The enumeration of extreme rigid honeycombs

The notion of a rigid honeycomb was introduced in [9] in relation to the Horn problem. It also played a central role in proving that the Horn inequalities hold in an arbitrary factor of type II₁ in [2], where it was shown that every rigid honeycomb arises as a special kind of overlay (more general than the clockwise overlays of [9]) of extreme rigid honeycombs. Extreme rigid honeycombs are scalar multiples of the rigid tree measures (which we now call rigid tree honeycombs) studied in [3].

In retrospect, the results of [2] depended on the study of these extreme rigid honeycombs, and one of the initial difficulties was the lack of examples of sufficiently complicated such honeycombs, which in turn hampered effective experimentation. (Producing just six such honeycombs of weight four required an extensive search; an intersection problem generated by one of these six examples is fully analyzed at the end of [2, Section 6].) Subsequently, several infinite families of extreme rigid honeycombs were constructed in [3].

Our purpose is to address the following two questions:

1. (a)

   Can one effectively enumerate all the rigid tree honeycombs of a given weight $\omega?$

2. (b)

   How can one generate rigid tree honeycombs of high weight from rigid tree honeycombs of smaller weight?

In answer to (a), we offer a method based on the study of certain combinatorial objects called locking patterns. This makes the enumeration of (types of) rigid tree honeycombs possible, though still tedious for large weights. For (b), we show that every rigid tree honeycomb of weight at least two can be constructed from an overlay of honeycombs with strictly lower weights. In principle, this allows one to use the enumeration of rigid tree honeycombs of weight less than a fixed $k\in\mathbb{N}$ to construct all possible such honeycombs of weight $k$. In practice, this is difficult to achieve because there are usually many overlays of two honeycombs of given type and these overlays are hard to construct directly. However, once an overlay of two honeycombs, or even a single honeycomb, is known, we show how to construct honeycombs of much greater complexity and weight. These constructions recover, for instance, some of the infinite families constructed in [3] and produce many additional infinite families of rigid tree honeycombs. The honeycombs constructed this way should be instrumental in the experimental study of related conjectures, such as those of [5].

The paper is organized as follows. Basic definitions about honeycombs, puzzles, and duality are covered in Sections 2 and 3. Locking patterns and their connection to the enumeration of rigid tree honeycombs are also covered in Section 3. (The term locking pattern was suggested by Ken Dykema before we really understood what these overlays are, and this is why the notion is not mentioned in [2].) In Section 4, we describe two constructions that help calculate the exit multiplicities of honeycombs supported in the edges of the puzzle of a given rigid honeycomb. One of these results yields the basic inductive construction of rigid tree honeycombs from other rigid tree honeycombs of smaller weights. Section 5 is dedicated to the study of degenerations of a rigid honeycomb, particularly simple degenerations. These simple degenerations, applied to a rigid tree honeycomb, produce either an extreme rigid honeycomb or an overlay of two extreme rigid honeycombs. Necessary conditions are derived on such overlays and Section 6 is dedicated to showing that these necessary conditions are sufficient as well. The arguments are based on a new construction that starts with a puzzle and a compatible honeycomb and produces another puzzle and compatible honeycomb using an inflation and a partial deflation. The constructions produce infinite sequences of rigid tree honeycombs whose weights increase rapidly (like the powers of a number $>1$) but with a fixed number of nonzero exit multiplicities. Some (but not all) of the infinite families produced in [3] are shown to arise from this construction and some new infinite families are produced.

We review the definition of a honeycomb as presented in [8]. Fix vectors $u_{1},u_{2},u_{3}$ of length $\sqrt{3}/3$ in the plane, arranged in counterclockwise order such that $u_{1}+u_{2}+u_{3}=0$ (see Figure 2.1), and define unit vectors

$$w_{1}=u_{3}-u_{2},\quad w_{2}=u_{1}-u_{3},\quad w_{3}=u_{2}-u_{1}.$$

Every point $X\in\mathbb{R}^{2}$ can be written uniquely as $X=x_{1}u_{1}+x_{2}u_{2}+x_{3}u_{3}$ such that $x_{1},x_{2},x_{3}\in\mathbb{R}$ and $x_{1}+x_{2}+x_{3}=0$. We view the numbers $x_{j}$ as the coordinates of $x$. Given a fixed $c\in\mathbb{R}$, the equation $x_{j}=c$ represents a line parallel to $w_{j}$.

