Reflexive coloring complexes for 3-edge-colorings of cubic graphs
RReflexive coloring complexes for 3-edge-colorings ofcubic graphs
Fiachra Knox ∗ Bojan Mohar †‡ Nathan SingerDepartment of MathematicsSimon Fraser UniversityBurnaby, BC V5A 1S6, CanadaApril 16, 2020
Abstract
Given a 3-colorable graph X , the 3-coloring complex B ( X ) is the graph whosevertices are all the independent sets which occur as color classes in some 3-coloringof X . Two color classes C, D ∈ V ( B ( X )) are joined by an edge if C and D appeartogether in a 3-coloring of X . The graph B ( X ) is 3-colorable. Graphs for which B ( B ( X )) is isomorphic to X are termed reflexive graphs. In this paper, we consider3-edge-colorings of cubic graphs for which we allow half-edges. Then we consider the3-coloring complexes of their line graphs. The main result of this paper is a surprisingoutcome that the line graph of any connected cubic triangle-free outerplanar graph isreflexive. We also exhibit some other interesting classes of reflexive line graphs. In [6], Tutte examined how many connected components the 4-coloring complex (see thedefinition below) of a triangulation of the plane could have. This was a question he hadexamined since the times of the Four Color Conjecture [5].Edge-colorings of cubic graphs appear naturally in this setting. By the well-knowncoloring-flow duality, 4-colorings of triangulations are in bijective correspondence with 3-edge-colorings of their dual cubic graphs. Nevertheless, the coloring complex correspondingto 4-colorings of a triangulation and the coloring complex of 3-edge-colorings of the dual ∗ Supported in part by a PIMS Postdoctoral Fellowship † Supported in part by the NSERC Discovery Grant R611450 (Canada), by the Canada Research Chairsprogram, and by the Research Project J1-8130 of ARRS (Slovenia). ‡ On leave from IMFM, Department of Mathematics, University of Ljubljana. a r X i v : . [ m a t h . C O ] A p r ubic graph may be very different. For example, the coloring complex of the icosahedronis connected, while all ten different 3-edge-colorings of the dual dodecahedron give separatecomponents in the corresponding coloring complex.Early on, this kind of questions raised attention, and Biggs [1] showed that the Coxetergraph, which has 28 vertices, has twice as many 3-edge-colorings, and the correspondingcoloring complex is the line graph of a 2-arc-transitive cubic graph B of order 56. Fromthe list of known 2-arc-transitive graphs, Biggs concluded that this graph B is the famousKlein map of genus 3 and of type { , } . It was only later that Fisk [3] actually discoveredthat there was an error in this conclusion, because the coloring complex in question isdisconnected, and is in fact isomorphic to two copies of the line graph of the Coxeter graph(which by itself is an interesting phenomenon).In his attempts to answer questions that were posed by Tutte and others, Fisk [3] intro-duced the notion of a reflexive coloring complex (see Section 2 for definition). Furthermore,he established in [3] that the 3-coloring complexes of line graphs of “cubic” cycles and pathsare reflexive. By cubic cycles and paths we mean cubic graphs obtained from these by addinghalf-edges. In his monograph [4], Fisk provided a number of further developments in thisarea.This paper continues the work started by Fisk. Our main result shows that the line graphof a connected cubic outerplanar graph G (with added half-edges) is reflexive if and only if G is triangle-free. This result, which appears as Theorem 9, gives a large infinite class ofreflexive line graphs. In the second part of the paper we discover several infinite families ofadditional, interesting examples of reflexive line graphs.Our results may be just the tip of an iceberg. The real question that remains open iswhy there are so many reflexive graphs (and why there are any at all). Our results andextensive computational evidence say that graphs with a large number of colorings tend tobe reflexive, although this appears counterintuitive from another perspective: when we havemany colorings, the coloring complex is large, so it may have too many colorings for theoriginal graph to be reflexive. Let X be a k -colorable graph and let us consider a k -coloring of X , which we treat as apartition { V , V , . . . , V k } of V ( X ) into k independent sets V i (1 ≤ i ≤ k ). Here, some of theparts of the partition may be empty, but in all considered cases we will have the propertythat X contains a ( k − V i arising in a k -coloring of X are called the color classes of thecoloring. The k-coloring complex B ( X ) is the graph whose vertices are the color classes ofall k -colorings of X . Two vertices C, D ∈ V ( B ( X )) are joined by an edge if the color classes C and D appear together in a k -coloring of X . It was Tutte [5] who used the word “complex” for B ( X ). The reason for this terminology is that B ( X )can be viewed as a simplicial complex in which each k -clique corresponding to a k -coloring of X is made intoa simplex of dimension k −
2t is not hard to see that the graph B ( X ) is k -colorable under the following simplecondition. Lemma 1 (Fisk [3]) . Let X be a k -colorable graph that contains a ( k − -clique. Then B ( X ) is k -colorable.Proof. Let Q be a ( k − X . For i ∈ [ k − C i be the set of all the color classesin k -colorings of X which contain the i th vertex of Q . Then {C i : i ∈ [ k − } ∪ { ( V ( B ( X )) \∪ k − i =1 C i ) } is a k -coloring of B ( X ).When B ( X ) is k -colorable, we can consider its k -coloring complex B ( X ) = B ( B ( X )).It appears surprising that X and B ( X ) are closely related. Lemma 2 (Fisk [3]) . Let X be a k -colorable graph without isolated vertices in which eachedge is contained in a ( k − -clique. Then the mapping φ X : v (cid:55)→ { C ∈ V ( B ( X )) | x ∈ C } ( v ∈ V ( X )) is a graph homomorphism X → B ( X ) . The graph homomorphism outlined in Lemma 2, φ X : X → B ( X )maps a vertex v ∈ V ( X ) to the set of all color classes of X containing v . We will referto it as a canonical homomorphism . Graphs for which the canonical homomorphism is anisomorphism are termed reflexive graphs and they will be our main concern in this paper.There are no obvious reasons why any graph would be reflexive, yet there are interestinginfinite classes. Our main goal is trying to understand this phenomenon.The following is a necessary condition for φ X to be reflexive. We say that the graph X is colorful (for k -colorings) if for any two vertices x, y there exists a k -coloring, which has x and y in different color classes. Let us add the following basic observation: Observation 3.
A graph X is colorful for k -colorings if and only if the mapping φ X isinjective. Let us now assume that k = 3. (This assumption will be kept throughout the paper.)The 3-coloring complex B ( X ) of a graph X is composed of triangles, one triangle for every3-coloring of X . The following result shows that B ( X ) has no other triangles and also showsthat the triangles are edge-disjoint. Lemma 4.
Let X be a 3-chromatic graph without isolated vertices. Then any triangle in B ( X ) represents a 3-coloring of X . Consequently, each edge of B ( X ) is contained in preciselyone triangle. Proof.
