Nordhaus-Gaddum for Treewidth

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📝 Original Info

  • Title: Nordhaus-Gaddum for Treewidth
  • ArXiv ID: 1109.1602
  • Date: 2013-06-18
  • Authors: Gwena’el Joret and David R. Wood

📝 Abstract

We prove that for every graph $G$ with $n$ vertices, the treewidth of $G$ plus the treewidth of the complement of $G$ is at least $n-2$. This bound is tight.

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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 [1,4,5,8,10,13,14] for a few. Our main result is the following Nordhaus-Gaddum-type theorem for treewidth 1 , which is a graph parameter 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,

The following lemma is the key to the proof of Theorem 1. 1 While treewidth is normally defined in terms of tree decompositions (see [3]), it can also be defined as follows.

Then the treewidth of a graph G is the minimum integer k such that G is a spanning subgraph of a k-tree. See [2,11] for surveys on treewidth.

Let G be a graph. Two subsets of vertices A and B in G touch 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.

Proof of Theorem 1. Let k := tw(G). Let H be a k-tree that contains G has a spanning subgraph. Thus H has no induced 4-cycle (it is chordal) and has no (k + 2)-clique. By Lemma 2 and since

Lemma 2 immediately implies the following result of independent interest. Theorem 3. For every graph G with girth at least 5, we have tw(G) ≥ n -3.

For k-trees we have the following precise result, which proves that the bound in Theorem 1 is tight. Let Q k n be the k-tree consisting of a k-clique C with nk vertices adjacent only to C.

Theorem 4. For every k-tree G,

By the definition of k-tree, V (G) can be labelled v 1 , . . . , v n such that {v 1 , . . . , v k+1 } is a clique, and for j ∈ {k + 2, . . . , n}, the neighbourhood of

Thus C j = C k+2 for some minimum integer j. Observe that each vertex in C j has a neighbour outside of C j . Arbitrarily label C j = {x 1 , . . . , x k+1 }, and let y i be a neighbour of each x i outside of C j .

We now describe an (nk -2)-tree H that contains G. Let A := V (G) \ C j be the starting (n-k-1)-clique of H. Add each vertex x i to H adjacent to A{y i }. Observe that H is an (n-k-2)tree and G is a spanning subgraph of H. Thus tw(G) ≤ nk -2 and tw(G) + tw(G) ≤ n -2, with equality by Theorem 1.

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 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( 1 n ) in the sense of Erdős and Rényi, then asymptotically almost surely tw(G) = no(n); see [6,7] for related results. Setting p = 1 2 , it follows that asymptotically almost surely, tw(G) = no(n) and tw(G) = no(n), and hence tw(G) + tw(G) = 2no(n).

Theorems 1 and 4 can be reinterpreted as follows.

Proposition 5. For all graphs G 1 and G 2 , the union G 1 ∪ G 2 contains no clique on tw(G 1 ) + tw(G 2 ) + 3 vertices. Conversely, there exist graphs G 1 and G 2 such that G 1 ∪ G 2 contains a clique on tw(G 1 ) + tw(G 2 ) + 2 vertices.

Proof. For the first claim, we may assume that V (

Thus |S| ≤ tw(G 1 ) + tw(G 2 ) + 2 as desired. The converse claim follows from Theorem 4. Proposition 5 suggests studying G 1 ∪G 2 further. For example, what is the maximum of χ(G 1 ∪G 2 ) taken over all graphs G 1 and G 2 with tw(G 1 ) ≤ k and tw(G 2 ) ≤ k? By Proposition 5 the answer is at least 2k + 2. A minimum-degree greedy algorithm proves that χ(G 1 ∪ G 2 ) ≤ 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.

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