A honeycomb $\mu$ is described by two elements, namely,

1. (1)

   the support of $\mu$. This consists of a finite union $I_{1}\cup\cdots\cup I_{k}\subset\mathbb{R}^{2}$ where

   1. (a)

      each $I_{j}$ is a closed segment, a closed ray, or a line,

   2. (b)

      each $I_{j}$ is parallel to one of the vectors $w_{1},w_{2},w_{3}$, and

   3. (c)

      if $i\neq j$ then $I_{i}$ and $I_{j}$ have at most one point in common that is an endpoint of both $I_{i}$ and $I_{j}$.

2. (2)

   the multiplicities of the honeycomb. These are (not necessarily integer) numbers $\mu(I_{1}),\dots,\mu(I_{k})\in(0,+\infty)$.

It is convenient to consider that a honeycomb $\mu$ assigns zero multiplicity to segments parallel to some $w_{i}$ that intersect the support of $\mu$ in finitely many points, and that $\mu$ assigns multiplicity $\mu(I_{j})$ to all subintervals of $I_{j}$. A branch point of a honeycomb is a point where at least three of the sets $I_{j}$ meet. The segments $I_{j}$ are called edges of $\mu$. The multiplicities of a honeycomb are subject to the following balance condition: Suppose that that $J_{1},J_{2},J_{3},J_{4},J_{5},J_{6}$ are six segments, each parallel to some $w_{i}$, containing no branch points in their interior, having a common endpoint $X$ and arranged clockwise around $X$. Then

(2.1)

$$\mu(J_{1})-\mu(J_{4})=\mu(J_{5})-\mu(J_{2})=\mu(J_{3})-\mu(J_{6}).$$

It may be easier to think of a honeycomb as a Borel measure in the plane that assigns $\mu(I_{j})\times\text{length}(A)$ to each Borel subset $A\subset I_{j}$. This allows us to add honeycombs, thus giving the collection of all honeycombs the structure of a convex cone.

In this work, we do not distinguish between two honeycombs that are simply translates of each other, with the translation preserving multiplicities.

Suppose that $\mu$ is a honeycomb with support $I_{1}\cup\cdots\cup I_{k}$. It is easy to see that any endpoint of some $I_{j}$ must also be an endpoint of another $I_{\ell}$. The support of a honeycomb $\mu$ must contain lines or rays. Each ray points in the direction of one of the vectors $\pm w_{j}$, $j=1,2,3$. Figure 2.2 shows a few examples of honeycombs. All edge multiplicities, with one exception indicated by a thicker line, are equal to $1$.

We denote by $\mathcal{M}$ (respectively, $\mathcal{M}_{*}$) the collection of all honeycombs with the property that the rays in the support of $\mu$ point in the direction of $w_{1},w_{2},$ or $w_{3}$ (respectively $-w_{1},-w_{2}$, or $-w_{3}$). The second and third examples in Figure 2.2 represent a honeycomb in $\mathcal{M}$ and its mirror image (relative to a vertical line) in $\mathcal{M}_{*}$. Suppose that $\mu$ is a honeycomb in $\mathcal{M}$, $c\in\mathbb{R}$, and $j\in\{1,2,3\}.$ If the support of $\mu$ contains a ray $I\subset\{x_{j}=c\}$, we define

(2.2)

$$\mu^{(j)}(c)=\mu(I).$$

Otherwise, we set $\mu^{(j)}(c)=0.$ The numbers $\mu^{(j)}(c)$ are called the exit multiplicities of $\mu$. (The ordered collection of the nonzero exit multiplicities will later be called the exit pattern of $\mu.)$ Of course, $\mu$ only has finitely many nonzero exit multiplicities and the balance condition (2.1) is easily seen (see, for instance, [8]) to imply the identity

