How to Solve Fillomino Puzzles
Fillomino, invented by Japanese puzzlemaker Suranta, was first published in 1994 in the puzzle magazine Nikoli. Your goal is to divide the grid into polyomino regions by filling every cell with a number.
- Orthogonally contiguous cells with the same number form a region.
- Two regions of the same size may not be orthogonally adjacent, although diagonally is allowed.
- There may be regions that do not have a clue number given at the start.
Starting with small numbers is often useful. You can simply draw a region boundary around any given 1 immediately. Any adjacent pair of 2s would form a domino and could also be marked.
At the top, we see another useful thing to look for. The 3 has no orthogonally adjacent 3s, and only one open cell beside it, so it must use it as part of its triomino. We can fill it in with a 3, which will connect to the other 3 below it. That gives us a complete set, so we can mark the region.
One important thing to note about the process of solving a Fillomino puzzle is not to make assumptions about which numbers go together.
Here is one example, where you might think that because of their proximity, you only need to add one more 4 and mark an I-tetromino region. However, here is just one other possible option that would also be legal at this stage of the puzzle. Remember that the only clue numbers that are definitely part of the same region are those that are orthogonally adjacent.
Take your time and deduce what must be true, rather than using intuition of guess what could be true.
With large clue numbers, it is easier to determine if some clusters are part of the same region. Look for an area where there aren’t many empty cells around a large clue. For example, this 9 only has four empty spaces around it. I’m using red dots in this tutorial for hypothetical spaces, not solved ones.
Not only could it not form a region of 9 by itself, but in the process, it wouldn’t be able to avoid connecting to the pair of 9s below it, so now we know they are definitely part of the same polyomino.
However, because we can’t be certain of the exact cells that will be used in the final region, we will only mark the highlighted cell with a 9, as it is the bare minimum needed to connect the group.
Here, we have the same situation again. The 9 can’t possibly fill the cells around it by itself, and if it did, it wouldn’t be able to avoid connecting to the group above it.
So they must be part of the same region, and the yellow cell is the bare minimum that must connect them. By filling it, we’ve definitely discovered 6 out of the 9 cells we will need for the final region.
Since that’s all we can deduce immediately about the 9-region, let’s look at another large number.
Both of these 7s must be part of the same group, because for each of them, no matter which cell you fill out next to it, the other 7 will be connected.
So let’s look at where they would be trapped if they spread in that direction (the red dots). That only gives them a cluster of 4 cells, and the 3 highlighted cells would be the only escape path. This means that at the minimum, the yellow cells must contain 7s, which gives us 5 cells solved for this region.
This technique works for smaller numbers, too. Here, you can see that the 4 cannot fill adjacent cells without connecting to the other 4 above it, so they must be part of the same group, and the yellow cell definitely has to be a 4.
Most of what’s required to solve a Fillomino puzzle is looking for where a group of numbers is forced to spread, and slowly piecing together the shape of their polyomino.
This 3 can only grow in two directions. If it goes down, then it must also go left (red dots). If it only goes left, then there would have to be another 3 above that (blue dots).
Since both possibilities use the yellow cell, we know that it must contain a 3.
If this 5 grows to the right, it would still need 2 more cells to complete a pentomino, so it also must spread left, which will connect it to the other 5.
Note also that by filling the yellow cell with a 5, we now also know the correct placement of the triomino below it.
Another useful technique is when you can prove where a certain number can’t be. This can work especially well with small numbers like 2 and 3, which have a limited number of possible shapes they can form.
In this example, the 2 has only two possible positions – either the second one is to the left, or it’s below. If it was placed to the left, the 3 would have to be completed with the positions of the blue dots.
This traps the 4 without enough cells to form a complete region. Now we know the domino has to be vertical.
For the same reason, we also know the 3 can’t form an L-shape using the position of the red dot. That leaves either an L-shape using both blue dots, or an I-shape using the red dot and the blue dot above the 3. In either case, we know the blue dot above the 3 must also be a 3.
Now that we know the direction of the domino, here’s another example. The triomino can only be in two possible configurations.
If we use its L-shape with the red dot, it will force the tetromino for the 4 into an L-shape as well, with the blue dots.
Doing this traps the 9s with only 8 possible cells, proving that the 3 must be in an I-shape, instead.
Then the 4 cannot create a full tetromino without connecting to the pair of 4s above it, so now we know that is also an I-shaped tetromino.
Boxing in that group of 9s can be a useful strategy here. This 3 at the bottom can be part of only 4 possible triominoes: an L to the left (blue dots), an L to the right (red dots), or two different I-shapes on the bottom (green or orange dots).
We can rule out the blue dots because if we used both of those cells, there would only be 8 spaces available for the required group of 9s.
The other three options all have a cell in the space highlighted to the right of the known 3. Therefore, we know there must be a 3 in that cell, at least.
Once more, this time with this 2. If the second 2 were to be placed where the red dot is, we’re limiting the 9s to only 8 cells.
This means that the domino must expand to the left.
Speaking of dominoes, completing this region will force several moves.
Rather than simply letting this domino be forced by another region, I wanted to take the opportunity to demonstrate another situation to look for.
This domino can only be in two orientations, but notice that the red dot would be adjacent to another region that contains 2s. Because orthogonal adjacency is not allowed, the yellow cell must be the correct one to place the other 2.
By making that placement, we also create the pentomino region for the group of 5s.
Because of its position, we can see that the 3 in the bottom right corner has to connect with the 3 that is diagonally adjacent to it.
However, the red dot can’t possibly be correct, because it would create an adjacency with the I-triomino from earlier. So the yellow cell must contain the last 3.
This demonstrates that sometimes, you will discover a region that didn’t have a clue number given in the initial puzzle. The red dot is the position os a region of 1 cell, so we will fill it with a 1.
Finally, we can see the correct cells for that group of 9s we were picking away at when we began the puzzle.
Completing the region of 6s also forces a vertical position for a domino, on the right.
This is an interesting trick you can use to see if there is a hidden region.
Here, we have an area of exactly 14 cells. The visible clue numbers are 2, 3, 4, and 5.
If we add them together: 2+3+4+5=14, which means the correct placement of these polyominoes will completely fill the region. It also proves that those 3s are definitely part of the same region, so we can already place that triomino.
Because we know there are no hidden regions here, we must expand the pair of 5s to the right, because all cells in the grid must be used. Then we expand upward as shown, and we’ve completed the pentomino.
With the last 2 cells remaining in this part of the grid, the correct placements of the domino and tetromino are obvious.
This pair of 4s must form either a T-shape (red dots), or an S-shape (blue dots) tetromino.
If we try to place the S-shape, then the 5s will have to expand as shown by the green dots. This traps the group of 7s into an area with only 6 cells, so this placement cannot be correct, and the tetromino must be the T-shape.
We can easily see there are only 12 cells remaining, and 5+7=12, so we know there are no hidden regions left. So we know the cell with a red dot has to contain a 5, because otherwise, that cell would be isolated.
Then, the only possible final placement possible for a 5 to complete the region is the highlighted cell shown.
After, that, we can place the last two 7s to complete the last region.
The completed puzzle.
