#130 documented
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The document provides step-by-step logic to solve a Sudoku puzzle. It eliminates candidate numbers from cells based on constraints like unique digit rules, sequence rules, and logic deductions. Over many lines of reasoning, it arrives at a fully solved Sudoku grid with a single possible number in each cell.
![The solver very soon stops here:
The 9 in C3 can be eliminated (not in sequence with C4). 89 can be eliminated from
C69 (stranded by the 7 in C1). C69 must contain [34], and these numbers can be
eliminated from C34. 25 can be eliminated from C3 (not in sequence with C4).
123 can be eliminated from G14 (stranded by the 4 in G9). The 9 in G7 can be
eliminated (not in sequence with G6). G14 must be [5678] or [6789] and must
contain [678], which can be eliminated from G67. Both 5s and the 9 in G67 can also
be eliminated (each is not in sequence with the other compartment cell).
12 can be eliminated from F1 (not in sequence with G1). FG1 must be [45] or [56]
and must contain a 5, which can be eliminated from AB1, along with the 6s (stranded
by the necessary 5 in FG1). 14 can be eliminated from A1 (not in sequence with A2).
If A15 contained a 7, it would have to be [34567], but there is only one 3 and nothing
higher than 3 in A1 and A2, so 7 can be eliminated from A15. 13 can be eliminated
from A5 (not in sequence with B5). 8 can be eliminated from B5 (not in sequence
with A5).
2 can be eliminated from E5 (if E5 were 2, E6 would have to be 3 but there is no
candidate 3 in E6).
5 can be eliminated from C9 (not in sequence with D9). 89 can be eliminated from
D9 (not in sequence with C9).
Every row except G must have a 4. Row G does not have a 4. Column 9 does not
have a 4, so every other column must have a 4. Thus columns 1, 2 and 8 must each
have a 4. In column 2, as DH2 must have a 4, it cannot have a 9 (out of sequence
with 4).
BF8 must be [45678] or [56789] and must contain an 8, solving B8 for 8 (the only
candidate 8 in BF8) and eliminating the other 8s in the row.
BF8 must contain a 7, and E8 is the only candidate, solving it for 7 and eliminating
the other 7s in the row.](https://image.slidesharecdn.com/130documented-121217102454-phpapp02/85/130-documented-1-320.jpg)
![J5 cannot be 3 (if it were, HJ5 would have to be 23, implying that DF5 would be [456]
and all candidates would thus be eliminated from A5). CJ79 is thus an x-wing on 3,
eliminating all other 3 in columns 7 and 9.
2 can be eliminated from H5 (not in sequence with J5). If H5 were 5, J5 would have
to be 6, DF5 would be [234] and all candidates would thus be eliminated from A5, so
5 can be eliminated from H5. Similarly, if J5 were 5, H5 would have to be 6, DF5
would be [234] and all candidates would thus be eliminated from A5, so 5 can be
eliminated from J5.
There must be a 7 in column 5, else {AB5 would be [12] and HJ5 would be [56]} or
{AB5 would be [56] and HJ5 would be [12]}, eliminating all candidates from F5.
There must accordingly be at least 7 columns containing 7. There are two rows that
cannot contain 7, so there must be 7 rows and 7 columns that must contain 7, and 2
rows and 2 columns that cannot contain 7. Row J and thus J59 must therefore
contain a 7, and 2 can be eliminated from J59, solving, in turn, J8 for 4, C8 for 5
(eliminating the 1 from C69) and F8 for 9, eliminating the 1 from F7.
After eliminating the candidates inconsistent with the solved squares, 6 can be
eliminated from C3 (not in sequence with C4). 1 can be eliminated from H5 (not in
sequence with J5) and from H9 (not in sequence with J9).
B5 cannot be 7, else A5 would be 6 and all candidates would be eliminated from J5.
B18 must contain a 7, solving B7 for 7 (the only candidate 7 in the compartment) and
eliminating all other 7s in the column.
GH24 is an x-wing on 7, eliminating the other two 7s in row H. The 6 in J5 can be
eliminated (not in sequence with H5) solving J5 for 7 and eliminating 7 from J9 and 8
from H9. HJ9 is thus [23] or [56]. CD9 must also be [23] or [56] (it cannot be [12]
because if it were, C7 would be 3, J7 would not be 3 and J9 would be 3, whereas for
CD9 to be [12] we know that HJ9 must be [56] and J9 cannot be 3.