(2.3)

$$\sum_{c\in\mathbb{R}}\mu^{(1)}(c)=\sum_{c\in\mathbb{R}}\mu^{(2)}(c)=\sum_{c\in\mathbb{R}}\mu^{(3)}(c).$$

This common value is called the weight of $\mu$ and is denoted

(2.4)

$$\omega(\mu)=\sum_{c\in\mathbb{R}}\mu^{(1)}(c).$$

The exit multiplicities also satisfy the equality

(2.5)

$$\sum_{c\in\mathbb{R}}c(\mu^{(1)}(c)+\mu^{(2)}(c)+\mu^{(3)}(c))=0,$$

sometimes known as the trace identity on account of its connection with linear algebra.

A honeycomb $\mu\in\mathcal{M}$ (or $\mu\in\mathcal{M}_{*}$) is said to be rigid if there is no other honeycomb that has the same exit multiplicities as $\mu$. It is easy to see that an arbitrary honeycomb is not rigid if it has a branch point adjacent to six edges in its support; see Figure 2.3.

In fact, one can determine whether a honeycomb $\mu$ is rigid just by looking at its support. Let $\mu$ be a honeycomb and let $A,B,C$ be branch points of $\mu$ such that the segments $AB$ and $BC$ are edges of $\mu$. We say that $ABC$ is an evil turn [2, p. 1592] if one of the following situations occurs:

1. (1)

   $C=A$, and the support of $\mu$ contains edges $BX,BY,BZ$ that are $120^{\circ},180^{\circ}$, and $240^{\circ}$ clockwise from $AB$.

2. (2)

   $BC$ is $120^{\circ}$ clockwise from $AB$.

3. (3)

   $C\neq A$ and $A,B,C$ are collinear.

4. (4)

   $BC$ is $120^{\circ}$ counterclockwise from $AB$ and the support of $\mu$ contains an edge $BX$ that is $120^{\circ}$ clockwise from $AB$.

5. (5)

   $BC$ is $60^{\circ}$ counterclockwise from $AB$ and the support of $\mu$ contains edges $BX,BY$ which are $120^{\circ}$ and $180^{\circ}$ clockwise from $AB$.

The possible evil turns are illustrated in Figure 2.4 where the edges that must be in the support of $\mu$ are labeled, but the support may contain all six edges incident to $B$. A sequence $A_{1},A_{2},\dots,A_{n},A_{n+1}=A_{1}$ of branch points of $\mu$ is called an evil loop if $A_{j-1}A_{j}A_{j+1}$ is an evil turn for every $j=1,\dots,n$. Each of the first three configurations in Figure 2.5 contains exactly one evil loop (up to cyclic permutations) but the fourth one contains none.

The following result is proved as in [2, Proposition 2.2]. (The proof uses a characterization of rigidity via puzzles; see [9, Theorems 4 and 7].)

Suppose that $e=AB$ and $f=BC$ are edges of a rigid honeycomb $\mu$. As in [2], we write $e\to_{\mu}f$ (or $e\to f$ when $\mu$ is understood; see Figure 2.6) if one of the following two situations arises:

1. (1)

   $\varangle ABC=120^{\circ}$ and the segment $BX$ opposite $AB$ is not in the support of $\mu$.

2. (2)

   $e$ and $f$ are opposite and there exists a segment $BX$ not in the support of $\mu$ such that $\varangle CBX=60^{\circ}$.

