Sudoku Very Hard Solution
This is a solution to the VERY HARD Sudoku Puzzle. You
should start with the puzzle as given, then examine the image
below, and read the narrative that follows it. Each number
found is shown with a dash and two-digit sequence value after
it to show when each was found (the order they were found).
||=====|=====|=====||=====|=====|=====||=====|=====|=====||
||1-40 |2-37 |9-04 ||3-06 | |5-23 ||4-19 |7-39 | ||
|| | | || | 8 | || | | 6 ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
||4-20 | | ||7-09 |1-22 | || | |5-21 ||
|| | 6 | 3 || | | 2 || 8 | 9 | ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
||5-41 |7-38 | || |9-05 |6-08 || |1-42 |2-36 ||
|| | | 8 || 4 | | || 3 | | ||
||=====|=====|=====||=====|=====|=====||=====|=====|=====||
||3-46 |4-52 |2-53 ||1-11 |6-34 | ||9-02 |5-48 |8-44 ||
|| | | || | | 7 || | | ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
|| |9-03 | || |5-29 | || |4-30 | ||
|| 7 | | 6 || 8 | | 3 || 2 | | 1 ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
||8-45 |1-17 |5-49 || |2-35 |4-28 ||6-25 |3-47 |7-43 ||
|| | | || 9 | | || | | ||
||=====|=====|=====||=====|=====|=====||=====|=====|=====||
||6-13 |8-14 | ||5-15 |3-07 | || |2-32 |4-31 ||
|| | | 1 || | | 9 || 7 | | ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
||2-12 | | || |4-27 |1-26 || | |9-01 ||
|| | 3 | 7 || 6 | | || 5 | 8 | ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
|| |5-51 |4-50 ||2-16 | |8-10 ||1-24 |6-33 |3-18 ||
|| 9 | | || | 7 | || | | ||
||=====|=====|=====||=====|=====|=====||=====|=====|=====||
Order
(R8,C9) = 9 by isolation. 01
(R4,C7) = 9 by isolation. 02
(R5,C2) = 9 by isolation. 03
(R1,C3) = 9 by isolation. 04
(R3,C5) = 9 by isolation. 05
(R1,C4) = 3 by isolation. 06
(R7,C5) = 3 by isolation. 07
(R3,C6) = 6 by isolation. 08
(R2,C4) = 7 by isolation. 09
(R9,C6) = 8 by isolation. 10
(R4,C4) = 1 because... 11
R7's 1 and C9's 1 force 1 to R9 (C7 or C8) in B9,
so (R9,C4) and (R7,C4) can't be 1, thus (R4,C4) is 1.
That leaves 2,5 in C4 (B8), and 1,4 pair in R8 of B8.
So...
(R8,C1) = 2 by what's left in R8 (outside B8). 12
(R7,C1) = 6 by isolation. 13
(R7,C2) = 8 by isolation. 14
(R7,C4) = 5 by exclusion from B9 (and 2,4,5 in R7). 15
(R9,C4) = 2 by what's left in C4. 16
B7 is only missing 4,5. C1 has 2,7 in B7,B4, and along
with the phantom 2 in C3 (B4), another 2,7 are in B4,B7.
So 2,7 must be in C2 of B1. That leaves 1,4,5 in C2 in
B4,B7. 1 is blocked in C2 for R4 and R9. So...
(R6,C2) = 1 by exclusion. 17
(R9,C9) = 3 by exclusion (1,6 in C9). 18
Note the 4,5 pair in B7, and 1,6 pair in B9 remain in R9.
At this point we are stuck. There are no exclusions or
isolations to act as clues. But there are several phantoms
that help. This is tricky logic, so read it careful.
