Twinning by replacement of pieces 3


This is the third special example file for the 24th TT CCM C 10.10.2008.

This time the focus is on the direct problems, among them four twomovers and two fivemovers - three orthodox and three fairy. Especially orthodox twomovers by the grandmaster György Bakcsi are delicious.
György Bakcsi
2nd Prize Pesti Hirlap 1980

a) 1.Bxe6! th. 2.Qa7#
1...Qh6 2.Se4#
1...Sd7 2.Sxd7#

b) 1.Se5! th. 2.Qa7#
1...Qh6 2.Sd3#
1...Sd7 2.Sexd7#

c) 1.f8=S! th. 2.Qa7#
1...Qh6 2.Sxe6#
1...Sd7 2.S8xd7#

The departure of twinned piece from f7 creates threat 2.Qa7#. There are two defences for Black: pinning of wQ by bQ and gate closing by bS. They lead to change of two mates in three phases (Z-32-26) with unguarding as harmful motive in all variations.









#2 (7+9)
b) wSf7
c) wpf7

Johannes J. Burbach
Jugendschach 1986

a) 1.Sc3! th. 2.Sb5#

b) 1.Be4! R~ 2.e3#

c) 1.f4! B~ 2.e3#

d) 1.Bf5! S~ 2.e3#

In diagram position is Black stalemated and White has to let him move by d-pawn. In other positions Black could check White, therefore White has to disable checking possibility for Black. Quite strong keys, not according to the standard twomover canons, but well, you can try more subtle keys in your own works...









#2 (6+3)
No captures
b) bRe3
c) bBe3
d) bSe3

Vyacheslav Pilchenko
Mat-Pat 1989

a)
1.Sb2! th. 2.Qd4,Qxb5#
1...Rxb2 2.Qd4#
1...Bxb2 2.Qxb5#

b)
1.Bb2! th. 2.Qd4,Qxb5#
1...Rxb2 2.Qd4#
1...Bxb2 2.Qxb5#

c)
1.Rb2! th. 2.Qd4,Qxb5#
1...Rxb2 2.Qd4#
1...Bxb2 2.Qxb5#

Cycle of types of pieces in the role of key piece and the one standing on e8: RS-SB-BR.









#2 (8+11)
b) wSe8
c) wBe8

György Bakcsi
1st Prize Sakkelet 1990

a) 1.Se3! th. 2.Qxe4#
1...R~ 2.Qf5#
1...Rf4! 2.gxf4#
1...Rxe3! 2.Qxe3#

b) 1.Sgf4! th. 2.Qxe4#
1...B~ 2.Qf5#
1...Bxd3 2.Sxd3#
1...Bxg6 2.Sxg6#

c) 1.Sh4! th. 2.Qxe4#
1...S~ 2.Qf5#
1...Sxd6! 2.Qxd6#
1...Sxg3! 2.Qxg3#

Any jump away by Sg2 threats Qxe4, random defence by a piece standing on e4 allows Qf5#. And then it becomes really interesting: there are two black corrections in each phase, leading to the set of mates. But what about key? What motivates selection of the key move? With

  • bRe4 - it is necessary to shield Se2,
  • bBe4 - it is necessary to prepare two mates on d3 and g6,
  • bSe4 - it is necessary to reguard f5 against possible 1...Sxf6, but d3-g3 line should not be cut.
Also pure change of type Z-32-66 (and with some reservations also Z-33-79).









#2 (12+10)
b) bBe4
c) bSe4

Fernand Joseph
F028 Probleemblad 1997

a) 1.Ke2{Ig2}! Sb3{Ih4} 2.h8Q{Ih5} Sa1{Ig3} 3.Qg8{If3} Sc2{Ih4} 4.Qd5{Ie1} Kc5{Id2} 5.Ke1{Id1}#

b) 1.h8Q{If3}! Bc3{Ih5} 2.Qh4{Ih1}+ Kc5{Ig2}+ 3.Qg3{If1} Be5{Ih3} 4.Qg4{Ih4} Bb2{Ie1} 5.Qc4{Ia1}#

c) 1.h8R{If3}! Ra6{If8} 2.Rh1{If1} Ra8{If3} 3.Rg1{Ie3} Rd8{Ih3} 4.Rxg5{Ih7} Rd5{Ih4} 5.Rg4{Ih3}#

No big strategy, rather precise play by White who manipulates bK into mating net in a) and b) and bR to blocking square in c). Moreover it is necessary to keep bS and bB in bay.









#5 (2+5+1)
Maximummer
imitator f2
b) bBa1
c) bRa1

Kari Valtonen
on occassion of Henry Tanner 50
70 Suomen Tehtäväniekat 2/2005

a) 1.Be3! Gf3 2.Bf4 Gf5 3.Bg5 Gh5 4.Bh6 Gh7 5.Bg7#

b) 1.Sd3! Ge3 2.Se5 Ge6 3.Sg4 Gh3 4.Sh6 Gh7 5.Sf7#

c) 1.SHc3! Gd3 2.SHf4 Gd7 3.SHg7 Gh7 4.SHf4 d5 5.SHe7#

While previous problem is really hard to grasp, these manoeuvres are understandable. White has to pull bG to h7 and the way depends on the piece standing on f2 in the initial position. Clean position, the only slight blemish being bp needed only in the camel position.









#5 (2+3)
Maximummer
grasshopper b3
b) wSf2
c) white camel f2

Comments to Juraj Lörinc.
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