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Pentomino Nurikabe is still elusive, but here is another Tetromino Nurikabe! I'm not sure if the 4s are actually socially distanced enough. (The ones on the right and bottom are doing better than the others). Nevertheless, they are further apart than before.

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. The new name for this puzzle genre can be blamed on Bubbler

CSV:

,,,,,,
,,4,,,,4
4,,,,,,
,,,4,,,
,,,,,,
,,,,,,
,,,,4,,
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  • $\begingroup$ Really neat puzzle, with a nice tough solve :-) At least for me... $\endgroup$ – Jeremy Dover Nov 10 '20 at 2:07
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The completed grid:

enter image description here

Reasoning:

By distance, no unshaded region can reach either the lower left corner, or the square diagonally northeast of it, so the bottom left 2x2 must contain an unshaded square from the I tetromino along either the left or bottom. But note that the 4 on the bottom row is the only one that can reach the bottom right 2x2, so the I has to be along the left side. Next, note that the 2x2 in R67,C23 also needs an unshaded square. The 4 on the bottom row can reach it, but it also needs to cover the bottom right 2x2, which would require it to use an I tetromino. But that's already been used, so it is the 4 in the middle that has to cover that 2x2. We can't be sure of the positioning, but it has to be either the L or S tetromino. The grid thus far:

enter image description here

Next:

Continuing with this tetromino, note that the 2x2 R45,C23 has to be covered by this same tetromino. This forces R5C3 to be unshaded, since shading it would force us to have at least 5 unshaded squares.

On the right, the overlapping 2x2s R34,C56 and R34,C67 both need to be covered by tetrominos. Only the 4 on the right side can cover either. One possibility is to use an L from the 4 down to R4C6, but that would require R12,C56 to be covered from the upper left 4, forcing it to use an I tetromino to cover this 2x2 and the upper left corner. Similarly, using an L to cover R3C5 leaves R34,C67 uncovered, so R3C6 must be in this tetromino. We still need to cover R12,C56, which forces R2C6 to be in this tetromino as well. The grid thus far:

enter image description here

The bottom right:

The 2x2 R56,C67 needs to be covered, and this can only be done by the bottom 4. If we avoid using R6C6, then this has to be done with an L around it to either R5C6 or R6C7. The former case forces the middle tetromino to also be an L, while the latter case forces the 2x2 R45,C67 to be covered by the right tetromino, which then must also be an L. So R6C6 must be unshaded.

Closing in:

The 2x2 R45,C67 is the key. Whichever tetromino covers it must be either an L or an S, forcing the middle tetromino to be the other. So the upper left 4 must be an O or a T. If it were an O, the four would have to be the bottom right corner to cover the upper left 2x2, but then the shaded R2C1 would be isolated. So this must be the T. This means R2C5 must be shaded (otherwise the upper right tetromino would be a T), and then R3C5 must also be shaded, since unshading would force the upper right tetromino to be an S but not cover R45,C67.Similar logic shows that R1C7 must be shaded. With some simple deductions, we have:

enter image description here

Finally:

The upper left tetromino must cover R2C2, since a T cannot reach anywhere else in the upper left corner. This forces R2C1 to escape around the top of the T. This yields:

enter image description here

And finally the shading at bottom right must escape over the bottom 4, focing its tetromino to be the L. The rest is simple deduction.

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