The document provides step-by-step logic to solve a Sudoku puzzle. Key deductions include: - Compartment AE3 must contain 567, eliminating those numbers from other areas. - Naked pairs and triples are identified and used to eliminate numbers from rows and columns. - Setti's rule is applied several times to determine where certain numbers must or must not appear. - A "deadly pattern" is avoided by determining one compartment must contain a 3. - The logic chains through multiple linked deductions to arrive at a fully solved puzzle.

1. The solver stops at this position:
Compartment AE3 must contain 567, so these numbers (along with stranded digits
89) can be eliminated from GJ3. GJ3 must contain a 3 so that number can be
eliminated from AE3.
Compartment AE4 has a 4 and a 6 so it must have a 5. The only possibility is D4
which is thus solved for 5, eliminating 5 from D7.
Compartment BD6 must contain a 2. The only possibility is B6 which is thus solved
for 2.
Compartment AD7 must contain 345, and these (along with stranded digit 2) can be
eliminated from GJ7. As AD7 must contain a 5 and C7 is the only candidate, C7 is
solved for 5. Similarly, AD7 must contain a 4 and D7 is the only candidate, it is
solved for 4, thus solving D6 for 3 and C6 for 1.
The 1 at C6 eliminates the 9 from row C.
Stranded digits 78 can be removed from G13, so G13 must contain 23, which can be
eliminated from the remainder of row G.
67 can be removed from F1 (not in sequence with G1) leaving a naked pair 34 in
FG1, eliminating 4 from A1, solving A1 for 6, and thus eliminating 6 from D1.

2. 67 can be removed from F2 (not in sequence with F1).
The naked quad 3678 in column 5 (rows CDE & G) reduces H5 and J5 to 12 each.
Compartment AE4 must contain a 7, so 7 can be eliminated from HJ4.
AE3 must contain a 5. E3 is the only possibility and is thus solved for 5.
9 can be eliminated from row J (not in sequence with J5, and 8 can be eliminated
from H4 (not in sequence with J4).
4 can be eliminated from E9 (not in sequence with E8).
3 can be eliminated from E2 (no 4 elsewhere in the compartment).
GJ3 must contain a 3, which can thus be eliminated from AE3.
Counting 4s, there are 8 rows containing 4 (ABCDEHJ and either F or G because of
the naked 34 pair in FG1) and 1 row that is unknown whether it contains a 4 (the
other of rows F and G). There are 6 columns containing 4 (124579), one column that
does not contain a 4 (6) and 2 columns that are unknown whether they contain a 4
(38). Since column 6 does not contain a 4, the unknown row cannot contain a 4
(applying Setti's rule). Thus one of rows F and G contains a 4 and the other does
not.
Because FG1 contains (and is limited to) 34, and because one of rows F and G
contains a 4 and the other does not, then if F1 is 4 there would be no 4 in row G, and
if F1 is 3 then G1 would be 4 and there would be no other 4 in row G. Thus 4 can be
eliminated from row G except for G1. Similarly, 4 can be eliminated from row F
except for F1, as can the stranded 3 in F9.
Again applying Setti's rule, since 8 rows contain a 4 and 1 row does not, 8 columns
must contain a 4 and 1 column (which we know to be column 6) must not. Thus
columns 3 and 8 must each contain a 4. The 4 in column 8 must be in compartment
EJ8, and 9 can thus be eliminated from that compartment (not in sequence with 4).
If A7 is 2 then B7 is 3, B8 is 1 and A8 is 3. If A7 is 3 then B7 is 6, A8 is 1 and B8 is
3. In either case, there is a 3 in row A and another in row B so we can eliminate the
other 3s in those rows.
The 12 naked pair at A9 and D9 eliminates those numbers from the rest of column 9.
The elimination of 1 from B9 solves B8 for 1 (the only possible 1 in the row), and thus
A8 for 3, A7 for 2, A9 for 1, D9 for 2 and B7 for 3.
The 78 naked pair at A4 and B4 eliminates the 8 in J4 and thus the stranded 9 in H4.
If HJ4 does not contain a 3 then H4, H5, J4 and J5 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, HJ4 must contain a 3, and therefore 1 can
be eliminated from it.
Every row has, or must have, a 6. Therefore every column (including column 7) must
have a 6, so 9 can be eliminated from GJ7.

3. Naked pair of 78 at G5 and G7 eliminates those numbers from G59, solving G8 for 5
and G9 for 9.
There are 4 rows containing 9 (BDFG) and 5 rows not containing 9 (ACEHJ). There
are 3 columns containing 9 (259), 5 columns not containing 9 (14678), and one
column not known whether it contains a 9 or not (3). By Setti's rule, column 3 must
contain a 9, and 4 can be eliminated from A3 (not in sequence with 9). The only
possible candidate is B3, which is solved for 9, and thus A2 for 4.
The stranded 3 can be eliminated from E5, and the naked pair 78 in E25 solves E8
for 4 and E9 for 3.
C5 is solved for 3 (the only possible candidate).
The naked triple 678 in B2, C2 and E2 eliminates those numbers from H2.
The naked pair 78 at E5 and G5 solves D5 for 6. The naked pair 78 at A3 and D3
solves C3 for 6. The naked pair 78 at C2 and E2 solves B2 for 6. The naked pair 78
at B9 and C9 eliminates those numbers from the rest of column 9.
If H5 is 1, J5 is 2, J4 is 3 and H4 is 2and H2 is 5. If H5 is 2, H2 is still 5. So H2 is
solved for 5, and thus H6 for 7, J6 for 5, H9 for 4, J9 for 6, H7 for 6, H8 for 8, J8 for 7,
F8 for 6, F9 for 5, J7 for 8, G7 for 7, G5 for 8, E5 for 7, E2 for 8, C2 for 7, C9 for 8,
B9 for 7, B4 for 8, A4 for 7, A3 for 8, D3 for 7 and D1 for 8.
J3 is solved for 4 (the only candidate in the row).
J5 is solved for 2 (1 would be out of sequence with 8), H5 for 1, H4 for 2, H3 for 3,
G3 for 2, G2 for 3, G1 for 4, F1 for 3 and F2 for 2.
The solution: