2
$\begingroup$

From Martin Gardner's 'From Penrose Tiles to Trapdoor Ciphers'

From page 14, Chapter 1; https://www.maa.org/sites/default/files/pdf/pubs/focus/Gardner_PenroseTilings1-1977.pdf

"Any spoke of the infinite cartwheel pattern can be turned side to side (or, what amounts to the same thing, each of its bow ties can be rotated end for end), and the spoke will still fit all surrounding tiles except for those inside the central cartwheel. There are 10 spokes; thus there are 2^10 = 1024 combinations of states. After eliminating rotations and re- flections, however, there are only 62 distinct combinations. Each combi- nation leaves inside the cartwheel a region that Conway has named a "decapod." Decapods are shapes of enlarged half darts. The decapods with maximum symmetry are the buzzsaw and the starfish shown in Figure 12. Like a worm, each triangle can be turned. As before, ignoring rotations and reflections, we get 62 decapods. Imagine the convex vertexes on the perimeter of each decapod to be labeled T and the concave vertexes labeled H. To continue tiling, these H's and T's must be matched to the heads and tails of the tiles in the usual manner. When the spokes are arranged the way they are in the infinite cart- wheel pattern shown, a decapod called Batman is formed at the center. Batman (shown in dark gray) is the only decapod that can legally be tiled. (No finite region can have more than one legal tiling.) Batman does not, however, force the infinite cartwheel pattern. It merely allows it. Indeed, no finite portion of a legal tiling can force an entire pattern, because the finite portion is contained in every tiling.'

The problem I am having is getting the 62. I thought that putting 10 triangles meeting at a vertex with an angle of 2pi/10 where the triangles can be oriented in 2 ways would be the same as a 2 colour bracelet with 10 beads.

Here are some OEIS sequences;

A006245 rhombus tiling of 2n-gon A(3)=2 A(4)=8 A(5)=62 A(6)=908 A(7)=24698

A000031 2 colour bracelets A(10)=46

A000029 2 color necklaces A(10)=60

Now A006245 gives the correct number for the term,but this is the number of tilings of a 2n-gon , not the number of Decapods.

I would have thought A000031 was the correct object but a(10)= 46, not 62.

So I must be misunderstanding, what I am doing wrong?

Stuart

$\endgroup$
  • $\begingroup$ I think the issue is that when you reflect a bracelet the colours don't change, but when you reflect an arrangement of triangles, the types change. $\endgroup$ – Anthony Quas Jun 14 '16 at 17:56
2
$\begingroup$

I get a total of $62$ by appealling to Burnside's lemma. In particular, note that the objects we are counting are 2-colourings of the vertices of a decagon up to:

  • Rotations;
  • Reflections composed with colour reversal.

Now let's consider the $20$ elements of this group:

  • $1$ identity fixes all $2^{10}$ colourings;
  • $1$ rotation fixes $2^5$ colourings;
  • $4$ rotations each fix $2^2$ colourings;
  • $4$ rotations each fix $2^1$ colourings;
  • $5$ reflections each fix $2^5$ colourings;
  • $5$ reflections each fix no colourings.

So the total number of colourings up to symmetry is given by:

$$ \frac{1}{20}(2^{10} + 6 \times 2^5 + 4 \times 2^2 + 4 \times 2^1) = 62 $$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

I always thought it should be a member of A000013. If the necklace colors are swapped, it is making a mirror image of the decapod, but reversing the necklace is not allowed as that would reverse the order of the aces on the border of the decapod, but not their orientation. This seems to coincide with some code I wrote to catalog them, but for this sequence a(10) = 56, not 62 as claimed.

If I am right, the 1024 decapods can be grouped into 2 (x2), 6 (x10), and 48 (x20) equivalence classes. Batman is one of the 6 decapods that takes 10 of the 1024 as it is mirror symmetric, the buzzsaw is one of the pair that takes only 2 of the 1024 since rotating it produces the original decapod, and only a mirror image will produce one of the other 1024.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Nevermind, I just realized what I was doing wrong. I was reversing the direction of each ace on the boundary, but a reflection also requires the order of all those aces to be reversed. Fixing my code to do this tells me that this is the 10th member of A053656. The classes are then 2(x2), 18(x10), and 42(x20) $\endgroup$ – Jon Nov 19 '16 at 23:39

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.