
1...d4! (guarding by opening of bBc4 line)

1.Bc1? th. 2.d3#
1...d4! (unpinning of Bc4)

1.Bb2? th. 2.Sxg5#
1...d4! (prepares creation of e5 flight in aticipation)

1.Sb5? th. 2.Sc3#
1...d4! (direct guarding of c3)

1.Sc8? zz
1...d4! (the only move)

1.Kxh7? th. 2.Qg6#
1...d4! (unblocks d5)

1.Sc6! th. 2.Sxg5#
1...h6 2.Qg6#
1...dxc6 2.Qe6#

That is the twomover mentioned above. Simple move by pawn and how varyingly strong!

#2 (11+7)

Hrvoje Bartolovic
1st Prize The Problemist 1972

1.Bf8! th. 2.Qc8#
1...Rd~ 2.Se3#
1...Rd6+! 2.Sxd6#
1...Rd4! 2.cxd4#
1...Re~ 2.Sd6#
1...Re6+! 2.Qxe6#
1...Re4! 2.Qxe4#
(1...Bb5 2.Qxb5#)

Two systems of black correction. Random moves allow mates using pin of Bf4 and halfpin mechanism. One correction unpins bishop, the other checks.

#2 (11+10)

Hrvoje Bartolovic
Die Schwalbe 1961

a) 1.Kc4 e8S 2.Kb5 Sd6#

b) 1.Kd5 Kd8 2.Kxe6 Ba2#

c) 1.Ke4 Bg3 2.Kf3 Sd2#

d) 1.Ke4 Rh4+ 2.Kf3 Rf4#

e) 1.Kc4 Ba5 2.Kb5 Bxe2#

f) 1.Kd5 Kd8 2.Kxe6 Sf4#

g) 1.Kd5 exf7 2.Ke6 fxg8Q#

h) 1.Ke4 Bg3 2.Kf3 Bxc6#

i) 1.Kc4 Ba5 2.Kb5 Sd6#

Very bold example of Forsberg type twinning with additional cyclic theme. There is complete cycle of black moves compared to the white piece present in the twin position on b1, h5 and e8. The repetition of wBc7 moves is however regrettable.

h#2 (9+10)
b) wBb1
c) wSb1
d) b1 -» h5
e) = d) + wBh5
f) = d) + wSh5
g) b1 -» e8
h) = g) + wBe8
i) = g) + wSe8

Hrvoje Bartolovic
Die Welt 1969

1.d5 a4 2.d4 axb5 3.d3 bxc6 4.d2 cxd7 5.d1=S d8=S#

Not far from now well known and probably impossible 100$ theme (in orthodox chess). It asks for h#5 with excelsior by both sides ending in the knight promotion. Here the motivation is standard for Black (just temping by pawn, promotion being forced by check and wp promotion square guarded), white pawn just eats his way to d8.

h#5 (2+11)

Hrvoje Bartolovic
2nd Prize e.a. feenschach 1976

a) 1.f4! zz
1...Kxc6 2.Rb5=
1...Ke4 2.Rxf5=
1...Kxc4 2.Rd3=
1...Ke6 2.Rd7=

b) 1.Sa5! zz
1...Kc5 2.Rd7=
1...Ke5 2.Rd3=
1...Kd4 2.Rf5=
1...Kd6 2.Rb5=

Stalemate in two resembling Latzel's similar mate twomover. In a) position black king makes star, in b) he finishes cross. What is new and allowed just by stalemate aim, there are the same stalemating moves in two phases, thus we see 4 transferred stalemates (Z-24-84).

=2 (12+5)
b) f5 -» e3

Hrvoje Bartolovic
Milivoj Slezinger

1st HM 2nd TT Sahovski vjesnik 1950-51

The last move was c7xQd8S+. Why?

Black king is in check, so White must have moved the last. There is no place from which he might have jumped to d8. So there is one last possibility, capture promotion from c7. What did the pawn capture? Not bishop nor knight as there would be no possible last move for Black (here the role of Rf7 is multiplied - it not only guards f6, f5 and blocks for retro moves f7, but also makes quick f6xe5 impossible as the rook would be trapped there with no chance to escape in retro play). And why it could not have been rook, the piece captured on d8? Because there is other configuration that stays on the place from the beginnning of the game - pb7, Bc8, pd7, pe7, Bf8, pg7. Black rook would have no chance to enter e8, d8 - or quit these squares in retro play. So what remains? Just black queen - and it is the piece that was captured on d8.

Last move? (3+9)

Comments to Juraj Lörinc.
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