Some results on multithreshold graphs

TL;DR

This paper investigates the structure of multithreshold graphs, proving that for three thresholds, different threshold choices define infinitely many distinct graph classes, thus answering a longstanding open question.

Contribution

The paper demonstrates that for three thresholds, the classes of graphs are infinitely many and distinct, resolving Jamison's question about the equivalence of threshold choices.

Findings

Infinitely many distinct classes of 3-threshold graphs exist.

Different threshold choices can define different graph classes for k ≥ 3.

Some open problems in multithreshold graphs are identified.

Abstract

Jamison and Sprague defined a graph $G$ to be a $k$-threshold graph with thresholds $\theta_1 , \ldots, \theta_k$ (strictly increasing) if one can assign real numbers $(r_v)_{v \in V(G)}$, called ranks, such that for every pair of vertices $v,w$, we have $vw \in E(G)$ if and only if the inequality $\theta_i \leq r_v + r_w$ holds for an odd number of indices $i$. When $k=1$ or $k=2$, the precise choice of thresholds $\theta_1, \ldots, \theta_k$ does not matter, as a suitable transformation of the ranks transforms a representation with one choice of thresholds into a representation with any other choice of thresholds. Jamison asked whether this remained true for $k \geq 3$ or whether different thresholds define different classes of graphs for such $k$, offering \$50 for a solution of the problem. Letting $C_t$ for $t > 1$ denote the class of $3$-threshold graphs with thresholds $-1, 1, t$, we prove that there are infinitely many distinct classes $C_t$, answering Jamison's question. We also consider some other problems on multithreshold graphs, some of which remain open.

Equations

$$a_k+b_j\ge a_j+b_j\ge -1,$$

$$a_k+b_j\le a_k+b_k<1,$$

$$a_j+b_k\ge a_k+b_k\ge -1,$$

$$a_k+b_j\le a_k+b_k<1,$$

$$(b_j-a_j)-(b_{j-1}-a_{j-1})=(a_{j-1}+b_j)-(a_j+b_{j-1}),$$

$$b_j-a_j\ge (b_{j-1}-a_{j-1})+2\ge 2(j-2)+2=2(j-1).$$

$$b_j-a_j$$

$$b_j+a_j$$

$${\theta }_{\alpha }\le x_1+x_2\le x_1+y_2\le y_1+y_2<{\theta }_{\alpha +1},$$

$${\theta }_{\beta }\le y_2+y_3\le y_2+x_3<x_2+x_3<{\theta }_{\beta +1},$$

$${\theta }_{\gamma }\le x_1+x_3\le y_1+x_3<y_2+y_3<{\theta }_{\gamma +1},$$

Full text

1. Introduction

Multithreshold graphs were introduced by Jamison and Sprague [3] as a generalization of the well-studied threshold graphs, first introduced by Chvátal and Hammer [1]. Given real numbers $\theta_{1},\ldots,\theta_{k}$ with $\theta_{1}<\theta_{2}<\cdots<\theta_{k}$, we say that a simple graph $G$ is a $k$-threshold graph with thresholds $\theta_{1},\ldots,\theta_{k}$ if there exist real numbers $(r_{v})_{v\in V(G)}$, called ranks, such that for every pair of distinct vertices $v,w\in V(G)$, we have $vw\in E(G)$ if and only if the inequality $\theta_{i}\leq r_{v}+r_{w}$ holds for an odd number of indices $i$. (Equivalently, adopting the convention that $\theta_{k+1}=\infty$, we want $vw\in E(G)$ if and only if $r_{v}+r_{w}\in[\theta_{2i-1},\theta_{2i})$ for some $i$.) In this case, we call $r$ a $(\theta_{1},\ldots,\theta_{k})$-representation of $G$.

We will abbreviate this notation by saying that $G$ is $(\theta_{1},\ldots,\theta_{k})$-threshold to mean that $G$ is $k$-threshold with thresholds $\theta_{1},\ldots,\theta_{k}$. When $k=1$, we obtain the classical threshold graphs.

In the case of the classical threshold graphs, it is clear that the exact choice of threshold does not matter: by appropriately rescaling the vertex ranks, any $\theta$-threshold graph is seen to also be a $\theta^{\prime}$-threshold graph. The same observation holds for $k=2$: any ranks witnessing that $G$ is $(\theta_{1},\theta_{2})$-threshold can be transformed, via an appropriate affine transformation, into ranks witnessing that $G$ is $(\theta^{\prime}_{1},\theta^{\prime}_{2})$-threshold.

