An exact upper bound on the size of minimal clique covers

TL;DR

This paper determines the exact maximum size of the minimal clique cover for any graph with given vertices and edges, which relates to the minimal alphabet size in indeterminate string representations.

Contribution

It provides an exact upper bound on the size of minimal clique covers for graphs with specified vertices and edges, addressing a problem from prior research.

Findings

Computed the exact maximum minimal clique cover size for given graph parameters

Established a precise upper bound related to graph structure and clique covers

Linked the bound to minimal alphabet size in indeterminate string representations

Abstract

Indeterminate strings have received considerable attention in the recent past; see for example Christodoulakis et al 2015 and Helling et al 2017. This attention is due to their applicability in bioinformatics, and to the natural correspondence with undirected graphs. One aspect of this correspondence is the fact that the minimal alphabet size of indeterminates representing any given undirected graph corresponds to the size of the minimal clique cover of this graph. This paper solves a related problem proposed in Helling et al 2017: compute $\Theta_n(m)$, which is the size of the largest possible minimal clique cover (i.e., an exact upper bound), and hence alphabet size of the corresponding indeterminate, of any graph on $n$ vertices and $m$ edges.

Equations

$$\hat{m}=m-\left(\genfrac{}{}{0ex}{}{s(G)}{2}\right)-s(G)(c(G)-s(G))=m-\frac{s(G)(2c(G)-s(G)-1)}{2}.$$

$${\Theta }_n(m)=$$

Full text

1 Introduction

Given an undirected graph $G=(V,E)$, we say that $C=\{C_{1},C_{2},\ldots,C_{k}\}$ is a clique cover of $G$ of size $k$ if each $C_{i}$ is a set of vertices comprising a clique, and $\cup C=V$, and furthermore, given any edge $(u,v)\in E$, there is a $C_{i}$ that contains both $u$ and $v$. We denote $\theta(G)$ as the size of the smallest clique cover of $G$ ([6]).

Let $\mathcal{G}_{n}(m)$ be the set of all undirected graphs on $n$ vertices and $m$ edges; of course, $0\leq m\leq{n\choose 2}$. Let $\Theta_{n}(m)$ be the largest possible $\theta(G)$ for $G\in\mathcal{G}_{n}(m)$. In [3, Problem 11] the authors pose the following problem: describe the function $\Theta_{n}(m)$ for every given $n$, and they provide as an example a graph for $\Theta_{7}(m)$, where $m$ ranges over $\{0,1,\ldots,21\}$, $21={7\choose 2}$ (see [3, Fig. 3]). Given the fact that for $n>7$ the number of graphs quickly becomes unwieldy, it is desirable to compute $\Theta_{n}(m)$ analytically, as $\Theta_{n}(m)$ provides an exact bound on the alphabet size of indeterminates obtained from a graph on $n$ vertices and $m$ edges.

We already know from [3] that for each $n$, the global maximum is reached at $m=\lfloor n^{2}/4\rfloor$. The reason for this is that $\lfloor n^{2}/4\rfloor$ is the largest number of edges that can fit in a graph on $n$ vertices without forcing any triangles; note that such a graph is simply a complete bipartite graph. On the other hand, if a graph has no triangles and no singletons, the only possible clique cover for such a graph consists of all the edges (more precisely, the cover consists of all pairs $\{u,v\}$ where $e=(u,v)$ is an edge in the graph).

We aim to characterize $\Theta_{n}(m)$ in our primary result, Theorem 20. We also establish Algorithm 1, which computes $\Theta_{n}(m)$ in linear time. Our motivation comes from [3], where $\Theta_{n}(m)$ would be used as an upper bound for the size of a minimal alphabet for an indeterminate string based on the edge and vertex counts of said string’s corresponding undirected graph. The hope, of course, is that exploration of the structural causes behind the upper bound of $\theta(G)$ will help us to better understand the problem of finding a minimal or near-minimal clique cover, which corresponds directly to finding a small alphabet for an indeterminate string. While the motivation comes from string processing, our results are primarily in extremal graph theory. We’ll apply theorems provided by Mantel [5] (Theorem 1) and Lovász [4] (Theorem 2) to prove that $\Theta_{n}(m)$ has many recursive properties, which we will then use to characterize it.

Theorem 1** (Mantel)**

If a graph on $n$ vertices contains no triangle, then it contains at most $\lfloor n^{2}/4\rfloor$ edges.

The expression $\lfloor n^{2}/4\rfloor$ will be used frequently throughout the paper, and so we abbreviate it as $\overline{n}:=\lfloor n^{2}/4\rfloor$. In general, for any expression exp, we let $\overline{\text{exp}}=\lfloor\text{exp}^{2}/4\rfloor$.

Theorem 2** (Lovász)**

Given $G\in\mathcal{G}_{n}(m)$, let $k$ be the number of missing edges (i.e. $k={n\choose 2}-m$), and let $t$ be the largest natural number such that $t^{2}-t\leq k$. Then $\theta(G)\leq k+t$. Moreover, this bound is exact for $k=t^{2}$ or $k=t^{2}-t$.

Of course, as Lovàsz’s bound relies solely on the number of missing edges, it also relies on the assumption that the vertex and edge counts are arbitrarily large; the bound is exact at the specified values of $k$, as long as that $m\geq\overline{n}$.

Clearly, $t^{2}=\overline{2t}$, and $t^{2}-t=\overline{2t-1}$ (or identically $t^{2}+t=\overline{2t+1}$), so Theorem 2 can be restated:

Given $G\in\mathcal{G}_{n}(m)$, define $k$ as above and let $t$ be the largest natural number such that $\overline{2t-1}\leq k$. $\theta(G)\leq k+t$. Assuming $m\geq\overline{n}$, this bound is sharp if $k=\overline{2t-1}$ or $k=\overline{2t}$. That is, $\Theta_{n}(m)=\overline{2t-1}+t=\overline{2t}$ if $m={n\choose 2}-\overline{2t-1}\geq\overline{n}$ and $\Theta_{n}(m)=\overline{2t}+t=\overline{2t+1}$ if $m={n\choose 2}-\overline{2t}\geq\overline{n}$.

We propose an improvement to Theorem 2 in Conjecture 17. In lieu of proof for all $m$, we provide Lemmas 16, 18 and 19, which prove that Conjecture 17 is true for some $m$. The conjecture reads:

Given $m\geq\overline{n}$, define $k$ as above and let $t$ be the largest natural number such that $\overline{t}\leq k$. $\Theta_{n}(m)=\overline{t+1}$

We use $i(G)$ to denote the number of singletons (or isolated vertices) in $G$, and $c(G)$ to denote the number of non-isolated vertices in $G$. Of course, for any graph $G$ on $n$ vertices we have $i(G)+c(G)=n$. Let $I(G)$ denote the subgraph of $G$ consisting of the isolated points in $G$ ($i(G)=|I(G)|$), and $C(G)$ denote the subgraph of $G$ consisting of all of the non-isolated points in $G$ ($c(G)=|C(G)|$). Let $S(G)$ denote those vertices in $C(G)$ which are connected by an edge to every other vertex in $C(G)$, i.e., $S(G)=\{v\in C(G):\forall u\in C(G)-\{v\},(v,u)\in E\}$. We call such vertices stars. Let $s(G)=|S(G)|$. Finally, $\hat{G}$ will denote the subgraph of $G$ which results from removing all vertices in $S(G)$, along with their edges, but with one exception: if $C(G)$ is a clique, it is simply replaced with a new singleton vertex $v_{C(G)}$ in $\hat{G}$.

As discussed in [3, §2], based on the results of Mantel and Erdös ([5, 2]), $\Theta_{n}(m)$ achieves its global maximum at precisely $\overline{n}$, and $\Theta_{n}(m)$ is non-decreasing for $m\leq\overline{n}$ and non-increasing for $m\geq\overline{n}$, and $\Theta_{n}(\overline{n})=\overline{n}$. This fixed point corresponds to the situation where the number of edges is as large as possible without forcing any triangles, i.e., $m=\overline{n}$, and it is precisely at this point when the best cover can be forced to include all $\overline{n}$ of the edges. See Figure 1 for $\Theta_{8}(m)$, where $\overline{n}=\overline{8}=\lfloor 8^{2}/4\rfloor=16$.

We describe $\Theta_{n}(m)$ in two sections: Section 2 for $m\leq\overline{n}$, which relies primarily on Theorem 1, and Section 3, for $m\geq\overline{n}$, based on Theorem 2. Note that we assume throughout that $n\geq 4$, thus when we say “for all $n$,” we mean “for all $n\geq 4$.”

