Extremal Graphs of Girth Five

David K. Garnick Y.H. Harris Kwong Felix Lazebnik

Research Report 93-1 February 1993

Department of Mathematical Sciences University of Delaware Newark, Delaware 19716 Abstract

We derive bounds for f(v), the maximum number of edges in a graph on v vertices that contains neither three-cycles nor four-cycles. Also, we give the exact value of f(v) for all v up to 24 and constructive lower bounds for all v up to 200. Extremal Graphs of Girth Five

David K. Garnick1 Department of Computer Science Bowdoin College Brunswick, Maine 04011 Y.H. Harris Kwong Department of Mathematics and Computer Science SUNY College at Fredonia Fredonia, New York 14063 Felix Lazebnik Department of Mathematical Sciences University of Delaware Newark, Delaware 19716

February 1993

1Research supported by the National Science Foundation under Grant CCR-8909357. Contents

1 Introduction 1

2 Theoretical Results 3

3 Values of f(v) for v ≤ 24 7

4 Constructive Lower Bounds on f(v) for v ≤ 200 16

5 Closing Remarks 18

A Alternate Proofs of f(16) = 28 and f(23) = 50 22

i Chapter 1

Introduction

This paper investigates the values of f(v), the maximum number of edges in a graph of order v and girth at least 5. For small values of v we also enumerate the set of extremal graphs. This problem has been mentioned several times by P. Erd˝os(for instance, in [7]) who conjectured that f(v) = (1/2 + o(1))3/2v3/2. We begin with some basic definitions; undefined terms can be found in any standard textbook on graph theory (e.g. [14]). Let G be a simple graph with vertex-set V (G) and edge-set E(G). The values |V (G)| = v and |E(G)| = e are called the order and size of G, respectively. Let d(x) = dG(x) denote the degree of a vertex x of G, and let δ(G) and ∆(G) be the minimum and maximum degrees of vertices of G. The argument G will be left off when the intended graph is obvious. Given x ∈ V , N(x) = NG(x) and N[x] = NG[x] = N(x) ∪ {x} denote the (open) neighborhood 0 0 S 0 and closed neighborhood, respectively, of x. For V ⊂ V , define N(V ) = x∈V 0 N(x) and hV i as the subgraph induced by V 0. By Cn, n ≥ 3, we denote an n-cycle, i.e. a cycle with n vertices. We will also refer to a 3-cycle as a triangle and to a 4-cycle as a quadrilateral. Pn denotes a path of n vertices. The girth g(G) of a graph G is the length of the shortest cycle in G. If G has no cycles (that is, G is a forest), then g(G) is defined as ∞. Given graphs G1,G2,...,Gk, let ex(v; G1,G2,...,Gk) denote the greatest size of a graph of order v which contains no subgraph isomorphic to some Gi, 1 ≤ i ≤ k. One of the main classes of problems in extremal graph theory, known as Turán-type problems, is for given v, G1,G2,...,Gk to determine explicitly the function ex(v; G1,G2,...,Gk), or to find its asymptotic behavior for large values of v. Thus, the problem we consider in this paper, that of finding the maximum size of a graph of girth at least 5, can be stated as finding the value of ex(v; C3,C4). Exact results for all values of v are known for only a few instances of Turán-type problems. Sometimes the exact value is computed for particular values of v, but typically some lower and upper bounds or asymptotic results are all that is known. Asymptotic results are quite satisfactory if each graph Gi, 1 ≤ i ≤ k, has chromatic number χ(Gi) at least 3. Let this be the case, and let χ = min{χ(Gi) : 1 ≤ i ≤ k}. Then Ã ! 1 v ex(v; G ,G ,...,G ) ∼ (1 − ) , v → ∞. (1.1) 1 2 k χ − 1 2

If at least one of the Gi’s is bipartite, then χ = 2, and there is no general result similar to (1.1). Excellent references on the subject are [1] and [22] (pp. 161–200).

1 2 It is well known that ex(v; C3) = bv /4c, and the extremal graph is the complete bipartite graph Kbv/2c,dv/2e. 2 The exact value of ex(v; C4) is known for all values of v of the form v = q + q + 1, where q is a power of 2 [9], or a prime power exceeding 13 [10], and it is equal to q(q + 1)2/2. The extremal graphs for these values of v are known, and were constructed in [3, 8]. For 1 ≤ v ≤ 21, the values of ex(v; C4) and the corresponding extremal graphs can be found in [4]. It is well known that 3/2 ex(v; C4) = (1/2 + o(1))v (see [1, 8]). It is important to note that attempts to construct extremal graphs for ex(v; C3,C4) by destroying all 4-cycles in the extremal graphs for ex(v; C3), or by destroying all 3-cycles in the extremal graphs for ex(v; C4), fail; neither method leads to graphs of order v with f(v) edges. In Chapter 2 we derive upper and lower bounds on the value of f(v), and provide some remarks on the structure of extremal graphs. In Chapter 3 we present the exact values of f(v) for all v up to 24; we also enumerate all of the extremal graphs of order less than 11. In Chapter 4 we provide a table of constructive lower bounds for f(v), 25 ≤ v ≤ 200, which are better than the theoretical lower bounds.

2 Chapter 2

Theoretical Results

In this chapter we present some theoretical results about f(v) and the structure of the extremal graphs. Many of them will be used in the subsequent chapters. We call a graph G of order v extremal if g(G) ≥ 5 and e = e(G) = f(v). The following statement gives some simple facts about extremal graphs.

Proposition 2.1 Let G be an extremal graph of order v. Then

1. The diameter of G is at most 3. 2. If d(x) = δ(G) = 1, then the graph G − {x} has diameter at most 2.

Proof: Let x, y be two vertices of G of distance at least 4. Since xy is not an edge of G we can add this edge to E(G). The obtained graph G0 = G + xy has girth g(G0) ≥ 5 and one more edge than G. This contradicts the assumption that G is extremal and proves part 1. Let a, b be two vertices of G − {x} of distance at least 3. By deleting the only edge of G incident to x and introducing two new edges xa and xb, we obtain a graph G00 of order v having one more edge than G and girth at least 5. This contradicts the assumption that G is extremal and proves part 2.

