1 Introduction
A cycle is a graph each vertex of which has an even degree. A circuit is a regular connected graph. A cycle cover of a graph is a multiset of cycles from such that each edge of is contained in at least one of the cycles. As every cycle is decomposable into circuits, a circuit cover can be obtained in a straightforward manner from a cycle cover by decomposing the cycles into circuits. The length of a cycle is the number of its edges, the length of a cycle cover is the sum of the lengths of all cycles in the cover. If has a bridge (a bridge is an edge whose removal increases the number of components of the graph), then the bridge does not belong to any cycle of . Thus only bridgeless graphs are of interest regarding cycle covers. In this paper we are interested in finding cycle covers with small length.
Conjecture 1.1 (Short Cycle Cover Conjecture).
Every bridgeless graph with edges has a cycle cover of length at most .
A shortest cycle cover of the Petersen graph has length and consists of four cycles. The upper bound given by Conjecture 1.1 is, therefore, tight and the constant cannot be improved. The Short Cycle Cover Conjecture is surprisingly strong as, among others, it implies [9] the famous Cycle Double Cover Conjecture by Szekeres [14] and Seymour [13].
The best general result on short cycle covers is due to Alon and Tarsi [1] and Bermond, Jackson, and Jaeger [2] who independently proved that every bridgeless graph with edges has a cycle cover of length at most . Despite numerous results on short cycle covers for special classes of graphs (see e.g. Chapter 8 of the book [15]) no progress was made for the general case. Significant attention has been devoted to cubic graphs and graphs with minimal degree three.
Let be a bridgeless graph on edges. Among the recent results, Kaiser et. al. [10] proved that has a cycle cover of length at most for cubic and for loopless without vertices of degree two. Fan [6] proved that has a cycle cover of length at most for cubic, for loopless without vertices of degree two, and for without vertices of degree two. Candráková and Lukoťka [3] proved the bound for cubic, for cubic without intersecting circuits, and for cubic without circuits.
In this paper we show how to overcome the problems in the proof from [3] arising from circuits that intersect and obtain the following result.
Theorem 1.
Let be a bridgeless cubic graph on edges. Then has a cycle cover consisting of three cycles of length at most .
This result makes the biggest improvement of the bound on the length of shortest cycle cover of cubic graphs made by a single paper up to date (see e.g. [3] for previous best bounds) and the proof is simpler than in [3]. Besides the idea concerning intersecting circuits, the structure of covers follows that of [10] while we use several optimisations presented in [3]. The ideas of Fan [6] make many steps in our proof easier, which was crucial for us to examine the intersections of circuits in the necessary depth.
As a byproduct we obtain a stronger result for cubic graphs that are cyclically edgeconnected.
Theorem 2.
Let be a cyclically edgeconnected cubic graph on edges. Then has a cycle cover consisting of three cycles of length at most .
The paper is organised as follows. In Section 2 we show how one can modify flows. We prove all flow manipulating lemmas used in the rest of the paper. Given a fixed factor, to perform flow manipulating operations we have to process some of the circuits of the factor in a specific order, while some others have to be processed in pairs. In Section 3 we specify the required order and pairs. In Section 4 we use the lemmas from Section 2 to construct a flow that has special properties with respect to the factor and how the pairs and order of circuits were chosen. In Section 5 we define two cycle covers based on the factor, on the way how the pairs and the order of circuits were chosen, and on the flow. We bound the length of the cycle covers by using these properties. We combine the two bounds to obtain a new bound that depends only on the number of edges in the graph and the number of circuits in the factor that do not intersect other circuits in the graph. Finally, in Section 6 we show that there exists a factor with few circuits that do not intersect other circuits in the graph, which completes the proofs of Theorems 1 and 2.
2 Chain extensions and modifications
Let be a graph and be a vertex of . Let be the set of edges with exactly one end in . For a set , let be the set of edges with exactly one end in . For a subgraph of let be the multiset of edges from that contains each edge as many times as is the number of vertices from incident with . Moreover, if is clear from context, we omit from the notation. To work with multisets that arise from the definition of we use the following operations. In the operation (), the first operand may be a multiset, while the second operand is always an ordinary set. By () we denote the multiset that contains elements from that are in (that are not in ) the same number of times as they are in .
