Maximal Chordal Subgraphs

A chordal graph is a graph with no induced cycles of length at least $4$. Let $f(n,m)$ be the maximal integer such that every graph with $n$ vertices and $m$ edges has a chordal subgraph with at least $f(n,m)$ edges. In 1985 Erd\H{o}s and Laskar posed the problem of estimating $f(n,m)$. In the late '80s, Erd\H{o}s, Gy\'arf\'as, Ordman and Zalcstein determined the value of $f(n,n^2/4+1)$ and made a conjecture on the value of $f(n,n^2/3+1)$. In this paper we prove this conjecture and answer the question of Erd\H{o}s and Laskar, determining $f(n,m)$ asymptotically for all $m$ and exactly for $m \leq n^2/3+1$.


Introduction
One of the central questions in extremal combinatorics can be formulated as follows. Given a graph G and a property P, what is the maximal subgraph of G one can find which satisfies this property. The study of this problem goes back to the work of Turán in 1941, whose theorem states that the largest subgraph of the n-vertex complete graph with no clique of size k + 1 is the complete k-partite graph with sides as equal as possible. This graph is called the Turán graph. We denote it by T k (n) and its size by t k (n). Turán's theorem is the starting point of extremal graph theory and has inspired extensive research. One such research direction studies which other (more elaborate) structures must appear in a graph with more than t k (n) edges. For example, a series of works determined how many (k + 1)-cliques must exist in a graph with t k (n) + a edges (for a suitable range of a) [7,11,16]. Other examples are results on finding many (k + 1)-cliques which share one or more vertices [5,7,8,15], and results on finding (k + 1)-cliques with large degree sum [1,3,4,12].
In this paper we study the Turán type problem for chordal graphs. A graph is called chordal if it contains no induced cycle of length at least 4. Chordal graphs are one of the most studied classes in graph theory and have numerous applications, for example in semidefinite optimisation (see the survey [17]) and evolutionary trees (see [6]). In 1985, Erdős and Laskar [10] asked to determine the maximum integer f (n, m) such that every graph with n vertices and m edges contains a chordal subgraph with at least f (n, m) edges. To put this question under the umbrella of classical extremal graph theory, one needs to consider equivalent definitions of chordal graphs. It is well known that a graph is chordal if and only if it can be constructed from a single-vertex graph by repeatedly adding a vertex and connecting it to a clique of the current graph 1 (this is called a perfect elimination ordering), see [13,Chapter 4]. So if G is a triangle-free graph, then every chordal subgraph of G must be a forest. More generally, if G has no cliques of size k + 1, then every chordal subgraph of G has at most (k − 1)(n − k + 1) + k−1 2 = (k − 1)n − k 2 edges. In particular, this bound applies to k-partite graphs. Another way of proving this bound for kpartite graphs is to observe that if G is k-partite with parts V 1 , . . . , V k and H is a chordal subgraph of G, then e H (V i , V j ) ≤ |V i | + |V j | − 1 for every i < j (because a chordal subgraph of a bipartite graph must be a forest). Hence, e(H) ≤ i<j (|V i | + |V j | − 1) = (k − 1)n − k 2 . The above discussion shows that if m ≤ t k (n) then f (n, m) ≤ (k − 1)n − k 2 . It is natural to guess that the value of f (n, m) "jumps" as m increases from t k (n) to m = t k (n) + 1, because at this point the graph must contain (k + 1)-cliques. Erdős and Laskar [10] proved that this is indeed the case for k = 2, showing that f (n, t 2 (n) + 1) ≥ (1 + ε)n. In the late 80's, Erdős, Gyárfás, Ordman and Zalcstein [9] determined the value of f (n, t 2 (n) + 1) exactly for even n, showing that f n, n 2 4 + 1 = 3n 2 − 1. This bound is achieved by the graph T 2 (n) + e, obtained by adding an edge to the Turán graph T 2 (n). It is natural to conjecture that for every k and n, the value of f (n, t k (n) + 1) is determined by T k (n) + e, which is the graph obtained by adding an edge to a largest class of T k (n). It is not hard to check that the largest chordal subgraph of T k (n) + e has kn − n k + 2 − k+1 2 edges. So we get the following conjecture.
