1 Introduction
One of the unchallenged milestones in planar graph theory is the result by Tutte [11] that every 4connected planar graph on vertices is Hamiltonian, i.e. has circumference , where the circumference of a graph is the length of a longest cycle of . However, decreasing the connectivity assumption from 4 to 3 reveals infinitely many planar graphs that do not have long cycles: Moon and Moser [9] showed that there are infinitely many 3connected planar (i.e., polyhedral) graphs that have circumference at most for , and this upper bound is best possible up to constant factors, as there is a constant such that every polyhedral graph contains a cycle of length at least .
One of the biggest remaining open problems in this area ever since is to characterize the connectivity properties between connectivity 3 and 4 that imply long cycles. Essential connectivity is such a property and will be a focus of this paper.
Indeed, essentially connected graphs have been thoroughly investigated throughout literature for this purpose. An upper bound for the circumference of essentially 4connected planar graphs was given by an infinite family of such graphs on vertices in which every graph satisfies [2]; the graphs in this family are in addition maximal planar. Regarding lower bounds, Jackson and Wormald [8] proved that for every essentially 4connected planar graph on vertices. Fabrici, Harant and Jendroľ [2] improved this lower bound to ; this result in turn was recently strengthened to [4], and then further to [5]. For the restricted case of maximal planar essentially 4connected graphs, the matching lower bound was proven in [3]; however, the method used there is specific to maximal planar graphs. For the general polyhedral case, it is still an open conjecture that every essentially connected planar graph on vertices satisfies and thus the upper bound; while this conjecture has been an active research topic at workshops for over a decade^{1}^{1}1personal communication with Jochen Harant, it was only recently explicitly stated [3, Conjecture 2].
Here, we show that for every essentially 4connected planar graph; this matches the upper bound up to the summand . Moreover, this is only an implication of a much more general result about polyhedral graphs, which we present here. While one part of the proof scheme follows the established approach of using Tutte cycles in combination with the discharging method, we contribute an intricate intersection argument on the weight distribution between groups of neighboring faces, which we call tunnels. This methods differs substantially from the result in [3] (which exploits the inherent structure of maximal planar graphs) and is, unlike the previous results, able to harness the dynamics of extending cycles in polyhedral graphs. In particular, we will discharge weights along an unbounded number of faces, which was an obstacle that was not needed to solve for the weaker bounds.
In fact, we will prove this tool in the following variant that provides also a method to prove many different cycle lengths. Let a cycle of a graph be extendable if contains a larger isolating cycle such that and , where is the number of faces of size five in .
Lemma 1 (Isolation Lemma).
Every isolating cycle of length in a polyhedral graph on vertices is extendable.
The assumption is redundant if and only if . The Isolation Lemma may be seen as a polyhedral relative of Woodall’s Hopping Lemma [13] that allows cycle extensions through common neighbors of cycle vertex pairs even when none of these pairs have distance two in . Despite this correlation, the Isolation Lemma makes inherently use of planarity; in fact, it fails hard for nonplanar graphs, as the graphs for any and any (isolating) cycle of length in these show. We state some immediate corollaries.
Corollary 2.
If a polyhedral graph contains an isolating cycle of length , contains cycles of at least different lengths in .
Corollary 3.
If a bipartite polyhedral graph contains an isolating cycle of length , contains a cycle of length for every even .
The conclusion of Corollary 3 holds even when the polyhedral graph does not have faces of size 5. In view of the sheer number of results in Hamiltonicity studies that use subgraphs involving faces of size five (see e.g. the Tutte Fragment or the one of Faulkner and Younger), this provides evidence that these faces are indeed key to a small circumference. Another corollary is that polyhedral graphs on vertices, in which all cycles have length less than , do not contain any isolating cycle; for example, this holds for the sufficiently large MoonMoser graphs [9] and the 18 graph classes of [7, Theorem 1] that have shortness exponent less than 1.
Finally, the Isolation Lemma implies also the following theorem.
Theorem 4.
Every essentially 4connected planar graph on vertices contains an isolating Tutte cycle of length at least and such a cycle can be computed in time .
