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This puzzle is a combination between a Slitherlink and a Fillomino.

You should put a number in each square and draw a single closed loop over the grid such that the following properties are satisfied:

  • All the shaded squares should contain 1, 2 or 3, and the number should equal the number of sides of the square that are part of the loop.
  • The region bounded by the curve should form a valid Fillomino puzzle, that means: the inside region is divided into several connected groups such that any square in such a group contains the size of the group, and such that no groups with the same size border each other.
  • Similarly the region outside the loop should form a separate valid Fillomino puzzle.

An example of a solved puzzle:
enter image description here

And here is the puzzle for you to solve:
enter image description here

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  • $\begingroup$ Oh so the outside needn't be connected. $\endgroup$ – Arnaud Mortier Aug 22 at 20:16
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Solution:

The two adjacent shaded 1s must be split by the path, because otherwise we have an illegal fillomino region. Therefore their other edges must not be part of the path, which we can therefore continue a little further.
enter image description here
The 2 next to the bit of path we have just drawn must now be part of a block extending upward:
enter image description here
The 4 nearby is part of a fillomino block of size 4. It can't extend into any of the nearby shaded squares (clearly a shaded 4 is impossible). There's only one way to do this, and since it results in a block adjacent to another 4 we must put a bit of path in between the two:
enter image description here
Now we can fill in the rest of the other 4-block:
enter image description here
The shaded 2 near the top right must have the path on either its N and E, or the S and W, edges. The latter is impossible because it results in a size-1 fillomino block labelled 2, so it must be the former. We can then immediately draw in another path segment there:
enter image description here
The shaded square near the bit of path we have just drawn can't be numbered 2, because both the 2-blocks it might join with are already of size 2. Nor can it be numbered 4, because shaded squares never can. Nor 3, because none of its neighbours is numbered 3. So it's a 1, and the path runs between it and the 1 below it. Filling this in, a number of other path segments and non-path segments follow quickly:
enter image description here
The remaining un-filled shaded square near the top right can't contain a 4 (of course), nor a 2 (it's on the same side of the path as the block of 2s above it), nor a 3 (because that would make the path bend around it and cut it off from the blank squares that might contain other 3s). So it's a 1. We can draw a bit more path and non-path (including one segment over on the left that could have gone in earlier):
enter image description here
The path can't go around the 4s at top right, because then it would form a loop isolated from other bits of path we know are present. So it goes the other way.
enter image description here
There's a shaded square near the top left with two path segments around it. So its number is at least 2 and (as usual) at most 3. It can't be a 3 because then the path would curve around it and block it from anything else that might be part of the same 3-block. So it's a 2. Its neighbour on the left is therefore also a 2, and the path must pass above that neighbour to separate it from the 2 above it -- which must therefore have another 2 above it.
enter image description here
The square in the middle of the left-hand side can't contain a 1 or 2 because of already-complete adjoining blocks. Can it be a 4? No, because there's only one way to arrange the resulting 4-block, which then requires a 3-block at bottom left which would include those two adjacent shaded squares, which would then have to form an illegal 2-square loop. So, that middle square is a 3 and so therefore its its neighbour below. The square below that is also a 3, but can't be part of the same block, because that would require the path to do something impossible. So we have this instead:
enter image description here
If the bottom-left square contains a 2 then the path is forced to cut it off from any possible neighbour, so it's a 1. That pushes the path rightward and tells us where the last 3 from the block above it goes.
enter image description here
The empty square nearest the bottom left corner isn't a 1 (because there's another 1 next to it) nor a 3 or 4 (which would force two 3s or 4s into the shaded squares adjoining it), so it's a 2, which therefore extends into the adjacent shaded square.
enter image description here
The blank square two up-and-in from the bottom left corner can't contain a 1, so the square to its right is on the same side of the path. That lets us fill in a few more non-path segments and another bit of path.
enter image description here
We now see that that square can't contain a 2, because that would leave nothing to fill in above the right-hand square of that block. Nor of course can it contain a 3. So it's a 4 and there's only one way to complete the block.
enter image description here
The shaded square in the middle of the lower side can't be a 1 (it already has two adjacent path segments) or a 4 (duh) or a 3 (because that would force the path to cut it off from possible neighbours). So it's a 2, forcing the path to run above it to separate it from that other 2.
enter image description here
There's then only one thing to do with the nearby 3.
enter image description here
The rest is sufficiently straightforward that I shan't split it up further.
enter image description here

