This is a Fillomino puzzle. Divide the grid into orthogonally connected regions so that no two regions of the same size share an edge, and every number inside a region indicates the size of its region. A region can contain any number of numbers, including none at all.
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1 Answer
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4
I think I've got it but don't know how to use Penpa's solution checking function.
Solution path:
Notice the central 14 and 18 must use all the squares in their enclosed set and the single width path between them forces a unique assignment:
The 15s must connect and we have a pair of squares which need to be removed to get the right number
Some pleasant deductions to fill out the top. Notice the blank squares nearby require some thinking to sort out:
Now we get to the tricky part. We notice that the 12s at the bottom must be separate islands. Some trial and error was used to find an appropriate arrangement.
The title describes the difficulty solvers will have dealing with all (but one) indicated number being more than ten and perhaps
the elegant "fingers" which run out of the center around the upper left and also the step determining which narrow finger escapes the bottom right. It was fun to look at the possibility of 13 escaping over the top before finding the 12 and 16 before another run of fingers.
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$\begingroup$ If you draw all the edges it will tell you when it is correct. $\endgroup$Florian F– Florian F2026-03-28 14:07:13 +00:00Commented Mar 28 at 14:07
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$\begingroup$ I'll try to do that although edge drawing is awkward on the phone $\endgroup$n1000– n10002026-03-28 14:14:49 +00:00Commented Mar 28 at 14:14
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1$\begingroup$ The solution is correct. But the interesting is not the solution, it is how you solved it. $\endgroup$Florian F– Florian F2026-03-28 14:19:45 +00:00Commented Mar 28 at 14:19
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2$\begingroup$ @FlorianF I added some details. Let me know if anything calls for expansion $\endgroup$n1000– n10002026-03-28 15:17:46 +00:00Commented Mar 28 at 15:17