Bo Waldemar Lindgren (26021927  04062011) Swedish composer and Grandmaster

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 twopart 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 198688 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 sers#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.