Bipartizing fullerenes^{†}^{†}thanks: Supported by a CZSL bilateral project MEB 091037 and BICZ/1011004.
Abstract
A fullerene graph is a cubic bridgeless planar graph with twelve 5faces such that all other faces are 6faces. We show that any fullerene graph on vertices can be bipartized by removing edges. This bound is asymptotically optimal.
Keywords: Fullerene graph; Fullerene stability; Bipartite spanning subgraph
1 Introduction
Fullerenes are carboncage molecules comprised of carbon atoms that are arranged on a sphere with pentagonal and hexagonal faces. The icosahedral , wellknown as Buckminsterfullerene was found by Kroto et al. [10], and later confirmed by experiments by Krätchmer et al. [9] and Taylor et al. [12]. Since the discovery of the first fullerene molecule, the fullerenes have been objects of interest to scientists all over the world.
From the graph theoretical point of view, the fullerenes can be viewed as cubic 3connected graphs embedded into a sphere with face lengths being or . Euler’s formula implies that each fullerene contains exactly twelve pentagons, but provides no restriction on the number of hexagons. In fact, it is not difficult to see that mathematical models of fullerenes with precisely hexagons exist for all values of with the sole exception of . See [3, 5, 6, 11] for more information on chemical, physical, and mathematical properties of fullerenes.
The question of stability of fullerene molecules receives a lot of attention. The goal is to obtain a graphtheoretical property whose value influences the stability. Different properties, like the number of perfect matchings [7] or the independence number [4] were considered. The property investigated in this paper is how far the graph is from being a bipartite graph, which was suggested by Došlić [1] and further considered in [2]. Despite of the effort none of the so far considered parameters works in all cases. Hence more research is still needed.
For a plane graph , let be the set of the faces of . Let be a fullerene graph, and let be the weighted complete graph whose vertices correspond to the faces of and the weight of the edge joining two faces and is equal to the distance from to in the dual of . Let be the size of the minimum set such that is bipartite. Došlić and Vukičević [2] proved the following:
Theorem 1.
If is a fullerene graph, then is equal to the minimum weight of a perfect matching in .
A corollary of the above theorem is a polynomialtime algorithm for finding a set of edges whose removal makes the graph bipartite.
Došlić and Vukičević [2] conjectured that . In fact, they gave the following stronger conjecture.
Conjecture 2.
If is a fullerene graph with vertices, then .
The main result of this paper is an upper bound on , confirming the weaker version of the conjecture.
Theorem 3.
If is a fullerene graph with vertices, then .
2 Proof of Theorem 3
Let be a fullerene graph. A patch with boundary is a connected subgraph such that (usually, we consider to be the outer face of ) and (but it is possible for the boundary to be also a face of ). Let be a vertex incident with . If , then is a vertex (with respect to ), otherwise is a vertex (with respect to ). An edge incident with is a edge (resp. a edge) if both vertices incident with are vertices (resp. vertices) with respect to . If is neither a edge nor a edge, then it is a edge. The description of the boundary is the cyclic sequence in that represents a edge, represents a edge, and a maximal consecutive segment of edges is represented by the integer giving its length. For example, the boundary of the patch consisting of a face and a face sharing an edge is described as .
Let and be the numbers of edges and edges of , respectively, and let be the number of pairs of consecutive edges of . Let be the number of faces of distinct from . The following lemma relates the number of  and edges; a similar relation was derived by Kardoš and Škrekovski [8].
Lemma 4.
If is a patch with the boundary , then .
Proof.
Suppose that the length of is . Let , and let be the number of faces of . Since each edge of is incident with two faces,
i.e., . Note that the number of vertices is , which can be easily seen from the modification of the boundary by adding and deleting 3vertices so that there is no 33edge or 22edge. Thus
Substituting for , we obtain
By Euler’s formula, , thus and the claim of the lemma follows. ∎
A patch with the boundary is a fat worm if , the subgraph of induced by is a path , and the edges of incident with each two consecutive inner vertices of are not incident to a common face of . See Figure 1(a). Note that in this case, the description of is

if has length and

if has length .
We consider the patch with exactly one vertex not incident with (and boundary ) to be a fat worm as well (in this case, has length ). The patch is a slim worm if , and . Geometrically, it is a straight line of hexagons, see Figure 1(b). Note that for some (or , when and is a face). The patch is a worm if it is a fat worm or a slim worm. The shell is the patch with boundary such that and (having internal vertices). See Figure 1(c).
An chord of a cycle in a patch is a path of length with distinct endvertices belonging to such that the inner vertices and edges of the path do not belong to . We say a chord instead of a chord. Consider an chord of the boundary of a patch . Let and be the two patches into that splits (i.e., the subgraphs such that , and ), and and their boundaries. We say that splits off a face if or . The patch is decomposable if it contains a simplifying cut, that is

an chord of with such that , or

two chords and such that , and are edges of . See Figure 2.
Otherwise, we call indecomposable. We say that is a normal patch if is indecomposable, no face of distinct from shares an edge with and is neither a worm nor a shell.
Lemma 5.
Let be a normal patch with boundary and an chord of , with . Then and splits off a face. Furthermore, the number of edges incident with the endvertices of is most .
Proof.
Let and with boundaries and , respectively, be the patches to that splits . Let and .
Suppose first that . Since is not a slim worm, there exists an edge that either is a edge of or is incident with a vertex in . Let us choose the chord and the patches and so that and is minimal. As is indecomposable, each edge of is also a edge in or . It follows that and are vertices, and since the internal face incident with has length six, and must be adjacent. Since , we have ; hence, is a chord of . The chord splits to patches and with . However, this contradicts the choice of , since it is easy to see that . We conclude that is an induced cycle.
Suppose now that . By symmetry between and , we may assume that is a vertex in . Since , we have that and are vertices, and it follows that is a face split off by .
Finally, suppose that . Suppose first that both and are vertices in , and thus is a edge with respect to . Since , at least one of and (say ) is a vertex. Thus, are all incident with a common face, which is only possible if and consists of a single face. It follows that splits off a face.
The case that both and are vertices in is symmetrical. Hence, without loss of generality, we assume that is a vertex and is a vertex in . As , we infer that both and are vertices. Let be the third neighbor of . Observe that

if , then both and split off a face (for the former, note that the edge joining with a neighbor of is not a chord, since we already proved that is an induced cycle). See Figure 3(a).

if , then and are adjacent, is a face and we may apply the same observations to the chord . See Figure 3(b).
By symmetry, this argument also holds for . Hence by repeating the argument we conclude that is a fat worm, contradicting the assumption that is a normal patch.
Furthermore, note that if splits off a face, then , where is the number of edges incident with or . Since is indecomposable, it follows that . ∎
For a patch with boundary , let be the subgraph consisting of the outer layer of the faces of ; that is, is an edge of if and only if it is incident with a face that shares an edge with . Let be the set of vertices that have at least two neighbors in . Let . See Figure 4(a).
Lemma 6.
If is a normal patch with boundary , then is a cycle, and the patch bounded by satisfies , and . Furthermore, .
Proof.
Since is not a fat worm, we have . If two vertices of were adjacent, then by Lemma 5 and would be a fat worm, thus is an independent set and is not empty. Lemma 5 also implies that is connected, and since only contains vertices whose degree in is one, is connected as well.
Suppose that a vertex of is adjacent to more than one vertex of . Since is not the shell, is adjacent to exactly two vertices in ; let be the neighbor of not in . Since is not a fat worm, we have . Let and be the neighbors of distinct from ; since , we may assume that . Let be the face of incident with , and , and let be the neighbor of in distinct from . Note that is incident with a neighbor of that belongs to , and thus shares an edge with . Hence is a face and we have . If , then the chord contradicts Lemma 5. Otherwise, by a symmetric argument we conclude that a face incident with , and is also a face sharing an edge with , see Figure 4(b). However, forms a simplifying cut (a pair of chords) in , which is a contradiction. Therefore, each vertex of has at most one neighbor in .
By Lemma 5, no vertex of has a neighbor both in and in , since at least one of the two resulting chords would not split off a face. If is a vertex of that has a neighbor in or , then has two neighbors in , and thus has at least three vertices.
Suppose that contains a bridge . Note that both faces and of incident with share an edge with . As , these two vertices do not lie on chords. Note that contains an chord of such that and , where . Similarly, let be an chord of such that and . As neither nor splits off a face, Lemma 5 implies that . Since and are faces, we conclude that and are chords. Lemma 5 further implies that and are middle vertices of and , thus and is a pair of chords forming a simplifying cut. This is a contradiction; therefore, is edgeconnected. Since , every edge of is incident with a face that shares an edge with . We conclude that is a cycle.
Consider now a edge in and let be the incident face. Lemma 5 implies that each of and has only one neighbor in , as otherwise one of them would belong to a chord whose endpoint is incident with a edge . Therefore, and is a part of , and is a edge with respect to . It follows that . On the other hand, consider a edge of , and let be a part of the boundary of . As and are vertices in , there exists a face in , and is a edge in . Hence, we have and by Lemma 4, .
Similarly, consider a part of , where and are edges. The common neighbor of and belongs to , and its neighbor distinct from and belongs to . As we observed before, both neighbors and of distinct from belong to . Furthermore, by Lemma 5, the endpoints of the chord are incident with no edges, thus and are vertices. It follows that both and have a neighbor in , and and are edges with respect to . Hence, we conclude that .
In fact, can be obtained from in the following way: Add between each two consecutive letters in . Since endvertices of a chord of are not incident with edges, if appears in the resulting sequence, then it is as a part of a subsequence , where . We construct by

for each subsequence, decreasing each of and by ,

for each not contained in such a subsequence, decreasing each of the neighboring integers by ,

for each , increasing each of the neighboring integers by , and

suppressing any zeros.
Note that the increases/decreases are cumulative, e.g., if contains a subsequence , then the sequence contains a subsequence (or after suppressing zeros). By Lemma 4, , and the formula for the length of follows:
∎
Consider a patch with boundary . A sequence of cycles , , …, (with ) is called an uninterrupted peeling if for , the subpatch of bounded by is normal and .
Lemma 7.
Let be the boundary of a patch such that . If , , …, is an uninterrupted peeling, then the number of vertices of outside of (and not including) is at least .
Proof.
Let be the largest index such that and let be the smallest index such that . Note that if the whole sequence is decreasing then and similarly if the whole sequence is increasing then , hence in all the cases. We compute a lower bound on the the number of vertices of outside of as
First, we deal with the middle term. Let . Suppose that . In this case, we have ; let . By Lemma 6, for . Since , we conclude that . It follows that contains a subsequence , where by Lemma 5. By Lemma 4, . As , we conclude that , and thus . By symmetry, assume that . As observed in the proof of Lemma 6, contains a subsequence , where (the equality is achieved if is adjacent to in ). The same observation applies to , …, . In the normal patch , the integers adjacent to are greater or equal to three, thus and . It follows that . In the case that , we have , since .
Now we deal with the other terms of the sum. If , then the sequence dominates the arithmetic sequence with the first element and step due to Lemma 6 and the fact that for . Hence . If then .
Similarly, the sequence dominates the arithmetic sequence with the last element and step , hence .
Note that . Summing these inequalities, we obtain
where the lower bound in the last inequality is achieved for , and . Since all the cycles , …, are strictly outside of , the claim follows. ∎
Lemma 8.
Let be a fullerene with vertices and a face of . There exist at least five faces distinct from whose distance to in the dual of is at most .
Proof.
We define a rooted tree with each vertex of corresponding to a patch such that and . Furthermore, we assign a weight to each edge of . The root of is the patch whose boundary is the cycle bounding , i.e., . Suppose that a patch with boundary is a vertex of . Let us note that is neither a worm nor the shell, since . The sons of in the tree are defined as follows:

If and shares an edge with a face of , then is a leaf of .

If is a normal patch, then has a single son , equal to the last element of the maximal uninterrupted peeling starting with . The weight of the the edge joining with is equal to the length (number of patches) of the uninterrupted peeling. Note that is not a normal patch.

If has a simplifying cut, then let and be the boundaries of the two patches and to that it splits . Note that . The patch (with ) is a son of if and . In that case, the edge between and has weight . Since and , has at least one son.

Finally, if is indecomposable, and shares an edge with a face , note that there exists an chord (with ) splitting off (otherwise would be incident with a chord and a chord and both of them would witness the decomposability of ). We let the son of with boundary be the patch obtained from by removing edges incident to both and and by removing isolated vertices. We let the edge of between and have weight .
The type of is defined according to the rule ((a) to (d)) in that its sons are described.
Observe that at least five faces distinct from share edges with boundaries of the patches forming the vertices of of type (a) or (d). Indeed, either all faces are reachable in this way, or there are exactly six potentially unreachable faces contained in a single patch that is a leaf of , or split off by a simplifying cut from an internal vertex of . Let be a subtree of of smallest possible depth that contains five vertices of type (a) or (d). We choose to be minimal, i.e., all leaves of are of type (a) or (d).
Consider a vertex with a son in , and let and be the boundaries of these patches. If is of type (b), then and by Lemma 6. If is of type (c), then and . If is of type (d), then and .
Let be the path in joining the root with a leaf whose boundary is incident with a face . Observe that the distance between and in the dual of is at most the sum of the weights of the edges of , plus . Let and be the boundaries of and , respectively. Let , and be the numbers of vertices of types (b), (c) and (d) in distinct from , respectively. By the observations in the previous paragraph, we have . By the choice of , we have , and since is nonnegative, . Therefore,
Let , , …, be the sequence of the weights of all edges of such that is a normal patch; by the construction of , is not a normal patch in this case, hence . Using Lemma 7, we obtain
Therefore, the total weight of these edges is at most , and the distance between and is at most . ∎
Lemma 9.
Every graph on vertices with minimum degree such that has a perfect matching.
Proof.
If does not have a perfect matching, then there exists a set such that has more than components of odd size. Consider such a set , and observe that . As , is either (and thus has a perfect matching) or is connected. Therefore, .
If , then since , no component of may consist of a single vertex, and hence has at most three odd components and . Since , each component of has size at least . However, must have at least two components of odd size, thus it would have exactly two components of size . However, then , which is not smaller than the number of odd components.
Therefore, and has at least components of odd size. However, this is only possible if each component of consists of a single vertex, and hence . ∎
Proof of Theorem 3.
Let be the subgraph of consisting of edges with weight at most . By Lemma 8, , and thus either has a perfect matching or as a subgraph, by Lemma 9. In the former case, the weight of each perfect matching in (and thus of the minimumweight perfect matching in ) is at most . In the latter case, note that the weights in satisfy the triangle inequality, thus the weight of any edge in is at most , and we conclude that has a perfect matching of weight at most . By Theorem 1, . ∎
The multiplicative constant is likely to be far from the best possible. Indeed, it can be somewhat improved by a more complicated analysis of our argument (e.g., observing that not all faces can appear in on the lowest possible level, indicating that some of the edges of are much shorter than we estimated). Nevertheless, we could not improve it enough to approach the best known lower bound of of Došlić and Vukičević [2].
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