Disjoint G-Designs and the Intersection Problem for Some Seven Edge Graphs. Daniel Hollis

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1 Disjoint G-Designs and the Intersection Problem for Some Seven Edge Graphs by Daniel Hollis A dissertation submitted to the Graduate Faculty of Auburn University in partial fulfillment of the requirements for the Degree of Doctor of Philosophy Auburn, Alabama May, 0 Keywords: Intersection Problem, G-Designs, Graph Decompositions, Dragon Graphs Copyright 0 by Daniel Hollis Approved by Dean G. Hoffman, Chair, Professor of Mathematics and Statistics Chris A. Rodger, Don Logan Endowed Chair in Mathematics Peter D. Johnson, Professor of Mathematics and Statistics Jessica McDonald, Professor of Mathematics and Statistics

2 Abstract A G-Design of order n is a decomposition of the edges of K n (the complete graph on n vertices) into subgraphs of K n isomorphic to G, called blocks, in which each edge of K n appears in exactly one block. If B and B are arbitrary block sets corresponding to two G-designs of order n, one might ask under what circumstances can a G-design B of order n be found such that B and B are isomorphic and B and B are disjoint. Luc Teirlinck found that if B and B are K -designs of order n, then such a design B can be found for all n. In the first part of this dissertation, a similar result is shown for all graphs G with the exception of four small graphs (i.e. some graphs with less than vertices and edges). That is, if B and B is a pair of arbitrary G-designs of order n, then there is a G-design B such that B and B are isomorphic and B and B are disjoint provided n is sufficiently large. The remainder of the dissertation focuses on the intersection problem for some selected graphs G. The intersection problem is concerned with determining the positive integer pairs n,k for which there are G-designs B and B of order n such that B B = k. In a landmark result, C.C. Lindner and A. Rosa solved the intersection problem for Steiner Triple Systems (which are K -designs). Since then the intersection problem has been solved for various combinatorial designs among which are cycle systems where the cycles have length less than 0, star designs, and various other simple connected graphs with no more than edges. The results presented here solve the intersection problem for bipartite graphs each of which has exactly vertices and edges. ii

3 Acknowledgments First and foremost, I want to thank my advisor, Dean G. Hoffman, for without his guidance and encouragement this dissertation would have never come to fruition. I would also like to thank my advisory committee consisting of Peter D. Johnson, Jessica McDonald, and Chris A. Rodger whose helpful critiques and comments on my work have been instrumental in bringing this document to completion. To my parents, Wayne and Beverly Hollis, I am grateful for your unconditional love and support throughout my many years of formal and informal education. Without you, I would have never made it this far. Last but not least, I want to thank my many friends, family, mentors, and colleagues past and present who have influenced me, making me the person I am today. iii

4 Table of Contents Abstract Acknowledgments ii iii List of Figures v List of Tables ix Preliminaries G-Designs Group Actions, Orbits, and Stabilizers Disjoint G-Designs The Intersection Problem for Graphs with Vertices and Edges A Pair of Dragons with Edges A Graph Containing a -Cycle, Pendant Edges on Vertex in the Cycle, and a Single Pendant Edge on an Adjacent Vertex in the Cycle The Starship Enterprise Graph Containing Vertices and a -Cycle..... A Graph Containing a -Cycle with a Pendant Edge on One Vertex and a Path of Length Two on the Opposite Vertex in the Cycle A Graph Containing a -Cycle with a Pendant Edge on One Vertex and a Path of Length Two on an Adjacent Vertex in the Cycle The Viper Graph Containing a -Cycle and Edges Comments on the Remaining Bipartite Graphs with Seven Edges Summary of Results and Discussion Results for the Intersection Problem Further Inquiries into the Intersection Problem A Intersections of Designs on K, iv

5 List of Figures. Disjoint K -designs of Order {K n, K n,n }-Decomposition of K nt with t {K n+, K n,n }-Decomposition of K nt+ with t G-Design of Order nt with t G-Design of Order nt + with t The Dragon D () B, a D ()-Design of Order B 8, a D ()-Design of Order B,, a Cyclic D ()-Design on K, A Trade of Volume and a Mate in a D ()-Design of Order The Dragon D () B, a D ()-Design of Order B 8, a D ()-Design of Order B,, a Cyclic D ()-Design on K, A Trade of Volume and a Mate in a D ()-Design of Order v