It turns out that the extremal graphs of diameter 2 are very rare. In fact, it was claimed in [2] that the only graphs of order v with no 4-cycles and of diameter 2 are:

1. The star K1,v−1;

2. Moore graphs: C5, Petersen graph (the only 3-regular graph of order 10, diameter 2 and girth 5, see G10 in Figure 3.1), Hoﬀman-Singleton graph [16] (the only 7-regular graph of order 50, diameter 2 and girth 5), and a 57-regular graph of order 3250, diameter 2 and girth 5 if such exists (its existence is still an open problem [16]);

3. Polarity graphs (see [3, 8]).

Remark: The only graphs from the list above which in addition contain no triangles and are regular are the Moore graphs. It is also known that a graph of diameter k ≥ 1 and girth 2k + 1 must be regular [23]. P We now derive an upper bound on f(v). Let d¯ = d¯ = 1 d(x) be the average degree P G v x∈V (G) 2 ¯ 2 of G, and let σ = x∈V (G)[d − d(x)] . By Di = Di(G), i = 1, 2, 3, we denote the number of unordered pairs of vertices of G of distance i apart.

3 Theorem 2.2 Let G be an extremal graph of order v and size e. Then 1q 1 √ f(v) = e = v2(v − 1) − vσ2 − 2vD ≤ v v − 1 (2.1) 2 3 2 The inequality in (2.1) becomes an equality if and only if G is an isolated vertex or G is regular and of diameter 2, i.e., G is a Moore graph.

Proof: According to Proposition 2.1 G is connected and of diameter at most 3. Since G is {C3,C4}- P ¡d(x)¢ free, we have D2 = x∈V (G) 2 . Therefore Ã ! Ã ! v X d(x) = D + D + D = e + + D . (2.2) 2 1 2 3 2 3 x∈V (G) P P ¯ 2 2 2 Using the standard equalities 2e = x∈V (G) d(x), d = 2e/v and x∈V (G) d(x) = σ + 4e /v, we can rewrite (2.2) as Ã ! v = σ2/2 + 2e2/v + D . (2.3) 2 3 Solving (2.3) with respect to e we obtain (2.1). The statement about the equality sign in the inequality (2.1) follows immediately from the proof above and the Remark following Proposition 2.1.

Remark: The inequality (2.1) was found independently in [5], its proof appeared in [6].

Corollary 2.3 Let G be an extremal graph of order v, size e, diameter 3. If G is not regular, then r 1 5 f(v) = e ≤ v2(v − 1) − v. (2.4) 2 2 If, in addition, the average degree of G is an integer, then 1q f(v) = e ≤ v2(v − 1) − 4v. (2.5) 2

Proof: Since G is not regular, then at least one vertex of G has degree strictly greater than the average degree d = d¯G, and another vertex has degree strictly less than d. We want to show that 2 1 in this case σ ≥ 2 . If for at least one vertex x, |d − d(x)| ≥ 1, then it is obvious. Therefore we can assume that d is not an integer and |d − d(x)| < 1 for all vertices of G. Let m be the number of vertices whose degrees are equal to bdc, and let n be the number of vertices whose degrees are dde. Then m + n = v, m, n ≥ 1. Let a = d − bdc and b = dde − d. Then a + b = 1, and 2 2 2 2 2 2 1 σ = ma + nb ≥ a + b ≥ (a + b) /2 = 2 . Since D3 ≥ 1, the inequality (2.4) follows from (2.1). If the average degree d is an integer, but G is not regular, then σ2 ≥ 2 and (2.1) implies (2.5).

2 Now we derive a lower bound for f(v). Let q be a prime power, and let vq = q + q + 1, and eq = (q + 1)vq. By Bq we denote the point-line incidence bipartite graph of the projective plane P GL(2, q). More precisely, the partite sets of Bq represent the set of points and the set of lines of P GL(2, q), and the edges of Bq correspond to the pairs of incident points and lines. Then Bq is a (q + 1)-regular bipartite graph of order 2vq and size eq. Also g(Bq) ≥ 6 since bipartite Bq has neither 3-cycles nor 5-cycles, and the existence of a 4-cycle in Bq would mean that in P GL(2, q) there are two distinct lines passing through two distinct points. It is easy to show that g(Bq) = 6.

4 Theorem 2.4 Let G be an extremal graph of order v and size e. Let q be the largest prime power such that 2vq ≤ v. Then

f(v) = e ≥ eq + 2(v − 2vq) = 2v + (q − 3)vq. (2.6)

Proof: Consider a graph obtained from Bq by joining to V (Bq) a set of v − 2vq isolated vertices and connecting each of these isolated vertex to two nonadjacent vertices of Bq taken from diﬀerent partite sets (such two vertices always exist, since Bq is not a complete bipartite graph). If the chosen pairs of vertices of Bq are distinct for distinct isolated vertices, then the obtained graph H 2 of order v and size eq + 2(v − 2vq) will have girth at least 5. There are q vq pairs of disjoint vertices 2 of Bq in which two vertices in the pair belong to diﬀerent partite sets. We claim that q vq > v−2vq. 2 Indeed, if it is not the case and q > 2, then v ≥ (q +2)vq > 8vq. According to Bertrand’s Postulate [15], for any integer n ≥ 1, there is at least one prime number p such that n < p ≤ 2n. Let n = q, and let p be a prime satisfying the inequality q < p ≤ 2q. Then

2 2vq < 2vp ≤ 8q + 4q + 2 < 8vq < v, which contradicts our choice of q in the statement of the theorem. For q = 2, 2v2 = 14 ≤ v < 2 2v3 = 26 implies that v − 2v2 = v − 14 < 12 < 28 = 2 v2. Therefore, for all prime powers q, 2 q vq > v − 2vq, and the construction of H described above is possible.

