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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.)


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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.
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  • $\begingroup$ What numbers can be used? 1-4? 1-5? $\endgroup$ – npkllr Sep 27 at 14:51
  • $\begingroup$ @npkllr it can be any numbers, say 1-100.. $\endgroup$ – athin Sep 27 at 14:59
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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:

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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:

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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.

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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!

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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.

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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.)

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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...

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

by an unclued size-5 region.

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

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