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I enjoy combining polyominos with grid-deductions. My current plan is to create a Pentomino Nurikabe. But that sounds hard, so I made this Tetromino Nurikabe as practice first. I think it came out well, so here goes!

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

I've included all available tetrominoes as a reference.

CSV:

,,,4,,,
,,,,,,
,,4,,,,
,,,,4,,
,,,,,,
,,,4,,,
,,4,,,,
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Completed grid:

Step-by-step solution:

The number at R6C4 must extend to its right, otherwise the wall on its left is isolated (and both regions starting from R6C4 and R7C3 will be L's). If the cell R6C7 is unshaded, it is only reachable by R6C4, but it would isolate the bottom wall.

So the lower right region is an L, which means the lower left region must be a T.

I must be the one on the top, placed horizontally. Neither O nor S covering R3C3 can touch both the squares R12C12 and R45C12, so the I must start at R1C1 or R1C2. Then the three cells at the top right are unreachable.

Therefore the middle right is an S and the remaining one is an O. Completing the grid by plain Nurikabe deduction gives the following result:

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