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I've been doing a bunch of shape-placing puzzles lately, and I found some spare parts leftover. I figured I'd make a Statue Park out of them! The grid:

Grid

Solver Note

I know there are a lot of squares given...part of that is to keep the mirror symmetry, but part is to get a unique solution without an unconscionable amount of case-bashing (a consequence of including the L-triomino). There are still some fun deductions in there, but all in all, not too difficult.

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The solution is:

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Step 1:

There is only 1 place where the 'X' can fit. We can mark as unshaded (beige) a couple of other spaces that must be blank for connectivity (R1C4 and R9C9). Note also that the space R3C2 (in between the 2 shaded spaces) must be blank. If it was shaded, there would be no legal piece in the shape bank to which the shaded space in the top-left corner can belong.

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Step 2:

Only 1 piece fits the shaded space in the bottom right – the ‘v’.

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Step 3:

Now only the ‘W’ can go on the right-hand side.

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Step 4:

'L' cannot fit in the bottom section without cutting the board into two disconnected halves or rendering the shaded space in R7C4 'pieceless', so must go top-left (only other place where it fits), though we don’t know yet which way it is oriented.

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Step 5:

Now only one place for ‘I’.

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Step 6:

Then only one place for the ‘i’ and ‘Y’ pieces.

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Step 7:

Regardless of which of ‘s’ and ‘o’ goes on the left, we need the unshaded spaces to loop up the left side and around the top to connect to those in the top-right section of the grid; this helps us position the 'L' at last.

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Step 8:

Now only the ‘o’ can fit on the left, and there is only way to position the ‘s’ and leave all unshaded spaces connected. Solved!

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  • 2
    $\begingroup$ Great job! I was hoping to make it bit more difficult to place the "v", but everything I tried ended up not having a unique solution. But stay tuned for my next logic grid (prob Friday unless I get overexcited)...I think you're going to like it a lot! $\endgroup$ – Jeremy Dover Oct 19 at 14:28

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