    # Density of Binary Disc Packings: Playing with Stoichiometry

We consider the packings in the plane of discs of radius 1 and √(2)-1 when the proportions of each type of disc are fixed. The maximal density is determined and the densest packings are described. A phase separation phenomenon appears when there is an excess of small discs.

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## 1 Statement of the results

A disc packing (or circle packing) is a set of interior-disjoint discs in the Euclidean plane. Its density is the proportion of the plane covered by the discs:

 δ:=limsupk→∞area of the square [−k,k]2 % covered by discsarea of the square [−k,k]2.

We are here interested in packings by two discs in well-defined proportions. Specifically, we consider the packings by discs of radius and , simply called ”packing” afterwards. The packings with a proportion of large discs (radius ) will be called -packings. We denote by the supremum of the densities of -packings. On the one hand, it was proven in [FT43]

 δ(0)=δ(1)=π2√3≈0.9069.

On the other hand, it was proven in [Hep00] (see also [Hep03, BF20])

 δ(x)≤δ(12)=π2+√2≈0.9202.

These densities are reached by the periodic packings depicted in Fig. 1. Here, we extend this to any stoichiometry .

###### Theorem 1

For , the maximal density of -packings is (Fig. 2):

 δ(x≤12)=π(x+(1−x)r2)4x+2(1−2x)r2√3,δ(x≥12)=π(x+(1−x)r2)4(1−x)+2(2x−1)√3.

Moreover, for any , there exists a packing with density . Figure 2: The maximal density as a function of the proportion of large discs.

Further, we would like to describe the set of all the densest -packings. One difficulty that arises is that, whatever packing we consider, there are continuously many packings with the same density that are different, even though they look pretty much the same (Fig. 3). We need to formalize this similiarity. Figure 3: Removing a disc in a packing does not affect the density (left). This holds for infinitely many discs, as long as they are in negligible proportion (center). Remaining discs can then be slightly moved (even arbitrarily far ones) without affecting the density (right).

Given a disc packing, define the cell of a disc as the set points of the plane which are closer to this disc than to any other. These cells form a partition of the plane whose dual is a triangulation, referred to as the FM-triangulation of the packing. Introduced in [FTM58] (see also [FT64]), FM-triangulations are also known as additively weighted Delaunay triangulations. For example, the FM-triangulations of the leftmost and rightmost packings in Fig. 1 are triangular grids (the one of the central packing is a so-called tetrakis square tiling). In what follows, we will ignore small disc displacements as long as they do not modify the isomorphism class of the FM-triangulation.

Now, let us call neighborhood of a disc the sequence (up to reversal and circular permutation) of its neighbor discs, where two discs are neighbor if the corresponding vertices are connected in the FM-triangulation. The neighborhood of a disc is denoted by the word over the alphabet which encodes the sequence of radii of the neighbors. For example, in Fig. 1, each small disc has the neighborhood rrrrrr in the leftmost packing or 1111 in the central packing, while each large discs has the neighborhood 1r1r1r1r in the central packing or 111111 in the rightmost packing. This allows to state out second result:

###### Theorem 2

Consider an -packing with density . If , then almost any small (resp. large) disc has a neighborhood 1111 or rrrrrr (resp. 1r1r1r1r). If , then almost any small (resp. large) disc has a neighborhood 1111 (resp. 1r1r1r1r, 1111r1r, 111r11r or 111111).

For , this yields a good insight into the look of the densest packings. Condider indeed a small disc. If its neighborhood is rrrrrr, then each of these small neighbor discs has itself three small neighbor discs, hence a neighborhood rrrrrr. Continuing from neighbour to neighbour unveils discs packed as in Fig. 1, left. If, on the contrary, its neighborhood is 1111, then each of the small neighbor discs has itself neighborhood 1111, and each of the large neighbor discs has neighborhood 1r1r1r1r. Continuing from neighbour to neighbour this times unveils discs packed as in Fig. 1, center. Both cases can actually coexist in the same packing, because Theorem 2 deals only with almost any disc: the negligible proportion of discs with other neighborhood can play the role of a ”joint” between large regions of near-periodically packed discs. The packing then looks like what is called in materials science a twinned crystal, i.e., an aggregate of different crystalline domains (Fig. 4). Figure 4: Typical look of a densest packing with an excess of small discs.

