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Here is another Greener Grasses puzzle.

Also, happy happy birthday to my fellow Indonesian puzzler, Nusi! I'm sorry that I realize the puzzle was broken in the last-minute checking so a single letter on N was then removed.. XD


Rules of Greener Grasses:

  1. Divide the white cells into regions of the indicated size (top-right of the grid.)
  2. Whenever two regions share a side, each region has something the other does not: a symbol that the other region has less of. (For example, a region containing AA and AB may share a side: the AA region has more A's than the other, and the AB region has more B's than the other. But AB and B may not share a side; the B region doesn't have anything the AB region doesn't have.)

enter image description here

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I decided to post my answer on PSE because I think it's already been unanswered for a long time here :)

Please keep in mind that I'm not really good at explaining (same reason as why I'm not trying to be active here), but I'll try my best. Also, please tell me if we can resize the spoiler box.

Take a look into the R on row/column 12/2. Let it not connected with the R on 12/3. As a consequence, the cell on 13/1 should be connected with R on 12/3. However, it'll break the rules. So, the RR should be connected. They should be connected through 12/3-12/2-12/1-13/1, else the H on 9/2 will have nowhere to go. Similar logic applied with B on 3/2 as well.
enter image description here

Take a look into A on 6/2. It'll have no another alphabet connected with it, or else some of column 1 cell will left empty. Also, you had no choice but make 5/1, 5/2, 6/1, and 6/2 as tetromino. Do some trivial stuff after this, and you'll have something like this.
enter image description here

Take a look into 10/3. A should have no connected character, else 9/3 will empty or there will be a contradict on neighboring on A-AX or D-DX. After that, it'll easily conclude on how 11/4 works by contradicting the rules on 12/7 for some pattern. So, it'll be like this.
enter image description here

Now everything is trickier than ever. If you take a look into 6/3, you know it should not be connected with H on 5/5. Our final choice is it's either connected with B on 4/4 or D on 8/4. However, if you connect it with the B, you'll have single empty cell on some upper cell which we should avoid. So, it must be D. Then, allowing similar logic with the AA on 6/2 and 8/2, we can do similar thing with BB on 6/2 and 2/8, which lead us into this.
enter image description here

Take a look into B on 2/8. it have no choice but make L shaped, else some upper cell will be empty. The I on 2/13 also should be on square shaped, else it'll neighboring with I on 4/12. But we don't know yet if it should fill the lefty or righty one. So, I'll leave it for now. Now let's go back to down side. Take a look at T on 12/12. If you decided to connect it up, it'll give some empty cell on the last column, and we don't want that. Applying this, we'll have some great break.
enter image description here

We want to fill the 10/12, but we have two possibility. It's either connected with H on 9/10 or A on 8/11. Let it connected with H, then you'll have some trouble with filling the centre part because the AB will be a neighbor. So, H should be stretch to left and let A fill the 10/12 and make it like this.
enter image description here

Applying green grasses rules on 8/9 (really hard to spot this one, but satisfying after I found it), we will have
enter image description here

Final answer:

Then just some nice filling with the squared I I mentioned above, it should be 'righty', or some cell on mid-left can't be divided as tetromino.
enter image description here

Thanks for the puzzle! :)

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    $\begingroup$ You can make images smaller by putting an "m" at the end of the alphanumeric string in the image link, as I've done for you. If you meant something else feel free to revert. $\endgroup$ – bobble Jan 13 at 18:50
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    $\begingroup$ Aww, happy birthday once again Nusi! XD This is a nice explanation tho, btw the tricky part on the fourth spoiler might be easier to tackle if you consider the 1/10 first :) Also, welcome to PSE! ^^ $\endgroup$ – athin Jan 15 at 0:50
  • $\begingroup$ Thanks bobble, that's exactly what I meant. I'll keep that in mind for my future answer. Also, thanks for the warm welcome, athin :) $\endgroup$ – Nusi Jan 15 at 10:34

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