6 . The Graph R (, ) B, an R (, )-Design of Order B 8, an R (, )-Design of Order B,, a Cyclic R (, )-Design on K, A Trade of Volume and a Mate in an R (, )-Design of Order The Starship Enterprise Graph SE (, ) B, an SE (, )-Design of Order B 8, an SE (, )-Design of Order B,, a Cyclic SE (, )-Design on K, A Trade of Volume and a Mate in an SE (, )-Design of Order The Graph T (, ) B, a T (, )-Design of Order B 8, a T (, )-Design of Order B,, a Cyclic T (, )-Design on K, A Trade of Volume and a Mate in a T (, )-Design of Order The Graph U(, ) B, a U(, )-Design of Order B 8, a U(, )-Design of Order vi

7 . B,, a Cyclic U(, )-Design on K, A Trade of Volume and a Mate in a U(, )-Design of Order A Trade of Volume and a Mate in a U(, )-Design of Order The Viper Graph V () B, a V ()-Design of Order B 8, a V ()-Design of Order B,, a Cyclic V ()-Design on K, A Trade of Volume and a Mate in a V ()-Design of Order Bipartite Graphs with Vertices and Edges Unsolved Bipartite Graphs with Vertices and Edges Unsolved Bipartite Graphs with Edges and No More Than Vertices..... A. B,, a D ()-Design on K, A. B,, a D ()-Design on K, with Exactly One Block in Common with B,... 8 A. B,, a D ()-Design on K, A. B,, a D ()-Design on K, with Exactly Two Blocks in Common with B,.. 8 A. B,, an R (, )-Design on K, A. B,, an R (, )-Design on K, with Exactly Two Blocks in Common with B, 8 A. B,, an SE (, )-Design on K, vii

8 A.8 B,, an SE (, )-Design on K, with Exactly Two Blocks in Common with B, 8 A.9 B,, a T (, )-Design on K, A.0 B,, a T (, )-Design on K, with Exactly Three Blocks in Common with B,. 8 A. B,, a U(, )-Design on K, A. B,, a U(, )-Design on K, with Exactly Two Blocks in Common with B,. 8 A. B,, a V ()-Design on K, A. B,, a V ()-Design on K, with Exactly Four Blocks in Common with B,.. 8 viii

9 List of Tables. Summary of -Vertex Graphs Summary of Intersection Problem Results for -Edge Graphs ix

10 Chapter Preliminaries. G-Designs Let us begin by establishing some notation that will be in effect for the remainder of discussion unless stated otherwise. For a graph G with vertex set V (G) and edge set E(G), v G = V (G) and e G = E(G). Definition.. For a graph G with v G = n, the automorphism group of G, denoted Aut(G), is the subgroup of S n, the group of permutations on V (G), such that for each edge {u, v} E(G) and σ Aut(G), {σ(u), σ(v)} E(G). Let a G = Aut(G). For everything else, the notation and definitions of [8] will be used unless specifically stated otherwise. Moreover, all graphs are simple, finite, and nontrivial, that is, they have at least one edge, unless stated otherwise, and n will refer to a positive integer unless stated otherwise. Definition.. If G and H are graphs, then a G-design on H is an ordered pair (V, B) such that V is the vertex set of H and B is a collection of subgraphs of H, each isomorphic to G, called blocks, where each edge in H is in exactly one block of B. In practice, a G-design on H is named by the set of blocks B; that is, B is referred to as a G-design on H. If B is a G-design on K n, then B is called a G-design of order n. For future reference, the vertex set of the complete graph on n vertices is V (K n ) = {,,..., n} unless stated otherwise.

11 Observation. Some obvious necessary conditions for the existence of a nontrivial G-design B on H are. v G v H. e G e H since the number of blocks B = e H e G. d H 0 (mod d G ) where d G is the greatest common divisor of the degrees in G and d H is the greatest common divisor of the degrees in H. Definition.. Letting V = {,..., n} be the vertex set for a graph H, two G-designs B and B on H are isomorphic if B = σb for some permutation σ Aut(H) with σb = {σb b B } where σb is the graph obtained by applying sigma to the vertices in b. If two G-designs B and B on H are isomorphic, we denote it B = B. Definition.. Two G-designs B and B on a graph H are disjoint if B B =. Example.. Figure. shows disjoint K -designs of order 9. Figure.: Disjoint K -designs of Order 9 K 9 B B Definition.. The spectrum of G, denoted Spec(G), is the set of all n for which there is a G-design of order n. Note that Spec(G) vacuously for any graph G.