For , we also get aggregates of large regions in which only the coronas mentioned in Theorem 2 appear. But these regions may not be near-periodically packed anymore. They actually look like so-called square-triangle tilings. These are coverings of the plane by interior disjoint squares and regular triangles with edges of length , with edges being moreover allowed to intersect only in a single point or a on a whole edge (Fig. 5, left). Any square-triangle tiling can indeed be transformed into a disc packing (Fig. 5, right). A simple computation shows that if there is a proportion of squares, then we get an -packing of density . Conversely, the possible disc neighborhoods in an -packing of density for ensure it can be transformed into a square-triangle tiling (Fig. 6). However, one could argue about how much this correspondence really sheds light on the set of densest packings. Indeed, the set of square-triangle tilings with a given proportions of squares and tiles is itself quite complicated, as it will be illustrated by Proposition 1. Figure 5: A square-triangle tiling (left) can be transformed into a packing by discs of radius 1 and r=√2−1 (right) by putting a large disc on each vertex and a small disc at the center of each square. Figure 6: A disc packing with only suitable disc neighborhoods (from left to right: 1111, 1r1r1r1r, 1111r1r, 111r11r and 111111) can be transformed back into a square-triangle tiling.
###### Proposition 1

Call pattern of size of a square-triangle tiling its restriction to its tiles which lies into some ball of radius . Then, for any given , the number of different patterns which appear in the square-triangle tilings with squares for triangles grows exponentially in .

In other words, the set of square-triangle tiling with fixed proportions of tiles has positive entropy. Determining rigorously the value of the base of the growth exponent is an open problem (it is conjectured to be maximal for the tilings with a ratio of squares for large triangles, known as -fold quasicrystals).

Though not discussed here, we conjecture that the proof of both Th. 1 and Th. 2 can be readily adapted to each of the other ratios of disc sizes which allow so-called compact packings, i.e., packings whose contact graph is a triangulation [Ken06].

The rest of the paper is organized as follows. In Section 2, we describe -packings with density for any , relying on some basic word combinatorics. In Section 3, we prove that -packings have density at most . This is a computer-assisted proof which is adapted from the proof in [BF20] with only minor changes. Recalling the whole proof strategy of [BF20] would be like cutting and pasting several pages of [BF20]. We therefore assume that the reader is already acquainted with [BF20] and only sketch the strategy and point out the changes. This paper is thus not self-contained. Theorem 1 will follows from Sections 2 and 3. We then prove Theorem 2 in Section 4. This is, again, a modification of the previous computer-assisted proof. We also prove Prop. 1, which just relies on a ”tiling widget”.

## 2 Lower bound for the density

We here explicitly build an -packing with density . We rely on elementary combinatoric on words. For , denote by the sequence of whose -th letter is defined by

 uk=0⇔kαmod1∈[0,1−α).

For example (the bold letter has index ):

 u(13) = ⋯0100100100100100100100100100100100100100⋯ u(√2−1) = ⋯1001010100101001010100101001010010101001⋯

The case corresponds to a so-called Sturmian word introduced in [MH40]. words. Here, we will only use the fact that has a proportion of letter .

For , we use Sturmian words to mix the central and rightmost packings depicted in Fig. 1. Let . We associate with the square-triangle tiling made of vertical columns of either squares of triangles, with the -th column being made of squares if and only if the the -th letter of is (Fig. 7). This tiling has squares for triangles. Putting a large disc on each vertex and a small disc in the center of each square (recall Fig. 5) yields a packing with large discs for small discs, i.e., an -packing. Its density is

 α(π+πr2)+2(1−α)π24α+2(1−α)√3=π(x+(1−x)r2)4(1−x)+2(2x−1)√3=δ(x).

