1 Decomposition into bipartite graphs with minimum degree 1. Raphael Yuster.

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1 Decomposition into bipartite graphs with minimum degree 1. Raphael Yuster

Transcript of 1 Decomposition into bipartite graphs with minimum degree 1. Raphael Yuster.

Page 1: 1 Decomposition into bipartite graphs with minimum degree 1. Raphael Yuster.

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Decomposition into bipartite graphs with minimum degree 1.

Raphael Yuster

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Definitions, notations and background

• Let H be a fixed connected graph. A graph G has an H-decomposition if there is an edge coloring of G such that each color class induces a copy of H.

• Trivial necessary conditions: e(H) divides e(G)gcd(H) divides gcd(G)

These conditions are not always sufficient.

Example: G=K4 and H=K1,3.

• A seminal result of Wilson (1975): if n is sufficiently large and G=Kn then these conditions are also sufficient.

• The decomposition problem is trivial when H=K1,1 and almost trivial when H=K1,2.

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• For any fixed graph H with at least 3 edges the corresponding decision problem is NP-Complete (Dor & Tarsi).

• As usual in such cases, the focus of research turns towards extremal graph theory:

Problem: Determine the function fH(n) that is the minimum possible integer such that for every graph G with n vertices and δ(G) fH(n), then the necessary conditions also suffice.– Wilson’s result shows that fH(n) n-1 for n

sufficiently large.

– fH(n) n/2-1 because of connectivity.

– Hence:

0.5 0.5 liminf liminf ffHH((nn)/)/nn limsup limsup ffHH((nn)/)/nn 1 1

• Until 1991, it was not known for any fixed graph H other than the two trivial ones, whether liminf fH(n)/n < 1.

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• In 1991, T. Gustavsson proved that for every graph H, there exists =(H) such that liminf fH(n)/n < 1- .Unfortunately, this is only an existential result, since is very small. Even for the simplest cases, K3 or K1,3 , < 10-24. Thus, it is probably very far from the correct bound for any graph H.

• Conjecture (Nash-Williams) for K3

liminf fH(n)/n=0.75 (lower bound is easy).

• My original interest was to find at least one nontrivial fixed graph for which liminf fH(n)/n can be determined.

• Theorem (Y. 1998) lim fH(n)/n = 0.5 for every fixed tree H.This leads to conjecture the following:

• Conjecture (Y) lim fH(n)/n=0.5 if and only if H is bipartite.

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The main result• The difficult direction in the last

conjecture is far from being solved. We can, however, prove it for an important class of graphs:

• Theorem (Y. 2001) lim fH(n)/n = 0.5 for

every bipartite graph with δ(H)=1.

• Although this extends the result for trees, the proof is completely different.

BB

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Outline of proof• Let H be a fixed bipartite graph with h

edges and assume δ(H)=1. Let a0 be a vertex of degree one (a leaf) and let b1 be its unique neighbor. We call (a0,b1) the leaf edge and we call b1 the pivot.Denote H’=H \ {a0} and let A’={a1,…,ar} B={b1,…,bs} be the vertex classes of H’.Example:

a0 b1

b1 b2

a2

a3

a1

H h=7

H’

pivot

leaf edge

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The naive approach (that would fail)

• G=(V,E) is a graph with n vertices – |E|=h•m– δ(G) > 0.5n(1+o(1)).

We must show that there are m edge disjoint copies of H in G.

• Use the fact that H’ has a small Turán Number o(n2), and find m edge disjoint copies of H’ in G using a greedy algorithm. This succeeds since hm-(h-1)m = m = (n2) > T(n,H’).

• The remaining m edges must play the role of the leaf. Hence, we need to show that there is a perfect matching between the m copies of H’ and the m remaining edges, where each remaining edge (x,y) may be matched to a copy whose pivot is x (and y is not present at all) or whose pivot is y (and x is not present at all).

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• The reasons for possible failure are many:– The vertex x appears as pivot qx times, but is

an endpoint only of px remaining edges, and qx > px.

– The vertices x and y never appear as pivots, but the edge (x,y) needs to be matched.

– Whenever x appears in a copy of H’, so does y.

– These problems can be magnified to sets of vertices. Namely, Hall’s conditions for the existence of a perfect matching do not hold.

• Three essential obstacles need to be addressed in order to solve these problems– The possible ugliness of H’. In other words,

its automorphism group may be trivial so only one specific vertex can play the role of the pivot in any copy. We would like all vertices to be pivots many times. We have no control over this if H’ is very ugly.

– Every vertex x should appear in a significant number of copies of H’ as a pivot, and any other vertex y should not appear together with x in a significant number of these copies.

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– The number of remaining edges incident with x should be about 1/h of its original degree (in order to guarantee the expansion properties needed in Hall’s Theorem). Similar requirement should hold for subsets of vertices.

• These obstacles are solved as follows:1. We create a (lager, but finite and depends

only on H’) graph H* with the following properties:

– H* is bipartite with 8(r+s)5 vertices in each class.

– H* has 64(h-1)(r+s)6 edges. (H*) 4(h-1)(r+s)2.– There is an H’-decomposition D of H*

with |D|=64(r+s)6, such that for any two distinct vertices v,u of H*, there is an element of D in which v plays the role of the pivot, and u does not appear at all.

We use several probabilistic arguments in order to construct H*.

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2. We initially select a random subset of m-T(n,H’) =m(1-o(1)) edges of G. This random selection guarantees that any m edges containing all of them would have “nice” expansion properties.

3. In the remaining m(h-1)+T(n,H’) edges we find edge-disjoint copies of H* one by one, such that whenever we need a new copy we select one that contains the vertex that appeared the least number of times in previous copies. We cannot do it until we absorb all edges, but we can do it sufficiently many times to guarantee that all vertices appear in at least, say, n/800(r+s)9 copies H*.

4. We now take each H* and decompose it, in turn, to H’, using the properties of these decompositions of H* that are guaranteed to exist.

5. The union of these copies of H’ from all the decompositions can now be matched to the remaining edges, as we can show (very technical) that Halls’ condition holds.