Proof.
We only consider existence and will give an algorithm at the end of this paper. It is wellknown that every 3connected plane graph on at most 10 vertices is Hamiltonian [1]. Since these graphs contain in particular the essentially 4connected planar ones, this implies the theorem if ; we therefore assume . For , it was shown in [2, Lemma 4(i)+(ii)] that contains an isolating Tutte cycle of length at least . If , is already long enough, since ; otherwise, applying iteratively the Isolation Lemma to gives the claim and preserves a Tutte cycle, as no vertices of are deleted. ∎
2 Preliminaries
We use standard graphtheoretic terminology and consider only graphs that are finite, simple and undirected. For a vertex of a graph , denote by the degree of in . We omit subscripts if the graph is clear from the context. Two edges and are adjacent if they share at least one end vertex. The distance of two edges in a connected graph is the length of a shortest path that contains both. We denote a path of that visits the vertices in the given order by .
A separator of a graph is a subset of such that is disconnected; we call a separator if . Let a cycle of a graph be isolating if every component of is a single vertex (see Figure 0(a) for a example). We do not require that these single vertices have degree three (this differs e.g. from [2, 4, 5]). A chord of a cycle is an edge for which and are in . According to Whitney [12], every 3connected planar graph has a unique embedding into the plane (up to flipping and the choice of the outer face). Hence, we assume in the following that such graphs are equipped with a fixed planar embedding, i.e., are plane. Let be the set of faces of a plane graph .
3 Proof of the Isolation Lemma
Let be a 3connected plane graph on vertices, and let be an isolating cycle of of length . We assume to the contrary that is not extendable.
Let be the subset of that is contained in the maximal pathconnected open set (i.e. region) of that is bounded (hence, strictly inside ), and . Without loss of generality, we assume . Since and , we have . Let be the plane graph obtained from by deleting either all chords of if or otherwise all chords of whose interior point set is contained in the bounded region of (see Figure 0(a)). Let and .
For a face of or of , the edges of that are incident with are called edges of and their number is denoted by . A vertex of is a vertex that is incident to a edge of ; a vertex of is extremal if it is incident to at most one edge of and nonextremal otherwise. A face is called face if . If
has an odd number of
vertices or edges, we call their unique vertex or edge in the middle the middle vertex or edge of .Let two faces and of be opposite if and have a common edge. If has an edge , let the opposite face of be the face of that is different from and incident to . Let a face of be thin if and is in ; otherwise, let be thick. A face of is called minor if it is either thick and incident to exactly one vertex of or thin and incident to exactly one edge that is not in ; otherwise, is called major. Let and be the sets of minor faces of and , respectively. For a minor thick face of , let be the unique vertex of incident to . A 2sandwiched vertex is the middle vertex of a thick minor 2face of .
Construct a graph (see Figure 0(b)) obtained from with vertex set and the following edgeset. First, for every face , add the edge to . Second, for every major face of (in arbitrary order), fix any vertex that is incident to and add the edge to for every vertex that is incident to . We first prove that is a tree.
Lemma 5.
is a tree with inner vertex set , leaf set and no vertex of degree two.
Proof.
Consider two faces and of that are incident to a common edge . Since does not contain any chord of , is not a chord of , so that and are incident to a common vertex . By construction of , all inner vertices of that are incident to or (in particular, ) are connected in . Hence, is connected. As is isolating, every two faces of are incident to at most one common vertex of . Hence, the union of the acyclic graphs that are constructed for every major face of , and thus itself, is acyclic. We conclude that is a tree with inner vertex set and leaf set .
∎
Note that may contain vertices of unbounded degree even when every vertex of has degree three in (for example, in Figure 0(b)). If , Lemma 5 holds by symmetry also for the tree that is constructed from in the same way as from .
Lemma 6.
We have . If , .
Proof.
Since , has at least four vertices by Lemma 5. It is wellknown that every tree on at least two vertices has exactly leaves, where is the number of vertices of degree at least 3 in . Since has no vertex of degree two, this gives the first claim by Lemma 5. The second claim follows from by the same argument applied to . ∎