12 Remark. When discussing G-designs, we only consider the positive integers n Spec(G) unless stated otherwise. The following gives some possible conditions on the graph G: T For n Spec(G), there is a pair B, B of G- designs of order n with B B =. T Given a G-design B of order n, there is a G-design B of order n such that B B =. T Given G-designs B and B of order n, there is a G- design B, isomorphic to B, such that B B =. For the sake of simplicity, if the statement about G in T i (i {,, }) holds for a particular integer n, we say G is T i for order n. Notice that the conditions ascend in order of strength; that is, G is T for order n G is T for order n G is T for order n. Condition T is a generalization to G-designs of a problem posed in [] concerning Steiner triple systems. It was shown in [] that K is T for all orders n. Incidentally, the T i naming scheme for the conditions on a graph G is for Luc Teirlinck whose work on K inspired further investigation into other graphs. Consequently, graphs that satisfy condition T for all sufficiently large n may also be referred to as Teirlinck graphs. Interestingly, each graph G (with the exception of a few small graphs) is T for all sufficiently large n. In order to show this, we need some results from group theory.. Group Actions, Orbits, and Stabilizers Much of the following discussion about group actions follows the development and proof techniques of [] except notation has been changed and some additional results are included.

13 Although many of the definitions and theorems apply when the group or set is infinite, we assume that all groups and sets are finite unless explicitly stated otherwise. Definition.. Let (Γ, ) be a group and let X be a set. Γ acts on X if for each g Γ and each x X, there is defined an element g. x X such that: i) For each x X, ι Γ. x = x, where ι Γ is the identity element of Γ. ii) For all g, g Γ and each x X, g.(g. x) = (g g ). x. If a group acts on a set, then it is called a group action. Remark. When it is clear that two elements of a group Γ are being combined under the group s binary operation, it is customary to omit the symbol for the binary operation. This practice will be adhered to from now on. Example.. Let X be the set of all subgraphs of K isomorphic to K, i.e. X= {,,, } The permutation group S acts on X by applying σ S as follows: σ.( a b c) = σ(b) σ(a) σ(c) for a < b < c. Remark. If G is a graph such that v G n and X is the set of all subgraphs of K n isomorphic to G, then the permutation group S n acts on X by applying the permutations in S n to the vertices of K n similarly to Example.. Observation. Let Γ be a group that acts on a set X. Define the relation Γ on X by x Γ y if and only if there is some g Γ such that g. x = y for all x, y X.

14 Using the properties of groups and group actions, it is easily shown that Γ is an equivalence relation on X. Thus Γ partitions the elements of X into disjoint equivalence classes. Definition.. Let Γ be a group that acts on a set X. The equivalence classes of the relation Γ are called orbits, and for each x X, Orb(x) is the orbit to which x belongs. Alternatively, Orb(x) = {g. x g Γ}. Definition.8. Let Γ be a group that acts on a set X. The group action of Γ on X is called transitive if for each pair x, y X there is some g Γ such that g. x = y. If the group action of Γ on X is transitive, we say that Γ acts transitively on X. Theorem.. A group Γ acts transitively on a set X if and only if Orb(x) = X for each x X. Example.. The group action given in Example. is transitive. In fact, the action of S n on the set X of all subgraphs of K n isomorphic to a graph G is transitive. Definition.9. Let Γ be a group that acts on a set X. For each x X, the stabilizer of x is the set of elements of Γ that fix x. We denote the stabilizer of x by Stab(x); thus, Stab(x) = {g Γ g. x = x}. Lemma.. Suppose G is a group that acts on a set X. Then for each x X, Stab(x) is a subgroup of G. Proof. The proof of this lemma follows easily by using the properties of group actions to check that Stab(x) satisfies the conditions for a subgroup of G.

