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Eight Doubled Tetrominoes Make a Tetronogram

This puzzle is part of the "Piece de Resistance" series. Go back to Part 1 (Ace) for the story.
Ace Two Three Four Five Six Seven Eight ...

Time for Another Tetronogram! (named by @Feeds aka MrPie)

  • The puzzle is made of a grid like a nonogram.
  • Notations are along the axes like a classic nonogram but numbers are replaced by the names of the tetromino.
  • The names are I, L, T, O, and S.
  • The tetrominoes can be flipped and rotated, therefore a J/Z tile would have the L/S notation.
  • A notation of ‘L’ means there is a part of the L tile on that row or column. It can be 1 tile, 2 tiles, or 3. Same theory for other tiles.
  • Most steps can be deduced by logic alone. There is only one solution, but only one make sense.
  • Not all grids have to be filled.
  • Different from other nonograms, there need not be a space between two tetrominoes.
  • Gray squares are shaded for you. There will not be any parts of a tetromino in those grids.

enter image description here

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Solution(s)

I've solved the puzzle, but there seem to be two very slightly different possibilities for the final answer, differing only in what happens with two L-tetrominoes at the top in the middle:

first option

second option

Edit: now that the question has been slightly edited to add another grey square, there's only one possibility left, which is the second of the two above. As @Stiv notes, this means the grey squares

spell the word ODE.

Reasoning

The first and easiest things to fill in are the

I-tetrominoes, because there are only two columns and four rows which contain I cells. Then in the 4th row, all cells between the two I-tetrominoes must be empty (grey). In the bottom three rows, there's a T and an L between the two I-tetrominoes. The T can't go into the 4th column, and the L must be below the T in the 3rd column, so we get:

bottom left done

Next we fill in the

T-tetromino in the top left, which is easy because it can't go into the 3rd column. For the L above it in the 2nd column, we can just make a start because we don't know whether it's three wide and two high or two wide and three high. (Edit: actually with the updated puzzle we know this for sure, but I'm going to continue with my original solution.)

Turns out that

T-tetrominoes are also easy to solve. We've got part of one in the bottom row, which can only be in the 9th column. That T-tetromino can't go anywhere to the left, so there's only one possible orientation. In the 10th column we've got part of an S below that T, so there's only one possible orientation for this S-tetromino. Then the remainder of the bottom three rows is easy to fill:

bottom finished

Considering again T-tetrominoes:

there must be part of one in the last column, which can only be in the 4th row, and then there's only one possible orientation for that T-tetromino. The next thing to the left of that must be an S-tetromino, which can't now be in the 10th or 11th columns, so we shade a few cells grey. Then a couple of L-tetrominoes are easy to place and we've finished the top right:

top right done

Now consider

S-tetrominoes. There must be part of one in the 9th column. The 8th column must contain two different S-tetrominoes, one of which extends to the right and the other to the left. The 7th column only contains one S-tetromino, so the one that's in the 9th column must stop at the 8th. Then there's only one possible orientation for both S-tetrominoes:

almost finished

Then there's just two L-tetrominoes left to finish, which gives the two different possibilities shown at the top.

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  • $\begingroup$ I imagine the second is the intended answer since the blankspace seems to spell a 3-letter word... $\endgroup$ – Stiv Sep 11 at 13:18

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