# Unique colorability and clique minors

###### Abstract

For a graph , let denote the largest such that has pairwise disjoint pairwise adjacent connected nonempty subgraphs, and let denote the largest such that has pairwise disjoint pairwise adjacent connected subgraphs of size or . Hadwiger’s conjecture states that , where is the chromatic number of . Seymour conjectured for all graphs without antitriangles, i. e. three pairwise nonadjacent vertices. Here we concentrate on graphs with exactly one -coloring. We prove generalizations of (i) if and has exactly one -coloring then , where the proof does not use the four-color-theorem, and (ii) if has no antitriangles and has exactly one -coloring then .

AMS classification: 05c15, 05c40.

Keywords: coloring, clique minor, Kempe-coloring, Hadwiger conjecture.

## 1 Introduction

All graphs throughout are assumed to be finite, simple, and undirected unless otherwise stated, and for terminology not defined here we refer to [2] or [3]. An anticlique of a graph is a set of pairwise nonadjacent vertices of , a coloring of is a partition of into anticliques, and denotes the chromatic number of , that is, the smallest integer such that admits a coloring of size . A clique minor of is a set of pairwise disjoint pairwise adjacent connected nonempty subsets of , where two subsets of are adjacent if there exists an edge containing a vertex of each of them, and a subset of is connected if it induces a connected subgraph of . A clique minor is shallow if all its members have size or . Let be the largest such that admits a clique minor of size , and let denote the largest such that admits a shallow clique minor of size .

Hadwiger conjectured for all graphs [5]. So far, this is known for graphs with , where the statements restricted to or to are equivalent to the four-color-theorem, respectively [8]. Seymour conjectured for all graphs without an antitriangle, that is, an anticlique of size (see [1]). Here, we target these conjectures under the additional assumption of unique colorability, that is, there exists exactly one -coloring for the graphs under consideration. In the case of Hadwiger’s conjecture, we do not get new facts but show that the four-color-theorem is not essential:

###### Theorem 1

If and admits exactly one coloring of size then , and the proof of this statement does not rely on the four-color-theorem.

Instead of assuming unique colorability, we look at a more general coloring concept: A coloring of is a Kempe-coloring if any two distinct members from induce a connected subgraph of . It is far from being true that a graph has a Kempe-coloring of any size at all; however, if has only one coloring of size then is a Kempe-coloring, for if, for distinct from , had more than one component, then we take one, say , and observe that is a coloring of size distinct from , contradiction. is obtained from by “exchanging colors” along the “Kempe-chain” , an absolutely classic process in graph coloring theory. The term Kempe-coloring thus simply indicates that we cannot apply it as to obtain new colorings of the same size. Now it is clear that Theorem 1 is a consequence of the following.

###### Theorem 2

If and admits a Kempe-coloring of size then , and the proof of this statement does not rely on the four-color-theorem.

Whereas I could not relax in Theorem 1 or Theorem 2, the situation improves for Seymour’s conjecture.

###### Theorem 3

If has no antitriangle and exactly one coloring of size then ; in particular, .

It is quite obvious that the colorings of size of a graph without antitriangles correspond to the maximum matchings of its complementary graph , so that matching theory is naturally involved. I could not relax the condition of having exactly one coloring of size in Theorem 3 to the condition of just having a Kempe-coloring of size . However, there is the following, more general “rooted version” of Theorem 3. Recall that a transversal of a set of sets is a set with for every . If is a transversal of then we also say that is traversed by .

###### Theorem 4

Suppose that has no antitriangles and exactly one coloring of size . Then, for every transversal of , there exists a shallow clique minor of size traversed by .

## 2 Unique colorability and Hadwiger’s conjecture

The main ingredience for the proof of Theorem 2 is the following result of Fabila-Monroy and Wood on “rooted -minors” [4].

###### Theorem 5

[4, Theorem 8, (1.)(3.)] Let be a graph and be a set of four vertices of . Then there exists a clique minor of size four traversed by if and only if for any two from there exists a path in such that the two vertices from are in the same component of .

The proof of Theorem 5 in [4] is almost self-contained and, in particular, does not use the four-color-theorem. Only little progress has been made so far on generalizing Theorem 5 to “rooted -minors”; it is not clear if there is a set of reasonable “linkage conditions” as there at all [10].

Proof of Theorem 2. We first prove the statement for . Let be a Kempe-coloring of size . is connected and has size at least . There exists a vertex such that is connected and nonempty (take a leaf of any spanning tree of ). Without loss of generality, we may assume . is connected and nonempty, too, and adjacent to . Observe that, for from , every vertex from must have at least one neighbor in . It follows that has a neighbor in for every . Let from , and let be the two indices in . There is a -path in and a -path in . Clearly, are disjoint. By Theorem 5, has a clique minor with in for all . Consequently, is a clique minor, and, since has a neighbor in , is a clique minor, too. This proves . —

For , let be obtained from by adding a set of new vertices and connecting every to every other vertex from by a new edge. (Later we will refer to the vertices of as apex vertices.) Then is a Kempe-coloring of of size so that, by what we have just proved, has a clique minor of size . Since every vertex from is contained in at most one member of , is a clique minor of size at least of . Since the -color-theorem is neither used in the proof of Theorem 5 nor in the preceeding arguments, this proves Theorem 2.

## 3 Unique colorability and rooted shallow clique minors

For the proof of Theorem 4, we need the following result from matching theory by Kotzig [6]. (For a short proof of a more general result let me refer to [11].) Recall that an edge is a bridge of a graph , if is disconnected for some component of .

###### Theorem 6

[6] If is the only perfect matching of some nonempty graph then contains a bridge of .

Proof of Theorem 4. We do induction on . Let , , and be as in the statement, and set . Observe that is a Kempe-coloring of size , and all members of have order or . If some member of consisted of a single vertex, say, , only, then and must be a star with center for all ; it follows that is adjacent to all other vertices in . It is easy to see that , and that is the unique coloring of size of . Clearly, is a transversal of so that, by induction, admits a shallow clique minor traversed by . But then one readily checks that is a shallow clique minor of traversed by .

Hence all members of have size , and, thus, they correspond to the edges of a perfect matching of the complementary graph of . Conversely, if had another perfect matching , then the edges of corresponded to the classes of a coloring of size of distinct from , a contradiction. Therefore, is the only perfect matching of .

We now construct a descending sequence of subgraphs of with unique perfect matchings as follows. Set , suppose that have been constructed, and let be the unique perfect matching of . By Theorem 6, contains a bridge of . Let be a component of containing exactly one vertex of such that , and let be the vertices of , where . Observe that for all . It is easy to see that is the unique perfect matching of (which is possibly empty), and that is the unique perfect matching of ; by the size condition to , the latter matching contains at least as many edges as , and we choose a subset of size exactly from it. Set (so is the unique perfect matching of ). Set ; then is the unique perfect matching of . We iterate until is empty.

If then we can construct a set of cliques of size of traversed by such that each contains both a member of and a member of : Let be the edges of and be those of , where and are from ; if then set , , , and if, otherwise, then set (see Figure 1 for an example of the first kind).

Otherwise, , so that consists of only; if then we set , and in the other case we call the index special and set . Observe that each is a shallow clique minor of traversed by .

We first apply induction to ( , with the unique coloring of size corresponding to the edges of , traversed by ), and find a shallow clique minor of traversed by .

Suppose that is not a special index. Since every member of contains a vertex from and all of these are adjacent to all of , is a shallow clique minor of traversed by , and we are done. Hence we may assume that is a special index. If is adjacent to all members from in then is a shallow clique minor of traversed by . Hence we may assume that is nonadjacent to at least one vertex . There exists a vertex such that ; clearly, is not in as is in , and is adjacent to as contains no antitriangles. There exists a unique such that , and we choose (and ) such that is as large as possible. If then let be the vertex from distinct from ; otherwise, there exists a set such that , and we let be the vertex in . Observe that, in both cases, follows. By induction, applied to , there exists a shallow clique minor of size traversed by .

Now let be the special indices smaller than in increasing order (see Figure 2 for an example).