All flows in this article are in . We use colour symbols R, G, B for nonzero elements of and for . As has only involutions we use the simplified definition of the notions chain and flow, omitting orientations. Let be a graph. A chain on is a function . A vertex is a zerosum vertex in if . A flow is a chain whose all vertices are zerosum in . For a chain , let the zeroes of , denoted by , be the set of all edges such that . A nowherezero flow is a flow such that . Let be a connected subgraph of and let be a chain on . The subgraph has a nowherezero boundary in if . The chain is extensible if each is zerosum in . We call a flow on an extension of the extensible chain if for each .
The Parity Lemma restricts the flow values on the edges from in an extensible chain.
Lemma 3.
Let be a graph, let be a connected subgraph of and let be a extensible chain on . Then for each
The following lemma allows us extend a extensible chain on into a flow.
Lemma 4.
Let be a graph, let be a subgraph of and let be an extensible chain on . Then there exists a flow on that is an extension of .
Proof.
Proof by induction on . If , then is an isolated vertex, we denote it . As has only involutions,
Thus is the flow required by this lemma.
If and is not a tree, then let be an edge of a circuit of . The induction hypothesis for the subgraph guarantees the existence of the required flow.
If and is a tree, then let be a leaf of and let be the edge of incident with . We set to be the flow value of and we remove from . The assumptions of the induction hypothesis are satisfied and the conclusion guarantees the existence of the required flow. ∎
Let be a graph, let be a connected subgraph of , and let be an extensible chain. An extensible chain is an modification of if . A simple way to make a modification is by using switches of Kempe chains. The following two lemmas help us to perform such modifications and track the changes of the flow values on .
Lemma 5.
Let be a graph, let be a connected subgraph of and let be a extensible chain on . Let and be two distinct edges from and let and be two distinct nonzero elements of such that . Assume that and have a common endvertex from . Then there exists an extensible chain that is an modification of such that for and for .
Proof.
Let for and for . As both and are in , both and are nonzero, thus is the modification of . For each vertex , is a zerosum vertex in , even for the common endvertex of and as we add the same value to both and . Thus is an exensible chain. It follows that is an modification of with all properties required by this lemma. ∎
Lemma 6.
Let be a graph, let be a connected subgraph of and let be an extensible chain on . Let be an edge from and let and be two distinct nonzero elements of such that . Then there exists an edge with such that the multiset is a submultiset of the multiset , and an extensible chain that is an modification of such that for and for .
Proof.
If has both endvertices on , then we set and define and for edges other than . This chain fulfills the conditions of the lemma.
Thus assume that has one endvertex in and the other endvertex outside . Let be the subgraph of induced by all the edges with . By Lemma 3, the component of that contains has even. Therefore there is an edge other than with . Let we the endvertex of in . The second endvertex must be in . Let we a path between and in .
We define as follows: , for , and , for . As the edges have , also and thus is nonzero on these edges. Moreover, each is zerosum in as compared to the same value from is added to exactly two edges from . Thus we conclude that is an modification of required by this lemma. ∎
Let be a connected subgraph of a graph an let be an extensible chain on . To facilitate the discussion on chain extensions we order the edges from the multiset (edges occurring twice in the multiset will occur twice in the ordering). We call such ordering an boundaryordering. If vertices of have degree three in and degree two or three in , then we can give an bounderyordering by a sequence of vertices of degree two in —if we have such a sequence of vertices , then we obtain the boundaryordering by setting, for each , to be the unique edge outside incident with . Given an boundaryordering and an extensible chain , we define to be the string of elements from such that, for each , . If is a circuit then a natural boundaryordering is an boundaryordering defined by a vertexsequence where the vertices of are in a consecutive order in the circuit.
The following lemma allows us to extend a chain to some circuits without introducing a new zero value.
Lemma 7.
Let be a graph, let be a circuit of that contains only vertices of degree three in , let be a extensible chain on such that has a nowherezero boundary in , let be a natural boundaryordering, and let . If , for some with indices taken modulo , then there exist a flow on that is a extension of such that .