In both equalities of Lemma 6, the last summand is nonnegative but may be zero, as a longest cycle of the graph obtained from the octahedron by inserting a vertex of degree three in every face shows (see also the graph obtained from Figure 0(b) by deleting ).
Consider a minor 1face of with edge ; since there are no thin minor 1faces, is thick. Then the cycle obtained from by replacing with the path shows that is extendable, which contradicts our assumption. We conclude that has no minor 1face. Since is isolating and has no chords in , has no minor 0face. To summarize our assumptions so far, we know that is not extendable, , and every minor face of satisfies .
For the final contradiction, we aim to prove
(1)  
(2) 
Assume . By Lemma 6, and . Since , Inequality 1 implies and thus the claim . In the special case , we have , so that Inequality 2 implies . Since every minor face of has a nonextremal vertex (since has no minor 1faces), minimum degree 3 in implies that we have at least one chord of in , so that . This gives (hence, we are only off by in the case ).
In order to prove Inequalities (1) and (2), we will charge every face of with weight ; hence, the total charge has weight . Then we discharge (i.e., move) these weights to minor faces such that no face has negative weight. We will prove that after the discharging every minor face of has sufficiently high weight to satisfy the inequalities.
3.1 Arches and Tunnels
For a face of , a path of is an arch of if is minor and is either the maximal path in all of whose edges are incident to (in this case we say that is proper; then has length one or two depending on whether is thin or thick) or a chord of whose inner point set is contained in and that does not join the two extremal vertices of (see Figure 0(a)). Hence, an arch is proper if and only if , so that every minor face has exactly one proper arch. The face of an arch is the minor face of that contains the inner point set of .
Let the archway of an arch be the path in between the extremal vertices of whose edges are incident to in . Since and its archway close a face in the graph , we define , thickness, and the vertices and edges of just as the corresponding terms for the face ; in particular, denotes the number of edges of the archway of , and is thick if is thick. Then every arch has exactly two extremal edges, as has no minor 1face and, by the last condition of the definition of arches, two arches of have never the same pair of extremal edges (this prevents that two arches of the same face “overlap”). We call an arch a arch if . If and are arches or faces of , let be the number of edges that and have in common; then and are opposite if . An arch of an arch is an arch of such that every edge of is a edge of ; we also say that has arch .
Consider the 3arches in Figure 3.1 and assume that every is thick and proper, so that every is a minor 3face. Since every receives only initial weight 3 and is unbounded, every local method of transferring weights to reach weight 4 per minor face is bound to fail. Unfortunately, 3faces are not the only example where nonlocal methods are needed: in fact, there is an unbounded number of faces in which weights must be transferred nonlocally. We will therefore design the upcoming discharging rule in such a way that weight transfers are not dependent on minor face but instead on arches; this will reduce all structures that have to be handled nonlocally to one common structure (called tunnel), which is the one in Figure 3.1.
Let two 3arches and be consecutive if . The reflexive and transitive closure of this symmetric relation partitions the set of 3arches; we call the sets of this partition tunnels (see Figure 3.1). Since is plane, imposes a notion of clockwise and counterclockwise on ; in the following, both directions always refer to . The counterclockwise track of a tunnel (which will transfer weights counterclockwise around ) is the sequence of all 3arches of such that is the clockwise consecutive successor of for every . We call a tunnel track and its tunnel cyclic if and and are consecutive, and acyclic otherwise.
The exit pair of a counterclockwise track consists of the counterclockwise extremal edge of and the opposite face of . Clockwise tracks and exit pairs are defined analogously. The exit pairs and of a tunnel are then the two exit pairs of the counterclockwise and clockwise tracks of ; we call and exit faces of ). Clearly, we have if and only if is cyclic, and if so, and are opposite faces, so that . Hence, the exit pairs of every acyclic tunnel are different, while the exit faces may be identical.
In order to describe the weight transfers through tunnels, we define the following reflexive and symmetric relation for faces and of and extremal edges and of (not necessarily different) 3arches of an acyclic tunnel such that and are incident to and , respectively. Let be ontrack with if the following statements are equivalent (see Figure 3.1).