Combining the results of Theorems 2.2 and Theorem 2.4, we obtain Corollary 2.5 1 f(v) f(v) 1 √ ≤ lim inf ≤ lim sup ≤ . (2.7) 2 2 v→∞ v3/2 v→∞ v3/2 2

When v = 2q2 another lower bound can be derived from graphs constructed by Murty [19]:

f(2q2) ≥ q2(q + 2)

This gives the same asymptotic bound as in Corollary 2.5. We next deﬁne a restricted type of tree; many of the proofs in Chapter 3 rely on the presence of these trees in the extremal graphs. Consider a vertex x of maximum degree ∆ in a {C3,C4}-free graph G. Let the neighborhood of x be N(x) = {x1, x2, . . . , x∆}. Clearly, N(x) is an independent set of vertices. Furthermore, the sets of vertices N(xi) − {x}, 1 ≤ i ≤ ∆, are pairwise disjoint; otherwise there would be a quadrilateral in G. This motivates the notion of an (m, n)-star Sm,n, which is deﬁned to be the tree in which the root (center) has m children, and each of the root’s children has n children, all of which are leaves. The subtree containing a child of the root and all its n children is called a branch of Sm,n. We will make use of the following easily established facts:

1. |V (Sm,n)| = 1 + m + mn and |E(Sm,n)| = m + mn.

2. Every {C3,C4}-free graph G with at least 5 vertices, ∆(G) = ∆, and δ(G) = δ, contains S∆,δ−1.

3. For any nonadjacent vertices t and u in Sm,n that are not both leaves, Sm,n + tu contains a C3 or C4.

5 4. In any {C3,C4}-free graph G containing an (m, n)-star S, no vertex in G − S can be adjacent to two siblings in S. In other words, every set of siblings has a unique common neighbor, namely, their parent.

2 Proposition 2.6 For all {C3,C4}-free graphs G, v ≥ 1 + ∆δ ≥ 1 + δ .

2 Proof: Since G contains S∆,δ−1, we have v ≥ |V (S∆,δ−1)| = 1 + ∆δ ≥ 1 + δ .

The lower bounds of ∆ and δ of any extremal graph will play important roles in our discussion.

Proposition 2.7 For all {C3,C4}-free graphs G on v vertices and e edges, δ ≥ e − f(v − 1) and ∆ ≥ d2e/ve.

Proof: Suppose there exists a vertex x in G such that d(x) < e − f(v − 1). Then G − x would have more than f(v − 1) edges, contradicting the deﬁnition of f(v − 1). This proves the ﬁrst inequality. The second one follows from the pigeonhole principle and it simply asserts that G contains at least one vertex of degree greater or equal to its average degree.

The next result will be used later to obtain upper bounds of f(v).

Proposition 2.8 For any {C3,C4}-free graph G of order v, v ≥ 1 + d2f(v)/ve(f(v) − f(v − 1)). Proof: The result holds for v ≤ 4 from direct computation. For v ≥ 5, the result follows immediately from Propositions 2.6 and 2.7.

6 Chapter 3

Values of f(v) for v ≤ 24

√ Theorem 2.2 states√ that f(v) ≤ bv v − 1/2c. For 1 ≤ v ≤ 10, we have constructed {C3,C4}-free graphs with bv v − 1/2c edges. These graphs are shown in Figure 3.1. We summarize this result

G G G G G G G 1 2 3 4a 4b 5 6a

G G G G G 6b 7 8 9 10

Figure 3.1: Extremal graphs with v ≤ 10 in the following theorem. √ Theorem 3.1 For 1 ≤ v ≤ 10, f(v) = bv v − 1/2c. This yields the following values for f(v):

v 1 2 3 4 5 6 7 8 9 10 f(v) 0 1 2 3 5 6 8 10 12 15

Next, we investigate the number of extremal graphs on v ≤ 10 vertices; Figure 3.1 shows all ∗ such extremal graphs. Let Fv denote the set of {C3,C4}-free graphs of order v, and let Fv denote ∗ the set of extremal graphs of order v. Define F (v) = |Fv |. We adopt the following notation. Given a rooted tree T of height 2, denote its root by r, the children of r by r1, r2,... , and the children of ri (that is, the leaves of T in the i-th branch) by r , r ,... . Denote by R the set of leaves in the i-th branch, that is, R = N(r ) − {r}. Define i,1 Si,2 i i i R = i Ri as the set of leaves in T . Suppose a {C3,C4}-free graph G contains an (m, n)-star S and a vertex z 6∈ V (S). Then either (i) z is adjacent to only one vertex in S, or (ii) z is adjacent to two leaves from different branches

7 in S. If riz ∈ E(G), then we will also consider z as a vertex in S; that is, z is regarded as a leaf in Ri. Under this convention, every child of the root in any Sm,n has at least n children of its own.

Theorem 3.2 For 1 ≤ v ≤ 10, we have the following values for F (v):

v 1 2 3 4 5 6 7 8 9 10 F (v) 1 1 1 2 1 2 1 1 1 1

Proof: The theorem is trivial when v ≤ 5. For v ≥ 6, we use Propositions 2.6–2.7 to obtain the following possible values of δ and ∆: v (δ, ∆) 6 (1,2), (1,3), (1,4), (1,5), (2,2) 7 (2,3) 8 (2,3) 9 (2,3), (2,4) 10 (3,3)

