DeepAI

Decreasing maximum average degree by deleting independent set or d-degenerate subgraph

The maximum average degree mad(G) of a graph G is the maximum average degree over all subgraphs of G. In this paper we prove that for every G and positive integer k such that mad(G) > k there exists S ⊆ V(G) such that mad(G - S) <mad(G) - k and G[S] is (k-1)-degenerate. Moreover, such S can be computed in polynomial time. In particular there exists an independent set I in G such that mad(G-I) <mad(G)-1 and an induced forest F such that mad(G-F) <mad(G) - 2.

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1 Introduction

The maximum average degree of a graph is a heavily studied notion. There are many results in a form of “if  satisfies some inequality then we are able to partition its vertex set into some parts such that some property holds” including [12, 6, 7, 2]. Various researchers approached problems of this form as partitioning of the edges instead of partitioning of the vertices [4, 1, 11], however these results are less relevant to our work. Some of referenced papers talk about planar graphs with girth that is at least some number, however bounding girth from below and bounding maximum average degree from above are tightly linked through inequality that holds in planar graphs.

Up to our knowledge this is the first theorem of a kind where we are given a graph with bounded value of its where we partition its vertex set into some parts so that their values of are smaller, however they need not be bounded by absolute constant. This is opposed to all results where every resulting part induces a forest or is an independent set or has maximum degree etc.

Our results can be applied as a tool for direct deriving many results for some specific sparse graph classes, for example planar graphs with constraints on girth. It seems that our results do not show as much expressive power as it is possible to get on such restrictive graph classes which is a price for deriving them from a more general theorem, where arguments specifically adjusted to the researched restricted graph classes can be used. However our results can be seen as a nice way of unifying these results.

Our results imply a positive answer for the open problem presented in [8], namely problem 2 from final remarks what in turn implies sub-exponential bound on the diameter of reconfiguration graphs of colourings for graphs with any bounded maximum average degree.

Rest of the paper is organized as follows. In Section 2 we introduce a few useful notions, in Section 3 we state our main result and provide its proof, in Section 4 we talk about conclusions.

2 Preliminaries

Theorems proved in this paper will be about simple undirected graphs, however multiple directed graphs will show up throughout the proofs.

Undirected edge between vertices and will be denoted as . If we are dealing with directed edges and we are given and edge from to we denote it as .

If is a graph and is a subset of its vertices then by we denote subgraph of induced on vertices of . Length of the shortest cycle in a graph will be denoted as . If is a forest we assume that . denotes maximum degree of a vertex.

The maximum average degree of a given graph is defined as follows:

We assume that of a graph with an empty vertex set is .

We say that undirected graph is -degenerate if every its subgraph contains a vertex of degree at most . Degeneracy of a graph is the smallest value of such that this graph is -degenerate.

Let us note that -degenerate graphs are exactly the same graphs as graphs with , because both are just edgeless graphs. Moreover -degenerate graphs are exactly the same graphs as graphs with , because both are just forests.

3 Main result

The following theorem is a main result of this paper.

Theorem 1

For every undirected simple graph and a positive integer there exists such that is -degenerate and . Moreover such can be computed in polynomial time.

In order to prove it we are going to investigate flow network that allows to determine value of in polynomial time. Example of such network can be found in [10], however we are going to use one adjusted to our own use which we are going to describe here.

Let us define flow network for given undirected graph and any nonnegative real number . Network will consist of one node for each , one node for each denoted as and two special nodes and , respectively source and sink. There will be three layers of directed edges in :

• The first layer – Edges of capacity one from to each node .

• The second layer – Edges of infinite capacity from each where to and to .

• The third layer – Edges of capacity from each to .

Lemma 2

For any graph and any real number , maximum flow between and in is equal to if and only if .

Proof:

By max-flow min-cut theorem we know that maximum flow in a graph is equal to the minimum cut, so we are going to investigate structure of cuts in this graph. We refer to cuts as sets of edges. Since edges in second layer have infinite capacity they surely do not belong to any minimum cut. If no edge from third layer belongs to the cut then all edges from first layer must belong to it and this is a cut of weight , so if maximum flow is smaller than then there exists a minimum cut with some edges in third layer. Let us fix some minimal cut and let us assume that is a nonempty subset of all vertices of such that edges for all belong to . Let . If then has to belong to . All mentioned edges, that is for and for actually already form a cut. Its weight is . If this value is less then then we know that maximum flow in this graph is less than . However if for any this value is not smaller than then we know that maxflow in this graph is .

We get that maxflow in this graph is smaller than if and only if there exists such that . Maximum value of equals , so we get that maxflow in is equal to is and only if , as desired.

Let us note that by using this lemma and observing that for some and and knowing that we can compute maximum flow in polynomial time we can conclude that can be computed in polynomial time.

Let us fix any graph and denote . Let us define graph for given flow in of capacity by directing some of edges from and discarding the rest. Flow routes one unit of flow through each . Node has two outgoing edges to and to . If sends more than unit of flow to then in we put directed edge , similarly if sends more than unit of flow to we put edge . Otherwise if sends exactly unit to both and we simply discard this edge.

Lemma 3

There exists flow of capacity in such that is acyclic. Moreover, it can be determined in polynomial time.