Proof.
Let us denote the vertices of so that may be defined by the vertexsequence . Due to the symmetry of we may without loss of generality assume . As has automorphisms between nonzero elements, we can without loss of generality assume . Then by Lemma 3 . As even with element R fixed there is an automorphism between G and B in , we may, again, without loss of generality assume and . We define the flow as follows. Let if , , , , and . ∎
While Lemma 7 allows us to extend a chain into a flow on a circuit provided that the edges from have right chain values, the following lemma allows us to modify a chain into a flow regardless of the chain values on . On the other hand, we require the existence of a special vertex next to the circuit.
Lemma 8.
Let be a bridgeless graph and let be a circuit of that contains only vertices of degree three in . Assume that there is a vertex that is neighbouring with at least two vertices of . Let be an extensible chain on such that has a nowherezero boundary in . There exists a flow on that is a modification of such that .
Proof.
Let . Let be the boundaryordering given by the vertex sequence , , , , . We may without loss of generality assume that is a neighbour of and either or is a neighbour of . Let . By Lemma 3, each element from
is used odd number of times in
. Due to automorphisms of we may without loss of generality assume that R is used three times in and G is before B in . Moreover, we may assume that , otherwise Lemma 7 gives the desired modification of . We conclude the proof by showing that in each case it is possible to make a modification of so that we can use Lemma 7 to get the modification of required by this lemma.Assume first that and are neighbors of . Due to the symmetries of , considering only automorphisms stabilizing the set , it suffices to assume .
If , then by Lemma 5, with and as the two edges and R and G as the two elements of , there exist a modification of such that . The chain satisfies the conditions of Lemma 7 as required.
If , then by Lemma 6, with as the edge and R and B as the two elements of there exist a modification of such that . If , then is the sought modification of , satisfying the conditions of Lemma 7. If , then we proceed according to the case when .
If , then by Lemma 5, with and as the two edges and R and G as the two elements of , there exist a modification of such that which is the sought modification of , satisfying the conditions of Lemma 7.
On the other hand, assume and are neighbors of . Due to symmetries of , considering only automorphisms stabilizing the set , it suffices to assume .
If , then by Lemma 5, with and as the two edges and R and G as the two elements of , there exist a modification of such that , which is the sought modification of , satisfying the conditions of Lemma 7.
The argumentation where we describe how the desired modification is constructed using Lemmas 5 and 6 as written in the proof of Lemma 8 is very repetitive. To avoid repetition, we introduce a way how to certify that a modification of a chain satisfying some property can always be made.
Let be a graph, be a subgraph of , be an boundaryordering and let be an extensible chain. If Lemma 5 is used with and as the two edges, with and as the two elements of , and Lemma 5 produces a chain , then we write
with th and th element of underlined.
If Lemma 6 is used
with as the edge, with and as the two elements of ,
and Lemma 6 produces a chain ,
then we do not know exactly what is, but
Lemma 6 gives us a list of possibilities
, , such that for some ,
.
We record this as follows
, 
with th element of underlined. Note that we know which lemma was used according to the number of underlined edge colours. We finish each line by indicating the reason why the resulting chain modification has the desired property. If this last entry in a line is a sequence in brackets, then it indicates that we complete the modification as in case (of course, we have to avoid cyclic references). It is possible that differs from the value before : an automorphism of or/and an automorphism of might be applied, provided that the autoromphisms preserve the desired property and preserve when Lemma 5 can be applied (e.g. in the proof of Lemma 8 we could use only automorphisms stabilising the set of vertices incident with the two edges from with the common neighbour).
As an example, we present how the modifications of from the proof of Lemma 8 in case and have a common neighbour outside are recorded.
RGRRB  GRRRB  Lemma 7  

RGRBR  BGRRR  Lemma 7  
RGBRR  Lemma 7  
RGRRB  [RGRRB]  
RRGRB  GGGRB  Lemma 7 
In case and have a common neighbour outside the modifications from Lemma 8 are recorded as follows.