Suppose, for a contradiction, that B ( X ) contains a triangle C C C which does notrepresent a 3-coloring of X . Each edge of C C C must be in a 3-coloring of X . Let C ij bethe third color class of the 3-coloring containing C i and C j , for each i (cid:54) = j in { , , } .Now, C ∩ C = C ∩ C = C ∩ C = ∅ . Hence, C ⊆ C , C ⊆ C and C ⊆ C .Let H = V ( X ) \ ( C ∪ C ∪ C ). Then C = V ( X ) \ ( C ∪ C ) = C ∪ H , C = C ∪ H and C = C ∪ H . If H = ∅ , then C = C and C C C is a 3-coloring of X , which is acontradiction. Thus, H (cid:54) = ∅ ; let us take a vertex v ∈ H . Since X has no isolated verticesand H is an independent set in X , v has a neighbour u / ∈ H . Without loss of generality,suppose that u ∈ C . Then, as C ⊆ C , u, v ∈ C . However, this is a contradiction, since u and v are neighbours, while C is an independent set.Let us now consider an edge AC in B ( X ). By the above, any triangle containing AC corresponds to a 3-coloring with color classes A and C . But the third color class is just thecomplement of A ∪ C , so there is only one such 3-coloring.As we will primarily concern ourselves with 3-edge-colorings in this article, we add a fewrelated definitions. Firstly, given a 3-edge-colorable graph G , we define the of G as the 3-coloring complex of the line graph L ( G ) of G . Furthermore, we saythat a 3-edge-colorable graph G is edge-reflexive if L ( G ) is a reflexive graph, and that G is edge-colorful if L ( G ) is colorful.When we refer to a cubic graph , we allow half-edges . These are edges which are incidentwith only one vertex. In this way, we can treat every graph of maximum degree three asa cubic graph by adding half-edges to the vertices of smaller degrees. (On the other hand,we do not allow double edges, since cubic graphs containing double edges cannot be edge-colorful, unless we have a triple edge.) With this understanding, we will in particular speakof cubic paths , cubic cycles and cubic trees . We refer to Figure 1 for some examples.The following observation (combined with Observation 3) shows that cubic graphs thatcontain triangles cannot be edge-reflexive. Observation 5.
A cubic graph containing a triangle is not edge-colorful. roof. Let abc be a triangle in G and let e be the third edge incident with the vertex a .Then the edges e and bc have the same color in every 3-edge coloring of G .We also have the following lemma. Lemma 6.
Let G be a cubic edge-colorful graph of order n and let X = L ( G ) . The followingstatements are equivalent: (i) G is edge-reflexive. (ii) B ( X ) has precisely n -colorings. (iii) For every -coloring {A , B , C} of B ( X ) , there is a vertex in G with incident edges e, f, g such that {A , B , C} = { φ X ( e ) , φ X ( f ) , φ X ( g ) } .Proof. As argued in the proof of Lemma 1, all partitions of V ( B ( X )) of the form { φ X ( e ) ,φ X ( f ) , φ X ( g ) } are 3-colorings of B ( X ). This shows that (ii) and (iii) are equivalent.Next, observe that φ X : X → B ( X ) is injective and that every triangle in X correspondsto a vertex of G . Thus X has precisely n triangles. This yields equivalence of (i) and (ii).Finally, let us add a necessary condition for reflexivity. Lemma 7.
Let X be a graph that is reflexive for 3-colorings. If v is a vertex of degree d in X , then d = 2 t , where t is the number of components of the bipartite graph B ( X ) − φ X ( v ) .Proof. The set φ X ( v ) is the color class for 3-colorings of B ( X ), and since B ( B ( X )) is iso-morphic to X , the number of 3-colorings of B ( X ) that have φ X ( v ) as one of the color classesis equal to d/ d/ B ( X ) − φ X ( v ). In particular, this subgraph is bipartite,and it is clear that the number of 2-colorings is equal to 2 t − . Thus, d = 2 · t − = 2 t .When applied to edge-colorings, we obtain the following corollary. Corollary 8.
Let G be a cubic edge-reflexive graph and let X = L ( G ) . For each edge e of G , the graph B ( X ) − φ X ( e ) is bipartite, and its either connected (when e is a half-edge), orhas precisely two connected components (when e is a full edge). Throughout the remainder of this article, we will assume that cubic graphs G satisfy thehypotheses of Theorem 9 below: they are always connected, and they may have half-edges.Recall that a graph G is outerplanar if G has a planar drawing in which all of its vertices ap-pear on the unbounded face of the drawing. We will consider cubic triangle-free outerplanargraphs. Let us observe that all such graphs must have at least four half-edges.5iorini [2] proved that every cubic outerplanar graph is 3-edge-colorable. The same proofshows that these graphs are edge-colorful whenever they are triangle-free.The main result of this paper is the following somewhat surprising theorem, whose proofoccupies the rest of this section. Theorem 9.
Let G be a connected cubic outerplanar graph. Then G is edge-reflexive if andonly if it is triangle-free. The nontrivial direction of Theorem 9 will be proved in three steps. First, we will reducethe problem to 2-connected graphs. Then we will exhibit two operations which are used toconstruct all 2-connected cubic triangle-free outerplanar graphs from a 4-cycle. Finally, wewill show that edge-reflexivity is preserved under both of these operations, completing ourargument that any connected, triangle-free, outerplanar graph G is edge-reflexive.Before proceeding, we will need to establish a brief lemma, for which we need to introducea definition. Suppose that S ⊆ V ( X ) and that we have a 3-coloring Γ = { A, B, C } of X [ S ].A coloring { A (cid:48) , B (cid:48) , C (cid:48) } of X is an extension of the coloring Γ if A ⊆ A (cid:48) , B ⊆ B (cid:48) , and C ⊆ C (cid:48) (or some permutation of color classes satisfies the same inclusion relations). If v is a vertexand for every extension of Γ, v is in the color class containing the same color class in Γ, thenwe say that the color of v is determined by Γ (or that the coloring Γ determines the color of v ). If colors of all vertices are determined by Γ and Γ has an extension, we say that Γ determines the extension (or that Γ extends uniquely to X ). Lemma 10.
Let F be a graph with a 3-coloring whose color classes are A , B and C . Supposethat F has no isolated vertices, that every edge of F is contained in exactly one triangle, andthat F [ B ∪ C ] is connected. Then any -coloring of F that uses at least two colors on A isdetermined by its restriction to A .Proof. Let x and y be two vertices of A which have different colors, and let v, w ∈ B beneighbours of x and y , respectively. Let P = v v . . . v n be a path in F [ B ∪ C ] with v = v and v n = w . For each i ∈ [ n − x i ∈ A be the common neighbour of v i and v i +1 .Relabel y as x n . Let j ∈ [ n ] be such that x j and x j − have different colors. Such an indexexists because x and x n have different colors. Then v j is adjacent to both of these vertices,and hence its color is determined. Relabel v j as z .Now let u be any vertex in B ∪ C . Let Q = u u . . . u m be a path in F [ B ∪ C ] with z = u and u = u m . The color of u is determined, and whenever the color of u i is determined, sois the color of u i +1 (since it is in a triangle with u i and a vertex of A , both of whose colorsare determined). By induction the color of u = u m is determined. Since u was arbitrary, theentire coloring is determined. In this section, we will prove that a cubic graph G is edge-reflexive whenever each blockof G is edge-reflexive. Thus, we will be able to reduce edge-reflexivity questions to 2-edge-connected graphs. 6et G be a cubic graph with a cutedge e joining vertices a and b . By cutting the edge e we obtain two cubic graphs H and K which are obtained from G − e by adding a half-edge to a and b , respectively. The added half-edge will be considered to be the same as the removededge e , so that we can consider E ( H ) ⊆ E ( G ), E ( K ) ⊆ E ( G ), and E ( H ) ∩ E ( K ) = { e } . Lemma 11.
Let G be a cubic graph with a cutedge e . Let H and K be cubic graphs obtainedfrom G by cutting the edge e . If H and K are edge-reflexive, then G is edge-reflexive, too.Proof. We let X = L ( H ), Y = L ( K ), and X (cid:48) = L ( G ). Let w and x be the neighbours of e in X , and let y and z be the neighbours of e in Y . Before proceeding further, let us firstobserve that since X and Y are reflexive, they are also colorful. This implies that X (cid:48) is alsocolorful. Thus, φ X (cid:48) is an injective homomorphism X (cid:48) → B ( X (cid:48) ), and therefore it suffices toshow that X (cid:48) and B ( X (cid:48) ) have the same number of triangles. Triangles in X (cid:48) correspond tovertices in V ( G ) = V ( H ) ∪ V ( K ), while the triangles in B ( X (cid:48) ) correspond to 3-colorings of B ( X (cid:48) ).Now, for any color class C ∈ V ( B ( X )), let F C be the set of color classes of X (cid:48) whichcoincide with C on the vertices of X . If e ∈ C , we also write F eC to denote the same set ofcolor classes. Observe that when e ∈ C and D ∈ F C , we have y, z / ∈ D . On the other hand,if e / ∈ C then exactly one of y and z is in D . In this case, we partition F C into the subset F yC consisting of color classes which contain y , and the subset F zC of color classes containing z . We refer to the set F eC , or the sets F yC and F zC as the clusters of B ( X (cid:48) ) corresponding to C . Each cluster is an independent set in B ( X (cid:48) ). We say that two clusters corresponding to C, C (cid:48) ∈ V ( B ( X )) are adjacent if CC (cid:48) ∈ E ( B ( X )).Claim 1: For any triangle ACD in B ( X ) with e ∈ A , the induced subgraph of B ( X (cid:48) ) on F eA ∪ F yC ∪ F zD is isomorphic to B ( Y ). The same is true for F eA ∪ F zC ∪ F yD .Proof: To prove the claim, consider the map ψ : F eA ∪ F yC ∪ F zD → B ( Y ) which takes a colorclass B of X (cid:48) to the intersection of B with V ( Y ). Since the restriction of B to V ( X ) isidentical to precisely one of A, C or D it is easy to see that ψ is injective. Further, we canform a color class of X (cid:48) from any color class of Y by combining it with one of A , C and D ,provided that they agree on containment of v ; hence ψ is surjective.It remains to show that, for B, B (cid:48) ∈ F eA ∪ F yC ∪ F zD , B is adjacent to B (cid:48) if and only if ψ ( B )is adjacent to ψ ( B (cid:48) ). Suppose that B ∈ F eA and B (cid:48) ∈ F yC . If B and B (cid:48) are adjacent in B ( X (cid:48) )then there is a color class F ∈ B ( X (cid:48) ) such that BB (cid:48) F is a triangle in B ( X (cid:48) ). Note that F ∈ F zD . The image of this triangle under ψ is also a triangle, and hence ψ ( B ) is adjacent to ψ ( B (cid:48) ). On the other hand, if ψ ( B ) and ψ ( B (cid:48) ) are adjacent in B ( Y ), then there is a triangle ψ ( B ) ψ ( B (cid:48) ) J ⊆ B ( Y ) since all edges in B ( Y ) come from 3-colorings of Y . Now B, B (cid:48) and D ∪ J form a coloring of X (cid:48) , and hence B and B (cid:48) are adjacent in B ( X (cid:48) ). The proofs for thecases when B ∈ F eA , B (cid:48) ∈ F zD and B ∈ F yC , B (cid:48) ∈ F zD are similar. This proves the claim.Claim 2: The subgraph of B ( X (cid:48) ) on color classes in F zD ∪ F yC is connected and bipartite, andis isomorphic to B ( Y ) − φ Y ( e ). Further, the graph B ∗ obtained from B ( Y ) by deleting theedges of B ( Y ) − φ Y ( e ) is connected and isomorphic to each of the bipartite graphs between F eA and F C , as well as between F eA and F yC ∪ F zD .7roof: B ( Y ) − φ Y ( e ) is connected by Corollary 8. Furthermore, every edge of B ( Y ) − φ Y ( e )is contained in a triangle in B ( Y ) using no other edge of B ( Y ) − φ Y ( e ). Therefore, B ∗ isalso connected since the edges of such triangles can be used in B ∗ to replace any of theremoved the edges. Now, suppose that ACD is a triangle in B ( X ), where e ∈ A . Thenthe bipartite graph between F zD and F yC is isomorphic to B ( Y ) − φ Y ( e ), while the bipartitegraphs between F eA and F C and between F eA and F yC ∪ F zD are each isomorphic to B ∗ , in thelatter case, by Claim 1. Hence both of these are also connected, establishing the claim.We say that a coloring is constant on a set S of vertices if S is contained in a color classof the coloring. If a coloring χ of B ( X (cid:48) ) is constant on each of F yC and F zC , we say that χ is near-constant on F C .Claim 3: Let ACD be a triangle in B ( X ), where e ∈ A , and let χ be a coloring of B ( X (cid:48) )that is constant on one of F C or F D . Then χ is constant on each of F eA , F C and F D .Proof: If χ is constant on F C , then F eA and F D form a connected bipartite graph on which χ uses only two colors; hence χ is constant on each of F eA and F D . The case where χ isconstant on F D is similar. This proves the claim.Claim 4: If χ is constant on F eA , then it is near-constant on F C and F D . If χ is near-constanton one of F C and F D , then it is also near-constant on the other, together using only twocolors on F C ∪ F D , and is constant on F eA , using the third color.Proof: To prove the claim, we first observe that if χ is constant on F eA , then F yC and F zD forma connected bipartite graph on which χ uses only two colors. Hence, χ is constant on eachof F yC and F zD (and, similarly, F zC and F yD ). It follows immediately that χ is near-constanton F C and F D .Suppose now that χ is near-constant on F C . By Claim 3, we may assume that χ is notconstant on F C , so it uses different colors on F yC and F zD . Then there is only one color leftfor the whole of F eA (noting from Claim 2 that every vertex of F eA is adjacent to some vertexin each of F yC and F zC ). So χ is constant on F eA , and hence near-constant on F D . Again, thecase where χ is near-constant on F D is similar. This completes the proof of the claim.Our next goal is to count 3-colorings of B ( X (cid:48) ). All such colorings are partitioned intothree classes.Class 1: Colorings that are constant on F C for some C ∈ V ( B ( X )) \ φ X ( e ). By Corollary8, B ( X ) − φ X ( e ) is connected. This fact, repeatedly combined with Claim 3, implies thatsuch a coloring χ is constant on F eA for every A ∈ φ X ( e ). Note that every coloring of B ( X )gives rise to a coloring of B ( X (cid:48) ) in the obvious way: If D ∈ B ( X ) is a color class in acoloring of B ( X ), we let D (cid:48) = (cid:83) D ∈D F D ∈ B ( X (cid:48) ). This correspondence yields a bijectionbetween colorings of B ( X ) and the colorings of B ( X (cid:48) ) that are of Class 1. In particular, thenumber of colorings of Class 1 is equal to the number of triangles of X , since X is reflexive.Of course, this is equal to the number of vertices of H .Class 2: Colorings that are constant on F eA for some A ∈ φ X ( e ), but not on F C for any C ∈ V ( B ( X )) \ φ X ( e ). In this case, Claim 4 implies that any such coloring is near-constanton F C for each C ∈ V ( B ( X )) \ φ X ( e ). Since B ( X ) − φ X ( e ) is connected, the same two colors8re used for every F C and the color of each F yC and F zC are determined by the choice of anyone of them. We can therefore construct exactly one coloring of B ( X (cid:48) ) in this way.Class 3: Colorings that are non-constant on each F eA , A ∈ φ X ( e ). In this case, wewill show that any such coloring is completely determined by its restriction to an arbitrarilychosen triangle of clusters F eA , F yC , F zD . The vertices of these clusters can be colored accordingto any coloring of B ( Y ), except for the coloring in which the clusters are themselves colorclasses. Therefore, the number of colorings covered by this case is one less than the numberof triangles in Y .Suppose now a coloring χ of B ( X (cid:48) ) of class 3 and consider its restriction to F eA ∪ F yC ∪ F zD .Our aim is to prove that this restriction determines χ . To see this, we employ an inductiveargument. First observe, by Lemma 10, that χ is determined on F yD and F zC , since theseclusters form a connected bipartite graph. Moreover, by Claim 4, neither F C nor F D maybe near-constant. Similarly, χ is determined on any clusters adjacent to F eA . Again, byLemma 10, noting that χ is not near-constant on F C , χ is also determined on F eB for any B ∈ φ X ( e ) which is adjacent to C , since the bipartite graph between F eB and F F is connected,where F is the third vertex of the triangle in B ( X ) containing B and C . Similarly to theargument for F eA , χ is also determined on F F . Repeating this argument, since B ( X ) isconnected, we conclude that we have uniquely determined the whole of the coloring χ .By the preceding cases, in total, the number of colorings of B ( X (cid:48) ), and therefore thenumber of triangles in B ( X (cid:48) ), is equal to the sum of the number of triangles of X and Y . Since X (cid:48) also has this many triangles, we conclude by Lemma 6 that the injection φ X (cid:48) : X (cid:48) → B ( X (cid:48) ) is in fact an isomorphism.As a direct corollary of Lemma 11 we obtain a theorem of Fisk [4]. Corollary 12 (Fisk [4]) . Every cubic tree is edge-reflexive.
