Hadwiger’s Conjecture for Dense Strongly Regular Graphs

Abstract

The famous Four-Color Conjecture (now Theorem) states that any planar graph could be colored using four colors. Hadwiger’s conjecture strengthens the Four-Color Conjecture by asserting that every graph with chromatic number t contains a complete minor of order t. In this paper we investigate Hadwiger’s conjecture for the complements of the Petersen graph and the Clebsch graph; both are strongly regular graphs with independence number two (hence dense graphs). We confirm Hajós’ conjecture, hence Hadwiger’s conjecture, for these graphs. Moreover, we show that for each of these graphs the exact hadwiger number is strictly greater its chromatic number.

1. Introduction

A minor of a simple graph G is a graph gained from G by a succession of deleting vertices, deleting edges and contracting edges. Equivalently, a minor is a graph isomorphic to H obtained from a subgraph of G by contracting edges, and such a minor is called an H-minor. In particular, if H is isomorphic to the complete graph on t vertices, it is called a $K_t$-minor. The hadwiger number $h\left(G\right)$ of G is the maximum integer t such that G contains a $K_t$-minor. A $K_t$-minor can also be viewed as t pairwise vertex-disjoint connected subgraphs in G, such that there is at least one edge joining every pair of distinct subgraphs. We call each of these subgraphs a branch set. For all other terminology we follow [1].

The Four-Color Conjecture (now Theorem), which states that any map (planar graph) could be colored using four colors, has been one central topics for a long time in combinatorics and has motivated considerable research findings and nice mathematics. In 1943, Hadwiger [2] proposed the following conjecture, which generalizes the Four-Color Conjecture, and is widely thought to be one of the most important problems in graph theory:

Hadwiger’s Conjecture. For every graph G, $h\left(G\right)\ge \chi \left(G\right)$.

Hadwiger’s conjecture asserts that every graph that is not $(t-1)$-colourable contains a $K_t$-minor. Equivalently, the hadwiger number $h\left(G\right)$ of G is bounded below by the chromatic number $\chi \left(G\right)$. It has been established [3] for graphs with $\chi \left(G\right)\le 6$, and remains open for graphs with $\chi \left(G\right)\ge 7$. Researchers also confirm this conjecture for certain specific graph classes, including powers of cycles and their complements [4], proper circular arc graphs [5], line graphs [6], quasi-line graphs [7], 3-arc graphs [8], complements of Kneser graphs [9] and 2-trees [10]. In [11], one Hadwiger-type conjecture with respect to path-chromatic number is proved. See [12,13] for two survey works

A stronger version of Hadwiger’s conjecture owing to Hajós states that every graph G with $\chi \left(G\right)\ge t$ contains a subdivision of $K_t$. A graph G is said to be an H-subdivision if it can obtained from H by a sequence of operations of edge subdividing. It is [14] proved that Hajós’ conjecture is incorrect for every $t\ge 7$. Obviously, if Hadwiger’s conjecture fails, then counterexamples must be detected within those examples counter to Hajós’ conjecture [15]. Another important direction for pushing research on this topic forward is to investigate dense graphs for Hadwiger’s conjecture [16]. Since dense graphs usually have very large chromatic number, intuitively, proving the existence of larger $K_t$-minor is believed to be not easy. In addition, there are also very few results have been obtained on the broad class of symmetric graphs. Motivated by these observations, we will examine a well-known family of symmetric graphs, namely, strongly regular graphs with high density.

In this paper, we concentrate on dense strongly regular graphs, that is, strongly regular graphs having small independence number. Recall that the independence number, often denoted by $\alpha \left(G\right)$, is the size of a maximum subset of pairwise nonadjacent vertices (namely, independent set) in G. A graph G on v vertices is called strongly k-regular if there are integers k, $\lambda$, and $\mu$ such that every vertex has degree k, every two adjoining vertices have $\lambda$ neighbors in common, and every two nonadjacent vertices have $\mu$ common neighbors. It can be observed that if G is strongly regular with parameters $(v,k,\lambda ,\mu )$, then the complement $\overline{G}$ of G is strongly regular with parameters $(v,\overline{k},\overline{\lambda },\overline{\mu })$, where $\overline{k}=v-k-1$, $\overline{\lambda }=v-2k+\mu -2$ and $\overline{\mu }=v-2k+\lambda$. Here, the complement $\overline{G}$ of G is a graph on the same vertices as G such that two distinct vertices of $\overline{G}$ are adjacent if and only if they are not adjacent in G.

