Bo Waldemar Lindgren (26-02-1927 - 04-06-2011) Swedish composer and Grandmaster
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Bo Lindgren and Norman Macleod
Benidorm 1990
[Wikipedia] |
Bo Lindgren was the son of Fritjof Lindgren (see
February 23rd). International Judge since 1966 and Grand Master since 1980, he was the editor of the problem columns of
Skakbladet from 1952 until 1957 and collaborated with
Stella Polaris from 1966 until 1970.
A collection of 197 of his best problems was published in 1978 by Walther Jørgensen, titled "
Maskrosor" (Dandelions).
John Rice wrote an inspiring two-part article "The Lindgren Legacy" selecting 52 of Lindgren's problems in The Problemist November 2011 and January 2012: a worthy homage to the Swedish GM. The problems below were also quoted by John Rice.
Let's start with a longmover:
Lindgren, Bo Waldemar
Probleemblad, 2000
2
nd Prize
Show Solution1.Rc2? ZZ Kd3!
1.Kc2? ZZ Ke4!
1.Bc2! ZZ Kc4 2.Bf5+ Kd4 3.Kc2 Kc4 4.Kd1+ Kd4 5.Bc2 Kc4 6.Bb1+ Kd4 7.Rc2 Ke4 8.Rc5+ Kd4 9.Kc2 Ke4 10.Kc3+ Kxf4 11.Rxd5 Rxd2 12.Rf5#
Many successive interferences at c2.
Critical and anticritical play, combined with pendulum manoeuvres.
Lindgren, Bo Waldemar
Mat (Beograd) 1984
Show Solution1.g1=S 2.Se2 3.Kg2 4.Kg1 5.Sc1 6.a1=Q 7.Qa8 8.Qh1 Bd4#
1.g1=B 2.Be3 3.Bc1 4.a1=R 5.Ra4 6.Bf4 7.Ke3 8.Re4 Rb3#
Allumwandlung with only 6 pieces.
One of Bo Lindgren's most impressive achievements was the following
series selfmate:
Lindgren, Bo Waldemar
feenschach, 1987
Dedicated to Walter Jörgensen for his 70
th birthday
1
st Prize
Show Solution1. e8=S 2. Sxf6 3. Sd5 4. f6 5. f7 6. f8=Q 7. Qxf3 8. Qg2 9. f4 10. f5 11. f6 12. f7 13. f8=B 14. Bxh6 15. Be3 16. h6 17. h7 18. h8=R 19. Rxh4 20. Re4 21. h4 22. h5 23. h6 24. h7 25. h8=Q 26. Qxb8 27. Qe5 28. b8=B 29. Bxa7 30. Bc5 31. a7 32. a8=R 33. Rxa4 34. Rb4 35. a4 36. a5 37. a6 38. a7 39. a8=S 40. Sb6 41. Sc4 42. Qxe2+ Sxe2#
Double white Allumwandlung ! An incredible achievement in series selfmate. Unsurprisingly it was awarded 12 points (out of 12) in the 1986-88 FIDE Album.
However, it was cooked by Arnold Beine in 2022: : 1.e8=S 2.Sxf6 3.Se4 4.Sd2 5.f6 6.f7 7.f8=Q 8.Qxf3 9.Qxe2 10.f4 11.f5 12.f6 13.f7 14.f8=B 15.Bxh6 16.Be3 17.h6 18.h7 19.h8=R 20.Rxh4 21.Re4 22.h4 23.h5 24.h6 25.h7 26.h8=Q 27.Qxb8 28.Qe5 29.b8=B 30.Bxa7 31.Bc5 32.a7 33.a8=R 34.Rxa4 35.Rb4 36.a4 37.a5 38.a6 39.a7 40.a8=B 41.Bd5 42.Sc4+ Sxe2#
For the achievement of the double white Allumwandlung, please have a look at
this ser-s#37 problem by G.P. Sphicas, 1st Prize StrateGems 2013.
Lindgren, Bo Waldemar
Ujcs, 1951
1
st Prize
Show Solution1.Qf2 ! ZZ
1...e2 2.Rd4+ exd4 3.Qf5+ Ke3 4.Bb8 d3 5.Be5 d2#
1...exd2+ 2.Kd1 Kd3 3.Qf4 exf4 (3...e4 4.Sh4/Rf2/Rg2) 4.Re2 f3 5.Bf2 fxe2#
1...exf2+ 2.Kf1 Kf3 3.Re2 e4 4.Rd2 e3 5.Rd4 e2#
An amazing key is followed by good play in three lines.