Conjecture 1.1. f (n, t k (n) + 1) = kn − n k + 2 − k+1 2 . The authors of [9] only studied Conjecture 1.1 in the cases k = 2, 3, although they very likely had the full conjecture in mind. For k = 3, they proved that f (n, t 3 (n) + 1) ≥ 7n/3 − 6 and asked to determine f (n, t 3 (n) + 1). This question was later mentioned again in the problem survey of Gyárfás [14]. Answering this question, we resolve Conjecture 1.1 for the case k = 3. Theorem 1.2. f (n, t 3 (n) + 1) = 3n − n 3 − 4. Our next result proves Conjecture 1.1 asymptotically for every k. In fact, we go a step further and determine f (n, m) asymptotically for every value of m, answering the question of Erdős and Laskar. Theorem 1.3. Let k, n ≥ 1 and t k (n) + 1 ≤ m ≤ t k+1 (n). Set a = m − t k (n). Then f (n, m) = (k − 1/k)n + 2(k + 1)a/k − k The construction giving the upper bound in Theorem 1.3 is to take an (unbalanced) complete k-partite graph with k − 1 smaller classes of the same size and one bigger class, and to add a balanced complete bipartite graph inside the bigger class. One then needs to optimise the sizes of the classes and the size of the complete bipartite graph so as to minimise the size of chordal subgraphs. It is best to take the k − 1 smaller classes of size n−r k , the bigger class of size n+(k−1)r k and the complete bipartite of size r × r, where r := 2ka k+1 . See Figure 1, and see Section 2 for the details. For k = 1, 2, we can go a step further and determine f (n, m) exactly. This is done in the following two theorems. Theorem 1.4. Let n ≥ 1 and m ≤ t 2 (n). Then f (n, m) = min{r : t 2 (r) ≥ m} − 1.
The extremal construction for Theorem 1.5 is given by taking a t × (n − t) complete bipartite graph and placing a complete bipartite graph with r vertices inside the side of size t.

Proof ideas
Recall that a graph is chordal if and only if it can be obtained from the one-vertex graph by repeatedly adding simplicial vertices, that is, vertices whose neighbourhood is a clique. In particular, adding simplicial vertices to a chordal graph keeps it chordal. We will often use this fact (implicitly) to claim that certain graphs are chordal.
Let us first recall the argument used by Erdős, Gyárfás, Ordman and Zalcstein [9] to prove Conjecture 1.1 for k = 2 (and n even). Let G be a graph with n vertices and n 2 /4 + 1 edges, and let x, y, z be a triangle in G. We need to show that G has a chordal subgraph H with at least 3n/2 − 1 edges. If d(x) + d(y) + d(z) ≥ 3n/2 + 2, then take H to be the subgraph consisting of all edges touching x, y, z. Suppose now that d(x) + d(y) + d(z) ≤ 3n/2 + 1. Then by averaging, we can assume without loss of generality that d(x) + d(y) ≤ n. Deleting x, y, we get a graph with at least n 2 /4 + 1 − (n − 1) ≥ (n − 2) 2 /4 + 1 edges. By induction, this graph contains a chordal subgraph H with at least 3(n − 2)/2 − 1 edges. Adding the edges xy, xz, yz gives the required chordal subgraph H.
Our proof of Theorem 1.5 is also based on this inductive argument, but with two key differences. First, we need a relation between g 2 (n, m) and g 2 (n , m ) (for n = n − 2, say), so that the induction can be carried through when deleting vertices. And second, it turns out that the induction scheme of deleting two vertices does not work to give the correct bound on f (n, m) for all m in the range of Theorem 1.5. Instead, we sometimes need to delete just one vertex and then add two edges when adding the vertex back. To this end, we need to know that the deleted vertex has two neighbours which form an edge in the chordal subgraph H that we find using induction. To guarantee this, we strengthen the induction hypothesis to say that not only does G contain a chordal subgraph with the correct number of edges, but that any given triangle in G can be included in such a chordal subgraph.
The idea of strengthening the induction hypothesis is also used in the proof of Theorem 1.2. Here we show that every K 4 can be included in a chordal subgraph with the correct number of edges. This proof has a more involved case analysis. It would be interesting to find a shorter proof.
