# Skeleton sudoku, the second

I've already set one skeleton Sudoku. Here's a rather harder one. (Nonetheless, it should be within the realms of what can be done by hand; I solved it manually myself to verify that there was only one solution. Actually, I got pretty good at solving these, because it took me something like 5 or 6 attempts to find a puzzle whose solution was unique, learning more each time about what sort of patterns would allow multiple solutions to exist…)

Here's the grid: In case you can't see images: I shaded a1, b1, e1, d2, c3, e3, and c4.

And here are the rules (basically the same as before, but hopefully stated a bit more clearly):

• Following the grid lines, separate this puzzle into five contiguous regions of five squares each, and also place a digit from 0 to 4 inclusive in each square, so that:
• (Generalised) Sudoku property:
• Each horizontal row must contain one each of 0, 1, 2, 3, 4;
• Each vertical column must contain one each of 0, 1, 2, 3, 4;
• Each of your regions must contain one each of 0, 1, 2, 3, 4.
• Each shaded cell must contain a number equal to the number of edges of that cell which are region borders. (This is the same meaning that numbers have in a Slitherlink puzzle.)
• Equivalently, the number on each shaded cell must be the number of adjacent cells that are either in a different region, or outside the map.

Looking at the first row, we can see that there are three blue squares, and blue squares on the edges are either 1, 2 or 3, so they must be 1 2 3 in some order. Furthermore, the middle one can be 1, so we have 2 1 3 or 3 1 2. Assuming 3 1 2, we eventually get here by placing forced numbers/walls: Here we can notice it is impossible to settle the zeros in the bottom-left corner, as they would have to go at least two in the same region. So 3 1 2 is bad and we need 2 1 3: A bit of "cheating" is now in order: we know this has a unique solution, so one of the blue squares is a 0 (otherwise, 0s and 4s would be completely interchangeable in the resulting sudoku). The only blue squares that can be a 0 are the ones not in the edges, and putting a 0 in the top two causes the top-right corner to be blocked, so the bottom most blue square is a 0. The 0 determines that entire region's shape, from where putting all of the other regions is now simple: • I was aware of the shortcut where you assume the answer has a unique solution, and using it here is legitimate, so I'm marking this answer as correct. (The puzzle is solvable without, though.) – ais523 May 12 '17 at 19:02
• Just so it's there, the other way to solve (without the shortcut) is after filling as much of the grid in as possible, to notice that the square that ends up 0 can only be 1 or 0, as the square above it must have 3 and the one above that must have 2. The center-right blue square also must be 2, so the 2s in the bottom rows must be in columns 2 and 4. If the blue square that turns out to be 0 were a 1, the pentomino containing it would have one square in each of columns 2 and 4, but not opposite each other, so it would either need to contain both 2s or neither 2. – P... Nov 8 '17 at 20:08