The lemma also enables us to restrict ourselves to 2-edge-connected cubic graphs.
Corollary 13.
Suppose that all graphs obtained from a connected cubic graph G by cuttingall cutedges of G are edge-reflexive. Then G is edge-reflexive. Corollary 13 shows that in order to prove Theorem 9, it suffices to prove the following.
Lemma 14.
Every 2-connected cubic triangle-free outerplanar graph is edge-reflexive.
It is well-known (and easy to see) that any 2-connected, triangle-free, cubic, outerplanargraph G can be constructed from a cubic 4-cycle by repeatedly applying the following twooperations:1. Adding a 4-cycle : Given an edge e = v v in a cubic graph H , incident with half-edges e and e (respectively), as well as the full edges f and f , add two new vertices v , v and9orm a 4-cycle v v v v , where e joins v and v , e joins v and v , and e joins v and v .Finally, add half-edges e v and e v incident with v and v , respectively.2. Subdividing an edge : Given an edge e = v v in a cubic graph H , where v and v are incident with half-edges e and e (respectively), as well as with full edges f and f ,subdivide the edge e into two edges e (cid:48) and e (cid:48)(cid:48) by inserting a new vertex v . Then add ahalf-edge g incident with v , in order to form a new cubic graph.The two operations with the corresponding notation that will be used when speakingabout them are depicted in Figure 2. ee e f f v v H e (cid:48) e e f f v v Gee e f f v v G e (cid:48)(cid:48) ge e v e v v v Figure 2: Adding a 4-cycle and subdividing an edgeIn order to prove Lemma 14, it suffices to show that these two operations preservereflexivity.
Lemma 15.
Operation 1 preserves edge-reflexivity.Proof.
Let G be the graph obtained from H by adding a 4-cycle, let X = L ( G ), Y = L ( H ),and assume that Y is reflexive. We have to show that the homomorphism φ X : X → B ( X )is a bijection. As before, we will establish injectivity and then count the number of coloringsof B ( X ).Claim 1: The graph X is colorful.Proof: We are given that Y is colorful. Since every 3-coloring of Y extends to X , it immedi-ately follows that, for any pair of vertices u, v ∈ V ( X ) \ { e , e v , e v } , there exists a coloringof X in which u and v are colored differently. Moreover, as there is a coloring of Y whichcolors f and e differently, there is a coloring of X in which e and e have the same color.Now, by performing an { e , e v , e v } Kempe change, if needed, we can ensure that for any u ∈ V ( X ) \ { e , e v , e v } and v ∈ { e , e v , e v } , there exists a coloring in which u and v arecolored differently. Finally, as there exists a coloring in Y which colors e and e differently,we can extend this coloring to X in order to obtain a coloring in which e v and e v are coloreddifferently. This completes the claim.Now, we partition the vertices of B ( X ) into the seven sets shown in Figure 3, where wedenote by C x x ...x t the set of color classes of X which contain the vertices x , x , . . . , x t ∈ e, e , e , e , f , f } . In other words, C x x ...x t = t (cid:92) i =1 φ X ( x i ) . Such sets will be referred to as clusters . In Figure 3 we also have the cluster C (cid:98) e f f , where (cid:98) e indicates that the color classes in this cluster do not contain e . Note that C e f f ∪ C (cid:98) e f f is a partition of C f f .The 3-colorings of X fall into three types as indicated by triangles in Figure 3. Thesubgraph of B ( X ) consisting of all triangles of type i ∈ { , , } will be denoted by T i . Notethat T i contains only those vertices from the corresponding three clusters that appear ascolor classes in T i . Thus, T i is obtained from the induced subgraph on the three clusters byremoving the isolated vertices (which must participate in colorings of the neighboring T j ,but not in T i ). We also denote the subgraph of T i induced by the union of the two adjacentclusters C x ...x t and C w ...w s by C x ...x t C w ...w s .Claim 2: T ∪ T is isomorphic to B ( Y ). Under this isomorphism, C ee is mapped onto φ Y ( e ).Proof: Consider the map ψ : T ∪ T → B ( Y ) which takes a color class C of X to itsintersection with V ( Y ). As every coloring c of Y extends uniquely to a coloring c (cid:48) of X with c (cid:48) ( e ) = c (cid:48) ( e ), ψ is a bijection. Moreover, this unique extension also establishes that ψ is a graph homomorphism, since it follows that ABC is a coloring of T ∪ T if and only if ψ ( A ) ψ ( B ) ψ ( C ) is a coloring of Y . Thus, our claim is established. C e C e f f C e e C (cid:98) e f f C ee C e f C e f T T T Figure 3: Partitioning B ( X ) into clustersSince Y is reflexive, Corollary 8 shows that B ( Y ) − φ Y ( e ) has precisely two components.Hence, it follows from Claim 2 that C e f C e f and C e e C (cid:98) e f f are bipartite and connectedsubgraphs of T ∪ T .Claim 3: T is isomorphic to T and C e C e f f is isomorphic to C ee C (cid:98) e f f .Proof: Note that the colorings of X forming T and T have e and e colored the same.Consider the map ψ : T → T which takes a color class C ∈ V ( T ) to the color class of X in T which results from performing a Kempe change on { e , e v , e v } . Observe that thismap is well-defined since e v and e v are colored the same and hence the colors of e and e remain the same. As ψ and ψ − are both invertible (as Kempe changes are reversible),11 is a bijection. Similarly, ψ is a graph homomorphism, since it follows from the existenceand reversablility of the Kempe change discussed that ABC is a triangle of T if and only if ψ ( A ) ψ ( B ) ψ ( C ) is a triangle of T . These observations confirm our claim.Claim 4: C (cid:98) e f f C ee is isomorphic to C (cid:98) e f f C e e .Proof: Consider the map ψ : T → T which takes a color class C ∈ V ( T ) to the color classof X in T which results from performing a Kempe change on { e, e , e , e } . Observe thatthis map is well-defined, as such a Kempe change always exists and leaves e and e thesame color, which remains different from the shared color of e and e . As Kempe changesare reversible, ψ is a bijection. Similarly, ψ is a graph homomorphism, since it follows fromthe existence and reversibility of the Kempe change discussed that ABC is a coloring of T if and only if ψ ( A ) ψ ( B ) ψ ( C ) is a coloring of T .Claim 4 implies that C ee C (cid:98) e f f is bipartite and connected. By Claim 3 it follows imme-diately that C e C e f f is bipartite and connected.Now, let χ be a coloring of B ( X ). Suppose first that χ is constant on C ee ∩ T . Weclaim that χ is constant on the whole cluster C ee . For a contradiction, suppose that thisclaim is false. As χ is constant on C ee ∩ T , we know that χ is constant on all threeclusters of T . Now, as Y is reflexive and T ∪ T is isomorphic to B ( Y ) (by Claim 2),the color class of T ∪ T ∼ = B ( Y ) containing C ee ∩ T must be of the form φ Y ( v ), forsome vertex v ∈ V ( Y ). If v = e or v = e , we are done, so we may assume otherwise.As C (cid:98) e f f and C e e ∩ T are both nonempty, the color class containing C ee ∩ T in B ( Y )cannot be any of φ Y ( f ), φ Y ( f ), φ Y ( e ), φ Y ( e ), φ Y ( e v ) or φ Y ( e v ). However, for anyvertex v ∈ V ( Y ) \ { e, e , e , e , f , f , e v , e v } , by Claim 4, if φ Y ( v ) ∩ C ee ∩ T (cid:54) = ∅ , then φ Y ( v ) ∩ C e e ∩ T (cid:54) = ∅ . But this contradicts the fact that χ is constant on C ee ∩ T ,completing our claim.Consequently, if χ is constant on C ee ∩ T , then χ is constant on the whole cluster C ee ,and hence, on all the clusters of B ( X ). There are four such colorings of B ( X ), compared totwo such colorings of B ( Y ).Let us now consider the case when χ is not constant on C ee ∩ T . In this case, byLemma 10 and Claim 4, χ is determined and non-constant on C e e ∩ T and determined on C (cid:98) e f f . By Claim 3 and Lemma 10, it then follows that χ is completely determined on T .Thus, the coloring χ of B ( X ) is completely determined by its restriction to T ∪ T , which isisomorphic to B ( Y ). Each such coloring is determined uniquely by a coloring of B ( Y ) whichis not constant on all the clusters of B ( X ). Since two colorings of B ( Y ) correspond to case1, we get two fewer colorings in case 2, so we obtain two less colorings than the number ofcolorings of B ( Y ).Thus, in total, B ( X ) has two more colorings than B ( Y ) had, which is precisely thenumber of additional triangles in X . Consequently, X ∼ = B ( X ), as required. Lemma 16.
Operation 2 preserves edge-reflexivity.Proof.
Again, let X = L ( G ), Y = L ( H ), and we assume that Y is reflexive. As in ourprevious arguments, we will demonstrate that the homomorphism φ X : X → B ( X ) is a12ijection by establishing injectivity and then counting the number of colorings of B ( X ). Wealso find it useful to define graphs H e (cid:48) and H e (cid:48)(cid:48) as shown in Figure 4, as well as their linegraphs X e (cid:48) and X e (cid:48)(cid:48) , respectively. Note that H e (cid:48) and H e (cid:48)(cid:48) are both isomorphic to H , buthave some of their edges labelled differently. Through this labeling we can view E ( H e (cid:48) ) and E ( H e (cid:48)(cid:48) ) as subsets of E ( G ).Figure 4 also shows the correspondence of 3-edge-colorings of H e (cid:48) and H e (cid:48)(cid:48) with certain3-edge-colorings of G . Observe that H e (cid:48) ( H e (cid:48)(cid:48) ) has precisely the 3-edge colorings of G inwhich the color of e (cid:48) is equal to the color of e (the color of e (cid:48)(cid:48) is equal to the color of e )and H e (cid:48) ∼ = H ∼ = H e (cid:48)(cid:48) H . Gf f e e ge (cid:48) e (cid:48)(cid:48) f f e e (cid:48) e (cid:48)(cid:48) f f e e (cid:48) e (cid:48)(cid:48) H e (cid:48)(cid:48) H e (cid:48) χ ( e (cid:48) ) = χ ( e ) χ ( e (cid:48)(cid:48) ) = χ ( e ) Figure 4: Two graphs isomorphic to H whose 3-edge-colorings correspond to certainsubsets of the 3-edge-colorings of G Claim 1: G is edge-colorful.Proof: Let u, v ∈ V ( X ). We examine four cases.Case 1: If u, v / ∈ { e (cid:48) , e (cid:48)(cid:48) , e , e , g } , then a 3-coloring c e (cid:48) of X e (cid:48) with c e (cid:48) ( u ) (cid:54) = c e (cid:48) ( v ) can beextended to a coloring of X in which c ( u ) (cid:54) = c ( v ). As X e (cid:48) is colorful, this establishes case 1.Case 2: u, v ∈ { e (cid:48) , e (cid:48)(cid:48) , e , e , g } : As X e (cid:48) is colorful, there exists a 3-edge coloring of G with c ( f ) (cid:54) = c ( f ) and c ( e (cid:48) ) = c ( e ). Since c ( e (cid:48) ) is different from c ( f ) and c ( f ), we havethat c ( e (cid:48)(cid:48) ) (cid:54) = c ( e ), c ( e ) (cid:54) = c ( e ) and c ( e ) (cid:54) = c ( g ). By the same argument on X e (cid:48)(cid:48) , we alsoestablish that there is a coloring distinguishing e (cid:48) , e and e , g .Case 3: If u ∈ { e (cid:48) , e (cid:48)(cid:48) , e , e } and v / ∈ { e (cid:48) , e (cid:48)(cid:48) , e , e , g } , then we may always arrange acoloring of X in which c ( f ) = c ( f ), as X e (cid:48) is colorful. In the event that c ( u ) = c ( v ), we canthen perform a Kempe change on the vertices of the set { e (cid:48) , e (cid:48)(cid:48) , e , e } in order to arrange for c ( u ) (cid:54) = c ( v ). This establishes case 3.Case 4: u = g and v / ∈ { e (cid:48) , e (cid:48)(cid:48) , e , e , g } : In case 2, we showed that there is a 3-edge coloringof G with c ( e (cid:48) ) = c ( f ). So, for each vertex v / ∈ { g, e } we can arrange for c ( g ) (cid:54) = c ( v ) byperforming a Kempe change on the path g, e (cid:48)(cid:48) , e (if needed). This completes the claim.Now, as X is colorful (and thus φ X : X → B ( X ) is injective), it suffices to establishthat B ( X ) has precisely one more 3-coloring than B ( Y ). In order to prove this fact, we willagain partition B ( X ) into clusters of the form C ab = φ X ( a ) ∩ φ X ( b ), where a, b ∈ V ( X ). Wewill consider the partition into clusters as depicted in Figure 5.13 T T T C e (cid:48) e C e (cid:48)(cid:48) e C e (cid:48) f C e (cid:48)(cid:48) f C e f C f f C e f C e e Figure 5: Partitioning B ( X ) into clustersAs in the proof of Lemma 15, let T i (1 ≤ i ≤