At the 2019 Spring Sectional AMS Meeting in Auburn, Jamison asked whether this phenomenon continues for higher $k$, and specifically whether it still holds when $k=3$. Observing that an affine transformation of the vertex ranks still uses up two “degrees of freedom” and let us express any $(\theta_{1},\theta_{2},\theta_{3})$-threshold graph as a $(-1,1,t)$-threshold graph for some $t$, his question can be phrased as follows.

Question 1* (Jamison).*

Do there exist real numbers $t,t^{\prime}>1$ such that the class of $(-1,1,t)$-threshold graphs and the class of $(-1,1,t^{\prime})$-threshold graphs differ?

Jamison offered a $50 bounty for an answer to this question. In this paper, we answer the question in the affirmative: letting$C_{t}$denote the class of$(-1,1,t)-threshold graphs, we prove in Section [2](#S2) that there are infinitely many distinct classes C_{t}$.

We also study some other questions involving multithreshold graphs. Say that $G$ is a $k$-threshold graph if there exist real numbers $\theta_{1}<\cdots<\theta_{k}$ such that $G$ is a $(\theta_{1},\ldots,\theta_{k})$-threshold graph. Jamison and Sprague [3] proved that for every graph $G$, there is some $k$ such that $G$ is a $k$-threshold graph. Thus, we may define the threshold number $\Theta(G)$ of a graph $G$ to be the smallest nonnegative $k$ such that $G$ is a $k$-threshold graph.

It is natural to compare the parameter $\Theta(G)$ to other graph parameters involving threshold graphs. Cozzens and Leibowitz [2] define the threshold dimension $t(G)$ of a graph $G$ to be the smallest nonnegative integer $k$ such that $G$ can be expressed as the union of $k$ threshold graphs. Since the complement of a threshold graph is a threshold graph, we can also view $t(\overline{G})$ as the smallest nonnegative $k$ such that $G$ can be expressed as the intersection of $k$ threshold graphs.

Doignon observed, in a personal communication with the authors of [4], that any $2$-threshold graph is the intersection of two threshold graphs, hence $t(\overline{G})\leq 2$ whenever $\Theta(G)\leq 2$. This observation suggests a possible converse:

Question 2* (Jamison).*

Replacing $t(\overline{G})$ with $t(G)$, does $\Theta(G)\leq 2$ imply any bound on $t(G)$?

Question 3* (Jamison).*

Is $\Theta(G)$ bounded by any function of $t(\overline{G})$ or of $t(G)$?

Question 2 has a brief answer. For any graph $G$ and positive integer $p$, let $pG$ be the disjoint union of $p$ copies of $G$. The graph $pK_{2}$ evidently has $t(G)=p$, since $2K_{2}$ is a forbidden induced subgraph for a threshold graph; on the other hand, $pK_{2}$ is a $(-1,1)$-threshold graph, as witnessed by giving the endpoints $u_{i}$ and $v_{i}$ of the $i$th edge ranks $r(u_{i})=-2i$ and $r(v_{i})=2i$. Hence there are graphs with $\Theta(G)=2$ for which $t(G)$ is arbitrarily large.

In Section 3, we partially answer by proving that there are graphs with $t(\overline{G})=3$ for which $\Theta(G)$ is arbitrarily large. Finally, in Section 4, we discuss some remaining open problems about multithreshold graphs, along the lines of the questions considered in this paper.

2. Distinct families of $(-1,1,t)$-threshold graphs

To facilitate proofs about multithreshold graphs, we introduce some notational conventions. Given a multithreshold representation of a graph $G$, the weight of an edge or non-edge $uv$ is the sum of the ranks of $u$ and $v$. When it is understood which multithreshold representation we are working with, we will omit the function $r$ and simply write $v$ to stand for the rank of the vertex $v$. (Hence, the weight of an edge $uv$ will simply be written as $u+v$.)

For positive integers $p$, let $G_{p}=pK_{2}$.

Lemma 1**.**

For any $p\geq 2$ and any $t>2p-3$, the graph $G_{p}$ is a $(-1,1,t)$-threshold graph.

Proof.

