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This is a Statue Park puzzle.

Rules of Statue Park: (shamelessly stolen from an earlier puzzle by @Deusovi)

  • Shade some cells of the grid to form the given set of pieces. Pieces may be rotated or reflected.
  • Pieces cannot be orthogonally adjacent (though they can touch at a corner).
  • All unshaded cells must be (orthogonally) connected.
  • Any cells with black circles must be shaded; any cells with white circles must be unshaded.

the puzzle

The piece bank is a set of bobblies, which (long story short) are little crowns with variable number of points. There are 4 no-point bobblies, 3 one-point bobblies, 2 two-point bobblies, and 1 three-point bobblie. I've labeled each with the number of cells they take up. The numbers and backstory have no effect on the puzzle.


Transcription of puzzle for those who have trouble with images

CSV:

,,,,w,,,,,b
,b,,,b,,,w,w,
,,b,,,,,,w,
w,,,w,,,,b,,w
,,,,b,,,,,
,,,,,b,,,,
b,,w,,,,b,,,b
,b,,,,,,b,,
,w,b,,,b,,,b,w
w,,,,,w,,,w,

There are 4 dominoes, 3 T-shaped tetrominoes, 2 C-shaped hexominoes, and 1 E-shaped octomino in the piece bank.

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  • 2
    $\begingroup$ +1 for bobblies :-) $\endgroup$ – Jeremy Dover Oct 26 at 18:52
  • 2
    $\begingroup$ Bobblies! Nice puzzle :) $\endgroup$ – Prince North Læraðr Oct 26 at 18:56
  • $\begingroup$ Nice! Did it take you longer to build this than it took Deusovi to answer it? (Difficulty-benchmark.) $\endgroup$ – BmyGuest Oct 26 at 19:29
  • $\begingroup$ Yep! I spent maybe ~10 hours constructing it, another hour checking, then another hour fixing a mistake, and another hour making the question & picture. $\endgroup$ – bobble Oct 26 at 19:32
  • $\begingroup$ Well, Deusovi spent less than 45 minutes answering it :):) $\endgroup$ – Voldemort's Wrath Oct 26 at 19:38
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To start,

take a look at the lower right black cell. It must be part of a 4-piece; any other piece would rub up against a different black dot. And given the placement of two nearby black dots, there's only one way to actually fit it in.
enter image description here

And we can do the same thing again:

Now the next two cells up the diagonal chain have the same deduction! We can't tell exactly how the piece will be placed yet, but it still has to be a 4 piece, or it'll run into a problem with the next black dot.
enter image description here

A big global deduction can be made:

enter image description here
Consider the highlighted dots here. There are 7 red cells; each of them must be part of a separate shape. There are 4 more blue cells; we can only merge one pair of them (either the top two or the bottom two; if we do both the shapes will touch). So this accounts for all 10 of our shapes. In other words, each shape must cover at least one of these cells.

Now, which dots can be 6s? The only available 6 dots are A, D/E, and F: any other placement would either block off the "cubby" of the 6 shape, or brush up against a different known-shaded cell.
And what about 8s? C must have the 8: it can either pair with B, or go vertically by itself.
If it goes vertically, we have a problem: F's spot for the 6 is blocked off, and now we can't fit a piece in B while still pairing E-D. So the 8 goes horizontally, covering B and C.

And now the rest of the pieces fall into place:

enter image description here
E must be a 2. I and J must also be 2s...
enter image description here
And now we need to place one more 4-piece -- if G and H are both ⊣-shaped, then D can't be extended without blocking off the path. So D is the other 4-piece, and the puzzle is solved!
enter image description here

| improve this answer | |
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  • $\begingroup$ I had a slightly different solving path to you, but not enough for a separate answer. My very first step (after trivial shading) was to try and place the 8 and two 6's. There are only two spots anywhere on the initial grid where the 8 can go, and only 3 for the two 6's. One of the 8-positions would block two 6's, permitting only one 6 to be placed. Thus the 8 must go where you placed it. One of the 6-spots then rubs up against the 8, so use the other two for the 6's - then continue. I do agree with an earlier comment of yours on this puzzle type - many interesting deductions! :) $\endgroup$ – Stiv Oct 26 at 20:16

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