We focus on strongly regular graphs G with $\alpha \left(G\right)=2$, in which no independent set of size three exists. That is, for every three distinct vertices of G, among which there is at least one adjacent pair. In this sense edges are dense in such G. We can quickly locate strongly regular graphs G with $\alpha \left(G\right)=2$ by the following observation:

Observation. A graph G of order v is strongly regular with parameters $(v,k,\lambda ,\mu )$ and $\alpha \left(G\right)=2$ if and only if the parameters $(v,\overline{k},\overline{\lambda },\overline{\mu })$ associated to $\overline{G}$ satisfy $\overline{\lambda }=v-2k+\mu -2=0$.

Proof. 

Let G be a strongly regular graph with parameters $(v,k,\lambda ,\mu )$ and $\overline{G}$ be its complement associated with $(v,\overline{k},\overline{\lambda },\overline{\mu })$. Since an independent set in G corresponds to a cliques in $\overline{G}$, the condition $\overline{\lambda }=0$ means that every two adjacent vertices in $\overline{G}$ (corresponding to an independent set in G) have $\overline{\lambda }=0$ neighbors in common. That is, $\overline{G}$ contains no triangle. Therefore, $\alpha \left(G\right)=2$. □

From this observation, we turn attention to two families dense strongly regular graphs: the complement of the Petersen graph and the complement of the Clebsch graph. As the Petersen graph is strongly regular with parameters $(10,3,0,1)$ and the Clebsch graph is strongly regular with parameters $(16,5,0,2)$, both with $\lambda =0$. It is known that the Petersen graph is one of the most famous graphs, which possesses interesting properties and often provides a counterexample to many graph-theoretic statements. The Petersen graph and relevant graphs have been received concentration on various optimization parameters, e.g., modeling texture images [17], chromatic Index [18], domination variants [19] and so on. While the Clebsch graph is actually the 5-folded cube graph, which can be obtained from the 5-hypercube graph $Q_5$ by identifying its antipodal vertices. Note that hypercubes and related graphs are classical tools in modelling interconnection networks because of their ideal topological properties [20].

The paper is organized as follows. In Section 2 we present a stronger result that the complement of the Petersen graph satisfies Hajós’ conjecture, which implies immediately that it satisfies Hadwiger’s conjecture as well. In Section 3 we prove similar results on complement of the Clebsch graph. In Section 4 we conclude the paper by proposing two open questions.

2. Petersen Graph

In this section, we focus on the complement of the well-known Petersen graph. We first show that the complement of the Petersen graph contains a subdivision of $K_5$ thus satisfies the Hajós’ conjecture, hence, satisfies Hadwiger’s conjecture. Then, we proceed further determining that the Hadwiger number of the complement of the Petersen graph is six, exactly its chromatic number plus one.

Let $PG$ be the Petersen graph with vertex set $U\cup V$, where $U=\{u_1,u_2,\dots ,u_5\}$ and $V=\{v_1,v_2,\dots ,v_5\}$. The adjacency relation is defined as usual and is depicted in Figure 1.

Theorem 1.

Hajós’ conjecture holds for the complement of $PG$.

Proof. 

Let $\overline{PG}$ be the complement of $PG$. It suffices to show that $\overline{PG}$ contains a $K_5$-subdivision. Now consider in $\overline{PG}$ the subgraph H induced on the five vertices $\{v_4,u_3,v_2,u_4,u_1\}$. Note that $\overline{PG}$ is the complement of $PG$, then two vertices are adjacent in $\overline{PG}$ if and only if they are non-adjacent in $PG$. Therefore, in H there are edges connecting $u_1$ to each of $\{v_4,u_3,v_2,u_4\}$, as well as edges $v_4{u}_3$, $u_3{v}_2$ and $v_2{u}_4$ (see solid lines in Figure 2). That is, all vertices in H pairwise adjacent (i.e., isomorphic to $K_5$) except the following three pairs $\{v_4,u_4\}$, $\{v_4,v_2\}$, and $\{u_4,u_3\}$ (see dashed lines in Figure 2). Hence, H is isomorphic to $K_5$ with three specific edges deleted.

