Dense graphs with a large triangle cover have a large triangle packing Raphael Yuster SIAM DM10 2 The problem, formulation and definitions Triangle (edge) cover: a set of edges meeting all triangles. Triangle (edge) packing: a set of
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Dense graphs with a large triangle cover have a large triangle packing Raphael Yuster SIAM DM’10<br>
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2 The problem, formulation and definitions Triangle (edge) cover: a set of edges meeting all triangles.
Triangle (edge) packing: a set of pairwise edge-disjoint triangles.
(G) - minimum triangle cover.
(G) - maximum triangle packing.
Obviously: (G) (G) 3(G) .
Long-standing Conjecture of Tuza: (G) 2(G) .<br>
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3 If true, this is best possible (e.g. G = {K4 , K5} ) .
Known: (G) 2.87 (G) [Haxell ’99] .
An important setting, where, asymptotically, Tuza's conjecture is known to hold is the dense graph setting.
To derive this, one considers fractional relaxations:
Fractional triangle cover: assigns nonnegative weights to the edges so that the weight sum on each triangle is 1.
Dually:
Fractional triangle packing: assigns nonnegative weights to triangles so that the weight sum on each edge is 1.<br>
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4 *(G) - min. fractional triangle cover. *(G) (G) .
*(G) - max. fractional triangle packing. *(G) (G) .
Linear programming duality: *(G) = *(G) .
[Krivelevich’95] proved a mixed fractional-integral version of Tuza’s conjecture:
(G) 2*(G) .
*(G) 2(G) .
[Haxell-Rödl ’2001, Y. ’2005] proved that integer and fractional packing are asymptotically the same in dense graphs. For triangles it implies:
*(G) (G) + o(n2) .<br>
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(G) 2(G) + o(n2) . 5 Theorem 1 Combining these two results we immediately obtain: Is this (the constant 2) best possible?
Clearly the question is interesting in dense graphs with (G) = θ(n2) . Perhaps most interesting when (G) is as large as we can expect it to be:
For any graph with m edges we have: (G) m/2 – o(m) .<br>
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6 But for many classes of graphs we have tightness:
(G) m/2 – o(m) .
We call such graphs Hard to make Δ-free.
Random graphs, complete graphs, as well as many other combinations of such are hard to make Δ-free.
More formally we define: (1-δ)-hard to make Δ-free as (G) (1-δ)m/2. Dense graphs that are 1-o(1) -hard to make Δ-free have
(G) ~ m/4 . Corollary of Theorem 1<br>
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7 Dense graphs that are 1-o(1) -hard to make Δ-free have
(G) ~ m/3 . Conjecture 1 In other word, if a dense graph is hard to make triangle-free then it has an almost perfect triangle packing.
Formally (with quantifiers): For ε, β > 0 there exist δ > 0 so that for large graphs with m βn2 edges that are (1-δ)-hard to make Δ-free:
(G) (1-ε)m/3 . Conjecture 1 – formal statement<br>
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8 There exists an absolute > 0 so that dense graphs that are 1-o(1) -hard to make Δ-free have
(G) ~ (1+)m/4 . Conjecture 2 There exists > 0 so that for all β > 0 there exist δ > 0 so that for large graphs with m βn2 edges that are (1-δ)-hard to make Δ-free:
(G) (1+)m/4 . Conjecture 2 – formal statement Strictly better than the Tuza bound by a fraction that is independent of the density.<br>
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9 β-dense graphs that are 1-o(1) -hard to make Δ-free have
(G) ~ (1+ f(β))m/4 . Conjecture 3 For all β > 0 there exist δ > 0 so that for large graphs with m βn2 edges that are (1-δ)-hard to make Δ-free:
(G) (1+ β4)m/4 . Theorem 3 – formal statement Strictly better than the Tuza bound by a fraction that depends on the density.<br>
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10 Proof of main result Since *(G) = *(G) (G) + o(n2) it is equivalent to prove that: For β > 0 there exists δ > 0 so that for large graphs with m βn2 edges that are (1-δ)-hard to make Δ-free:
*(G) (1+ β4)m/4 . Assume w.l.o.g. that m = βn2.<br>
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11 Take a minimum fractional cover f: E(G) [0,1]
Take a maximum fractional packing g: T(G) [0,1]
Let F0 E(G) be F0 = { e | f(e) = 0 }.
Let F1 E(G) be F1 = { e | f(e) = 1 }.
There are three cases to consider:
F1 is relatively large.
F0 is relatively small.
Neither (F1 relatively small and F0 relatively large).<br>
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12 Case 1 - |F1| > (δ+β4)m/2 Define G1 = G – F1 .
Notice that:
(G1) (G) - |F1| (we deleted |F1| edges).
*(G1) *(G) - |F1| (the total deleted weight is |F1|).
*(G) *(G1) + |F1|
½ (G1) + |F1|
½ ((G)- |F1|) + |F1|
= ½ (G) + |F1|/2
½ (1- δ)m/2 + (δ+β4)m/4
= (1+ β4)m/4 .<br>
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13 Case 2 - |F0| < (1-3β4)m/4
Let us recall that f and g are a minimum fractional cover and a maximum fractional packing respectively.
From linear programming duality we have the complementary slackness condition:
f(e) > 0 implies ∑et g(t) =1
This means that *(G) = *(G) | E(G) – F0 |/3 .
*(G) | E(G) – F0 |/3
(m-(1-3β4)m/4)/3
= (1+ β4)m/4 .<br>
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14 Case 3
|F0| (1-3β4)m/4 m/5.
|F1| (δ+β4)m/2 β4m.
Consider the graph H = G[F0] .
It is a dense triangle free graph.
We will prove that it contains an:
induced bipartite subgraph
with high density
whose vertex classes are partial neighborhoods.<br>
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15 Proof:
H has m/5= βn2/5 edges so if we delete vertices with degree less than βn/10 we still remain with subgraph H’ having m/10 edges and minimum degree βn/10.
Consider the following list coloring problem on H’: - The list of a vertex is the set of its neighbors.
Each list has size βn/10 and the colors are {1…n} H has an induced bipartite subgraph (A B, F*) with |F*| 2β4m . Furthermore, A has a common neighbor and B has a common neighbor. Lemma<br>
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16 A random set of (10/β) ln (20/ β) colors is expected to hit all but nβ/20 lists.
So let’s fix such a set C of “colors”.
The nβ/20 vertices corresponding to un-hit lists are incident with at most n2β/20=m/20 edges, so the graph H’’ without them contains m/10-m/20=m/20 edges.
Color each vertex of H’’ with an arbitrary color of C appearing in its list.
This partitions the vertices of H’’ into C independent sets.
The C2 pairs of parts contain all the m/20 edges. So on average, there is a pair with m/20C2 2β4m .<br>
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17 What do we gain from the lemma:
Let’s go back to G.
|E(A,B)| |F*| 2β4m .
E(A) and E(B) contain only edges of F1 .
|E(A,B)| - |E(A) E(B) | 2β4m - |F1| β4m . H has an induced bipartite subgraph (A B, F*) with |F*| 2β4m . Furthermore, A has a common neighbor and B has a common neighbor. < 1 A<br>
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18 Reaching a contradiction
Split the vertices of V-A-B into two parts X,Y at random:
The cut (AX , BY) contains an expected number of:
|E(A,B)| + ½( m - |E(A,B)| - |E(A)| - |E(B)| )
(1+ β4)m/2
Implying that (G) (1- β4)m/2 < (1- δ)m/2 . A B X Y<br>