The proof of Theorem 1.3 is based on induction as well. Here, instead of deleting only a few vertices, we delete a large number of vertices. To give the general idea, we sketch first the proof in the case m = t 3 (n) + 1. So let G be a graph with n vertices t 3 (n) + 1 edges. We need to show that G has a chordal subgraph H with at least 8n 3 − 6 − C √ n edges. Let us assume first that e(G) ≥ t 3 (n) + 2n. By a theorem of Faudree [12] (see also [1,3]), there is a triangle n for each i. Delete y, z, w 1 , . . . , w 7 to get a graph G on n − 9 vertices. It is easy to see that e(G ) is well above t 3 (n − 9) + 1. So by the induction hypothesis, there exists a chordal subgraph H of G with e(H ) ≥ 8(n−9) 3 − 6 − C √ n edges. Now add back the vertices y, z, w 1 , . . . , w 7 , and add to H the edges of the triangle x, y, z and the edges between x, y, z and w 1 , . . . , w 7 . This is a total of 24 edges. So e(H) = e(H ) + 24 ≥ 8n 3 − 6 − C √ n, as required. It is also easy to see that H is chordal (if we add the new vertices in the order y, z, w 1 , . . . , w 7 , then we always add a simplicial vertex). The number 7 was chosen here so that the number of edges added would be large enough for the induction to carry through. But the key point is that such a number must exist. Indeed, each w i contributes 3 edges to H. On the other hand, the term 8n 3 suggests that it is enough to add 8 3 edges per vertex on average. So by adding 3 edges per vertex, we are gaining over the required bound.
It now remains to handle the case that e(G) ≤ t 3 (n) + 2n. Here we proceed as follows. If the minimum degree of the graph is at least 2n 3 − √ n, then take a 4-clique x, y, z, w and take H to be the subgraph consisting of edges touching x, y, z, w. Else, delete a vertex of minimum degree and continue with the remaining graph. After O( √ n) steps, we get a graph with n = n − O( √ n) vertices and at least t 3 (n ) + 2n edges, so we can apply the first case.
To prove the general case of Theorem 1.3 we find a (k − 1)-clique x 1 , . . . , x k−1 and a forest F inside N(x 1 , . . . , x k−1 ) such that F has few components. We delete V(F) and x 2 , . . . , x k−1 and apply induction to find a chordal subgraph H . We then add to H the edges of the clique x 1 , . . . , x k−1 , the edges of F, and the edges between V(F) and x 1 , . . . , x k−1 . Note that when adding back the vertices of F one by one, most vertices contribute k edges: one edge in F and k − 1 edges to x 1 , . . . , x k−1 (this fails once for each connected component of F, and this is why we want the number of components to be small). On the other hand, the main term in Theorem 1.3 is (k − 1/k)n, which suggests that each vertex adds k − 1/k edges on average. So again we are gaining over the required bound (at least if we ignore the second term 2(k + 1)a/k for the moment). A somewhat lengthy calculation shows that this argument indeed works for any value of a.
The rest of this short paper is organised as follows. Theorem 1.3 is proved in Section 2, Theorems 1.4-1.5 in Section 3 and Theorem 1.2 in Section 4.

Proof of Theorem 1.3
In this section we prove Theorem 1.3. We begin with the upper bound. Here we use the following construction. For simplicity, assume that n is divisible by k, k + 1. For general n the construction is essentially the same (and, since we are only interested in an approximate result, we are allowed a small error due to divisibility issues). Fix k ≥ 1 and m ≤ t k+1 (n) = kn 2 2(k+1) , so that a := m − t k (n) ≤ kn 2 2(k+1) − (k−1)n 2 2k = n 2 2k(k+1) . Set r := 2ka k+1 ≤ n k+1 . Consider a complete k-partite graph with sides X, Y 1 , . . . , Y k−1 such that |X| = n+(k−1)r k and |Y i | = n−r k for every 1 ≤ i ≤ k − 1. Place an r × r complete bipartite graph with sides A, B inside X. This is possible as 2r ≤ n+(k−1)r k . The resulting graph G has Let H be a chordal subgraph of G. We have e H (A, because each of these bipartite graphs is induced in G, so its intersection with H is a forest. So giving the upper bound on f (n, m) for Theorem 1.3. We now prove the lower bound, which we restate for convenience as follows.
such that the following holds. Let n, a ≥ 1, and let G be a graph with n vertices and at least t k (n) + a edges. Then G has a chordal subgraph with at least For the proof of Theorem 2.1 we need two lemmas. The following lemma uses an argument originally used by Edwards [3,4] and Faudree [12] (see also [1]) to find cliques with a large degree sum.