Every column must have a 5 so row G (and thus G14) must have a 5, eliminating the
9s from G14 and arriving at the following position, which is very similar to that posted
by Kolumbus at de.slideshare.net/kolumbus7/extreme-str8ts-130 (the only difference
is that he still has a 2 in E5):](https://image.slidesharecdn.com/130documented-121217102454-phpapp02/85/130-documented-2-320.jpg)
![If C3=8 then C4=9, D4=[25], G3=56, G4=7, G2=8 and D2=[35]. But the naked pair
25 in row D would eliminate the 5 from D2 solving it for 3, which is out of sequence
with the 8 in G2. Thus 8 can be eliminated from C3 and 9 from C4. After eliminating
the other 1s and 2s inconsistent with C34, the naked triples 457 in row D solves D4
for 9, eliminating the 19 from D7. The 1 in C3 eliminates the 9s in B3 and E3.
GH67 is an x-wing on 1, eliminating the 1 in B6.
If B6 were 9, G6 would be 2, G7 would be 1, H7 would not be 1 and G7 would be 1,
which would be out of sequence with the assumed 9 in B6, so 9 can be eliminated
from B6. The naked triple 456 in row 6 eliminates those numbers from H6.
If E7 were 9, G7 would be 2 (1 is out of sequence with 9 in BJ7), G6 would be 1 and
H6 would be 2. There must be a 1 in H29 and H7 would be the only 1 candidate, but
it would be out of sequence with the assumed 9 in E7. Thus 9 can be eliminated E7.
Every row except J must have a 2. Row J cannot have a 2. Column 8 cannot have a
2 so every other column must have a 2, solving A1 for 2, B1 for 1 (eliminating the 1
from B5 and the 2 from A5, leaving a naked pair 56 and solving E5 for 3, F5 for 2 and
H5 for 8), and eliminating the 2 from A3.
D7 is now solved for 8 (the only candidate 8 in BJ7) and after the other 8s in the row
have been eliminated, G3 is solved for 8 (the only candidate 8 in AH3). The 8 in G2
is therefore eliminated, and DH2 must be [34567], eliminating the 3 from A2, solving
it for 1 and eliminating the 6s from A15, solving A5 for 5, B5 for 6, A4 for 4, and A3
for 3. The naked triple 245 in column 3 solves H3 for 6. Eliminating the other 6s in
the row eliminates 5 from J9.
The naked 25s in row D solves D2 for 3 and the rest is routine, leading to the
following solution:](https://image.slidesharecdn.com/130documented-121217102454-phpapp02/85/130-documented-3-320.jpg)
#130 documented
- 1. The solver very soon stops here: The 9 in C3 can be eliminated (not in sequence with C4). 89 can be eliminated from C69 (stranded by the 7 in C1). C69 must contain [34], and these numbers can be eliminated from C34. 25 can be eliminated from C3 (not in sequence with C4). 123 can be eliminated from G14 (stranded by the 4 in G9). The 9 in G7 can be eliminated (not in sequence with G6). G14 must be [5678] or [6789] and must contain [678], which can be eliminated from G67. Both 5s and the 9 in G67 can also be eliminated (each is not in sequence with the other compartment cell). 12 can be eliminated from F1 (not in sequence with G1). FG1 must be [45] or [56] and must contain a 5, which can be eliminated from AB1, along with the 6s (stranded by the necessary 5 in FG1). 14 can be eliminated from A1 (not in sequence with A2). If A15 contained a 7, it would have to be [34567], but there is only one 3 and nothing higher than 3 in A1 and A2, so 7 can be eliminated from A15. 13 can be eliminated from A5 (not in sequence with B5). 8 can be eliminated from B5 (not in sequence with A5). 2 can be eliminated from E5 (if E5 were 2, E6 would have to be 3 but there is no candidate 3 in E6). 5 can be eliminated from C9 (not in sequence with D9). 89 can be eliminated from D9 (not in sequence with C9). Every row except G must have a 4. Row G does not have a 4. Column 9 does not have a 4, so every other column must have a 4. Thus columns 1, 2 and 8 must each have a 4. In column 2, as DH2 must have a 4, it cannot have a 9 (out of sequence with 4). BF8 must be [45678] or [56789] and must contain an 8, solving B8 for 8 (the only candidate 8 in BF8) and eliminating the other 8s in the row. BF8 must contain a 7, and E8 is the only candidate, solving it for 7 and eliminating the other 7s in the row.