The balance condition shows that $\mu(e)\leq\mu(f)$ whenever $e\to f$. In fact, $\mu(f)$ equals $\mu(e)$ if $e\to f$ and $f\to e$, and it equals $\mu(e)+\mu(e^{\prime})$ when $e\to f$, $e^{\prime}\to f$, and $e\neq e^{\prime}$. The transitive closure $\Rightarrow$ of the relation $\to$ was called descendant in [2]. Thus, $e\Rightarrow f$ if there exists a descendance path, that is, a sequence $e_{1}\dots e_{n}$ of edges such that $e_{1}=e,e_{n}=f,$ and $e_{j}\to e_{j+1}$, $j=1,\dots,n-1$. Given such a path, $e$ is called an ancestor of $f$ and $f$ is called a descendant of $e$. We write $e\Leftrightarrow f$, and we say that $e$ and $f$ are equivalent, if either $e=f$ or both $e\Rightarrow f$ and $f\Rightarrow e$ hold. An edge $e$ of $\mu$ is said to be a root edge if the relation $f\Rightarrow e$ implies $e\Leftrightarrow f$. The above observations show that the multiplicity of an arbitrary edge of $\mu$ can be written as a linear combination with positive integer coefficients of the multiplicities of root edges. More precisely, suppose that $\mu$ is a rigid honeycomb and that $e_{1},\dots,e_{n}$ is a maximal sequence of pairwise inequivalent root edges. Then an arbitrary edge $f$ of $\mu$ satisfies

$$\mu(f)=\sum_{j=1}^{n}d_{j}\mu(e_{j}),$$

where $d_{j}\geq 0$ is the number of distinct descendance paths from $e_{j}$ to $f$. Another way of formulating this is to say that

$$\mu=\sum_{j=1}^{n}\mu(e_{j})\mu_{j},$$

where each $\mu_{j}$ is itself a rigid honeycomb (see [2, Section 3]). The summands $\mu_{j}$ belong to extreme rays in the cone $\mathcal{M}$. Extreme rigid honeycombs are characterized by the fact that they have a unique equivalence class of root edges. An extreme rigid honeycomb is a tree honeycomb precisely when the multiplicity assigned to its root edges is equal to $1$. Alternatively, rigid tree honeycombs are obtained as immersions of a special kind of tree that we now define.

The trees relevant to the construction of rigid honeycombs are finite trees whose edges are labeled $1,2,$ or $3$, and the following conditions are satisfied:

1. (1)

   each vertex has order $1$ or $3$, and

2. (2)

   the three edges adjacent to a vertex of order $3$ have distinct labels.

An immersion of a tree is a map $f$ that associates to each vertex $v$ of order 3 of a tree a point $f(v)$ in the plane such that the following conditions are satisfied:

1. (i)

   if $a$ and $b$ are joined by an edge labeled $j$, then $f(a)\neq f(b)$ and the segment $f(a)f(b)$ is parallel to $w_{j}$, and

2. (ii)

   if $xa,xb$, and $xc$ are three edges adjacent to a vertex $x$ then the angle between any two of the segments $f(x)f(a),f(x)f(b)$, and $f(x)f(c)$ equals $120^{\circ}$.

Given a tree $T$ that has at least two vertices of order $3$, and an immersion $f$ of $T$, we define a honeycomb $\mu_{f}$ as follows. Let $ab$ be an edge of $T$. If $a$ and $b$ have order $3$, we add the segment $f(a)f(b)$ to the support of $\mu_{f}$ and we add $1$ to the multiplicity of this segment. If $a$ is of order $1$, $ab$ is labeled $j$, and $c,d$ are the other vertices connected to $b$, we add to the support of $\mu_{f}$ a ray $h$ with endpoint $f(b)$ and parallel to $w_{j}$ and we add $1$ to the multiplicity of this ray. The ray is chosen such that the angles between any two of $h,f(b)f(c)$, and $f(b)f(d)$ equal $120^{\circ}$. The segments $f(a)f(b)$ and the rays constructed above can be viewed as images under the immersion $f$ of the edges of $T$. The honeycomb $\mu_{f}$ is simply determined by adding all of the multiplicities just defined. If $T$ has exactly one vertex of order $3$, there are two possible honeycombs associated to an immersion $f$, one in $\mathcal{M}$ and the other in $\mathcal{M}_{*}$. The honeycombs obtained from immersions of trees are called tree honeycombs.

Not every tree honeycomb is rigid. Figure 2.7 represents a tree $T$ and two immersions of $T$, the second of which yields a rigid honeycomb.