Lemma 2.2. Let k, n, a ≥ 1 and let G be a graph with n vertices and at least
Consider the following process: for i = 1, 2, . . . , take x i to be a vertex of maximum degree among all vertices in N(x 1 , . . . , x i−1 ). Then this process continues for at least k steps, and N(x 1 , . . . , x k−1 ) contains at least a edges.
Proof. We prove the lemma by induction on k. The base case k = 1 is trivial. Let k ≥ 2. By the induction hypothesis, the process continues for at least k − 1 steps. It remains to show that N(x 1 , . . . , x k−1 ) contains at least a edges, because this would also imply that N(x 1 , . . . , x k−1 ) = ∅ and hence the process continues for at least k steps. For 1 ≤ i ≤ k − 1, let S i be the set of vertices which are adjacent to x 1 , . . . , x i−1 but not adjacent to x i . In particular, S 1 is just the set of vertices not adjacent to x 1 and Since e(G) = e(N) + e(S) + e(N, S), we have 2e(S) = 2e(G) − 2e(N) − 2e(N, S). Plugging this into (1) and rearranging, we get We have e(N, S) ≤ |N| · |S| = (n − s)s. Also, by Cauchy-Schwarz, Plugging this into (2) gives .
≥ a, as required.

Lemma 2.3.
Let G be a graph with n vertices and a edges. Let s ≥ 1 and suppose that a ≥ 2s 2 . Then G contains a forest F with s vertices and at least s − 1 − sn a edges. Proof. Let C 1 , . . . , C m be the connected components of G with |C 1 | ≥ · · · ≥ |C m |. Let ≥ 1 be the minimal integer satisfying |C 1 | + · · · + |C | ≥ s. If ≤ 1 + sn a then take F to be a forest contained in C 1 ∪ · · · ∪ C having s vertices and connected components. Suppose now by contradiction that > 1 + sn a . Set r = |C 1 | + · · · + |C −1 |. Then r < s and where the last inequality uses a ≥ 2s 2 . We got a contradiction to e(G) = a.
We are now ready to prove Theorem 2.
1. An overview of the proof can be found in Section 1.1.
Proof of Theorem 2.1. The proof is by induction on n. Fix constants k c c 1 C, to be chosen implicitly later. Suppose first that a ≤ (ck + 1) 2 n. In this case we proceed as follows. If and take H to consist of all edges that touch x 1 , . . . , x k+1 . Then H is chordal and where the last inequality holds as C c 1 , c and a ≤ (ck By the induction hypothesis with parameter a = a + c 1 √ n, G contains a chordal subgraph H with Squaring and plugging in the value of a , we get Cancelling the term 2(k + 1)a/k from both sides and rearranging, we see that it is enough to have which holds because a ≤ (ck + 1) 2 n and c 1 c.
For the rest of the proof we assume that a ≥ (ck Note that d 0 ≥ (k−1)n k by our assumption that a ≥ (ck + 1) 2 n.

Claim 2.4. If the statement of the theorem does not hold, then G[N] contains a forest F with v(F)
Proof. We consider two cases. Suppose first that there is common neighbours, where the last inequality holds by the assumption a ≥ (ck + 1) 2 n. Take F to be the star whose centre is x k and whose leaves are and then the subgraph consisting of all edges touching edges, so the assertion of the theorem holds. Hence, we may assume that d(y) ≤ d 0 for every y ∈ N(x 1 , . . . , x k ). This means that as required by the claim. Also, F has the right number of edges, as e(F) = v(F) − 1.