∗ Let G ∈ F6 . If G is 2-regular, then it must be C6. Otherwise, let x be a pendant vertex of G. ∼ ∼ + + Since f(5) = 5 and F (5) = 1, we have G − x = C5 or G = C5 , where C5 is obtained from C5 by joining an additional vertex to one of its 5 vertices. ∗ Every G ∈ F7 contains S3,1, which has 7 vertices and 6 edges. The remaining two edges must connect the leaves, thus yielding G7 (see Figure 3.1). Assume r2,1 is adjacent to both r1,1 and r3,1 in G7. ∗ Let z1 be a vertex of degree 2 in G ∈ F8 . Since |E(G)| = 10, f(7) = 8 and F (7) = 1, we have ∼ ∼ G − z1 = G7 and G = G8, where G8 (Figure 3.1) is obtained from G7 by joining z1 to r1,1 and r3. ∗ There exists a vertex z2 of degree 2 in every G ∈ F9 . Since |E(G)| = 12, f(8) = 10 and ∼ F (8) = 1, we have G − z2 = G8. Suppose z1z2 6∈ E(G), then z2 must be adjacent to r1 and r3,1; 0 yielding G9 in Figure 3.1. If z1z2 ∈ E(G), then z2 is also adjacent to r2, yielding G9. By considering 0 the (3,1)-star in G9 rooted at r1,1, with children {r2,1, r1, z1} and grandchildren {r2, r, r3}, it is clear 0 ∼ that G9 = G9. Thus F (9) = 1. ∗ ∼ Finally, if G ∈ F10, then G is 3-regular. So G − x = G9 for every vertex x ∈ V . Note that δ(G9) = 2, ∆(G9) = 3 and G9 has exactly 3 vertices of degree 2. Thus x must be adjacent to these 2 3 vertices. Therefore F (10) = 1. Since σ = D3 = 0, G has diameter 2; The only 3-regular graph on 10 vertices with diameter 2 is the Petersen graph, which is displayed as G10 in Figure 3.1.

We now provide the exact value of f(v), 11 ≤ v ≤ 20, with a case-by-case proof based on the values of v. To show that f(v) = m, we use the following strategy. First, with the aid of computer programs (see Chapter 4 and [12]), we generate a graph Gv ∈ Fv with m edges to show that f(v) ≥ m. Next, assume that there is a graph G ∈ Fv with more than m edges. Using Propositions 2.6–2.8, we compute the possible values of δ and ∆. Finally, the adjacency of the vertices in G − N[r], where r is the root of a S∆,δ−1 contained in G, leads to a contradiction. Theorem 3.3 The values of f(v) for 11 ≤ v ≤ 20 are as follows:

v 11 12 13 14 15 16 17 18 19 20 f(v) 16 18 21 23 26 28 31 34 38 41

8 Proof: Following are the cases based on the values of v. f(11) = 16. The graph G11 with 16 edges (in Figure 3.2) belongs to F11. Suppose f(11) ≥ 17. It follows from Proposition 2.7 and Theorem 3.1 that for all graphs G ∈ F11 with 17 edges, δ ≥ 2 and ∆ ≥ 4. Proposition 2.6 implies that δ = 2. So there exists a vertex x with d(x) = 2. From the proof of Theorem 3.2, G − x is isomorphic to the Petersen graph, which has diameter 2. Thus the vertices in N[x] belong to a triangle or quadrilateral. Therefore f(11) < 17.

G G G 11 12 13

Figure 3.2: Extremal graphs on 11, 12, and 13 vertices

f(12) = 18 and f(13) = 21. Since G12 ∈ F12 (see Figure 3.2), f(12) ≥ 18. Since f(12) ≥ 19 would contradict Proposition 2.8, we must have f(12) = 18. Similarly, the graph G13 in Figure 3.2 shows that f(13) ≥ 21; and Proposition 2.8 implies that f(13) < 22. f(14) = 23. G14 in Figure 3.3 demonstrates that f(14) ≥ 23. Assume there exists G ∈ F14 with 24 edges. Propositions 2.6 and 2.7 imply that δ = 3 and ∆ = 4. Thus G has 8 vertices of degree 3 and 6 vertices of degree 4. Further, G cannot contain a pair of adjacent vertices x and y each having degree 3, since G − {x, y} would have 19 edges, contradicting f(12) = 18. Together with X d(z) = 24, z∈V d(z)=3 we conclude that every edge of G connects a degree 3 vertex to a degree 4 vertex. If for each degree 3 vertex we represent its neighborhood as a triple, then the existence of G implies the existence of a design of 8 triples on 6 elements (the six vertices of degree 4), where each element occurs in 4 triples and each distinct pair of elements occurs in at most 1 triple. But since 6 elements constitute 15 distinct pairs, and each triple speciﬁes 3 pairs, such 8 triples will specify 8 × 3 = 24 pairs, and therefore cannot exist. Thus f(14) < 24. f(15) = 26. The graph G15 in Figure 3.3 implies that f(15) ≥ 26. On the other hand, f(15) < 27 follows from Proposition 2.8. Hence, f(15) = 26.

9 G G G 14 15 16

Figure 3.3: Extremal graphs on 14, 15, and 16 vertices

f(16) = 28. We established f(16) ≥ 28 by construction; see G16 in Figure 3.3. Suppose there exists G ∈ F16 with e = 29; then δ ≥ 3 and ∆ ≥ 4. First we establish that G cannot contain a vertex of degree 5. If there is a degree 5 vertex, x, then it is at the center of the (5,2) star which has 16 vertices. Hence, the vertices in N(x) must all be of degree 3. Since all possible degree sequences for G which include at least 1 vertex of degree 5 also include at least 7 vertices of degree 3, then some of the leaves are also of degree 3. This implies that there will be an adjacent pair of degree 3 vertices. But adjacent vertices of degree 3 implies that f(14) > 23, which is a contradiction. Thus, there are no degree 5 vertices in G, implying there are 6 vertices of degree 3, and 10 of degree 4. We now look at the placement of the degree 3 and degree 4 vertices in a star. Having 10 vertices of degree 4, there must be at least 1 degree 4 vertex, x, that is adjacent to 3 other degree 4 vertices. Further, x must have a degree 3 neighbor; otherwise G would contain S4,3 which has 17 vertices. By the same argument, each branch with 3 leaves must contain at least one vertex of degree 3. In Figure 3.4, where each vertex is labeled with its degree, we call the left branch b1, and proceed clockwise so that the branch with 2 leaves is b4. Since one of the branches must have at least 2 leaves of degree 3, we now have two cases to consider depending on whether there is a branch that has 3 leaves of degree 3. Case 1, where there is no branch with 3 leaves of degree 3, is depicted in Figure 3.4a. The 4-leaves in b3 are each adjacent to a leaf in each other branch. Since they are adjacent to both of the leaves in b4, the 3-leaf in b3 must be adjacent to the 4-leaves in b1 and b2. Therefore, the 4-leaves in b3 must be adjacent to the 3-leaves in b1 and b2. Now it is not possible to place the remaining edges amongst the leaves without creating C3 or C4; to see this, ﬁrst add the edges extending from the 4-leaf in b2 to a leaf in each of b1 and b4, and then attempt to identify the remaining neighbors of the leaves in b4. Case 2, where there are 3 leaves of degree 3 in b1, is depicted in Figure 3.4b. Since each 4-leaf in a branch bi is adjacent to exactly one leaf in each branch other than bi, the degree 4 leaves in b2 and b3 are adjacent to the leaves in b4. Further, the leaves in b4 are also adjacent to leaves in b1. Since all of the neighbors of the leaves in b4 have been accounted for, and since degree 3 vertices cannot be adjacent, then it is not possible to ﬁnd 2 more neighbors for the 3-leaf in b2 without completing C3 or C4.

