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Fractional decompositions Fractional decompositions of dense hypergraphs of dense hypergraphs Raphael Yuster Raphael Yuster University of Haifa University of Haifa

Fractional decompositions of dense hypergraphs

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Fractional decompositions of dense hypergraphs. Raphael Yuster University of Haifa. Definitions, notations and background. The Rödl nibble The edges of the complete r -graph K ( n , r ) can be packed, almost completely, with copies of K ( k , r ) , where k is fixed. - PowerPoint PPT Presentation

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Page 1: Fractional decompositions of dense hypergraphs

Fractional decompositions of Fractional decompositions of dense hypergraphsdense hypergraphs

Raphael YusterRaphael YusterUniversity of HaifaUniversity of Haifa

Page 2: Fractional decompositions of dense hypergraphs

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Definitions, notations and background

The Rödl nibble

The edges of the complete r-graph K(n,r) can be packed, almost completely, with copies of K(k,r), where k is fixed.

This result is considered one of the most fruitful applications of the probabilistic method.

It was not known whether the same result also holds in a dense hypergraph setting.

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Example: Packing 7 copies of K(3,2) in a K(7,2)

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Definitions, notations and background

Let H0 be a fixed hypergraph.A fractional H0-decomposition of a hypergraph H is an assignment of nonnegative real weights to the copies of H0 in H such that for each e E(H) the sum of the weights of copies of H0 containing e is 1.

Example:K4 has a fractional K3-decomposition.Each triangle receives a weight of ½.

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Main result

There exists a positive constant α=α(k,r) so that everyn-vertex r-graph in which every (r-1)-set is contained in at least (n-r+1)(1-α) edges has a fractionalK(k,r)-decomposition.

• The proof is algorithmic.

• K(n,r) is simply the case α=0 since every r-1 set is contained in n-r+1 edges.

• In fact we obtain: α(k,r) > 6-kr

α(k,2) > 0.1k -10

α(3,2) > 10-4

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From fractional to integral• Combined with the following result of Rödl, Schacht,

Siggers and Tokushige our result has consequences for integral packing.

• Let ν(H0,H) denote the maximum number of edge-disjoint copies of H0 in H (the H0-packing number of H).

• Let ν*(H0,H) denote the fractional relaxation.

• Trivially, ν*(H0,H) ≥ ν(H0,H).

• They proved that if H is an r-graph with n vertices then ν*(H0,H) < ν(H0,H) + o(nr).

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Corollaries for graphs

• If G is a graph with n vertices, andδ(G) > (1- 1/10k10)n then G has an asymptotically optimal Kk-packing.

• Same theorem holds for k-vertex graphs.• For triangles (k=3), δ(G) > 0.9999n suffices.

• The previously best known bound (for the missing degree) in the triangles case was 10-24 (Gustavsson).The previously best known bound for Kk was 10-37k-94 (Gustavsson).

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Corollary for 3-graphs

• If H is a 3-graph with n vertices andminimum co-degree (1-216-k)n then H has an asymptotically optimal K(k,3)-packing.

• Same theorem holds for k-vertex 3-graphs.

• The previously best known bound (for the missing co-degree) was 0 (The Rödl nibble).

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Tools used in the proof• Some linear algebra.

• Kahn’s Theorem:For every r* > 1 and every γ > 0 there exists a positive constant ρ=ρ(r*,γ) such that the following statement is true:

If U is an r*-graph with:(i) maxdeg < D(ii) maxcodeg < ρDthen there is a proper coloring of the edges of U with at most (1+γ)D colors.

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Tools used in the proof

• Several probabilistic arguments

large deviation,

local lemma.

• Hall’s Theorem for hypergraphs (by Aharoni and Haxell. Has a topological proof):

Let U={U1,…,Um} be a family of p-graphs. If for every W U there is a matching in UU WU of size greater than p(|W|-1) then U has an SDR.

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The proofRecall our goal:

There exists a positive constant α=α(k,r) so that every r-graph in which every (r-1)-set is contained in at least n(1-α) edges has a fractional K(k,r)-decomposition.

• Let t=k(r+1) and consider the family of the 3 r-graphs:F(k,r) = { K(k,r), K(t,r), H(t,r) }

where H(t,r) is a K(t,r) missing one edge.

• Lemma: K(k,r) fractionally decomposes each element of F(k,r). (To show that K(k,r) fractionally decomposes H(t,r) requires some work. We use linear algebra here.)

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The proof

For r=2 (graphs) it suffices to take t=2k-1 and the lemma is easy.