4) be the subgraph of B ( X ) on all colorsets participating in 3-colorings of X whose edges are between the three clusters of T i , asshown in Figure 5. We will refer to the subgraph consisting of all edges in B ( X ) joining theclusters C uv and C xy as the edge C uv C xy of our cluster partition. When C uv C xy is nonempty,it is contained in precisely one of the subgraphs T i , i ∈ { , , , } .Claim 2: The following subgraphs of B ( X ) are isomorphic: T ∪ T ∼ = B ( X e (cid:48) ) ∼ = B ( Y ) ∼ = B ( X e (cid:48)(cid:48) ) ∼ = T ∪ T .Proof: That B ( X e (cid:48) ) ∼ = B ( Y ) ∼ = B ( X e (cid:48)(cid:48) ) follows immediately from their definitions. So, itsuffices to prove that the maps ψ : T ∪ T → B ( X e (cid:48) ) and ψ : T ∪ T → B ( X e (cid:48)(cid:48) ), each ofwhich takes a color class C of X to its restriction as indicated in Figure 4, are isomorphisms.As these two arguments are identical, we will only establish the claim for the map ψ .As every coloring c of X e (cid:48) extends uniquely to a coloring of X and T ∪ T includes thewhole C e (cid:48) e , ψ is a bijection. Moreover, this unique extension also establishes that ψ is agraph homomorphism, since it follows that a coloring ABC of X is in T ∪ T if and only if ψ ( A ) ψ ( B ) ψ ( C ) is a coloring of X e (cid:48) . Thus, Claim 2 is resolved.Claim 3: The edges of the cluster partition in Figure 5 representing C e f C e (cid:48)(cid:48) f , C f f C e (cid:48)(cid:48) e , C e (cid:48) e C f f and C e (cid:48) f C e f all represent connected, bipartite subgraphs of B ( X ). Moreover, C f f C e (cid:48)(cid:48) e ∼ = C e (cid:48) e C f f .Proof: Recall that, by Claim 2, T ∪ T ∼ = B ( X e (cid:48) ) ∼ = B ( Y ) ∼ = B ( X e (cid:48)(cid:48) ) ∼ = T ∪ T . Since X e (cid:48) is reflexive and e (cid:48) is in two triangles of X e (cid:48) , φ X e (cid:48) ( e (cid:48) ) must be in precisely two triangles of B ( X e (cid:48) ). Through the isomorphism T ∪ T ∼ = B ( X e (cid:48) ), φ X e (cid:48) ( e ) corresponds to the cluster C e (cid:48) e in T ∪ T . Therefore, C e f C e (cid:48)(cid:48) f and C f f C e (cid:48)(cid:48) e are bipartite and connected. Similarly, as φ X e (cid:48)(cid:48) ( e (cid:48)(cid:48) ) must be in precisely two triangles of B ( X e (cid:48)(cid:48) ), C e (cid:48) e C f f and C e (cid:48) f C e f are bipartiteand connected. 14ow, consider the map ψ : T → T , which takes a color class C ∈ V ( T ) to the color class C (cid:48) which results from performing a Kempe change on { e , e (cid:48) , e (cid:48)(cid:48) , e } . Observe that this mapis well-defined, as such a Kempe change always exists and leaves e and e (cid:48)(cid:48) the same color,which remains different from the shared color of e (cid:48) and e . Moreover, as Kempe changes arereversible, ψ is a bijection, and ABC is a triangle in T if and only if ψ ( A ) ψ ( B ) ψ ( C ) is atriangle in T . Consequently, ψ is a graph isomorphism. It remains to note that ψ maps C f f C e (cid:48)(cid:48) e onto C e (cid:48) e C f f . This establishes Claim 3.We will now discuss the structure of 3-colorings of B ( X ). If we consider our clusterpartition as an 8-vertex graph, each 3-coloring of that graph determines a 3-coloring of B ( X )in which each cluster is monochromatic (the coloring is constant on the cluster ). There areother 3-colorings of B ( X ). To understand them, we first show that each such coloring isdetermined by its restriction to certain subgraphs of B ( X ).Claim 4: For every 3-coloring of B ( X ), its restriction to T ∪ T determines the coloring on T .Proof: Observe that deleting either e (cid:48) and f or e (cid:48)(cid:48) and f separates X into two components,one of which only contains vertices in { e , e , e (cid:48) , e (cid:48)(cid:48) , g } . Consequently, we can obtain fromany coloring of X represented by a triangle in T a triangle in T through a Kempe changeon { e , e (cid:48) , g } and a triangle in T through a Kempe change on { e , e (cid:48)(cid:48) , g } . This shows that C e (cid:48)(cid:48) f = T ∩ T and C e (cid:48) f = T ∩ T . Thus, the coloring of C e (cid:48)(cid:48) f is determined by the coloringof C e (cid:48)(cid:48) f ∩ T , and the coloring of C e (cid:48) f is completely determined by the coloring of C e (cid:48) f ∩ T .Now, every vertex in C e e is in a triangle in T and hence also its color is determined.Claim 5: If a coloring χ of X is constant on C e (cid:48)(cid:48) e ∩ T , then χ is constant on the whole cluster C e (cid:48)(cid:48) e . If χ is constant on C e (cid:48) e ∩ T , then it is constant on C e (cid:48) e .Proof: For a contradiction, suppose that this claim is false. If χ is constant on C e (cid:48)(cid:48) e ∩ T ,then (by Claim 3) it is constant on each of C e (cid:48) e ∩ T and C f f . Now, as T ∪ T ∼ = B ( X e (cid:48)(cid:48) )is reflexive, the color class containing C e (cid:48)(cid:48) e ∩ T in B ( X e (cid:48)(cid:48) ) must be of the form φ X e (cid:48)(cid:48) ( v ), forsome vertex v ∈ V ( X e (cid:48)(cid:48) ). If v = e (cid:48)(cid:48) , we are done, so we may assume otherwise. As C f f and C e (cid:48) e ∩ T are both nonempty, the color class containing C e (cid:48)(cid:48) e ∩ T in B ( X e (cid:48)(cid:48) ) cannotbe any of φ X e (cid:48)(cid:48) ( f ), φ X e (cid:48)(cid:48) ( f ), φ X e (cid:48)(cid:48) ( e (cid:48)(cid:48) ), φ X e (cid:48)(cid:48) ( e ) or φ X e (cid:48)(cid:48) ( e (cid:48) ). However, when the vertex v ∈ V ( X e (cid:48)(cid:48) ) \ { e , e (cid:48) , e (cid:48)(cid:48) , f , f } , then φ X e (cid:48)(cid:48) ( v ) ∩ C e (cid:48) e ∩ T (cid:54) = ∅ by Claim 3. But this contradictsthe fact that χ is constant on C e (cid:48)(cid:48) e ∩ T , completing our claim. The proof of the secondstatement is the same.Claim 6: The subgraphs of B ( X ) corresponding to T and T are isomorphic, and C e (cid:48)(cid:48) f C e (cid:48) f is bipartite and connected.Proof: As in the proof of Claim 3, we consider the map ψ : T → T induced on B ( X ) byperforming a Kempe change in X on { g, e (cid:48)(cid:48) , e } . This map is a bijection between T and T .Moreover, ABC is a triangle in T if and only if ψ ( A ) ψ ( B ) ψ ( C ) is a triangle in T , so thisis indeed a graph isomorphism.Claim 7: If a 3-coloring χ of B ( X ) is non-constant on C e (cid:48)(cid:48) e ∩ T , then the restriction of χ on( C e (cid:48)(cid:48) e ∩ T ) ∪ ( C e (cid:48) f ∩ T ) determines χ on the whole of T .15roof: Suppose that A ∈ ( C e (cid:48)(cid:48) e ∩ T ) \ ( C e (cid:48)(cid:48) e ∩ T ). As T ∪ T ∼ = B ( X e (cid:48)(cid:48) ) and B ( X e (cid:48)(cid:48) ) \ φ X e (cid:48)(cid:48) ( e (cid:48) )is bipartite and connected (by Corollary 8), there exists a path in C e (cid:48)(cid:48) e