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This is a Pentominous puzzle.

Rules of Pentominous:

  • Dissect the grid into regions of 5 cells each.
  • No two regions of the same shape can be adjacent (but they may touch at a corner). Rotations and reflections count as the same shape.
  • Each given letter must correctly label the shape it is in. (An image of all the possible pentominoes and their corresponding letters has been given with the puzzle for your convenience. A region may contain any number of letters, including none at all.)

enter image description here

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The final grid:

enter image description here

An explanation of the path I took: (Thanks to Beastly Gerbil and Deusovi for helping with eliminating the brute forcing!)

We start by looking at the leftmost I in NIFTILY, in combination with the V-clue which is given below it. This V should extend at least 2 squares to the right, and this would be impossible if the I extended 4 squares downward. So it can extend at most 3 squares downward. However, if the I would extend 3 squares downward, we get the following situation:
enter image description here
Now the leftmost N cannot be drawn anymore without leaving some infillable cells. So we see that the I can occupy at most 2 squares below the clue, and has to occupy at least 2 cells above the clue:
enter image description here
Next assume that the right I also extends at least 2 cells up. Now the F and the T do not fit at the top in between these I's anymore, and there is only one way left to draw them:
enter image description here
The I's are forced up now, and they enclose an area which is not fillable anymore, so we get a contradiction. From this contradiction we learn that the right I has to extend at least 3 cells downward:
enter image description here
If we look at the F and the T now, we notice that they cannot both fit above the row of clues, and however we try to fit them in, there is always one of them which occupies the square directly below the left I. This implies that this I is forced completely upwards:
enter image description here
Next the N on the left can only be drawn in one way without enclosing an area with a number of cells not divisible by 5. After this, the V pentomino is also uniquely determined:
enter image description here

As a next step, we observe

that the F pentomino in the middle has only two possible postions left where it can go, one below NIFITLY, and one above. First assume that it takes the lower position:
enter image description here
Now the T and the U cannot be drawn anymore without locking in some empty cells. So apparently the Y needs to go in the upper position:
enter image description here
This immedately allows us to make some other deductions: enter image description here
The position of the bottom Y is also fixed now:
enter image description here

Next,

we look at the region in the bottom left, here marked red:
enter image description here
Note that this area has exactly 10 squares, so any pentomino crossing its border wouldd make the remaining part of the region unfillable. So it has to be filled with exactly two pentominos, and it turns out that the only way of doing this without having equal pentominos touch is the following:
enter image description here

Let's look at the top right corner now.

First, look at the L pentomino which is clued by NIFTILY. This L cannot be completely below this word: the only way of doing this would be by putting it horizontally below the Y and the N, but then the enclosed area in the top right contains a number of cells not divisible by 5. So apparently the cell below the letter L is occupied by this pentomino. The L pentomino also cannot be placed completely above the word: then the straight part of the L would need to go directly on top of the Y and the N, which does not leave enough room for the Y. Using these two deductions, the long part of the L can only go in one way:
enter image description here
Next we look at the Y. The long part of the Y has to be vertical in order to fit, we however do not know yet whether this part passes through the Y-clue, or whether it goes to the right of this clue. Assume for now that this long part goes to the right of the clue, there are only two ways to do this, and in both these cases there is only one way left to fit in the N. The Y and N together always occupy all the cells in the following region:
enter image description here
However, now we see that the cell immediately below the Y-clue cannot be covered anymore. We conclude that the long vertical part of the Y needs to pass through the Y-clue; in particular it has to occupy the cell immediately above this clue, as follows:
enter image description here Now there is only one way left to draw the L, and then the positions of the Y and N can be determined (for the N, count the number of isolated cells again):
enter image description here
The space below the Y and N can only be filled by an X or an F, however using an F will get you into trouble with the F which should still be drawn below. So we need to use an X here. The rest of the puzzle is fairly straightforward to fill in:
enter image description here

And then we are finally done!

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  • 3
    $\begingroup$ I got about 70% done before you posted, you can skip a lot of your first explanation by noticing that the 'V' on the left must go at least three to the right, meaning the first 'I' has to go one up. From there you can deduct the only possibilities for F and T $\endgroup$ – Beastly Gerbil Sep 26 at 21:01
  • $\begingroup$ Well done on solving though, brute force or not! This was very tricky $\endgroup$ – Beastly Gerbil Sep 26 at 21:01
  • $\begingroup$ Nice job solving it! There are definitely cleaner paths through a lot of the sections you casebashed, though - see if you can find them! (BG pointed out the intended one that skips most of your initial casebashing. For the second, consider how the L pentomino can go - instead of considering whole positions at once, it helps to consider individual squares as being occupied or not.) $\endgroup$ – Deusovi Sep 26 at 21:55
  • $\begingroup$ I indeed expected there to be some nicer deductions to be made, instead of the brute forcing. I will try to find some smarter trick for the final part, and implement the trick of @BeastlyGerbil tomorrow! $\endgroup$ – Reinier Sep 26 at 22:06
  • $\begingroup$ @BeastlyGerbil I am probably missing something obvious, but it seems that even after the deduction that the first I has to go one cell up, there are still quite some possibilities for the F and the T, for example like this or this. $\endgroup$ – Reinier Sep 27 at 11:29

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