Suppose that $\mu\in\mathcal{M}$ is a rigid honeycomb with a unique equivalence class of root edges, all of which have multiplicity $1$. We construct a tree $T_{0}$ as follows. Fix a root edge $e_{0}=XY$ of $\mu$. The vertices of $T_{0}$ are $x,y$, and the descendance paths $e_{0}\dots e_{n}$. Given such a descendance path, the vertices $v=e_{0}\dots e_{n-1}$ and $v^{\prime}=e_{0}\dots e_{n}$ are joined by an edge of $T_{0}$ labeled $j$ if $e_{n}$ is parallel to $w_{j}$. In addition, $x$ is joined to $y$ by an edge labeled $j$ if $XY$ is parallel to $w_{j}$, $x$ is joined to $e_{0}e_{1}$ if $X$ is a an endpoint of $e_{1}$, and $y$ is joined to $e_{0}e_{1}$ if $Y$ an endpoint of $e_{1}$. The same labeling rule applies to the last two kinds of edges. The tree $T_{0}$ obtained this way may have some vertices of order $2$. To obtain a tree $T$ satisfying condition (1) we simply remove the vertices of order $2$ as follows: suppose that $v_{0}v_{1}\dots v_{n}$ is a path in $T_{0}$ such that $v_{0}$ and $v_{n}$ have order $1$ or $3$ and $v_{1},\dots,v_{n-1}$ have order $2$. Then $v_{1},\dots,v_{n-1}$ are removed from $T_{0}$ and an edge joining $v_{0}$ and $v_{n}$, carrying the label of $v_{0}v_{1}$, is added instead. Clearly, we have $\mu=\mu_{f}$ for some immersion $f$ of $T$. The construction of $T_{0}$ and $T$ is illustrated in Figure 2.8 for a particular rigid tree honeycomb.

Suppose that $T$ is a tree, that $f$ is an immersion of $T$, and that $\mu=\mu_{f}$ belongs to $\mathcal{M}$. Suppose also that there exists an edge $xy$ of $T$ such that $\mu$ assigns unit multiplicity to part of $f(x)f(y)$. Orient the edges of $T$ other than $xy$ away from $xy$. We say that the immersion $f$ is coherent if the following condition is satisfied: if a segment $I$ is contained in the images under $f$ of two different edges of $T$, then the two orientations induced on $I$ are the same. The following result is [4, Theorem 4.1].

Having obtained a description of rigid tree honeycombs, it is important to determine which linear combinations $c_{1}\mu_{1}+\cdots+c_{n}\mu_{n}$, $c_{1},\dots,c_{n}>0$, of rigid tree honeycombs are rigid. Such combinations are called rigid overlays and their description involves a bilinear map [3] defined on honeycombs in $\mathcal{M}$ as follows: given $\mu,\nu\in\mathcal{M}$, we set

(2.6)

$$\Sigma_{\nu}(\mu)=\sum_{c<d}[\mu^{(1)}(c)\nu^{(1)}(d)+\mu^{(2)}(c)\nu^{(2)}(d)+\mu^{(3)}(c)\nu^{(3)}(d)]-\omega(\mu)\omega(\nu),$$

where $\mu^{(j)}(c)$ is as defined above (2.2). It was seen in [3, end of Section 2] that

(2.7)

$$\Sigma_{\mu}(\mu)=\frac{1}{2}\left[\omega(\mu)^{2}-\sum_{c\in\mathbb{R}}[\mu^{(1)}(c)^{2}+\mu^{(2)}(c)^{2}+\mu^{(3)}(c)^{2}]\right].$$

The following result is [3, Theorem 4.2].

It is possible that $\Sigma_{\nu}(\mu)>0$ and $\Sigma_{\mu}(\nu)>0$ for two rigid tree honeycombs $\mu$ and $\nu$, see Figure 2.10.

It follows from (2.7) that the second assertion in Theorem 2.3 can be written as

(2.8)

$$\sum_{c\in\mathbb{R}}[\mu^{(1)}(c)^{2}+\mu^{(2)}(c)^{2}+\mu^{(3)}(c)^{2}]=\omega(\mu)^{2}+2.$$

The following characterization of rigid overlays is [3, Corollary 4.5] (see also [2, Theorem 3.8]).