For , we want to use alike Sturmian words to mix the leftmost and central packings depicted in Fig. 1. The situation is however more complicated because these packings do not ”mix well” anymore: we have to combine two types of regions and deal with the ”joint” (recall Fig. 4). For this purpose, introduce the following transformation on sequences. If , then denotes the sequence obtained by replacing the -th letter of by the same letter repeated times. For example (the bold letter has index ):

 ˆu(13) = ⋯0000001111100000001100011100000000011111⋯ ˆu(√2−1) = ⋯0000011111000011100100011100001111100000⋯
###### Lemma 1

The transformation does not modify letter proportions.

Proof. Consider a factor of and a letter which has frequency in . We assume it has only positive indices (the case of negative indices is symmetric, and if it has both positive and negative indices we break it into two parts). It can be written

 w=pui+1iui+2i+2⋯ui+k+1i+ks,

where the prefix (resp. the suffix ) has length (resp. ). This can be rewritten

 w=p(ui⋯ui+k)i+1(ui+1⋯ui+k)(ui+2⋯ui+k)⋯ui+ks.

Fix and let grow. Since has length of order , the prefix and the suffix do not affect the proportion of in . By hypothesis, the proportion of in - hence in - tends towards . The Stolz-Cesàro theorem ensures that the proportion of in tends towards . The proportion of in thus also tends towards

We then associate with any sequence a disc packing as follows (Fig. 8). It is made of vertical column of identical adjacent discs. The columns alternate horizontally as the letters in , with a colum of large discs for a letter or a column of small discs for a letter . Consecutive columns of small discs are disposed so that small discs form an hexagonal compact packing as in the leftmost packing depicted in Fig. 1. Consecutive columns of large discs form a square grid and a small disc is inserted in each hole between four large discs, as in the central packing depicted in Fig. 1.

Let . The sequence has a proportion of . The packing associated with has large discs for small disc, i.e., it is an -packing. However, each factor or in a sequence corresponds in the associated packing to a ”joint” between two regions which locally decreases the density. In , there is way too much and (two for each small disc) in order to reach the density . This is where the above transformation comes in. Lemma 1 ensures that yields an -packing as well as , but it has moreover a negligible proportion of factors and (a factor of length contains factors and ). The two types of region thus respectively cover a proportion and of the plane. The density is

 απ+πr24α+2(1−α)r2√3=π(x+(1−x)r2)4x+2(1−2x)r2√3=δ(x).

## 3 Upper bound for the density

As mentionned in the introduction, we follows the strategy of [BF20] that is just sketched here. Fix a proportion of large discs. Fix an -packing and an FM-triangulation of the center of its discs. Define the excess of any triangle by

 E(T):=δ(x)×area(T)−cov(T),

where is the area of , is the area of inside the discs centered on the vertices and is the density defined in Th. 1. To prove that our packing has density at most , we will prove

 ∑T∈TE(T)≥0.

For this, we will define a potential on triangles satisfying the global inequality

 ∑T∈TU(T)≥0, (1)

and, for any triangle , the local inequality

 E(T)≥U(T). (2)

The potential will be the sum of a vertex potential and an edge potential . The edge potential shall satisfy the inequality

 ∑T∈T|e∈TUe(T)≥0. (3)

The vertex potential must satisfy a modified version of the vertex inequality of [BF20], namely

 ∑T∈T|v∈TUv(T)≥αq, (4)

where is the radius of the disc of center and the two real numbers and satisfy

 xα1+(1−x)αr=0. (5)

Let us stress that this is the one and only difference with the strategy of [BF20]! The global inequality (1) then follows:

 ∑T∈TU(T)=∑e∈T∑T∋eUe(T)≥0+∑v∈Tq=1∑T∋vUv(T)≥α1+∑v∈Tq=r∑T∋vUv(T)≥αr≥0 by disc proportions% and Eq.~{}(???)≥0.

Let us now define the potential and the constant and . Both the edge potential and the vertex potential will be defined as in [BF20]. We shall here only define the values of the base vertex potentials and of the parameters and .

