Let's Play Tetris!

Question

Fill the whole grid with Tetrominos.

Every cell must be part of exactly one Tetromino.

Tetrominos can be rotated as necessary.

The colored clue cells must be part of the indicated type of Tetromino.

As an additional rule, Tetrominos of the same type must not be orthogonally connected.

(Obviously, aside from the Tetromino shapes, usual Tetris rules do NOT apply.)

Comments are very appreciated!

EDIT: As pointed out by @isaacg a small region at the bottom right does have multiple possible solutions.

Answer 1

Boom, Icosaetris for Zilvarro!

This is only the final result; full write-up is below. I think the solution is unique (at least if I made no mistakes (EDIT: turns out I did.)), because the deduction chains needed were of reasonable length, as long as I picked the appropriate variations to rule out. (Which was the hard part).

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Edit: Here's the promised write-up on how I got there.

Step 1: The low hanging fruit: (clicky will unsmallify)

* The top left Z cannot be vertical, because it would isolate the corner square
* If the O piece at the bottom were one space to the right, there would be no way to fill the bottom right corner square
* Given the O piece at the bottom, there are only two ways to place the adjacent L. The other way would force the Z on top of the T piece.

This is basically what happens all the way: Some pieces are forced, some are impossible. Here are some heuristics that will get repeatedly used:

• Try the "tight spots" first (the fewer options there are, the more likely it is to find something definite)
• The "no adjacent blocks of same colour rule" often rules out almost every piece
• When completely cutting off an area, the number of squares must be divisible by four on both sides

Using these, we quickly get these pieces in place:

And then we need to start really thinking for the first time. Happily, we can make progress on multiple fronts:

• At the top left, the square in the armpit of the Z piece must belong to a J piece by elimination: I and S would instantly break the adjacency rule, an L or a T piece would do the same with the next piece, and neither Z or O can fit.
• At the bottom left, the nook between the T and the Z can be filled with an I or with an S. Either choice soon fills the spot next to the given L square, which forces the two nearby I blocks vertical. (The leftmost one because it won't fit horizontally, and the other one because horizontally it would cut off an area of 22 unsolved squares at the bottom right)

After these, we complete the unfinished T near the top, which gives us a couple of pieces on the left side for free:

Then there's the question posed in the comments above: can the two yellow Os belong to the same O piece? Turns out we can solve it now.

If they belonged to different squares, their places would be fixed, and they would wall off an area. Depending on the orientation of the L piece at border, the size of that area would be 15, 17 or 18 squares, none of them divisible by four! So there can only be one yellow square there.

This forces the neighbouring S piece in place, and a bit surprisingly, also the L-piece on the right edge: no matter how you place it, it cuts off the top right corner, and there's only one way to do it so that an even number of squares gets cut off. (NOTE: LOGIC FLAW SPOTTED. If you place the L one spot lower, it doesn't cut off the corner. Let's see if this bug cancels out later, it's quite possible I just got lucky here. All the other pieces are still provably correct though. RE-EDIT: logic flaw dissolved: placing the L one spot lower would necessitate two adjacent L pieces, as kindly pointed out by OP in the comments.)

This makes the top left easy to solve and gives a free piece on the left too.

Then it's time to look at the bottom again.

Now that there's a square on the left side, it's easy to see that the previously mentioned nook between the T and Z pieces cannot be filled with an I piece, so it must be an S, which also fixes the L above. The remaining two empty squares along the side cannot be taken by a single piece (The J is the only option, and that would force the nearby Z to block of an area of 7 squares.) So we get

The S at the bottom right has three possible placements. The vertical ones won't work: the topmost one would cause the J to wall of an improperly sized area, and the other would need two adjacent T blocks to fill the bottom right corner. So the S must be horizontal, and with some simple deductions (EDIT: .. that involved the typical mistake: confusing the handedness of the pieces. Without that mistake it should be obvious the L and J pieces can fill the same space in more than one way. So the solution's not completely unique after all.) we get the corner filled right up to L piece on the side.

We left a gap near the I pieces at the center in the previous step, it's easily fillable too:

From here, the clues get scarce, but since there's very little left to solve, it's not going to be too bad. I got to the final solution by trying all the possible positions and orientations of the clued-in S piece; there are many, but most of them are very easily ruled out.

Answer 2

I solved the puzzle, and found that the solution is not quite unique:

In the bottom right corner, there are three possible arrangements of three pieces: Either J, L, O, or L, J, O or J, Z, L. These are indicated by the top left dots, bottom dots, and top right dots, respectively.

The main techniques I used were:

• Connecting squares to each other, one at a time.

• Using the no orthogonal condition.

• Trying out different orientations of crucial pieces to reach contradictions

• Counting unfilled squares up to multiples of 4.

I started by solving the top left corner, which wasn't too complicated:

Next, I did a case analysis of the L in the top right. The only possibility was the horizontal arrangement with the marked square on the corner. That allowed me to solve the top right corner, and the beginning of the bottom left was pretty straightforward:

Next, I solved the bottom right, I found it to be non-unique. The crucial region was the interaction of the L and J at the top of the bottom right. There were two possible orientations of the L, but only one of the J.

The next part of the bottom left came from a case analysis of the Z around there. It turns out, the only possible orientations are the two vertical right-aligned ones. Playing out those two possibilities means that there must be a border as drawn, though it's not yet clear what the pieces within that border are.

Next, I moved to the middle right. I did a case analysis of the Z in the mid-right, with a sub-case analysis of the S over there. I identified a unique Z possibility, and with it a unique S possibility. Many possibilities were eliminated due to square counting or a conflict with the O just above the S. With that, I solved the middle right.

With that taken care of, I could finish off the bottom left, and then the rest of the puzzle.