Circe problems without captures 4

This is the fourth special example file for the 26th TT CCM C 11.11.2009.

The deadline is near, nevertheless, there is still a vast space for originality as is proved by flow of originals for the tourney. To stimulate other authors and perhaps to turn their attention somewhere else, here is the selection of stalemate problems with Circe variants and with no captures, in the spirit of the tourney.

Alain Goset
1st-5th Prize Europe Echecs 1969

1.Bh1 c6 2.Qg2 c7 3.Qa8 c8=R 4.Bb7 Rc7=

Three black pieces free in the diagram position are immobilized in three different ways. Black queen is incarcerated in the corner, Bb7 is pinned and Bb8 cannot capture white rook due to the Circe self-protection. Although largely unrelated, Turton manoeuvre reminded me of #4 by Heinz Curth.

h=4 (2+4)

Zvonimir Hernitz
feenschach 1980

1.Rh7! Kg5 2.Sf7+ Kg4 Rh3=
    2...Kg6 3.Ke6=

Nothing deep, just an example of two different stalemates with Circe self-protected rook.

=3 (3+1)

Romeo Bedoni
Phénix 1994

1.Rg3 2.Qg2 3.Bg1 4.Qh2 5.Rg2 6.g3 Sf3=

Exchange of place by bQ and bB, but the key point of the problem lies in the final position. Although seemingly free, bK cannot move anywhere as thanks to the Circe Assassin all square around him are unsafe:

  • d8 due to Sxh2(Qd8),
  • d7 due to Bxd3(Pd7),
  • e7 due to Kxe6(pe7),
  • f7 due to Kxf4(pf7),
  • f8 due to Sxg1(Bf8),
always with removal of bK.

ser-h=6 (4+11)
Circe Assassin

Christian Poisson
2nd Place 15th TT Problemkiste 1995

1.Bc6? Kc7! 2.b8Q Kc8!

1.Rc7! Kc8 2.b8Q Kb7 3.Bc6 Kb6 4.Qb7 Kc5 5.Qb6 Kb4 6.Qc5 Kc3 7.Qb4 Kb2 8.Qc3=
(3...Kc8? 4.Bb7=)

The most unexpected stipulation of this file. With very limited resources White manages to tame Black king in Mars Circe. It is possible due to the fact that wR and wQ limit strictly the movement possibilities for bK to two columns. Add to that diagonal activity of wB from f1 and wQ from d1 and it works. The try fails to the tempo win of Black, impossible in the solution thanks to the secondary stalemate on the other side of the board.

=8 (3+1)
Mars Circe

Manfred Rittirsch
1st Prize diagrammes 1999

1.d1nG f8nR+ 2.h1nQ Be1=

1.h1nG f8nQ+ 2.d1nR Gg1=

Fantastic idea - three promoted neutral pieces pin each other cyclically in both solutions by standing on rebirth squares from the White's viewpoint. Moreover, the nature of promoted pieces is cycled too thanks to the simple trick (simple once you find it...).

h=2 (5+1+3)
Anticirce type Cheylan
grasshopper g7

Klaus Wenda
Problemkiste 2002

1.Kh5 2.Kg4 3.Kf3 4.Ke2 5.Ge1 6.Kd1 Qd3=

This stalemate position simply asks for echo, e.g. by change of Anticirce to Mirror Anticirce or some other kind...

ser-h=6 (1+3)
Anticirce type Cheylan
0+2 grasshopper

Jorge J. Lois
Jorge M. Kapros

3rd HM Shakhmatna Misl 2004

1.b1Q g8B 2.Qe1 Bf7 3.Kf1 c8R+ 4.g1S Re8=

Another AUW based on the peculiar properties of the final position. wRe8 pins bSg1 and blocks d8. bQe1 is pinned by wK. Black cannot promote bishop on g1 due to the check from f8 and rook at b1 due to the check from a8, while White needs wB to guard g2.

h=4 (4+5)
Mars Circe

Lennart Werner
Comm Springaren 2005

Black stalemated: 1.g1=B Kb7 2.Ka8 f8=R+ 3.Ba7 Rb8=

White stalemated: 1.f8=S Kb7 2.Sd7 g1=Q+ 3.Sb8 Qa7=

In both solutions pinned king is at a8, while the other blocks b7. Another analogy might be found in the play of promoted pieces - bQ blocks a7 and pins white knight on b8 (from d8), wR blocks b8 and pins black bishop on a7 (from a1).

h=3 (5+4)
Mars Circe

Horst Backer
Problemkiste 2007

1.h4! Bh2 2.h5 Bg3 3.h6 Bh4 4.h7 Bg5 5.h8=B Bh6 6.Bg7 Bg5 7.Bh6 Bh4 8.Bg5 Bg3 9.Bh4 Bh2 10.Bg3 Bg1 11.Bh2=

Nothing especially exciting, unfortunately Black has no place to lose tempo , otherwise the play might have been much richer. Bishop promotion is noteworthy, but straightforward.

=11 (3+8)

Comments to Juraj Lörinc.
Back to main page of Chess Composition Microweb.