This is a Sunday Fillomino puzzle.

Puzzle 171

The largest polyominoes in this puzzle are as large as the other givens allow them to be. You can use that to avoid some tedious counting.

I like how this one came out. I think I’m going to stop saying that I’m still learning how to make Fillomino.

*Thanks to glmathgrant for testsolving the puzzle.*

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Tags: game, logic, nikoli, problem, puzzle

This entry was posted on January 24, 2010 at 5:00 am and is filed under Fillomino, [4] Sunday. You can follow any responses to this entry through the RSS 2.0 feed.
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January 24, 2010 at 3:35 pm |

I like it as well ๐ A very nice solve too, I finished with almost no serious struggles. I did do the tedious counting, though, just to be sure.

January 24, 2010 at 4:59 pm |

You certainly are more forgiving than I am when it comes to counting in Fillomino. So far, all of your 50-ominoes and 128-ominoes have been as large as the surrounding givens allow. I believe that only three of the puzzles on my blog have givens greater than 12:

http://mathgrant.blogspot.com/2008/09/puzzle-69-polyominous-8.html (tons of 69’s)

http://mathgrant.blogspot.com/2008/10/puzzle-93-polyominous-12.html (14’s and a 21)

http://mathgrant.blogspot.com/2009/05/evil-zinger-5-polyominous.html (some 76’s)

Both of the first two puzzles have at least one instance of a region larger than 12 in size not being as large as the other givens allow, requiring the solver to actually count cells. I’m evil. ๐

January 24, 2010 at 10:38 pm |

I confess I don’t get any help from “as large as the other givens allow them to be”. Every region eventually fills the empty space around it, running into its neighbors. Isn’t this just equivalent to saying “every cell must be used”?

January 24, 2010 at 10:43 pm |

For one, you don’t have to suspect the presence of hidden polyominoes in the various branches. But there is one part where it saves a lot of time (and I’m guessing you did this without realizing it): just to the right of the bottom center with the line of three 4s, you eventually have two options of where to put the fourth 4. Both are valid, but one allows the 128 to reach one space more and so that is the one you have to write down. I don’t believe you can do that part without counting or using the hint I gave.

January 25, 2010 at 8:47 am |

Yeah, I realised that but still decided I don’t want to take your word for it just yet ๐ No big deal, though, we’re all big boys, we can count to a couple hundred at least ๐

BTW, I was a bit disappointed to find 3s in there but at least there’s exactly 3 (sets) of them…

January 25, 2010 at 8:29 pm

Eh, I thought it was neat that other numbers showed up in the solution. ๐

January 25, 2010 at 12:54 am |

Yep, those 3 4’s were one of the things I resolved by counting the 128. I also did an earlier count, to prove that there wasn’t room for 2 separate 128’s, at which point the available space (the count of all squares that might plausibly contain a 128) was slightly under 256. Knowing that all the 128’s must grow together was helpful. The final count, in addition to resolving those 3 4’s, was necessary to figure out the area along top edge, which had a few 8’s and 4’s and enough empty space that I couldn’t figure out which 8’s needed to connect to which other 8’s until I determined that I needed one more 128 in the area, which separated the 8’s and made the rest easy. So what your shortcut means, I guess, is that the Big Region is not adjacent to any unclued regions.

January 25, 2010 at 5:42 am |

This deduction is actually just a different way of saying we should trust you there is only one solution.

In puzzles like this one, where there are many locations of various sizes which can be filled the large poyomino, the large polyomino can either expand to fill everything, expand to no place at all. For example, were the row of 3 4s to expand up from the middle, there would be another solution of expanding it up from the left, and removing a square from one of the other edges.

In this puzzle itself, you can also “prove” it expands, and is not minimal, because the 7 free squares in the right middle can be 7 or 1+6, the 5 free squares on the top left can be 5 or 1+4, and the 5 free squares on the bottom right can be 5 or 4+1 or 3+2 (two ways) or 2+1+2.

Of course this assumption of uniqueness is more “pure” than regular deductions based on assuming uniqueness, if you leave them to the end of the solution. In such a case the deduction is available, but the solver is lazy and knows what the deduction is. In regular cases, the solver knows what the solution is, but does not see the logical deduction process.

January 25, 2010 at 10:19 am |

It’s not true that expanding the 3 4’s up from the middle would necessarily imply an alternate solution. Imagine that the 128 had its last 2 squares removed from that branch, leaving 2 empty squares. Now you have the row of 3 given 4’s, with a row of 3 empty squares above it. You can’t fit a pair of 2’s in that row of 3 empty squares, because 2 of the empties are adjacent to 2’s already. So the row of empties has to be a 4 and two 1’s, and the 4 has to go in the middle to separate the 1’s from each other. Still a unique solution, as far as you can tell from anything local. Only the 128 count makes it wrong.

January 25, 2010 at 8:37 pm

But then you can add the extra two squares there, and remove them somewhere else. I didn’t actually check, but I suspect that within this puzzle, any number between the take all and take minimally would allow multiple solutions. If you are referring to an example of taking the minimum possible, I mentioned the other multiple solutions that allows. I never said that there will always be multiple solutions.

January 26, 2010 at 6:28 am |

Um, I’m seeing an ambiguity in the 7-10th rows and the 5-10th columns.

January 26, 2010 at 6:42 am |

Do you mean multiple solutions? If so: besides the testsolver listed above and I, someone else sent me their full solution path and all three of us got the same thing. You can email me what you have if you can’t see anything wrong.

July 6, 2010 at 10:38 pm |

I just discovered your site and I love challenging puzzles so started with this Sunday Filomino. I had a slight scare when I got done and didn’t have room for all the squares in a 128 region. Fortunately I discovered where I had extended a 4 a little too far (with 5 squares). That blocked off a region where 128 could flow and take up 3 more squares. Phew!!!

I’ll confirm that there is a solution and there is no ambiguity.

December 2, 2010 at 9:57 am |

The best Fillomino I know.

On december, 1st it was published at janko.at. There is a great java-applet to solve the puzzles online.

Greetings from Germany

Jรผrgen

December 2, 2010 at 10:47 am |

Yeah, the java applet is pretty nice, even for a puzzle with a big number like this one, although not perfect. Glad you enjoyed this one; I was certainly proud of it, as you can read here.

Actually my favorite among my own Fillomino puzzles (not sure what my favorite of everyone’s would be) is a more recent one, puzzle 259. I was actually a little surprised by the lack of response to that one, as I thought it was one of my best puzzles to date. But it’s not the first time that’s happened.

October 3, 2011 at 1:39 am |

Trying to go back and solve some of the ones I hadn’t yet. I *almost* admitted defeat because I kept counting 130 with no way to know where the two extraneous bits were, but I finally noticed I had only used 3 numbers for a 4-piece. A little mistake-erasing, and I was done! Phew!