#129 documented
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The solver progresses through a Sudoku puzzle by eliminating inconsistent candidates based on logic deductions at each step. It determines that A1 must be 3, solving the rest of the puzzle, as setting A1 to 6 would lead to a contradiction in cell E49.
![The solver stops at this position:
Compartment C79 must contain 5, so C9=5.
6 can be removed from D89 (neither 6 is in sequence with the other cell).
E12 must be [23] and E49 must be [456789], so inconsistent candidates can be
eliminated.
F49 must contain [78] so 789 can be eliminated from F12.
The 3 can be eliminated from F12 (unique rectangle). EF12 is an x-wing on 2 and
eliminates the other 2s from columns 1 and 2.
In column C, if D3 is not 5 than AD3 must be [6789] and G3 must be 5. Therefore
every column must have a 5, so every row must also have a 5 (Setti's rule). Thus
D13 must have a 5 and must be [345] or [456] (7 is barred by D7), and all
inconsistent candidates can be eliminated from D13. As D13 must contain 4, this
can be eliminated from D89.
BF9 must contain 4, so 4 can be eliminated from HJ9.
3 can be eliminated from J8 (not in sequence with H8) and from J9 (not in sequence
with H9). 7 can be eliminated from H8 (not in sequence with J8).
The lowest candidate in C3 is 6, so AD3 cannot contain a cell lower than 3,
eliminating the 2 from A3.
Columns 3 and 6 do not contain a 2. Every other column must have a 2. Rows C
and G do not have a 2, so all other rows must have a 2 (Setti's rule). Thus row D
must have a 2 and 89 can therefore be eliminated from D89.
The highest D9 can be is 3, so no cell in BF9 can be higher than 7, eliminating the 8s
in the compartment. Similarly, the highest D8 can be is 3, so no cell in AF8 can be
higher than 8, eliminating the 9s in the compartment.
BF9 must be [12345], [23456] or [34567], and must contain 3, which can be
eliminated from H9.](https://image.slidesharecdn.com/129documented-121209115520-phpapp01/85/129-documented-1-320.jpg)
![AD3 must be [3456], [4567], [5678] or [6789], and must contain a 6, which can thus
be eliminated from GJ3.
GJ3 can only be [345] or [789]. If GJ3 is [345] then G3 must be 5 which can be
eliminated from the rest of the cells in the compartment. If GJ3 is [789] then H3 must
be 8, J3 must be 7, and thus G3 must be 9 and 7 can be eliminated from G3 and H3,
and 9 from H3.
AC5 can only be [123], [234], [678] or [789]. If AC5 is [123] or [234], A5 must be 2
and 34 can be eliminated from A5.
Every row must contain a 4, so every column must also column 4 (Setti's rule). Thus
EJ4 must contain a 4, eliminating 9 from EJ4.
AF8 must contain a 3 so 3 can be eliminated from H8.
C13 must contain an 8, which can be eliminated from C5.
BF9 must be [12345], [23456] or [34567]. If it is [23456] or [34567], D9 cannot be 1.
If BF9 is [12345], E9 must be 4, F9 must be 3, B9 must be 1 and D9 must be 2, so in
no case can D9 be 1, which can thus be eliminated from D9.
AD3 can only be [3456], [4567], [5678] or [6789]. If AD3 is [3456], [4567] or [5678],
D3 must be 5 (the only 5 in the compartment) and D3 is not 3 or 4. If AD3 is [6789],
then D3 is not 3 or 4 either, so 34 can be eliminated from D3. Since D13 must be
[345] or [456], it must contain a 4, solving D2 for 4 (the only 4 in the compartment).
AC5 can only be [123], [234], [678] or [789]. If AC5 is [123], [234] or [678], B5 cannot
be 9. If AC5 is [789], C5 must be 9 and again B5 cannot be 9, so 9 can be
eliminated from B5. If AC5 is [123], [234] or [789], A5 cannot be 6. If AC5 is [678],
C5 must be 6 and again A5 cannot be 6, so 6 can be eliminated from A5.
AF8 must be [123456] or [345678]. If it is [123456], D8 would be 1, and 2 can thus
be eliminated from D8.
The 3 can be eliminated from D9 (not in sequence with D8), solving D9 for 2,
eliminating 2 for J9 and 1 from J8. The 2 in D9 means that BF9 cannot contain a cell
higher than 6, eliminating the 7s from BF9.