Gerhardus Johannes Gerard Smits (26-02-1950) Dutch composer and International Master
Gerard Smits composes in all genres: direct mates, helpmates, selfmates and fairy problems, and he has won prizes in all. He enjoys cyclic themes, but his cycles are always elegant and expressive.
Smits, Gerhardus Johannes Gerard
Probleemblad, 1983
1
st Prize
Show Solution1.Qxc6 ! (2.e5+ Be4 3.Rf1#)
1...Rb6 2.Qc5 (3.Qe3#) [2...Qxa7?]
1...Rb8 2.Qd7 (3.Qg4#) [2...Qc8??]
1...Rxc7 2.Qd6 (3.Qf4#) [2...Bxc7??]
1...Rxa7 2.Qxb5 (3.Qe2#); 1...Qxa7 2.Qd7 (3.Qg4#)
Self-obstructions by the black Rook.
A fairy helpmate:
Smits, Gerhardus Johannes Gerard
The Problemist, 1985
2
nd Prize
Show Solutiona) 1. Kxd5[+wPd2] fxe5[+bBf8] 2. Kxe5[+wPe2] e4 3. Kd4 Ge3 4. Bc5 d3#
b) 1. Kd4 dxe6[+bBc8] 2. Bxe6[+wPe2] Gf3 3. Bd5 Ge1 4. Ke4 e3#
The mates are specific: in a) 5.Kxe3 is illegal because the white Grasshopper is reborn on e8 with check.
Chameleon echo mates.
Smits, Gerhardus Johannes Gerard
Thema Danicum 1989
1
st Prize
h#2 | Circe
b) bPh6->h5
c) bPh4->h5 | 4 + 13 |
Show Solutiona) 1. Bxc2 Bxd4[+bPd7] 2. Bxb2[+wQd1] Qxg4[+bPg7]#
b) 1. Bxb2 Rc6 2. Sxf2[+wBc1] Bxf4[+bPf7]#
c) 1. Sxf2 Qb6 2. Bxc2[+wRh1] Rxh5[+bPh7]#
Black cycle of moves and captures of white pieces. Cyclic
Zilahi.
The aim of Black's first move is to free the rebirth square of the white piece that will be captured at the 2nd move and that will give mate.
Jaroslav Štúň (26-02-1951 - 24-02-2021) Slovak composer
Jaroslav Štúň composed orthodox problems at the beginning of his career. He has focused on fairy problems and has been more successful at them. He proposes a wide collection of his works on
his website. Some of his problems show extreme fairy exotism. His use of ultra-economical positions and combinations of fairy conditions and pieces manages to surprise solvers, composers and sometimes also, unfortunately, tourney judges.
Short announcement of his death in
PataMat 115.
Štúň, Jaroslav
Východoslovenské noviny 1986
1
st Prize
Show Solution1... Re2 2. Bxd5 Sf5#
1... Bf2 2. Rxd5 Sc4#
1. Bxd5 Sc4+ 2. Ke6 Bxg4#
1. Rxd5 Sf5+ 2. Kc5 Rc2#
Štúň, Jaroslav
Šachová skladba 2007
h=2.5 2 sol. | Sentinelles
Neutral Royal Bishop | 1 + 0 + 1n |
Show Solutiona) 1...nrBf8[+wPe7]+ 2.nrBg7 e8=B 3.nrBh8[+bPg7] Bg6 =
1...c8=S+ 2.nrBf6[+bPe7]+ Sxe7 3.nrBh8[+bPf6] Sf5[+wPe7] =
b) 1...c8=Q+ 2.nrBf7 nrBg6[+wPf7] 3.nrBh7[+bPg6] Qc2 =
1...c8=R+ 2.nrBf7 Rc6 3.nrBe8[+bPf7] Rf6[+wPc6] =
Allumwandlung.
The problem was placed 2nd-3rd ex-aequo in the award of the
Wenigsteiner Jahrespreis 2007.
