... by: newby

4B4
6D5, 2E5>contradiction
therefore 6F5
2E5, 1(F4, F6)>contradiction
therefore 2D5
1F6, 2E7> contradiction
1F6, 2E9, 9E7>contradiction
1F6, 2E9, 7E7, 3(E5, F4)> contradiction
therefore 1F4
2E9> contradiction
therefore 2E7 solves

... by: Fab

Starting from where the solver is stuck (without Bowman).
column forcing on the three possibilities in column 7 leads, through basic strategies only, to the solutions 5A3 4D1 5D9 5F1 7G1 6H1
After this the sudoku becomes solvable (extreme)

... by: JC Van Hay

... UR(35)BH56=8B5 ...

... by: JC Van Hay

FWIW, instead of using the straightforward UR(35)BH56=8A5 in #4 :

#4. J5=7->UP57; Broken Wing 3AE1+3F24+3A4 :=> J6=7; UP36; NP(89)D46; UP81

... by: ray

Get working board using basic strategies.Choose cellG4{178} - bi-location cell for for value 1. Forcing G4 = 1 or 7 gives an inconclusive result, G4 = 8 gives a solution. Using this solutuon(with basic strategies), other 1-cell solutions are: J3 = 8 ,J6 = 7 , J8 = 3, J9 = 2, E5 = 7, A7 = 9, A8 = 2, A4 = 7; somme 2-cell solutions are: F12 = (5,8), C23 = (6,4),DE9 = (5,8).

... by: JC Van Hay

#1. B4=4; UP24; HP(45)FJ7; NT(359)H456; LC(456)
#2. Kite(7C67.AF8) :=> -7F6
#3. J2=7->UP40; Chain[5] : 9A7=9E7-9E3=HP(59-6)AD3=6C3-6B2=6B9 :=> A7=9;->Contradiction(Singles); G1=7
#4. J5=7->UP57; B5=3or5or8->Contradiction(Singles) :=> J6=7; UP36; NP(89)D46; UP81