2 $\Theta_{n}(m)$ for $m\leq\overline{n}$

We prove a sequence of auxiliary results that will help us characterize the graph of $\Theta_{n}(m)$ for $m\leq\overline{n}$. The forthcoming material is rather technical, but the reader will find it easier to follow by keeping the graph in Figure 1 in mind.

Claim 3

$\Theta_{n}(0)=\Theta_{n}(4)=n$* and $\Theta_{n}(1)=\Theta_{n}(2)=\Theta_{n}(3)=n-1$.*

Claim 3 is trivial, as it can be shown very quickly through enumeration of all possible arrangements of 0, 1, 2, 3 and 4 edges.

Claim 4

$\Theta_{n}(m+1)\leq\Theta_{n}(m)+1$.

Proof

Consider $G\in\mathcal{G}_{n}(m+1)$. Choose any edge $e$ in $G$, and remove it (while keeping its end-points) to obtain $G_{0}\in\mathcal{G}_{n}(m)$. Let $C_{0}$ be the smallest clique cover of $G_{0}$, and so $|C_{0}|\leq\Theta_{n}(m)$. Let $C=C_{0}\cup\{e\}$. Since $C_{0}$ covered all of $G$ except for $e$, $C$ by extension covers $G$, and $|C|=|C_{0}|+1\leq\Theta_{n}(m)+1$. We have found a cover for $G$ with cardinality of at most $\Theta_{n}(m)+1$. ∎

Lemma 5

$m\leq\overline{n}\wedge G\in\mathcal{G}_{n}(m)\wedge\theta(G)=\Theta_{n}(m)\implies G$* is triangle-free.*

Proof

We will prove this Lemma by contrapositive. That is, we will show that if $G\in\mathcal{G}_{n}(m)$, $m\leq\overline{n}$, and $G$ contains triangles, then $\theta(G)\neq\Theta_{n}(m)$.

Let $G\in\mathcal{G}_{n}(m)$, and assume $G$ has at least one triangle. 3 edges can be covered with 1 clique, so $\theta(G)\leq m-2+i(G)$.

Case 1: $m>\overline{c(G)}$. We must first note that $i(G)\neq 0$, because if $i(G)=0$ then $c(G)=n$, so $m>\overline{n}$; this directly contradicts the assumptions of this Lemma.

$\theta(G)\leq\overline{c(G)}+i(G)$; because the largest clique cover of any graph on $c(G)$ vertices is $\overline{c(G)}$, we need only add the singletons to this bound to get an upper bound for $\theta(G)$. Note that $c(G)\geq 3$, since $G$ has a triangle. Consider a graph $G_{1}\in\mathcal{G}_{n}(m)$ such that $c(G_{1})=c(G)+1$ and $i(G_{1})=i(G)-1$. Such a graph can be constructed; we have more edges than can fit on $c(G)$ vertices without triangles, so we can simply choose 1 edge $\{u,v\}$ from a triangle in $G$, remove it, and replace it with $\{u,s\}$, where $s$ is a singleton in $G$. We can write a similar bound for this graph: $\theta(G_{1})\leq\overline{c(G_{1})}+i(G_{1})$. Let’s compare the two bounds.

If $c(G)$ is even, then $\overline{(c(G)+1)}-\overline{c(G)}={c(G)/2}$. Since $c(G)$ is even and $\geq 3$, $c(G)\geq 4$. So $\overline{(c(G)+1)}-\overline{c(G)}\geq 2$.

If $c(G)$ is odd, $\overline{(c(G)+1)}-\overline{c(G)}={(c(G)+1)}/{2}$. Given that $c(G)\geq 3$, this difference is, again, $\geq 2$.

As a result, our upper bound for $\theta(G_{1})$ is at least 1 more than that for $\theta(G)$. Of course, this isn’t enough to prove that $\theta(G_{1})\geq\theta(G)$. We can, however, repeat this process to get increasing upper bounds for $\theta(G_{2})$ on $c(G_{1})+1$ non-isolated vertices and $i(G_{1})-1$ singletons, and so on until we reach a $G_{\alpha}$ such that $\overline{c(G_{\alpha})}\geq m$. Note that since $m\leq\overline{n}$, this will necessarily happen before or when we run out of singletons.

If $\alpha=1$, then $\overline{c(G)+1}\geq m$, so we can construct triangle-free $H\in\mathcal{G}_{c(G)+1}(m)$. Let $I$ be a graph of $i(G)-1$ singletons, and let $G^{\prime}=H\cup I$. $G^{\prime}\in\mathcal{G}_{n}(m)$, and $\theta(G^{\prime})=m-1+i(G)$. Recall that $\theta(G)\leq m-2+i(G)$, so $\theta(G^{\prime})>\theta(G)$.

If $\alpha>1$, then we can construct triangle-free $H\in\mathcal{G}_{c(G_{\alpha})}(m)$. Let $I$ be the graph of $i(G_{\alpha})$ singletons, and let $G^{\prime}=H\cup I$. $G^{\prime}\in\mathcal{G}_{n}(m)$ and $\theta(G^{\prime})=m+i(G^{\prime})$ by construction. Let $B_{\beta}$ denote the previously established upper bound for $\theta(G_{\beta})$ ($1\leq\beta\leq\alpha$), and let $B$ denote the upper bound established for $G$. Note that $m\geq\overline{c(G_{\alpha-1})}+1$ or we would have stopped prior to $G_{\alpha}$, and that $i(G^{\prime})=i(G_{\alpha-1})-1$. Thus, $\theta(G^{\prime})\geq B_{\alpha-1}>B\geq\theta(G)$, so $\theta(G^{\prime})>\theta(G)$.

Regardless of $\alpha$’s value, we have found $G^{\prime}\in\mathcal{G}_{n}(m)$ such that $\theta(G^{\prime})>\theta(G)$, so $\Theta_{n}(m)\neq\theta(G)$.

Case 2: $m\leq\overline{c(G)}$. We can construct a triangle-free graph $H\in\mathcal{G}_{c(G)}(m)$. Since $H$ has no triangles, $\theta(H)\geq m$. Let $I$ be the graph of $i(G)$ singletons, and let $G^{\prime}=H\cup I$. Then $G^{\prime}\in\mathcal{G}_{n}(m)$ and $\theta(G^{\prime})\geq m+i(G)>m+i(G)-2\geq\theta(G)$. Thus, $\theta(G)\neq\Theta_{n}(m)$.

In either case, we have shown that if $m\leq\overline{n}$, $G\in\mathcal{G}_{n}(m)$, and $G$ contains at least 1 triangle, then $\theta(G)\neq\Theta_{n}(m)$. Thus, if $m\leq\overline{n}$ and $G\in\mathcal{G}_{n}(m)$, then $\theta(G)=\Theta_{n}(m)$ implies that $G$ is triangle-free. ∎

Lemma 6

If $m\leq\overline{n}$ and $m=p^{2}$ or $m=p(p+1)$ for some positive integer $p$, then $\Theta_{n}(m)=m+n-2p$ or $\Theta_{n}(m)=m+n-2p-1$, respectively.

Proof

Case 1: $m=p^{2}$. Consider the complete bipartite graph $K_{p,p}$. Since it has no triangles or singletons, $\theta(K_{p,p})=m$. Let $I$ be a graph consisting of $n-2p$ singleton vertices, and let $G=K_{p,p}\cup I$. $K_{p,p}$ has $2p$ vertices, and $I$ has $n-2p$ vertices, so $G$ has $n$ vertices. Similarly, $K_{p,p}$ has $m$ edges and $I$ has [math] edges, so $G$ has $m$ edges. Thus, $G\in\mathcal{G}_{n}(m)$. Moreover, since $G$ is triangle-free, $\theta(G)=m+n-2p$; that is, $G$’s smallest clique cover is equal to its edge count plus its number of singleton vertices.

Let $H\in\mathcal{G}_{n}(m)$ such that $H$ is triangle-free, and let $H_{0}$ be $H$ without its singletons. Then $H_{0}$ is also triangle-free, and $H_{0}\in\mathcal{G}_{c(H)}(m)$. Mantel’s Theorem shows that $m\leq\frac{c(H)^{2}}{4}$, or identically that $c(H)\geq 2\sqrt{m}$. Since $m=p^{2}$, this in turn is equivalent to $c(H)\geq 2p$. $H$ is not necessarily bipartite, but it can be covered by one clique for each edge (i.e., a clique consisting of the edge’s incident vertices) plus one clique for each singleton vertex. That is, there is a cover of $H$ with cardinality $m+n-c(H)\leq m+n-2p$. Thus, $\theta(H)\leq m+n-2p$. In conclusion, if $m=p^{2}$ and $m\leq\overline{n}$ then $\Theta_{n}(m)=m+n-2p$.