Proof:

From Lemma 2 we know that there exists at least one flow between and of capacity . Let us take such that number of edges in is as small as possible. Suppose there is a cycle in on vertices respectively. Denote as we are dealing with a cycle. Let be minimum of amounts of flow that sends through some edge for some valid . From definition of we deduce that . Let us define by decreasing flow on edges and increasing it on edges by . Amount of flow leaving and entering each vertex remains unchanged hence still restricts capacities. Flow for at least one vertex sends exactly unit of flow through both edges outgoing from it, so at least one edge on the cycle is no longer present in and edges outside the cycle remain unchanged when compared to . It contradicts the assumption that has the smallest possible number of edges what concludes proof of existence of such .

In order to compute such in polynomial time let us take any of capacity in (let us remind that we can determine value of in polynomial time which is used in construction of ). If contains some cycle we can detect this cycle, determine value of and adjust values of units that flow sends through mentioned edges accordingly. Number of edges in is strictly smaller than in , so we will not do this more than times, what gives us algorithm performing polynomial number of operations. In order to omit dealing with rational numbers we can multiply all capacities in by , where for some coprime integers . That concludes description of polynomial algorithm determining desired .

Let us fix from above lemma. We will present an algorithm in which the routine returns any vertex from directed acyclic graph which has no incoming edges (as the graph is acyclic there always exists at least one such vertex) and the routine given graph , subset of its vertices and integer returns set of all vertices from outside of adjacent to at least vertices from .

Theorem 4

Function will return -degenerate set such that .

Proof:

First we argue the algorithm will return -degenerate set. In each iteration picked by algorithm is adjacent to at most already picked vertices. So is -degenerate indeed.

To show that we just have to find flow in graph of capacity thanks to lemma 2. Observe that is a subgraph of with capacities of edges on the third layer reduced by . Flow has to saturate all edges from the first layer to have capacity . On the second layer we define using , for each edge from the second layer of flow will send exactly the same amount of flow as on corresponding edge in . Now we just have to argue that amount of flow sent by to any node between second and third layer in is bounded by i.e. capacity of edge going from that node to sink. Each such node corresponds to vertex from , so let us take arbitrary vertex . During execution of the algorithm vertex has been removed from as incident to some vertices already picked to . Denote them and let us consider arbitrary . When the algorithm picked from , there were no incoming edges to . In particular in there was no edge . At that time still belonged to , so there was no edge even in . Since and are adjacent in , there was either an edge in which means that flow sends more than unit of flow from to in or there was no and which means that flow sends exactly unit of flow from to in . Through node in flow sends at most units of flow and for every at least unit of flow comes from to . Therefore flow going through is decreased by at least unit per each in what implies that sends at most units of flow to vertex in .

What is more, procedure can be trivially implemented in a polynomial time. Theorem 1 directly follows from Theorem 4.

As a two notable special cases we mention following corollaries:

Corollary 1

For every undirected simple graph there exists such that is an independent set and . Moreover such can be computed in polynomial time.

Corollary 2

For every undirected simple graph there exists such that is a forest and . Moreover such can be computed in polynomial time.

4 Conclusions and open problems

As already mentioned, our result is a positive answer for the open problem presented in [8], namely problem 2 from final remarks which in turn implies sub-exponential bound on the diameter of reconfiguration graphs of colourings for graphs with any bounded maximum average degree. However this bound has already been improved in [9] to the polynomial bound depending on value of .

Our results imply many results for some specific classes of graphs as a direct consequence and here we mention a few of them.

Following folklore fact will come handy in deriving some of them:

Fact 1

For every planar graph we have .

Corollary 3

For every planar graph its vertex set can be partitioned into such that are forests.

Proof:

Every planar graph satisfies , so using Corollary 2 we can partition into and such that and and then using Corollary 2 again we can partition into and such that and . Hence are forests.

Corollary 4

For every planar graph without triangles its vertex set can be partitioned into such that are forests.

Proof:

Based on Fact 1 we know that if has no triangles then . Therefore using Corollary 2 we deduce that there exist such that and are forests.

Corollary 5

For every planar graph without cycles of length and its vertex set can be partitioned into such that is a forest and .

Proof:

Since we deduce that , so based on Corollary 2 we get that there exist and such that and and . Therefore is a forest and , because if contains a vertex with degree then this vertex together with its two neighbours induce a graph with at least .

Corollary 6

For every planar graph with its vertex set can be partitioned into such that is a forest and is an independent set.

Proof:

Since we deduce that , so based on either Corollary 2 or Corollary 1 we get that there exist and such that and and . Therefore is a forest and is an independent set.

However, all of these corollaries have already been proven and even improved before. Corollary 3 has been proven in [5] and later improved in [13]. Improved version of corollary 4 has been proven in [14]. Improved version of both corollaries 5 and 6 has been proven in [3].

Corollary 7

For every planar graph with its vertex set can be partitioned into such that is an independent set and is a forest where every connected component has less than vertices.

Proof:

Since we deduce that , so based on Corollary 1 we get that there exist and such that is an independent set and . It can be readily verified that graphs with are graphs which are forests with connected components of size less than .

As a main open problem in this area we recall the following one:

Conjecture 1

For every graph and positive real numbers if then there exists a partition of vertex set such that and .

Our results show that this conjecture is true for . Moreover since for positive we have that -degenerate graphs fulfill we can deduce that for every integer and a graph that satisfies there exists a partition of vertex set such that and .

Acknowledgements

Majority of this research has been conducted during Structural Graph Theory workshop in Gułtowy, Poland, in June 2019. Thanks to the organizers and to the other workshop participants for creating a productive working atmosphere. We would also like to thank Bartosz Walczak for discussions on this topic.

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