15 Theorem. (The Orbit-Stabilizer Theorem). Let Γ be a group that acts on a set X. Then for each x X, Orb(x) = [Γ : Stab(x)] or Γ = Orb(x) Stab(x). Proof. Let x X and C = {g Stab(x) g Γ} be the collection of left cosets of Stab(x) in Γ. Then by definition C = [Γ : Stab(x)]. Let us define a function ϕ : C Orb(x) by ϕ(g Stab(x)) = g. x for each g Γ. To see that ϕ is well defined, suppose g, g Γ with g Stab(x) = g Stab(x). Then by left cancellation we have Stab(x) = g g Stab(x) which means g g Stab(x). Thus g. x = g.(g g. x) = g.[g.(g. x)] = g g.(g. x) = ι Γ.(g. x) = g. x. We now want to show that ϕ is bijective. Suppose g. x Orb(x). Then ϕ(g Stab(x)) = g. x; hence, ϕ is an onto function. Furthermore, suppose ϕ(g Stab(x)) = ϕ(g Stab(x)) for some g, g Γ. Then g. x = g. x, and g g x = x which means g g Stab(x) (or g g Stab(x) = Stab(x)). Consequently, g Stab(x) = g Stab(x) implying that ϕ is a one-to-one function. Therefore, C = Orb(x) proving the desired result. Lagrange s Theorem shows that Γ [Γ : Stab(x)] = which means Stab(x) Orb(x) = Γ Stab(x).

16 Lemma.. Let Γ be a group that acts on a set X, and define [x y] = {g Γ g. x = y} for any pair x, y X. If g [x y], then [x y] = g Stab(x). Proof. Suppose g [x y] for some pair x, y X and g Stab(x) is the left coset of Stab(x) in Γ containing g. By definition, h [x y] h. x = y = g. x (g h). x = x g h Stab(x) h g Stab(x). Therefore [x y] = g Stab(x). Definition.0. Let Γ be a group that acts on a set X. For each g Γ and S X define gs = {g. s s S}. Lemma.. If Γ is a group that acts transitively on a set X, A X, and B = {b} X, then A gb = A Stab(b). g Γ Proof. Suppose A = n, A = {a,..., a n } and for simplicity A i = {a i } for i n. Then for i n, there is a g i Γ such that g i. b = a i (or g i B = A i ), and [b a i ] = g i Stab(b) by Lemma.. Consequently, for i n, if g g i Stab(b) A i gb = 0 otherwise.

17 Thus A i gb = g i Stab(b) = Stab(b) for each i {,..., n}. Moreover, since g Γ [b a i ] [b a j ] = for i < j n, A gb = ( n ) A i gb = n (A i gb) g Γ g Γ i= g Γ i= ( ) n n = A i gb = Stab(b) = n Stab(b) i= g Γ i= = A Stab(b). Theorem.. Suppose Γ is a group that acts transitively on a nonempty set X. If A, B X, then Γ A gb = g Γ A B. X Proof. Suppose B = m, B = {b,..., b m }, and for ease of notation B i = {b i } for each i {,..., m}. For each g Γ, (A gb i ) (A gb j ) = for i < j m. Hence A gb = ( ) m A g B i g Γ g Γ i= = ( m ) A gb i g Γ i= = m (A gb i ) g Γ i= ( ) m = A gb i = = i= g Γ m ( A Stab(b i ) ) by Lemma. i= m i= ( A Γ ) X by the Orbit-Stabilizer Theorem 8

18 = m A X Γ = A B Γ. X Therefore, Γ A gb = g Γ A B. X Corollary.. Suppose Γ is a group that acts transitively on a nonempty set X. If A, B X such that A B < X, then there is some g Γ such that A gb =. Proof. From Theorem., g Γ A B A gb = Γ X < Γ. Hence, for some g Γ, A gb = ; otherwise, the sum on the left side of the inequality is too large. 9

19 Chapter Disjoint G-Designs From the necessary conditions mentioned previously, a G-design of order n, say B, exists for a graph G only if v G n B = e K n e G = ( n ) e G = n(n ) e G. For a graph G, let X n G be the set of all subgraphs of K n that are isomorphic to G. Lemma.. For a graph G, X n G = n(n ) (n v G + ) a G. Proof. The number of ways of labelling the vertices of G with vertices from V (K n ) is n(n ) (n v G + ) because there are n choices for the first vertex, n choices for the second vertex, and continuing on n v G + choices for the v G th vertex. These labellings will represent distinct subgraphs of K n except when there is a permutation of the vertices of G that gives the same labelling of G; that is, the permutation has the same adjacencies between vertices as the original labelling. The set of these permutations is Aut(G). Therefore X n G = n(n ) (n v G + ) a G. 0