10 3 3 3 4 4 4

4 4 3 3 4 3

3 4 4 4 4 3 4 4 4 4

3 3 4 3 3 4

4 4 4 4

a) Case 1 b) Case 2

Figure 3.4: Degrees of vertices in the star resulting from 29 edges on 16 vertices

Since it is impossible to have suﬃcient edges amongst the leaves in the star, we conclude that f(16) < 29. f(17) = 31 and f(18) = 34. From G17 in Figure 3.5, f(17) ≥ 31. Suppose f(17) ≥ 32. ∗ Propositions 2.6 and 2.7 imply that all G ∈ F17 have ∆ = δ = 4. This in turn implies that f(17) = |E(G)| = 34, which contradicts Proposition 2.8. Again, by construction (G18 in Figure 3.5), ∗ f(18) ≥ 34. If f(18) ≥ 35, then δ ≥ 4. Thus all G ∈ F18 contain at least 36 edges, which contradicts Proposition 2.8. f(19) = 38 and f(20) = 41. From G19 in Figure 3.6, f(19) ≥ 38. Since the outermost 3 vertices (the vertices outside the dodecagon) are mutually distance 3 apart, we can construct G20 ∈ F20 by adding a vertex to G19 that is adjacent to the 3 outer vertices. (In Figure 3.6 dashed lines are used to show the edges from the 20th vertex to G19.) Thus, f(20) ≥ |E(G)| = 41. The upper bounds are obtained by applying Proposition 2.8.

Computation of F (v) is more involved when v > 10. Values of F (v) for 11 ≤ v ≤ 21 can be found in [13]. However, it is easy to show that F (19) = F (20) = 1 based on the order of the (4, 5)-cage. An (r, g)-cage is deﬁned as an r-regular graph with girth g of the smallest order. See [25] for more about cages. The (4, 5)-cage was discovered by Robertson [20]; it has 19 vertices and 38 edges and is known to be unique. The Robertson graph has a unique set of three vertices that are mutually distance 3 apart. These are the vertices outside the dodecagon in G19 (Figure 3.6). By Propositions 2.6 and 2.7, δ = 3 for any extremal graph G of order 20. So G has a degree 3 vertex x such that G − x is isomorphic to the Robertson graph. In addition, N(x) consists of the three vertices (in the Robertson graph) that are pairwise distance 3 apart. Hence F (20) = 1. We conclude this chapter by determining the exact value of f(v) for 21 ≤ v ≤ 24. The proof will be based in part on the structure of the Robertson graph.

Theorem 3.4 The values of f(v) for 21 ≤ v ≤ 24 are as follows:

11 G G 17 18

Figure 3.5: Extremal graphs on 17 and 18 vertices

G G 19 20

Figure 3.6: The unique extremal graphs on 19 and 20 vertices

12 v 21 22 23 24 f(v) 44 47 50 54

Proof: Once again, we prove the theorem on a case-by-case basis. f(21) = 44. See G21 in Figure 3.7 for f(21) ≥ 44. Assume there exists G ∈ F21 with 45 edges. Propositions 2.6 and 2.7 imply that δ = 4 and ∆ = 5. Thus there are 6 vertices of degree 5 and 15 vertices of degree 4 in G. The (5,3)-star S it contains will have 5 leaves of degree 5 and 10 leaves of degree 4; therefore, there must be at least 3 branches in S with at least 2 leaves of degree 4. On any such branch, the two degree 4 leaves, together with their parent, form a P3 of degree 4 vertices. The existence of such a P3 of degree 4 vertices implies that the subgraph induced by the remaining 18 vertices will have 35 edges, contradicting f(18) = 34. Therefore, f(21) < 45.

G G 21 22

Figure 3.7: Extremal graphs on 21 and 22 vertices

f(22) = 47. By construction, f(22) ≥ 47 (Figure 3.7). Assume there exists a triangle- and quadrilateral-free graph G with v = 22, e = 48. By Proposition 2.7, δ ≥ 4 and ∆ ≥ 5. If there exists a vertex x in G with degree greater than 5, then there is a star in G with more than 22 vertices; therefore, G contains 14 vertices of degree 4, and 8 vertices of degree 5. This in turn implies that there are at least 8 edges amongst the degree 4 vertices. Since any graph with 14 vertices and at least 8 edges will contain a path of length 3, that there must be at least one degree 4 vertex adjacent to two other degree 4 vertices; thus there is a P3 in G where all vertices are of degree 4. However, if there is a path P on 3 vertices of degree 4, then G − V (P ) is the unique graph in ∗ F19, which is the Robertson graph. Each of the pendant vertices in P , x and y, has 3 neighbors in the Robertson graph that are mutually distance 3 apart. But the Robertson graph has only one such set of 3 vertices. Therefore, G cannot contain a P3 of degree 4 vertices, thus contradicting the above statement. Therefore f(22) < 48.

13 f(23) = 50. By construction, f(23) ≥ 50 (Figure 3.8). Suppose there exists G ∈ F23 with 51 edges, then δ = 4, ∆ = 5 by Propositions 2.6 and 2.7. Let Q4 and Q5 denote the sets of vertices in G having degree 4 and 5, respectively. We have |Q4| = 13 and |Q5| = 10.

G G 23 24

Figure 3.8: Extremal graphs on 23 and 24 vertices

If G contains a subgraph G0 which is either a path or a star on four degree 4 vertices, then G − V (G0) is the Robertson graph because f(19) = 38 and F (19) = 1. Any pendant vertex x in G0 has 3 neighbors that are mutually distance 3 apart. The Robertson graph has only one such set of 3 vertices, and G0 has at least two pendant vertices; thus, G contains neither a path nor a star on four degree 4 vertices. Consider the (5,3)-star S rooted at any r ∈ Q5. Deﬁne W = V − N[r] and H = hW i. There are at most 2 vertices not in S, thus |Q5 ∩ N(r)| ≤ 2. We may assume d(ri) = 4 for 1 ≤ i ≤ 3. Note that if d(ri) = 4 then |Q5 ∩ Ri| ≥ 1, since G does not contain any K1,3 of degree 4 vertices only. Suppose |Q ∩ N(r)| = 2 for some r ∈ Q . Then d(r ) = 5 and |Q ∩ R | ≥ 3 for i = 4, 5; thus 5 P 5 i 4 i |Q4 ∩ (R1 ∪ R2 ∪ R3)| ≥ 2. Since x∈Ri dH (x) ≥ 12 for i = 4, 5, there exists a path of the form r5,ar4,brc,drc or r4,ar5,brc,drc for some 1 ≤ c ≤ 3 such that V (P ) ⊂ Q4. But such path cannot exist, so |Q5 ∩ N(r)| ≤ 1 for all r ∈ Q5. Suppose |Q5 ∩ N(r)| = 1 for some r ∈ Q5. Let z be the vertex not in S. Assume d(r4) = 4 and d(r ) = 5, then R ⊂ Q . For each i 6= 5, R has only one degree 4 vertex, otherwise since P 5 5 4 i x∈R5 dH (x) = 12, there would be a path of the form r5,ari,briri,c consisting of degree 4 vertices only. Thus d(z) = 4. In addition, the two degree 5 vertices in Ri, i 6= 5, are adjacent to 6 vertices in W − Ri; they are z, the degree 4 vertex in each of Rj, j 6= i, 5, and two vertices in R5. This accounts for all the edges incident to Q4 −R5, but only 8 edges incident to R5. Hence hR5i contains 4 edges, and a C3 or C4 will be formed.

14 Therefore N(r) ⊂ Q4 for all r ∈ Q5. There are eight degree 4 vertices in W , and each degree 5 vertex in Ri is adjacent to four degree 4 vertices in W −Ri. So we have |Q5 ∩Ri| = 1 for all i, which in turn implies that |Q5| ≤ 8. This contradiction shows that G does not exist. Hence f(23) = 50. f(24) = 54. By construction, f(24) ≥ 54 (G24 in Figure 3.8). Proposition 2.8 implies that f(24) < 55. Thus f(24) = 54.

Remark: The proofs of f(16) = 28 and f(23) = 50 are substantially more complicated than other cases. In fact, we found alternate proofs of both values. They are included in Appendix A.

15 Chapter 4

Constructive Lower Bounds on f(v) for v ≤ 200

We have developed and implemented algorithms, combining hill-climbing [24] and backtracking techniques, that attempt to ﬁnd maximal graphs without triangles or quadrilaterals; the algorithms are described in [12]. We have succeeded in generating graphs, with sizes greater than the lower bound presented in Chapter 2, for all orders from 25 to 200. We present these results in Table 4.1.

f(v) 0 1 2 3 4 5 6 7 8 9 0 0 0 1 2 3 5 6 8 10 12 10 15 16 18 21 23 26 28 31 34 38 20 41 44 47 50 54 57 61 65 68 72 30 76 80 85 87 90 94 99 104 109 114 40 120 124 129 134 139 144 150 156 162 168 50 175 176 178 181 185 188 192 195 199 203 60 207 212 216 221 226 231 235 240 245 250 70 255 260 265 270 275 280 285 291 296 301 80 306 311 317 323 329 334 340 346 352 357 90 363 368 374 379 385 391 398 404 410 416 100 422 428 434 440 446 452 458 464 470 476 110 483 489 495 501 508 514 520 526 532 538 120 544 551 558 565 571 578 584 590 596 603 130 610 617 623 630 637 644 651 658 665 672 140 679 686 693 700 707 714 721 728 735 742 150 749 756 763 770 777 784 791 798 805 812 160 819 826 834 841 849 856 863 871 878 886 170 893 901 909 917 925 933 941 948 956 963 180 971 979 986 994 1001 1009 1017 1025 1033 1041 190 1049 1057 1065 1073 1081 1089 1097 1105 1113 1121 200 1129

Table 4.1: Constructive lower bounds on f(v)

16 Adjacency lists for the graphs appear in [11]. For the sake of completeness, we include in Table 4.1 the exact values of f(v) where they are known; those values are in bold face. The values of f(v) for 25 ≤ v ≤ 30 are determined in [13]. Note that f(50) = 175; the extremal graph of order 50 is the Hoﬀman-Singleton graph [16], the size of which attains the upper bound from Chapter 2. The Hoﬀman-Singleton graph is the only 7-regular Moore graph having diameter 2. The uniqueness was shown in [16, 17] and recently by Cong and Schwenk [21].

17 Chapter 5

Closing Remarks

In this paper, theoretical and computational bounds of f(v) were determined, and exact values were derived for small v. Figure 5.1 compares the computational and theoretical bounds. The upper bound is attained for v = 50, but after that point our computational results fall away from the upper bound. It is not clear whether the upper or lower bound seems to be closer to the

1500 Theoretical upper bound

Computational lower bound 1000 Theoretical lower bound

f(v)

500

100 50 100 150 200 v

Figure 5.1: Computational and theoretical bounds on f(v) asymptotic value of f(v); the upper bound might be attained on 3250 vertices. Erd˝os’conjecture that f(v) = (1/2 + o(1))3/2v3/2 remains unsolved. We hope this paper has shed some light on the problem. To improve the theoretical bounds of f(v), more properties of the extremal graphs need to be recognized. The extremal graphs seem to have a highly symmetric structure; it may be worthwhile to study the groups associated with them. Another interesting topic concerns the values of F (v). In particular, what are the necessary and/or suﬃcient conditions for F (v) = 1?

