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This is a Statue Park puzzle.

Rules of Statue Park: (adapted 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.

enter image description here

The piece bank is a double set of the 5 tetrominoes, which have been given the standard lettering for your convenience – the letters themselves have no effect on the puzzle.

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    $\begingroup$ The tetrominoes labelled L and S are not double; they are mirror image versions of each other. $\endgroup$ – Acccumulation Oct 16 at 17:05
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    $\begingroup$ @Acccumulation Since the rules stipulate that "Pieces may be rotated or reflected", to all intents and purposes similar-lettered tetrominoes are entirely interchangeable. Their depiction is purely for symmetry and aesthetics :) $\endgroup$ – Stiv Oct 16 at 17:11
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My answer:

enter image description here

First steps:

first step
Some initial non-shaded deductions to force connectivity and to prevent any groups of more than 4 tiles from being shaded
next step
That forces some shaded squares to extend out in order to have enough space for tetrominoes; the upper left one is forced to be a T so as to not trap an unshaded square in the corner

A tricky deduction:

next step
The orange squares must be part of an L tetromino (they can't make anything else legally). If the blue square is unshaded, that forces the green squares to be shaded to avoid making an illegal 3rd L. However, now R2C7 can only make Ls and Ts, which are all used up elsewhere. So the blue square must be shaded.

Working on the right side:

next step
We can now set several squares around the newly-made 3-block to unshaded to avoid another L
next step
The newly-made 3-block must be the final T, so the other 3-block must be an S

Another tricky deduction

next step
If R3C8 is shaded, that forces all the green squares to be shaded and the blue square to be unshaded. Now the orange square can only form Ls, Ts, and Ss, all of which are used up (the 3-block to its immediate right must be an S). Therefore R3C8 must be unshaded

Working on the middle:

next step
Basic extensions from unshading R3C8. Note that the orange 2-block must be an S (no Ls or Ts are left)
next step
The 2-block that was part of the i cannot be an S (none are left), so it must connect with R2C9 above it to form an I

Using up the remaining shapes:

next step
If the orange S has its second part to the left, it will isolate some unshaded cells. Therefore its second part is to its right.
next step
R5C11 can't be a T, L, or S so it must be an O
next step
The right side must have an O and an I, and there is only one way to fit them
next step
The final L must be turned to the left so as to not isolate the unshaded squares in the upper right, and then we can set all remaining squares to unshaded

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  • $\begingroup$ Well, you got the start and end logic but not the middle (you know that already), which houses the more interesting logical steps... See if you can find that! :) $\endgroup$ – Stiv Oct 14 at 23:19
  • $\begingroup$ Hey bobble, at your invalid deduction step, rot13(lbh pna pbapyhqr gung gur grgebzvab pbagnvavat gur hccre yrsg oenapu bs gur I arrqf gb or na Y). Still looking to see if that matters. $\endgroup$ – Jeremy Dover Oct 14 at 23:46
  • $\begingroup$ That's now pretty much bang on the intended logic :) Note one typo in the 'A tricky deduction' spoiler - you meant 'illegal 3rd L', not T... (Interestingly, my logic in that step involved focussing on the cell 2 to the left instead, which results in an identical conclusion - and in 'Another tricky deduction' my attention was on the gap in the 'i' clues, 1 to the right... again with an identical conclusion) Fix that typo and the checkmark is yours :) Well done! $\endgroup$ – Stiv Oct 15 at 8:29
  • $\begingroup$ Checkmark awarded for being first to answer correctly and following my intended logical solution path too :) I highly suggest any readers also check out Deusovi's answer, as his answer reaches the same solution with a very interesting deduction that I had not considered...! $\endgroup$ – Stiv Oct 15 at 21:11
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A third answer might seem a bit redundant, but the path I used was significantly different from the other two. There's a step in mine that eliminates a lot of the "bashy" hypothetical logic (and is, at least in my opinion, more interesting).

So, some basic deductions get you here:

enter image description here

And now you're presented with a question:

enter image description here

We have at least nine pieces accounted for already. (The 8 group may be broken up, if we happen to have all ten. Also, the top left cell of the 6 group could technically go with the 7.)

So, where do the I pieces go?

Group 5 must be an L. So 6 can't have the other I piece, or it would make an L with the 7. And none of the other groups can form an I... except for group 7. So the two I pieces are group 7, and our mysterious missing piece.

(And this means that group 8 is indeed one group as well!)

enter image description here

And the rest of the puzzle can be finished off with similar logic:

We've already used up both Ls (though one hasn't been fully decided yet), so the group in the upper middle must be a T.
enter image description here
That uses up both Ts, so the left-side group must be an S, and then the one next to it is forced to be S as well. And then the two right-hand groups must be the two Os...
enter image description here

The S must bend right in order to not block off an area; the mysteriously missing I piece now has a single place to go, wedged in the lower right corner; and then the top-right L is finally resolved!
enter image description here

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  • $\begingroup$ Just to reiterate on the post itself after our discussion in chat, I really like this approach. It's a perfectly valid alternative logical process and displays a really useful technique for solvers to learn - namely, "where can piece X go?" rather than merely "what colour must this cell be?" and "which piece can go here?" With this logic you reach the solution very smoothly - a fantastic alternate answer! $\endgroup$ – Stiv Oct 15 at 9:55
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Similar to but maybe a bit simpler than @bobble's deduction:

enter image description here
First thing to observe that we can use up the L's and then the T's pretty much immediately. This only leaves an S at the indicated position.

![enter image description here
if the second S were at the indicated position the there would be no solution at the arrow.

enter image description here
This only leaves an I and also forces us to us spend the second S.

enter image description here
With an I and two O's left the rest is easy.

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    $\begingroup$ +1 for the solve and a nice solution. I'm not convinced that all of the logic behind your first spoiler block is quite as straightforward as you claim - particularly the identification of the second T, which requires some conditional logic. Nice use of colour though, and well done for cracking it :) $\endgroup$ – Stiv Oct 15 at 8:33
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    $\begingroup$ @Stiv Hm, unless I'm missing something it is straightforward: focus on f5 (the black spot surrounded by 3 whites. SInce at this point no L's are available it has to extend to the right, so it has to connect with g6 which leaves only T, no? $\endgroup$ – Paul Panzer Oct 15 at 9:19
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    $\begingroup$ That's precisely what I'm saying though - you're using the conditional logic of not having any L's left, which isn't necessarily a trivial observation in itself. Your solution is correct, absolutely - just I felt that looking at your first diagram it might not be immediately obvious to everybody why exactly that should be the case. Not a criticism, just to point out that not all readers would necessarily make the same deductions at the same pace :) $\endgroup$ – Stiv Oct 15 at 9:28

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