Exact square coloring of subcubic planar graphs
Florent Foucaud, Hervé Hocquard, Suchismita Mishra, Narayanan Narayanan, Reza Naserasr, Éric Sopena, Petru Valicov
aa r X i v : . [ c s . D M ] S e p Exact square coloring of subcubic planar graphs ∗ Florent Foucaud † Herv´e Hocquard † Suchismita Mishra ‡ Narayanan Narayanan ‡ Reza Naserasr § ´Eric Sopena † Petru Valicov ¶ k
September 17, 2020
Abstract
We study the exact square chromatic number of subcubic planar graphs. An exact squarecoloring of a graph G is a vertex-coloring in which any two vertices at distance exactly 2receive distinct colors. The smallest number of colors used in such a coloring of G is its exactsquare chromatic number, denoted χ [ ♯ ( G ). This notion is related to other types of distance-based colorings, as well as to injective coloring. Indeed, for triangle-free graphs, exact squarecoloring and injective coloring coincide. We prove tight bounds on special subclasses of planargraphs: subcubic bipartite planar graphs and subcubic K -minor-free graphs have exact squarechromatic number at most 4. We then turn our attention to the class of fullerene graphs, whichare cubic planar graphs with face sizes 5 and 6. We characterize fullerene graphs with exactsquare chromatic number 3. Furthermore, supporting a conjecture of Chen, Hahn, Raspaud andWang (that all subcubic planar graphs are injectively 5-colorable) we prove that any inducedsubgraph of a fullerene graph has exact square chromatic number at most 5. This is doneby first proving that a minimum counterexample has to be on at most 80 vertices and thencomputationally verifying the claim for all such graphs. In this paper, we study exact distance coloring problems for graphs. The celebrated Hadwiger-Nelson problem asking for the “chromatic number of the plane” falls into this category of problems:there, one wishes to assign a color to each point of the Euclidean plane, such that two points atdistance exactly 1 receive distinct colors. A recent breakthrough result on this problem, showingthat at least five colors are necessary, appeared in [22] (it is long known that seven colors suf-fice [24]). Similar problems are studied for other metric spaces, see [30]. In the graph setting, fora positive integer p , an exact p -distance coloring of a graph G is an assignment of colors to thevertices of G , such that two vertices at distance exactly p receive distinct colors [37, Section 11.9].The exact p -distance chromatic number of G , denoted χ [ ♯p ] ( G ), is the smallest number of colorsin an exact p -distance coloring of G and the study of this parameter is gaining growing attention,see [8, 28, 30, 40]. Denote G [ ♯p ] , the exact distance- p -power of G , which is a graph obtained bytaking vertices of G and adding an edge between any two distinct vertices at distance exactly p . ∗ This research was supported by the IFCAM project “Applications of graph homomorphisms”(MA/IFCAM/18/39), by the ANR project HOSIGRA (ANR-17-CE40-0022) and by the ANR project DISTAN-CIA (ANR-17-CE40-0015) † Univ. Bordeaux, CNRS, Bordeaux INP, LaBRI, UMR5800, F-33400 Talence, France ‡ Department of Mathematics, IIT Madras, Chennai 600036, India. § Universit´e de Paris, CNRS, IRIF, F-75006, Paris, France ¶ Aix-Marseille Univ, Universit´e de Toulon, CNRS, LIS, Marseille, France k LIRMM, CNRS, Universit´e de Montpellier, France
1e have χ [ ♯p ] ( G ) = χ ( G [ ♯p ] ). Thus, for p = 1 this notion coincides with the usual chromatic num-ber. Similarly denote ω [ ♯p ] ( G ) and α [ ♯p ] ( G ), respectively the maximum clique and the maximumindependent set in G [ ♯p ] , that is ω [ ♯p ] ( G ) = ω ( G [ ♯p ] ) and α [ ♯p ] ( G ) = α ( G [ ♯p ] ).Exact distance p -powers of graphs were first studied by Simi´c [42], see also [4, 9] for morerecent works. Exact p -distance colorings have first been studied for graphs of bounded expansion.For a fixed graph class C of bounded expansion (for example, the class of planar graphs), theexact p -distance chromatic number is bounded by an absolute constant for graphs in C when p isodd [28, 37], and by a linear function of the maximum degree when p is even [28]. Exact p -distancecolorings have been studied in specific graph classes: trees [8], graphs of bounded tree-width [28],chordal graphs [40], graphs of bounded genus [28, 40].Distance-based colorings of graphs have been extensively studied since the first papers on thesubject were published in the 1960s by Kramer and Kramer [32, 33]. In their setting, for a positiveinteger p , a p -distance coloring of a graph G is an assignment of colors to the vertices of G , suchthat two vertices at distance at most p receive distinct colors. The smallest possible number ofcolors of such a coloring is denoted by χ p ( G ) (see [1, 27, 43, 44] for some important results). Theproblem of 2-distance coloring subcubic and cubic planar graphs is already far from trivial, andthe focus of many research works: see [7, 20, 26, 27, 43]. Notably, a special case of a conjectureby Wegner [44], recently solved in [43], states that χ ( G ) ≤ G .The goal of the present paper is to study exact p -distance colorings with a focus on the specialcase p = 2 and for subclasses of planar graphs. Given a graph G , the graph G [ ♯ is called the exactsquare of G . Similarly, an exact 2-distance coloring of G is called an exact square coloring of G ,and χ [ ♯ ( G ) is the exact square chromatic number of G .What makes the study of exact distance coloring rather more difficult, is that χ [ ♯p ] is notnecessarily monotone with respect to taking subgraphs. Indeed K n is exact p -distance 1-colorablefor every p ≥
2. However, if restricted to exact square coloring ( p = 2), then χ [ ♯ is monotone withrespect to taking induced subgraphs. On the other hand, the parameter χ [ ♯ is unbounded evenfor trees, indeed for the star K ,t with t leaves, ( K ,t ) [ ♯ is isomorphic to the disjoint union of thecomplete graphs K t and K , and hence χ [ ♯ ( K ,t ) = t . Thus, it is natural to restrict the study ofexact square colorings to graphs with no induced K ,t . For triangle-free graphs, this turns out tobe the same as bounding the maximum degree. Thus, the class of subcubic graphs will be a focusof this study.A related notion is the one of injective coloring , introduced in [25] and well-studied since then,see for example [10, 15, 35]. An injective coloring of a graph G is a vertex-coloring where anytwo vertices receive distinct colors whenever they are joined by a path of length 2. The smallestnumber of colors of an injective coloring of G is its injective chromatic number , denoted χ i ( G ).From these definitions, we have for any graph G : χ [ ♯ ( G ) ≤ χ i ( G ) ≤ χ ( G ) . Moreover, whenever G is triangle-free, we have χ [ ♯ ( G ) = χ i ( G ), since in G , vertices joinedby a path of length 2 must also be at distance 2. Since many of our results are for triangle-freegraphs, they can be re-interepreted as results on injective colorings. For a subcubic graph G , themaximum degree of G [ ♯ is at most 6 and thus χ [ ♯ ( G ) ≤
7. In fact, the bound χ i ( G ) ≤ G is the Heawood graph [15, 25], andusing the arguments of [15, 25], the same holds for the exact square chromatic number. Thus, forevery subcubic planar graph G , we have χ i ( G ) ≤