18 Acknowledgments

We would like to express our gratitude to Professor Simonovits for his valuable information and provision of many references, to Professor F¨uredifor giving us his preprint ([10]) on C4-free graphs, and to Professors Dutton and Brigham for providing their preprint ([6]) on upper bounds on the size of graphs of a given girth. Special thanks are due to Dr. Werman, with whom the last two authors enjoyed many stimulating discussions during the early stage of the research.

19 Bibliography

[1] B. Bollob´as, Extremal Graph Theory, Academic Press, New York, 1978.

[2] J. A. Bondy, P. Erd˝osand S. Fajtlowicz, Graphs of diameter two with no 4-cycles, Research Report CORR 79/11, Dept. of Combinatorics & Optimization, Univ. of Waterloo, Waterloo, Ontario, Canada, March 29, 1979.

[3] W. G. Brown, On graphs that do not contain a Thomsen graph, Canad. Math. Bull. 9 (1966), 281–285.

[4] C. R. J. Clapham, A. Flockhart and J. Sheehan, Graphs without four-cycles, J. Graph Theory 13 (1989), 29–47.

[5] R. D. Dutton and R. C. Brigham, Bounds on some graph invariants as a function of girth, Congr. Numer. 59 (1987), 13–22.

[6] R. D. Dutton and R. C. Brigham, Edges in graphs with large girth, Graphs Combin. 7 (1991), 315–321.

[7] P. Erd˝os,Some recent progress on extremal problems in graph theory, Congr. Numer. 14 (1975), 3–14.

[8] P. Erd˝os,A. R´enyi and V. T. S´os,On a problem of graph theory, Studia Sci. Math. Hungar. 1 (1966), 215–235.

[9] Z. F¨uredi,Graphs without quadrilaterals, J. Combin. Theory Ser. B 34 (1983), 187–190.

[10] Z. F¨uredi,Quadrilateral-free graphs with maximum number of edges, preprint.

[11] D. K. Garnick, Y. H. Kwong, and F. Lazebnik, Graphs without three-cycles or four-cycles for each order from 25 to 200, Computer Science Research Report 91-2, Bowdoin College, Brunswick, ME, 1991.

[12] D. K. Garnick, Y. H. Kwong, and F. Lazebnik, Algorithmic search for extremal graphs of girth at least ﬁve, preprint.

[13] D. K. Garnick and N. A. Nieuwejaar, Non-isomorphic extremal graphs without three-cycles or four-cycles, J. Combin. Math. Combin. Comput. 12 (1992), 33–56.

[14] F. Harary, Graph Theory, Addison-Wesley, Reading, MA, 1969.

[15] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford University Press, Oxford, 1979.

20 [16] A. J. Hoﬀman and R. R. Singleton, On Moore graphs with diameters 2 and 3, IBM J. Res. Develop. 4 (1960), 497–504.

[17] L. O. James, A combinatorial proof that the Moore (7,2) graph is unique, Utilitas Math. 5 (1974), 79–84.

[18] B. D. McKay, Isomorph-free exhaustive generation, preprint.

[19] U. S. R. Murty, A generalization of the Hoﬀman-Singleton graph, Ars Combinatoria 7 (1979), 191–193.

[20] N. Robertson, The smallest graph of girth 5 and valency 4, Bull. Amer. Math. Soc. 70 (1964), 824–825.

[21] A. J. Schwenk, Structure of the Hoﬀman-Singleton graph, Abstracts Amer. Math. Soc. 13 (1992), #876-05-198, p. 423.

[22] M. Simonovits, Extremal graph theory, in Selected Topics in Graph Theory Vol. 2, L. W. Beineke and R. R. Wilson, eds., Academic Press, New York, 1983.

[23] R. Singleton, There is no irregular Moore graph, Amer. Math. Monthly 75 (1968), 42–43.

[24] D. Stinson, Hill-climbing algorithms for the construction of combinatorial designs, Ann. Dis- crete Math. 26 (1985), 321–334.

[25] P. K. Wong, Cages — a survey, J. Graph Theory 6 (1982), 1–22.

21 Appendix A

Alternate Proofs of f(16) = 28 and f(23) = 50

We provide here two proofs that are alternate versions of the two corresponding proofs in the main paper. The alternate version of f(16) = 28 is more compact than the one in the paper, and the alternate version of f(23) = 50 is more expansive than the one in the paper. f(16) = 28 (alternate version)

