Covering perforated hexagon with triminoes

[witzar 18]

This puzzle is inspired by posted by bonanova.

Again we work on a hexagonal tiling of a plane, and the question is about
possibility of covering some shape with triminoes.
Trimino is a "triangle" formed by three unit hexagons sharing common vertex.

The shape to cover is defined as follows:
Let's pick a unit hexagon and call it H₁.
Now we recursively define H_n+1 as a sum of H_n and all unit hexagons adjacent to H_n.
So basically H_n is a "hexagon" with side of length n (unit hexagons).

Let D_n be H_n with one unit hexagon at it's center removed.
So, can you cover D₂₀₁₅ with triminoes?

Edited November 14, 2014 by rookie1ja

[k-man 26]

Nice puzzle!

 

I don't have a solid proof yet, but I came to an interesting conclusion.

the only D

_n that can be filled with triminoes is D₄, so the answer to the OP is No.

 

[k-man 26]

I found a counterexample for my previous conclusion. This puzzle is even tougher than I thought.

[witzar 18]

Recall a classic: When a chessboard with two squares removed can be covered with 2x1 dominoes? The same approach works here.

[witzar 18]

Paint unit hexagons with three colors so that no two hexagons sharing side have the same color (like

this)
and observe then no matter how you place a trimino, it will cover one unit hexagon of each color.

[gavinksong 11]

I just saw this puzzle today.

The coloring trick immediately came to mind, but I couldn't figure out how to count the colors...

Just now, I reasoned that you could split any D(n) into six triangle shaped pieces that look like this after coloring (each triangle will only look the same as two others, but the color pattern will be the same for all six):

R

GB

BRG

RGBR

GBRGB

BRGBRG

...

After the first three rows, the later rows repeat except with some sets of three different colors appended to the end. In other words, the regex for every third row after the first row would be R(GBR)*, the regex for every third row after the second row would be GB(RGB)*, and the regex for every third row after the third row would be BRG(BRG)*.

Since D(n) has a side length of n, the triangles each have n-1 rows. Since the colors repeat, we can just look at the first n-1 (mod 3) rows and count the colors.

From this, I gleaned that there is only an odd number of colors if n (mod 3) = 2. So for D(2014), there is an equal number of each color.

So after all this, all I can say is that it's not necessarily impossible to tile D(2014) with triminoes...

Edited December 3, 2014 by gavinksong

[gavinksong 11]

Wait, I just realized that the question is asking about D(2015), not D(2014).

Using my reasoning above (which was not very elegant), D(2015) has an odd number of colors, namely three extra of each color that isn't what the center tile would have been. Thus, a complete tiling is impossible. Yay!

[witzar 18]

Well done, gavinksong.

At first counting colors of D

_n looks hard for big n. The trick is to realize that all that matters is n mod 3.

[gavinksong 11]

Well done, gavinksong.

At first counting colors of D

_n looks hard for big n. The trick is to realize that all that matters is n mod 3.

Is there a more elegant reasoning to show why that is?

[witzar 18]

Is there a more elegant reasoning to show why that is?

I was analyzing D

_n+1-D_n instead of dividing D_n into 6 triangles, but I don't think that it was more elegant

[gavinksong 11]

Is there a more elegant reasoning to show why that is?

I was analyzing D

_n+1-D_n instead of dividing D_n into 6 triangles, but I don't think that it was more elegant

I see, so you were looking at each ring.

Anyways, that was an excellent puzzle, witzar! Well done!