6. A stronger bound was conjectured as follows.
Conjecture 1 ([15]) . If G is a subcubic planar graph, then χ i ( G ) ≤ . Note that there is a subcubic planar graph G (with triangles) satisfying χ i ( G ) = 5 [15], so iftrue, the conjectured bound would be tight. However, we do not know of any subcubic planar2raph with exact square chromatic number 5. Conjecture 1 was generalized to arbitrary valuesof the maximum degree [15, 34], in the spirit of a well-studied conjecture by Wegner [44] on the(non-exact) square chromatic number. Conjecture 1 was proved for K -minor-free graphs [15] (andit follows from [6] that χ [ ♯ ( G ) ≤ K -minor-free graph G ). Furthermore, the bound wasimproved to four colors for outerplanar subcubic graphs [36]. If G is a subcubic graph with girthat least 19 (resp. 10) then χ i ( G ) ≤ χ i ( G ) ≤
4) [35]. If G has girth at least 6, then χ i ( G ) ≤ χ ( G ), see [7, 20].Exact square colorings also appear in another, more general, context: the one of L ( p, q )-labelings. Given two non-negative integers p and q , an L ( p, q )-labeling of a graph G is an assignment ℓ of non-negative integers to the vertices of G , such that for any two vertices u and v , we have | ℓ ( u ) − ℓ ( v ) | ≥ p if u and v are adjacent, and | ℓ ( u ) − ℓ ( v ) | ≥ q if they are at distance 2. The first(and most) studied case is when p = 2 and q = 1 [23]; see the survey [14]. Thus, for any graph G , there is a 1-to-1 correspondence between: (a) L (1 , G , (b) L (1 , G , and (c) L (0 , G . However, it seems that L (0 , G be a subcubic graph. We show that χ [ ♯ ( G ) ≤ G is K -minor-free. We also show that if G is planar and bipartite, then its exact square is planar, thus χ i ( G ) = χ [ ♯ ( G ) ≤
4. Moreover these bounds are tight: there exist bipartite subcubic K -minor-free graphs with exact square chromatic number 4. In passing we also show that the exact squareof every subcubic bipartite outerplanar graph G is outerplanar, thus χ i ( G ) = χ [ ♯ ( G ) ≤
3. Thisis tight since for every tree T with a vertex of degree 3, χ [ ♯ ( T ) ≥ , -nanotubes [2, 29] which we call drums . Then, in Section 3.2, we prove Conjecture 1 for fullerenegraphs. The proof is computer-assisted: we first consider a potential minimum counterexample,and prove that it cannot have more than 80 vertices. We then use a computer program to check thelist of fullerene graphs of order up to 80, which is available online and certified complete [11, 12, 21].We conclude in Section 4. We now recall some facts from the literature and prove a few new results.
Given a vertex v , we denote by d ( v ) the degree of v and we say that v is a k -vertex if d ( v ) = k .For a planar graph with a given planar embedding, we call a k -face a face of length k .A thread in a graph G is a path all of whose internal vertices are 2-vertices. We will use thefollowing lemma. 3 emma 2 ([38]) . Any planar graph G of girth at least d + 1 ( d ≥ ) contains a vertex of degree or a thread with at least d internal vertices. Observation 3.
For any graph G , G [ ♯ is a subgraph of G (the complement of G ). If G hasdiameter , then G [ ♯ is isomorphic to G . The two following results were formulated in the context of injective coloring [15, 25], but thesame arguments hold for exact squares.
Theorem 4 ([15, 25]) . If G is a connected graph of maximum degree ∆ = k + 1 , then ω [ ♯ ( G ) ≤ k + k + 1 . Moreover, equality may only happen if G is the incidence graph of a projective geometryof order k , in which case G [ ♯ is isomorphic to two copies of K k + k +1 . Using Brook’s theorem, the following can be deduced for the special case k = 2. Corollary 5 ([15]) . Let G be a connected subcubic graph. Then, χ [ ♯ ( G ) ≤ , unless G is theHeawood graph (in which case G [ ♯ is isomorphic to two disjoint copies of K ). It is not difficult to find general subcubic graphs whose exact square contains a 5-clique (e.g.the Petersen graph ) or a 6-clique (e.g. the triplex graph, a cubic graph of girth 5 and order 12,depicted in Figure 1a).Another interesting graph G with χ [ ♯ ( G ) = 6 is the subcubic bipartite graph of order 22 builtin a similar fashion as the Heawood graph, see Figure 1c. This graph is also the incidence graphof the 11 geometric configuration number 31 described in [39]. Each connected componenentof its exact square is isomorphic to the (non-exact) 3-distance power of the 11-cycle C . It hasmaximum degree 6, clique number 4, chromatic number 6 and is thus an extremal example forReed’s conjecture stating that χ ( G ) ≤ l ω ( G )+∆( G )+12 m , for every graph G [41]. (a) The triplex graph with ω [ ♯ ( G ) = χ [ ♯ ( G ) = 6 (b) The Heawood graph : ω [ ♯ ( G ) = χ [ ♯ ( G ) = 7 (c) A bipartite graph with ω [ ♯ ( G ) = 4 and χ [ ♯ ( G ) = 6 Figure 1: Some extremal examples of cubic graphs having high exact square chromatic number.The following lemma is interesting to observe.