First, here is what the puzzle looks like now:
||=====|=====|=====||=====|=====|=====||=====|=====|=====||
|| | | 9 || 3 | | 15 || 14 | | ||
|| | | || | 8 | || | | 6 ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
|| 145 | | || 7 | 15 | || | | 45 || 1 4 5
|| | 6 | 3 || | | 2 || 8 | 9 | ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
|| | | || | 9 | 6 || | | || 1 2 5 7
|| | | 8 || 4 | | || 3 | | ||
||=====|=====|=====||=====|=====|=====||=====|=====|=====||
|| | | 245 || 1 | 26 | || 9 | | ||
|| | | || | | 7 || | | ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
|| | 9 | || | 45 | || | 45 | || 4 5
|| 7 | | 6 || 8 | | 3 || 2 | | 1 ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
|| | 1 | 245 || | 26 | 45 || 46 | | ||
|| | | || 9 | | || | | ||
||=====|=====|=====||=====|=====|=====||=====|=====|=====||
|| 6 | 8 | || 5 | 3 | || | 24 | 24 || 2 4
|| | | 1 || | | 9 || 7 | | ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
|| 2 | | || | 14 | 14 || | | 9 || 1 4
|| | 3 | 7 || 6 | | || 5 | 8 | ||
||-----|-----|-----||-----|-----|-----||-----|-----|-----||
|| | 45 | 45 || 2 | | 8 || 16 | 16 | 3 || 4 5 : 1 6
|| 9 | | || | 7 | || | | ||
||=====|=====|=====||=====|=====|=====||=====|=====|=====||
2 2 1 1
4 4 4 4
5 5 5 6
7
You can see the what's left lists, and something new. I've
filled in a few "empty" cells with "naked pairs", which are
the pairs of numbers that can only go in those cells. Such
pairs can assist in making decisions. There also are some
"naked triples", like those in R2 and C3. (All in red.)
I generally don't like cluttering up a puzzle like this,
but sometimes, desperate measures are needed to continue.
Notice how many 4's are missing. If we could find one of
them, we might find more, get past this point, and solve it.
C7 has 1,4,6 left. B3 has a 6, so R1,C7 can only be 1,4.
B6 has a 1, so R6,C7 can only be 4,6. We already know that
R9,C7 can only be 1,6 because of the phantom 4 in R9 of B7.
Therefore, we can only put 4 in R1,C7 or R6,C7, not R9,C7.
Note that B2 is missing 1,5 in R1,C6 and R2,C5. Likewise,
4,5 is missing in R5,C5 and R5,C8, and 2,6 can't appear
in R6,C6 because they already occur in C6, so 2,6 must
appear in C5 of B5. Since (R5,C5) is either 4 or 5, the
2,6 values must occur in R4,C5 and R6,C5. That leaves
R6,C6 which must be another 4,5 possibility.
If we put 4 in R6,C7, then R1,C7 must be 1, and R6,C6
must be 5. Since only 1,5 can go in R1,C6, that's not
possible if R1,C7 is a 1 and R6,C6 is 5. Logically,
R6,C7 can't be a 4, and we can't put 4 in R9,C7, so
the only place left in C7 for 4 is:
(R1,C7) = 4 by the logic described above. 19
(R2,C1) = 4 by isolation. 20
(R2,C9) = 5 because 1 is excluded. 21
(R2,C5) = 1 by what's left in R2. 22
(R1,C6) = 5 by what's left in B2. 23
(R9,C7) = 1 by exclusion from R6,C7. 24
(R6,C7) = 6 by what's left in C7. 25
(R8,C6) = 1 by isolation. 26
(R8,C5) = 4 by what's left in R8. 27
(R6,C6) = 4 by what's left in C6. 28
(R5,C5) = 5 by exclusion. 29
(R5,C8) = 4 by what's left in R5. 30
(R7,C9) = 4 by isolation. 31
(R7,C8) = 2 by what's left in R7. 32
(R9,C8) = 6 by what's left in B9. 33
(R4,C5) = 6 by isolation. 34
(R6,C5) = 2 by what's left in C5. 35
(R3,C9) = 2 by isolation. 36
(R1,C2) = 2 by isolation. 37
(R3,C2) = 7 by what's left in C2 of B1. 38
(R1,C8) = 7 by isolation. 39
(R1,C1) = 1 by what's left in R1. 40
(R3,C1) = 5 by what's left in B1. 41
(R3,C8) = 1 by what's left in R3. 42
(R6,C9) = 7 by isolation. 43
(R4,C9) = 8 by what's left in C9. 44
(R6,C1) = 8 by isolation. 45
(R4,C1) = 3 by what's left in C1. 46
(R6,C8) = 3 by isolation. 47
(R4,C8) = 5 by isolation. 48
(R6,C3) = 5 by what's left in R6. 49
(R9,C3) = 4 by exclusion. 50
(R9,C2) = 5 by what's left in R9. 51
(R4,C2) = 4 by what's left in C2. 52
(R4,C3) = 2 by what's left in C3. 53