The base vertex potentials , , , , and ,as well as the real number and are defined by the eighth following independent equations. The first four equations ensure that the sum of the base vertex potentials in any triangle is equal to the excess of this triangle (as in [BF20]):

 3V111 = E111, 3Vrrr = Errr, 2V11r+V1r1 = E11r, 2V1rr+Vr1r = E1rr.

The two following equations ensure Ineq. (4) around small and large discs of the central packing in Fig. 1:

 8V11r = α1, 4V1r1 = αr.

The seventh equation depends on :

 6Vqqq = αq,

with if or otherwise. It ensures Ineq. (4) around small (resp. large) discs in the leftmost (resp. rightmost) packing in Fig. 1. The eighth and last equation also depends on . It assigns an arbitrary value to (the way this value has been chosen is discussed at the end of this section):

 V11r={865269x3−985063x2+101139x−42831if x<12,0.006otherwise.
###### Lemma 2

For , the equation 5 is satisfied.

Proof. First, consider the case . The density has been defined in Section 2 as the density of a packing that corresponds to a square tiling with square for triangles. Each square corresponds in the FM-triangulation to triangles, each with two large discs and one small disc, and each triangle corresponds to one triangle with three large discs. Since the total excess of this packing is zero, one has

 0=4αE11r+(1−α)E111=4(1−x)E11r+2(2x−1)E111.

We then rely on the equations that define the ’s and ’s:

 0 = 4(1−x)(2V11r+V1r1)+2(2x−1)3V111 = (1−x)8V11r+(1−x)4V1r1+(2x−1)6V111 = (1−x)α1+(1−x)αr+(2x−1)α1 = xα1+(1−x)αr.

Then, consider the case . The density has been defined in Section 2 as the density of a packing whose FM-triangulation has triangles with two large discs and one small disc for triangles with three small squares (and a negligible proportion of other triangles), with . We then proceed as above:

 0 = 4αE11r+2(1−α)Errr = 4xE11r+2(1−2x)Errr = 4x(2V11r+V1r1)+2(1−2x)3Vrrr = x8V11r+x4V1r1+(1−2x)6Vrrr = xα1+xαr+(1−2x)αr = xα1+(1−x)αr.

The equation 5 is thus satisfied for any .

We further proceed as in [BF20]. First, one finds coefficients and of the angle deviation in the vertex potential which ensures Ineq. 4 around any vertex of any FM-triangulation. We then compute the ceiling and . The vertex potential is fully defined. The edge potential is defined by the parameters given in Tab. 1. One then finds such that the local inequality (2) follows from the mean value theorem for any triangle of the FM-triangulation with distance at most between aby two of its discs. We finally check with a computer program the local inequality (2) on all the other possible triangles by dichotomy, using interval arithmetic.

Last but not least, we have to check that is maximal not only for a given proportion , but for any proportion in . As usually, we use interval arithmetic: the interval is divided into intervals that are small enough to perform the above check with each of these intervals as a value for .

We first wrote a SageMath/Python program to perform the above check. It works quite well for a given value of but is too slow to check all the possible proportions. We thus wrote a C++ program and used on the Boost interval arithmetic library. Cutting in intervals of length appeared to be sufficient to check both the local and global inequalities for any proportion . Fig. 9 gives an idea of the number of triangles on which, for each interval of proportions, the local inequality has been checked by dichotomy (this is the most time-consuming part of the check). The complete checking took about minutes on our laptop, an Intel Core i5-7300U with cores at GHz and Go RAM. Figure 9: Variation with the % of large discs of the number of checked triangles. Each bar corresponds to an interval of 1%. The first bar is quite high, but dividing the interval in 10 yields 10 bars of at most 60000 triangles each: there is a trade-off between size and number of intervals.

To conclude this section, let us explain how the values of have been chosen. For a given proportion , we tried different random values of . Some values yield an error (i.e., an FM-triangle which does not satisfy the local inequality) and some other values yield an infinite loop-recursion when trying to refine too far the precision while checking the local inequality. But many values allow to succesfully check the local inequality. We take this value for . We do this for different proportions

, then interpolate the obtained values by a polynomials to define

for any .