RGRRB  GGGRB  Lemma 7  

GRBRR  RGBRR  Lemma 7  
GGBGR  [RRGRB]  
GGBRG  Lemma 7  
RRGRB  GRRRB  Lemma 7 
We prove two lemmas of similar flavor, one concerning circuits and the second one concerning a special subgraph containing three circuits.
Lemma 9.
Let be bridgeless graph, let be a circuit of that contains only vertices of degree three in and let be a extensible chain on such that has nowherezero boundary in . Let be a natural boundary ordering. There exists a extensible chain on that is a modification of such that satisfies at least one of the conditions from Figure 1.
No.  Condition 

1  , , 
2  , , 
3  , , 
4  , , 
5  , , 
6  , , 
Proof.
We may without loss of generality assume that . By Lemma 3, . Note that the set of conditions in Figure 1 is preserved by automorphisms of (the lines separating the conditions on Figure 1 separate conditions in different orbits under action of the automorphism group of ). Thus among the symmetric cases arising from different consecutive vertex orders we consider only the cases where is alphabetically maximal. In what follows we show that in all these cases we can satisfy one of the conditions in Figure 1. The values of are sorted by and then by the reverse alphabetical order. Note that and if then no colour can occour at two consecutive positions .
RRRRRR  Condition 1  

RRRRGG  Condition 1  
RRRGRG  BRRGBG  [RRGBGB]  
RBRGBG  [RGRBGB]  
RRBGBG  [RRGBGB]  
RRGRRG  Condition 3  
RRGGBB  Condition 1  
RRGBGB  RRBGGB  Condition 5  
RRBBBB  Condition 1  
RRBBGG  Condition 1  
RRGBBG  Condition 5  
RGRBGB  BGBBGB  Condition 4  
BGRRGB  Condition 4  
BGRBGR  Condition 6  
RGBRGB  Condition 6 
∎
If is a subgraph of isomorphic to graph from Figure 2, then a natural boundaryordering is an ordering of given by the vertexsequence , , , , , , where vertices of are the isomorphic images the vertices , , , , , from , respectively.
Lemma 10.
Let be bridgeless graph, let be a subgraph of isomorphic to graph from Figure 2 that contains only vertices of degree three in and let be an extensible chain on such that has a nowherezero boundary in . Let be a natural boundaryordering. There exists an extensible chain on that is as an modification of such that satisfies at least one of the conditions from Figure 2.
No.  Condition 

1  
2  
3  
4  
5  
6  
7  
8 
Proof.
We may without loss of generality assume that . By Lemma 3, . Note that the set of conditions in Figure 2 is preserved by automorphisms of (besides identity, the only other automorphism takes to and to ; the lines separating the conditions on Figure 1 separate conditions in different orbits under action of the automorphism group of ). Among the symmetric cases arising from different vertex orders we consider only the cases where is alphabetically maximal. In what follows we show that in all these cases we can satisfy one of the conditions from Figure 2. The values of are sorted by and then by the reverse alphabetical order. Note that if , then we can require that is alphabetically behind . Also note that the case when is symmetric to the case when and the case when is symmetric to the case when .
RRRRRR  Condition 1  
RRRRGG  Condition 1  
RRRGRG  Condition 3  
RRGRRG  Condition 6  
RRGRGR  Condition 5  
RRGGRR  Condition 1  
RGRRRG  Condition 5  
RGRRGR  GGRRRR  Condition 1  
RRRRRR  Condition 1  
RGGRRR  Condition 2  
RGRGRR  Condition 6  
RGRRRG  Condition 5  
RGGRRR  Condition 1  
GRGRRR  RGGRRR  Condition 4  
GGRRRR  Condition 1  
GGGGRR  Condition 1  
GGGRGR  Condition 3  
GGGRRG  Condition 2  
RRGGBB  Condition 1  
RRGBGB  Condition 3  
RRGBBG  RRBGBG  Condition 3  
RRBBGG  Condition 1  
RRBBBB  Condition 1  
RGRGBB  GRRGBB  [RRGBBG]  
GGRRBB  Condition 1  
GGGGBB  Condition 1  
RGRBGB  GRRBGB  Condition 4  
GGRBRB  Condition 3  
GGGRGR  Condition 3  
RGBRGB  GGBRRB  [RRGBBG]  
RRBRRB  Condition 6  
RGBGRB  Condition 6  
RGBRBG  Condition 5  
RGBGBR  Condition 8  
RGBBGR  GRBBGR  Condition 7  
RRBBRR  Condition 1  
RRBBGG  Condition 1 
∎
We conclude this section by the following reformulation of a results of Fan [6].