Notice that in $\overline{PG}$, $u_2$ is adjacent to each of $\{v_4,u_4\}$, $u_5$ is adjacent to each of $\{v_4,v_2\}$ and $v_1$ is adjacent to each of $\{u_4,u_3\}$. Then we add to H the three vertices $u_2$, $u_5$ and $v_1$ to connect the three non-adjacent pairs $\{v_4,u_4\}$, $\{v_4,v_2\}$, and $\{u_4,u_3\}$, respectively. Now the subgraph H together with the three paths of length two $v_4{u}_2{u}_4$, $v_4{u}_5{v}_2$ and $u_4{v}_1{u}_3$ forms a graph isomorphic to a subdivision of $K_5$. □

Our next result determines the exact Hadwiger number of the complement of $PG$.

Theorem 2.

Let $\overline{PG}$ be the complement of the Petersen graph $PG$. Then $h\left(\overline{PG}\right)=6$.

Proof. 

We first show $h\left(\overline{PG}\right)\ge 6$ by constructing explicitly a $K_6$-minor in $\overline{PG}$. Consider the following six vertex-disjoint branch sets: $B_1:=\left\{u_1\right\}$, $B_2:=\{u_2,v_5\}$, $B_3:=\left\{u_3\right\}$, $B_4:=\{u_4,v_1\}$, $B_5:=\{u_5,v_4\}$ and $B_6:=\{v_2,v_3\}$. Then each of these branch sets induces a connected subgraph in $\overline{PG}$. For distinct i, j $\in \{1,2,\dots ,6\}$, there always exists one vertex in $B_i$, which is not adjacent to certain vertex in $B_j$ in $PG$. Hence, branches $B_i$ and $B_j$ are connected by at least one edge in $\overline{PG}$. Therefore, $B_1,B_2,\dots ,B_6$ provide a $K_6$-minor in $\overline{PG}$, which implies $h\left(\overline{PG}\right)\ge 6$. For instance, vertices $u_2\in B_2$ and $u_4\in B_4$ are not adjacent in $PG$, then in $\overline{PG}$ branches $B_2$ and $B_4$ are connected by the edge $u_2{u}_4$.

Next, we show $h\left(\overline{PG}\right)\le 6$ by contradiction. Here the inspiration comes from the structural properties of $\overline{PG}$: it has 10 vertices and each of degree 6, it contains no triangles. Thus, finding 7 disjoint but mutually adjacent branch sets among these 10 vertices without dense local clustering (triangle-free) is not easy. Suppose on the contrary that $\overline{PG}$ contains a $K_7$-minor say H. Without loss of generality assume that H is a $K_7$-minor with maximum singleton branches (branch containing unique vertex). Then, the collection of singleton branches induce in $\overline{PG}$ a complete subgraph, which is one-to-one corresponding to an independent set in $PG$. Since $\alpha \left(PG\right)=4$, there are at most 4 singleton branches in H. Now we can observe that the largest size of a branch set in H is at most 2. Otherwise, assume there is a branch set containing three vertices. Then the remaining seven vertices (ten vertices in total in $PG$) will be required to form the other six branch sets, which is impossible since the number of singleton branches is at most four. Consequently, each branch set is either a singleton or contains exactly a pair of vertices. Denote by t the number of singleton branches in H. Then, each of the remaining $7-t$ branch sets contains precisely two vertices. Thus, by counting the total number of H we have the inequality