Lemma 6.
Let G be a subcubic planar graph and let I be an independent set of G . Then, G [ ♯ [ I ] is planar.Proof. Let G ′ be the graph build on I as follows: for each vertex x ∈ V ( G ) \ I , delete x and joinall its neighbors in I . Observe that, as x has at most three neighbors in G , and by considering If G has diameter 2, then G [ ♯ is isomorphic to G and if G has girth 5, then α ( G [ ♯ ) = 2. G , the new graph G ′ is planar. However this graph may have multiedges(when two vertices of I are in a 4-cycle). By removing all but one edge from any group of paralleledges we get G [ ♯ [ I ].We deduce the following from Lemma 6. Proposition 7.
For any subcubic planar graph G , we have ω [ ♯ ( G ) ≤ .Proof. Let K be a clique in G [ ♯ . In G , K is an independent set. Thus, by Lemma 6, G [ ♯ [ K ] isa planar complete graph, which implies that | K | ≤
4, as desired.The bound of Proposition 7 is sharp, as we see next.
Proposition 8.
There exist bipartite subcubic K -minor-free graphs G of girth with ω [ ♯ ( G ) = 4 .Proof. Consider the graph G consisting of two vertices connected with three vertex-disjoint pathsof length 3. Then G [ ♯ contains two disjoint copies of K . We deduce the following from Lemma 6.
Theorem 9.
Let G be a connected bipartite planar subcubic graph. Then, G [ ♯ consists of twoplanar connected components, each induced by one part of the bipartition of G . Moreover, if G isouterplanar, then these two components are outerplanar.Proof. The first part of the statement directly follows from Lemma 6.For the second part, suppose that G is outerplanar with bipartition ( X, Y ). Recall that G [ ♯ consists of two connected components: the one induced by X and the one induced by Y . It sufficesto prove that each biconnected component of G [ ♯ is outerplanar.If G has a bridge, then the two vertices of this bridge are cut-vertices in G [ ♯ . Since they belongto two different connected components of G [ ♯ , each component contains a cut-vertex. Since weconsider biconnected components of G [ ♯ , we may assume in the following that G has no bridge.Since G is cubic, we conclude that G is biconnected. Consider an outerplanar embedding of G and let C be the facial cycle of this embedding. Consider any part of the bipartition of G , say X ,ordered in the same way as in C ; it suffices to show that G [ ♯ [ X ] is outerplanar. In G [ ♯ , the orderof X induced by C also induces a cycle C X . We claim that the natural embedding of C X inducesan outerplanar embedding of G [ ♯ [ X ]. Indeed, the only edges of G [ ♯ [ X ] not in C X are createdalong edges that are chords of C in G . Let xy be such a chord, and assume that x , x , y , y arethe neighbors (in G ) of x and y , respectively (with x ∈ X and y ∈ Y ). Then, in the component G [ ♯ [ X ], we obtain the edges xy , xy and y y . Now, we easily build an outerplanar embeddingof G [ ♯ [ X ] from the one of G : the coordinates of the vertices stay the same. The edges of C X follow the embedding of C . The two additional edges corresponding to the chord xy can be drawnalong the position of the edge xy in the embedding of G . In this way we obtain an outerplanarembedding of G [ ♯ [ X ].Clearly, Theorem 9 implies that the square of bipartite subcubic outerplanar graphs can becolored with four colors (by using the Four Color Theorem).5 .3 K -minor-free graphs A nested ear decomposition of a graph G is a partition of E ( G ) into paths E , . . . , E k of G (theears), such that the following conditions hold.(i) For every ear E i , only the two end-vertices might be the same (thus E i induces a path or acycle).(ii) For every ear E i with i >
1, there is an ear E j with j < i such that the two endpoints of E i belong to E j (we say that E i is nested in E j , and the sub-path of E j between the twoendpoints of E i is the nest interval of E i ).(iii) Apart from the endpoints of E i , for every j < i , no other vertex of E i belongs to E j .(iv) If two ears E i and E i ′ are both nested in E j , then their nest intervals are either disjoint orone is contained in the other.This concept was defined in [19]. It is known that in any K -minor-free graph, every biconnectedcomponent has a nested ear decomposition. Indeed, a graph is K -minor-free if and only if ev-ery biconnected component is two-terminal series-parallel [5], and every biconnected two-terminalseries-parallel graph has a nested ear decomposition [19].Figure 2: An outerplanar graph with ∆ = 3 and χ [ ♯ ( G ) = 4 [36]. Theorem 10. If G is a subcubic K -minor-free graph, then χ [ ♯ ( G ) ≤ . Furthermore, this boundis tight on subcubic bipartite K -minor free graphs of girth at least 6, and on outerplanar graphs.Proof. Let G be a minimum counterexample, that is, G is subcubic and K -minor-free, χ [ ♯ ( G ) > K -minor-free graph is 4-colorable. Claim 10.A. G is -connected.Proof of claim. Suppose that G has a cut-vertex v . Since G is subcubic, then it has a bridge xy . The case where one of the two vertices (say x ) has degree 1 is easy, since any exact square4-coloring of G − x would extend to G .So assume that both x and y are vertices of degree at least 2. We remove the edge xy from G :this creates two components. We add a degree 1 neighbor y ′ to x and a degree 1 neighbor x ′ to y .Let G x and G y be the components containing x and y , respectively. Now, let φ be a k -coloring of G [ ♯ x and G [ ♯ y .Suppose first that we have φ ( x ) = φ ( y ′ ) and φ ( x ′ ) = φ ( y ). Then, we interchange the colors φ ( x ) and φ ( y ) in G y . This new coloring induces a valid k -coloring of G [ ♯ .6imilarly, if φ ( x ) = φ ( y ′ ) and φ ( x ′ ) = φ ( y ), we permute the colors of vertices of G y so that φ ( x ) = φ ( x ′ ) and φ ( y ) = φ ( y ′ ). Again this yields a valid k -coloring of G [ ♯ .Finally, suppose that φ ( x ) = φ ( y ′ ) but φ ( x ′ ) = φ ( y ) (the symmetric case is handled similarly).Since k ≥