We let , and for , and . Since and are in different components of and , is a clique. Since and are in different components of and , is a clique for . By construction, is a clique. All of are pairwise disjoint, disjoint from any clique in , and disjoint from any clique in with the only exception that meets the unique clique of containing (which is either or ) if is not special. Observe that has size if is not special, and size if is special. Therefore, consists of exactly many disjoint cliques of size or . Moreover, is traversed by , as , each , and are traversed by , respectively.

We accomplish the proof by showing that is a clique minor. To this end, it suffices to prove for all that if contains some vertex from and contains some vertex from then are adjacent. This is true for , as are members of the clique minor in this case.

Consider and from with and . Suppose, to the contrary, that are not adjacent in . If is contained in then it contains a vertex from , which is adjacent in to all vertices from , implying that unless for some . If then and ; since is adjacent to all of , it is adjacent to all of , contradiction. Hence . If then and , so that is not traversed by , contradiction. Thus, has size and contains a vertex from , which is adjacent in to all vertices from , so that . Since are nonadjacent in , or follows. We infer (as all of have size ), so that does not contain a member of , contradiction. Therefore, must be among .

If for some then contains from , and contains from if or from if . In both cases, , and from is adjacent to all of , and, thus, to all of , contradiction. If, finally, then . By choice of , is nonadjacent to all vertices from and, hence, adjacent to all of , implying that . Since every vertex from , , is adjacent to , we know that is adjacent to all from and to all of . It follows that . But then has a vertex in and a vertex in ; if then it is adjacent to all of , and if then it is adjacent to all of , so that in either case, is adjacent to .

## 4 Open problems

So far, I could neither generalize Theorem 2 to the case , nor Theorem 4 under the weaker assumption that there is a Kempe-coloring of order instead of a unique coloring of size , not even if the conclusion asks for just any clique minor of size instead of a shallow one. The following, being a common generalization of these two projects, is, therefore, perhaps a little bit too optimistic.

###### Conjecture 1

Suppose that has a Kempe-coloring of size . Then, for every transveral of , there exists a clique minor traversed by .

For , this follows from Theorem 5, as in the proof of Theorem 2 in Section 2, and the statement inherits to smaller values of by augmenting to as above (add only instead of apex vertices and augment to ). It suffices to prove the Conjecture for the case that is (edge-) minimal with the property that is a Kempe-coloring of size . This is equivalent to saying that any two distinct members of induce a tree, which is in turn equivalent to saying that — independent from the actual Kempe-coloring of size — the graph has exactly edges. Since, for fixed , the latter is an increasing function in as long as , it follows easily that the minimality condition implies that has no Kempe-coloring of size larger than (proofs are left to the reader).

In general, an affirmative answer to Conjecture 1 could give us a clique minor of much larger size than we actually need to verify Hadwiger’s conjecture for the graphs under consideration, because the difference of the size of a (largest) Kempe-coloring (if any) and may be arbitrarily large: Consider the graph on , , where are connected if and only if and and (. Consider . For , induces a path on six vertices (see Figure 3), which implies that is a Kempe-coloring of size and is minimal with that property.

(By the remark above, would even be a largest Kempe-coloring of .) On the other hand, is a coloring of size , and the triangle formed by , , shows . Adding apex vertices to as in the proof of Theorem 2 yields a graph with a Kempe-coloring of size and chromatic number .

The proof of Theorem 4 (and also the much easier proof of the unrouted version, where special indices can be avoided) depends heavily on Kotzig’s Theorem, Theorem 5, and the latter does not generalize to the situation of a Kempe-coloring because there are infinitely many graphs without antitriangles admitting a Kempe-coloring of size whose complementary graph is bridgeless and, in fact, -connected: Let be a -connected tournament on vertices. We construct from by introducing a pair for every vertex from and connect and if or there is a directed edge from to in . It is easy to see that is a -connected bipartite graph on vertices. The perfect matching of corresponds to a coloring of size in , and, since distinct members of are connected by only one edge in , any pair of members of induces a connected graph (in fact: a path of length ) in . Therefore, is a Kempe-coloring of (even a largest one). — An answer to the following question could lead further:

###### Problem 1

Let be a triangle free graph with a perfect matching such that any two edges of are connected by at most one edge (out of the four possible edges). Does there exists an with such that every -cycle of contains two nonadjacent vertices from or two nonadjacent vertices from ?