Suppose that $\mu\in\mathcal{M}$. The puzzle of $\mu$ is obtained by a process of inflation that we describe next. Cut the plane along the edges of $\mu$ to obtain a collection of white puzzle pieces and translate these pieces away from each other in such a way that, given an edge $AB$ of $\mu$, the parallelogram formed by the two translates of $AB$ has two sides of length $\mu(AB)$ that are $60^{\circ}$ clockwise from $AB$. We call this parallelogram the inflation of $AB$ and we color it dark gray. The balance condition for $\mu$ implies that these white and dark gray pieces fit together and leave a space corresponding to each branch point of $\mu$. These extra spaces are colored light gray. The light gray piece associated to a branch point $X$ of $\mu$ is a convex polygon with as many sides as there are edges incident to $X$. One can recover the honeycomb $\mu$, up to a translation, from its puzzle by deflation. This process consists simply of moving the white pieces together and replacing each dark gray parallelogram (or half-infinite strip) by a segment parallel to its white sides and with multiplicity equal to the length of its light gray side. There are similar processes of $*$-inflation and $*$-deflation which are the same as inflation and deflation except that ‘clockwise’ is replaced by ‘counterclockwise’ in the construction of the $*$-inflation.

Starting with a nonzero honeycomb $\mu\in\mathcal{M}$, one can construct the puzzle of $\mu$ and apply $*$-deflation to this puzzle. The only difficulty is that the segments arising from dark gray infinite strips would be assigned infinite multiplicity. To overcome this difficulty, choose a closed equilateral triangle $\Delta$ containing all the branch points of $\mu$ whose sides are $60^{\circ}$ clockwise from those rays in the support of $\mu$ that they intersect. Apply now the process of inflation only to the part of the support of $\mu$ that is contained in $\Delta$ and apply $*$-deflation to the resulting puzzle. The result of these operations is the part of a honeycomb $\mu_{\Delta}^{*}\in\mathcal{M}_{*}$ contained in an equilateral triangle $\Delta_{*}$ with sides equal to $\omega(\mu)$. (A simple example is illustrated in Figure 3.1.)

Choosing a different triangle $\Delta$ changes the multiplicities of $\mu_{\Delta}^{*}$ only on the boundary of $\Delta_{*}$. Denote by $\Delta_{0}$ the smallest triangle satisfying the requirements above. We define $\mu^{*}$ to be equal to $\mu_{\Delta_{0}}^{*}$. The honeycomb $\mu^{*}$ has the property that each side of $\Delta_{*}$ contains a segment with zero multiplicity.

In general, we observe that the sides of $\Delta_{*}$ are divided by the rays in the support of $\mu^{*}$ into segments with lengths equal to the exit multiplicities of $\mu$. (The triangle $\Delta_{*}$ is partitioned by the support of $\mu^{*}$ into light gray pieces that form the degeneracy graph of $\mu$ introduced in [8].) If $\mu$ has only one branch point, we define $\mu^{*}=0$.

It is easily seen that a honeycomb $\mu\in\mathcal{M}$ is rigid if and only if $\mu^{*}\in\mathcal{M}_{*}$ is rigid (see [2, p. 1591]). Given a rigid honeycomb $\mu\in\mathcal{M}$, we denote by ${\rm exit}(\mu)$ the number of nonzero exit multiplicities of $\mu$, and we denote by ${\rm root}(\mu)$ the number of equivalence classes of root edges in the support of $\mu$. As noted earlier, $\mu$ is a linear combination with positive coefficients of ${\rm root}(\mu)$ distinct rigid tree honeycombs. The following result is equivalent to [1, Theorem 1.1]. (To see the equivalence, observe that the number of attachment points of $\mu$ relative to the minimal triangle $\Delta_{0}$ considered above is equal to ${\rm exit}(\mu)-3$.)

In order to study the exit multiplicities of a rigid tree honeycomb, we require an auxiliary result of convex analysis.