and are contained in the same region of

the distance between and in the union of the edges of (measured by the length of a path that does not exceed ) is a multiple of .
Clearly, this relation is an equivalence relation. Moreover, if is an extremal edge of a 3arch of a tunnel , is ontrack with exactly one exit pair of . Tunnels will serve as objects through which we can pull weight over long distances. We will later prove that tunnels transfer weights only oneway, i.e. use (the ontrack pairs of) at most one of their tracks. Based on the structure of , weight may not be transferred through the whole tunnel track; the following definition restricts the parts where weight transfers may occur.
Let be an acyclic tunnel track, an extremal edge of an arch of such that is ontrack with the exit pair of , the extremal edge of different from , and the opposite face of (see Figure 3.1; informally, is the ontrack pair in that precedes ). Recursively, we define that is a transfer pair of if is either the exit pair of or a transfer pair, and

the opposite face of is minor and ,

is either an extremal edge of or adjacent to that edge, and in the latter case the middle edge of is incident to or a major face, and

the middle edge of is not incident to (in particular, ).
For a tunnel track , an arch of that has an extremal edge such that is a transfer pair of is called transfer arch of . Note that a transfer arch of is not necessarily a transfer arch of the other tunnel track of .
3.2 Discharging Rule
By saying that a face pulls weight over its edge for a positive weight , we mean that is added to and subtracted from the opposite face of ; we sometimes omit if the precise value is not important, but positive.
Definition 7 (Discharging Rule).
For every minor face of and every edge of (both in arbitrary order), pulls weight 1 over from the opposite face of for every of the following conditions that is satisfied (see Figure 2).

is major

is minor, and is a thick 2face

is minor and , is the middle edge of a 3arch of and not an extremal edge of a 3arch of , either is a 3face or an extremal edge of is an extremal edge of and the other extremal edge of is incident to an extremal vertex of and not incident to a major face such that, if is incident to , has a 4arch that has

is minor, is a nonextremal edge of a 4arch of such that the extremal edge of that is adjacent to is an extremal edge of and the other extremal edge of is incident to a thick minor 2face , is not the middle edge of a 3arch of , and

is minor, is a nonextremal edge of a 4arch of such that the extremal edge of that is adjacent to is an extremal edge of and the other extremal edge of is an extremal edge of a 3arch of a face , is a transfer pair, no 2arch has extremal vertex , there is no 3arch of , is not an extremal edge of a 3arch of , and

is minor, is a nonextremal edge of a thick minor 4face such that the extremal edge of that is not adjacent to is the middle edge of a 3arch of a face , and is not the middle edge of a 3arch of .

