# Kropki Sudoku – Sudoku based on the relation between neighbouring numbers

This puzzle is named Kropki Sudoku.

• 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.

• A white dot indicates that the adjacent numbers differ by 1.

• A black dot indicates that one of the adjacent numbers is twice the other.

• The dot between 1 and 2 can be either black or white.

• All applicable black and white dots are shown: if two numbers do not have a white dot between them, then they do not differ by 1, and similarly for black dots. • I was going to ask, why the coloured backgrounds, then I realised you need it to show white and black dots properly :-D – Rand al'Thor Apr 10 at 20:50
• That was fun! And I think not as hard as your usual :-P The hardest part was to start off, to get any numbers in the grid at all; after that it was a lot easier to do the rest. – Rand al'Thor Apr 10 at 23:35

General note: the only possibilities for two cells with a black dot between (a double-pair) are $$\{1,2\},\{2,4\},\{3,6\},\{4,8\}$$.

First consider the top brown box, specifically the double-pair, one of which forms an end of a string of three consecutive numbers. Recall there's also another string of three consecutive numbers in the box.

That can't be $$2$$ adjoining $$1,2,3$$ or $$4$$ adjoining $$2,3,4$$ or $$2$$ adjoining $$4,3,2$$, obviously.
It can't be $$3$$ adjoining $$6,7,8$$, because then there's no space for the other string of three.
It can't be $$3$$ adjoining $$6,5,4$$, because then the other string of three must be $$7,8,9$$ leaving $$2$$ next to either $$3$$ or $$4$$.
It can't be $$4$$ adjoining $$8,7,6$$, because then the other string of three must be $$1,2,3$$ leaving $$5$$ next to either $$4$$ or $$6$$.
It can't be $$2$$ adjoining $$4,5,6$$, because then the other string of three must be $$7,8,9$$ leaving $$1$$ or $$3$$ next to $$2$$.
It can't be $$8$$ adjoining $$4,5,6$$, because then the other string of three must be $$1,2,3$$ respectively, leaving $$7$$ or $$9$$ next to $$8$$.

So it must be
EITHER $$8$$ adjoining $$4,3,2$$, with the other string of three as $$5,6,7$$ ($$1$$ above $$8$$ and $$9$$ in the corner),
OR $$1$$ adjoining $$2,3,4$$, with the other string of three as 5,6,7 or 6,7,8 are impossible $$7,8,9$$ ($$5$$ above $$1$$ and $$6$$ in the corner),
OR $$6$$ adjoining $$3,4,5$$ with the other string of three as $$7,8,9$$ ($$1$$ and $$2$$ either way round),
OR $$6$$ adjoining $$3,2,1$$ with the other string of three as $$7,8,9$$ ($$4$$ above $$6$$ and $$5$$ in the corner).

9 6 5 6 8 8 5 8
2 7 1 4 5 5 1 4
3 4 8 3 2 1 4 3 6 2 3 6

The first of these four possibilities can't be right, by considering the top left blue box:

The one next to $$9$$ must be $$8$$, and the numbers $$3$$ and $$4$$ can't be either on the bottom row or the right column, so they're both involved in the string of four. But that can't be $$3,4,5,6$$ (puts $$3$$ next to $$6$$) or $$1,2,3,4$$ or $$2,3,4,5$$ (leaves no possibilities for the double-pair on the right of the top row).

The third of those four possibilities also can't be right:

If top left in that brown box is $$1$$, then left of it is $$2$$, and again that string of four to the left must involve both $$3$$ and $$4$$ which turns out to be impossible.
If top left in that brown box is $$2$$ and next to it is $$3$$, then that string of four to the left must involve $$4$$ and not $$3$$, so it must be $$4,5,6,7$$ which puts $$3$$ next to $$6$$.
If top left in that brown box is $$2$$ and next to it is $$1$$, then consider the top right blue box. That double-pair must be $$\{3,6\}$$ in some order, so the $$1$$ must be in the bottom left and the $$2$$ above it, which leaves no possibilities for a string of three ending in one of a double-pair.

The last of those four possibilities also can't be right:

In the top left blue box, $$2$$ must be in the string of four, but $$1,2,3,4$$ and $$2,3,4,5$$ are both impossible, because $$1,4$$ can't be in the second row and $$5$$ can't be in the top row.

So we can solve the top brown box (almost) completely, and fill in a bunch more cells with only two possibilities, including the string of four in the top left:

Now consider the top right square, which contains a string of three ending in one of a double-pair.

That end cell can't be $$1,2$$ so it must be one of $$4,6,8$$. If it's $$6$$ or $$8$$, then the string of three is $$6,7,8$$ in some order, so the two adjacent cells on the left must be $$3$$ (above) and $$4$$ (below), with the two on the right being $$5$$ (above) and $$9$$ (below). But then either $$3$$ is next to $$6$$ or $$9$$ is next to $$8$$, contradiction. So we have: Now consider the right brown box, which contains a string of four.

Turns out the only possibilities for that string are (from top to bottom) $$5,6,7,8$$ and $$6,7,8,9$$. So the one at the top right must be $$2$$, and the remaining four cells are $$1,3,4$$ and one of $$5,9$$. Note that $$4$$ can't be in the top left because it's the end of a string of three which can't be $$2,3,4$$ or $$6,5,4$$.
If the string of four is $$6,7,8,9$$, then $$4$$ must be in the bottom left (can't be next to $$8$$), but then one of $$3,5$$ must be next to it. Contradiction, so it's $$5,6,7,8$$.

Now consider the bottom right blue box, which contains a double-pair.

That double-pair can't be $$\{3,6\}$$ or $$\{4,8\}$$ or $$\{1,2\}$$ because there's a $$6$$ and an $$8$$ and one of $$1,2$$ in the same column. So it must be $$\{2,4\}$$. Neither $$1$$ nor $$5$$ can be above that, so it must be $$3$$; neither $$3$$ nor $$5$$ can be on the right, so we can fill in $$1,2,3,4$$ in this box and $$9$$ in the brown box above.
Going back to that brown box, we only have $$1,3,4$$ left to fill in on the left-hand side, so $$4$$ must be at the bottom, $$1$$ in the middle, and $$3$$ at the top giving a string $$5,4,3$$ and another $$3$$ below the $$4$$. In the fourth row, only $$1,6,7,9$$ are left.

So the two adjacent must be $$6$$ and $$7$$. They're part of a string of four which can't be $$7,6,5,4$$, so it must be $$6,7,8,9$$.

In the central blue box, $$1,2,6,7,8,9$$ are left, including two separate adjacent pairs.

The bottom right square can't be $$1,2,6,8$$, and the one next to it also can't be $$8$$, so it must be $$7$$ with $$6$$ next to it. The other pair can't involve $$1$$, so it must be $$8,9$$ and the $$8$$ must be on the left. So we have $$1$$ above and $$2$$ below.

That $$7$$ lets us finish off the top three rows:

Now consider the left brown box, specifically the double-pair.

It can't involve $$2$$ or $$6$$, so it must be $$4,8$$ and the $$4$$ must be at the top; next to that can't be $$3$$ so it must be $$5$$, and from the $$8$$ we also have a string $$8,7,6$$.

The rest of the grid easily follows.