The solver stops at the following position:




A36 must be [1234] eliminating 89 from the compartment, solving A6 for 4 and
eliminating the other 4s in the compartment.

The highest number in A3 is 3 so the highest in AD3 is 6 eliminating 789 from the
compartment.

The solver then progresses to the following position:




Each of AC4 and AC5 must contain 34 eliminating those numbers from the rest of
columns 4 and 5 along with the stranded 2 in GJ5.

The lowest number in J4 and J5 is 5, eliminating the 1 from J6 and J7.

D13 must be [234], [345] or [456] and must contain a 4, eliminating that number and
the stranded 23s in D69.

D7 is higher than 5, so everything lower than 5 is eliminated from BD7, and
everything higher than 5 from FJ7.
The solver stops at the following position:




GJ5 must contain 67 so those numbers can be eliminated from E5.

The lowest number in F3 and F4 is 5 eliminating 2 from F14. F14 must be [3456],
[4567] or [5678] and must contain 56 which can be eliminated from F79. F79 must
contain a 3 which can be eliminated from F14. The highest number in F9 is 4,
eliminating 9 from EH9.

J4 and J5 are each [56] so J6=3 and J7=4, eliminating 3s from column 6 and 4s from
column 7.

F7 cannot be 3 or GH67 would be an impossible pattern as the puzzle has a unique
solution (see notes on Unique Rectangles in the Sudoku solver).

D6 and D8 are each [56] so D69 must contain a 5, eliminating 9 from D7.

DH7 must be [12345] or [23456], and must contain 2345, eliminating those numbers
from B7.

Each of C3, C4 and C5 is [234], so C17 must contain a 2 and therefore cannot
contain a 9.

B4C5 is an x-wing on 4, eliminating the other 4s in rows B and C and thus solving D3
for 4.

The 4 in D3 reduces D1 and D2 to [23], and the maximum in the column 2 and
column 3 compartments is thus 7.

E1F2 is an x-wing on 4, eliminating the other 4s in rows E and F, thus solving F9 for
3 and eliminating the other 3s in row F and column 9.

The solver then stops at the following position:
As BF1 and BF2 both contain a 4, F14 must contain a 4 and therefore cannot contain
an 8.

The naked pair of 56 in FJ4 eliminates the 5 from B4, and the naked quad [2345] in
row B (columns 2 to 5) eliminates the 23 from B1 solving it for 1 and eliminating the 1
from B8. The 1 in B1 eliminates the 67 from F1 solving it for 4, F2 for 5, F4 for 6 and
F3 for 7. J4 is thus solved for 5 and J5 for 6, eliminating the other 6s in column 5.
The naked quad [2345] in row B (columns 2 to 5) also eliminates the 5 from B6,
solving B6 for 6, D6 for 5 and D8 for 6, thereby eliminating the 6 from B1 and the 6
from E8, solving E8 for 5 and E9 for 6, and eliminating the other 6s from column 9.

The 5 in F2 eliminates the 5 from B2, and the naked pair of [23] in B2 and B3 solve
B4 for 4 (eliminating the 4 from C4) and B5 for 5, which in turn solves C5 for 4, A5 for
3, A4 for 2, C4 for 3, C3 for 2, B3 for 3, B2 for 2, D2 for 3, D1 for 2, E1 for 3 and E2
for 4.

The 6 in DH8 eliminates the 1 from F8, solving it for 2 and F7 for 1, and eliminating
the other 1s and 2s from columns 7 and 8. Each of F4 and J4 must be [56], so FJ4
must contain a 5 and therefore cannot contain a 9, eliminating 9 from H4, thus
solving H4 for 8 and eliminating the other 8s in row H.

The solver does the rest.

#128 documented ver 2

  • 1.
    The solver stopsat the following position: A36 must be [1234] eliminating 89 from the compartment, solving A6 for 4 and eliminating the other 4s in the compartment. The highest number in A3 is 3 so the highest in AD3 is 6 eliminating 789 from the compartment. The solver then progresses to the following position: Each of AC4 and AC5 must contain 34 eliminating those numbers from the rest of columns 4 and 5 along with the stranded 2 in GJ5. The lowest number in J4 and J5 is 5, eliminating the 1 from J6 and J7. D13 must be [234], [345] or [456] and must contain a 4, eliminating that number and the stranded 23s in D69. D7 is higher than 5, so everything lower than 5 is eliminated from BD7, and everything higher than 5 from FJ7.
  • 2.
    The solver stopsat the following position: GJ5 must contain 67 so those numbers can be eliminated from E5. The lowest number in F3 and F4 is 5 eliminating 2 from F14. F14 must be [3456], [4567] or [5678] and must contain 56 which can be eliminated from F79. F79 must contain a 3 which can be eliminated from F14. The highest number in F9 is 4, eliminating 9 from EH9. J4 and J5 are each [56] so J6=3 and J7=4, eliminating 3s from column 6 and 4s from column 7. F7 cannot be 3 or GH67 would be an impossible pattern as the puzzle has a unique solution (see notes on Unique Rectangles in the Sudoku solver). D6 and D8 are each [56] so D69 must contain a 5, eliminating 9 from D7. DH7 must be [12345] or [23456], and must contain 2345, eliminating those numbers from B7. Each of C3, C4 and C5 is [234], so C17 must contain a 2 and therefore cannot contain a 9. B4C5 is an x-wing on 4, eliminating the other 4s in rows B and C and thus solving D3 for 4. The 4 in D3 reduces D1 and D2 to [23], and the maximum in the column 2 and column 3 compartments is thus 7. E1F2 is an x-wing on 4, eliminating the other 4s in rows E and F, thus solving F9 for 3 and eliminating the other 3s in row F and column 9. The solver then stops at the following position:
  • 3.
    As BF1 andBF2 both contain a 4, F14 must contain a 4 and therefore cannot contain an 8. The naked pair of 56 in FJ4 eliminates the 5 from B4, and the naked quad [2345] in row B (columns 2 to 5) eliminates the 23 from B1 solving it for 1 and eliminating the 1 from B8. The 1 in B1 eliminates the 67 from F1 solving it for 4, F2 for 5, F4 for 6 and F3 for 7. J4 is thus solved for 5 and J5 for 6, eliminating the other 6s in column 5. The naked quad [2345] in row B (columns 2 to 5) also eliminates the 5 from B6, solving B6 for 6, D6 for 5 and D8 for 6, thereby eliminating the 6 from B1 and the 6 from E8, solving E8 for 5 and E9 for 6, and eliminating the other 6s from column 9. The 5 in F2 eliminates the 5 from B2, and the naked pair of [23] in B2 and B3 solve B4 for 4 (eliminating the 4 from C4) and B5 for 5, which in turn solves C5 for 4, A5 for 3, A4 for 2, C4 for 3, C3 for 2, B3 for 3, B2 for 2, D2 for 3, D1 for 2, E1 for 3 and E2 for 4. The 6 in DH8 eliminates the 1 from F8, solving it for 2 and F7 for 1, and eliminating the other 1s and 2s from columns 7 and 8. Each of F4 and J4 must be [56], so FJ4 must contain a 5 and therefore cannot contain a 9, eliminating 9 from H4, thus solving H4 for 8 and eliminating the other 8s in row H. The solver does the rest.