$$t+2(7-t)\le 10.$$

Solving it gives $t\ge 4$. Thus $t=4$, that is, H contains four singleton branches. Moreover, the four singleton branches are those vertices of a maximum independent set of $PG$. By the symmetric property of $PG$, we can assume that $\left\{u_1\right\}$, $\left\{u_3\right\}$, $\left\{v_4\right\}$, and $\left\{v_5\right\}$ (whose union forms a maximum independent set in $PG$) are the singleton branches. Then the other three branch sets, each containing two vertices, form a partition of $V\left(PG\right)-\{u_1,u_3,v_4,v_5\}$. Let B be the branch set containing $u_2$. Then the other vertex in B must be from $V\left(PG\right)-\{u_1,u_3,v_4,v_5\}$ and non-adjacent to $u_2$. That is, the other vertex can only be chosen from $\{v_1,v_3,u_4,u_5\}$. Then we will see that each possible choice for the other vertex in $\{v_1,v_3,u_4,u_5\}$ produces a contradiction. Suppose first that $B=\{u_2,v_1\}$ or $B=\{u_2,u_5\}$. Then branch B and the singleton branch $\left\{u_1\right\}$ are not adjacent in H, a contradiction. Similarly, if $B=\{u_2,v_3\}$ or $B=\{u_2,u_4\}$, B will be non-adjacent to the singleton branch $\left\{u_3\right\}$. Therefore, the presumed $K_7$-minor does not exist. Hence, $h\left(\overline{PG}\right)\le 6$ and the result is established. □

In the proof of Theorem 2, the first step is to obtain a partition containing six parts from a subset of $U\cup V$ (that is, some vertices of $\overline{PG}$ are absent in constructing the desired minor). Then it is shown that the six parts of the partition are actually forming branches in a $K_6$ minor. After that, it is proved by contradiction that $\overline{PG}$ contains no $K_7$ minor.

Since the chromatic number of the complement of $PG$ is 5, we have

Corollary 1.

Hadwiger’s conjecture holds for the complement of $PG$.

Remark 1.

The result of Corollary 1 also follows from [3] as $\chi \left(\overline{PG}\right)=5\le 6$. Our results here confirm a stronger conjecture (namely, Hajós’ conjecture) for the graph $\overline{PG}$. In addition, we show that the hadwiger number of $\overline{PG}$ is strictly greater than its chromatic number.

3. Clebsch Graph

In this section, we focus on the Clebsch graph, which contains the Petersen graph as a subgraph and is also strongly regular with independence number two.

The Clebsch graph, also known as the Greenwood-Gleason graph, is the unique strongly regular graph with parameters $(v,k,\lambda ,\mu )=(16,5,0,2)$ [21]. The Clebsch graph has many other names and different representations. For instance, it is isomorphic to the 5-folded cube graph, which can be obtained from the 5-hypercube graph $Q_5$ by identifying antipodal points. It is also isomorphic to the 16-cyclotomic graph and 2-Keller graph.

Let S be a set and $A\subseteq S$. If the number of elements of A is even, we call A an even subset of S. The empty set ∅ is an even subset of every set.

Among different representations of the Clebsch graph, we use the following one: the vertex set consists of all even subsets of a 5-element set, say $\{1,2,3,4,5\}$, and two vertices (even subsets) are adjacent if their symmetric difference has cardinality 4. See Figure 3 for an illustration of the Clebsch graph, noting that it contains the Petersen graph as a subgraph (in blue/thick edges).

In the following, we first show that the complement of the Clebsch graph satisfies the Hajós’ conjecture. From which it follows immediately that Hadwiger’s conjecture holds for $\overline{CG}$.

Theorem 3.

Hajós’ conjecture holds for the complement of $CG$.

Proof. 

To prove the statement, we show that $\overline{CG}$ contains a $K_8$-subdivision. Consider the subgraph H induced by vertices $\{12,23,24,14,15,1234,1235,1245\}$ in $\overline{CG}$. Then, in H every pair of vertices are adjacent except the following pairs $\{23,15\}$, $\{23,14\}$, $\{24,15\}$, $\{1234,15\}$, $\{1235,14\}$, $\{1235,24\}$, and $\{1245,23\}$. (It is not hard to observe in H that vertex 12 is adjacent to every other vertex, 23 is connected to all other vertices but 15, 14, and 1245, and so on.)

Next for each non-adjacent pair above, we add one extra vertex to join them so that eventually the pair is connected by a path of length two. Specifically, for the non-adjacent pair $\{23,15\}$, we add vertex 35 as it adjoins both 23 and 15 in $\overline{CG}$. Analogously, we add 34, 45, 13, 1345, 2345, and 25 to join $\{23,14\}$, $\{24,15\}$, $\{1234,15\}$, $\{1235,14\}$, $\{1235,24\}$, and $\{1245,23\}$, respectively. Consequently, H together with these seven extra vertices (that is, seven paths of length two) gives rise to a $K_8$-subdivision. The result follows. □

From Theorem 3 it follows $h\left(\overline{CG}\right)\ge 8$. We next improve this result.