Write $t=(1+2\epsilon)(2p-3)$ with $\epsilon>0$. Letting $a_{i},b_{i}$ be the endpoints of the $i$th edge for $i=1,\ldots,p$, observe that the following ranks yield a $(-1,1,t)$-threshold representation of $G_{p}$:

• •

$a_{i}=-(1+\epsilon)(i-1)$ for $i=1,\ldots,p$,

• •

$b_{i}=(1+\epsilon)(i-1)$ for $i=1,\ldots,p$.

Evidently $a_{i}+b_{i}=0$ for every edge $a_{i}b_{i}$. On the other hand, any nonadjacent pair of vertices has a weight whose absolute value is at least $1+\epsilon$, hence does not fall into the interval $[-1,1)$, and whose value is at most $(1+\epsilon)(p-1)+(1+\epsilon)(p-2)=(1+\epsilon)(2p-3)<t$, hence does not fall into the interval $[t,\infty)$. Hence, this is a $(-1,1,t)$-representation of $G$. ∎

Computational experiments suggest that this bound is sharp: that $G_{p}$ is not $(-1,1,t)$-threshold for any $t\leq 2p-3$. Lacking a formal proof of this sharpness, we prove a weaker statement.

Lemma 2**.**

For integer $p\geq 4$, if $G_{p}$ is a $(-1,1,t)$-threshold graph then $t>2p-5$.

Proof.

View the edges whose weight lies in $[-1,1)$ as colored red and view the edges whose weight lies in $[t,\infty)$ as colored yellow. Since the yellow edges form a threshold graph and $2K_{2}$ is a forbidden induced subgraph for threshold graphs, there is at most one yellow edge in $G_{p}$. Let $a_{1}b_{1}$, $a_{2}b_{2}$, …, $a_{p}b_{p}$ be the edges of $G_{p}$.

By symmetry, we may assume that $a_{i}\leq b_{i}$ for each $i$ and that $b_{1}\leq\cdots\leq b_{p}$. This implies that $b_{p}$ has the largest rank of all vertices and, thus, if there is a yellow edge, then that edge is $a_{p}b_{p}$.

Let $q=p$ if $a_{p}b_{p}$ is red, and otherwise let $q=p-1$, so that all edges $a_{1}b_{1},\ldots,a_{q}b_{q}$ are red.

Claim 1: $a_{k}<a_{j}$ whenever $j<k\leq q$. If not, then there exist $j<k$ with $a_{k}\geq a_{j}$ and $b_{j}\leq b_{k}$. Hence

[TABLE]

and

[TABLE]

which contradicts the fact that the edge $a_{k}b_{j}$ is absent.

It follows that the intervals $[a_{i},b_{i}]$ are nested, with $[a_{1},b_{1}]\subset[a_{2},b_{2}]\subset\cdots\subset[a_{q},b_{q}]$.

Claim 2: $a_{j}+b_{k}\geq 1$ and $a_{k}+b_{j}<-1$ whenever $j<k\leq q$. Using the previous claim, we have

[TABLE]

hence $a_{j}+b_{k}\geq 1$ since otherwise the edge $a_{j}b_{k}$ should be present. Similarly, since $b_{j}\leq b_{k}$, we have

[TABLE]

hence $a_{k}+b_{j}<-1$ since otherwise the edge $a_{k}b_{j}$ should be present.

Claim 3: $b_{j}-a_{j}\geq 2(j-1)$ for all $j\in[q]$. We prove this by induction on $j$. When $j=1$ this is just the assumption that $b_{j}\geq a_{j}$. Assuming it holds for $j-1$, we prove that it holds for $j$. Observe that

[TABLE]

and by the previous claim we have $a_{j-1}+b_{j}\geq 1$ and $a_{j}+b_{j-1}\leq-1$, so that

[TABLE]

Claim 4: $b_{j}\geq j-3/2$ for all $j\in[q]$. This follows immediately from the inequalities

[TABLE]

Having established these claims, we now complete the proof. If $q=p$, then Claim 4 gives $b_{p-1}\geq p-5/2$ and $b_{p}\geq p-3/2$, so to avoid the unwanted edge $b_{p-1}b_{p}$, it is necessary that $b_{p-1}+b_{p}<t$, which requires $(p-5/2)+(p-3/2)=2p-4<t$.