Case 2: $m=p(p+1)$. Consider the complete bipartite graph $K_{p,p+1}$. Again, since it’s bipartite, $\theta(K_{p,p+1})=m$. $K_{p,p+1}$ has $2p+1$ vertices, so let $I$ be the graph consisting of $n-2p-1$ singleton vertices, and let $G=K_{p,p+1}\cup I$. $G\in\mathcal{G}_{n}(m)$, and $\theta(G)=m+n-2p-1$. Similar to the previous case, given triangle-free $H\in\mathcal{G}_{n}(m)$, we can bound the number of non-isolated vertices in $H$: $c(H)\geq 2\sqrt{p^{2}+p}$. Since $c(H)$ is an integer, this bound can be improved to $c(H)\geq\lceil 2\sqrt{p^{2}+p}\rceil$. Obviously, $2\sqrt{p^{2}}<2\sqrt{p^{2}+p}<2\sqrt{p^{2}+p+0.25}$. Identically, $2p<2\sqrt{p^{2}+p}<2p+1$. So $\lceil 2\sqrt{p^{2}+p}\rceil=2p+1$. Thus, $c(H)\geq 2p+1$. Again, since $H$ can be covered by its individual edges and singletons, $\theta(H)\leq m+n-2p-1$. We have shown that if $m=p(p+1)$ and $m\leq\overline{n}$ then $\Theta_{n}(m)=m+n-2p-1$. ∎

Lemma 7

If $m<\overline{n}$ and $m=p^{2}$ or $m=p(p+1)$, then $\Theta_{n}(m)=\Theta_{n}(m+1)$.

Proof

Case 1: $m=p^{2}$. Let $G\in\mathcal{G}_{n}(m+1)$ be triangle-free. Mantel’s Theorem grants $c(G)\geq 2\sqrt{p^{2}+1}$. Since $c(G)$ is an integer, $c(G)\geq\lceil 2\sqrt{p^{2}+1}\rceil$. Obviously $2\sqrt{p^{2}+1}>2p$, so $c(G)\geq 2p+1$; $G$ has $m+1$ edges and at most $n-2p-1$ singletons. Thus, $\theta(G)\leq m+1+n-2p-1=m+n-2p$; in other words, $\theta(G)\leq\Theta_{n}(m)$. A graph $G\in\mathcal{G}_{n}(m+1)$ for which $\theta(G)=\Theta_{n}(m)$ can be constructed easily. For example, let $I$ be the graph of $n-2p$ singletons, and let $G=K_{p,p}\cup I\cup\{e\}$, where $e$ is an edge with one incident vertex in $K_{p,p}$ and the other in $I$.

Case 2: $m=p(p+1)$. Let $G\in\mathcal{G}_{n}(m+1)$ be triangle-free. Mantel’s Theorem, combined with the fact that $c(G)$ is an integer, grants $c(G)\geq\lceil 2\sqrt{p^{2}+p+1}\rceil$. $2\sqrt{p^{2}+p+0.25}<2\sqrt{p^{2}+p+1}<2\sqrt{p^{2}+2p+1}$, so $2p+1<2\sqrt{p^{2}+p+1}<2p+2$. Therefore, $c(G)\geq 2p+2$. $G$ is a graph with $m+1$ edges and at most $n-2p-2$ singletons. As such, $\theta(G)\leq m+n-2p-1$; $\theta(G)\leq\Theta_{n}(m)$. A graph $G\in\mathcal{G}_{n}(m+1)$ for which $\theta(G)=\Theta_{n}(m)$ can be constructed in much the same way as above.

In both cases, we have shown that $\Theta_{n}(m+1)=\Theta_{n}(m)$. ∎

Lemma 8

$(\forall m\leq\overline{n})$* $\Theta_{n+1}(m)=\Theta_{n}(m)+1$.*

Proof

Let $G_{0}\in\mathcal{G}_{n}(m)$ such that $\theta(G_{0})=\Theta_{n}(m)$. Lemma 5 ensures that $G_{0}$ is triangle-free. Therefore, $\theta(G_{0})=m+i(G_{0})$; $G_{0}$ could be any triangle-free graph on $n$ vertices and $m$ edges which maximizes $i(G)$. Maximizing $i(G_{0})$ is identical to minimizing $c(G_{0})$, as $i(G_{0})=n-c(G_{0})$. So let $n^{\prime}$ be the smallest number of vertices which can contain $m$ edges without a triangle. $\theta(G_{0})=m+n-n^{\prime}$. Similarly, let $G_{1}\in\mathcal{G}_{n+1}(m)$ be a triangle-free graph such that $\theta(G_{1})=\Theta_{n+1}(m)$. Again, $c(G_{1})=n^{\prime}$, as the number of edges is the same, so $\Theta_{n+1}(m)=\theta(G_{1})=m+(n+1)-n^{\prime}=\theta(G_{0})+1=\Theta_{n}(m)+1$. ∎

We are now ready to describe $\Theta_{n}(m)$ for $m\leq\overline{n}$. From Lemma 8, we can first take the entirety of $\Theta$ for $n-1$ vertices (up to its maximum) and add 1 to every dependent value. This establishes the portion of $\Theta_{n}$ ranging from [math] to $\overline{(n-1)}$ edges, and gives us a current right-most point at $(\overline{(n-1)},\overline{(n-1)}+1)$. Then, Lemma 7 grants that the next point is $(\overline{(n-1)}+1,\overline{(n-1)}+1)$. From here, the plot must make it to $(\overline{n},\overline{n})$; the increase in cover size is equal to the increase in edge count. In other words, from this point on the cover size must increase by 1 for each edge, on average. This, combined with Claim 4, shows that it actually must increase by exactly 1 per additional edge up to $(\overline{n},\overline{n})$.

So, essentially, to get the left side of $\Theta_{n}$, simply take the left side of $\Theta_{n-1}$, shift it upward by 1, then add the portion of the line $m=n$ ranging from $(\overline{(n-1)}+1,\overline{(n-1)}+1)$ up to the new maximum at $(\overline{n},\overline{n})$. We need only find the horizontal length of this added segment, $d=\overline{n}-(\overline{(n-1)}+1)$, to determine the pattern.

If $n$ is even, then $n-1$ is odd. So $\overline{n}=\frac{n^{2}}{4}$, and $\overline{(n-1)}=\frac{(n-1)^{2}-1}{4}$. So $\overline{n}-(\overline{(n-1)}+1)=\frac{n^{2}}{4}-(\frac{(n-1)^{2}-1}{4}+1)$. Simplification grants $d=\frac{n}{2}-1=\lfloor\frac{n}{2}\rfloor-1$.

Similarly, if $n$ is odd then $n-1$ is even. As such, $\overline{n}=\frac{n^{2}-1}{4}$ and $\overline{(n-1)}=\frac{(n-1)^{2}}{4}$. Again, subtract and simplify to get $d=\frac{n-1}{2}-1=\lfloor\frac{n}{2}\rfloor-1$. Note that this is the same difference that was attained from the even number of vertices directly preceding this odd $n$, and that in both cases, increasing $n$ by two increases $d$ by 1.

Let $\delta_{d}$ denote the sequence of pairs $(\Delta m,\Delta\Theta_{n})$, comprised of the pair $(+1,+0)$ followed by $d$ pairs $(+1,+1)$. For example, $\delta_{2}$ would be $\{(+1,+0),(+1,+1),(+1,+1)\}$. Recall that the left-most points of $\Theta_{n}(m)$ are $(0,n)$, $(1,n-1)$, $(2,n-1)$, $(3,n-1)$, $(4,n)$. This is the entire left side of $\Theta_{4}$. To extend this to show the left side on 5 vertices, we need only add $\delta_{\lfloor\frac{5-1}{2}\rfloor-1}$ (or $\delta_{1}$) to the right of these points. $\delta_{1}=\{(+1,+0),(+1,+1)\}$, so we get 2 additional points. Our last point was $(4,n)$. Addition of $(+1,+0)$ grants our first new point, $(5,n)$. Then, addition of $(+1,+1)$ grants $(6,n+1)$. To extend this to be the left side of the graph on 6 vertices, we would then add $\delta_{2}$ since $\lfloor\frac{6}{2}\rfloor-1=2$. Then $\delta_{2}$ to get to 7 vertices, then $\delta_{3}$ twice to get to 8 and 9 vertices, $\delta_{4}$ twice to get 10 and 11 vertices, and so on.

The pattern is clearest if we start from the point $(2,n-1)$, after which we have the sequence of changes: $\delta_{1},\delta_{1},\delta_{2},\delta_{2},\delta_{3},\delta_{3},\ldots,\delta_{p},\delta_{p},\ldots$, until the maximum $(\overline{n},\overline{n})$ is reached.

