12
$\begingroup$

Tetris Puzzle

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.

$\endgroup$
  • 1
    $\begingroup$ You're sure this is logically deducible? It's looking pretty unconstrained to me at the moment... $\endgroup$ – Deusovi Mar 28 at 1:48
  • $\begingroup$ I did double check the logic of the puzzle, but might have missed something. If you want to, you can post where you are stuck. It might be necessary to think a few tiles ahead or use process of elimination. An important observation is that rot13(pybfrq bss nernf zhfg unir n zhygvcyr bs sbhe pryyf). $\endgroup$ – Zilvarro Mar 28 at 1:59
  • $\begingroup$ When there are two coloured cells of the same type near to one another, are we to assume that they belong to different tetrominoes, or could they be in the same one? $\endgroup$ – Gareth McCaughan Mar 28 at 2:07
  • $\begingroup$ Tetriminos of the same type must not be orthogonally connected. They may be diagonally adjacent though. If you are referring to the two O clues, they could be part of the same tetrimino but do not need to be. $\endgroup$ – Zilvarro Mar 28 at 2:08
  • $\begingroup$ I did indeed have the two O clues in mind. $\endgroup$ – Gareth McCaughan Mar 28 at 2:10
9
$\begingroup$

Boom, Icosaetris for Zilvarro!

enter image description here

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).


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

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

enter image description here
* 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:

enter image description here

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:

enter image description here

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.)

enter image description here

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

enter image description here

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:

enter image description here

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Nice job! Regarding uniqueness, I was halfway through when you posted and what I have agrees with yours. $\endgroup$ – Jens Mar 28 at 11:18
  • $\begingroup$ Great writeup so far, and pretty much the intended path. Having the L piece not fence off the upper right corner would force another L piece directly between it and the O piece, because the upper right corner must still have a size of 20. $\endgroup$ – Zilvarro Mar 28 at 13:26
  • $\begingroup$ I want to accept this answer, but as of now it ends too abruptly. No need to go over every single detail but I would really appreciate if you complete the answer. $\endgroup$ – Zilvarro Mar 29 at 12:06
3
$\begingroup$

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

enter image description here

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:

enter image description here

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:

enter image description here

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.

enter image description here

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.

enter image description here

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

| improve this answer | |
$\endgroup$
  • $\begingroup$ You are totally right. I am sorry for that, I really tried to make it unique but must have confused myself with L/J down there. $\endgroup$ – Zilvarro Mar 29 at 15:40
  • $\begingroup$ No worries, it was a really fun puzzle! $\endgroup$ – isaacg Mar 30 at 1:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.