is minor, is a transfer pair of an acyclic tunnel track , and the exit pair of satisfies (in the notation and ) at least one of the conditions –
Note that the weight transfers of this rule are solely dependent on and (and not on the current weight transfers). This holds in particular for the ones caused by , as these do not depend on (but instead of – on the exit pair). Since tunnels partition the set of 3arches, it suffices to evaluate once for the transfer pairs of each tunnel track.
After the discharging rule has been applied, effectively routes weight 1 through an extremal part of a tunnel track towards its exit face if this exit face pulls weight from by any other condition. By definition of –, the only faces that do not have an arch of a tunnel (i.e. reside “outside” ) and pull over a edge of such an arch are the exit faces of ; in this sense, weight may leave only through an exit face of .
3.3 Structure of Tunnels and Transfers
We give further insights into the structure of tunnels and the location of edges over which our discharging rule pulls (positive) weight.
Lemma 8.
Every tunnel track with satisfies . In particular, every tunnel is acyclic.
We remark that it is possible, but slightly more involved, to prove Lemma 8 solely by using the discharging rule of Definition 7. Using Lemma 8, we assume from now on that every tunnel is acyclic. We now show that does not contain the dotted edges of Figure 2 for the respective conditions; this sheds first light on the implications that are triggered by the assumption that is not extendable.
Lemma 9.
Proof.
We use the notation of Figure 2. Assume . If (or, by symmetry, ) in Figure 1(a) is a edge of a 2arch of , is an extremal vertex of , since is polyhedral. Then is extendable by the path replacement , which adds one or two new vertices to (depending on whether is thick). If (or, by symmetry, ) is the middle edge of a 3arch, we have , as , since is polyhedral; hence, neither nor is the middle edge of a 3arch. Using the same argument, has no 4arch with extremal edges and .
Assume . Then is not the middle edge of a 3arch of , as is polyhedral. By definition of C3, is not an extremal edge of a 3arch of .
Assume . Then the first argument for implies . In addition, , as otherwise is extendable by the path replacement . Hence, is not an extremal edge of a 3arch of . If is an extremal edge of a 3arch of , is extendable by the path replacement , as this adds at most three new vertices to . Note that we have in this replacement if is proper ( is thick because of ), as has no minor 1face (here, with edge ). In addition, is not the middle edge of a 3arch of , as otherwise would be a 2separator of by the previous claims.
Assume . By definition of , it only remains to prove that is not the middle edge of a 3arch of . If it is, is a 2separator of , since and are not contained in . This contradicts that is polyhedral.
Assume . Then , as otherwise is extendable by the replacement . Since is polyhedral, has degree at least three in , which implies as only remaining option. Then , as otherwise is extendable by the replacement . Sinec is polyhedral, this implies that there is no 2arch of with extremal vertices and . Assume that has a 2arch with extremal vertices and . Then is thick, as has a 2arch, but nonproper, as otherwise would be a minor 1face of . Hence, , so that is extendable by the replacement . This implies in addition that is not an extremal edge of a 3arch of , as would have degree two in . ∎
For a edge of a face of and a condition , let denote that is satisfied for and in Definition 7. For notational convenience throughout this paper, whenever pulls weight from a face , we denote by the extremal vertex of whose clockwise neighbor in is vertex of , and denote by the th vertex modulo in a clockwise traversal of starting at .
So far, a tunnel might transfer weights through both of its tracks simultaneously. The next lemma shows that this never happens.
Lemma 10.
Let and be the exit pairs of a tunnel such that pulls weight over . Then

is minor, and no other condition is satisfied for ,

every 2arch of an arch of has a edge such that is ontrack with ,

and there is no 2arch of that has edge ,

does not pull any weight over .

for every 4arch that has an arch of , the common extremal edge of and satisfies that is ontrack with ,

every arch of that is consecutive to two transfer arches of satisfies
By Lemma 104, all weight transfers that are caused within a tunnel by Condition C7 go oneway, i.e., use only one track of . As an immediate implication, the following lemma shows that all weight transfers strictly within a tunnel are solely dependent on the weight transfers on its exit pairs.
Lemma 11.
Let be a transfer pair of a tunnel track with exit pair such that the opposite face of is minor. Then pulls weight over if and only if pulls weight over (and if so, and ).
Proof.
Assume that pulls weight over . By Lemma 8, is acyclic. By Lemma 101, . Hence, is satisfied for , so that pulls weight over .
Assume to the contrary that for some and does not pull any weight over . The latter implies . Since is minor, . Since is a edge of a 3arch of with , . By planarity, . By Lemma 9, , which is a contradiction. ∎
An immediate implication of the discharging rule in Definition 7 is that every face pulls a nonnegative integer weight over every edge, as every satisfied condition adds 1 to that weight. We next prove that no two of the conditions C1–C7 are satisfied simultaneously for the same face and edge ; hence, pulls either weight 0 or 1 over . This is crucial for keeping the amount of upcoming arguments on a maintainable level; in fact, our conditions were designed that way.
Lemma 12.
The total weight pulled by a face of over its edge is either 0 or 1. If it is 1, the opposite face does not pull any weight over .
Proof.
Assume to the contrary that is incident to two faces and of such that and for conditions and and ; without loss of generality, we assume that is not stated before in Definition 7. In general, implies . If , is major, which implies and thus ; then is major, which contradicts . Hence, , so that both and are minor. If , Lemmas 104 and 11 imply that and is an extremal edge of two consecutive 3arches; then by Lemma 9 and by planarity, which is a contradiction. Hence , which implies .
We distinguish the remaining options for and in .
Assume ; by our notational convention, is the 2sandwiched vertex of . Then , which implies , as the remaining options for require . Since is thick, exists. By Lemma 9 (for C2), . Assume . If is not incident to any extremal edge of (see Figure 1(c)), contains neither nor by Lemma 9 (for C2), which contradicts . Otherwise, consider Figure 2(a). Then implies , which contradicts Lemma 9 (for C4). If , contradicts the assumption of that case. If , Lemma 9 implies that neither nor is contained in , which contradicts .
Assume . Then is the middle edge of a 3arch of , which does not satisfy and by definition of these conditions. We conclude . Then contradicts Lemma 9, and contradicts that is plane.
Assume . If , , as 2faces do not have 3arches, so let . Assume (see Figure 2(b) for ). By Lemma 9 (applied for and ), neither nor is adjacent to a vertex of in , so that is a 2separator of . If , we get a contradiction to planarity.
Assume . If , by planarity, so let . Assume . By Lemma 9 (applied for and ), neither nor is adjacent to a vertex of in , so that is a 2separator of . If , we get a contradiction to planarity.
Assume . Then and thus , which contradicts that is plane. ∎
By Lemma 12, we know that whenever weight 1 is pulled over some edge by some condition –, no other condition is satisfied on and 1 is the final amount of weight transferred over .
3.4 The Proof
Throughout this section, let denote the weight function on the set of faces of after our discharging rule has been applied. Clearly, still holds. For and the set of edges of a face of , let the (weight) contribution of to be (i.e. the initial weight these edges give to ) plus the sum of weights pulled by over edges in minus the sum of weights pulled by opposite faces of over edges in . The contribution of an arch to is the contribution of the edges of to ; in particular, every proper arch contributes weight to . Note that we have if a set contributes weight to , as may loose weight at most 1 for every of its edges that is not in by Lemma 12, which cancels the initial weight 1 given by this edge.
Lemma 13.
For two edges and of a minor face , let and such that and and are not contained in . Then

