Immobilization of hurdle 3

And again the examples for our 7th TT CCM... The start is humble, two older one-phase twomovers with grasshoppers only, but some other problems are stunning - e.g. #16 with grasshoppers or #13 with chinese pieces.
Milan R. Vukcevich
6th Place Rose d'Or 1973

1.Gd8! zz,
1...Gc5 2.Qd5#
1...Gb5 2.Qc5#
1...Gc7 2.Qd6#
1...Ge3 2.Qe4#
1...Ge2 2.Qe3#
1...Gg3 2.Qf4#

Battery G-Q mates 6 times over immobilized grasshopper.

#2 (10+14)
3+7 grasshopper

Fritz Hoffmann
1st Prize Chessics 1981

1.Gh4! th. 2.e8G#
1...Ge1, Re1 2.Ra8#, Ga8#
1...Gh3, Se4 2.Bxd1#, Gxd1#

R, G and B, G mate from the same square.

Thanks to Joost de Heer for pointing the position error.

#2 (8+12)
2+3 grasshopper

Vaclav Kotesovec
Phénix 1988

a) 1.Kc2! Ka2 2.Ga7+ Ka1 3.Gc7 Ka2 4.Gc1 Ka1 5.Gh1 Ka2 6.Ga8+ Ka1 7.Gc6 Ka2 8.Gd5 Ka1 9.Ge4 Ka2 10.Gf3 Ka1 11.Gb1+ Ka2 12.Gd3 Ka1 13.Gc3 Ka2 14.Gc1 Ka1 15.Kb3 a2 16.Ga3#

b) 1.Kc2! Ka2 2.Ga7+ Ka1 3.Gc7 Ka2 4.Gc3 Ka1 5.Gc8 Ka2 6.Gc6 Ka1 7.Gc5 Ka2 8.Gd5 Ka1 9.Gb5 Ka2 10.Gc4 Ka1 11.Gc3 Ka2 12.Gd3 Ka1 13.Gb3 a2 14.Ga3#

Echo diagonal-orthogonal and systematical movement - good for six pieces! Immobilization element is only a very small part of the whole problem.

#16 (4+2)
3+0 grasshopper
b) g1 -» c1, #14

Gerard Smits
3rd HM Probleemblad 1988

1.Gb6 Ge4 2.Gg1 Gc5 3.Gf3 a4#

1.Gg5 Ge2 2.Gg2 Ge1 3.Gf1 Kc8#

Other kind of echo diagonal-orthogonal, with more strategy. Note the fact that mates use in fact 3 paralyses each, one of them beeing thematical immobilization.

h#3 (9+10)
4+4 grasshopper

Hans Peter Rehm
Die Schwalbe 1991

1.Ke5! th. 2.Kf5+ PAg4 3.Kxg4 ~ 4.Kxf5#
1...PAg6 2.PAbg4+! PAf5 3.PAg5+ PAf5~ 4.PA1g4+ PAf5 5.PAf4#
1...VAf2 2.PAbg4+! VAf5 3.PAg6+ VAf5~ 4.PA1g4+ VAf5 5.PAe4#

Very good moremover showing analogical immobilization of black pao and vao in two variations. Although the second move is the same in both variations, it is spiced by choice of right pao to move.

#5 (6+10)
2+2 pao, 1+2 vao

Hans Peter Rehm
1st Prize 47th TT feenschach 1991
(P. Kniest JT)

1.VAd4? (th. 2.Sc2#) VAd6! 2.VAb6 (th. 3.Sc2#) VAf8 3.PAd4+ PAxb6!
...PAb2 must be decoyed...

1.PAh8? (th. 2.PAh3#) PAf2! 2.PAe8+ VAd6!! 3.PAd8 VAxb4! shows some vao-shaking must be done first...

1.VAd6! (th. 2.Sc2#) VAd4 2.VAe5! (th. 3.Sc2# and 2.PAe8#) VAc5 3.PAh8! (th. 4.PAh3#) PAf2 4.PAe8+ VAe7 5.VAf6+! (not 5.VAd6+ VAxb4!) VAd6! 6.VAe7+! (not 6.PAd8? VAe5!) VAe5 7.VAd6+ VAf6! 8.PAd8! (th. 9.Sc2#) VAd4 9.VAe5! (9.VAc5+? PAxc5! guarding c2) VAc5! 10.VAd4 (th. 11.Sc2#) VAd6 11.VAb6 (th. 12.Sc2#) VAf8 12.PAd4+ VAc5 13.PAd6#!

Immobilization of black vao on c5 with complicated logic of play and multiple echange of vao places. Really excellent moremover.

#13 (8+6)
1+2 pao, 2+1 vao

Michal Dragoun
Comm Kotesovec 40 C 29.7.1996

a) 1.LIgd1! (LIad1?) Rf7 2.Lxe2-f1 LIe1#

b) 1.LIad1! (LIgd1?) Be7 2.Lxb3-a3 LIa2#

So, here we have the case of very humble immobilization. Both thematical hurdles, Le3 in a) and Lc4 in b) have only one possible move in diagram position and this is removed by annihilation by Black in 2nd moves of solutions. As I said - humble, but analogical and with multiple reciprocal changes of functions.

h#2 (6+7)
1+2 lion, 0+4 locust
rook locust e2, bishop locust b3
b) h1 -» a8

Newman Guttman
Probleemblad 1999

1.Kg2 Kd1 2.Ge1 Ke2 3.G8g1 Gd1 4.Kh1 Kf3 5.Bh2 Gf1#

Well, given material requires immobilization as white has only grasshopper and he surely cannot mate over own king. But the play isn't entirely trivial and all pieces on board fly into the same corner to construct fine mating picture.

h#5 (2+4)
1+2 grasshopper

Newman Guttman
StrateGems 9 - Jan-Mar 2000

1.Bb1+! Kg1 2.Sf1+ Rf2 2.Gxg3! g4 4.Sxg4! Gh4 5.Gxe1+ Gxe1#

Multiple immobilization of Gd1 on 4 lines!

s#5 (9+6)
2+1 grasshopper

Hans Moser
10535v Die Schwalbe April 2001

1.Ge6! g4 2.Se3+ Kf2 3.Sxg4+ Ke2 4.Se3+ Kf2 5.Sf5+ Ke2 6.Gh1 Gg5 7.Se3+ Kf2 8.Sg4+ Ke2 9.Kg1 Gg3#

Mate over immobilized pawn g2 is naturally thematical for our 7th TT, but I like multiple switchbacks by wS too.

Thanks to Joost de Heer for pointing the position error.

What is worse, Arnold Beine have found some serious faults here: "Besides the dual 4.Ba4 Ga3 5.Gh1 Ga5 6.Kg1 Ga3 7.Sh2 Ga5 8.Sf3 Ga3 9.Bc2 Gg3# there are some other solutions, e.g. 1.Se3+! Kf2 2.Sg4+ Ke2 3.Bc5 Gd5 4.Rxd5 Ke1 5.Kg1 Ke2 6.Bd1+ Ke1 7.Sh2 g4 8.Ghh1 g3 9.Bf2+ gxf2#. The other solutions start with 1.Gf1! or 1.Ghh1!."

s#9 (9+4)
2+2 grasshopper

Comments to Juraj Lörinc.
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