GJ3 must be [345] or [789], and AD3 must be [6789] (if GJ3 is [345]) or [3456] (if GJ3
is [789]). If AD3 is [3456], D3 must be 5, C3 must be 6 and A3 cannot be 6. If AD3
is [6789], D3 must be 6 and again A3 cannot be 6, so 6 can be eliminated from A3.
If EJ5 contains a 3 it must be [12345] or [23456]. If EJ5 is [12345], H5 must be 1 and
J5 must be 2 and cannot be 3. If EJ5 is [23456], J5 must be 2 and cannot be 3. 3
can therefore be eliminated from J5.
EJ56 is an x-wing on 5, eliminating the other 5s in columns 5 and 6.
The 1 is eliminated from H5 as it cannot be in sequence with G5. EJ5 must contain a
6, eliminating 6 from C5.](https://image.slidesharecdn.com/129documented-121209115520-phpapp01/85/129-documented-2-320.jpg)
![HJ8 must be [12] or [78]. If it were [78], a chain leads to F5=71. However, if HJ8
were [78], AF8 would be [123456] and A8 would be 2 (the only 2 in the
compartment). That would eliminate the 2 in AC5 which would be [789] containing a
naked pair of 78s, and that would be inconsistent with F5=7, showing that HJ8 could
not have been [78] and must have been [12].
Setting HJ8 to [12], the solver progresses to the following position:
If A1 were 6, there would be a contradiction in E49.2 Accordingly A1 must be 3, and
the solver does the rest.
1
If HJ8 were [78], H9 would be 7, J9 would be 6, J6 would be 5, AF8 would be (in order) 253164, BF9
would be (in order) 15243, F49 would not contain a cell higher than 8 and F7 would be solved for 8,
E7 for 9, E6 for 7, E4 for 8, F6 for 6, E5 for 8 and F5 for 7.
2
A2 would be solved for 8, H1 for 3, A8 for 4, B2 for 9, H3 for 4, A6 for 3, B3 for 8, G2 for 6, J3 for 3,
B6 for 4, C3 for 9, G4 for 7, C1 for 8, G5 for 9, J1 for 9, F7 for 9, H6 for 9, J4 for 4, E7 for 8, E4 for 6
and E8 for 7, leaving E47 with only three numbers (568) for four cells.](https://image.slidesharecdn.com/129documented-121209115520-phpapp01/85/129-documented-3-320.jpg)
#129 documented
- 1. The solver stops at this position: Compartment C79 must contain 5, so C9=5. 6 can be removed from D89 (neither 6 is in sequence with the other cell). E12 must be [23] and E49 must be [456789], so inconsistent candidates can be eliminated. F49 must contain [78] so 789 can be eliminated from F12. The 3 can be eliminated from F12 (unique rectangle). EF12 is an x-wing on 2 and eliminates the other 2s from columns 1 and 2. In column C, if D3 is not 5 than AD3 must be [6789] and G3 must be 5. Therefore every column must have a 5, so every row must also have a 5 (Setti's rule). Thus D13 must have a 5 and must be [345] or [456] (7 is barred by D7), and all inconsistent candidates can be eliminated from D13. As D13 must contain 4, this can be eliminated from D89. BF9 must contain 4, so 4 can be eliminated from HJ9. 3 can be eliminated from J8 (not in sequence with H8) and from J9 (not in sequence with H9). 7 can be eliminated from H8 (not in sequence with J8). The lowest candidate in C3 is 6, so AD3 cannot contain a cell lower than 3, eliminating the 2 from A3. Columns 3 and 6 do not contain a 2. Every other column must have a 2. Rows C and G do not have a 2, so all other rows must have a 2 (Setti's rule). Thus row D must have a 2 and 89 can therefore be eliminated from D89. The highest D9 can be is 3, so no cell in BF9 can be higher than 7, eliminating the 8s in the compartment. Similarly, the highest D8 can be is 3, so no cell in AF8 can be higher than 8, eliminating the 9s in the compartment. BF9 must be [12345], [23456] or [34567], and must contain 3, which can be eliminated from H9.