Example: r=2, k=3 hence t=5 and F(3,2) = { K(3,2), K(5,2), H(5,2) }

1 2

a cb

H(5,2)

w(i,x,y)=1/2

i=1,2

x,y=a,b,c

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The proof

It suffices to prove the stronger theorem

There exists a positive constant α=α(k,r) so that everyr-graph in which every (r-1)-set is contained in at leastn(1-α) edges has an integral F(k,r)-decomposition.

• Let ε = ε(k,r) be chosen later.

• Let η = (2-H(ε)0.9)1/ε. H(ε) the entropy function.

• Let α = min{ (η/2)2 , ε2/(t24t+1) }

• Let γ satisfy (1-αt2t)(1-γ)/(1+γ)2 > 1-2αt2t

• Let r* =

• Let ρ = ρ(r*,γ) be the constant from Kahn’s theorem.

• Assume n is suff. large as a function of all these constants.

r

t

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The proof

• Let δd(H) and Δd(H) denote the min and maxd-degrees of H, 0 < d < r.

• Our r-graph H satisfies δd(H) >

• It is not difficult to prove (induction) that every edge of H lies on many K(t,r). In fact, if c(e) denotes the number of K(t,r) containing e then

)1(

dr

dn

)21()!(

)()!(

trtrt

trt

nec

rt

n

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The proof

• Color the edges of H randomly usingq=n1/(4r*-4) colors (that’s many colors!)

• Let Hi be the spanning r-graph colored with i.

• Easy (Chernoff): δd(Hi) very close to δd(H)/q

• Not so easy: we would also like to show thatci(e) is very close to its expectation c(e)n-1/4.

Note that two K(t,r) that contain e may share other edges as well – a lot of dependence.

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The proof

• By considering the dependency graph of the c(e) events we can show:

(1+γ)nt-r-1/4 > ci(e) (t-r)! > (1-γ)nt-r-1/4(1-αt2t)

• We fix the coloring with q colors satisfying the above.

• For each Hi we create another r*-graph Ui:- the vertices of Ui are the edges of Hi

- the edges of Ui are the copies of K(t,r) in Hi

• Notice that Δ(Ui) < D = (1+γ)nt-r-1/4 (t-r)!-1

Notice that Δ2(Ui) < nt-r-1 << ρD

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The proof• By Kahn’s theorem this means that the K(t,r) copies of Hi

can be partitioned into at most D(1+γ) packings.

• We pick one of these packings at random. Denote it Li.

• The set L=L1 U…U Lq is a K(t,r) packing of H.

• Let M denote the edges of H not belonging to any element of L.

• Let p = A p-subset {S1,…,Sp} of L is good fore M if we can select one edge from each Si such that, together with p, we have a K(k,r).

• We say that L is good if for each e M we can select a good p-subset, and all |M| selections are disjoint.

1

r

k

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Example: being good

ab

b

c

L M

a

c

S100

S700

{S100,S700} is good for (a,c)

k=3

r=2

So:

t=5

p=2

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The proof• Recall F(k,r) = { K(k,r), K(t,r), H(t,r) }.

Clearly:L is good → H has an Fk-decomposition.

• It remains to show that there exists a good L. We will show that with positive probability, the random selection of the q packings L1 U…U Lq yields a good L.

• We use Hall’s theorem for hypergraphs.

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The proof• Let M={e1,…,em}.

• Let U={U1,…,Um} be a family of p-graphs defined as follows:

• The vertex set of Ui is L (i.e, K(t,r) copies)

• The edges of Ui are the p-subsets of L good for ei

• U has an SDR → L is good.

• Thus, it suffices to show that the random selection of the q packings L1 U…U Lq guarantees that, with positive probability, for every W U there is a matching in UU WU of size greater than p(|W|-1).

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The proof

• It turns out that the only thing needed to guarantee this is to show that with positive probability, for all 1 < d < r:Δd(H[M]) < 2ε

• Once this is established, the remainder of the claim is deterministic, namely

• Δd(H[M]) < 2ε → U has an SDR.(purely combinatorial proof, but not so easy).

dr

dn

dr

dn

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Open problems• Determine the correct value of α(k,r).

• The simplest case is α(3,2) (triangles). We currently have α(3,2) > 10-4.

• A construction shows that α(3,2) ≤ ¼.

• More generally, a construction given in the paper shows that α(k,2) ≤ 1/(k+1).We conjecture α(k,2) = 1/(k+1).

• For hypergraphs we don’t even know what to conjecture.