3 $\Theta_{n}(m)$ for $m\geq\overline{n}$

As in the previous section, the material is technical, but the reader will find it easier to follows by keeping in mind Figure 2.

In the style of the $\delta_{d}$’s discussed at the end of the previous section, let $\gamma_{d}$ denote the change in coordinates $(-1,+d)$ followed by $d-1$ iterations of the change $(-1,+0)$. For example, $\gamma_{3}$ is $\{(-1,+3),(-1,+0),(-1,+0)\}$. If we start from $({n\choose 2}+1,0)$ and move left, we can construct the right side of the graph with the sequence of changes $\gamma_{1},\gamma_{1},\gamma_{2},\gamma_{2},\gamma_{3},\gamma_{3},\ldots,\gamma_{p},\gamma_{p},\ldots$ until the maximum at $(\overline{n},\overline{n})$ is reached (with the “$(-1,+0)$”s trailing the last $\gamma$ omitted).

Recall the definitions of $I(G),i(G),C(G),c(G),S(G),s(G)$ from the introduction: isolated vertices and their count, non-isolated vertices and their count, and stars and their count, respectively. Also recall that if $C(G)$ is not a clique, then $\hat{G}$ denotes the subgraph of $G$ which results from removing all stars, along with their edges, from $G$. If $C(G)$ is a clique, then $\hat{G}$ is the graph which results from replacing $C(G)$ with a single new vertex.

Additionally, in order to keep the notation as simple as possible, we will assume that the intersection or union of a vertex set and a graph includes edges whenever convenient.

We prove below that removing $S(G)$ does not change $\theta(G)$, but an identical proof works for any subset of $S(G)$.

Lemma 9

1. (i)

$\theta(\hat{G})=\theta(G)$. 2. (ii)

The removal of any subset of $S(G)$ does not decrease $\theta$. 3. (iii)

Due to the above, if $G\in\mathcal{G}_{n+1}(m+n)$ and $s(G)\neq 0$ then $\theta(G)\leq\Theta_{n}(m)$

Proof

First, note that $\theta(G)=\theta(C(G))+i(G)$, and similarly that every vertex in $I(G)$ is also a singleton in $\hat{G}$, so $\theta(\hat{G})=\theta(C(G)\cap\hat{G})+i(G)$ if $C(G)$ is not a clique, and $1+i(G)$ otherwise. As such, we can assume without loss of generality that $G$ has no singletons (i.e., $C(G)=G$).

If $C(G)$ is complete, then it can be covered by 1 clique, so replacing it with a singleton vertex has no effect on $\theta$; $\theta(\hat{G})=\theta(G)$.

So let $G$ be a graph with no singletons such that $C(G)$ is not complete, and let $C$ be a minimal clique cover of $G$ consisting entirely of maximal cliques. We know such a cover exists from [3]. Let $C_{0}=\bigcup_{c\in C}\{c\cap\hat{G}\}$. The elements of $C_{0}$ are still cliques in $\hat{G}$, as any pair of vertices in $\hat{G}$ which were connected in $G$ are still connected in $\hat{G}$. Moreover, $C_{0}$ covers $\hat{G}$; every edge and vertex of $\hat{G}$ was covered by $C$, and the only difference from $C$ to $C_{0}$ is the removal of the vertices and edges which were not included in $\hat{G}$. It remains to be seen that the elements of $C_{0}$ are nonempty and unique. Let $c_{0}$ be an element of $C_{0}$. Then there is a maximal $c\in C$ such that $c_{0}=c\cap\hat{G}=c-S(G)$. Moreover, since $C(G)$ is incomplete, there is a vertex in $C(G)$ which is not in $S(G)$, and since $S(G)$ is fully connected to any element of $C(G)$, $S(G)$ cannot be a maximal clique. Thus every $c\in C$ contains a vertex not in $S(G)$, so every $c_{0}\in C_{0}$ is nonempty. Moreover, $G$ has no singletons so every $c\in C$ is a subset of $C(G)$, and as such $(\forall c\in C)(c\supset S(G))$. So each $c_{0}\in C_{0}$ is the result of removing the entirety of $S(G)$, from a clique $c\in C$. Therefore, if two elements of $C_{0}$ are identical, then their corresponding elements from $C$ were identical, so $C$ is not a smallest clique cover, as a duplicate clique could be removed to find a smaller one. Thus, every element of $C_{0}$ is nonempty and unique. We have found a cover $C_{0}$ of $\hat{G}$ such that $|C_{0}|=|C|$.

Assume that there is a cover $C_{0}^{\prime}$ of $\hat{G}$, composed of maximal cliques, such that $|C_{0}^{\prime}|<|C_{0}|$. Let $C^{\prime}=\bigcup_{c\in C_{0}^{\prime}}\{c\cup S(G)\}$. Obviously, $C^{\prime}$ covers any edge or vertex in $\hat{G}$. So let $v$ be a vertex in $G$ that is not in $\hat{G}$. $v\in S(G)$, so $v$ is covered by every clique in $C^{\prime}$. Let $e$ be an edge in $G$ but not in $\hat{G}$. Then $e$ either has two incident vertices in $S(G)$ or one in $S(G)$ and the other in $\hat{G}$. If both incident vertices are in $S(G)$, then $e$ is covered by every clique in $C^{\prime}$, as $(\forall c\in C^{\prime})(c\supset S(G))$. Alternately, if one incident vertex is in $\hat{G}$, then there is a cover $c_{0}\in C_{0}^{\prime}$ which contains this vertex; thus there is a $c^{\prime}\in C^{\prime}$ such that $c^{\prime}=c_{0}\cup S(G)$ which covers $e$. So $C^{\prime}$ covers $G$, and $|C^{\prime}|=|C_{0}^{\prime}|<|C_{0}|=|C|$; $C$ is not a smallest cover of $G$, which directly contradicts its definition.

We have found a cover for $\hat{G}$ with the same cardinality as a minimal cover of $G$, and shown that a smaller cover for $\hat{G}$ cannot exist. Thus, $\theta(\hat{G})=\theta(G)$. ∎

Claim 10

If $G\in\mathcal{G}_{n}(m)$ and $\hat{G}\in\mathcal{G}_{\hat{n}}(\hat{m})$, then $\hat{n}=n-s(G)$, and

[TABLE]

Proof

$\hat{n}=n-s(G)$ follows directly from the definitions of $\hat{G}$ and $s(G)$.

In order to get $\hat{G}$ from $G$, we remove every vertex in $S(G)$ and their incident edges. There are two types of edges which are removed: those connecting vertices in $S(G)$ to each other, and those connecting vertices in $S(G)$ to vertices not in $S(G)$. $S(G)$ is a complete subgraph of $G$, so its removal results in the removal of ${s(G)\choose 2}$ edges. Every vertex in $S(G)$ is also fully connected to the remaining vertices in $C(G)$, of which there are $c(G)-s(G)$. Thus, removing these edges lowers the edge count by $s(G)(c(G)-s(G))$. This grants $\hat{m}=m-{s(G)\choose 2}-s(G)(c(G)-s(G))$. Simplification grants $\hat{m}=m-\frac{s(G)(2c(G)-s(G)-1)}{2}$. ∎

Claim 11

$c(\hat{G})\leq c(G)-s(G)$, and $i(\hat{G})\geq i(G)$

Proof

From the definitions of $S(G)$ and $\hat{G}$, it is clear that $C(\hat{G})\subseteq C(G)-S(G)$, so $c(\hat{G})\leq c(G)-s(G)$. Similarly, $I(\hat{G})\supseteq I(G)$, so $i(\hat{G})\geq i(G)$. Note that both inequalities result from fact that $\hat{G}$ may have singletons which were not isolated in $G$; specifically, if a vertex $v\in C(G)$ only has edges incident to vertices in $S(G)$, then $v\in I(\hat{G})$. ∎

Claim 12

If $C(G)$ is not complete, then $\theta(G)\leq\overline{c(G)-s(G)}+i(G)$. If $C(G)$ is complete, then $\theta(G)=1+i(G)$.

Proof

Consider $\hat{G}$ for any graph $G$ such that $C(G)$ is not complete. Every vertex in $I(G)$ is still a singleton in $\hat{G}$. Thus, all of the edges in $\hat{G}$ are confined to the remaining vertices: $c(G)-s(G)$ of them. So $\theta(\hat{G})\leq\overline{c(G)-s(G)}+i(G)$. Moreover, since $C(G)$ is not complete, we know from Lemma 9 that $\theta(G)=\theta(\hat{G})$, so $\theta(G)\leq\overline{c(G)-s(G)}+i(G)$. For any graph $G$ such that $C(G)$ is complete, $\theta(C(G))=1$, so $\theta(G)=1+i(G)$. ∎

Claim 13

If $G$ has a triangle, let $G_{0}$ be the graph which results from removing two of the edges in said triangle. $\theta(G_{0})\geq\theta(G)-1$.