Fix a positive number $r$ and consider an equilateral triangle $\Delta=ABC$ with side length $r$ and sides parallel to $w_{1},w_{2},$ and $w_{3}$. For simplicity, suppose that $A,B,$ and $C$ are arranged counterclockwise around $\Delta$ and $AB$ is parallel to $w_{1}$.

A pattern consists of a sequence $A_{0}=A,A_{1},\dots,A_{k_{1}-1},A_{k_{1}}=B=B_{0},B_{1},\dots,B_{k_{2}-1},B_{k_{2}}=C=C_{0},C_{1},\dots,C_{k_{3}-1},C_{k_{3}}=A$ of points arranged counterclockwise on the boundary of $\Delta$; see Figure 3.2. Alternatively, a pattern is determined by the distances $\alpha_{1}^{(1)},\dots,\alpha_{k_{1}}^{(1)}|\alpha_{1}^{(2)},\dots,\alpha_{k_{2}}^{(2)}|\alpha_{1}^{(3)},\dots,\alpha_{k_{3}}^{(3)}$ defined by

$$\alpha_{j}^{(1)}=\text{length}(A_{j-1},A_{j}),\ \alpha_{j}^{(2)}=\text{length}(B_{j-1},B_{j}),\ \alpha_{j}^{(3)}=\text{length}(C_{j-1},C_{j}).$$

These positive numbers are subject to

$$\sum_{j=1}^{k_{1}}\alpha_{j}^{(1)}=\sum_{j=1}^{k_{2}}\alpha_{j}^{(2)}=\sum_{j=1}^{k_{3}}\alpha_{j}^{(3)}=r.$$

It is easy to see that $\mathcal{M}$ can be replaced by $\mathcal{M}_{*}$ in the above definition. Moreover, the above condition need only be satisfied for honeycombs $\mu\in\mathcal{M}$ with rays intersecting $\partial\Delta$ in a subset of $\{A_{j}\}_{j=0}^{k_{1}-1}\cup\{B_{j}\}_{j=0}^{k_{2}-1}\cup\{C_{j}\}_{j=0}^{k_{3}-1}$; such honeycombs are allowed to have a ray meeting $\partial\Delta$ in $A_{0}(=C_{k_{3}})$ that is parallel to the other rays passing through some $A_{j}$ with $j\neq 0$ but not such a ray that is parallel to the ones passing through some $C_{j}$ for $j\neq k_{3}.$ A similar observation (with $A_{0},B_{0},C_{0}$ in place of $A_{k_{1}},B_{k_{2}},C_{k_{3}}$) applies to honeycombs in $\mathcal{M}_{*}$. It is also obvious that a pattern that has a locking refinement must itself be locking.

Suppose now that $\mu\in\mathcal{M}$ and the only nonzero exit multiplicities of $\mu$ are $\alpha_{j}^{(1)}=\mu^{(1)}(c_{j}^{(1)})$, $\alpha_{j}^{(2)}=\mu^{(2)}(c_{j}^{(2)})$, and $\alpha_{j}^{(3)}=\mu^{(3)}(c_{j}^{3})$, where $c_{1}^{(1)}<\cdots<c_{k_{1}}^{(1)}$, $c_{1}^{(2)}<\cdots<c_{k_{2}}^{(2)}$, and $c_{1}^{(3)}<\cdots<c_{k_{3}}^{(3)}$. These numbers define a pattern with $r=\omega(\mu)$ that we call the exit pattern of $\mu$.

The multiplicity pattern of a rigid tree honeycomb has an additional property that helps determine whether a given pattern arises from such a honeycomb. Suppose that an arbitrary pattern $A_{0},\dots,A_{k_{1}},B_{0},\dots,B_{k_{2}},C_{0},\dots C_{k_{3}}$ is given on a triangle $\Delta$ with sides parallel to $w_{1},w_{2},$ and $w_{3}$. A convex region $\Omega\subset\Delta$ is said to be flat relative to this pattern if every honeycomb $\mu\in\mathcal{M}$, whose rays intersect $\partial\Delta$ in a subset of $\{A_{j},B_{j},C_{j}\}$, assigns zero multiplicity to all segments contained in the interior of $\Omega$.