The second case is that d(y) ≤ d 0 for every y ∈ N. Since e(N) ≥ a 2 ≥ 2n, we can apply Lemma 2.3 to G[N] with parameters a 2 and s = √ n to obtain a forest F with √ n vertices and at least v(F) − 1 − 2n 3/2 a edges. All vertices in F have degree at most d 0 , so (3) holds. We continue with the proof of the theorem. Let F be the forest given by Claim 2.4. Let G be the graph obtained from G by deleting the t : where the last inequality uses that t ≥ √ n and c k, so that Let H be the subgraph of G obtained by adding to H the edges of the clique x 1 , . . . , x k−1 , the edges between x 1 , . . . , x k−1 and V(F), and the edges of F. Then H is chordal. To complete the proof, it suffices to verify that By the definition of H, we have Note that and therefore, For convenience, set h : So it remains to verify that the right-hand side of (4) is at least as large as Cancel the terms (k − 1/k)n, k+1 2 which appear in both expressions. Also, we may drop the terms C √ n − t, C √ n. After rearranging, we get the inequality By squaring and plugging in the value of a , we get: Both sides of the inequality (5) have the terms 2(k+1) k a and − 2t k · 2(k+1)a k . We can also drop the negative term 2th k on the right-hand side. After rearranging, we get the inequality We as n ≤ a ≤ n 2 . So the right-hand side of (6) is O k (n). On the other hand, the left-hand side is larger than cn 2 because t ≥ √ n ≥ √ n/2. So (6) holds because c k, as required.
, in contradiction to the choice of r. So g 1 (e(G )) ≥ r − 1. By the induction hypothesis, G contains a chordal subgraph H with at least r − 2 edges. Now, H + {xy} is a chordal subgraph of G with at least r − 1 edges, as required.
The rest of this section is dedicated to proving Theorem 1.5. Recall that for n ≥ 1 and m ≥ t 2 (n) + 1, we define g 2 (n, m) := min t,r (2n − t + r), where the minimum is taken over all integers t, r ≥ 0 satisfying t(n − t) + t 2 (r) ≥ m. We start by proving the upper bound in Theorem 1.5. First we claim that g 2 (n, m) ≤ 2n. Indeed, for t = r = 2n 3 we have t(n − t) + t 2 (r) = t 3 (n) ≥ m, so g 2 (n, m) ≤ 2n − t + r = 2n, as required. Now take t, r such that t(n − t) + t 2 (r) ≥ m and g 2 (n, m) = 2n − t + r. Since g 2 (n, m) ≤ 2n, we have t ≥ r. Take a complete bipartite graph with sides X of size t and Y of size n − t and add a copy of T 2 (r) with sides A, B inside X. The resulting graph G has t(n − t) + t 2 (r) ≥ m edges. Let H be a chordal subgraph of G. Then e H (A, B) ≤ |A| + |B| − 1 = r − 1, e H (A, Y) ≤ |A| + |Y| − 1 and e H (X \ A, Y) ≤ |X| − |A| + |Y| − 1, because each of these bipartite graphs is induced in G, so its intersection with H is a forest. Overall, we got that e(H) ≤ r − 1 + |X| + 2|Y| − 2 = 2n − t + r − 3 = g 2 (n, m) − 3. This shows that f (n, m) ≤ g 2 (n, m) − 3, as required. To prove the lower bound in Theorem 1.5, we prove the following stronger claim. Theorem 3.1. Let G be a graph with n vertices and m ≥ t 2 (n) + 1 edges, and let x, y, z be a triangle in G. Then G has a chordal subgraph with at least g 2 (n, m) − 3 edges which contains the edges xy, xz, yz.
We need the following facts about the numbers g 2 (n, m).
Proof of Theorem 3.1. The proof is by induction on n. The base cases n = 1, 2 are trivial because for these n there is no graph on n vertices with t 2 (n) + 1 edges. The case n = 3 is also easy to verify. So from now on let n ≥ 4. Let x, y, z be a triangle in G. We consider several cases. After dealing with each case, we will assume in all subsequent cases that this case does not hold. g 2 (n, m). In this case, take H to be the graph consisting of all edges touching x, y, z. This graph is chordal and clearly contains the edges of the triangle x, y, z. Also, We claim that if cases 1-2 do not hold then m ≥ t 2 (n) + 2. Indeed, suppose by contradiction that m = t 2 (n) + 1. We have g 2 (n, t 2 (n) contradicting that case 2 does not hold. So m ≥ t 2 (n) + 2.

