Killer Sudoku
One thing that can be done with the Sudoku spreadsheet is to use it as an aid
to solving Killer Sudoku problems. This is a less automated process than the
usual use for solving standard Sudolu problems. However some of the features of
the spreadsheet are designed for such cases.
What is a Killer Sudoku problem? Well, it's played on the same 9x9 grid, divided
into 3x3 squares, into which digits 1 to 9 are put with the same constraints as
in a Sudoku problem. However rather than pre-seeding the grid with some digits,
the grid starts empty. Instead the grid is divided up into connected sets of
cells, each of which is labelled with the sum of the digits that make it up. In
some Killer Sudoku problems the addituional constraint is added that the sets
also obey the Sudoku rule of no repeated digits. However in the case of the
example considered here, that constraint was neither specified nor needed to
solve the problem. However the solution turns out to have that property, so it
may have been intended.
To consider how to use the Sudoku spreadsheet to solve a Killer Sudoku problem,
an example is considered. The usual format for showing these is able to do this
in black and white, but I've used colour to show the sets in
this spreadsheet.
Taking the standard Sudoku spreadsheet, configure it on the Sudoku page to show
Solutions and (probably) Level 3, and on the Hints page to use Right hints. As
most time (all if preferring simply to us the spreadsheet for record taking)
will be spent on the Hints page, on the right hand grid,
this version of the spreadsheet makes those changes.
For this discussion, cells in the Sudoku spreadsheet will be labelled according
to the cell numbers in the coloured spreadsheet - A1 in the top left, I9 in the
bottom right and so on. Of course on the grid on which the work will take place
these are cells L2 to T10. Explanation will also use rows 1 to 9, columns A to
I, and 3x3 squares described as, for example, top left, centre right, and
centre.
The first easy steps (in a way to solve this problem, there are of course
others) are:
- Row 1 consists of sets 13, 20 and 8 and cell I1. As each row (and column
and square) sums to 45, cell I9 is a 4.
- The centre bottom square, plus cell F6, is covered by sets 20, 15 and 12.
Thus cell F6 is a 2.
- Cells H2 and H3 must be a 1 and a 2, in some order. So these cells
can be set to 12.
- These settings can be used to eliminate 4s in the other cells in row 1,
column I and the top right square, to eliminate 2s in row 6, column F and
the centre square, and to eliminate 1s and 2s in column H and the centre
square. So, for example, H1 can be set to 356789. (Note that a blank cell
is equivalent to 123456789.)
A spreadsheet with these settings is here. The two
known cells are also filled in on the Sudoku sheet.
The next steps are the other size two sets, so for example A1+B1 being 13
means that (taking into account that 4 is already excluded from both) that
these can be reduced to 5689 each. I2+I3 can be considered as a size two
set summing to 14, so these also can be 5678 each. The one asymmetric set is
that while G7 can be 2 (and hence G6 can be 9) G6 can't be 2, so G7 can't be
9. A sequence of eliminations is then possible:
- With 1, 2 and 4 excluded, G1 and H1 have to be 3 or 5 (i.e. set to 35)
which eliminates 3 and 5 elsewhere in row 1 and the top right square.
- This leaves A1 and A2 having to be 6 or 7, and these can be eliminated from
row 1 and the top left square. This also eliminates 4 and 5 from cell D2.
- Eliminating 6 from C3 eliminates 8 from C4.
- I2 and I3 are reduced to 6 and 8, which eliminates those from column I and
the top right square.
- G2 and G3 are reduced to 7 and 9, which eliminates those from column G
(there's nowhere left to eliminate in the top right square).
- This reduces G6 and G7 down to 3568 each.
With all the size two sets so handled, the hints are as shown
here. There are still no new cells with a definite
value. Can now proceed by:
- The only option left for cells I6, I7 and I8 (summing to 9, no 4s or 6s) is
1, 3 and 5. Eliminating those from column I leaves I4 and I5 as 2, 7 and 9.
But these can't be 7 and 9, as that would leave H4 and H5 summing to 4, but
each with a minimum value of 3. So these are 2 and 7 or 2 and 9,
eliminating 2 from I9 and elsewhere in the centre right square.
- If sets were assumed not to include repetitions, then H6, H7, H8 and G8
would be 6, 7, 8 and 9 in some order. But with G8 not able to be 9, and the
other constraints on the cells, it's possible for H7 or H8 to be 5, if H6
and G8 are both 8.
- Now the only possibility for a 2 in the lower right square is in cell G9.
This can be seen by looking at the right hand grid on the Sudoku page, or
observed for yourself. (The former is one reason you might use this
spreadsheet.). Now 2 can be eliminated elsewhere in row 9 (it's also out
of column G).
- Now the only 4 in the lower right square is in cell H9. 4 can now be
eliminated from column H and row 9. This also removes 9 from cell A8 and 8
from cells D9 and E9.
- This means that cell I9 has to be 9, from that the sum of G8, H8 and I9 is
15. This means that 9 can be eliminated from column I and row 9. Other
eliminations are then 4 from A8 and 3 from D9 and E9.
- Now D9 and E9 are 5 and 7, so 5 and 7 can be eliminated elsewhere in row 9
and the centre bottom square. This takes 6 and 8 out of cell A8.
- As cells I4 and I5 now must be 2 and 7, these can be eliminated elsewhere
in the centre right square.
There are some more deductions that can be made based on these changes, but
before considering them, the current state is here.
This includes setting the known cells on the Sudoku sheet. Continuing:
- Now cell H6 is the only possible 9 in its set, and there can't be a 5 in
that set (remove from H7 and H8). This then means removing 9 from column H,
row 6 and the centre right square. This also removes 1 from cell E5.
