Edge-disjoint induced subgraphs with given minimum degree Raphael Yuster 2012.

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Presentation transcript:

Edge-disjoint induced subgraphs with given minimum degree Raphael Yuster 2012

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. 2

Let f h (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, f 1 (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 f 2 (G) =O(m 2 /n 2 ). 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. 3 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 ≥ n 1+α edges has: cm 2 /(h 2 n 2 ) edge-disjoint induced subgraphs, with min. degree h. Any two subgraphs intersect in at most 1+h 2 n 3 /(cm 2 ) vertices.

We make no attempt to optimize c (it is polynomial in α). If m = Ω(n 3/2 ) then the intersection is only 1. The number of subgraphs cm 2 /(h 2 n 2 ) 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 n 3 /m 2 cannot be improved to (n 3 /m 2 ) 1-ε for any ε. 4 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 ≥ n 1+α edges has: cm 2 /(h 2 n 2 ) edge-disjoint induced subgraphs, with min. degree h. Any two subgraphs intersect in at most 1+h 2 n 3 /(cm 2 ) vertices.

Tightness of cardinality 5 Proposition 2. KxKx All x(n-x) edges In-xIn-x Choose: x(x-1)/2+x(n-x) ~ m x ≤ 2m/n

Tightness of intersection 6 Simple Combinatorial lemma: Let F be a family of subsets of {1,…,n}. Assume that |F| ~ n 2α and that | X | ≥ n 1-α for each X  F. Then there are two elements of F that intersect in ~ n 1-2α elements. We want to prove that there are graphs with m ~ n 1+α edges such that in any set of n 2α ~ m 2 /n 2 edge-disjoint induced subgraphs with min. degree at least h there are at two subgraphs having intersection n 1-2α ~ n 3 /m 2. Now use the random graph G(n,p) with p ~ n α-1. Next, prove that whp, for large enough (but constant) h, every subgraph with k ≤ n 1-α vertices has less than hk/2 edges. Thus, every subgraph with minimum degree h has to contain at least n 1-α vertices. Now, use the lemma with F being a maximum cardinality set of induced subgraphs with minimum degree at least h.

Balanced graphs 7 A graph is balanced if its average degree is not smaller than the average degree 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' vertices has 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

Reducing to the case of balanced graphs 8 Instead of proving the main result for all graphs with n ≥ N(α,h), let’s reduce 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 ≥ n 1+α edges. Let G’ be a subgraph with the least number of vertices n' < n and with 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 ≥ 2n α ≥ 2N α ≥ N* By the reduced theorem G’ has the required set of subgraphs of size c(m’) 2 /(h 2 (n’) 2 ) ≥ cm 2 /(h 2 n 2 )

A naïve attempt 9 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 ~ m 2 /n 2 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 ~ n 2 /m and thus, two subgraphs may have large intersection.

A coloring lemma for balanced graphs 10 Instead, 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. 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

Proof idea of lemma 11 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.

Projective planes and graph decomposition 12

Mapping independent sets into a projective plane 13 A bijection π from the points of PG(p) to the parts on an r -partite graph R (recall r = p 2 +p+1 ) defines a mapping between the lines of PG(p) and their corresponding full (p+1) -partite subgraphs. Equivalently, π defines a set L π of r full (p+1) -partite subgraphs of R, where any two subgraphs in L π are edge-disjoint. We call L π a projective decomposition of R. There are r! projective decompositions. Lemma 4. Let S p+1 be the set of all full (p+1) -partite subgraphs of an r -partite graph R with r = p 2 +p+1. Let L π be a projective decomposition of R chosen uniformly at random. Then, each element of S p+1 has the same probability of being an element of L π. In particular, a randomly chosen element of L π corresponds to a random element of S p+1.

Proof of Theorem 1 14

Proof of Theorem 1 (cont.) 15 Lemma 5 (vertices lemma). Lemma 6 (edges lemma).

Proof of Theorem 1 (cont.) 16

Concluding remarks 17 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 n 3 /m 2 for the intersection size cannot be improved to (n 3 /m 2 ) 1-ε for any ε, it may be to prove that it cannot be improved even by logarithmic factors.

Thanks