# Tic tac toe - Sudoku: A variation in which the centre box defines the layout of the other boxes

I stumbled across this interesting variation of Sudoku the other day and thought I'd give it a try, it has however baffled me and I've been unable to progress past the very early stages.

The puzzle comes from the "World Class Puzzles NL" set of daily puzzles, this one published on 2020-04-29 set by Richard Stolk.

The rules for this variation are as follows:

• The classic sudoku rules apply: Place numbers on the grid below such that each row, column and 3×3 box contain the numbers 1 to 9.
• The central box (3x3 box) serves as a map to the 9 corresponding 3x3 boxes.
• Wherever the central box contains an odd digit, the 3x3 box corresponding to that cell must contain at least one line (row, column or diagonal) of odd digits.
• Wherever the central box contains an even digit, the corresponding box must contain one line of even digits.
• No box can contain a line of 3 even digits and a line of 3 odd digits.

• @iBug In a 3x3 box, there can't be both a line of 3 odds and 3 evens, that would be an invalid box. If a box should have a line of odds it can't have a line of evens and vice versa. Does that make it clearer? I hope that edit clarifies things. May 4 '20 at 13:01
• I've made it! See my answer below.
– iBug
May 4 '20 at 14:14
• Note that rot13(gur prageny obk freirf nf n xrl gb vgfrys)! May 4 '20 at 14:35
• @shoover exactly, I'd hoped that was clear from the use of the language saying it was a key to the "9 boxes", is that not so? May 4 '20 at 14:47
• Ooh, actually I hadn't realised that. I'd thought the central cell of the central box didn't give us any information. That helps. May 4 '20 at 15:31

## Step by step deduction

First look at the right middle box, since that's got four cells filled already including two odd in a column and two even in a column.

If that's an even-line box, then the one between $$5$$ and $$9$$ must be even; it can't be $$2,4,8$$, so it must be $$6$$. But now the only even thing left to place is $$2$$; there must be an even number in the middle column and there must be one in the top row, so there's only one place $$2$$ can be, but then we don't have an even line in this box. Contradiction.

So it's an odd-line box, which means the one above $$8$$ and $$4$$ must be odd; it can't be $$1,3,5,9$$ so it must be $$7$$.

Also, from normal Sudoku rules,

the $$5$$ in the top right box must be on the right-hand side. Now that whole rightmost column is filled except for $$1,2,9$$ in the bottom three, of which $$2$$ must be on top.

Going back to the top right box, its bottom left cell can't be $$2,3,4,5,6,7,8,9$$, so it must be $$1$$, then the one to the right of that must be $$4$$.

Now look at the left middle box, which also has four cells filled already.

The empty cell on the right-hand side can't be $$2,4,6,8$$, so it must be odd and this is an odd-line box. In fact, the $$8$$ in this box must be in the middle column, which means the $$8$$ in the bottom left box is in the right column, while the $$4$$ can't be in the right column or bottom row. So the bottom left box is also an odd-line box. The corresponding cell in the centre box can't be $$3,5,7,9$$, so it must be $$1$$.

The $$1$$ in the right-middle box can now be filled by normal Sudoku deduction.

Important realisation:

if a box has an even line in it, it must be one of the diagonals. (Assume it's a row or column; then each one of the other two parallel rows/columns must contain at least one even number, making five even numbers in total, contradiction.)

In particular, this means the top middle box is

an odd-line box. Looking in the centre box now, the top-middle and left-middle cells are both odd and can't be $$1,5,7$$, so they must be $$3$$ and $$9$$ in some order. Now the top-left, top-right, and bottom-middle cells can't be $$5$$ or $$7$$, so they must be even. That means the top left box, top right box, and bottom middle box are all even-line boxes and therefore have all-even diagonals. We can fill in lots of stuff immediately there:

Thanks to @shoover's observation that the central box is a key to itself,

it must be an odd-line box, since there's not enough evens left to make an even diagonal. So the bottom right one must be an even-line box, which means its central cell must be even, therefore $$6$$. That means the $$6$$ in the right middle box is on the left, and now we can fill in lots more stuff with pure Sudoku logic:

Now we're almost done.

Top left box: only one place for $$9$$, then only one place for $$1$$, then the rest is easy.
Second column: placing $$5$$ and $$9$$ helps us to finish off the bottom right box.
Remember that the bottom middle box has an even diagonal, and the rest can all be done by pure Sudoku logic.

# Final State (Solution)

My progress (images that I saved "in the middle"):

• First few steps
• First (and only) time I made a guess (See the bottom for the way of avoiding this guess, making this answer completely logical)

The exact guess I made was

Choose 7 for the middle-right grid in the middle box (row 5, column 6).

After making that guess, continuation was pretty straightforward: You stare at the board, following regular Sudoku rules, picking the digit where it's the only option for the grid, or where it's the only valid grid for a particular digit in the row / column / box. Shouldn't be difficult to get to the next "checkpoint" (the image below).

• Fill more digits

For the above image, the key is

Note the bottom-left box: digits 4 and 8 aren't available at positions where this box could be an "even box", so this must be an "odd box", meaning the bottom-left grid of the middle box (row 6, column 4) must be 1.

And now,

The bottom-middle and bottom-right boxes must both be "even boxes", so the middle grid of the bottom-right box must be 6 (even).

You should now be at the following image. It's "trial and error" with standard Sudoku rules again.

• Fill even more digits

Not too much to talk about in this period. Go with standard sudoku rules.

• Almost there

This one is a bit interesting as I had to experiment a bit.

Try putting the 6 of the bottom box in either the middle-top grid (row 7, column 5) or the top-right grid (row 7, column 6), you'll find that in both cases, the 4 of the box goes in the middle (row 8, column 5) and the 3 goes in the bottom-right (row 9, column 6).

That's an interesting discovery. Now standard rules should bring you to the solution.

After all, there weren't many special strategies. Most of the time I was looking at the digits and thinking which one is available where.

Like, this method, you would call it "trial and errors".

To avoid the guess: Follow Rand al'Thor's answer to this point: Image. Use this key discovery:

An even box must have an even diagonal (and not row or column, since either would require two more even numbers to "break" the odd lines, making a requirement for at least 5 even numbers, which is impossible).

And this discovery:

Three even numbers have been placed in the middle box, at a position that's impossible to make this box itself an even box, so the middle grid of the middle box (row 5, column 5) must be an odd number, which must be 5. Then the right grid of the middle box (row 5, column 6) must be 7.

Now we've deducted that the guess is correct, go on as the original answer :)

• In the spoiler below your "And now," I think you mean "even boxes", correct? Otherwise this is a nice answer. I am however withholding from accepting it as it required a guess as opposed to a logical solution using the rules given. But +1 none the less and thank you very much for your time spent working on it and congratulations on achieving it! May 4 '20 at 15:17
• @Edlothiad Oh yes, I ran drousy while typing the answer.
– iBug
May 4 '20 at 15:21
• @Edlothiad I've "fixed" the guess part and now it's completely deducted.
– iBug
May 4 '20 at 16:02