Narayan Shankar Ram (26-02-1961) Indian composer and International Master
Narayan Shankar Ram seems to enjoy cycles of moves and letter-themes too. He won the 3rd and 4th WCCT fairy sections (one of them with Ukrainian cycle problem) and also many other prizes. He became IM in chess composition in 1992 at the age of 31 (he currently has 43.50 points in the FIDE albums). He is one of the world experts in reflexmates.
A clean helpmate to start with:
Shankar Ram, Narajan
The Problemist, 1984
1
st Prize
Show Solutiona) 1.Rg3 Bxd7 2.Qe4 Bc6 3.Be2 Rf7#
b) 1.Be2 Rxd7 2.Qf4 Rf7 3.Rg3 Bc6#
The same white piece is unpinned, pins itself, is directly unpinned by the bQ and finally pins the bQ. Interchanges of functions (wB/wR, bBa6/bRg8).
Something tougher:
Shankar Ram, Narajan
4
th WCCT G2, 1989
1
st Place, 1989-1992
Show Solution1.Ree1 ! (2.Rxa1 [3.Qd4#] b2 3.Rb1 [4.Qd4#])
1...b2 2.Re5+ [A] Bxe5 3.Sf4+ [B] Rxf4 4.Sf6# [C]
1...g3 2.Sf4+ [B] Rxf4 3.Sf6+ [C] Qxf6 4.Re5# [A]
1...Bf7 2.Sf6+ [C] Qxf6 3.Re5+ [A] Bxe5 4.Sf4# [B]
White cycle of moves ABC-BCA-CAB.
And three enjoyable reflexmates to conclude this rich selection:
Shankar Ram, Narajan
The Problemist, 1988
1
st Prize
Show Solution1.Sg1! (2.Sh3 Rxg4#)
1...Kf2/Kd4/Ke4/Kf4 2.Rh3/g5/Kg5/Rh6 Rxg4/Re4/Qd8/Rxh6#
Other variations: 1...Ra4~/Ra5/Rxg4+ 2.g5/Rh6/Kh3 Re4/Rxh6/Sf2#
1...Bxg4/Qd8+ 2.Kg5/Rg5 Qd8/Rh6#
4th degree correction of the bK. Correction elements:
Kf2 (vacation of 3rd rank allowing Rh3)
Kd4 (interferes with Ra4 but unpins g4)
Ke4 (blocks e4 but guards f5
Kf4 (stops Kg5, but also guards that square)
Shankar Ram, Narajan
The Problemist, 1987
1
st Prize, ex aequo
Show SolutionThree tries fail specifically due to reflexmate:
1.Qxh5/Qxg2/Qf4? O-O/O-O-O/Rd8! 2.Qh7/Qa8/Rxe6#
1.Qh3 ! (2.Sce4 Sf4#)
1...Bxc5 2.Se4 Sf4#
1...0-0-0 2.Sd7 Sf4#
1...0-0 2.Sb3 Sf4#
wSc5 unpins bSe6 on three lines.
Shankar Ram, Narajan
The Problemist, 1989
1
st Prize
Show Solution1.R8c6? (2.Rb2
[A] Qxc6/bxc6#)
1...b6/b5 2.Sxd4
[B], Qxe5
[C] Rxd4, Rxe5# but 1...Qb1!
1.R8c5? (2.Sxd4
[B] Rxd4#)
1...b6/b5 2.Rb2
[A], fxe4
[D] Qxc5, Bxe4# but 1...Rxf5!
1.R8c4 ! (2.Qxe5
[C] Rxe5#)
1...b6/b5 2.fxe4
[D], Rb2
[A] Bxe4, Qxc4#
1...Rxf3, Rxf5 2.Kxe4, Qxe5
Zagoruiko (1...b6/b5)
Reciprocal Change (2nd Try - Actual play)
Le Grand (1...b6 in tries, 1...b5 - 1st try & key)
L.A. Kuijers (26-02-1864 - 07-03-1930) Dutch composer
Kuijers, L. A.
Haagsche Post, 1917
2
nd Prize
Show Solution1.Rf3! (2.Qd3#)
1...Kxf3/Qe2+/Qxf3+ 2.Qe2#
1...Sxd7 2.Qxb7#
1...Ba6 2.Qd5#
1...Rxd7 2.Sg5#
1...Sxf4 2.Rxf4#
1...Qxd2+ 2.Sxd2#
1...Qxf1+ 2.Rxf1#
Flight-giving key.
The problem was reproduced in the FIDE Album.