- 2. AD3 must be [3456], [4567], [5678] or [6789], and must contain a 6, which can thus be eliminated from GJ3. GJ3 can only be [345] or [789]. If GJ3 is [345] then G3 must be 5 which can be eliminated from the rest of the cells in the compartment. If GJ3 is [789] then H3 must be 8, J3 must be 7, and thus G3 must be 9 and 7 can be eliminated from G3 and H3, and 9 from H3. AC5 can only be [123], [234], [678] or [789]. If AC5 is [123] or [234], A5 must be 2 and 34 can be eliminated from A5. Every row must contain a 4, so every column must also column 4 (Setti's rule). Thus EJ4 must contain a 4, eliminating 9 from EJ4. AF8 must contain a 3 so 3 can be eliminated from H8. C13 must contain an 8, which can be eliminated from C5. BF9 must be [12345], [23456] or [34567]. If it is [23456] or [34567], D9 cannot be 1. If BF9 is [12345], E9 must be 4, F9 must be 3, B9 must be 1 and D9 must be 2, so in no case can D9 be 1, which can thus be eliminated from D9. AD3 can only be [3456], [4567], [5678] or [6789]. If AD3 is [3456], [4567] or [5678], D3 must be 5 (the only 5 in the compartment) and D3 is not 3 or 4. If AD3 is [6789], then D3 is not 3 or 4 either, so 34 can be eliminated from D3. Since D13 must be [345] or [456], it must contain a 4, solving D2 for 4 (the only 4 in the compartment). AC5 can only be [123], [234], [678] or [789]. If AC5 is [123], [234] or [678], B5 cannot be 9. If AC5 is [789], C5 must be 9 and again B5 cannot be 9, so 9 can be eliminated from B5. If AC5 is [123], [234] or [789], A5 cannot be 6. If AC5 is [678], C5 must be 6 and again A5 cannot be 6, so 6 can be eliminated from A5. AF8 must be [123456] or [345678]. If it is [123456], D8 would be 1, and 2 can thus be eliminated from D8. The 3 can be eliminated from D9 (not in sequence with D8), solving D9 for 2, eliminating 2 for J9 and 1 from J8. The 2 in D9 means that BF9 cannot contain a cell higher than 6, eliminating the 7s from BF9. GJ3 must be [345] or [789], and AD3 must be [6789] (if GJ3 is [345]) or [3456] (if GJ3 is [789]). If AD3 is [3456], D3 must be 5, C3 must be 6 and A3 cannot be 6. If AD3 is [6789], D3 must be 6 and again A3 cannot be 6, so 6 can be eliminated from A3. If EJ5 contains a 3 it must be [12345] or [23456]. If EJ5 is [12345], H5 must be 1 and J5 must be 2 and cannot be 3. If EJ5 is [23456], J5 must be 2 and cannot be 3. 3 can therefore be eliminated from J5. EJ56 is an x-wing on 5, eliminating the other 5s in columns 5 and 6. The 1 is eliminated from H5 as it cannot be in sequence with G5. EJ5 must contain a 6, eliminating 6 from C5.
- 3. HJ8 must be [12] or [78]. If it were [78], a chain leads to F5=71. However, if HJ8 were [78], AF8 would be [123456] and A8 would be 2 (the only 2 in the compartment). That would eliminate the 2 in AC5 which would be [789] containing a naked pair of 78s, and that would be inconsistent with F5=7, showing that HJ8 could not have been [78] and must have been [12]. Setting HJ8 to [12], the solver progresses to the following position: If A1 were 6, there would be a contradiction in E49.2 Accordingly A1 must be 3, and the solver does the rest. 1 If HJ8 were [78], H9 would be 7, J9 would be 6, J6 would be 5, AF8 would be (in order) 253164, BF9 would be (in order) 15243, F49 would not contain a cell higher than 8 and F7 would be solved for 8, E7 for 9, E6 for 7, E4 for 8, F6 for 6, E5 for 8 and F5 for 7. 2 A2 would be solved for 8, H1 for 3, A8 for 4, B2 for 9, H3 for 4, A6 for 3, B3 for 8, G2 for 6, J3 for 3, B6 for 4, C3 for 9, G4 for 7, C1 for 8, G5 for 9, J1 for 9, F7 for 9, H6 for 9, J4 for 4, E7 for 8, E4 for 6 and E8 for 7, leaving E47 with only three numbers (568) for four cells.