4, there is one free color c f among the neighbors of x (in G ). We now permute the colorsin G y such that φ ( x ′ ) = φ ( x ) and φ ( y ) = c f . Again this yields a valid k -coloring of G [ ♯ . ( (cid:3) ) Claim 10.B. G contains no -vertex lying on a k -cycle, with k ∈ { , } .Proof of claim. Suppose v is a 2-vertex in G lying on a k -cycle ( k ∈ { , } ) and consider G ′ = G − v .Then χ [ ♯ ( G ′ ) ≤ G . Now, since v is part of a k -cycle in G , it is easy to seethat the distance in G ′ between any pair of vertices x, y is the same as their distance in G . Thusan exact square coloring of G ′ is valid in G . Since G is subcubic and v is lying on k -cycle, weconclude that v has degree at most 3 in G [ ♯ . Therefore v can be colored and we are done. ( (cid:3) ) Claim 10.C. G contains no pair of adjacent -vertices.Proof of claim. The proof is similar to the one of the previous claim. Suppose u, v are two adjacent2-vertices in G and consider G ′ = G − { u, v } . Then χ [ ♯ ( G ′ ) ≤ G . It is easyto see that if the exact distance between two vertices of V ( G ′ ) in G is 2, then this distance ispreserved in G ′ . Thus an exact square coloring of G ′ is valid in G and since u, v have degree atmost 3 in G [ ♯ , we are done. ( (cid:3) ) By Claim 10.A, G is 2-connected. Thus, as seen above, G has a nested ear decomposition E , . . . , E k . Moreover, by Claim 10.C, G has no pair of adjacent 2-vertices. Thus, there must beat least two ears in the decomposition. Consider the last ear E k , and let E i be the ear such that E k is nested in E i . Moreover, among all ears nested in E i , let E j be an ear whose nest intervaldoes not contain any other nest interval. Again by Claim 10.C, the length of E j is at most 2. Butthen the vertices of E j , together with the vertices of E i that belong to the nest interval of E j , formeither a triangle or a 4-cycle, one of whose vertices has degree 2. By Claim 10.B this contradictsthe minimality of G .For the tightness of the bound, in Proposition 8 we have presented a bipartite K -minor-freegraph of girth 6 whose exact square is the disjoint union of two K ’s. For the class of outerplanargraphs we have the example of Figure 2, given in [36] in the context of injective coloring ofouterplanar graphs. We now turn our attention to a special class of cubic planar graphs, namely fullerene graphs . Theyare the skeletons of cubic 3-dimensional convex polyhedra, each of whose faces is either a pentagonor a hexagon. As the skeleton of a convex polyhedron, a fullerene graph must be 3-connectedand most authors consider this condition as part of the definition. However, we rather define afullerene graph as a ”cubic plane graph each of whose faces is of size either 5 or 6”. The followingcan then be proved as an exercise.
Proposition 11 (Folklore) . Every fullerene graph has girth 5 and is 3-connected.
As any 3-connected graph admits a unique plane embedding, we may refer to a fullerene graphas a planar graph rather than a plane graph. Further results on their structure were proved, andwe give here a few of them which we will use in the proofs of this section. Recall that a graph G is cyclically k -edge-connected if any edge-cut separating two cycles of G has at least k edges. Doˇsli´cproved the following. 7 v v v v v v ′ v ′ v ′ v ′ v ′ v ′ u ′ u ′ u ′ u ′ u ′ u ′ u u u u u u (a) The 1-drum: the unique fullerene graph on 24vertices (b) The 3-drum D Figure 3: Examples of drums.
Theorem 12 ([18]) . Every fullerene graph is cyclically 5-edge-connected.
Using this theorem, one can easily classify all possible small cycles of fullerene graphs. Moreprecisley, we have the following (perhaps folklore) fact; we refer to [31, Lemma 4.1] for a proof.
Lemma 13.
Given a fullerene graph G , every non-facial cycle of G is of length at least 9. More-over, the only cycles of length 9, if any, are the cycles around a vertex incident with 5-faces only. Note that χ [ ♯ ( G ) ≥ G . We first characterize those fullerene graphswhose exact square is 3-colorable. Then, we prove χ [ ♯ ( G ) ≤ G , thusproving Conjecture 1 for this class of graphs. -colorable fullerene graphs We first need to define a special class of fullerene graphs, that we call drums . Definition 14.
A drum is a fullerene graph with two specific 6-faces F and F ′ , each of which isa neighbor with six 5-faces such that all these twelve 5-faces are distinct. A drum where F and F ′ are at facial distance k + 1 is called a k -drum. For an example, the 1-drum and 3-drum are depicted in Figure 3. In the literature, drumsare known as a specific type of so-called nanotubes , more precisely, following the terminologyfrom [2, 29], they are exactly one of the five types of (6 , -nanotubes .We will show next that there exists a unique k -drum up to isomorphism. Proposition 15.
Given a k -drum G , all faces at distance ℓ ≤ k from F are of a same length.Proof. We consider a planar embedding of G . Let v , v , . . . , v be the six vertices of F in thecyclic order of vertices, and let u , u , . . . , u be the six vertices of F ′ in the cyclic order of vertices,see Figure 3a for the labeling of the 1-drum. 8bserve that, since G is cubic, each vertex v i has a unique neighbor in G \ F , we call it v ′ i .Similarly, each neighbor of u i not in F ′ is called u ′ i . Observe that each pair of edges v i v ′ i and v i +1 v ′ i +1 (addition in indices here and in the rest of the proof are taken modulo 6) is a pair ofparallel edges of a 5-cycle.To prove the claim of the proposition, for ℓ = 1, by the definition, all faces at distance 1 from F are 5-faces. For ℓ = 2 we show that if one of the faces is a 5-face, then they are all 5-faces. Soassume a face f at distance 2 from F is a 5-face. Then, as there are only twelve 5-faces, f mustbe incident to F ′ and since the 5-faces incident to F ′ are distinct from those of F , the vertices of f furthest away from F form an edge of F ′ . Thus, we may label the vertices of f , without loss ofgenerality, u u ′ v ′ u ′ u . But then, the two faces incident to u u ′ and u u ′ are 5-faces next to F ′ ;by continuing this process, we conclude that all faces at distance 2 from F are the 5-faces incidentto F ′ .This completes the proof for ℓ = 1 , k , which is exhaustive for k ≤
2. For theremaining cases, we apply induction on k . Assume that the claim is true for k and all values of ℓ , ℓ ≤ k . Consider a ( k + 1)-drum ( k ≥ F are 6-faces. In eachof these 6-faces, one of the vertices is already labeled v ′ i , noting that different faces correspond todifferent v ′ i ’s. Label x i the common neighbor of v ′ i and v ′ i +1 (see Figure 4). Thus, x i − v ′ i x i formpart of a 6-face. On this face, label the neighbor of x i by y i . Finally, label the common neighborof y i and y i − by z i . Let G ′ be the graph obtained from G by deleting all the edges y i z i +1 andthen contracting edges z i y i and y i x i . We claim that G ′ is a ( k − ℓ from F in G are at distance ℓ − F in G ′ . This would complete the proof by induction. Tosee that G ′ is a fullerene graph, observe that from the construction it is 3-regular. Each face of G containing a path z i y i z i +1 becomes a face of the same size on the path x i v ′ i +1 x i +1 , and all otherfaces remain the same. Hence, we only have 5-faces and 6-faces in G ′ , so we are done. F v v v v v v v ′ v ′ v ′ v ′ v ′ v ′ x x x x x x y y y y y y z z z z z z Figure 4: The neighborhood of face F in a drum.Our goal in this section is to characterize drums as the only fullerene graphs which are exactsquare 3-colorable. To this end, in the next lemma, we present two planar subcubic graphs that arenot exact square 3-colorable, thus they cannot be induced subgraphs of an exact square 3-colorablegraph. Lemma 16.