C e f from A to somevertex D ∈ C e (cid:48)(cid:48) e ∩ T . Since D ∈ C e (cid:48)(cid:48) e ∩ T , its color is determined. Since the coloring on C e (cid:48) f ∩ T is determined, we conclude that χ is determined on the whole path from D to A .Thus, the color of A is determined. Now, as A was chosen arbitrarily, it follows that thecoloring is determined on C e (cid:48)(cid:48) e ∩ T , from which it follows that χ is determined on all of T .This proves the claim.Claim 8: If a 3-coloring χ of B ( X ) is non-constant on C e (cid:48)(cid:48) e ∩ T , then the restriction of χ on( C e (cid:48)(cid:48) e ∩ T ) ∪ ( C e (cid:48)(cid:48) f ∩ T ) determines χ on the whole of T .Proof: Firstly, let A ∈ C e (cid:48) f be a vertex of B ( X ) in a connected component K of C e (cid:48) f C e (cid:48)(cid:48) e .Since e is a half-edge in X e (cid:48)(cid:48) and X e (cid:48)(cid:48) is reflexive, B ( X e (cid:48)(cid:48) ) \ φ X e (cid:48)(cid:48) ( e ) is connected (byCorollary 8). Thus, there exists a path from A to some vertex D ∈ C e (cid:48)(cid:48) e ∩ T , which is alsoin the connected component K . Now, the color class of D is of the form φ X e (cid:48) ( x ), for some x ∈ V ( X e (cid:48) ) (because T ∪ T is isomorphic to X e (cid:48) ). The vertex D is also adjacent to somevertex D (cid:48) ∈ C e (cid:48) f , which is in turn adjacent (by the aforementioned isomorphism between T and T ) to a vertex D (cid:48)(cid:48) ∈ C e (cid:48)(cid:48) f whose color is determined.If D and D (cid:48)(cid:48) are colored with the same color, then they both belong to φ X e (cid:48) ( x ). Therefore, x ∈ D and x ∈ D (cid:48)(cid:48) . However, by Claim 6, unless x ∈ { g, e (cid:48)(cid:48) , e } , x ∈ D implies that x / ∈ D (cid:48)(cid:48) .Meanwhile, if x ∈ { e (cid:48)(cid:48) , e } , then it easily follows that χ is constant on C e (cid:48)(cid:48) e ∩ T . Finally, x cannot be equal to g . So, in each case, we have a contradiction. Thus, D and D (cid:48)(cid:48) must becolored differently.Since D and D (cid:48)(cid:48) are colored differently, the color of D (cid:48) is determined. Consequently, eachconnected component K of C e (cid:48) f C e (cid:48)(cid:48) e must contain some vertex D (cid:48) ∈ C e (cid:48) f whose color isdetermined.Now, we make an argument similar to that in the proof of Claim 7. Suppose that A ∈ C e (cid:48) f . By using the isomorphism between T and T , we see that, in the connectedcomponent K (cid:48) of C e (cid:48) f C e e containing A , there is a path from A to some vertex D (cid:48) ∈ C e (cid:48) f whose color is determined. Since the color of D (cid:48) and the colors of vertices in C e (cid:48)(cid:48) f ∩ T are determined, the neighbor of D (cid:48) on this path has its color determined. By iterating thisargument, we conclude that all the vertices on this path have their color determined. Now,as A was chosen arbitrarily, it follows that the coloring is determined on C e (cid:48) f , from which itfollows that the coloring on all of T is determined. This proves the claim.Now, consider an arbitrary 3-coloring χ of B ( X ) and its restriction χ (cid:48) on T ∪T ∼ = B ( X e (cid:48) ).We consider two cases.Firstly, if χ is constant on C e (cid:48)(cid:48) e ∩ T , then by Claim 3, χ is constant on each cluster of T . Thus, applying Claim 5 and then Claim 3 again, we observe that χ is constant on all theclusters of T , T and T . It then follows from Claim 4 that χ is constant on every cluster of B ( X ). There are three such colorings, compared to two colorings χ (cid:48) of T ∪ T which havethis form.Secondly, if χ is not constant on C e (cid:48)(cid:48) e ∩ T , we need to show that χ is uniquely determinedon all of B ( X ) by its restriction χ (cid:48) to T ∪T . So, we apply Claim 8. The 3-coloring χ of B ( X )is determined and non-constant on C e (cid:48)(cid:48) e ∩ T , and determined on C e (cid:48)(cid:48) f by the 3-coloring χ (cid:48) of16 ∪ T . Thus, the coloring is determined on T . Consequently, the 3-coloring is determinedand non-constant on C e (cid:48)(cid:48) e ∩ T , and determined on C e (cid:48) f , so by Claim 7, it is determined on T . Hence, all 3-colorings of T ∪ T ∼ = B ( Y ) uniquely extend to B ( X ) (except in the casewhen C e (cid:48)(cid:48) f and C e (cid:48)(cid:48) e are colored identically, in which case the coloring extends in two ways to B ( X )). This shows that B ( X ) has precisely one more 3-coloring than B ( Y ), as required. Proof of Lemma 14 and of Theorem 9.
By Observation 5, G cannot be edge-reflexive if itcontains a triangle. So let us assume that G is triangle-free. By applying Lemma 15 andLemma 16 repeatedly, we can construct any 2-connected, triangle-free, cubic, outerplanargraph from a cubic 4-cycle (which is edge-reflexive). This establishes Lemma 14. Then,taking this result together with Corollary 13, Theorem 9 follows immediately. Outerplanar triangle-free cubic graphs have at least four half-edges. It is therefore a naturalquestion whether there are some cubic graphs with less than four half-edges that are edge-reflexive. If there is just one half-edge, then the graph is not 3-edge-colorable and if thereare two half-edges, they receive the same color in every edge-coloring, so such a graph is notedge-colorful. Of course, the most interesting class is when no half-edges are present. Let usobserve that K , is an edge-reflexive graph (see Figure 6) without half-edges. X B ( X ) { , , } { , , }{ , , }{ , , } { , , }{ , , } B ( X ) Figure 6: The disjoint union of two triangles X , its 3-coloring complex B ( X ) (drawn in theprojective plane) which is isomorphic to L ( K , ), and the graph B ( X ). This shows that K , is edge-reflexive.But there are other classes of reflexive cubic graphs without half-edges. Let us start withthe prisms. The n -prism Π n is the cubic graph of order 2 n which is obtained by taking the17artesian product of an n cycle and K . If n is odd, the n -prism is not edge-colorful: inany 3-edge-coloring each pair of the corresponding cycle edges in the Cartesian product iscolored the same. This implies that B ( L (Π n )) is isomorphic to B ( L ( C (cid:48) n )), where C (cid:48) n is thecubic n -cycle with half-edges. However, prisms of even length are different. Theorem 17.
