The goal is to divide the grid into regions of exactly four cells (tetromino). Each region contains exactly two different symbols. Regions of the same shape must contain the same symbols. The tetrominoes may be rotated or mirrored.

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  • $\begingroup$ "Regions of the same shape must contain the same symbols." Also, if two regions contain the same symbols, must they be the same shape? $\endgroup$ – Rand al'Thor May 25 '20 at 20:32
  • $\begingroup$ Yes, same symbols, same shape, but the shapes can be rotated or mirrored $\endgroup$ – perayu May 25 '20 at 20:33
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    $\begingroup$ It's the evening here and I'm starting on a perayu puzzle ... I must be mad, I'll be up all night :-) $\endgroup$ – Rand al'Thor May 25 '20 at 20:36

First, some preliminary steps:

Let's figure out what combinations of shapes are allowed. There are five tetrominoes, and six possible pairs of the four shapes. So at least one pair must be unused.

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The bottom left diamond must connect to a square. So blue-green exists. The middle right triangle must connect to a circle. So yellow-red exists.
One of the two diamonds in the upper left must connect to a triangle. So blue-yellow exists.
One of the two shapes in the top right must form a red-blue pair. (If the red circle does not, it must connect to the green in the top right, and then the blue diamond must connect to the red circle just below it.) So blue-red exists.
The square in the center must connect to a circle. So green-red exists.

This accounts for all five pairings. So yellow and green are never connected.

Now, we begin the actual deductions.

The bottom left diamond must be an L shape, reaching the square. So this tells us that square+diamond = L tetromino.
Since green and yellow are never connected, this also lets us resolve a bunch of the bottom right corner, and deduce that circle + triangle = I tetromino. This deduction continues up the side of the grid.

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the top right can be resolved. The green squares must be part of an L shape: if that L shape includes the blue diamond on the left, this would block off the upper red circle. So the L shape must include the blue diamond on the right wall. This lets us deduce that red+blue makes a T shape.

If the lower red dot connects to the blue there, we can't match up the yellow triangle. So the lower red dot must go straight down.

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Next we look at a certain square:

The square two southwest of the top right corner. What can fill it? If the currently-unfinished T tetromino goes left, it can't be filled. So that one must go downwards, and then we can resolve more of that area.
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And now it's time to finish off the assignment:

Yellow-blue must be O or S, because those are the only remaining pieces. If it's O, then the upper left diamond must connect to the square, and then one just right of that must make an O. But then we have a contradiction.
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So now there's a definitive assignment: yellow+red is I, blue+green is L, green+red is O, yellow+blue is S, blue+red is T.

The top right can be resolved: first consider where the first square in the second row goes, then how the square above-right of it is accessed.

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Next, the top left: the key step is figuring out how this space can be filled.

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And then, we need to figure out how the upper part of the left wall can work. The alcove can be filled in two ways, each leading to a tall yellow-red piece -- but only one lets the green square escape with the diamond below it.
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And that's like this:
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More logic based on how certain squares can be filled takes care of the left wall:

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And next note that the yellow and blue near the bottom left can't go together, because it would block off one of the two squares left of them.

That last realization starts a chain that solves the rest of the puzzle:

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The blue in its own alcove near the lower left of the unfinished area must be a T. The yellow triangle above it must turn into an S, and the red circle that S overhangs must be the one to fill that newly created empty alcove. We can then resolve the top area, and that gives us more deductions that finally finish the puzzle.

The final answer:

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Final solution

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Step-by-step deduction

There are five possible tetrominoes: L, T, O, I, S. There are four different symbols, giving six possible pairs of distinct symbols (blue-green, blue-yellow, blue-red, green-yellow, green-red, yellow-red). So one of the pairs doesn't occur, and the other five have a one-to-one correspondence with the tetromino shapes.

Starting in the bottom left corner,

there's only one allowable tetromino including the blue diamond in the corner, so we have L-tetromino corresponding to blue-green and we can fill in some more:

starting off

(Some of these are only partial tetrominoes. E.g. in the top right, the green square in the corner can't be disconnected from the cell to its left, because then we'd have an L-tetromino just going up the right-hand side and it wouldn't fit.)

Looking at the top right and top left corners, it's clear that

blue-red and blue-yellow pairings exist. So the one pairing that doesn't exist must be among green, yellow, and red only. In total there are 17 yellow triangles, 37 green squares, and 24 red circles. Every one of the 48 blue diamonds must be paired with one of these, leaving just 16 tetrominoes to be made among green, yellow, and red only. Also, looking on the right-hand side, there must be yellow-red pairings, so the one that doesn't exist must be either green-yellow or green-red.

Some more deductions in the top right enable us to get

that the T-tetromino corresponds to blue-red. In detail: if the corner green square connects with the blue diamond to its left, then the top red circle must connect with the blue diamond to its right, which must then be an L-tetromino, contradiction. So the corner green square connects with the blue diamond below it. The top red circle can't connect with the green square to its left, as that would cut off the two blue diamonds above, so it must connect with the nearest blue diamond since it can't be in an L-tetromino. That connection can't be an O, I, or S tetromino, so it must be a T-tetromino.

Also, in the yellow-red group on the right-hand side (specifically the red on the far right between two yellows),

the yellow-red tetromino cannot be O. If it's S, then in the bottom right the green square in the corner must connect to the blue diamond (otherwise green-yellow S-tetromino), so the yellow triangle above it must connect to the red circle above that, which must be either an I-tetromino or an L-tetromino, contradiction. So the I-tetromino corresponds to yellow-red. Now we have:

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From the top left,

the blue-yellow shape can't be O, so the S-tetromino corresponds to blue-yellow. That enables us to fill in the bottom right corner and a bunch more stuff in the top left.

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Now finally we see that a green-red connection is necessary, so green-yellow never happens and by elimination the O-tetromino corresponds to green-red.

And the rest is easy deduction step by step, now that we know exactly which colours correspond to which shapes:

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  • $\begingroup$ "Regions of the same shape must contain the same symbols." I don't think this necessarily means that the same symbols always correspond to the same shape. The way I read it, it's possible to have a single pair of symbols that represents multiple shapes. (I'm trying to get a bound on that rigorously, by finding shapes that definitely must connect to certain other shapes.) $\endgroup$ – Deusovi May 25 '20 at 21:10
  • $\begingroup$ @Deusovi Check comments on the question - I asked the OP to be sure before starting to work on this. $\endgroup$ – Rand al'Thor May 25 '20 at 21:11
  • $\begingroup$ Oh, I see - good point. My bad. $\endgroup$ – Deusovi May 25 '20 at 21:12
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    $\begingroup$ Why can't the yellow-red tetromino be I? I'm working on the puzzle right now, and it seems perfectly fine if that's the case. $\endgroup$ – Deusovi May 25 '20 at 21:25
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    $\begingroup$ Welp, you're right. I was fooling myself with that red on the left being so surrounded by yellows. $\endgroup$ – Rand al'Thor May 25 '20 at 21:28

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