Download
1 / 19
190 likes | 322 Views
Dense graphs with a large triangle cover have a large triangle packing. Raphael Yuster. SIAM DM’10. 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.
E N D
Dense graphs with a large triangle cover have a large triangle packing Raphael Yuster SIAM DM’10
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) .
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.
*(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) .
Combining these two results we immediately obtain: Theorem 1 (G) 2(G) + o(n2) . 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) .
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. Corollary of Theorem 1 Dense graphs that are 1-o(1) -hard to make Δ-free have (G) ~ m/4 .
Conjecture 1 Dense graphs that are 1-o(1) -hard to make Δ-free have (G) ~ m/3. In other word, if a dense graph is hard to make triangle-free then it has an almost perfect triangle packing. Formally (with quantifiers): Conjecture 1 – formal statement For ε, β > 0 there exist δ > 0so that for large graphs with m βn2edges that are (1-δ)-hard to make Δ-free: (G) (1-ε)m/3 .
Conjecture 2 There exists an absolute > 0 so that dense graphs that are 1-o(1) -hard to make Δ-free have (G) ~ (1+)m/4. Strictly better than the Tuza bound by a fraction that is independent of the density. Conjecture 2 – formal statement There exists > 0 so that for all β> 0 there exist δ > 0so that for large graphs with m βn2edges that are (1-δ)-hard to make Δ-free: (G) (1+)m/4 .
Conjecture 3 β-dense graphs that are 1-o(1) -hard to make Δ-free have (G) ~ (1+f(β))m/4. Strictly better than the Tuza bound by a fraction that depends on the density. Theorem3 – formal statement For all β> 0 there exist δ > 0so that for large graphs with m βn2edges that are (1-δ)-hard to make Δ-free: (G) (1+β4)m/4 .
Proof of main result Since *(G)= *(G) (G) + o(n2) it is equivalent to prove that: For β> 0 there exists δ> 0so that for large graphs with m βn2edges that are (1-δ)-hard to make Δ-free: *(G) (1+ β4)m/4. • Assume w.l.o.g. that m= βn2.
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 (F1relatively small and F0 relatively large).
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 .
Case 2 - |F0|<(1-3β4)m/4 Let us recall that fand 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 ∑etg(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 .
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.
Lemma Hhas an induced bipartite subgraph (A B, F*) with |F*| 2β4m . Furthermore, A has a common neighbor and B has a common neighbor. • 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}
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 .
Hhas an induced bipartite subgraph (A B, F*) with |F*| 2β4m . Furthermore, A has a common neighbor and B has a common neighbor. • 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 . < 1 A 0 0 0 0 0 0
Reaching a contradiction Split the vertices of V-A-Binto 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 X B Y