If $q=p-1$, then Claim 4 gives $b_{p-1}\geq p-5/2$, and since $a_{p}+b_{p}\geq t$ with $a_{p}\leq b_{p}$, we have $b_{p}\geq t/2$. Hence, to avoid the unwanted edge $b_{p-1}b_{p}$ it is necessary that $(p-5/2)+t/2<t$, which implies $2p-5<t$. ∎

Corollary 3**.**

For each $k\geq 3$, the graph $G_{2^{k}}$ is $(-1,1,2^{k+1})$-threshold but not $(-1,1,t)$-threshold for any $t\leq 2^{k}$.

Corollary 4**.**

The classes $C_{2^{k}}$ for $k\geq 3$ are pairwise distinct.

Corollary 5**.**

For all $t>1$, there exist $2$-threshold graphs that are not $(-1,1,t)$-threshold graphs.

3. Threshold number versus threshold dimension

In this section, we partially answer Question 3 by proving that there exist graphs with $t(\overline{G})=3$ for which $\Theta(G)$ is arbitrarily large. We will require the following result of Cozzens and Leibowitz [2] concerning the threshold dimension of complete multipartite graphs:

Theorem 6** (Cozzens–Leibowitz [2]).**

For positive integers $m_{1}\leq\cdots\leq m_{p}$, the complete $p$-partite graph $K_{m_{1},\ldots m_{p}}$ has threshold dimension $t(K_{m_{1},\ldots,m_{p}})=m_{p-1}$.

Let $G=pK_{3}$. Applying Theorem 6 with all $m_{i}=3$ shows that $t(\overline{G})=3$. Therefore, to show that $\Theta(G)$ can be arbitrarily large for graphs with $t(\overline{G})=3$, it suffices to prove the following theorem.

Theorem 7**.**

If $pK_{3}$ is a $k$-threshold graph, then $p\leq{k+2\choose 3}$. In particular, $\Theta(pK_{3})\geq\frac{1}{2}(6p)^{1/3}$.

Note that the first part of the theorem is stronger than the second part, which is obtained using a crude lower bound on ${k+2\choose 3}$.

To prove this theorem, we will use a lemma stated in terms of edge colorings (not necessarily proper) induced by a threshold representation. Given a $(\theta_{1},\ldots,\theta_{k})$-representation of $pK_{3}$, we assign colors $1,\ldots,\left\lfloor k/2\right\rfloor$ to the edges of $pK_{3}$ by giving edge $e$ color $i$ if its weight lies in the interval $[\theta_{2i-1},\theta_{2i})$. (By the definition of a $(\theta_{1},\ldots,\theta_{k})$-representation, for each edge there is exactly one such $i$.)

Now, given an edge coloring, we can view each triangle as inducing a multiset of colors on its edges (for example, we consider “$2$ red edges and $1$ yellow edge” and “$1$ red edge and $2$ yellow edges” as different multisets, despite having the same underlying set).

Lemma 8**.**

In a $(\theta_{1},\ldots,\theta_{k})$-representation of $pK_{3}$, no two triangles have the same multiset of colors appearing on their edges.

Before proving the lemma, we show how the proof of the theorem follows immediately.

Proof of Theorem 7.

When $p=1$, both parts of the theorem clearly hold, since $k\geq 1$ is required. For $p\geq 2$, observe that $pK_{3}$ has an induced $2K_{2}$ and thus is not a threshold graph; thus, we may assume $k\geq 2$.

Assume $pK_{3}$ has a $(\theta_{1},\ldots,\theta_{k})$-representation. By Lemma 8, no two triangles have the same multiset of colors on their edges. Hence, by the pigeonhole principle, the number of triangles is at most the number of size-$3$ multisets from $\{1,\ldots,k\}$, which by the standard stars-and-bars argument is ${k+2\choose 3}$. Since $k\geq 2$, we have ${k+2\choose 3}\leq{2k\choose 3}\leq(2k)^{3}/6$. Rearranging $p\leq(2k)^{3}/6$ gives the desired inequality. ∎

Proof of Lemma 8.

Suppose to the contrary that two triangles $x_{1},x_{2},x_{3}$ and $y_{1},y_{2},y_{3}$ have the same multiset of colors on their edges. Without loss of generality, we may assume that:

• •

$x_{1}\leq x_{2}\leq x_{3}$,

• •

$y_{1}\leq y_{2}\leq y_{3}$, and

• •

$x_{1}\leq y_{1}$.

