Dense graphs with a large triangle cover have a large triangle packing Raphael Yuster SIAM DM’10

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).

3 If true, this is best possible (e.g. G = {K 4, K 5 } ). 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.

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(n 2 ).

 (G)  2 (G) + o(n 2 ). 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(n 2 ). 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).

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

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  βn 2 edges that are (1-δ)-hard to make Δ-free: (G)  (1-ε)m/3. Conjecture 1 – formal statement

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  βn 2 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.

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  βn 2 edges that are (1-δ)-hard to make Δ-free: (G)  (1+ β 4 )m/4. Theorem 3 – formal statement Strictly better than the Tuza bound by a fraction that depends on the density.

10 Proof of main result Since  *(G) = *(G)  (G) + o(n 2 ) it is equivalent to prove that: For β > 0 there exists δ > 0 so that for large graphs with m  βn 2 edges that are (1-δ)-hard to make Δ-free:  *(G)  (1+ β 4 )m/4. Assume w.l.o.g. that m = βn 2.

11 Take a minimum fractional cover f: E(G)  [0,1] Take a maximum fractional packing g: T(G)  [0,1] Let F 0  E(G) be F 0 = { e | f(e) = 0 }. Let F 1  E(G) be F 1 = { e | f(e) = 1 }. There are three cases to consider: 1. F 1 is relatively large. 2. F 0 is relatively small. 3. Neither (F 1 relatively small and F 0 relatively large).

12 Case 1 - |F 1 | > (δ+β 4 )m/2 Define G 1 = G – F 1. Notice that:  (G 1 )   (G) - |F 1 | (we deleted |F 1 | edges).  *(G 1 )   *(G) - |F 1 | (the total deleted weight is |F 1 |).  *(G)   *(G 1 ) + |F 1 |  ½  (G 1 ) + |F 1 |  ½ (  (G)- |F 1 |) + |F 1 | = ½  (G) + |F 1 |/2  ½ (1- δ)m/2 + (δ+β 4 )m/4 = (1+ β 4 )m/4.

13 Case 2 - |F 0 | < (1-3β 4 )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 ∑ e  t g(t) =1 This means that  *(G) = *(G)  | E(G) – F 0 |/3.  *(G)  | E(G) – F 0 |/3  (m-(1-3β 4 )m/4)/3 = (1+ β 4 )m/4.

14 Case 3 |F 0 |  (1-3β 4 )m/4  m/5. |F 1 |  (δ+β 4 )m/2  β 4 m. Consider the graph H = G[F 0 ]. 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.

15 Proof: H has m/5= βn 2 /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*|  2β 4 m. Furthermore, A has a common neighbor and B has a common neighbor. Lemma

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 n 2 β/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 C 2 pairs of parts contain all the m/20 edges. So on average, there is a pair with m/20C 2  2β 4 m.

17 What do we gain from the lemma: Let’s go back to G. |E(A,B)|  |F*|  2β 4 m. E(A) and E(B) contain only edges of F 1. |E(A,B)| - |E(A)  E(B) |  2β 4 m - |F 1 |  β 4 m. H has an induced bipartite subgraph (A  B, F*) with |F*|  2β 4 m. Furthermore, A has a common neighbor and B has a common neighbor < 1 A

18 Reaching a contradiction Split the vertices of V-A-B into two parts X,Y at random: The cut (A  X, B  Y) contains an expected number of: |E(A,B)| + ½( m - |E(A,B)| - |E(A)| - |E(B)| )  (1+ β 4 )m/2 Implying that  (G)  (1- β 4 )m/2 < (1- δ)m/2. A B X Y

19 Thanks