This claim doesn’t require proof; we could remove the whole triangle and still decrease $\theta$ by at most, as it cannot take more than 1 clique to cover a clique.

Lemma 14

If $m\geq\overline{n}$, $G\in\mathcal{G}_{n}(m)$, and $i(G)\neq 0$ then $\theta(G)\neq\Theta_{n}(m)$.

Proof

Let $m\geq\overline{n}$, and let $G\in\mathcal{G}_{n}(m)$ such that $i(G)\neq 0$. Then $m$ edges must fit on $n-i(G)$ vertices, so $m\leq{n-i(G)\choose 2}$.

Case 1: If $m={n-i(G)\choose 2}$, then $C(G)$ is complete, so $\theta(G)=1+i(G)$. Let $\{u,v\}$ be an edge in $C(G)$, and let $s$ be a singleton in $I(G)$. Remove the edge $\{u,v\}$, and replace it with the edge $\{u,s\}$ to attain graph $G^{\prime}$. Since $s$ is part of exactly 1 edge in $G^{\prime}$, this edge is a maximal clique, so it must be in any cover of $G^{\prime}$. To cover the remainder of $C(G^{\prime})$, exactly 2 cliques are necessary: one including $v$ and not $u$, and the other including $u$ and not $v$. Thus, $\theta(G^{\prime})=3+i(G^{\prime})=2+i(G)>\theta(G)$.

Case 2: Alternately, if $\overline{n}\leq m<{n-i(G)\choose 2}$, then $C(G)$ is not complete. Thus, there is a pair of vertices $u$ and $v$ in $C(G)$ such that $\{u,v\}$ is not an edge in $G$. Since $m\geq\overline{n}>\overline{n-1}$, there is at least 1 triangle in $G$. Let $\{x,y,z\}$ be a triangle in $G$. Remove the edges $\{x,y\}$ and $\{y,z\}$ to obtain graph $G_{0}$. We know from Lemma 13 that $\theta(G_{0})\geq\theta(G)-1$. Next, choose some singleton $s$ in $G_{0}$, and add to $G_{0}$ the edges $\{u,s\}$ and $\{v,s\}$ to get graph $G^{\prime}$. Since $s$ was isolated and $\{u,v\}$ was not an edge in $G_{0}$, these two new edges are maximal cliques. Thus, $\theta(G^{\prime})=\theta(G_{0})+2\geq\theta(G)+1$.

In both cases, we have found a graph $G^{\prime}\in\mathcal{G}_{n}(m)$ such that $\theta(G^{\prime})>\theta(G)$. Thus, $\theta(G)\neq\Theta_{n}(m)$. ∎

Lemma 15

For $G\in\mathcal{G}_{n}(m)$, if $1<m<{n\choose 2}$ and $C(G)$ is complete then $\theta(G)\neq\Theta_{n}(m)$.

Proof

Case 1: Let $1<m\leq\overline{n}$, and consider $G\in\mathcal{G}_{n}(m)$ such that $C(G)$ is complete. Since $m>1$, there are at least 3 vertices in $C(G)$, so $G$ contains a triangle. Thus, by Lemma 5, $\theta(G)\neq\Theta_{n}(m)$.

Case 2: Let $\overline{n}\leq m<{n\choose 2}$, and consider $G\in\mathcal{G}_{n}(m)$ such that $C(G)$ is complete. Then $c(G)<n$, because it would take ${n\choose 2}$ edges to make a complete graph on $n$ edges. As such, $i(G)\neq 0$, so by Lemma 14, $\theta(G)\neq\Theta_{n}(m)$. ∎

Lemma 16

$\Theta_{n}(\overline{n}+1)=\overline{n-1}$**

Proof

It can be shown quickly through enumeration of all arrangements of edges that this is true for 3 and 4 vertices. We will prove it for larger integers through induction.

Assume the Lemma holds for some even $n\geq 4$. Let $G\in\mathcal{G}_{n+1}(\overline{n+1}+1)$ such that $i(G)=0$. $G$ has degree sum $D=2\overline{n+1}+2$. Let $v$ be a vertex of minimum degree in $G$. For the sake of contradiction, assume $deg(v)>n/2$. Then $deg(v)\geq n/2+1$, so every vertex in $G$ has degree $\geq n/2+1$. Therefore, $D\geq(n+1)(n/2+1)=\frac{n^{2}+3n+2}{2}$. Moreover, since $n\geq 4$, $D\geq\frac{n^{2}+2n+6}{2}=\frac{n^{2}+2n+1}{2}+5/2>2\overline{n+1}+2=D$; that is, $D>D$. The assumption that $deg(v)>n/2$ led to a contradiction, so $deg(v)\leq n/2$. Remove $v$, all of its incident edges, and $n/2-deg(v)$ additional edges to construct graph $G^{\prime}$. Since $n$ is even, $\overline{n+1}-\overline{n}=\overline{n}-\overline{n-1}=n/2$. $G^{\prime}$ has $1$ fewer vertices and $n/2$ fewer edges than $G\in\mathcal{G}_{n+1}(\overline{n+1}+1)$, so $G^{\prime}\in\mathcal{G}_{n}(\overline{n}+1)$. By the hypothesis, $\theta(G^{\prime})\leq\overline{n-1}$. We will now reconstruct $G$ from $G^{\prime}$, taking note of any changes to $\theta$. As such, whenever we add an edge, it is an edge which had previously been removed from $G$ in the construction of $G^{\prime}$. First, add $v$ and any 1 of its incident edges; $v$ necessarily had at least 1 incident edge as $i(G)=0$. This addition increases $\theta$ by exactly 1, as the newly added edge comprises a maximal clique. There are now $n/2-1$ edges missing from $G$. Add them all back, noting that Claim 4 (or, more accurately, the same reasoning used to prove it), guarantees that each additional edges increases $\theta$ by at most 1. Thus, $\theta(G)\leq\theta(G^{\prime})+1+n/2-1\leq\overline{n-1}+n/2=\overline{n}$.

Assume the Lemma holds for some odd $n\geq 3$. Let $G\in\mathcal{G}_{n+1}(\overline{n+1}+1)$ such that $i(G)=0$. $G$ has degree sum $D=2\overline{n+1}+2$. As such, the average degree in $G$ is $\frac{n+1}{2}+\frac{2}{n+1}$. Thus, the minimum degree of any vertex in $G$ is at most $\frac{n+1}{2}$.

Case 1: If the minimum degree in $G$ is $\leq\frac{n-1}{2}$, then let $v$ be a vertex of degree $\leq\frac{n-1}{2}$. Remove $v$ and all of its incident edges, along with any additional edges necessary to remove $\frac{n-1}{2}$ total, to get graph $G^{\prime}$. Of course, $\theta(G)\leq\theta(G^{\prime})+\frac{n-1}{2}$, as we can simply add the $\frac{n-1}{2}$ edges to any cover of $G^{\prime}$ to obtain a cover of $G$. Moreover, $G^{\prime}\in\mathcal{G}_{n}(\overline{n}+2)$. Since $\Theta_{n}(m)$ is non-increasing for $m\geq\overline{n}$, $\theta(G^{\prime})\leq\Theta_{n}(\overline{n}+1)$. So, by the hypothesis, $\theta(G^{\prime})\leq\overline{n-1}$. Thus, $\theta(G)\leq\overline{n-1}+\frac{n-1}{2}$; identically, $\theta(G)\leq\overline{n}$.

Case 2: If the minimum degree in $G$ is $\frac{n+1}{2}$, then let $v$ be a vertex of degree $\frac{n+1}{2}$.

Subcase 1: If $v$ is in a triangle, remove it an all of its edges to get $G^{\prime}$. $G^{\prime}\in\mathcal{G}_{n}(\overline{n}+1)$, so $\theta(G^{\prime})\leq\overline{n-1}$ by the hypothesis. Since two of these edges were in a triangle, their removal reduced the clique cover size by at most 1. Removing the remaining $\frac{n-3}{2}$ edges reduced $\theta$ by at most $\frac{n-3}{2}$. Therefore, $\theta(G)\leq\theta(G^{\prime})+\frac{n-1}{2}\leq\overline{n-1}+\frac{n-1}{2}=\overline{n}$; that is, $\theta(G)\leq\overline{n}$.

