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Inspired by PSE Advent Calendar 2021 (Day 13): A Christmas Hokuro, I challenged myself to create a Christmas tree-shaped puzzle. Unfortunately I had to edit the shape a little bit because... reasons. Hope it still looks close enough.

Rule of LITSO:

Divide the grid into tetrominos so that two identical tetrominos never share an edge (but may touch diagonally).

The following is an example puzzle with its solution.

Now solve the following LITSO puzzle.

You can try this puzzle on Penpa.

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    $\begingroup$ It still looks like a tree! $\endgroup$
    – justhalf
    Dec 14, 2021 at 9:52
  • 1
    $\begingroup$ I attempted to make a LITSO puzzle several times before, but without success (it always ended up having no clean way to force deduction or simply having no solution). So I'm quite happy with how this one turned out. $\endgroup$
    – Bubbler
    Dec 14, 2021 at 12:55
  • $\begingroup$ (On the other hand, the example puzzle was rushed out and doesn't have a clean solving path. I just randomly placed dark squares and case-bashed to death, for the sole purpose of showing how the solution looks like.) $\endgroup$
    – Bubbler
    Dec 14, 2021 at 13:03

2 Answers 2

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Full solution

here

Step-by-step:

First, label the cells which have only a single connection so they must be a part of a single tetromino (note we don't yet know that the cell on the lower left of #7 is the part of #2 or not):
here
However, if it isn't (if it's the another tetromino labeled 8), we get a contradiction because both #2 and #8 must be L-type and share an edge:
enter image description here
So, it must be the part of #2. So, we get some more deduction and arrive to the following (we can label one cell of the #10 since it obviously does not belong to any of the shapes already labeled 1 to 9):
enter image description here
Now, we notice that the cell to the right from #1 cannot belong to it since we get the following contradiction (both #3 and #11 are Ls and share an edge):
enter image description here
So, it's nov obvious that #1 is I-type, and now we get the following:
enter image description here #10 now has the borders with #1 (I), #8 (S), and #9 (L), so it must be T since it's no place for O-type:
enter image description here
Note that newly labeled #12 cannot be I-type for the parity reasons (the 0-labeled cell cannot be assigned to #5 because there would be an odd number of cells remaining for the upper part): enter image description here
So, #12 is S-type, and the newly created #13 is T-type: enter image description here Now, it's only one way to fill the top, since the symmetrical variation won't work because #5 and #13 must be of different types:
enter image description here Finally, the remaining part can be easily filled (we arrived to the final solution, which is colored for the sake of clarity):
enter image description here

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Start by marking pairs of squares that must belong to the same piece: If the red pair at bottom-left does not connect to the purple pair right below it, it must extend to the right and then right or down, both of which lead to contradictions. Thus these two pairs are connected: Now if we put either red border in place, it is easy to see that they force an L to touch the existing L, so the three squares they connect are adjacent. This is easily completed into a second L, leaving a gap that can only be filled by an S; by similar logic we conclude that the bottom-centre purple pair below is connected. This last purple pair cannot be connected to the yellow pair that is the tree's trunk (L touching L), so the yellow pair is eventually forced to be an I. This also forces a T and an extension of the left-centre pair into a triplet:

Now assume the bottom purple pair doesn't extend upwards – we'll eventually force S touching S: Not extending rightwards, on the other hand, forces a T touching T: Thus: The purple triplet at left must now extend upwards, leaving a 16-square triangle on top and allowing more forcings:

There is now only one way to fill up the top without causing like to touch like: And the rest is easy to piece together:

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  • $\begingroup$ Both of you solved the puzzle nicely, with equally detailed solving path. I'm giving the checkmark to trolley813's because it is the first answer and explanation was added roughly at the time yours was posted. $\endgroup$
    – Bubbler
    Dec 14, 2021 at 10:13

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