Raphael Robinson in his work about the undecidability of the domino problem introduced a set of six tiles depicted in Figure 1. These tiles can be rotated and reflected, one tile can fit the other only in such a way that the arrowhead matches the arrow tail and each block must contain exactly one bumpy corner, the leftmost in Figure 1.
It is possible to tile the Euclidean plane with copies of these six tiles, but only in an aperiodic way. The key to this result is that any Robinson tiling has a hierarchical structure: for all , define a supertile of rank as shown in Figure 2. Bumpy corner tiles are said to be supertiles of the first rank. An increasing union of supertiles of rank is called an infinite order supertile. The Robinson tiling can be either made of only one infinite order supertile or contain two or four infinite order supertiles.
We will prove that the number of distinct blocks of size (with ) that appear in Robinson tiling made of one infinite order supertile is defined by the formula
cccl A(n) & = & 32n^2 + 72 n ⋅2^⌊log_2 n⌋ - 48 ⋅2^2⌊log_2 n⌋ .
Similar result for the number of factors in paper folding sequence has been obtained by Jean-Paul Allouche in [References].
2 Complexity of Robinson tiling
In this paper, we say that a tiling of an square is correct if it can appear in a Robinson tiling made of one infinite supertile. In this case, defining a correct block is equivalent to defining the hierarchy of intersecting squares, see Figures 3 and 4. This can be done by placing corner tiles corresponding to supertiles of all ranks, note that some of them may be outside of our block.
We have an square denoted by . We will mark all cells of with a symbol if it is possible to place a corner tile in this cell, with a symbol if we have chosen this cell to be a corner and with a black dot if it is impossible to place a corner tile. Initially, we have just an block with all cells marked with . The first step will be to choose places for the corner tiles of rank 1. There are only four variants to do so, example for step 1, , is in Figure 5. Then, at each step , we choose places for corner tiles corresponding to supertiles of rank . Let us remark that when we place a corner tile of rank , it fixes the orientation of the corner tiles of rank as in Figure 6. We stop when the number of vacant places is less than four.
At each step, the set of all possible correct tilings of is divided into four disjoint classes. Resulting classes by construction can have either one or two vacant places.
If two squares have the same number of vacant places for corner tiles, there is the same number of possibilities to complete them to a correct pattern.
First let us prove that once , if two squares and at the end of the procedure described earlier remain with the same number of vacant places (thus one of two), then there are exactly the same number of possibilities to complete both.
Suppose both squares have only one vacant place, then they have all the cells filled except for two rows and two columns. One of the ’crossroads’ has to be a corner tile (marked with ) and the other two we will mark with letters and as in Figure 7. Indeed, for any correct tiling of one square we can specify one tiling of the other simply by choosing the same tiles for cells marked with A, B, C, and V. Same holds true for the situation with two vacant places.
To conclude the proof, it remains to note that the number of classes with one (two) vacant places is completely defined by the initial number of vacant places . ∎
Using Lemma 1 we can obtain recurrence relations for the number of correct tilings. All we need to do is to count the number of vacant places after placing the corner tiles of the first and second ranks and then to find any smaller square with the same number of vacant places.
Denote by the number of correct tilings of with a corner tile in the position [1,1] and by with for the number of all correct tilings of . Denote by the number of correct tilings of with a corner tile in the position [1,2]. If , is a sum of all possible tilings with corner tile in positions [1,1], [2,2], [2,1] and [1,2].
If is an even number then after the first step we have exactly vacant places for all four possibilities to place first rank corner tiles. By using Lemma 1 we can state that once , the number of all correct tilings of with a corner tile in the position [1,1] is equal to the number of all correct tiling of . The same is true for any other choice at step one, which gives us the first recurrence relation:
In the case when
is odd number we can write the number of vacant places as follows :
corner tile in the position [1,1] : vacant places ;
corner tile in the position [2,2] : ;
corner tile in the position [1,2] : ;
corner tile in the position [2,1] : .
The number of possibilities for last the two options was already denoted by . Now we can write the second recurrence relation:
Now we need to find recurrence relations for . Again, by carefully examining all the possibilities for the second order corner tiles we can write the last two recurrence relations:
Values of and can be found by an exhaustive search:
A_1 & = & 56 ;
B_1 & = & 124. The solution for recurrence relations above can be written as
For any Robinson tiling made of one infinite order supertile, once , the number of distinct square blocks is given by
cccl A(n) & = & 32n^2 + 72 n ⋅2^⌊log_2 n⌋ - 48 ⋅2^2⌊log_2 n⌋.
3 Concluding Remarks
Franz Gähler and Johan Nilsson conjectured in [References] that the number of distinct blocks in 2D paper folding sequence is equal to
-  R. Robinson, Undecidability and nonperiodicity for tilings of the plane, Invent. Math. 12 (1971) 177–209.
-  J.-P. Allouche, The number of factors in a paperfolding sequence, Bull. Austral. Math. Soc. 46, 23–32 (1992).
-  F. Gähler, J. Nilsson , Substitution Rules for Higher-Dimensional Paperfolding Structures, Preprint arXiv:1408.4997, 2014.