Theorem 4.

Let $\overline{CG}$ be the complement of the Clebsch graph $CG$. Then $h\left(\overline{CG}\right)=9$.

Proof. 

We first show that $h\left(\overline{CG}\right)\ge 9$ by constructing a $K_9$-minor in $\overline{CG}$. Let

$$B_1:=\left\{13\right\},B_2:=\left\{23\right\},B_3:=\left\{1234\right\},B_4:=\left\{1235\right\},B_5:=\{12,24\},$$

$$B_6:=\{34,1345\},B_7:=\{35,2345\},B_8:=\{14,15,25\},B_9:=\{45,1245,\varnothing \}.$$

Then in $\overline{CG}$, each $B_j$, $j=1,\dots ,9$, induces a connected subgraph, and by a routine verification, we can get that each pair of distinct $B_i$ and $B_j$ are connected for $i,j\in \{1,\dots ,9\}$. For example, see Figure 4, $B_1$ is joined to each of the remaining eight branches as depicted; $B_2$ and $B_6$ are connected via edge $\{23,34\}$, and $B_5$, $B_6$ are connected via edge $\{24,34\}$, et cetera. Therefore, $B_1,\dots ,B_9$ give rise to a $K_9$-minor in $\overline{CG}$, $h\left(\overline{CG}\right)\ge 9$.

Next we prove that $h\left(\overline{CG}\right)\le 9$ by contradiction. Suppose on the contrary that $\overline{CG}$ contains a $K_{10}$-minor, say H. Let t be the number of singleton branches (branch containing single vertex) and s the number of non-singleton branches in H. Then $t+s=10$. Without loss of generality assume that H is a $K_{10}$-minor with t maximum. Then the collection of t singleton branches induce in $\overline{CG}$ a complete subgraph, which is one-to-one corresponding to an independent set in $CG$. Since $\alpha \left(CG\right)=5$, $t\le 5$. Moreover, we have

$$t+2s\le 16.$$

By noting that $t+s=10$, we have $t+2(10-t)\le 16$, which implies $t\ge 4$. That is, $t=4,5$.

We first show that $t\ne 5$. Suppose otherwise $t=5$. By symmetry of $CG$, assume without loss of generality that each vertex with cardinality 4 (i.e., the neighborhood of vertex ∅ in $CG$) forms a singleton branch. Then there are 11 vertices left to compose the remaining five branches. Since each of these five branches contains at least two vertices, no branch contains four or more vertices. Let $s_i$ be the number of branches containing i vertices, then $i=2,3$ only. Combining $s_2+s_3=5$ and $2s_2+3s_3\le 11$, we get $s_3\le 1$. That is, at most one branch consists of three vertices. Let B be an arbitrary branch containing two vertices. Then the two vertices in $\overline{CG}$ are adjacent, hence non-adjacent in $CG$. Suppose that vertex ∅ is included in B. In $\overline{CG}$, since B is required to neighbor every singleton branch, and vertex ∅ is not adjacent to any singleton branch, the other vertex in B must be adjacent to every singleton branch. Such a vertex does not exist, otherwise this vertex together with the five singletons forms an independent set with order 6, which is a contradiction. Suppose now that vertex $\varnothing \notin B$. Since the two vertices in B are of cardinality 2 and not adjacent in $CG$, the two vertices in B are of the form $ab,bc$, where $a,b,c,d,e\in \{1,2,3,4,5\}$. Now $B=\{ab,bc\}$ and the singleton branch $\left\{acde\right\}$ are not adjacent in $\overline{CG}$, again a contradiction.