G16 is displayed in Figure 3.3. The upper bound f(16) ≤ 29 follows from Proposition 2.8. Suppose ∗ there exists G ∈ F16 with 29 edges, then (δ, ∆) = (3, 4), (3, 5). Note that any two vertices x and y of degree 3 cannot be adjacent, otherwise G − {x, y} would contain 24 edges, which contradicts f(14) = 23. If ∆ = 5, then G contains S5,2 and d(ri) = 3 for each 1 ≤ i ≤ 5. Then d(x) ≥ 4 for all x ∈ R, so that G has a minimum degree sum of 60, which contradicts |E| = 28. Therefore ∆ = 4 and G has 6 vertices of degree 3 and 10 vertices of degree 4. Consider any (4,2)-star S, at least one of the ri’s has degree 3. Suppose N(r) contains 3 vertices of degree 3, then the removal of these vertices and r yields a graph on 18 edges, which contradicts f(12) = 18. Thus we may assume d(r1) = 3 and d(r3) = d(r4) = 4; then d(r1,1) = d(r1,2) = 4. Suppose d(r2) = 4. For each 2 ≤ i ≤ 4, since ri can be regarded as the root of a (4,2)-star, it follows that there is at least one degree 4 vertex in Ri. We may assume d(ri,3) = d(r4,2) = 4 for 2 ≤ i ≤ 4, and d(r2,j) = d(r3,j) = d(r4,1) = 3 for j = 1, 2. There exists a matching between R4 and each of Ri, 1 ≤ i ≤ 3. So we may assume these edges are r4,1r2,3, r4,1r3,3, r4,2ri,2, r4,3ri,1 for 1 ≤ i ≤ 3. The vertices r2,1 and r2,2 can only connect to R1 ∪ {r3,3}. Suppose r2,1r1,2, r2,2r1,1 ∈ E. One of r3,1 and r3,2 must also be incident to R1. Since r3,1r1,1, r3,2r1,2 6∈ E, either r3,1r1,2 ∈ E or r3,2r1,1 ∈ E, which will complete the 4-cycle r3,1r1,2r2,1r4,3r3,1 or r3,2r1,1r2,2r4,2r3,2, respectively. Therefore we may assume r2,1r3,3, r2,2r1,1 ∈ E. Similarly, one of r3,1 and r3,2 must be adjacent to r2,3, and the other adjacent to r1,1 or r1,2. Since T = {r2,3, r3,3} ⊂ N(r4,1), the remaining two edges incident to T form a matching between T and R1. Thus r3,2r1,1 6∈ E. But r3,2 cannot be adjacent to r1,2 either. So we must have r3,2r2,3, r3,1r1,2 ∈ E. Finally, r3,3 is adjacent to r1,1 or r1,2, which will create the 4-cycle r3,3r1,1r4,3r2,1r3,3 or r3,3r1,2r3,1r3r2,3, respectively. We have succeeded in showing that d(r2) = 3. In other words, every degree 4 vertex must be adjacent to two degree 3 vertices and two degree 4 vertices. Let z be the vertex external to S. By considering r3 and r4 as the roots of some (4,2)-stars, it follows that we may assume

22 d(ri,1) = d(ri,2) = 3 and d(ri,3) = d(z) = 4 for i = 3, 4. Among the vertices of degree 4 in W = V − N[r], each of z, r3,3 and r4,3 is adjacent to two degree 3 vertices, and every vertex in R1 ∪ R2 is adjacent to one degree 3 vertex. Thus there are 10 such edges in hW i. However, each of the 4 vertices in W of degree 3 is incident to two edges in hW i, giving a total of 8 such edges. Thus f(16) 6= 29. f(23) = 50 (alternate version)

From G23 in Figure 3.8, f(23) ≥ 50. Let us assume that there exists G ∈ F23 with 51 edges. From Propositions 2.6 and 2.7, G has 13 vertices of degree 4 and 10 vertices of degree 5. We now show that G cannot contain a P3 of degree 4 vertices. Note that if G contains a path P of 3 degree 4 vertices, then hG − P i has 20 vertices and 41 edges; therefore hG − P i is the unique extremal graph on 20 vertices, G20, shown in Figure 3.6. Since each of the two end vertices in P must be adjacent to 3 mutually distance 3 vertices in G20, and every such set of 3 vertices in G20 includes the vertex x external to the embedded Robertson graph, then hP +xi forms C4. Therefore, G does not contain a P3 of degree 4 vertices. We now establish the existence of a star in G that contains all 23 vertices. First we show that there must be at least one pair of adjacent vertices of degree 5. If there are no adjacent degree 5 vertices, then 50 edges extend from the 10 degree 5 vertices to vertices of degree 4, and there is exactly 1 edge incident on a pair of degree 4 vertices. Therefore, G contains S5,3 where the neighbors of the center all have degree 4, and at least 14 of the leaves in the star are of degree 5. But since G has only 10 vertices of degree 5, we conclude that there must be at least 1 pair of adjacent degree 5 vertices. Next, we establish by contradiction that G must contain a path of 3 degree 5 vertices. If G does not contain a path of 3 degree 5 vertices, then G does contain a star where the center is of degree 5, and 1 of the neighbors of the center has degree 5. Since we have assumed that there is no P3 of degree 5 vertices, then the remaining neighbors of the root all have degree 4. Also, all of the leaves in branch b, with the degree 5 root, have degree 4. Figure A.1a shows the vertices in G labeled with their degrees. Since there are 12 edges extending from the leaves in b to the set, V 0,

4 4 4 4 4 4 4 4 5 4 5 4 5 4 4 4 5 4 5 5 5

4 4 4 4 4 5 5 5 5 4

a) Assume no path of 3 degree 5 vertices b) The star with a path of 3 degree 5 vertices

Figure A.1: Degrees of vertices in graph with 51 edges on 23 vertices

23 of 13 unlabeled vertices in Figure A.1a, and 5 of the vertices in V 0 have degree 4, then at least 4 of the edges extending from the leaves of b are incident on degree 4 vertices in V 0. Thus, at most 20 edges extend from the degree 5 vertices in V 0 to the leaves in b and the degree 4 vertices in V 0. Now there are 6 edges amongst the 8 vertices of degree 5 in V 0. Therefore, there will be a path of degree 5 vertices in V 0, thus contradicting our assumption. We have established that G contains a (5, 3)-star where, if we label the top branch b1, and label the rest clockwise, then b1 and b2 both have 4 leaves. This star, which contains all 23 vertices in G, is shown in Figure A.1b. A 4-branch with at least 2 leaves of degree 4 contains a path of 3 degree 4 vertices, which we have already determined cannot exist in G; therefore, each 4-branch has at least 2 leaves of degree 5. Since this accounts for 9 of the 10 vertices of degree 5, we may, without loss of generality, label the leaves in the star with their degrees as in Figure A.1b; one of the unlabeled vertices has degree 4, and the other degree 5. Let branch b1 be the branch with 4 degree 4 leaves at the top of Figure A.1b. 12 edges extend from the leaves of b1 to leaves in other branches. Further, each of the 12 edges is incident on a distinct vertex from amongst the leaves not in b1. Therefore, an edge will extend from a leaf in b1 to a degree 4 leaf in a 4-branch, thus completing a path of 3 degree 4 vertices. Since this contradicts the proposition that G does not contain a path of 3 degree 4 vertices, we must conclude that our assumption, f(23) > 50, is false.