# Filling the plane with two colors

In this puzzle you must tile the plane with colored T-tetraminos. I will start by laying down 3 of them for you like so:

Your task will be to tile the entire rest of the plane meeting the following three conditions

1. My initial tiles will remain where they are.

2. No two tetraminos of the same color can share a side.

3. With a finite number of exceptions, all tetraminos should be one of two colors. That is to say that you should be able to draw a box such that only two colors of tetraminos appear outside of the box.

How can this be achieved?

• How many additional tetraminos can you use? Is there a limit? Commented Mar 6, 2020 at 4:22
• @ÉbeIsaac What do you mean by "additional". You must use an infinite number of tetraminos, since the plane is infinitely large. Commented Mar 6, 2020 at 12:15
• @ÉbeIsaac Maybe there is a misunderstanding. A green one should be able to fit just fine. Commented Mar 6, 2020 at 12:35
• @ÉbeIsaac "With a finite number of exceptions..." Commented Mar 6, 2020 at 12:42
• @KonstantinPfennig I'm not sure what you are asking. By box I meant closed curve, however every closed curve is also contained within a square or rectangle and vice versa, so if you substitute those words it does not change the meaning. Commented Mar 6, 2020 at 15:15

The black lines show boundaries between different "crystalline forms" which can be extended infinitely.

This can be extended infinitely in all directions - see my route to solving below for how.

First a detour to explain how I made a tool (which competing answers could also use, perhaps improving on the finite number of differently-coloured tiles) by which ideas can be quickly tried using an Excel spreadsheet:

1. Select all, and set column width to about 20 pixels so you get a square grid.
2. Enter 'R' in K9, 'D' in M10, 'L' in O9 (this will become the starting layout as in question)
3. Select a region from B2 to (e.g.) DZ100 (large enough to play with - in screenshot below, it seems I'd selected to CX63)
4. Open "Conditional formatting" => "Manage rules..." dialog.
5. Create 4 new rules. Each will have as its action setting one of the 4 borders of a cell. For each "Use a formula to determine which cells to format" In particular:
• =OR(B2="R",AND(B1<>"",B1<>"U"),AND(B3<>"",B3<>"D"),AND(C2<>"",C2<>"R")) => format set left border
• =OR(B2="L",AND(B1<>"",B1<>"U"),AND(A2<>"",A2<>"L"),AND(B3<>"",B3<>"D")) => format set right border
• =OR(B2="D",AND(B3<>"",B3<>"D"),AND(A2<>"",A2<>"L"),AND(C2<>"",C2<>"R")) => format set top border
• =OR(B2="U",AND(B1<>"",B1<>"U"),AND(A2<>"",A2<>"L"),AND(C2<>"",C2<>"R")) => format set bottom border
6. After this, the starting cells should have the right shapes around them. I then used (non-conditional) background formatting to highlight those starting pieces.
7. Play around by adding 'L', 'R', 'D' and 'U' to other cells. You can copy-paste regions around easily too...

In particular, portions of the layout that extend to infinity will likely need to follow a simple infinitely-repeating pattern such as the one shown in the answer given by PuzzlesAndSolutionsYT

After setting everything up as described above, it should look a bit like the following:

route to solving

One observation is that, within the infinitely-repeating section,

pairs of tetranimos must be "back-to-back" along their long edge, as if this were not the case, the long edge would form a 3-way junction with 2 other tiles. At least one other possible infinite tiling pattern exists.

If, outside the finite area,

there were regions of different "crystal structure", there would need to be a join that can be extended infinitely, still using only 2 colours. This is indeed possible, using a join along a diagonal. e.g. see below a join between a L/R crystalline area and a U/D crystalline area.

Using this knowledge I was able to put together the solution at the top of the post.

• There are ambient patterns other than the one PuzzlesAndSolutionsYT has shown. rot13("Fbzr bs juvpu ner irel hfrshy") Commented Mar 6, 2020 at 16:00
• Wow! VERY impressive!
– Avi
Commented Mar 6, 2020 at 18:01
• Nice solution, I've accepted it. My solution actually uses a different crystal, since I had discounted your crystal as less useful. It is very fun to see someone solve my puzzle in a way I had never expected. Commented Mar 6, 2020 at 20:49
• @SriotchilismO'Zaic I'd observed quite late that "your" crystal is the only one that fits in an infinite quadrant between two infinite diagonal lines (it's what my solution also uses in the 2-dimensionally infinite regions, even though I was originally starting by trying to build a different one), and had a sense that there would be a more efficient solution possible. I think if you'd left it open for improved answers a bit longer, someone (possibly me!) would have posted a more efficient solution by today. Commented Mar 7, 2020 at 12:01

## My own solution

Since this puzzle has been proper solved by Steve. I thought I would post my solution.

This solution is neat because it only uses 3 colors with 2 tiles of the third color.

1: Create a 4x4 block by placing one tetramino in the bottom right facing up and one of identical color in top left.

2: Fill in the rest of the space with the opposite color (2 tetraminos facing inwards from the sides)

3: to connect these blocks to neighboring blocks, simply create a 4x4 block of identical placement but with the two colors swapped out. place these blocks at all corners and keep repeating this process forever to create a plane.

Picture of solution:

• How can you do this from the given starting position, though? It seems like this solution doesn't actually include the starting tiles.
– Deusovi
Commented Mar 6, 2020 at 5:37
• @Deusovi It is simply not possible to solve it with the given starting position. I'm assuming he gave it as an example of how they look. it's not possible to solve it because if you look at the empty tile on the top it neighbors both colors therefore no color can fit into this space and its not solvable. Commented Mar 6, 2020 at 5:46
• The question states that you can use more than two colors, but only a finite number of tiles can be other colors besides your main two. Condition 2 states this pretty clearly. And you must start with three arranged as shown.
– Deusovi
Commented Mar 6, 2020 at 5:53
• Right, the question is how to fit the starting pattern into a 2-colored tiling, so that the space in between can also be filled with T tetrominoes. Might not be too hard, but it's definitely a necessary component of an answer.
– Deusovi
Commented Mar 6, 2020 at 7:02
• @Deusovi It may in fact be impossible to make a rectangle that doesn't slice through any tetrominos starting with the OP's original tiles. (Note that none of the valid solutions so far have done so.) That might be the basis for another puzzle... Commented Mar 6, 2020 at 22:06