The solver provides updates on compartments and candidates that can be eliminated from the sudoku puzzle. It realizes it missed that compartment C7 must be 7 and C9 must be 5, eliminating other candidates. It also missed that compartment AE7 limits candidates in H7 and H4. Further deductions are made by considering sequences, candidates in compartments, and Setti's rule until naked pairs and an X-wing technique solve more values.

1. The solver stops at:
Compartment F16 must be 4 to 9 and compartment F89 must be 2 and 3. Therefore
F8=2 and F9=3.
The solver proceeds to the following position:

2. Compartment AF6 must be 3 to 8 or 4 to 9, so 78 can be eliminated from H6 and 89
from J6.
Compartment H12 must be 12 or 78. If compartment H12 were 12, compartment
H49 must be 3 to 8, and if compartment H12 were 78, compartment H49 must be 1
to 6. But H6 is 12 so compartment H49 cannot be 3 to 8 and must be 1 to 6, and
compartment H12 must be 78. H2 must be 8, so H1=7.
The solver then stops at:
The solver has missed that C7=7 (the only 7 in compartment C59, which must
contain a 7). It follows that C9=5 (now the only 5 in compartment C59, which must
contain a 5), eliminating the other 5s in column 9.
The solver has also missed that compartment AE7 is 5 to 9, so that 5 can be
eliminated from H7, and thus H4=5 (the only 5 in compartment H49, which must
contain a 5).
Compartment E47 cannot contain a 2 (or compartment E12 would have to contain at
least one number higher than 6, which is impossible). Compartment E47 must
contain 56 so those numbers can be eliminated from the remainder of row E.
F3=7 (the other candidates are not in sequence with G3), eliminating the 7 from F4.
AC3 must contain 2 and 3, so those numbers can be eliminated from J3.
Compartment DJ4 cannot contain a 1, because then it would also have to contain a
2, but the only 1 and 2 candidates are in the same cell (J4). Thus compartment DJ4

3. must contain a 7, and 7 can be eliminated from B4. Compartment DJ4 must also
contain 456, which can be eliminated from compartment AB4.
AB4 must contain a 2, so 2 can be eliminated from J4.
E4=7 (the only candidate 7 in compartment DJ4, which must contain a 7), eliminating
the 7 from F6. Similarly B6=7 (the only candidate 7 in compartment AF6, which must
contain a 7). 59 can be eliminated from B7 (not in sequence with B6).
Every column except 3 must contain a 5, and column 3 does not contain a 5. Every
row except D and perhaps B must contain a 5. D does not contain a 5, so by Setti's
rule, B14 must contain a 5 and therefore cannot contain a 1.
Every row (other than perhaps B) contains a 6. Column 8 does not contain a 6
therefore row B does not contain a 6 (Setti's rule).
The solver stops here:
Compartment CJ5 cannot contain a 1. If it did, it would have to contain 123 but there
are only two cells with those candidates (H5 and J5), so 8 can be eliminated from A5
(compartment CJ5 must contain an 8) and 1 from H5 and from J5. Compartment
CJ5 must contain a 3 so J5=3 (the only candidate 3 in the compartment) and the 3 is
eliminated from J6 and the 8 from A5.
The naked pair 24 in row H solves H6=1, J6=2, J7=4, H7=2 and H5=4.
The solver then stops at:

4. If A6 were 3, E6 would be 4 (compartment AF6 must contain a 4 and E6 would be
the only remaining candidate), A9 would be 8 (the only 8 in the row) eliminating 8
from E9 which would thus be 1. The 1 in E9 and the 4 in E6 would reduce each of
E1 and E2 to 23. Compartments D12 and E12 would be a so-called "deadly pattern"
precluding a unique solution (see www.sudokuwiki.org/Unique_Rectangles). If the
puzzle is to have a unique solution, A6 cannot be 3.
There is now no 3 in column 6. Every row has a 3 except perhaps row E, so 3 can
(by Setti's rule) be eliminated from compartment E12, as can the stranded 4 in that
compartment.
There is now an X-wing on 2 in D1-E2, eliminating the remaining 2s in columns 1 and
2, and solving C3 for 2.
The solver completes the solution: