||=====|=====|=====||=====|=====|=====||=====|=====|=====||
|| | 2 | 7 || 8 | 3 | || | 6 | || 2 3 6 7
|| 5 | | || | | 9 || 1 | | 4 ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
|| 1 | | 9 || 7 | 4 | || 8 | | 5 || 1 5 7 8
|| | 3 | || | | 6 || | 2 | ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
|| 6 | 4 | || 1 | 5 | 2 || | 3 | 7 ||
|| | | 8 || | | || 9 | | ||
||=====|=====|=====||=====|=====|=====||=====|=====|=====||
|| 2 | 8 | 1 || | 6 | 4 || | 7 | ||
|| | | || 9 | | || 5 | | 3 ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
|| | 7 | 6 || 3 | | 5 || 4 | 1 | ||
|| 9 | | || | 8 | || | | 2 ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
|| | 5 | || 2 | 7 | || 6 | 9 | 8 || 2 5 7
|| 4 | | 3 || | | 1 || | | ||
||=====|=====|=====||=====|=====|=====||=====|=====|=====||
|| 3 | 1 | || 6 | 2 | 8 || | 5 | 9 || 1 2 3 6
|| | | 4 || | | || 7 | | ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
|| 7 | | 5 || | 1 | 3 || 2 | | 6 || 1 3 5 7
|| | 9 | || 4 | | || | 8 | ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
|| | 6 | || | 9 | 7 || 3 | 4 | || 6 7 9
|| 8 | | 2 || 5 | | || | | 1 ||
||=====|=====|=====||=====|=====|=====||=====|=====|=====||
1 3 1 5
5 4 3 6
6 6 6 7
7 8 7 9
Sudoku Class Puzzle
There are already 27 numbers given. Of those remaining,
we can find the first few numbers quickly, by Isolation...
(R3,C2) = 4 because of 4's in B3, B4, B7
(R2,C3) = 9 because of 9's in B3, B4, B7
(R8,C7) = 2 because of 2's in B3, B6, B7
This creates two what's left lists, one below C3 and one below C7.
C3 list: 1 5 6 7
C7 list: 3 4 6 8
The newly added 4 in B1 leads to another 4 in B2...
(R2,C5) = 4 because of 4's in B1, B3, B8
We now get a what's left list for R2...
R2 list: 1 5 7 8
Because C7 already has 1, 5, 7, we get this by Exclusion...
(R2,C7) = 8
This leads to a sequence of finds, all for number 8 ...
(R1,C4) = 8 because of 8's in B1, B3, B5
(R7,C6) = 8 because of 8's in B2, B5, B7, B9
(R6,C9) = 8 because of 8's in B3, B9; and occupation in C9
(R4,C2) = 8 because of 8's in B1, B5, B6, B7
That completes the 8's within the entire puzzle.
It also leads to new what's left lists for R1 and C9...
R1 list: 2 3 6 7
C9 list: 5 6 7 9
Since B4 has a 9, and R8 has a 9, the 9 must go here...
(R7,C9) = 9 leaving only 5, 6, 7 in the C9 list.
Because 3 can't go in B6, and C7 is occupied in B3 ...
(R9,C7) = 3 the only place left for 3 in that column.
That leaves us with 4, 6 in C7 (B6), and R6 has a 4 in B4 ...
(R5,C7) = 4 by isolation, and
(R6,C7) = 6 what's left. C7 is now complete.
(R6,C8) = 9 because of 9's in B3, B4, B5, B9
Returning to the what's left list for R2, we see there are
only three empty cells scattered across the top boxes. C1
and C4 both already have 5's, so by exclusion there's only
one place left for a 5 in R2...
(R2,C9) = 5
Now what's left in C9 is just 6 and 7 in empty cells of B3, B9.
But B9 already has a 7, so 7 is forced to B3, and 6 to B9...
(R3,C9) = 7
(R8,C9) = 6 and C9 is now complete.
There are only two numbers missing from B9: 4, 5 and because
of the 4 in R7 and R8, they are forced to go in these cells...
(R9,C8) = 4
(R7,C8) = 5 (what's left in B9).
We now get what's left lists for C8, R6, R7, R8, R9...
C8 list: 1 3 6 7
R6 list: 2 5 7
R7 list: 1 2 3 6
R8 list: 1 3 5 7
R9 list: 6 7 9
Things now fall into place very quickly....
(R9,C5) = 9 because of 9's in B2, B7, B9
That leaves us with 6, 7 in R9, but 6 already appears in C6, so...
(R9,C6) = 7
(R9,C2) = 6 (what's left in R9).
Both 2 and 6 must occur in R7 of B8 because they are blocked in R8
and R9 of both B7 and B9. That leaves 1 and 3 in R7 of B7, and 3
is already in C2, so...
(R7,C2) = 1
(R7,C1) = 3
(R8,C5) = 1 because of 1's in B5, B7, B9.
Since 5 and 7 both appear in B8, exclusion gives us....
(R8,C6) = 3
Only 5 and 7 are left in R8 (B7), and C1 already has a 5, so...
(R8,C1) = 7
(R8,C3) = 5 (what's left in R8).
(R5,C4) = 3 because of 3's in B4, B6, B8.
For a long time, we've ignore the what's left list for R2.
There's only 1 and 7 left, and C1 has been found to be 7, so...
(R2,C1) = 1 by exclusion of 7.
(R2,C4) = 7 (what's left in R2). R2 is now complete.
(R3,C4) = 1 because of 1's in B5, B8; and occupation in C4.
(R4,C6) = 4 because of 4's in B2, B6, B8.
That leaves just 2, 5 in C6; and R5 already has a 2, so...
(R5,C6) = 5
(R3,C6) = 2 (what's left in C6).
(R6,C2) = 5 because of 5's in B5 and B6; and occupation in R6.
(R6,C4) = 2 because of 7 in C4 and what's left in R6.
(R6,C5) = 7 (what's left in R6).
(R4,C5) = 6 (what's left in B5).
(R5,C3) = 6 because of 6's in B5, B6, B7.
(R7,C4) = 6 (what's left in C4).
(R7,C5) = 2 (what's left in B8).
In C2, all that's left are 2, 7; and R5 has already has 2, so...
(R5,C2) = 7 by exclusion of 2.
(R1,C2) = 2 (what's left in C2).
In C3, all that's left is 1, 7; and 1 is blocked in B1, so...
(R1,C3) = 7
(R4,C3) = 1
(R5,C8) = 1 (what's left in R5).
(R4,C8) = 7 (what's left in B6).
(R4,C1) = 2 (what's left in R4).
(R3,C1) = 6 (what's left in C1).
(R3,C5) = 5 by isolation.
(R1,C5) = 3 (what's left in B2).
(R1,C8) = 6 (what's left in R1).
(R3,C8) = 3 (what's left in B3).
The puzzle is solved!
Could numbers be found in a different order? Certainly, but
it may not make much difference. You have to find numbers one
at a time (or in pairs). Usually you find several numbers very
early, then you find them very slowly, and then you find them
rapidly again toward the end. That SLOW section in the middle
is most frustrating. You look and look and look, and nothing
seems to be found with absolute confidence. Then there's a
break-through...you find something by Phantom or Exclusion,
and then several numbers fall into place. If you're forced to
guess (usually between two numbers in two empty cells), then
the puzzle was badly designed. You should NEVER have to guess.