Józef Klemensiewicz (26-02-1874 - 12-06-1926) Polish composer
Józef Klemensiewicz was a poet and translator. He preferred to compose miniatures with sacrifices of white pieces.
Klemensiewicz, Józef
L'Echiquier, 1926
Show Solutiona) 1.Sc2 ! (2.Qf5#) 1...Kd5 2.Qc8 Ke5/Kxe4 3.Qf5#
b) 1.Kg5 ! ZZ 1...Kd4 2.Qg3 Kc4, Kxe4 3.Qc3, Qf4#
Mates in diagonal/orthogonal correspondence.
Alberto Batori (26-02-1884 - 05-11-1923) Italian player and composer
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Alberto Batori |
Alberto Batori is known as the composer of the task of 11 direct checks given by the same piece (bQ) as a defence in a twomover (see it
here). He is also remembered as the editor of
L'Italia Scacchistica, from 1914 until his death in 1923.
He composed some retroanalytical problems too.
Batori, Alberto
Good Companion, 1919
1
st Prize
Show Solution1.Qc4! ZZ
1...d6 2.Qe6#
1...d5/Rxc4/Re5 2.Re5#
1...Qa7/Qa8 2.Qf1#/Sd6#
1...Qb6/Qc6/Qe6/Qf6/Qg6/Rb5/b5 2.Qf1#
1...Qd6+/Qb5/Qxc4/b6 2.Sxd6#
1...Rc6/Rc7 2.Re5#/Qd5#
1...Rc8/Rd5 2.Qd5#
1...g6 2.Sxh6#
1...g5/Be7+ 2.Se7#
1...Bg5/Bf6/Bd8 2.Sxg3#
6 different interferences by black Pawns (d6, d5, b6, b5, g6, g5).
Viktoras Paliulionis (26-02-1965) Lithuanian composer and International Master
Viktoras Paliulionis composes mostly helpmates and recently seems to have focused on moremovers (winning for instance the FIDE World Cup 2013 for helpmates, award
here).
Here is a helpmate twomover composed at the age of 21:
Paliulionis, Viktoras
Lettonie-Lituanie-Leningrad, 1986
1
st-2
nd Place
Show Solutiona) 1. Qc5+ Sd5 2. Qa3 Sb6#
b) 1. Qe3+ Be4 2. Qa3 Bc6#
As a tempo move, the bQ gives a check and a selfpin follows.
Here is a recent longmover that is at least as remarkable as the FIDE World Cup problem:
Paliulionis, Viktoras
Kobulchess, 1
st Aug 2013 (225)
Show Solution1…Sd3 2.Qe8+! Kg7 3.Qh8+! Kf7 4.Bg8+! Ke8! 5.Ba2+ Kd7 6.Bb1 Kc6 7.Ka2 Kb5 8.Qa1 Ka4 9.Bb2 Sb4#
1...Sd3 is a very paradoxical move that closes the diagonal of bBh7 and forces it to reach b1 via g8 and a2 (peri-manoeuvre). The black Queen's Indian at h8 is also excellent strategy.
Константин Константинович Сухарев (26-02-1912 - 08-10-2004) Russian composer (Konstantin Konstantinovich Sukharev)
Konstantin Sukharev was a great developer of the land and chess in Novosibirsk. From 1963 to 1993 Sukharev led a column in "Vechernij Novosibirsk" ("Novosibirsk Evening") and organized a lot of solving and composing tourneys. Bondarenko published the results in western magazines and eventually led to contact of Sukharev with John Roycroft. (Information
from Siegfried Hornecker)
Сухарев, Константин Константинович
Шахматный листок 1930 (473)
Show Solution1. Bg3 Qxg3
{1... Qh3 2. Rd6+ Ke4 3. Sf2+;
1... Qg1 2. Sc3+ Kd4 3. Se2+}
2. Sc3+ Ke5 3. Rb5+ and a fork follows:
3... Kf6 / Kd6 4. Se6+; 3... Kd4, Kf4 4. Se2+
The study would also be correct without bPh6, as observed by Arpad Rusz and others:
1.Bg3! Qh6 (1...Qg3 2.Sc3+ Ke5 3.Rb5+ +-; 1...Qg2 2.Se3+-) 2.Sc3+ Kd4 3.Bf4! Qf8 (3...Qf4 4.Se2+-) 4.Rd6+.
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