Case 3:
There are distinct u, v ∈ {x, y, z} such that the edge uv is on exactly one triangle (namely the triangle x, y, z). Without loss of generality, suppose that u = x, v = y. Since case 2 does not hold, we have d(x) + d(y) ≥ n + 1. On the other hand, since xy is on exactly one triangle, it must be that Suppose now that z is not isolated in H , and let w ∈ V(G) \ {x, y, z} such that zw ∈ E(H ). As mentioned above, w is adjacent in G to either x or y; without loss of generality it is adjacent to x. Take H = H + {xw, xy, xz, yz}. By adding to H first x and then y, we always add a simplicial vertex. Hence, H is chordal.
Theorem 4.1. Let G be a graph with n vertices and t 3 (n) + 1 edges, and let X = {x 1 , x 2 , x 3 , x 4 } be a 4-clique in G. Then G has a chordal subgraph with at least g 3 (n) − 6 edges which contains the edges of the clique X.
We need some simple facts on the numbers t 3 (n) and g 3 (n).

Lemma 4.2. For every n
Proof. For i = 1, 2, 3, T 3 (n − i) is obtained from T 3 (n) by deleting one vertex from each of the i largest classes of T 3 (n). For i = 1, the deleted vertex has degree 2n 3 . For i = 2, the sum of degrees of the deleted vertices is 4n 3 , and these vertices are adjacent. For i = 3, the sum of degrees of the deleted vertices is 2n, and they form a triangle. Finally, Proof. g 3 (n) − g 3 (n − 1) = 3 − n 3 + n−1 3 , and it is easy to see that n 3 − n−1 3 equals 0 if n ≡ 0, 2 (mod 3) and equals 1 if n ≡ 1 (mod 3).
We also need the following simple lemma saying that in a graph with t 3 (n) + 1 edges, we can find two 4-cliques sharing 3 vertices. This fact is originally due to Dirac [2]. For completeness, we include a proof.
Proof. The proof is by induction on n. For the base case n = 5, take a 4-clique and notice that the remaining vertex must send at least 3 edges to this 4-clique as t 3 (5) + 1 = 9. Let n ≥ 6. Take a 4clique x 1 , . . . , x 4 . If there is a vertex outside of x 1 , . . . , x 4 which has three neighbours in x 1 , . . . , x 4 then we are done. Else, d( We consider three cases: Then d(x 1 ) + · · · + d(x 4 ) ≤ 2n + 3 by (11). Let G be the graph obtained from G by deleting x 2 , x 3 , x 4 and all edges touching In particular, n − 3 ≥ 4 (else G cannot have more than t 3 (n − 3) edges). By the induction hypothesis (applied to an arbitrary 4-clique in G ), G contains a chordal subgraph H with e(H ) ≥ g 3 (n − 3) − 6. Take z ∈ A 2 and assume without loss of generality that z is adjacent to Also, H is chordal: adding x 1 z to H keeps it chordal because x 1 is isolated in H , and then by adding x 2 , x 3 , x 4 in this order we always add a simplicial vertex. By pigeonhole, two of the y i 's have a common neighbour. Without loss of generality, we may assume that y 1 , y 2 are both adjacent to x k for some k ∈ [4]. Suppose that y 1 is also adjacent to x . Let Also, d(x 1 ) + · · · + d(x 4 ) ≤ n + 8 by (11). Let G be the graph obtained from G by deleting Then e(H) = e(H ) + 9 ≥ g 3 (n) − 6, and it is easy to check that H is chordal. Claim 4.6. We may assume that for every i ∈ [4], the following holds. If d(x i ) ≥ 2n 3 + 1, then there is y ∈ V(G) \ {x 1 , . . . , x 4 } such that y is adjacent to x i and to at least two vertices in X \ {x i }.
Proof of Claim 4.6. Without loss of generality, i = 1. Suppose that d(x 1 ) ≥ 2n 3 + 1 but there is no y as above. We will show that the assertion of Theorem 4.1 holds. For 0 ≤ j ≤ 3, let A j be the set of vertices z ∈ V(G) \ {x 2 , x 3 , x 4 } which are adjacent to exactly j of the vertices x 2 , x 3 , x 4 , and let a j = |A j |. Note that x 1 ∈ A 3 . By Claim 4.5, we may assume that there exists z ∈ V(G) \ {x 1 , . . . , x 4 } which is adjacent to three of the vertices x 1 , . . . , x 4 . By assumption, these three vertices cannot include x 1 , so we must have z ∈ A 3 . Hence, We have a 0 + a 1 + a 2 + a 3 = n − 3 and Pick a set of edges F as follows: • Suppose first that a 0 = 0. We claim that On the other hand, d(x 1 ) ≥ 2n 3 + 1 by assumption. So 2n 3 ≤ 2, which is false for every n ≥ 5. Now pick an arbitrary v ∈ A 1 ∪ A 2 , and suppose without loss of generality that v is adjacent to x 4 . Take F to be the set of all edges between x 4 and N( • If a 0 ≥ 1 then F is the set of all edges between x 4 and N( In both cases we have |F| ≤ d(x 4 ) − a 3 − 3 + a 0 . Let G be the graph obtained from G by deleting x 1 , x 2 , x 3 and all edges in F. So v(G ) = n − 3. To count the deleted edges, note that there are d(x 1 ) + d(x 2 ) + d(x 3 ) − 3 edges touching x 1 , x 2 , x 3 , and that the edges in F do not touch x 1 , x 2 , x 3 . So the number of deleted edges is where the last inequality uses (12). So e(G ) ≥ e(G) − 2n + 3 = t 3 (n) + 1 − 2n + 3 = t 3 (n − 3) + 1. By the induction hypothesis, G contains a chordal subgraph H with e(H ) ≥ g 3 (n − 3) − 6. Define a subgraph H of G as follows:
By Claim 4.5, we may assume that there is y 0 ∈ V(G) \ {x 1 , . . . , x 4 } which is adjacent to at least 3 of the vertices x 1 , . . . , x 4 ; say to x 2 , x 3 , x 4 . By Claim 4.7, we may assume that x i is not deletable for any i ∈ [4]. Since x 1 is not deletable, d(x 1 ) ≥ 2n 3 + 1. By Claim 4.6, there is z 0 ∈ V(G) \ {x 1 , . . . , x 4 } which is adjacent to x 1 and to at least two of the vertices x 2 , x 3 , x 4 ; without loss of generality, z 0 is adjacent to x 3 , x 4 . Since x 2 is not deletable, we have d(x 2 ) ≥ 2n 3 + 1. Also, we may assume that y 0 = z 0 because else we would have a 5-clique containing x 1 , . . . , x 4 . Claim 4.11. We may assume that d(x i ) ≤ 2n 3 for each i = 3, 4. Proof. By Claim 4.10, we may assume that d( If d(y 0 ) ≤ 2n 3 then we can delete y 0 , apply induction to the remaining graph to find a chordal subgraph H with at least g 3 (n − 1) − 6 edges which contains the edges of the clique x 1 , . . . , x 4 and then add the edges x 2 y 0 , x 3 y 0 , x 4 y 0 to H to conclude the proof. So suppose that d(y 0 ) ≥ 2n 3 + 1. Let v / ∈ {x 3 , x 4 } be a common neighbour of x 2 and y 0 . Such a vertex v exists because d(x 2 ) + d(y 0 ) ≥ 2 · 2n 3 + 1 ≥ n + 3, where the last inequality holds for every n ≥ 5. Note that v = x 1 because x 1 is not adjacent to y 0 , as otherwise x 1 , . . . , x 4 , y 0 would be a 5-clique. Similarly, v = z 0 because v is adjacent to x 2 and z 0 is adjacent to x 1 , x 3 , x 4 (so otherwise x 1 , . . . , x 4 , z 0 would be a 5-clique).
Suppose first that x 2 , y 0 , v have no common neighbour. Then d(

Concluding remarks
In this paper we study the maximal value f (n, m) such that every graph with n vertices and m edges has a chordal subgraph with at least f (n, m) edges. We determine this function asymptotically for all m and exactly for m ≤ t 3 (n) + 1. Our results suggest the following conjecture on the value of f (n, m) for general m. where the minimum is taken over all t, r ≥ 0 satisfying t k−1 (n − t) + t(n − t) + t 2 (r) ≥ m.
It seems very likely that some progress on this conjecture can be achieved using our techniques together with a careful case analysis, but it would be interesting to find a proof which avoids such a case analysis as much as possible.  (1) in the special case m = t k (n) + 1, proving that f (n, t k (n) + 1) = (k − 1/k)n − O k (1).