Neither of the two graphs of Figure 5 admits an exact square 3-coloring. roof. We will repeatedly use the fact that in a proper 3-coloring of a K − (that is, the com-plete graph on four vertices minus one edge), the nonadjacent vertices must receive a same color.Applying this observation to the exact square of the graphs of Figure 5, we conclude that in ahypothetical 3-coloring of each of them, vertices v , t , u and w must receive a same color (redin the figure). This is already a contradiction in the graph of Figure 5a, as vertices w and v areat distance 2.To complete the proof for the graph of Figure 5b, observe that for the same reason, vertices u , t and v must also get a same color, and this color must be distinct from red because u and v are at distance 2. We suppose this color to be green. Furthermore, vertices v and t are coloredby the third color (blue), because each of them sees both other colors at distance 2. Moreover,since w must receive the same color as t , we conclude that w must be colored blue.Now, we know that v sees both blue and green at distance 2, and thus it must be colored red.Repeating the K − argument, we conclude that t and then u must be colored red as well. Thisis a contradiction, as u is at distance 2 from the vertex v , colored red. v v v v v u u u u t t t t t t t u v t u v t v w w (a) v v v v v u u u u t t t t t t t u v t u v t v w w (b) Figure 5: The forced precolorings from the proof of Lemma 16.Similarly, we show in the next lemma, that the graph of Figure 6, while admitting an exactsquare 3-coloring, has limits on its possible 3-colorings.
Lemma 17.
In an exact square 3-coloring of the graph G of Figure 6, vertices x and y mustreceive distinct colors.Proof. By contradiction, suppose φ is an exact square 3-coloring of G with φ ( x ) = φ ( y ) = 1.Then, without loss of generality, we can assume that φ ( v ) = 2 and φ ( v ) = 3. Therefore, we have φ ( v ) = 2. But then, φ ( u ) = 2 because u sees v and y at distance 2. Similarly, φ ( u ) = 2. Thisis a contradiction since u and u see each other at distance 2. Lemma 18. If G is a fullerene graph which is not the 1-drum and contains the graph of Figure 6as a subgraph, then χ [ ♯ ( G ) ≥ .Proof. Let G be an exact square 3-colorable fullerene graph which contains the graph of Figure 6as a subgraph. Moreover, let φ be its exact square 3-coloring. By Lemma 17 and without lossof generality, we may assume that φ ( x ) = 1 and φ ( y ) = 2. Hence, φ ( u ) = φ ( v ) = 3 and10 u x y u v v v u v u v Figure 6: A 5-cycle surrounded by three consecutive 5-cycles. { φ ( u ) , φ ( v ) } = { , } and { φ ( u ) , φ ( v ) } = { , } . By the symmetry along the edge xy , we mayassume that φ ( u ) = 3, which then implies φ ( v ) = 1, φ ( u ) = 2 and φ ( v ) = 3. Therefore, wehave φ ( u ) = φ ( v ) = 2 and φ ( u ) = φ ( v ) = 1.Next, noting that G is a 3-regular graph, we consider the remaining neighbors of degree 2vertices of this subgraph. Let a, b, c, d, e, f be, respectively, the neighbors of u , u , u , v , v , v .The coloring extends uniquely to these six vertices as follows: φ ( a ) = φ ( b ) = 1, φ ( c ) = φ ( f ) = 3,and φ ( d ) = φ ( e ) = 2. Since G is planar and cubic, vertices a and b are lying on the same face.Moreover, since G can have only 5-faces and 6-faces, we conclude that a and b (which are coloredwith the same color) must be adjacent. Similarly, we conclude that e and d must be adjacent. Onthe other hand, b and c cannot be adjacent, as otherwise b and u would be at distance 2 while bothhaving the same color. Hence, b and c have a common neighbor, say b ′ , and thus we get a 6-face u u u bb ′ cu , which we name F . Similarly, vertices e and f have a common neighbor, say e ′ , thuswe get a 6-face v v v ee ′ f v and we name it F ′ . Now, vertices a and f must lie on the same face,which can be either a 5-face or a 6-face. Observe also that the third neighbor of a , say a ′ , distinctfrom u and b must be colored 3. Therefore, since f is colored 3, vertex a ′ cannot be at distance 2from f and thus we conclude that a and f are lying on a 5-face au v f a ′ a . Symmetrically, we getthat vertices c and d are lying on a 5-face as well: dv u cc ′ d (where c ′ is the common neighbor of c and d ).In summary, starting from the graph of Figure 6, with xy being the central edge, we concludedthat the neighboring structure is forced. But we now have other isomorphic copies of the graph ofFigure 6 inside G , for example one centered around u u and another one centered around v v .Thus, the same local neighborhood structures should exist around these edges. The 6-faces inthese structures are already given ( F and F ′ ). Thus, we conclude that e ′ , e, d, c ′ are lying on a5-face. Similarly, a ′ , a, b, b ′ are lying on a 5-face as well. This forces a graph where all but twovertices have degree 3, the other two vertices being of degree 2. To complete this to a 3-connectedcubic graph, we then join these two vertices and obtain the 1-drum. Lemma 19.
The graph of Figure 7 does not admit an exact square 3-coloring.Proof.
To prove the lemma, we claim that in any possible exact square 3-coloring of this graph,vertices x and y must receive a same color. Considering the symmetry of edges xy and xz , thesame argument then would apply to x and z . This would lead to a contradiction, since y and z cannot be colored the same. 11 s x y vt Figure 7: The graph of Lemma 19.To prove the claim, assume φ is an exact square 3-coloring of the graph of Figure 7, and withoutloss of generality, assume that φ ( y ) = 1, φ ( u ) = 2 and φ ( v ) = 3. Then, φ ( s ) = 3 and φ ( t ) = 2.This in turn implies φ ( x ) = 1, which is the color of y . s t z y z t s t z y x x y x y x u v Figure 8: The graph of Lemma 20.
Lemma 20.
The graph of Figure 8 does not admit an exact square 3-coloring.Proof.
By contradiction, suppose φ is a 3-coloring of the exact-square of this graph. We first claimthat φ ( x ) = φ ( y ). If not, then we may assume φ ( x ) = 1 and φ ( y ) = 2, then φ ( y ) = φ ( y ) = 3,which in turn implies that φ ( t ) = φ ( t ) = 1 but t and t are at distance 2. Note that this proofis based solely on the three faces around y , so similarly, if a vertex has two 5-faces and one 6-facearound it, then its color must be the same as the color of its neighbor on the 5-faces. Applyingthis to our graph we have: φ ( x ) = φ ( y ), φ ( z ) = φ ( y ).Suppose φ ( x ) = φ ( y ) = 2, then as y is at distance 2 from y , and by the symmetry of othercolors, we have φ ( x ) = φ ( y ) = 3, then φ ( z ) = 1 and since z is at distance 2 from both z and x we have φ ( z ) = φ ( y ) = 2. This in turn implies that φ ( t ) = 1, but then t sees all three colorsat distance 2, that is y for color 2, y for color 3 and t for color 1.We can now state the main theorem of this section. Theorem 21.
A fullerene graph is exact square -colorable if and only if it is a k -drum, for apositive integer k . roof. First, we show that every drum is exact square 3-colorable. Given a k -drum, considerits 6-face F surrounded by six 5-faces. Color the vertices of F with 1 , , , , , F . Observe that, by Proposition 15, for 2 ≤ l ≤ k , there are six faces of the samelength at distance l from F . If we consider the subgraph G l induced by faces at distance at most l from F , its outer face is a 12-cycle. The 3-coloring of F is then uniquely extended to a 3-coloring of G l , where the 12 vertices of the outer face are colored consecutively 1 , , F .Similarly, if we start from the face F ′ of the k -drum, and color the vertices of F ′ with 1 , , F ′ ), then the coloring extends uniquely to any subgraphobtained by faces at distance at most k − l from F ′ . In such a coloring, vertices of the outer 12-faceare colored consecutively 1 , , F ′ . As this orientationof the outer 12-face matches its counterclockwise orientation with respect to F , we can choose aproper rotation of colors on F ′ , so that we can merge the colorings of both parts of the drum.Notice that, except the vertices of the 12-cycle, all the vertices of one part of the drum are atdistance at least 3 from vertices of the other part.It remains to show that if a fullerene graph admits an exact square 3-coloring, then it mustbe a drum. Let G be an exact square 3-colorable fullerene graph. By Lemma 16, a 5-face cannothave three consecutive 6-faces. By Lemmas 18 and 19, a vertex cannot be incident to three 5-faces. This leaves us with one possibility: each 5-face C is neighbor with two other 5-faces throughtwo non-adjacent edges of C . Labelling the vertices of one of the 5-faces, in the cyclic order, x , y , z , y , x , we may assume that each of the edges x y and x y is incident to another 5-face,thus so far we have the subgraph of Figure 8 induced by vertices x i ’s, y i ’s and z i ’s, for 1 ≤ i ≤ x y z y x cannot be on the edge z y , thus it must be on y x . Completing this sequenceof 5-faces, we conclude that vertices x i form a face of G as they are already of full degree. Thisface then can only be a 6-face which is the face F of the drum. The face F ′ of drum is found thesame way by considering the remaining 5-faces. To prove the main result of this section, we first give two lemmas on proper coloring of somegraphs. Specifically, we show that the precoloring extension of two graphs (where vertices takecolors from lists of given sizes) can always be done when the lists are subsets of { , , , , } . Lemma 22.
Let abcdef be a 6-cycle and x be a vertex such that N ( x ) = { a, b, c, d, e, f } . Sup-pose these vertices have lists of available colors from the set { , , , , } satisfying the following: | L ( a ) | ≥ , | L ( b ) | ≥ , | L ( c ) | ≥ , | L ( d ) | ≥ , | L ( e ) | ≥ , | L ( f ) | ≥ and | L ( x ) | = 5 (see Figure 9).Then there exists a proper L -coloring of this graph.Proof. Assume to the contrary that there is no such coloring.Suppose that L ( a ) ∩ L ( c ) = ∅ and without loss of generality let 1 ∈ L ( a ) ∩ L ( c ). We color φ ( a ) = φ ( c ) = 1. We then greedily color vertices b, d, e, f in this order by observing that at eachstep there is an available color for the current vertex. Now, if there is an available color from L ( x )left for x , we are done. If not, then it is easy to see that b, d, e, f have each been colored with adistinct color from { , , , } , say φ ( b ) = 2 , φ ( d ) = 3 , φ ( e ) = 4 , φ ( f ) = 5 and L ( x ) = { , , , , } .Moreover, observe that L ( b ) = { , } , L ( d ) = { , } , L ( e ) = { , } and L ( f ) = { , , } , asotherwise one of these vertices could be recolored in order to gain a free color for vertex x . Butthen we uncolor all the vertices and give the following coloring procedure: assign color 1 to b , d and f , color vertices a, c, e independently, and finish by coloring vertex x . A contradiction.13 | L ( a ) | ≥ b | L ( b ) | ≥ c | L ( c ) | ≥ d | L ( d ) | ≥ e | L ( e ) | ≥ f | L ( f ) | ≥ x Figure 9: Graph of Lemma 22, where L ( x ) = { , , , , } .We conclude that L ( a ) ∩ L ( c ) = ∅ and therefore, by symmetry we have L ( c ) ∩ L ( e ) = ∅ . Sincethe colors are taken from the set { , , , , } , we conclude that L ( a ) ∩ L ( e ) = ∅ , and we fix withoutloss of generality φ ( a ) = φ ( e ) = 1. Then observe that 1 ∈ L ( b ), as otherwise we could color greedilythe remaining vertices d, c, b, x, f in this order. Symmetrically, we have 1 ∈ L ( d ). Therefore, weuncolor vertices a, e and fix φ ( b ) = φ ( d ) = 1. Since L ( a ) ∩ L ( c ) = ∅ , we know that 1 L ( c ). Thuswe color greedily the remaining vertices in the following order: a, e, f, x, c . Note that at each step,there exists an available color for the current vertex. Lemma 23.
The graph of Figure 10 with the given lower bounds on the sizes of lists of colorsfrom the set { , , , , } is L -colorable. b | L ( b ) | ≥ a | L ( a ) | ≥ c | L ( c ) | ≥ i | L ( i ) | ≥ yx d | L ( d ) | ≥ h | L ( h ) | ≥ zg | L ( g ) | ≥ e | L ( e ) | ≥ f | L ( f ) | ≥ Figure 10: Graph of Lemma 23, where L ( x ) = L ( y ) = L ( z ) = { , , , , } . Proof.
Let G be the graph from the statement of the lemma. We distinguish three cases:1. Suppose L ( f ) ∩ L ( d ) = ∅ and let α ∈ L ( f ) ∩ L ( d ). Then we give the following partial coloringof G : we color both f and d with color α and then greedily color e, c, b , in this order (at eachstep, there is at least one available color with respect to L ). Now the remaining uncoloredvertices a, y, z, g, h, i, x form the configuration of Lemma 22, so we are done.2. We have L ( f ) ∩ L ( d ) = ∅ . Now, suppose L ( f ) ∩ L ( e ) = ∅ . Then we color e with some color α ∈ L ( f ) ∩ L ( e ) and thus we still have | L ( d ) | ≥
3. We then color greedily f, g, h, i , in thisorder. The remaining uncolored vertices z, x, a, b, c, d, y form the configuration of Lemma 22,so we are done. 14. We have L ( f ) ∩ L ( d ) = ∅ and L ( f ) ∩ L ( e ) = ∅ . We claim that this case is impossible. Indeed,since the colors used by L are taken from the set { , , , , } , without loss of generality wecan assume that L ( f ) = { , } , { , } ⊆ L ( e ) and L ( d ) = { , , } . Moreover, by symmetrywe have L ( g ) ∩ L ( e ) = ∅ and thus L ( g ) = { , , } . Therefore | L ( g ) ∩ L ( d ) | = 1. Now note thatvertex a is symmetric to vertex d and to vertex g and thus must satisfy | L ( a ) ∩ L ( d ) | = 1 and | L ( a ) ∩ L ( g ) | = 1, which is impossible since the colors are taken from a set of five elements.Let C be the class of induced subgraphs of fullerene graphs. In order to show that everyfullerene graph is exact square 5-colorable, we will prove a stronger statement: that every graphin C is exact square 5-colorable.The main result of this section is the following theorem. Theorem 24.
Every graph G ∈ C is exact square -colorable.Proof. Let G be a graph of C that is a minimum counterexample to our claim, that is, it is ofsmallest order among those that are not exact square 5-colorable. Let H be a fullerene graph thatcontains G as an induced subgraph. In the remainder of the proof, G will be a regarded as a planegraph whose embedding is induced by the unique embedding of H . By Lemma 13, we know thatevery 5-cycle or 6-cycle of G is a face of both G and H . Furthermore, G has no other cycle oflength less than 9, and all cycles of length 9 are obtained from the symmetric differences of three5-faces sharing a common vertex. Further properties of G are as follows. Claim 24.A. G is -connected.Proof of claim. Note that the proof could be done in the same lines as for Claim 10.A for exactsquare 4-colorability of K -minor-free graphs. However, having five colors, we give a simpler proofhere.Suppose that G has a cut-vertex v . Since G is subcubic, it has a bridge uv . If one of u or v isof degree at most 2 (say, it is u ), then an exact square 5-coloring of G ′ = G − u extends to G byusing a proper permutation of colors in one of the connected components of G ′ .Therefore, both u and v are 3-vertices and the graph G ′ = G − uv has exactly two connectedcomponents G u and G v , containing u and v respectively. Let u and u (resp. v and v ) be theneighbors of u (resp. v ) in G u (resp. G v ). By minimality of G , graphs G u and G v are exactsquare 5-colorable independently. Take such a coloring φ of G u and let φ ( u ) = 1, φ ( u ) = 2and φ ( u ) = α . We show that an exact square 5-coloring φ of G v can be chosen to be compatiblewith φ in G . Without loss of generality we can fix φ ( v ) = 3 = α . Then by applying a properpermutation of colors of φ on G v , one can choose φ ( v ) and φ ( v ) such that α / ∈ { φ ( v ) , φ ( v ) } and we are done. ( (cid:3) ) Claim 24.B.
Any two -vertices of G are at distance at least .Proof of claim. The configurations of Figure 11 are reducible. Indeed, if one of these configurationsoccurs, then we remove from G vertices y, v (resp. y, z, v and y, z, t, v ) in the case of Configura-tion 11a (resp. 11b and 11c), in order to obtain a graph G ′ . The graph G ′ is in C and thereforehas an exact square 5-coloring which can be easily extended to G . ( (cid:3) ) Claim 24.C. G has no -face. y v w (a) distance 1 x y z v w (b) distance 2 x y z t v w (c) distance 3 Figure 11: Configurations where two 2-vertices are at distance at most 3. The neighborhood ofthe black vertices is exactly the one depicted in the figure.
Proof of claim.
By Lemma 13, every 9-face of G must contain three 2-vertices pairwise at distanceat most 3, contradicting Claim 24.B. ( (cid:3) ) For a face f of G , let ℓ ( f ) denote the length of f , and n ( f ) the number of 2-vertices on theboundary of f . Claim 24.D.
For every face f of G , we have then n ( f ) ≤ j ℓ ( f )4 k .Proof of claim. This follows directly from Claim 24.B. ( (cid:3) ) Claim 24.E. G is a fullerene graph.Proof of claim. Let F ( G ) denote the set of faces of G in its planar embedding. By Euler’s Formulawe have the following: X v ∈ V ( G ) (2 d ( v ) −
6) + X f ∈ F ( G ) ( ℓ ( f ) −
6) = −
12 (1)We assign to each vertex v the charge ω ( v ) = 2 d ( v ) − f the charge ω ( f ) = ℓ ( f ) −
6. We redistribute the charges by applying the following rule: every face f gives 1 to each2-vertex lying on its boundary.Note that after this redistribution of charges, the initial sum of charges is preserved. We analysethe new amount of charges of vertices and faces of G : • every 3-vertex has charge 0, • every 2-vertex has charge 0 since by Claim 24.A it lies on two faces, • every face f with ℓ ( f ) ≥
10 has strictly positive charge by Claim 24.D, • every face f with ℓ ( f ) = 6 has charge 0 by Claim 24.D, • every face f with ℓ ( f ) = 5 has charge − −
12 after the redistribution of charges, weconclude that G contains exactly twelve 5-faces and no other face of length at least 10. On theother hand, by definition of C , we know that G has no 7-faces, nor 8-faces. Also by Claim 24.C, G has no 9-faces. Thus we conclude that G is a fullerene graph. ( (cid:3) ) Claim 24.F.
Every -face of G is adjacent to at least one -face. roof of claim. Suppose to the contrary that G contains a 6-face adjacent to six 6-faces, as inFigure 12a. The graph G = G − { a, b, c, . . . , x } belongs to C and is exact square 5-colorable.By applying such a coloring to G , we get a valid partial exact square coloring of G . Observethat the remaining uncolored vertices induce two isomorphic connected components in G [ ♯ (seeFigure 12b). Thus, each can be (properly) colored independently. By counting the number ofremaining colors for each of the vertices of each of these connected components we obtain theconfiguration of Lemma 23, so we are done. ( (cid:3) ) a bc d e fg h i jk l m no p q rs t u vw x (a) A 6-face surrounded by 6-faces ac eh jk mp rs uw bd fg il no qt vx (b) Exact square Figure 12: A 6-face surrounded only by 6-faces and its exact square.
Claim 24.G.
Every -face of G is adjacent to at least two -faces.Proof of claim. Suppose to the contrary that G contains a 6-face adjacent to at least five 6-faces. ByClaim 24.F, this 6-face has exactly five adjacent 6-faces and one 5-face, as depicted in Figure 13a.The graph G = G − { a, b, c, . . . , q, s, . . . , x } belongs to C and is exact square 5-colorable. Byapplying such a coloring of G to G , we get a valid partial exact square coloring. Observe that theremaining uncolored vertices induce in G [ ♯ the graph depicted in Figure 13b. After counting thenumber of available colors for each uncolored vertex, we properly color this graph using Lemma 22:1. since | L ( v ) | ≥ | L ( t ) | ≥
3, by the pigeonhole principle assign to v and t the same color,2. color vertices j, e, a, c in this order,3. color vertices m, h, k, s, w, u, p by Lemma 22,4. color vertices x, o, g in this order,5. color vertices n, q, l, d, b, f, i by Lemma 22.Thus, the claim is proved. ( (cid:3) ) We can now deduce our last claim.
Claim 24.H. G has at most vertices.Proof of claim. Since there are exactly twelve 5-faces in G and each can be adjacent to at most five6-faces, there are at most sixty 6-faces in G . By Claims 24.F and 24.G, every 6-face is adjacent toat least two 5-faces. Thus there can be at most thirty 6-faces in G . Since each vertex belongs tothree faces, we conclude that G has at most · · = 80 vertices. ( (cid:3) ) bc d e fg h i jk l m no p qs t u vw x (a) A 6-face adjacent to exactly one 5-face ac eh jk mps uw bf di gn lq ov tx (b) Exact square Figure 13: A 6-face surrounded only by one 5-face and its exact square.To finish the proof, we have verified by computer using
SageMath [45] that all fullerene graphswith up to 80 vertices are exact square 5-colorable (in fact they are exact square 4-colorable). Thus G does not exist, and our theorem is proved. The list of these fullerene graphs is available online and was generated independently using the two computer programs fullgen [11] and buckygen [12,21] (these two different programs use two different methods, and thus this list is trusted to becomplete). We have studied the notion of coloring of exact square on some families of subcubic graphs. Thisfits into a larger frame of studying the chromatic number of exact distance d -power of graphs,which has recently got attention. However, this special case is also closely related to the wellstudied notion of injective coloring. While in exact square coloring the pair of vertices of an edgeare allowed to have a same color, in injective coloring this is only allowed for those edges that arenot in a triangle.A main question studied here is the maximum possible exact square chromatic number of theclass of subcubic planar graphs. Conjecture 1 would imply that this is at most 5. There areexamples of subcubic planar graphs which need five colors in any injective coloring, but all knownsuch examples contain triangles and are exact square 4-colorable. These examples are built using K − whose vertices must receive four different colors in an injective coloring, however the exactsquare of this graph has only one edge and can be colored by two colors only. Thus 4 is also apossible answer for our question.It can be easily checked that a minimum counterexample to the conjecture has no triangle.Thus, as a natural class, we considered cubic planar graphs each of whose faces is either a 5-cycleor 6-cycle. These are are the duals of planar triangulations having only vertices with degree 5or 6. Known as fullerene graphs they are well studied. For this class of graphs we characterizedthe ones admitting an exact square 3-coloring, and we proved that they all admit an exact square5-coloring.Further evidence that the upper bound of 5 might be replaced by 4 is the fact that the exactsquare of any bipartite subcubic planar graph is 4-colorable. Our proof of this fact uses the Four https://hog.grinvin.org/Fullerenes K -minor-free graphs. The upper bound of 5 is proved in [15] andthe upper bound of 4 on the class of triangle-free K -minor-free graphs follows from Theorem 10. Acknowledgements
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