The answer is ,,yes” if, for example, is bipartite (like the graphs above obtained from tournaments) or if the girth of is larger than . (In fact, I do not know a single triangle-free graph on an even number of vertices where the answer is ,,no”, but I should doubt that there aren’t any.) The point is, of course, that if is a graph without antitriangles providing a Kempe-coloring of size then meets the conditions in Problem 1 (where the matching of corresponds to the Kempe-coloring of ). Suppose that we get a set as in Problem 1. We may assume that is -connected (see below), so that, by Hall’s or Menger’s Theorem (see [3] or [2]), there exists a perfect matching from to in . If, for distinct from , all four potential edges between were absent in , then would induce a -cycle in without two nonadjacent vertices from or two nonadjacent vertices from , contradiction. So the edges of correspond to a shallow clique minor of .

Concerning the connectivity issue, let us prove the following Lemmas.

###### Lemma 1

Suppose that has a Kempe-coloring and let be a separator of . Then contains an element of all but at most one member of .

Proof. Suppose that there exists a such that . Then there exists a component of such that . Let be another component of . If then contains a member of every , since is connected. Otherwise, , and there exist and with . Since is connected, contains a member of . For any , has a neighbor . Since and is connected, contains a member of , too.

###### Lemma 2

Suppose that has no antitriangles and has a Kempe-coloring of size . Then is -connected or admits a shallow clique minor of size .

Proof. Suppose that is a minimal counterexample to the statement of the Lemma, and let be a Kempe-coloring of of size . Since is not a clique of size , had a separator of size by the previous Lemma. Among all separators of size , we choose and a component of such that is minimal (with respect to either set inclusion or size).

If some member of consisted of a single vertex then, as usual, would be adjacent to all other vertices, would be a separator of of size , and would be a Kempe-coloring of size of . By minimality of , had a shallow clique minor of size , so that would be a shallow clique minor of size of , contradiction.

Hence all members of have size . There exists a with . By Lemma 1, for all . Since has no antitriangles, has exactly one further component distinct from , and and induce cliques. We classify by and . If then there exists a as , but the vertex in cannot have a neighbor in , contradicting the fact that is connected. It follows that there exist unique vertices and .

Let . If there was no matching from into then, by Hall’s Theorem (see [2] or [3]), for some , so that contained the neighborhood of . Since is a nonempty (proper) subset of and , is a separator of of size less than , contradiction. Therefore, we find a matching from into .

Let . If there was no matching from into then, by Hall’s Theorem, for some , so that contained the neighborhood of . Since is a nonempty proper subset of and , is a separator of of size and contains a component of whose vertex set is properly contained in , contradicting the choice of . Therefore, we find a matching from into . (It may be that .)

The edges of are pairwise adjacent, since is a clique, and the edges of are pairwise adjacent, since is a clique. Now let and . Then there exists a and and a and . Since is connected and lacks the edge connecting the two vertices in , we know that . Therefore, are adjacent. Now is not an endvertex of any member of , but it is adjacent to all of as is a clique and to all of as, for every , must have a neighbor in which can only be the vertex of .

It follows that is a shallow clique minor of size in .

According to the above arguments, verifying Conjecture 1 for some value of would verify the statement of Theorem 2 for ; therefore, I think it would be already interesting to verify Conjecture 1 for the smallest open case of . In general, graphs on vertices with a Kempe-coloring of size must have at least edges; by a classic result of Thomassen [9], they admit not only a minor but even a subdivision of where one could, moreover, prescribe a single vertex in the interior of one of the subdivision paths. By a result of Mader [7] (answering a question of Dirac), edges suffice to guarantee a clique minor of size (implying that there is such a minor in any graph which has an edge such that has a Kempe-coloring of size ). Both results indicate that there is considerable freedom in choosing a clique minor of size provided that has a Kempe-coloring of size .

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Author’s Address.

Matthias Kriesell

Technische Universität Ilmenau

Weimarer Straße 25

D–98693 Ilmenau

Germany