Long fairy helpmates 7

You can sometimes find longer fairy helpmates in awards of tourneys. Here are some of them selected randomly from my (limited) sources. It is hard to generalize, but one can say they usually don't get prizes. Either they don't exploit fairy elements enough (compared to shorter helpmate forms) or the theme is not ambitious enough (compared to orthodox helpmate moremovers) or whatever you can imagine... Just look below.
Frantisek Sabol
3rd Comm Phénix Divers 1995

a) 1.c4 Rxc4(pc7) 2.Kd8 Rxd4(pd7) 3.f5 Rxe4(pe7) 4.fxe4(Rh1) Rh8#

b) 1.e3+ Kxf3(pf7) 2.d3 Kxe3(pe7) 3.Ke8 Kxd3(pd7) 4.bxc3(Ra1) Ra8#

Judge Hans Gruber: "Nice echo with change of active white piece."

h#4 (2+6)
b) f6 -» b4

Didier Innocenti
4th HM Phénix Tanagras 1990

1.Qb3R cxb3S(Ra8) 2.Ra1B Sxa1B(Bf8) 3.Bg7S Bxg7R(Sb8) 4.Sd7p Rxd7Q#

Judge Maryan Kerhuel: "Analogically to four promotions we can see this problem executes theme of four metamorphoses in Einstein chess in both directions."

h#4 (1+2)
Einstein chess, Circe

Václav Kotesovec
Prize for miniature Martin 1994-95

1.Kc6 Ge7 2.Sf4 Ka6 3.Sd5 Ge5 4.Gd6 Ge8#

1.Gd8 Ge8 2.Sd4 Gc6 3.d6 Ge6 4.Gd5 Ge7#

1.Kc4 Ge8 2.d5 Ge5 3.Sf4 Ka4 4.Sd3 Ge6#

Judge Ladislav Salai: "13 halfmoves out of 24 are done by grashoppers. Thanks to them we can enjoy 3 echo mates."

h#4 (3+4)
2+1 grasshopper

Peter Wong
1st HM Phénix Tanagras 1990

a) 1...Sb3 2.RRBc3 Sc1 3.RRBc5 Sd3+ 4.RRBf5 Se5 5.RRBg5 Sf3#

1...Sc2 2.RRBc3 Sa3 3.RRBe3 Sc4+ 4.RRBe6 Se5 5.RRBe7 Sc6#

c) 1...Sc2 2.RRBd2 Se3 3.RRBd4 Sg4 4.RRBa4 Se5 5.RRBb4 Sc6#

d) 1...Sb3 2.RRBc2 Sc1 3.RRBc3 Sb3 4.RRBd3 Sd4 5.RRBd2 Sf3#

Judge Maryan Kerhuel: "Four echo mates in which royal rook bouncer occupies 4 squares where the mate is possible with white knight on white square. Nice technical performance. Pity that twinning is not totally homogeneous and that the move 2.RRBc3 is repeated."

h#4,5 (1+1)
royal rook bouncer f3
a) 2 solutions
b) f3 -» d1, 1 solution
c) f3 -» f2, 1 solution

Petko Petkov
3rd Comm Phénix Tanagras 1988

a) 1.Kb7 Kd6 2.Ga7 Kd7 3.Ka8 Kc8 4.Bb3 Ga2+ 5.Ba4 Ga5#

b) 1.Ge5 Kc7 2.Gb8 Kb6 3.Bg6 Ka6 4.Bf7 Gg8+ 5.Be8 Gd8#

Judge Bo Lindgren: "Nice precise echo, but I'd prefer form of two solutions."

h#5 (2+3)
1+1 grasshopper
b) d5 -» b6

Václav Kotesovec
3rd HM Phénix Tanagras 1990

1...Kc2 2.(2,5)c6 Kd3 3.(2,5)h4 Ke4 4.(2,5)c2 (2,5)d3 5.(2,5)a7 (2,5)b8 6.(2,5)f5 (2,5)g6 7.(2,5)a3 (2,5)e1 8.(2,5)c8 Kf5 9.(2,5)e3 Kg6 10.(2,5)g8 (2,5)c6#

Judge Maryan Kerhuel: "A problem for solvers, but soft control of white king road also has some necessary artistical touch..."

h#9,5 (2+2)
1+1 (2,5)-leaper

Comments to Juraj Lörinc.
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