Edge-disjoint induced subgraphs with given minimum degree

Raphael Yuster

2012

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Problems concerning edge-disjoint subgraphs that share some specified property are extensively studied in graph theory.

Many fundamental problems can be formulated in this way:

• Proper edge coloring,

• Proper vertex coloring,

• H-packing, …

When we require the set of edge-disjoint subgraphs to be induced, any two subgraphs can only intersect in an independent set.

Our main goal: determine (asymptotically) the maximum number of edge-disjoint induced subgraphs that share the basic property of having a given minimum degree.

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Let fh(G) denote the maximum size of a set of edge-disjoint induced subgraphs of a graph G, each having minimum degree at least h.

• Trivially, f1(G) equals the number of edges of G.

• It is not difficult to construct examples of graphs with n vertices and m edges for which already f2(G) =O(m2/n2).

• Our main result proves that (for any fixed h) this bound is tight for all graphs that have a polynomial number of edges (otherwise there are some logarithmic factor), while, at the same time, keeping the intersection of any two subgraphs relatively small.

Theorem 1.

Let h be a positive integer and let α be a positive real.There exists c=c(α) and N=N(α,h) such that:

any graph with n ≥ N vertices and m ≥ n1+α edges has:• cm2/(h2n2) edge-disjoint induced subgraphs, with min. degree h.• Any two subgraphs intersect in at most 1+h2n3/(cm2) vertices.

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• We make no attempt to optimize c (it is polynomial in α).• If m = Ω(n3/2) then the intersection is only 1.• The number of subgraphs cm2/(h2n2) is actually optimal also w.r.t.

the dependence on h (so it is tight up to a constant factor).• The intersection size is optimal in the sense that n3/m2 cannot be

improved to (n3/m2)1-ε for any ε.

Theorem 1.

Let h be a positive integer and let α be a positive real.There exists c=c(α) and N=N(α,h) such that:

any graph with n ≥ N vertices and m ≥ n1+α edges has:• cm2/(h2n2) edge-disjoint induced subgraphs, with min. degree h.• Any two subgraphs intersect in at most 1+h2n3/(cm2) vertices.

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Tightness of cardinalityProposition 2.

For positive integers n, m, h, there are graphs with n vertices and m edges that do not contain

edge-disjoint induced subgraphs with minimum degree at least h.

KxAll x(n-x) edges In-x

Choose:x(x-1)/2+x(n-x) ~ m

x ≤ 2m/n

A subgraph with minimum degree h must contain at least h vertices from Kx and hence all its edges.

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Tightness of intersection

Simple Combinatorial lemma:Let F be a family of subsets of {1,…,n}. Assume that |F| ~ n2α and that | X | ≥ n1-α for each X F . Then there are two elements of F that intersect in ~ n1-2α elements.

We want to prove that there are graphs with m ~ n1+α edges such that inany set of n2α ~ m2/n2 edge-disjoint induced subgraphs with min. degreeat least h there are at two subgraphs having intersection n1-2α ~ n3/m2.

• Now use the random graph G(n,p) with p ~ nα-1.• Next, prove that whp, for large enough (but constant) h, every subgraph

with k ≤ n1-α vertices has less than hk/2 edges. • Thus, every subgraph with minimum degree h has to contain at least

n1-α vertices.• Now, use the lemma with F being a maximum cardinality set of induced

subgraphs with minimum degree at least h.

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Balanced graphsA graph is balanced if its average degree is not smaller than the averagedegree of any of its subgraphs.

• As the average degree of a graph with n vertices and m edges is 2m/n,a balanced graph has the property that any subgraph with n' verticeshas at most n'm/n edges.

• Some examples of balanced graphs are complete graphs,complete bipartite graphs, and trees.

Non balanced: 7/5 < 6/4 balanced: 9/6 = 6/4

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Reducing to the case of balanced graphsInstead of proving the main result for all graphs with n ≥ N(α,h) , let’sreduce to proving it for balanced graph with at least N*(α,h) vertices:

• We are given a graph G with n ≥ N(α,h) vertices and m ≥ n1+α edges.

• Let G’ be a subgraph with the least number of vertices n' < n andwith m' edges for which m'/n' ≥ m/n. By minimality, G’ is balanced.

• We claim that n' ≥ N* . (for the choice N=(N*)1/α )

• As m/n ≥ n α and since the average degree of any graph is less than its number of vertices, we must have

n' > 2m'/n' ≥ 2m/n ≥ 2nα ≥ 2Nα ≥ N*

• By the reduced theorem G’ has the required set of subgraphs of sizec(m’)2/(h2(n’)2) ≥ cm2/(h2n2)

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A naïve attempt• It is easy to see that a balanced graph is d-degenerate for d = 2m/n:

Indeed, as long as there is a vertex with degree at most d, delete it and continue. The process must end with the empty graph.

• As a d -degenerate graph is (d+1)-colorable, we have that a balanced graph can be colored with 2m/n + 1 colors.

• So, we have found ~ m2/n2 induced edge-disjoint (in fact bipartite) subgraphs that correspond to pairs of color classes.