- Now A8 and A9 have to be 6 and 7 or 5 and 8. But the former would
contradict A1 being 6 or 7, so A8 is 5 and A9 is 8. This can then be
followed up by eliminating 5 and 8 in column A, rows 8 and 9 respectively
and the lower left square. This also removes 1 from cell B2.
- Adding up all the sets that cover the lower right square, this also covers
cells G6, H6 and I6, and sums to 11+30+9+15=65, so G6+H6+I6=20, so
G6+I6=11, which reduces the options to G6 being 6 or 8 and I6 being 3 or 5
(given the existing possibilities). Then G7 must also be 3 or 5.
- Now in column G there are two 68s and two 35s, so these can be eliminated
from other cells in that column, putting 14 in G4 and G5.
- Cells D9 and E9 are 5 and 7 (or vice versa). Thus cell F7 can't be 5.
The current state is now here. Continuing:
- A4 to A7 summing to 14 doesn't allow any of them to be 9, so that option
can be removed. This leaves A3 as the only possible 9 in column A (again,
this can be seen from the Sudoku sheet). 9 can then be eliminated from
row 3 and the upper left square. This eliminates 5 from cell C4, 3 from
cell D4 and 2 from cell B4.
- Now cell G3 is 7, which makes cell G2 equal 9. This allows more elimination
in rows 2 and 3, and also 5 from cell D4 and 2 from cell C2.
- As the set A3, B3 and B4 sums to 17, B3 and B4 sum to 8, which leaves the
former as 1235 and the latter as 3567.
- As the set F3, G3 and G4 sums to 16, F3 and G4 sum to 9, and as G4 is 1 or
4, F3 is 5 or 8.
- As there are now two 58s in row 3, can remove 5 and 8 from it. This also
leaves cell D4 as 8 or 9, and removes 3 from cell B4.
- This leaves cell I3 as 6, and then cell I2 as 8. The elimination based on
those in those rows also removes 3 from cells C2 and D2.
Eliminations are about to come thick and fast, so the current state is now
here. Continuing:
- Now the only 3 in the top right square is in cell B3 (again from the Sudoku
sheet). This sets cell B4 to 5. Elimination sets cell D3 to 4 and then cell
D4 to 8. As well as all the direct elimination, this also eliminates 1 from
cell A2.
- Now as cells A2 and B2 are 2 and 4, these can be removed from row 2 and the
top left square. This sets (some following on from others) H2 to 1, H3 to
2, E3 to 1, C2 to 5, C3 to 8, C4 to 6, C1 to 1, H4 to 3, H1 to 5, G1 to 3,
G7 to 5, C9 to 3, F3 to 5, D2 to 6. The elimination also removes 9 from E5.
- There are also some other eliminations that if these above results have all
been entered in the Sudoku sheet will show up as warnings. These are a 9
from cells D5 and F5 (because there must be a 9 in row 5 in the centre left
square), a 3 from cell I6 (because either cell I7 or cell I8 must be a 3).
That leaves cell I6 as a 5, and more elimination.
The current state is now here. Continuing:
- Now sheet Sudoku gives two cells that can be set to 5, D5 and E9 (based on
only possibilities in columns). This also sets cell D9 to 7.
- Now based on the set H4, I4, H5 and I5, cell H5 can be set to 8. Then cell
G6 can be set to 6 and cell G8 to 8. The elimination also includes 4 from
cell E5.
- Now based on the set F3, G3 and G4, cell G4 can be set to 4, and then cell
G5 can be set to 1. More eliminations follow. Eliminate 1 from cell A7, as
the 1 in the centre left square is in column A.
- Now cell B6 is the only cell that can be an 8 in column B or row 6.
The current state is now here. Continuing:
- Considering the set consisting of cells E4, F4, F5 and G5, the 1 can be
eliminated from cell F4, as otherwise the cells can't sum to 20.
- This now means that cell F5 can be set to 3. Also as cells E4 and F4 are 7
and 9, those values can be eliminated from row 4 and the centre square.
- Elimination includes cell I4 can be set to 2, cell I5 to 7, cell A4 to 1,
cell E6 to 4, cell C6 to 7, cell A6 to 3, cell D6 to 1, cell F2 to 7, cell
E2 to 3, cell F4 to 9, cell E4 to 7, cell F1 to 8, cell E7 to 8 (only place
in column E), remove 24 from cell A7 (cells A2 and A5 are 2 and 4).
The current state is now here. Continuing:
- The set with cells A4, A5, A6 and A7 can only sum to 14 with A5 set to 4
and A7 set to 6.
- B9 is set to 1, F9 is set to 6, A2 is set to 2, B2 is set to 4, H7 is set
to 7, H8 is set to 6, A1 is set to 7, B1 is set to 6 and B8 is set to 7
(only option in column B or row 8).
The current state is now here. Concluding:
- The set of cells C6, C7 and B7 summing to 13, with C6 equal to 7 means that
B7 and C7 sum to 6, which, with the options remaining, means that B7 must
be 2 and C7 must be 2.
- Then the sollowing elimination gives C8 set to 9, F7 set to 1, F8 set to 4,
I7 set to 3, I8 set to 1, D7 set to 9, D1 set to 2, E1 set to 9, E8 set to
2, D8 set to 3, C5 set to 2, B5 set to 9.
The final state is now here.
Please send any feedback to:
christopher.dearlove@gmail.com
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Last modified 2nd October 2020: Updated contact information.