(and thus ) is either an extremal edge of or adjacent to that, and

and have distance at least three in .
Proof.
For Claim 1, assume to the contrary that is neither an extremal edge of nor adjacent to that. Then and . Since is minor, . Since the definition of requires an edge that is not incident to , . For , the edge in Figures 1(c)–1(e) is not incident to , which contradicts the choice of .
For Claim 2, assume to the contrary that and have distance at most two in . Since is minor, . Let and let be the 3arch of that has edge . Then the existence of and planarity imply , and Lemma 9 implies . Hence, . Then by planarity and the fact that, in the notation of the conditions , and , we have and no 3arch with edge . This is a contradiction. ∎
Lemma 14.
Let , be the opposite face of , be the arch of shown in Figure 2 (for , let be the proper arch of ), and be the set of common edges of and .

If and , contributes weight at least to .

If , contributes weight at least to .

If , contributes weight at least to .
Lemma 15.
For a minor face , let be an arch of with minimal such that a face pulls weight over a edge of by Condition . Then .
In particular, we may choose in Lemma 15 as the proper arch of . This implies the following helpful corollary.
Corollary 16.
Every minor face that has a edge over which an opposite face of pulls weight by or satisfies .
Lemma 17.
Let be a face of . Then , if is thick and minor, if is a thin minor 2face, and if is thin and minor such that .
Proof.
By Lemma 12, an opposite face of pulls weight at most weight one over any edge of . Since the initial weight of such an edge for is one, we have . In the remaining proof, let be minor. Assume that has a edge such that the opposite face of pulls weight over . Then by Lemma 15. We therefore assume that no opposite face of pulls weight over a edge of by Condition or .
Let . If is thick, Condition and Lemma 12 imply . If is thin, assume to the contrary that . Then has a edge such that for the opposite face of . Since , Lemma 13 implies that the edge of different from contributes weight one to . This contradicts .
Let . Assume to the contrary that has an edge such that for the opposite face of . Then, for every , Lemma 9 contradicts that is a edge of . Hence, . If is thin, this gives the claim, so let be thick. Let be the middle edge of . Then is not an extremal edge of a 3arch, as then is extendable by the replacement consisting of the two 3arches and their two common edges. Hence, , which gives the claim .
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