Subcase 2: If $v$ is not in a triangle, we need only find a vertex $v_{0}$, in a triangle, with degree $\frac{n+1}{2}$ for the previous subcase to apply. None of $v$’s $\frac{n+1}{2}$ neighbors are adjacent to each other. Let $V$ be the set of vertices adjacent to $v$, and let $V^{\prime}$ be the set of vertices in $G$ with are neither $v$ nor adjacent to $v$. Note that $|V|=\frac{n+1}{2}$ and $|V^{\prime}|=\frac{n-1}{2}$. The vertices in $V$ all have degree of at least $\frac{n+1}{2}$, as this is the minimum degree in $G$. As such, every vertex in $V$ is adjacent to at least $\frac{n-1}{2}$ vertices other than $v$; none of the vertices in $V$ are adjacent to each other, and there are only $\frac{n-1}{2}$ vertices (other than $v$) remaining. Thus, every vertex in $V$ is adjacent to all $\frac{n-1}{2}$ vertices in $V^{\prime}$. Let’s count edges: there are $\frac{n+1}{2}$ between $v$ and $V$ and another $\overline{n}$ between $V$ and $V^{\prime}$, and there are $\overline{n+1}+1$ total edges; 1 edge is unaccounted for. This edge must be between 2 vertices in $V^{\prime}$, as it cannot be in $V$, and every edge from $V$ to $V^{\prime}$ is already counted. So let $v_{0}$ be any vertex in $V$. $deg(v_{0})=\frac{n+1}{2}$ and $v_{0}$ is necessarily in a triangle, so apply the previous subcase.

We have shown that $\Theta_{n}(\overline{n}+1)=\overline{n-1}\implies\Theta_{n+1}(\overline{n+1}+1)\leq\overline{n}$. It is easy to construct a graph $G\in\mathcal{G}_{n+1}(\overline{n+1}+1)$ such that $\theta(G)=\overline{n}$; simply add an extra edge to $K_{\lfloor\frac{n}{2}\rfloor,\lceil\frac{n}{2}\rceil}$ between any two vertices in the larger partition (if $n$ is odd) or in either partition (if $n$ is even). Therefore, for all values $n$, $\Theta_{n}(\overline{n}+1)=\overline{n-1}$. ∎

Lemma 16 implies that $\Theta_{n}(m)=\overline{n-1}$, if $\overline{n}<m\leq{n\choose 2}-\overline{n-2}$, as $\Theta_{n}(m)$ is non-increasing for $m\geq\overline{n}$ and Theorem 2 implies that $\Theta_{n}({n\choose 2}-\overline{t})=\overline{t+1}$. This, shows that the following three statements are equivalent. We conject that they are true:

Conjecture 17

1. (i)

If $m\geq\overline{n}$, then $\Theta_{n+1}(m+n)=\Theta_{n}(m)$. 2. (ii)

If $m={n\choose 2}-\overline{t}+1>\overline{n}$, then $\Theta_{n}(m)=\overline{t}$. 3. (iii)

If $m\geq\overline{n}$, let $k={n\choose 2}-m$, and let $t$ be the largest natural number such that $\overline{t}\leq k$. Then $\Theta_{n}(m)=\overline{t+1}$.

Lemma 16 shows that (ii) is true when $t=n-1$. We will show in Lemma 19 that (ii) also holds when $t=n-2$, and in Lemma 18 that (i) is true when ${n\choose 2}-m\leq n/2$.

Lemma 18

Let $k={n\choose 2}-m$. If $k\leq n/2$, then $\Theta_{n+1}(m+n)=\Theta_{n}(m)$.

Proof

Let $G\in\mathcal{G}_{n+1}(m+n)$. ${n+1\choose 2}-(m+n)={n\choose 2}-m=k$, so $G$ is missing $k$ edges. $k\leq n/2$, so $k<\frac{n+1}{2}$. Every edge only has 2 endpoints, so there is at least 1 vertex $v$ which is not adjacent to any of the missing edges. $v$ must be in $S(G)$, as it is adjacent to every other vertex in $G$. Therefore, $\theta(G-\{v\})=\theta(G)$. $G-\{v\}$ has $n$ vertices and $m$ edges, so $\theta(G)\leq\Theta_{n}(m)$.

We can easily construct a graph $G\in\mathcal{G}_{n+1}(m+n)$ such that $\theta(G)=\Theta_{n}(m)$; simply add a star to a graph $G_{0}\in\mathcal{G}_{n}(m)$ such that $\theta(G_{0})=\Theta_{n}(m)$. ∎

Lemma 19

$\Theta_{n}({n\choose 2}-\overline{n-2}+1)\leq\overline{n-2}$**

Proof

It should first be noted that ${n\choose 2}-\overline{n-2}+1=\overline{n+1}$, so this Lemma can be reformulated as $\Theta_{n}(\overline{n+1})\leq\overline{n-2}$. We have shown through exhaustive search of all graphs on $\leq 8$ vertices that this is true for all $n\leq 8$.

Case 1: Even $n$

Assume $n$ is even and $n\geq 10$. Let $G\in\mathcal{G}_{n}(m)$, where $m={n\choose 2}-\overline{n-2}+1$. Moreover, assume that the Lemma is true for all $n_{0}<n$.

Since $n$ is even, ${n\choose 2}-\overline{n-2}+1=\frac{n^{2}+2n}{4}$. Thus, the degree sum of $G$ is $\frac{n^{2}+2n}{2}$, and as such the average degree in $G$ is $\frac{n}{2}+1$. Let $d=\min{\{\text{deg}(v)|v\in V\}}$. Based on the average degree, $d\leq\frac{n}{2}+1$.

Subcase 1: $d\leq\frac{n}{2}-1$

If $d\leq\frac{n}{2}-1$, let $v$ be a vertex with degree $\leq\frac{n}{2}-1$, and let $G^{\prime}$ be $G$ without $v$ or any of its edges. $G^{\prime}$ has $n-1$ vertices and $m-d$ edges. $m-d\geq m-\frac{n}{2}+1=\overline{n}+1\geq\overline{n}$. Thus, $\theta(G^{\prime})\leq\Theta_{n-1}(\overline{n})\leq\overline{n-3}$. The remaining edges (those adjacent to $v$) can be covered by at most $\frac{n}{2}-1$ cliques. Thus, $\theta(G)\leq\overline{n-3}+\frac{n}{2}-1=\overline{n-2}$.

Subcase 2: $d=\frac{n}{2}$

If $d=\frac{n}{2}$. Let $v$ be a vertex of degree $d$. Let $\mathcal{N}_{v}$ denote $v$’s neighborhood, and let $\mathcal{N}_{v}^{c}$ denote the set of vertices in $G$ which are neither $v$ nor in $\mathcal{N}_{v}$.

Assume $v$ is not in a triangle. $|\mathcal{N}_{v}|=\frac{n}{2}$, so $|\mathcal{N}_{v}^{c}|=\frac{n}{2}-1$ (as there are $n-1$ vertices other than $v$). Since $v$ is not in a triangle, $\mathcal{N}_{v}$ is pairwise disjoint. Every vertex must be adjacent to at least $\frac{n}{2}$ others, and there are only $\frac{n}{2}$ vertices not in $\mathcal{N}_{v}$, so every vertex in $\mathcal{N}_{v}$ must be adjacent to all of them. Moreover, $\mathcal{N}_{v}$ is pairwise disjoint, no more edges can be added to any vertex in it, so the degree of every vertex in $\mathcal{N}_{v}$ is $\frac{n}{2}$ exactly. Let’s count edges: there are $\frac{n}{2}$ between $v$ and $\mathcal{N}_{v}$, and another $\frac{n}{2}(\frac{n}{2}-1)$ between $\mathcal{N}_{v}$ and $\mathcal{N}_{v}^{c}$; that’s $\overline{n}$ total edges; $\frac{n}{2}$ edges are missing. These edges must be in $\mathcal{N}_{v}^{c}$, as $v$ is not in any triangles. Since every vertex in $\mathcal{N}_{v}$ is connected to every vertex in $\mathcal{N}_{v}^{c}$, a single edge in $\mathcal{N}_{v}^{c}$ is enough to imply that there is an element of $\mathcal{N}_{v}$ which is in a triangle. Since every element of $\mathcal{N}_{v}$ has degree $\frac{n}{2}$, there must be a vertex of degree $\frac{n}{2}$ which is in a triangle.

So it is safe to assume that there is a vertex of degree $\frac{n}{2}$ which is in a triangle, as such a vertex necessarily exists. Let $v$ be such a vertex. Let $G^{\prime}$ be $G$ without $v$ or its edges. $G^{\prime}$ has $n-1$ vertices and $\overline{n}$ edges, so $\theta(G^{\prime})\leq\Theta_{n-1}(\overline{n})$. This, by our hypothesis, implies that $\theta(G^{\prime})\leq\overline{n-3}$. In order to cover the rest of $G$, we need only cover the edges adjacent to $v$. There are $\frac{n}{2}$ of them, 2 of which can be covered by 1 triangle. Thus, $\theta(G)\leq\overline{n-3}+\frac{n}{2}-1=\overline{n-2}$.