Suppose now that $t=4$. By symmetry, without loss of generality, we assume that each of the four singleton branches contains unique vertex of cardinality 4, say $B_1=\left\{1234\right\}$, $B_2=\left\{1235\right\}$, $B_3=\left\{1245\right\}$ and $B_4=\left\{1345\right\}$. Then each of the remaining branches $B_5,\dots ,B_{10}$ contains a pair of vertices. Let the branch containing vertex 2345 be $B_5$. Then the other vertex of $B_5$ is not adjacent to 2345 in $CG$. Consider two vertices $34,35$, at least one of which is not included in $B_5$. If $34\notin B_5$, let $34\in B_j$, $j\in \{6,\dots ,10\}$. Then the other vertex in $B_j$ must be of the form $3X$ or $4Y$, where $X,Y\in \{1,2,5\}$. If $B_j=\{34,3X\}$, then branches $B_j$ and $B_3=\left\{1245\right\}$ are not adjacent. If $B_j=\{34,4Y\}$, then branches $B_j$ and $B_2=\left\{1235\right\}$ are not adjacent. In both cases, we have a contradiction. If $35\notin B_5$, the analysis is similar and again lead to a contradiction. Consequently, $h\left(\overline{CG}\right)\le 9$ and the result follows. □

The main idea of the proof of Theorem 4 is analogous to that of Theorem 2, that is, it first obtain a partition containing nine parts from $V\left(\overline{CG}\right)$ (but different here, all vertices of $\overline{CG}$ will be used in constructing the desired minor). Then it is shown that the nine parts are forming branches in a $K_9$ minor.

Since the complement of the Clebsch graph has chromatic number 8, we have

Corollary 2.

Hadwiger’s conjecture holds for the complement of $CG$.

4. Conclusions

We have proved that the complements of both Petersen graph and Clebsch graph satisfy Hajós’ conjecture, a strengthening version of Hadwiger’s conjecture. Moreover, we determine the exact hadwiger numbers of the complements of Petersen graph and Clebsch. In fact, for all nontrivial strongly regular graphs with order not greater than 10, their chromatic numbers, independence number and hadwiger number can be calculated, as shown in

Table 1. Values of $\chi \left(G\right)$, $\alpha \left(G\right)$, $h\left(G\right)$ and whether satisfying Hajós’ conjecture for small strongly regular graphs.

$(\mathit{v},\mathit{k},\mathit{\lambda },\mathit{\mu })$	Graph	$\mathit{\chi }\left(\mathit{G}\right)$	$\mathit{\alpha }\left(\mathit{G}\right)$	$\mathit{h}\left(\mathit{G}\right)$	Hajós’ Conjecture
$(4,2,0,2)$	cycle $C_4$	2	2	3	Yes
$(5,2,0,1)$	cycle $C_5$	3	2	3	Yes
$(6,3,0,3)$	$K_{3,3}$	2	3	4	Yes
$(6,4,2,4)$	$K_{2,2,2}$	3	2	4	Yes
$(8,4,0,4)$	$K_{4,4}$	2	4	5	Yes
$(8,6,4,6)$	$K_{2,2,2,2}$	4	2	5	Yes
$(9,4,1,2)$	9-Paley	3	3	5	Yes
$(9,6,3,6)$	$K_{3,3,3}$	3	3	5	Yes
$(10,3,0,1)$	$PG$	3	4	5	Yes
$(10,5,0,5)$	$K_{5,5}$	2	5	6	Yes
$(10,6,3,4)$	$\overline{PG}$	5	2	6	Yes
$(10,8,6,8)$	$K_{2,2,2,2,2}$	5	2	6	Yes

.

In view of the values in Table 1, we can observe first that each of these small strongly regular graphs satisfies Hajós’ conjecture. If, in addition, the graph G is complete m-partite, then its hadwiger number is strictly greater than its chromatic number. And this conclusion can be easily extended to all such graphs with arbitrary larger order. The constructions obtained in this paper exploit and rely heavily on specific combinatorial properties of the graphs discussed. It would be meaningful to extend the idea to investigate other strongly regular graphs with independence number two or higher. Intuitively, we believe that confirming the Hadwiger’s conjecture on strongly regular graphs with independence two is more difficult than on those of less density (higher independence number). From these facts, we conclude this paper by proposing the following question:

Question 1. Is every strongly regular graph satisfing Hajós’ conjecture?

Question 2. Is every strongly regular graph G with $\alpha \left(G\right)=2$ or 3 and order greater than 10 containing a $K_t$-minor with $t>\chi \left(G\right)$?