- 2. J5 cannot be 3 (if it were, HJ5 would have to be 23, implying that DF5 would be [456] and all candidates would thus be eliminated from A5). CJ79 is thus an x-wing on 3, eliminating all other 3 in columns 7 and 9. 2 can be eliminated from H5 (not in sequence with J5). If H5 were 5, J5 would have to be 6, DF5 would be [234] and all candidates would thus be eliminated from A5, so 5 can be eliminated from H5. Similarly, if J5 were 5, H5 would have to be 6, DF5 would be [234] and all candidates would thus be eliminated from A5, so 5 can be eliminated from J5. There must be a 7 in column 5, else {AB5 would be [12] and HJ5 would be [56]} or {AB5 would be [56] and HJ5 would be [12]}, eliminating all candidates from F5. There must accordingly be at least 7 columns containing 7. There are two rows that cannot contain 7, so there must be 7 rows and 7 columns that must contain 7, and 2 rows and 2 columns that cannot contain 7. Row J and thus J59 must therefore contain a 7, and 2 can be eliminated from J59, solving, in turn, J8 for 4, C8 for 5 (eliminating the 1 from C69) and F8 for 9, eliminating the 1 from F7. After eliminating the candidates inconsistent with the solved squares, 6 can be eliminated from C3 (not in sequence with C4). 1 can be eliminated from H5 (not in sequence with J5) and from H9 (not in sequence with J9). B5 cannot be 7, else A5 would be 6 and all candidates would be eliminated from J5. B18 must contain a 7, solving B7 for 7 (the only candidate 7 in the compartment) and eliminating all other 7s in the column. GH24 is an x-wing on 7, eliminating the other two 7s in row H. The 6 in J5 can be eliminated (not in sequence with H5) solving J5 for 7 and eliminating 7 from J9 and 8 from H9. HJ9 is thus [23] or [56]. CD9 must also be [23] or [56] (it cannot be [12] because if it were, C7 would be 3, J7 would not be 3 and J9 would be 3, whereas for CD9 to be [12] we know that HJ9 must be [56] and J9 cannot be 3. Every column must have a 5 so row G (and thus G14) must have a 5, eliminating the 9s from G14 and arriving at the following position, which is very similar to that posted by Kolumbus at de.slideshare.net/kolumbus7/extreme-str8ts-130 (the only difference is that he still has a 2 in E5):
- 3. If C3=8 then C4=9, D4=[25], G3=56, G4=7, G2=8 and D2=[35]. But the naked pair 25 in row D would eliminate the 5 from D2 solving it for 3, which is out of sequence with the 8 in G2. Thus 8 can be eliminated from C3 and 9 from C4. After eliminating the other 1s and 2s inconsistent with C34, the naked triples 457 in row D solves D4 for 9, eliminating the 19 from D7. The 1 in C3 eliminates the 9s in B3 and E3. GH67 is an x-wing on 1, eliminating the 1 in B6. If B6 were 9, G6 would be 2, G7 would be 1, H7 would not be 1 and G7 would be 1, which would be out of sequence with the assumed 9 in B6, so 9 can be eliminated from B6. The naked triple 456 in row 6 eliminates those numbers from H6. If E7 were 9, G7 would be 2 (1 is out of sequence with 9 in BJ7), G6 would be 1 and H6 would be 2. There must be a 1 in H29 and H7 would be the only 1 candidate, but it would be out of sequence with the assumed 9 in E7. Thus 9 can be eliminated E7. Every row except J must have a 2. Row J cannot have a 2. Column 8 cannot have a 2 so every other column must have a 2, solving A1 for 2, B1 for 1 (eliminating the 1 from B5 and the 2 from A5, leaving a naked pair 56 and solving E5 for 3, F5 for 2 and H5 for 8), and eliminating the 2 from A3. D7 is now solved for 8 (the only candidate 8 in BJ7) and after the other 8s in the row have been eliminated, G3 is solved for 8 (the only candidate 8 in AH3). The 8 in G2 is therefore eliminated, and DH2 must be [34567], eliminating the 3 from A2, solving it for 1 and eliminating the 6s from A15, solving A5 for 5, B5 for 6, A4 for 4, and A3 for 3. The naked triple 245 in column 3 solves H3 for 6. Eliminating the other 6s in the row eliminates 5 from J9. The naked 25s in row D solves D2 for 3 and the rest is routine, leading to the following solution: