Павел Павлович Бобров (29-08-1862 - 23-12-1911) Russian composer (Pavel Pavlovich Bobrov)
If you can read Russian, you will discover
details about Pavel Bobrov on Wikipedia.
While
this threemover deserves a look because of its presentation of two en-passant captures (maybe the first achievement of this task in a threemover) this endgame study has been selected:
Бобров, Павел Павлович
Шахматное обозрение 1904
Show Solution1. g4 ! hxg3
{1... Qxg4 2. Bc8+ wins the bQ}
2. Sf4+ Kd6 3. Sxh5 gxh5 4. Kg2 !
{4. Bc8? g2+ 5. Kxg2 Bf4 -}
4... h4 5. Bc8 Kc5 6. Bd7 Kb6 7. Bf5 +-
Steffen Kamstra (29-08-1863 - 1930) Dutch composer
Steffen Kamstra was well-known at the beginning of the 20th century.
Kamstra, Steffen
Tijdschrift vd NSB, 1926
1
st Prize
Show Solution1.Kf3-g2 ! ZZ
1…Sxa5 2.Rf2 (3.Rc2#)
1…Sc1 2.Rf5 (3.Qxd5#)
1…Sxd4 2.Rf3 (3.Rc3#) Se2,Sb5,Sxf3 3.Qxd5#
Duel between bS and wR.
1…Sd2 2.Rc1+ Kd3 3.Rc3#
1…Bf8,Bh6,Bh8 2.Qg3 (3.Qxb3,Qc3,Qd3#) Rg5/Rxd4 3.Sxb3/Qg8#
1…f5 2.Qb8 (3.Qxb3,Qb4,Qb5#) Rxd4 3.Qg8#
(1…Sa1/Sxc5 2.Qe6,Rf3,Rf5/c8=Q,c8=R)
This problem was published in the FIDE Album.
Jens Otto [Oluf] Pedersen (29-08-1880 - 15-04-1968) Danish composer
Jens Pedersen composed such elegant problems as this aristocratic selfmate:
Pedersen, Jens Otto
Problemnoter, 1956 (547)
4
th HM
Show Solution1. Sd1+ !
1. ... cxd1=Q+ 2. Qd4+ Qxd4#
1. ... cxd1=R+ 2. Qd2+ Rxd2#
1. ... cxd1=B 2. Qf3+ Bxf3#
1. ... cxd1=S 2. Qe3+ Sxe3#
Allumwandlung in a pawnless position.
Johan Rudolf Jessurun Lobo (29-08-1893 - 16-12-1978) Dutch composer
This s#5 is unfortunately cooked but
this helpmate in 2 with Nightriders was good and here is another good fairy helpmate:
Jessurun Lobo, Johan Rudolf
Probleemblad, 1958 (v)
Show Solutiona) 1. c1=S Sd5 2. Sb3 Bb4 3. Sc5 Ge6 4. Sd7 Be7#
b) 1. c1=B Gxd4 2. Ba3 Sb5 3. Bf8 Sd6 4. Be7 Ba5#
c) 1. Ke8 Gf4 2. c1=R Sd5 3. Rc8 Bb4 4. Rd8 Sf6#
The black promoted units goes a long way to selfblock the black King.
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