Lemma 11.
Let be a bridgeless graph and let be a circuit of that contains only vertices of degree three in . Let be an extensible chain on .

There exists a flow on that is a extension of such that and

If , then there exists a flow on that is a modification of such that .
Proof.
By Lemma 4, can be extended into a flow on . By pigeonhole principle there is an element such that . We set for and otherwise. The resulting flow satisfies the first part of the lemma.
Lemma 3.1 and Theorem 3.3 of [6] guarantee the existence of a flow which is a modification of (which is also a modification of ) satisfying the second statement of this lemma^{1}^{1}1Lemma 3.1 and Theorem 3.3 in [6] are stated for flows. However, well known results of Tutte (see e.g. Corollary 6.3.2 and Theorem 6.3.3. of [4]) guarantee that we can convert a flow into a flow and vice versa without changing which edges have zero values assigned.. ∎
3 Intersecting circuits
Let be a bridgeless cubic graph. Let be the set of circuits of that intersect some other circuit of in exactly one edge or in exactly two adjacent edges ( stands for “intersecting”). Let be the set of all circuits on that do not intersect other circuit from ( stands for “nonintersecting”). In case is clear from the context, we sometimes omit in and and in several derived notations used further in the paper.
Although, not all circuits are in or , the circuits in the factors of our interests are either from or from . By we denote the graph created by contracting in , that is by contracting all the edges from in . A graph is oddedgeconnected if for each .
Lemma 12.
Let be a bridgeless cubic graph and let be a factor of such that is oddedgeconnected. All circuits in are either from or .
Proof.
Let be a circuit of that is not in . Thus intersects a circuit of . If , then (or if has two chords). This contradicts the oddedgeconnectivity of . If are three edges consecutive on , then there is exactly one vertex in . This vertex has two neighbours in , which contradicts with being a factor. If are three nonconsecutive edges, then , a contradiction with being bridgeless. If are two nonconsecutive edges, then the unique vertex in has two neighbours in , which contradicts with that is in a factor. In all other cases is in as required by the lemma statement. ∎
Let be a factor of . Let us denote the set of circuits from that are in and the set of circuits from that are in . Let . Consider the set of all circuits of other than such that there exists a circuit in that intersects both and . Due to the definition of , .
In Section 4, we will create a flow on by contracting circuits of and then we undo these contractions one by one. When we restore a circuit from , we would like to use Lemma 8. The condition of Lemma 8 requiring two boundary edges towards sharing a vertex is satisfied when some circuit in was not restored before restoring . Thus it is critical to define order in which circuits from are restored. Unfortunately, it is not possible to define an ordering that would allow us to use Lemma 8 in every case. However, for the remaining cases it turns out that we can form disjouint pairs of such circuits so that Lemma 10 is applicable.
We fix a complete strict order on . For we define as follows. If , then we set to be an arbitrary circuit from . Otherwise we set to be the smallest element of in . Let be the set of circuits such that and ( stands for “paired”). As on the function is an involution without fixed point thus it naturally partitions the circuits of into pairs. Let . Let ( stands for “unpaired”). We call circuits that are in the same set from paired.
Newt we show that paired circuits are in some graph isomorphic to the graph from Figure 2, so that Lemma 10 is applicable.
Lemma 13.
Let be a bridgeless cubic graph and let be a factor of such that is oddedgeconnected. Let be an ordering of the circuits in . Then is a partition of the set into two element subsets such that for each two element subset the two circuits in this subset are contained in a subgraph isomorphic to the graph from Figure 2.
Proof.
Due to the definition of , is in thus the sets of contain only elements from . By the definition of
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