Choose indices $\alpha,\beta,\gamma\in\{1,\ldots,k\}$ so that $x_{1}x_{2}$ has weight in $[\theta_{\alpha},\theta_{\alpha+1})$, $x_{1}x_{2}$ has weight in $[\theta_{\beta},\theta_{\beta+1})$, and $x_{2}x_{3}$ has weight in $[\theta_{\gamma},\theta_{\gamma+1})$. (For convenience, we will adopt the convention that $\theta_{k+1}=\infty$.) Observe that $x_{1}\leq x_{2}\leq x_{3}$ forces $\alpha\leq\beta\leq\gamma$.

Say an edge with weight in $[\theta_{\alpha},\theta_{\alpha+1})$ is red, an edge with weight in $[\theta_{\beta},\theta_{\beta+1})$ is yellow, and an edge with weight in $[\theta_{\gamma},\theta_{\gamma+1})$ is pink. (It is possible that some of these thresholds coincide, in which case an edge may be, say, both red and yellow.)

Since $y_{1}\leq y_{2}\leq y_{3}$ and the $y$-edges have the same multiset of colors as the $x$-edges, the colors of the $y$-edges must agree with the colors of the corresponding $x$-edges:

• •

$x_{1}x_{2}$ and $y_{1}y_{2}$ are red,

• •

$x_{1}x_{3}$ and $y_{1}y_{3}$ are yellow,

• •

$x_{2}x_{3}$ and $y_{2}y_{3}$ are pink.

Now we will derive our contradiction using the absence of the $x_{i}y_{j}$-edges.

Claim 1: $y_{2}<x_{2}$. If instead $x_{2}\leq y_{2}$, then we have

[TABLE]

forcing a red $x_{1}y_{2}$-edge, a contradiction.

Claim 2: $x_{3}<y_{3}$. If instead $y_{3}\leq x_{3}$, then since $y_{2}<x_{2}$, we have

[TABLE]

forcing a pink $y_{2}x_{3}$-edge, a contradiction.

Now since $x_{3}<y_{3}$, we have

[TABLE]

forcing a yellow $y_{1}x_{3}$-edge, again a contradiction. This completes the proof. ∎

4. Remarks and Open Questions

After being informed of a preliminary version of the results in Section 2, Jamison (personal communication) suggested studying the class $D=\bigcap_{t>1}C_{t}$, where $C_{t}$ is the class of $(-1,1,t)$-threshold graphs.

Intuition suggests that perhaps $D$ is related somehow to the class of $2$-threshold graphs. Since $G_{p}$ is a $2$-threshold graph for all $p$, Lemma 2 implies that not all $2$-threshold graphs lie in the class $D$. On the other hand, since all $2$-threshold graphs satisfy $t(\overline{G})\leq 2$, and since Theorem 6 implies that $t(\overline{2K_{3}})=3$, we see that $2K_{3}$ is not a $2$-threshold graph; however, $2K_{3}\in D$, as the ranking in Figure 1 can easily be verified to be a $(-1,1,t)$-representation for $2K_{3}$ whenever $\epsilon$ is sufficiently small (in terms of $t$). Thus, $2K_{3}\in D$ but $2K_{3}$ is not $2$-threshold; the two classes are incomparable.

Open Question 1*.*

Is there a nice characterization of the class $D$?

While the results in Section 4 imply that there are at least countably many distinct classes $C_{t}$, it is not clear whether there are countably many distinct classes or uncountably many distinct classes. Indeed, it seems plausible that $C_{t}\neq C_{t^{\prime}}$ whenever $t,t^{\prime}$ are distinct real numbers exceeding $1$.

Open Question 2*.*

Are there uncountably many distinct classes $C_{t}$?

Open Question 3*.*

Are there distinct real numbers $t,t^{\prime}>1$ such that $C_{t}=C_{t^{\prime}}$?

Theorem 7 only partially answers Question 3, which seeks a bound of $\Theta(G)$ in terms of $t(\overline{G})$ or $t(G)$. In particular, the following questions remain open:

Open Question 4*.*

Is $\Theta(G)$ bounded on the class of graphs $G$ with $t(\overline{G})=2$?

Open Question 5*.*

Is $\Theta(G)$ bounded by any function of $t(G)$?

5. Acknowledgments

I thank Robert E. Jamison both for posing the original Question 1 that motivated this paper as well as the thought-provoking followup questions 2 and 3 after receiving a preliminary version of these results.