Subcase 3: $d=\frac{n}{2}+1$

Finally, if $d=\frac{n}{2}+1$, then every vertex has degree of exactly $d$. Since $m>\overline{n}$, there is a triangle $T=(u,v,w)$ in $G$. Let $G^{\prime}$ be $G$ without $u$, $v$, $w$ or their edges. $G^{\prime}$ has $n-3$ vertices and $\overline{n-2}-1$ edges, so by Lemma 16, $\theta(G^{\prime})\leq\overline{n-4}$. Each vertex in $T$ has $\frac{n}{2}-1$ neighbors in $G^{\prime}$ (as its other 2 neighbors are in $T$. This gives us a lot of information. First, since there are $n-3$ vertices in $G^{\prime}$ and $2(\frac{n}{2}-1)=n-2$, each pair of vertices in $T$ has at least 1 common neighbor in $G^{\prime}$. This means that all three edges in $G^{\prime}$ can be covered by triangles containing 2 elements of $T$ and 1 element of $G^{\prime}$. Moreover, $3(\frac{n}{2}-1)=\frac{n}{2}+n-3$; if we were to list all of the neighbors (in $G^{\prime}$) of the vertices in $T$, there would be at least $\frac{n}{2}$ repeats (if a vertex is listed twice, it’s been repeated once, and if it’s been listed three times it’s been repeated twice…). Let $x$ be a repeated vertex. If $x$ was only repeated once, it is adjacent to two elements of $T$. As such, two edges connecting $T$ to $G^{\prime}$ can be covered with 1 triangle. As such, this repeat allows us to cover at least 1 “extra” edge with 1 clique. Alternately, if a vertex is repeated twice, it is adjacent to all 3 elements of $T$, so 3 edges between $T$ and $G^{\prime}$ can be covered with 1 clique, so we’ve covered 2 “extra” edges. Moreover, these triangles and 4-cliques do not repeat edges (other than those in $T$, as they each have a single unique vertex in $G^{\prime}$. So these $\frac{n}{2}$ repetitions equate to at least $\frac{n}{2}$ fewer cliques needed to cover the edges between $T$ and $G^{\prime}$, while simultaneously covering all 3 edges in $T$. As such, we can cover all edges in $T$ and all edges between $T$ and $G^{\prime}$ with at most $3(\frac{n}{2}-1)-\frac{n}{2}=n-3$.

So $\theta(G)\leq\theta(G^{\prime})+n-3\leq\overline{n-4}+n-3=\overline{n-2}$.

Case 2: Odd $n$

Assume $n$ is odd and $\geq 9$. Let $G\in\mathcal{G}_{n}(m)$, where $m=\overline{n+1}$. Moreover, like the previous case, assume that the Lemma is true for all $n_{0}<n$.

$m=\overline{n+1}=\frac{n^{2}+2n+1}{4}$, so the degree sum is $\frac{n^{2}+2n+1}{2}$. As such, the average degree is $\frac{n+1}{2}+\frac{n+1}{2n}$, so $d\leq\frac{n+1}{2}$ (as $\frac{n+1}{2}$ is obviously less than 1).

Subcase 1: $d\leq\frac{n-3}{2}$

If $d\leq\frac{n-3}{2}$, let $v$ be a vertex with degree $d$. Let $G^{\prime}$ be $G$ without $v$ or its edges. $G^{\prime}$ has $n-1$ vertices and at least $\overline{n}+2$ edges. Thus, $\theta(G^{\prime})\leq\overline{n-3}$. We can cover the edges adjacent to $v$ with at most $\frac{n-3}{2}$ cliques, so $\theta(G)\leq\overline{n-3}+\frac{n-3}{2}=\frac{n^{2}-4n+3}{4}=\overline{n-2}$.

Subcase 2: $d=\frac{n-1}{2}$

Let $v$ be a vertex of degree $d$. There are $\frac{n-1}{2}$ vertices in $\mathcal{N}_{v}$.

Assume $v$ is not in a triangle. Every vertex in $\mathcal{N}_{v}$ is adjacent to at least $\geq\frac{n-3}{2}$ vertices and $\leq\frac{n-1}{2}$ in $\mathcal{N}_{v}^{c}$. Thus, there are at most $(\frac{n-1}{2})^{2}=\overline{n-1}$ edges between $\mathcal{N}_{v}$ and $\mathcal{N}_{v}^{c}$. We have accounted for $\overline{n-1}+\frac{n-1}{2}=\overline{n}$ edges; there are at least $\frac{n-1}{2}$ edges remaining, all of which must be in $\mathcal{N}_{v}^{c}$.

If there is a vertex$v_{0}$ in $\mathcal{N}_{v}$ with degree $\frac{n-1}{2}$, this vertex is adjacent to $\frac{n-3}{2}$ of the $\frac{n-1}{2}$ vertices in $\mathcal{N}_{v}^{c}$; let the vertex in $\mathcal{N}_{v}^{c}$ to which it is nonadjacent be $w$. Given that there are at least $\frac{n-1}{2}$ edges in $\mathcal{N}_{v}^{c}$, which is strictly more than $\frac{n-3}{2}$, at least one of the edges in $\mathcal{N}_{v}^{c}$ does not include $w$, and therefore forms a triangle with its two adjacent vertices ($u_{1}$ and $u_{2}$) and $v_{0}$. We have found $v_{0}$, a vertex with degree $\frac{n-1}{2}$ which is also in a triangle. Let $G^{\prime}$ be $G$ without $v_{0}$ or its edges. $G^{\prime}$ has $n-1$ vertices and least $\overline{n}+1$ edges, so $\theta(G^{\prime})\leq\Theta_{n-1}(\overline{n}+1)\leq\overline{n-3}$. The remaining edges in $G$ are those adjacent to $v_{0}$; there are $\frac{n-1}{2}$ of them, and 2 can be covered by a triangle, so $\theta(G)\leq\overline{n-3}+\frac{n-3}{2}=\overline{n-2}$. Also note that this works with any vertex of degree $\frac{n-1}{2}$ which is in a triangle, so the case in which $v$ is in a triangle has been proven in the process.

If there are no vertices of degree $\frac{n-1}{2}$ in $\mathcal{N}_{v}$, then every vertex in $\mathcal{N}_{v}$ has degree $\frac{n+1}{2}$ (as there are only $\frac{n+1}{2}$ vertices not in $\mathcal{N}_{v}$). Therefore, every vertex in $\mathcal{N}_{v}$ is adjacent to every vertex in $\mathcal{N}_{v}^{c}$. So there are $(\frac{n-1}{2})^{2}=\overline{n-1}$ edges between $\mathcal{N}_{v}$ and $\mathcal{N}_{v}^{c}$, and $\frac{n-1}{2}$ between $v$ and $\mathcal{N}_{v}$ for a total of $\overline{n}$; there are $\frac{n}{2}$ edges in $\mathcal{N}_{v}^{c}$.

Let $E$ be the set of edges in $\mathcal{N}_{v}^{c}$. Assume, for contradiction, that none of the elements of $E$ are pairwise disjoint. That is, assume every pair of edges in $E$ has a common vertex. Then $\mathcal{N}_{v}^{c}$ contains no triangles, as any edge in $E$ (not in the triangle) could contain at most 1 of the triangles vertices and would therefore be disjoint with at least 1 of the triangles edges.

Choose any $2$ edges in $E$. These edges share a vertex, $a$. Choose a third edge in $E$. If it is not adjacent to $a$, then it must form a triangle with the previous 2 edges in order to share a vertex with each of them. Therefore, the third edge must be adjacent to $a$. As must the fourth…until there are $\frac{n-3}{2}$ edges connecting $a$ to every other vertex in $\mathcal{N}_{v}^{c}$. There must be another edge in $\mathcal{N}_{v}^{c}$, and it cannot be adjacent to $a$; label this edge $(c,d)$. Since $n\geq 9$, there are $\geq 4$ vertices in $\mathcal{N}_{v}^{c}$, so there is at least $1$ more vertex, $b$. Since $a$ is adjacent to everything in $\mathcal{N}_{v}$, $(a,b)$ is also an edge. We have found a pair of disjoint edges in $E$, so obviously our assumption that it is not pairwise disjoint was false; there are two disjoint edges $(a,b)$ and $(c,d)$ in $\mathcal{N}_{v}^{c}$.