• But there are problems:

a) Most of them may be too sparse and not contain subgraphs with the required minimum degree.

b) Color classes may be huge (in fact, the average size of a color class is already ~ n2/m and thus, two subgraphs may have large intersection.

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A coloring lemma for balanced graphsInstead, we require a more “balanced” coloring of a balanced graph where we do have control over the density of edges between a color class and the other vertices.

Lemma 3.

Let δ and α be positive reals. The following holds for all n suff. large:

Let G be balanced graph with n vertices and m ≥ n1+α edges.Then G has a proper coloring with the following properties:

1. The number of colors is at most

2. the size of each color class is at most

3. any color class is incident with at most n edges.

Color classes will be grouped to form subgraphs

No two color classes will be in the same group.

Important for controlling dependencies in a later probabilistic argument

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Proof idea of lemma• We will partition the vertex set into many (but constant number of)

parts according to their degrees:

Smaller parts will contain high degree vertices while larger parts will contain smaller degree vertices.

We make sure that the induced subgraph in each part has relatively small maximum degree.

We properly color each part using an equitable coloring (this is very important).

A proper vertex coloring is equitable if the numbers of vertices in any two color classes differ by at most one.

A fundamental theorem of Hajnal & Szemerédi:

Any graph with maximum degree d has an equitable coloring with d +1 colors.

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Projective planes and graph decompositionRecall: for any prime power p, there exists a projective plane of order p, denoted by PG(p).

• In graph-theoretic terms, if r = p2+p+1, the complete graph Kr can be decomposed into r pairwise edge-disjoint cliques of order p+1.

• The vertices of Kr correspond to the points of PG(p) and the cliques in the partition correspond to the lines of PG(p).

• For example, the Fano plane (the case p = 2) corresponds to a decomposition of K7 into 7 pairwise edge-disjoint triangles.

For an r-partite graph R and 1 ≤ t ≤ r, an induced subgraph of R consisting of precisely t parts is called a full t-partite subgraph of R.

• An r-partite graph has precisely full t-partite subgraphs,• We allow some parts to be empty sets, so under this assumption,

empty parts are considered distinct.

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Mapping independent sets into a projective planeA bijection π from the points of PG(p) to the parts on an r-partite graph R (recall r = p2+p+1) defines a mapping between the lines of PG(p) and their corresponding full (p+1)-partite subgraphs.

• Equivalently, π defines a set Lπ of r full (p+1)-partite subgraphs of R, where any two subgraphs in Lπ are edge-disjoint.

• We call Lπ a projective decomposition of R.• There are r! projective decompositions.

Lemma 4.

Let Sp+1 be the set of all full (p+1)-partite subgraphs of an r-partite graph R with r = p2+p+1. Let Lπ be a projective decomposition of R chosen uniformly at random.Then, each element of Sp+1 has the same probability of being an element of Lπ .In particular, a randomly chosen element of Lπ corresponds to a random element of Sp+1.

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Proof of Theorem 1We outline the case m = O(n3/2) (the denser case is a bit simpler).

• Recall that by Lemma 3, we have colored our balanced graph G with r colors where r ~ .

• We may assume that r = p2+p+1 as otherwise we can add a few empty color classes and still p ~ (Bertrand’s postulate).

• Denote the color classes by C1,…,Cr so by Lemma 3 we also have:

a) e(Ci, V(G)-Ci) n

• By Lemma 4, what remains is to study the properties of a random element of Sp+1 (a randomly chosen full (p+1)-partite subgraph),

namely - the edge density of a random selection of p+1 color classes.

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Proof of Theorem 1 (cont.)

• This is easy because p ~ and because each | So we only need Markov’s inequality.

Lemma 5 (vertices lemma).

The probability that a randomly selected element of Sp+1has at most

vertices is high (say, > 11/12).

Lemma 6 (edges lemma).

The probability that a randomly selected element of Sp+1has at least

edges is relatively high (say, > 3/10).

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Proof of Theorem 1 (cont.)

• The proof of this lemma is somewhat involved and requires careful analysis of the dependencies between pairs of edges whose endpoints share a color class.This is where we need to use the property that e(Ci, V(G)-Ci) n .

By Lemma 5 and Lemma 6 we have that with high probability (> 1/5), a randomly selected element of Sp+1 is sufficiently dense:

• By our choice of it has average degree at least 2h.• So the expected number of elements is a random projective

decomposition (i.e. set of edge-disjoint induced subgraphs) that have a subgraph with minimum degree at least h is at least

~ .• Any two subgraphs intersect in at most one color class, so the

intersection is at most .

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Concluding remarks• The proof of Theorem 1 is algorithmic.

a projective plane of order p is elementary constructed from the addition and multiplication tables of a field with p elements.

The major algorithmic component is an implementation of Lemma 3 (balanced graph coloring lemma). The coloring constructed there requires a recent result of [KKMS-2010] (algorithmic version of the Hajnal-Szemerédi Theorem).

• If m is very close to linear (say, m = n log n) the proof of Lemma 3 introduces a logarithmic factor. It may be interesting to determine if this is essential.

• Although we proved that the term n3/m2 for the intersection size cannot be improved to (n3/m2)1-ε for any ε, it may be to prove that it cannot be improved even by logarithmic factors.

Thanks