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This puzzle is dedicated to Sciborg. Copying the dear gentleperson, some of the 4s are hiding in the corners.

Rules: (Nurikabe section shamelessly stolen from an earlier puzzle by @jafe)

  • Numbered cells are unshaded.
  • Unshaded cells are divided into regions, all of which contain exactly one number. The number indicates how many unshaded cells there are in that region.
  • SPECIAL RULE: the regions will form a tetromino set, with rotation and reflection allowed.
  • Regions of unshaded cells cannot be (orthogonally) adjacent to one another, but they may touch at a corner.
  • All shaded cells must be connected.
  • There are no groups of shaded cells that form a 2 × 2 square anywhere in the grid.

the puzzle - see transcription below

I've included all available tetrominoes as a reference.

A playable version of this puzzle can be found here. The link leads to a puzz.link editor. Note that this editor won't force you to use the tetromino rule, and it has a timer.

The first answer with a fully-explained logical solution path will get the checkmark. I welcome multiple answers, if later ones can show a better-explained or more elegant path.

CSV:

,,,,,,4
,,,,,,
,,,4,,,
,,4,,4,,
,,,,,,
,,,,,,
4,,,,,,
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  • $\begingroup$ Should each tetromino be used exactly once? Also, this seems quite easy even to brute force since the shape is restricted. But good puzzle! $\endgroup$
    – justhalf
    Commented Dec 2, 2020 at 4:10
  • $\begingroup$ Yes, each tetromino should be used exactly once, that's what is meant by a "tetromino set". And of course you could brute-force it due to the small size, but you can do that with any small puzzle. The fun is supposed to be finding the elegant solution. $\endgroup$
    – bobble
    Commented Dec 2, 2020 at 4:12

2 Answers 2

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@Bubbler and others solved this before me, but I figured I would share my solve path too, since I love that this puzzle was dedicated to me!

So first, I filled in the obvious squares to give me a starting point:

enter image description here

Then I saw that there were two 2x2 regions that needed to be filled with island, since we can't have any 2x2 oceans. Those were these regions here:

enter image description here

Then I realized those 2x2 regions could only be reached in specific ways - that is, I needed to have the bottom right piece reach downwards, and a piece reaching up to the top left corner. So I knew that I had to place the L and the S pieces in those two spots, although I wasn't sure yet which was which.

I filled in some oceans. And, since I knew the top piece had to reach upwards:

enter image description here enter image description here

From here it was clear to me that the L piece had to go in this spot, since the S-piece wouldn't fit. So now I had placed a tetromino, and I knew the S piece had to go in the other spot in the only orientation that made sense.

enter image description here enter image description here

Now I looked at my grid again. Having placed the L and S, it was clear to me that the top right corner must be the T piece. If it was the O piece, there would be a 2x2 region left unfilled, and there wasn't enough room for it to be the I piece.

So I placed the T:

enter image description here

And from there, the final grid was clear:

enter image description here

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Apparently too late to the game, but anyway here it goes. Hope this one is the intended solving path. (I think the existing two answers have at least some logical leaps.)

Step 1:

Start by marking walls between the crammed fours at the center. Looking at top left and bottom right 2x2 corners, the only cell that can be occupied by a tetromino is the inner cell (R2C2 and R6C6 respectively).

Step 2:

R2C2 must be part of a 4 starting from either R3C4 or R4C3. That piece is an L either way. R6C6 must share the area with R4C5, and it can't be L, so it must be an S.

Step 3:

In order to avoid 2x2 wall at R6-7C4-5, the only way is to place an I horizontally at the bottom. (Placing an L starting from R4C3 to cover R6C4 doesn't work because L must contain R2C2.)

Finally:

Placing L on the left side makes problems, so L should go right and cover R3C4. Then it is straightforward to see that the middle left must be an O and the upper right corner must be a T.

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  • $\begingroup$ This is indeed the intended solve path. The exact reason that placing an L on the left side causes problems is that R5C3 must be unshaded to prevent a 2x2 in R56C34. $\endgroup$
    – bobble
    Commented Dec 2, 2020 at 4:56
  • $\begingroup$ Oh yes, that works too. $\endgroup$
    – Bubbler
    Commented Dec 2, 2020 at 4:58

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