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  • $\begingroup$ Ah whoops, didn't notice you had posted an answer already. Threw a +1 your way. $\endgroup$ – Deusovi Aug 22 at 20:37
  • $\begingroup$ No problem! Your solution takes some different paths from mine, so there's no harm in having both. $\endgroup$ – Gareth McCaughan Aug 22 at 20:42
  • $\begingroup$ This is indeed the intended solution, and also a beautiful explanation! I will accept your answer since you were slightly earlier than Deusovi. $\endgroup$ – Reinier Aug 23 at 9:26
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In this answer, I use "RxCy" for the cell in the x-th row (from the top) and the y-th column (from the left).


First:

The two 1s must be separated by the loop. Since they're both Slitherlink clues, this gives a few loop segments:

enter image description here

Second:

The 2 in R3C3 must extend upwards. Then the 4 in R2C2 must extend upwards and to the right (because it cannot use any Slitherlink clue cells); this gives another loop segment to the right of that, because otherwise it would touch the other 4 clue.
Then, the other 4 clue can be extended right and downwards.

enter image description here

Following directly from that,

take a look at the 2 in R2C6. To not leave an endpoint stranded in the top right, that 2 needs to have its edges at either the top and right, or the bottom and left. But if the bottom and left are used, that traps the Fillomino region. So it must be the top and right, and therefore that Fillomino region extends leftwards.

enter image description here

And then:

The cell in R2C4 must be a 1, since it cannot be part of any of the 4 groups around it. This gives a couple more loop segments - specifically, the endpoint above the 1s from the start must go leftwards. This tells us that the unknown clue in R2C3 is at least a 2, and so it extends left in Fillomino. And this tells us that the clue is exactly 2, and so we can resolve the top left area.

enter image description here

Now that the top left is resolved,

which way can the 3 in R6C1 extend? It can't go upwards, since the left endpoint would be trapped. It can't go downwards, since the clue below would be a 3 as well, and that would force a tiny loop in the bottom left. So it must go right. And then R4C1 must be a 3 as well.

Also, the segment near the 1 in R3C4 can be extended as well (with two more regions partially decided).

enter image description here

Now we move to the top right:

The clue in R3C6 cannot be a 2, because it touches a 2. It cannot be a 3, because that would make a U shape, blocking off the Fillomino region. So it must be a 1, giving more of the loop.

Then, the B region cannot extend to the right, because it would complete a small loop in the center. So it must extend downwards.

enter image description here

Finally, we've reached the bottom!

What can the clue in R7C3 be? It can't be a 3, since it would trap an endpoint in the bottom left. It can't be a 1, because R7C2 would be unfillable. So it must be a 2. By similar logic, so must R7C4, and that forces B to be 4.

enter image description here

And finally:

The 2 in R6C5 cannot go upwards, since it would force region A to also be 2. So it must go right. This pushes the 3 in R7C6 right and up, and lets us complete the loop in the bottom right. Finally, the rest of the loop can be finished, and the other regions can be resolved.

The solution:

enter image description here

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  • 2
    $\begingroup$ When you take the time to put such a nice answer together, it's worth posting even if there is one already. $\endgroup$ – Arnaud Mortier Aug 22 at 20:41
  • $\begingroup$ You can make out the shape of an alien-like creature, especially due to the black square/eyeball just above centre-right. $\endgroup$ – TheSimpliFire Aug 23 at 16:59

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