## 4 Densest packings

For lighter wording, let us call bad neighborhood the following neighborhood:

• if , the neighborhood of a small (resp. large) disc other than 1111 or rrrrrr (resp. 1r1r1r1r);

• if , the neighborhood of a small (resp. large) disc other than 1111 (resp. 1r1r1r1r, 1111r1r, 111r11r or 111111).

In other words, these are the neighborhoods which are negligible in a densest packing, according to Theorem 2.

###### Lemma 3

Assume that there is such that, in an FM-triangulation of any -packing, the vertex inequality (4) can be improved as follows around the center of a disc of radius with a bad neighborhood:

 ∑T∈T|v∈TUv(T)≥αq+η. (6)

Then, the proportion of bad neighborhoods in an -packing with density is at most .

Proof. Consider an -packing with density . Fix an FM-triangulation of its disc centers. Denote by the restriction of to the discs within distance from the origin. On the one hand, the definition of the density yields

 ∑T∈Tncov(T)=δ∑T∈Tnarea(T)+O(n).

Since , this rewrites

 ∑T∈TnE(T)=(δ(x)−δ)∑T∈Tnarea(T)+O(n).

On the other hand, Ineq. (2) and (3) first yield

 ∑T∈TnE(T)≥∑T∈TnU(T)≥∑T∈Tn∑v∈TUv(T)+O(n).

Then, the hypothesis (6), Eq. (5) and the fact that we have an -packing yield

 ∑T∈Tn∑v∈TUv(T)+O(n)≥b(n)η+O(n),

where is the number of vertices of with a bad neighborhoods. Hence

 b(n)≤δ(x)−δη∑T∈Tnarea(T)+O(n).

Dividing both sides by the sum of the areas of triangle in , which grows as , yields the proportion of bad neighborhoods on the left hand side and the claimed bound on the right hand side.

Theorem 2 will be a corollary of the above lemma for . The actual value of has no importance for Theorem 2, though it provides an information on how fast bad neighborhoods disappear when the density approaches the optimal density. The following proposition provides a suitable (though not optimal) value :

###### Proposition 2

The improved vertex (6) inequality holds for .

Proof. We simply replace by in the code use to prove Th. 1 and proceed similarly, except to compute suitable coefficients and : we do not take into account the good neighborhoods (around which only the original vertex inequality (4) is theoretically ensured by the way vertex potential have been chosen). The obtained coefficients are (slightly) larger, and must be small enough so that the local inequality (2) still holds for any suitable triangle. The chose value of suits for any interval of length for and only slightly increase the computation time (3min 30s instead of 3min). This value is not optimal: it can be increased, especially if we make it depend on and take a smaller interval for (but the computation time increases as well).

Theorem 2 is thus proven. We conclude this section by proving Proposition 1:

Proof. Let . There are two different ways to tile a regular dodecagon by squares and regular triangles (Fig. 10). Any square-triangle tiling with a positive frequency of dodecagons will thus yields a number of different patterns of size which grows exponentially in (by tiling independently each dodecagon). We shall define such a tiling with squares for large triangles.

The idea is to replace, in a square-triangle tiling associated with a sequence as explained in Section 2), each square and triangle by respectively, the building blocks and depicted in Fig. 11. The point is to find suitable and . The numbers of squares and of triangle in (resp. and in ) are

 s□n = (n+1)2+4n2+64=n2+4n+52 s△n = 4(2n+1)+124=8n+7, t□n = 3n2+63=32n+2, t△n = 1+3+…+(2n+3)−3+123=n2+4n+5.

The ratio of squares and triangles in the resulting tiling is

 f(β,n):=βs□n+(1−β)t□nβs△n+(1−β)t△n.

In particular

 limn→∞f(0,n) = limn→∞32n+2n2+4n+5=0, limn→∞f(1,n) = limn→∞n2+4n+528n+7=+∞.

Since is continuous, this ensures that for any , i.e., for any , for large enough there exists such that . For such and , replacing the squares and triangles of the tiling associated with by the building blocks and yields a tiling with squares for triangles.