For every even n ≥ , the n -prism Π n is edge-reflexive.Proof. Let A and B be the perfect matchings of the first n -cycle S in Π n and let A , B be the corresponding perfect matchings in the second n -cycle S . Also, let M be the perfectmatching in Π n consisting of all edges joining the two cycles.Since n is even, there are two 3-edge-colorings of Π n containing M as a color class: { A ∪ A , B ∪ B , M } and { A ∪ B , B ∪ A , M } . The latter one is called the mixedcoloring . Further, any other 3-edge-coloring { A, B, C } of the cubic n -cycle S ∪ M (wherethe edges in M are treated as half-edges) that does not contain the whole M as a color classextends uniquely to a 3-edge-colouring of Π n by adding in each color class all edges in S that are copies of the edges of S in the color class. This shows that B ( L (Π n )) is isomorphicto B ( L ( C (cid:48) n )) with one added triangle corresponding to the mixed coloring. That triangleshares the color class M with the rest of the coloring complex.It is easy to see that Π n is edge-colorful. Thus it suffices to show, by Lemma 6, that B ( L (Π n )) has precisely 2 n C (cid:48) n is edge-reflexive (by Theorem 9), B ( L ( C (cid:48) n ))has precisely n B ( L (Π n ))extends in two ways to the whole B ( L (Π n )) since we have two ways to color the vertices of B ( L (Π n )) corresponding to color classes A ∪ B and B ∪ A of the mixed coloring.In Lemma 16 we showed that, under certain circumstances, the graph G (cid:48) , which we ob-tain from an edge-reflexive cubic graph G by subdividing an edge e of G , is edge-reflexive.However, there exist edge-reflexive graphs G where, regardless of how many times we sub-divide one of its edges, the result will never be edge-reflexive. One such example is K , . Infact, there is a more general family. Proposition 18.
Suppose that G is an edge-reflexive cubic graph without half-edges. Thenno graph H which results from subdividing a single edge of G k times ( k ≥ is edge-reflexive. In the proof we will employ the well-known Parity Lemma.
Lemma 19 (Parity Lemma) . Suppose that a cubic graph G is edge-colored. Let n , n and n be the number of half-edges of G in each of the three color classes. Then n , n and n are congruent modulo 2.Proof. The number of half-edges in a color class is equal to the number n of vertices in G ,minus twice the number of full edges in the same color class. Thus, n ≡ n ≡ n ≡ n (mod 2). Proof of Proposition 18.
Let the graph H be obtained from G by subdividing e = uv k ≥ H (cid:48) obtained from H − e by adding twohalf-edges shows that H is not 3-edge-colorable when k = 1 and that it is not edge-colorfulif k ≥ H (cid:48) , the half-edges are colored the same.18roposition 18 shows that subdividing a single edge in K , yields a graph that is notedge-reflexive. Of course, this is still not the full story, as some subdivisions of K , areedge-reflexive. For example, if we subdivide each edge of K , once, the resulting graph isedge-reflexive. At this time, we do not fully understand the relation between subdividingedges and edge-reflexivity. However, we can still use Lemma 16 to help identify additionalinfinite families of edge-reflexive graphs. For example, this lemma is instrumental in provingour next result.We construct the cubic theta graph T k,l,m ( k, l, m ≥
1) as follows. Begin with threepaths of lengths k , l and m , respectively. Label their vertices u , u , . . . , u k , v , v , . . . , v l and w , w , . . . , w m . Then identify the vertices u , v and w , as well as the vertices u k , v l and w m . Finally, add half-edges to make the graph cubic. Observe that T k,l,m ∼ = T l,k,m ∼ = T k,m,l ,and hence we may assume that k ≤ l ≤ m .A number of small theta graphs are not edge-reflexive. In particular, T , ,m is not edge-reflexive for any m ≥ T , ,m is not edge-reflexive for any m ≥
1, since it contains a triangle. Additionally, using a computer, we found that T , , , T , , , T , , , T , , , T , , and T , , are not edge-reflexive. However, all other theta graphs areedge-reflexive. Theorem 20.
The cubic theta graphs T , ,m , T , ,m ( m ≥ T , , , T , , , T , , , T , , , T , , and T , , are not edge-reflexive. All other cubic theta graphs are edge-reflexive.Proof. As mentioned above, the graphs listed in the statement of the theorem are not edge-reflexive. To show that all other cubic theta graphs are edge-reflexive, we have verified byusing computer that T , , , T , , , T , , , T , , and T , , are edge-reflexive. Since any othercubic theta graph can be obtained from one of these by subdividing edges, Lemma 16 impliesthat they are all edge-reflexive. As we saw in the previous section, many non-outerplanar graphs exist which are edge-reflexive. A theta graph can have a large number of vertices which do not appear on theouter face of any drawing, while K , is a non-planar graph. Nonetheless, in both of thesecases (with the exception of a few small theta graphs) we obtain edge-reflexive graphs.Another class of potentially edge-reflexive cubic graphs we consider particularly interest-ing are the fusenes (also known as hexagonal graphs ). We say that G is a fusene if G is a2-connected plane graph, in which every interior face is a hexagon, all vertices of G havedegree three (after adding half-edges) and only vertices on the boundary of the outer faceare permitted to be incident with half-edges. Question 21.
Do there exist any fusenes that are not edge-reflexive?
We are not aware of any examples, and have confirmed through a lengthy computationthat none exist with nine or fewer hexagonal faces. Given any edge-reflexive fusene graph, we19btain an infinite family of edge-reflexive fusenes by using the operation of adding a 4-cyclefollowed by two subdivisions. However, not all fusenes are obtained this way.We have also uncovered a generalization of the theta graphs, which may yield anotherinfinite family of edge-reflexive graphs. We will call a graph theta ladder
T L ( l, m, n ) if it isconstructed as follows. Begin with two 6-cycles a a a a a a and b b b b b b . Now, join theedge a a to the edge b b with a ladder of length k . Similarly, the edges a a and b b areconnected by a ladder of length l , while a a and b b are connected by a ladder of length m . An example given in Figure 7 is the theta ladder T L (3 , , T L (3 , ,
3) constructed by attaching three ladders to a pair of6-cycles. Many graphs of this form are edge-reflexive.We have confirmed, using a computer, that
T L (1 , , T L (1 , , T L (1 , , T L (1 , , T L (1 , , T L (3 , , T L (1 , , T L (1 , , T L (3 , , T L (3 , ,
5) and
T L (3 , ,
7) areedge-reflexive. For those
T L ( l, m, n ) where at least one parameter is even and l + m + n ≤ T L (1 , ,
3) (the onlyodd-odd-odd exception). Based on this evidence, we ask the following question.
Question 22. (a)
Do there exist any theta ladder graphs
T L ( l, m, n ) , where l, m, n ≥ areall odd, that are not edge-reflexive? (b) Do there exist any theta ladder graphs
T L ( l, m, n ) , where l is even, that are edge-reflexive? The graph
T L ( l, m, n ) is not edge-colorful if one parameter is even and the other two areodd. (We leave the proof of this fact as an exercise.) Thus, in this case T L ( l, m, n ) is notreflexive. We were not able to establish a similar result for other cases of Question 22(b). References [1] Norman Biggs. Pictures. In
Combinatorics (Proc. Conf. Combinatorial Math., Math.Inst., Oxford, 1972) , pages 1–17, 1972. 202] Stanley Fiorini. On the chromatic index of outerplanar graphs.
Journal of CombinatorialTheory, Series B , 18(1):35–38, 1975.[3] Steve Fisk. Cobordism and functoriality of colorings.
Advances in Mathematics , 27:177–211, 1980.[4] Steve Fisk.
Coloring Theories . American Mathematical Society, Providence, R.I., 1989.[5] W. T. Tutte. Even and odd 4-colorings. In
Proof Techniques in Graph Theory , pages161–169. Academic Press, New York, 1969.[6] William T. Tutte. Some comments on the four-colour problem. Available at: