1 Introduction
Finding long cycles in graphs of low degree but high connectivity has been a fruitful and interesting line of research. Relevant to our work is the result by Bondy and Entringer [2] which shows that every 2edgeconnected cubic graph on vertices has a cycle of length . Lang and Walther showed that this result is essentially tight [5]. However, to the best of our knowledge, the question about the longest path has not been explicitly answered. One might be tempted to conjecture that the answer should be similar; that the best lower bound we can find is of the order of . Surprisingly, we show that there always exists a path of the order of (ignoring factors of order ). Using a simple construction we show that this is tight up to lowerorder factors, even for the case of planar cubic graphs.
For our lower bounds we decompose the graph along edgecuts of size two and use the resulting recursive structure to construct a long path. The recurrence takes advantage the important fact that a 3connected cubic graph always contains a cycle of length at least , where , which was first shown by Jackson [4], proving a conjecture by Bondy and Simonovits [3], and was later improved by Bilinski [1] to and by Liu [6] to . Together these result show that, up to lowerorder factors, every cubic connected graph contains a path of order and we cannot expect a longer path in general.
2 Preliminaries
For an integer we use the notation to denote the set , and for integers we use the notation to denote the set .
All graphs considered in this paper will be finite, undirected, and loopless; but may contain parallel edges. For sets and set members we will use the notation and . For vertex sets we will frequently use the notation for its complement.
For a graph we use the shorthand for the number of vertices in the graph, and we write to denote that the edge is contained in . A graph is cubic if every vertex has degree exactly three, and it is edge connected if the removal of any edges does not disconnect the graph.
We define a cut of a graph to be any bipartition of . The cutset of a cut is the set of edges with one endpoint in and the other in . The size of a cut is the size of its cutset. We call the start and endvertex of a path its endpoints and all other vertices on it the internal vertices. For , an path in is a path with endpoints and . A path avoids an edge (vertex) if the edge (vertex) is not a part of the path.
3 Upper bound
In this section we define a family of graphs in which every path has length . Every graph in the family consists of two isomorphic binary trees whose leaves are identified, plus an edge between the roots of the trees and some edges between their leaves, see Figure 1. Formally, for every integer , we define with vertex set
and edge set which contains

the edges and , for every and ,

the edges and , for every and ,

the edge between the roots of the trees, and

the edges , for every , between leaves of the trees.
For every and , we let be the induced subgraph of all the vertices in the subtree of that is rooted in and the subtree of that is rooted in . Formally, a is the induced subgraph of on the set of vertices
We call the subgraphs the blocks of .
Lemma 3.1.
is regular, planar, edgeconnected and has vertices.
Proof.
By construction, is regular and, for every , there are labelled vertices . Thus, the cardinality of is indeed .
We prove that is edgeconnected by showing that for any pair of vertices and of , there exists a cycle in that includes both and . First assume that and lie inside the same tree, let’s say . Let be the least common ancestor of and in . Choose two paths , starting both at and going through and , respectively, down towards some leaves of . If we mirror and into we obtain a cycle that contains both and .
Now assume that and lie in different trees, say and . Let be the mirror image of in (so if then ). We construct the same cycle as in the previous case and simply note that it contains and therefore, by construction, also . We remark that this also works when .
A planar embedding is exemplified in Figure 1 if we move the edge either into the middle of the figure or route it through the outer face. ∎
To analyse the length of paths in we introduce the following terminology. Let be a path of
. We classify every vertex
of as follows:
a boundary if
is odd,

a leaf if and is even

a turn if , and in it is adjacent to both vertices , , where ,

a way if , and it is not a turn.
We write just boundary, turn or way, when the relevant path is clear from context. See Figure 2 for a depiction of the vertex types.
Observation 3.2.
If is a path of , then every one of its vertices is either a leaf, a boundary, a turn, or a way, and the number of boundary vertices is or .
Proof.
The observation follows from the fact that, as a standalone graph, is Eulerian. ∎
For the remainder of this section let be an arbitrary path in . Recall that our goal is to upper bound the number of edges in . To do so we show that every edge in which is not between leaf vertices can be charged to a specific segment of . A segment is a minimum subpath of with one endpoint being either at a turn or a boundary and the other either a boundary or a leaf. We further on bound the number of segments in . This is sufficient for bounding the path’s length because the number of edges between leaves is bounded from above by the number of such segments.
We note that since there are at most two boundary vertices each of degree , the set of specific segments contains at most six paths with one of their ends being a boundary vertex. The length of every one of these paths is bounded above by . Thus, for our bound we need to focus on the paths between a turn vertex and a leaf. We note that every turn vertex , is an endpoint of at most two segments. Consequently, to prove this section’s main result, we show that has at most two turn vertices with the same first index.
Lemma 3.3.
Let be a path of and . There exist at most turn vertices of such that .
Proof.
Assume for the sake of contradiction that there exist three distinct turn vertices , and such that and , for every .
If any two of and are both equal to some , then two of the three vertices are labelled and . Since and are turn vertices, all the edges of are contained in , because the only edges between vertices in and the vertices not in are adjacent to and and are not in , since and are turn vertices. Therefore, there can be no other turn vertex whose first index has an absolute value of , contradicting our assumption on , and .
Now assume that each of the vertices , and are contained in a distinct block, e.g. all three indices are distinct. We next prove that this implies that every one of these blocks has a boundary vertex, contradicting Observation 3.2.
Fix . We show that has a boundary vertex. Assume towards a contradiction it does not. Thus, in particular, neither one of and is a boundary vertex. Since is connected, one of and is a way and the other is a boundary vertex or a way. Let be subgraph consisting of the edges common to and . We observe that is an vertex of and has degree in . Thus, by Observation 3.2, has another vertex of odd degree in . Since this vertex is different from and , it also has odd degree in . Thus, has a boundary vertex in contradiction to the assumption that it does not.
Hence our assumption on the existence of , and implies that has three boundary vertices, contradicting Observation 3.2. This proves the claim. ∎
Theorem 3.4.
Every path in is of length at most .
Proof.
Fix some path in , we show that it has length at most . With every turn we associate segments in , both between and either a boundary vertex or a leaf vertex. One of the segments contains the vertex and the other , where and has the same sign as . Every internal vertex in either segment is a way vertex, so the length of both paths is bounded above by . Thus, by Lemma 3.3, the total sum of segmentlengths associated with every turn vertex on is bounded above by .
Edges of may reside on a segment that includes both and . Such segment is either between two boundary vertices, or a boundary vertex and a leaf vertex. The length of such a path is bounded above by .
All the edges which are not part of the segments above are as follows: (i) adjacent to leaf vertices, or (ii) in paths that do not include and and are between a boundary vertex and either a boundary vertex or a leaf vertex.
There are at most edges as in (ii). Every edge as described in (i) is adjacent to at most distinct segments, each one between a boundary or turn vertex and a leaf vertex. By Observation 3.2 and Lemma 3.3, there are at most such paths. Thus, over all the number of edges as in (i) is bounded above by . Consequently, the number of edges in is bounded above by . ∎
4 Lower bound
We prove here that every connected cubic graph has a path of length . For the simplicity of the presentation we have not optimized the constants involved. We will need the following two known lemmas.
Lemma 4.1 (Based on Bondy [2]).
Every connected graph with maximum degree has a cycle of length at least .
Lemma 4.2 (Jackson [4]).
Let be a connected graph on vertices and let . Then and are contained in a cycle of of length at least .
We will further need the following simple observation:
Observation 4.3.
A connected cubic graph is either simple or it is the multigraph on two vertices with three edges.
Proof.
Assume there exists a graph with parallel edges between on more than two vertices. Then and must each have one further edge that leaves the set . But then is a cut of size two, contradiction. ∎
We will also need the following simple proposition regarding connected cubic graphs.
Proposition 4.4.
Let be a connected cubic multigraph and , edges in the graph. There exists a cycle in containing both and . Furthermore, if is contained in a cycle of length then is contained in a cycle of length at least .
Proof.
Since is connected, every pair of vertices incident to the same edge have two edgedisjoint paths between them. Therefore, has a cycle containing and a cycle containing . Let be the length of . If is in , then the proposition holds. So, assume that is not in .
Suppose that and have common edges. There exists a shortest path in that contains and is between two vertices in . Using this path and the rest of , we get a cycle that has and as edges, or possibly just , but has length at least ( may be in parallel to an edge of ), which implies the theorem.
We note that if and have a common vertex, then they also share at least one edge, because is cubic. Hence we only need to deal with the case that and are vertexdisjoint. Let be a vertex in and a vertex in . Since is connected, there exist two edgedisjoint paths between and . These paths each have a minimal subpath that contains a vertex from and from but none of their edges. Using these paths and and , we get a cycle that has and as edges, or possibly just , but has length at least ( may be in parallel to an edge of ), which proves the theorem. ∎
For the remainder of this section we fix to be an arbitrary but sufficiently large connected graph on vertices. We next give three more definitions and one proof that are essential for this section’s main result. Afterwards we explain how these notions work together.
Definition 4.5 (Tombolo, tombolocut).
A cutset of size exactly is called a tombolo. A tombolocut of a graph is a cut whose cutset is a tombolo.
Proposition 4.6.
For every tombolocut of the edges of its tombolo are vertex disjoint.
Proof.
Let be a tombolocut and its tombolo, where . Assume, for the sake of contradiction that . This implies that is a cut of size one, contradicting that is connected. Hence, and, by a symmetric argument, . ∎
The following definition of a virtual subgraph is somewhat similar to the more general idea of a torso in a graph decomposition: we take a subgraph and encode the external connectivity by adding (virtual) edges.
Definition 4.7 (Virtual subgraph, virtual edges and real edges).
A pair of vertices is called a portpair if they are in the same set of the tombolocut and incident to the cut’s edges. A virtual subgraph of a multigraph is a multigraph such that and is obtained by taking

all the edges of the subgraph of induced on , referred to as the real edges of , and

an edge (possibly parallel), called a virtual edge, for every and that are the vertices of a portpair of a tombolo whose edges are both not in .
For the sake of clarity, we will call a tombolocut in a virtual subgraph a virtual tombolocut. The cut itself might or might not actually use virtual edges.
Definition 4.8 (Peninsula).
A virtual subgraph is called a peninsula, if is a tombolocut or if .
The proof of this section’s main result is based on an induction over the number of vertices in a peninsula. The next proposition enables us to use Lemma 4.1 on peninsulas. After that we provide an extra definition that enables us to formally explain how peninsulas connect with virtual graphs and how they can be used for constructing paths.
Proposition 4.9.
Let be a peninsula. If , then it has exactly one virtual edge and the rest of its edges are real edges. The graph obtained from by replacing parallel edges with a single edge is connected and has maximum degree .
Proof.
If , then the proposition holds, by the assumption on , so we assume that .
By construction, the only vertices that are not adjacent to the same vertices in and in , are the ones that are adjacent to the tombolocut separating from the rest of the graph. We note that, by Proposition 4.6, that these are a pair of distinct vertices . Thus, by the definition of a virtual subgraph, the single virtual edge is added to create from . Since is cubic, all vertices in have degree exactly three. The simple graph obtained from by removing potential parallel edges between and has therefore only vertices of degree and .
It is left to show that this graph is indeed connected. If then the statement clearly holds, so assume otherwise. Consider any pair of vertices , since is connected there exist two edgedisjoint paths connecting them. Since is cubic, at most one of them can use an edge between and , thus the same paths exist in and we conclude that is connected. ∎
Definition 4.10 (Internal port and external port).
The single virtual edge of a peninsula, that is not the whole graph, is call the peninsula’s internal port. If an arbitrary edge is fixed to be its internal port. The virtual edge created by a peninsula in a virtual subgraph (other than ) is called its external port.
Lemma 4.11.
Let be the external port of a peninsula . There exists an path in whose internal vertices lie entirely in and that has length at least .
Proof.
Let and be the vertices adjacent to ’s internal port. By Proposition 4.9, the graph that we get by replacing every parallel edge in with a single edge is connected and has maximum degree . It also still has an edge . Thus, by Lemma 4.1, has cycle of length at least . This implies, by Proposition 4.4, that has a cycle of length at least that includes . By construction, the same holds for . Using this cycle and the tombolo separating from the rest of the graph, we conclude the existence of the path required for the lemma to hold. ∎
Peninsulas by themselves are not sufficient for our goal. We need a virtual subgraph that will enable us to use Lemma 4.2. We next give an algorithm that provides us with such a virtual graph.
The core of a peninsula .
The core a peninsula , with internal port , is a virtual subgraph obtained using the following algorithm:
Algorithm 1 always terminates since at each the new graph has strictly less vertices and the initial number of vertices is finite. For simplicity, we will still speak of the core of a peninsula with the understanding that any virtual subgraph constructed by the above algorithm will work^{1}^{1}1Cores are unique but we do not need that fact and hence do not prove it here..
Lemma 4.12.
Let be a peninsula with internal port and core . Then .
Proof.
Let and let , be the vertices adjacent to ’s external port, hence and . Furthermore, let be the sequences of virtual subgraphs constructed by Algorithm 1.
For a core and a peninsula whose external port lies in we say that and are linked or linked via . The following lemma proves that all peninsulas linked to a core are disjoint and separated by the core from each other.
Now we are ready to explain how the induction we use for the main result works. We have a peninsula and its core . Our induction assumption is that for every peninsula with internal port and every vertex adjacent to , has a long enough path that avoids and has as an end point. The idea is to find a path with the required properties in and turn it into a path in , by replacing external edges in this path, with paths in their peninsulas. The following lemmas provide the means to show that this can be done.
Lemma 4.13.
Let be a peninsula with internal port and core . Then is cubic, connected and for every pair of distinct peninsulas and linked to it holds that and do not have edges between them and that they do not share vertices with each other or with .
Proof.
Let with and be the sequence of virtual subgraphs constructed by Algorithm 1. We prove, by induction on , that the above statement holds for all and therefore in particular for . We further prove that together with all peninsulas linked to it forms a partition of .
Let us begin with . By construction and by Proposition 4.9, is cubic, connected and only has as a virtual edge. If then the trivially partitions . Otherwise, is a peninsula linked to another peninsula via and both together partition via the tombolocut between them. The claimed properties therefore hold trivially.
This concludes the induction base and we now assume that the lemma statement holds for and prove it for . Let be the cut with a cutset of size that was used in order to construct . Assume without loss of generality that (otherwise rename the cut). Let be the vertices that are adjacent to the edges of the cutset of .
Let contain and all vertices of peninsulas linked to . In the following, we will refer to these peninsulas as islands. By induction, contains and the vertices of all peninsulas which are linked to but not to . We will refer to these peninsulas as islands. () Note that, by construction, there can be edges from islands into but there cannot be any edges from islands into .
Assume first that is a tombolocut in . Then and are adjacent to the tombolo edges and lie inside . Accordingly, there exists a virtual edge in which does not exist in . Since is, by induction, cubic and connected, it follows that is, too, due to this additional virtual edge. Therefore it suffices to show that indeed is a tombolocut in order to proof that cubic and connected.
To that end, we show that the cutset of in is at most as large as the cutset of in , which contains edges. Since is connected, this implies that is a tombolocut.
Consider any edge from the cutset of in with and . By the induction assumption, cannot connect two distinct peninsulas linked to . Since every peninsula linked to is, by construction, contained completely in either or , we further conclude that cannot lie entirely within a single island or island. As observed above (), islands have no edges into and therefore cannot lie inside an island, which implies that .
If , then is an edge in and therefore a real edge in , hence it is also contained in the cut in . This leaves the case in which is contained in a island . Let be the external port of in . Since is a island, must lie in and, by construction of , is a virtual edge in . We charge the virtual to the real edge . We conclude that every distinct edge in the cutset of implies a distinct edge in the cutset of .
Finally, let us proof that all peninsulas linked to are pairwise vertexdisjoint, not connected to each other and disjoint from . We proved above that the cut is actually a tombolocut. Accordingly, all islands together with form a single large peninsula linked to via while all islands are linked to the same way they were linked to . Thus the claim still holds among all islands and we have to only consider cases involving the newly linked peninsula . We already observed that islands have no edges into and, by induction, no edges into islands, thus they have no edge into and are disjoint from it. The peninsula is, by construction via a cut in , disjoint from and the claim that together with and all islands partitions follows from the same construction. ∎
Lemma 4.14.
Let be a peninsula with internal port and core , where . Then is either connected and does not have parallel edges, or is connected with a twocut that contains in its cutset and in which the virtual subgraphs induced by and are both cubic and connected.
Proof.
Since Algorithm 1 iterates as long as there is a virtual tombolocut that does not contain it is clear that it terminates if either is connected or if every virtual tombolocut contains . Let us first show that there is actually only a single such cut.
Assume towards a contradiction that there exist multiple virtual tombolocuts with in them. Note that the respective other edge is a bridge in the graph . Let be bridges in which both have an endpoint in some connected component of . Then is is a tombolocut which does not contain in its tombolo—a contradiction. We conclude is the only virtual tombolocut in and we let be its tombolo.
Since every other cut of must have size three or larger, its directly follows that the virtual subgraphs induced by (with the additional virtual edge ) and (with the additional virtual edge ) are cubic and connected. ∎
Lemma 4.15.
Let be a peninsula, its internal port and its core. For every vertex adjacent to there exists a path in of length at least that avoids and has as one of its endpoints.
Proof.
By Lemma 4.14, the core is either connected or has a special cut. Suppose first that is connected. If is the multigraph with two vertices and three edges, then there is trivially a cycle of length at least that includes . Otherwise, by Observation 4.3, is simple and by Lemma 4.2 has a cycle of length at least that includes . In either case, the respective cycle shows that for every vertex adjacent to there exists a path as required in the lemma.
Now assume that is not connected but instead can be partitioned into two sets and such that the virtual subgraphs induced by these sets are both connected. At least one of the sets has vertices, without loss of generality assume that this is (otherwise rename the sets accordingly). Let be the virtual subgraph whose vertices are the set and let be the virtual edge in that includes a vertex adjacent to . If is simple, then by Lemma 4.2 it has a cycle of length at least that includes ; if has only two vertices then trivially it has a cycle of length at least that includes . According to the construction of , this cycle implies the existence of a cycle in that includes and has length at least , which in turn implies the existence of a path as stated by the lemma. ∎
Lemma 4.16.
Let be a peninsula, its internal port, its core and an edge in . For every vertex adjacent to there exists a path in that starts in , avoids and has as its last edge.
Proof.
Since is cubic and connected, by Proposition 4.4, and are in some cycle in . Consequently, for every vertex adjacent to there is a path in this cycle that does not include and can be seen as starting in and having as its last edge. ∎
Lemma 4.17.
Let be a peninsula, its internal port, its core and , distinct edges in . For every vertex adjacent to , there exists a path in that avoids , starts in , contains and , and ends in either or .
Proof.
Since is cubic and connected and, by Proposition 4.4, and are in some cycle in . If is also in then for every vertex adjacent to there exists a path in this cycle that avoids and starts in , contains both and and ends in either or .
Thus assume that is not on the cycle and fix one of the endpoints of as . Since is connected there exist two edgedisjoint paths from to some arbitrary vertex on the cycle and at least one of these two paths avoids . Therefore there exists a path starting at , avoiding , containing both and and ending in either or , as claimed. ∎
Definition 4.18 (power).
Let be a peninsula, its internal port. We denote by the length of the longest path in that starts in and avoids and we write
The following lemma now relates the existence of a long path in the core of a peninsula—which might include virtual edges—to the existance of a long path in the peninsula itself. The latter path, by definition, only consists of real edges. We express this fact by proving recurrent inequalities for .
Lemma 4.19.
Let be a peninsula, its internal port, its core. Let further be a path in that avoids but starts in a vertex that is incident to and has length at least .
Assume contains two virtual edges which link the peninsulas to , respectively. Assume is the edge closer to on . Then
If contains only one virtual edge which links to , then  
Proof.
Note that concerns paths of the peninsulas , hence in order to use to find lower bounds on we first need to argue that we can replace all virtual edges on by paths through peninsulas that contain only real edges. Let be the collection of all peninsulas linked to . By Lemma 4.13, all members of are pairwise disjoint, not connected by edges and also disjoint from . Therefore we can apply Lemma 4.11 to each virtual edge which links to and replace it a real path of length at least .
It is left to show that we can make the above claimed guarantees on . In the first case, we replace as described above by a path of length , however, we replace by crossing the tombolo and finding a path of length inside of , without resurfacing through the tombolo again (hence all edges of after are lost). Note that such a path exists by the definition of , proving the first inequality.
Similarly, if contains only one virtual edge , we replace it by a path of length without resurfacing into . The constructed path contains at least one edge more than since we count the tomboloedge that leads into . This proves the second inequality. ∎
Theorem 4.20.
Let be connected and cubic. Then has a path of length at least .
Proof.
We prove that for every peninsula on vertices , e.g. has a path of length at least . Since is also a peninsula this implies the theorem.
Let set . Suppose first that . Then, we only need to show that . The function is negative for , thus the inequality trivially holds. Therefore assume that and hence we may assume that is sufficiently large for the asymptotic inequalities we use in the sequel. By induction, assume that for every peninsula with . We next prove the induction step.
Let be ’s internal port, a vertex adjacent to and let be the core of . By Lemma 4.14, is either connected or can be partitioned into two set such that the virtual subgraphs induced by and are both connected. If , then by Lemma 4.15 there exists a path in which starts in , avoids and has length at least . Consequently, by Lemma 4.19, there exists a path in that includes only real edges, avoids , has as an endpoint and has length at least and the theorem statement holds. Thus assume that .
Since , has at least one virtual edge that is not . Suppose that there exists a peninsula linked to via such that . Let be a vertex adjacent to . By Lemma 4.16, there exists path in that starts in , avoids and has as its last edge. Consequently, by Lemma 4.19,
We now show that the above implies :
It remains to show that in case where no such big peninsula exists.
Assume that every peninsula linked to via an edge other than contains less than vertices. Since is cubic, which means that there are at most peninsulas linked to .
Let be the two largest peninsulas attached to via , respectively, with . By averaging we conclude that . However, by our prior assumption, . Therefore we can assert that
Let be a vertex adjacent to . By, Lemma 4.17 there exists a path in which starts in , avoids and contains both and . Consequently, by Lemma 4.19,
where is a lower bound on the sizes of both and . The following computation implies that :
This concludes the proof. ∎
References
 [1] Mark Bilinski, Bill Jackson, Jie Ma, and Xingxing Yu. Circumference of 3connected clawfree graphs and large eulerian subgraphs of 3edgeconnected graphs. Journal of Combinatorial Theory, Series B, 101(4):214–236, 2011.
 [2] John Adrian Bondy and RC Entringer. Longest cycles in 2connected graphs with prescribed maximum degree. Canadian Journal of Mathematics, 32(6):1325–1332, 1980.
 [3] John Adrian Bondy and M Simonovits. Longest cycles in 3connected 3regular graphs. Canadian Journal of Mathematics, 32(4):987–992, 1980.
 [4] Bill Jackson. Longest cycles in 3connected cubic graphs. Journal of Combinatorial Theory, Series B, 41(1):17–26, 1986.
 [5] Rainer Lang and Hansjoachim Walther. Über längste kreise in regulären graphen. Beiträge zur Graphentheorie. Teubner Verlagsgesellschaft, pages 91–98, 1968.
 [6] Qinghai Liu, Xingxing Yu, and Zhao Zhang. Circumference of 3connected cubic graphs. Journal of Combinatorial Theory, Series B, 128:134–159, 2018.
Comments
There are no comments yet.