I'm stuck as usual. But also want to see if there a Swordfish here for the twos:
R4C2 to R4C5
R7C1 to R7C6
R8C2 to R8C4
Is this right? And if so, does t his allow me to rid the 2s in R5C4 & R6C4?
(5)(7)(9) | (4)(2)(8) | (3)(1)(6)
(8)(4)(2) | (3)(16)(16) | (9)(7)(5)
(3)(1)(6) | (9)(7)(5) | (2)(8)(4)
(4)(26)(17) | (1267)(5)(3) | (167)(9)(8)
(26)(3)(8) | (1267)(9)(27) | (5)(4)(12)
(9)(5)(17) | (1267)(8)(4) | (167)(36)(123)
(26)(9)(3) | (8)(16)(27) | (4)(5)(17)
(1)(26)(5) | (27)(4)(9) | (8)(36)(37)
(7)(8)(4) | (5)(3)(16) | (16)(2)(9)
If so. The puzzle solves itself from this point.