Let $v_{0}$ be a vertex in $\mathcal{N}_{v}$. Let $G^{\prime}$ be $G$ without $v_{0}$ or its edges. $G^{\prime}$ has $n-1$ vertices and $\overline{n+1}-\frac{n+1}{2}=\overline{n}$. Thus, $\theta(G)\leq\Theta_{n-1}(\overline{n})\leq\overline{n-3}$. We can cover the remaining $\frac{n+1}{2}$ edges adjacent to $v_{0}$ with at most $\frac{n-3}{2}$ cliques, because $4$ of these edges can be covered by the triangles $(a,b,v)$ and $(c,d,v)$. Thus, $\theta(G)\leq\overline{n-3}+\frac{n-3}{2}=\overline{n-2}$. Note that this works with any vertex of degree $\frac{n+1}{2}$ which is in two otherwise disjoint triangles.

Subcase 3 $d=\frac{n+1}{2}$

Let $V$ be the set of vertices in $G$ with degree $\frac{n+1}{2}$, and let $V^{\prime}$ be the vertices with degree $\geq\frac{n+3}{2}$. Assume, for contradiction, that $|V|\leq\frac{n-3}{2}$. Then the minimum degree sum of $G$ is $(\frac{n-3}{2})(\frac{n+1}{2})+(\frac{n+3}{2})(\frac{n+3}{2})=\frac{n^{2}+2n+3}{2}$. Recall that the degree sum of $G$ is $\frac{n^{2}+2n+1}{2}$, so we’ve found our contradiction; we now know that $|V|\geq\frac{n-1}{2}$.

Assume, for contradiction, that $|V^{\prime}|\leq 1$. Then the maximum degree sum of $G$ is $(n-1)(\frac{n+1}{2})+1(\frac{n+3}{2})=\frac{n^{2}+n+2}{2}<\frac{n^{2}+2n+1}{2}$ (recall $n\geq 4$). This contradicts the known degree sum of $G$, so it must be false; $|V^{\prime}|\geq 2$.

Once more, assume for contradiction that there is no element of $V$ which is adjacent to two distinct elements of $V^{\prime}$. Every vertex in $V$ has $\frac{n+1}{2}$ neighbors. At most one of these neighbors is in $V^{\prime}$, so at least $\frac{n-1}{2}$ neighbors must be in $V$. Therefore $|V|\geq\frac{n+1}{2}$. As such, each there are at most $\frac{n-1}{2}$ vertices in $V^{\prime}$, so each vertex in $V^{\prime}$ must be adjacent to at least four vertices in $V$. Moreover, since no vertex in $V$ can be adjacent to two in $V^{\prime}$, we now know that $|V|\geq 4|V^{\prime}|$; in other words, $|V^{\prime}|\leq\lfloor\frac{n}{5}\rfloor$.

Choose two vertices in $V^{\prime}$. They must each be adjacent to at least $\frac{n+1}{2}$ vertices (other than each other). There are only $n-2$ other vertices, so they have at least 3 neighbors in common. These common neighbors are adjacent to two vertices in $V^{\prime}$, so they cannot be in $V$ and must be in $V^{\prime}$; $|V^{\prime}|\geq 5$. Choose five vertices in $V^{\prime}$. They are each adjacent to at least $\frac{n-3}{2}$ vertices (again, other than each other), of which there are $n-5$. $5(\frac{n-3}{2})-(n-5)=\frac{3n-5}{2}$, so there are at least $\frac{3n-5}{2}$ repeats. A vertex can be repeated at most four times (five vertices in $V^{\prime}$, first isn’t a repeat), so there are at least $\frac{3n-5}{8}$ repeated vertices; that is, there are at least $\frac{3n-5}{8}$ vertices (other than the give we already had) in adjacent to two elements of $V^{\prime}$, and therefore in $V^{\prime}$. Thus, $|V^{\prime}|\geq\frac{3n-5}{8}+5$.

We know that $\frac{3n-5}{8}+5\leq|V^{\prime}|\leq\frac{n}{5}$, so if $\frac{3n-5}{8}+5>\frac{n}{5}$, we have a contradiction. This turns out to be true for any value of $n$ which is $\geq-25$, which $n$ obviously is, so at long last we’ve arrived at a contradiction. Therefore, there must be an element $v$ of $V$ which is adjacent to two elements $a$ and $b$ of $V^{\prime}$.

Assume $a$ and $b$ are adjacent. $v$ is adjacent to $\frac{n-3}{2}$ vertices other than $a$ and $b$. $a$ is adjacent to at least $\frac{n-1}{2}$ vertices other than $b$ and $v$. There are $n-3$ total other vertices, so $a$ and $v$ have at least one neighbor $a_{1}\neq b$ in common. Similarly, $b$ and $v$ have a neighbor $b_{1}\neq a$ in common. If $b_{1}=a_{1}$, then $a_{1}$ is adjacent to $b$, so $(v,a,b,a_{1})$ is a clique. This clique can cover three edges between $v$ and $\mathcal{N}_{v}$. If $a_{1}\neq b_{1}$, then the two triangles $(v,a,a_{1})$ and $(v,b,b_{1})$ cover four edges between $v$ and $\mathcal{N}_{v}$. In either case, every edge between $v$ and its neighborhood can be covered with at most $\frac{n-3}{2}$ cliques.

If $a$ and $b$ are not adjacent, then $a$ is adjacent to at least $\frac{n+1}{2}$ vertices other than $v$. $v$ is still adjacent to $\frac{n-3}{2}$ vertices other than $a$ and $b$, so $a$ and $v$ have at least two neighbors $a_{1},a_{2}$ in common (and since $a$ and $b$ are not adjacent, these common neighbors cannot be $b$). Similarly, $b$’s and $v$’s neighborhoods have at least $b_{1},b_{2}$ in common. At least one of $\{b_{1},b_{2}\}$ is not $a_{1}$, so we can assume without loss of generality that $a_{1}\neq b_{1}$. The two triangles $(v,a,a_{1})$ and $(v,b,b_{1})$ cover four edges between $v$ and $\mathcal{N}_{v}$, so the edges adjacent to $v$ can be covered with at most $\frac{n-3}{2}$ cliques.

So, whether or not $a$ and $b$ are adjacent, let $G^{\prime}$ be $G$ without $v$ or its edges. $G^{\prime}$ has $n-1$ vertices and $\overline{n}$ edges, so $\theta(G)\leq\Theta_{n-1}(\overline{n})\leq\overline{n-3}$. We’ve shown that the remaining edges can be covered with at most $\frac{n-3}{2}$ cliques, so $\theta(G)\leq\overline{n-3}+\frac{n-3}{2}=\overline{n-2}$. ∎

We haven’t proven that Conjecture 17 is true for all $m$. The results of Lemma 16 and 19 and the non-decreasing nature of $\Theta_{n}(m)$ past $\overline{n}$ show that Conjecture 17(iii) is true for all $m\in[\overline{n},{n\choose 2}-\overline{n-3}]$. Similarly, Lemma 18 shows that Conjecture 17(i) is true if $m\geq{n\choose 2}-n/2$.

4 Summary of results

Putting together the results in Section 2, the behavior of $\Theta_{n}(m)$ for $m\leq\overline{n}$, and Section 3, the behavior of $\Theta_{n}(m)$ for $m\geq\overline{n}$, we can now state the conclusion of the paper succinctly as Theorem 20.

Theorem 20

Let $\Theta_{n}(m)$, for $1\leq m\leq{n\choose 2}$, be the size of the largest minimal clique-cover for any graph on $n$ vertices and $m$ edges. Then, for $m\leq\overline{n}$, we have:

[TABLE]

Where $k$ and $t$ are defined as in Theorem 2.

The recursive definition can be easily unwound, and the values of $\Theta_{n}(m)$ computed explicitly with Algorithm 1 which runs in linear time in the length of the binary encoding of $n$.

Proof

(1a) follows from Lemma 8. (1b) follows from several sources: the left bound, i.e., $m=\overline{n-1}+1$ is Lemma 7, the right bound, i.e., $m=\overline{n}$ is Mantel (Theorem 1), and the value, which consists of a subset of the line $y=x$, follows from Claim 4 which limits the growth, and therefore imposes an increase of 1 at each step. (1c) and (1d) follow from Lemmas 16 and 19 combined with Lovász (Theorem 2). (1e) is the bound provided by Lovász. ∎

Given proof of Conjecture 17 for all $m$, we can improve Theorem 20 to:

[TABLE]

Algorithm 1 (below), reflects the second version of Theorem 20 (i.e. the one directly above), because this version grants an exact $\Theta_{n}(m)$, as opposed to an upper bound. It could be adjusted to reflect the original theorem (roughly, omitting the improvements made in Lemmas 16 and 19) by replacing lines 18-21 with the single line “return $k+p$”.