Nordhaus-Gaddum for Treewidth

We prove that, for every n-vertex graph G, the treewidth of G plus the treewidth of the complement of G is at least n − 2. This bound is tight. © 2012 Gwenaël Joret and David R. Wood. Published by Elsevier Ltd. All rights reserved. Nordhaus–Gaddum-type theorems establish bounds on f (G) + f (G) for some graph parameter f , where G is the complement of a graph G. The literature has numerous examples; see [3,8,4,6,13,14,11] for a few. Ourmain result is the following Nordhaus–Gaddum-type theorem for treewidth,1which is a graphparameter of particular importance in structural and algorithmic graph theory. Let tw(G)denote the treewidth of a graph G. Theorem 1. For every graph G with n vertices, tw(G) + tw(G) ≥ n − 2. The following lemma is the key to the proof of Theorem 1. Lemma 2. For every n-vertex graph G with no induced 4-cycle and no k-clique, tw(G) ≥ n − k. E-mail addresses: [email protected] (G. Joret), [email protected] (D.R. Wood). 1 While treewidth is normally defined in terms of tree decompositions (see [2]), it can also be defined as follows. A graph G is a k-tree if G ∼= Kk+1 or G − v is a k-tree for some vertex v whose neighbours induce a k-clique. Then the treewidth of a graph G is the minimum integer k such that G is a spanning subgraph of a k-tree. See [1,10] for surveys on treewidth. Let G be a graph. Two subsets of vertices A and B in Gtouch if A ∩ B ≠ ∅, or some edge of G has one endpoint in A and the other endpoint in B. A bramble in G is a set of subsets of V (G) that induce connected subgraphs and pairwise touch. A set S of vertices in G is a hitting set of a bramble B if S intersects every element of B. The order of B is the minimum size of a hitting set. Seymour and Thomas [12] proved the Treewidth Duality Theorem, which says that a graph G has treewidth at least k if and only if G contains a bramble of order at least k + 1. 0195-6698/$ – see front matter© 2012 Gwenaël Joret and David R. Wood. Published by Elsevier Ltd. All rights reserved. doi:10.1016/j.ejc.2011.10.005 G. Joret, D.R. Wood / European Journal of Combinatorics 33 (2012) 488–490 489 Proof. Let B := {{v, w} : vw ∈ E(G)}. If {v, w} and {x, y} do not touch for some vw, xy ∈ E(G), then the four endpoints are distinct and (v, x, w, y) is an induced 4-cycle in G, which is a contradiction. ThusB is a bramble in G. Let S be a hitting set forB. Thus no edge in G has both endpoints in V (G)\ S. Hence V (G) \ S is a clique in G. Therefore n − |S| ≤ k − 1 and |S| ≥ n − k + 1. That is, the order of B is at least n − k + 1. By the Treewidth Duality Theorem, tw(G) ≥ n − k, as desired. Proof of Theorem 1. Let k := tw(G). Let H be a k-tree that contains G as a spanning subgraph. Thus H has no induced 4-cycle (it is chordal) and has no (k + 2)-clique. By Lemma 2, and since G ⊇ H , we have tw(G) ≥ tw(H) ≥ n − k − 2. Therefore tw(G) + tw(G) ≥ n − 2. Lemma 2 immediately implies the following result of independent interest. Theorem 3. For every n-vertex graph G with girth at least 5, tw(G) ≥ n − 3. We now show that Theorem 1 is tight. Lemma 4. Let G be a graph with treewidth k that contains a (k + 1)-clique C such that each vertex in C has a neighbour outside of C. Then tw(G) + tw(G) = n − 2. Proof. We describe an (n−k−2)-treeH that contains G. Let A := V (G)\C be the starting (n−k−1)clique of H . For each vertex x ∈ C , add x to H adjacent to A \ {y}, where y is a neighbour of x outside of C . Observe that H is an (n − k − 2)-tree and G is a spanning subgraph of H . Thus tw(G) ≤ n − k − 2 and tw(G) + tw(G) ≤ n − 2, with equality by Theorem 1. For k-trees, we have the following precise result. Let Q k n be the k-tree consisting of a k-clique C with n − k vertices adjacent only to C . Theorem 5. For every n-vertex k-tree G, tw(G) + tw(G) =  n − 1 if G ∼= Q k n n − 2 otherwise. Proof. First, suppose that G ∼= Q k n . Then G consists of Kn−k and k isolated vertices. Thus tw(G) = n − k − 1, and tw(G) + tw(G) = n − 1. Now assume that G ≁= Q k n . By the definition of a k-tree, V (G) can be labelled v1, . . . , vn such that {v1, . . . , vk+1} is a clique, and, for j ∈ {k + 2, . . . , n}, the neighbourhood of vj in G[{v1, . . . , vj−1}] is a k-clique Cj. If Ck+2, . . . , Cn are all equal, then G ∼= Q k n . Thus Cj ≠ Ck+2 for some minimum integer j. Observe that each vertex in Cj has a neighbour outside of Cj. The result follows from Lemma 4. In view of Theorem 1, it is natural to also consider how large tw(G) + tw(G) can be. Every n-vertex graph G satisfies tw(G) ≤ n − 1, implying that tw(G) + tw(G) ≤ 2n − 2. It turns out that this trivial upper bound is tight up to lower-order terms. Indeed, Perarnau and Serra [9] proved that, if G ∈ G(n, p) is a random n-vertex graph with edge probability p = ω( n ) in the sense of Erdős and Rényi, then asymptotically almost surely tw(G) = n−o(n); see [5,7] for related results. Setting p = 1 2 , it follows that, asymptotically almost surely, tw(G) = n − o(n) and tw(G) = n − o(n), and hence tw(G) + tw(G) = 2n − o(n). Theorems 1 and 5 can be reinterpreted as follows, where, for graphs G1 and G2, the union G1 ∪ G2 is the graph with vertex set V (G1) ∪ V (G2) and edge set E(G1) ∪ E(G2) (where G1 and G2 may intersect). Proposition 6. For all graphs G1 and G2, the union G1 ∪ G2 contains no clique on tw(G1) + tw(G2) + 3 vertices. This result is sharp, since there exist graphs G1 and G2 such that G1 ∪ G2 contains a clique on tw(G1) + tw(G2) + 2 vertices. 490 G. Joret, D.R. Wood / European Journal of Combinatorics 33 (2012) 488–490 Proof. For the first claim,wemay assume that V (G1) = V (G2) and E(G1)∩E(G2) = ∅. Let S be a clique in G1 ∪ G2. Thus G1[S] = G2[S]. By Theorem 1, tw(G1) + tw(G2) ≥ tw(G1[S]) + tw(G2[S]) ≥ |S| − 2. Thus |S| ≤ tw(G1) + tw(G2) + 2 as desired. The sharpness example follows from Theorem 5. Proposition 6 suggests studying G1 ∪ G2 further. For example, what is the maximum of χ(G1 ∪ G2) taken over all graphs G1 and G2 with tw(G1) ≤ k and tw(G2) ≤ k? By Proposition 6, the answer is at least 2k + 2. A minimum-degree greedy algorithm shows that χ(G1 ∪ G2) ≤ 4k. This question is somewhat similar to Ringel’s earth–moon problem, which asks for the maximum chromatic number of the union of two planar graphs.