Helpmates with 5 thematical phases 3

As the first reaction to the announcement of 15th TT CCM I got an e-mail from Chris J. Feather (CJF), who shared with me his knowledge about ancient and perhaps the first intentional helpmates with 5 phases. I write intentional as it is possible that some composer before Pauly composed some helpmate with less phases and intended and just required missing number of cooks...
Wolfgang Pauly
1564 Chess Amateur 01/1930

1...Sd5 2.Kc8 Sb6#
1...Sd5 2.Ke8 Sf6#
1...Sd3 2.Ke6 Sc5#
1...Sd3 2.Kc6 Se5#

1.Ke6 d5+ 2.Ke5 Sc6#

CJF: "It was interesting to read about your new TT. Five h# phases is not so uncommon. There are well over 200 examples in my h#2 collection. I believe the earliest example (h#2, but probably also the earliest h#n with 5 phases) is Pauly's no.1564 in 'The Chess Amateur' January 1930 (PDB P0513828). However this problem does not have 5 genuine phases, but only one solution and a 4x-branching set play."

h#2* (8+3)

Wolfgang Pauly
dedicated to Josif Schlarko
277 Magyar Sakkvilag 10/1930

1.Sd6 Qb3 2.Se4 Rd5#

1.Sd6 Qf3 2.Sc4 Rd5#

1.Kc5 Qc8+ 2.Kb6 Ra6#

1.Kc4 Qxa3 2.Sd4 Rc5#

1.Ke4 Qg3 2.Sd4 Re5#

CJF: "From the same year and by the same composer however, there is no.277 in 'Magyar Sakkvilag' (October 1930; P0550530) which is of the form + I have Pauly's own diagram of this before me as I write (he sent it to Dawson for his records), and it is inscribed, in a mixture of English and French: 'Helpmate in 2 Type Neumann 5 échos'."

h#2 (3+3) +

Lajos Bukovinszky
Romanian tourney 1962

a) 1.Kd3 Qf1+ 2.Ke4 Sf6#

b) 1.Kd5 Sf4+ 2.Kd6 Bc5#

c) 1.Kf3 Bb3 2.e4 Bd1#

d) 1.Kd5 Kb5 2.e4 Sf4#

e) 1.Kf5 Sf4 2.Kf6 f8Q#

Compare this problem to h#2 by Henry Forsberg with the almost the same twin mechanism - here the twins are made by changing white unit instead of black one. Pity that some black moves are repeated, 1.Kd5 and 2.e4...

h#2 (5+3)
b) wRf7, c) wBf7
d) wSf7, e) wpf7

Ricardo Vieira
Mario Figueiredo
Jose Figueiredo

1st Prize Sredba na Solidarnosta 1977

1.Sxc2 Ba4 2.Kc4 Bb3#

1.Sxb5 Rd2+ 2.Kc6 Sd8#

1.Sxe6 Rc3 2.Kd4 Rd3#

1.Sxf5 Rd2+ 2.Kxe6 Bc4#

1.Sxf3 Rc3 2.Ke4 Bc6#

All the first black moves are made by Sd4, all second by bK. Furthermore, note that in the first triplet of solutions there is a cyclical Zilahi there!

h#2 (12+9)

László Anyos
6056 Die Schwalbe 108 - XII 1987

1.Be3 Rxf6 2.Kf4 Bb8#

1.Be3 Rc5+ 2.Kd4 c3#

1.Bd5+ Rc4 2.Kd6 Bb8#

1.Qg6 Be6 2.Kf6 Bd4#

1.Sd5 Bxg4 2.Qf4 Re6#

The same combination of solutions as in h#2 by Wolfgang Pauly.

h#2 (6+6) +

Hilding Fröberg
Christer Jonsson

8632 Springaren 1997

a) 1.Kd3 d7 2.Kc2 d8Q 3.Kxb1 Qd1#

b) 1.Kd5 b6 2.Kc6 d7 3.Kb7 d8S#

c) 1.Kxf3 g7 2.Kg4 g8B 3.f3 Be6#

d) 1.Kf5 g7 2.Kg6 g8R+ 3.Kh7 Rg7#

e) 1.Kxf3 d7 2.Kg2 dxc8Q 3.Kh1 Qh3#

Position a), b), d), e) form big star of bK and there are 5 different promotions. Compare to h#3 by Fadil Abdurahmanovic. Here in the solution not fitting to big star bK moves - but he is mated on the different square. Also there 4 different promotions making full AUW.

h#3 (7+10)
a) a2 -» a3
b) b1 -» a4
c) b1 -» h3
d) b1 -» f7
e) b1 -» e2

Comments to Juraj Lörinc.
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