# Spiral Stumper Series: Fillomino

Spiral Stumper Series is a $$5$$-puzzles series taken from the Final Round of a local (national) contest, KPK, which has been ended recently and authored by me. The theme is spiral and each puzzle is standalone (there will be no meta, etc.)

Fillomino (taken from Nikoli)

• Fill in all empty cells with numbers under the following rules.
• Divide all of the board into blocks. Fill each block with the same number horizontally or vertically.
• Each block contains as many cells as the number in the block.
• Same sized blocks cannot touch each other, horizontally or vertically.
• What numbers can be used? 1-4? 1-5? – npkllr Sep 27 at 14:51
• @npkllr it can be any numbers, say 1-100.. – athin Sep 27 at 14:59

This is a pretty difficult puzzle! My solution here has a lot of 'usual' deductions left unmentioned.

So first of all, all the 1s can be closed off:

Now, there's one cell that must be focused on:

The top left cell cannot be a 1, so it must extend rightwards. It then cannot be a 2, so the 2 near the top must go down. The same logic can be applied again to get this far:

Now, check

the 2 in the top middle. Can it extend downwards? If it does, then R3C5 must go left to make a size-3 flat region that touches the 3 clue. That's a problem -- so the 2 must go upwards. This forces the top left region to be a 5.

Now, an important question appears:

What's going on in the center? There can't be any 1s in there. There can't be any 2s either. And we can't put a size-3 region in there: so it must be filled by a 4!

Now, we can finish the upper right corner, and start on the lower right:

There must be a single isolated 1 in the upper right, and one of the two potential places is adjacent to an already-existing 1.

Meanwhile, R5C8 must be a 4, completing a 4 region. This lets us continue down the right-hand side.

Now, I make an assumption for sake of contradiction. (This part of the puzzle seems like it should have a 'cleaner' path, but I couldn't find one.)

Assume that the 3 clue near the center does not go down. It must then go left, and then either left or down: either way, the remaining empty cells in column 3 cannot be satisfied. (If left, R6C3 cannot be a 3 or a 1; if down, R8C3 cannot be a 1, 2, or 3, but it only has 3 cells of space to expand into.)

This allows us to finish off the bottom right corner

by realizing that R8C7 must be a 1. Finally, the unresolved 2 clue cannot go right, or R7C4 would be unfillable. This leads to this situation...

and the only way to resolve this (completing the puzzle) is

by an unclued size-5 region.

• Exactly right, very well done! – athin Sep 28 at 0:34