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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.

enter image description here

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    $\begingroup$ I was going to ask, why the coloured backgrounds, then I realised you need it to show white and black dots properly :-D $\endgroup$ – Rand al'Thor Apr 10 at 20:50
  • $\begingroup$ 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. $\endgroup$ – Rand al'Thor Apr 10 at 23:35
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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:

enter image description here

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:

enter image description here

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 $4$.

enter image description here

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:

enter image description here

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.

enter image description here

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