20 Theorem.. Let G be a graph and G / {,,, } Then there is a positive integer N G such that G is T for n N G. Proof. For a graph G to have at least one edge and not be in the set of excluded graphs, v G. If v G = and G is not in the set of excluded graphs, then G is K. This graph was shown to be T for each n in []. For a pair B and B of G-designs of order n, we have B, B XG n, B = B = n(n ) e G, and B B X n G = a G n (n ) e G n(n ) (n v G + ). To proceed, we consider two cases for a graph with at least vertices: (i) Suppose v G = and B and B are G-designs of order n. Then B B a G n(n ) lim n XG n = lim n e G (n )(n ) = a G. e G As summarized in Table., set of excluded graphs. a G e G < for all -vertex graphs except for the one in the (ii) Suppose v G and B and B are G-designs of order n. Then B B a G n(n ) lim n XG n = lim n e G (n )(n ) (n v G + ) = 0. In either case, lim n B B X n G < which means there is a positive integer N G such that for all n N G and n Spec(G), B B < X n G. From Theorem., there is a σ S n such that B σb =.

21 The above proof does not show that the excluded graphs are not T for all orders n such that a design exists. However, there is exactly one way to decompose a graph into its component edges which means K is clearly not T for any order n. As it turns out, all of the excluded graphs except possibly the path of length are not T for any order n. Example.. Let G =, and let B and B be the following G-designs of order n : B = {i B = {i j + j j i + i < j n} { i i n } { n } i + i < j n } { i n i n } {n n } Notice that in B every edge of K n not incident on vertex is in a block where is the isolated vertex. In B, each vertex of K n is the isolated vertex in some block of the design. For any σ S n, there is an i {,,..., n} such that σ(i) =. By construction there must be some block in B such that i is the isolated vertex along with some edge {j, k}. Applying σ to that block gives σ.( j i k) = σ(j) σ(k) B. Thus B σb for each σ S n which means G is not T for any order n. Example.. Let G =, and let B and B be the following G-designs of order n : B = { i j B = { i i + i < j n} { i i + i + i n} { i i + j, i i n} {,, }. i + i + j + i n, j n}. (The addition in B is done modulo n with n replacing 0.) Each edge of K n not incident on vertex or is in a block in B in which the isolated vertices are and. In B, each pair of vertices in K n is in some block as the isolated vertex pair; that is, for each i, j V (K n ) (i j), there is some block

22 i j B. k l Suppose σ S n. Then there is a pair i, j {,,..., n} such that σ(i) = and σ(j) =. Applying σ to the block in B that contains i and j as isolated vertices gives σ.( k i j ) = σ(k) l σ(l) B. Hence B σb for each σ S n meaning G is not T for any order n. As shown in Theorem., if G is a graph with v G, e G, and n Spec(G) is sufficiently large, then we can find a pair of disjoint G-designs of order n. Moreover, if G is a graph with v G and n Spec(G) is sufficiently large, then we can find k pairwise disjoint G-designs of order n for any integer k. Theorem.. Let G be a graph with v G, and let k be an integer. There is a positive integer N G,k such that if n N G,k and B, B,..., B k are G-designs of order n, then there are G-designs B, B,..., B k of order n such that B i = B i for i k and B i B j = for i < j k. Proof. Suppose G is a graph such that v G. The proof proceeds by induction on the number of G-designs k. (i) Suppose k =. By Theorem. there is an integer N G such that for all n N G, if B and B are G-designs of order n, then for some σ Aut(K n ), B σb =. Let B = B and B = σb. (ii) Suppose for r < k that there is an integer N G,r such that for all n N G,r if B, B,..., B r are G-designs of order n, then there are G-designs B, B,..., B r such that B i = B i for i r and B i B j = for i < j r.

23 (iii) Suppose k. By the induction hypothesis there is a positive integer N G,k such that for all n N G,k if B, B,..., B k are G-designs of order n, then there are G-designs B, B,..., B k such that B i = B i for i k and B i B j = for i < j k. Since B, B,..., B k are pairwise disjoint G-designs of order n, k i= B i = (k )n(n ) e G. Furthermore, lim n k i= B i X n G and B k X n G and k B i i= B k X n G = lim n a G (k )n(n ) e G (n )(n ) (n v G + ) = 0. Thus there is a positive integer N G,k such that k B i i= B k n Spec(G). By Theorem., there is a σ S n such that B k = σb k which completes the proof. X n G < for all n N G,k with k i= B i σb k =. Let For a graph G with v G =, the limit in the proof of Theorem. is a G (k )n(n ) lim n e G (n )(n ) = a G(k ). e G Provided the limit is less than, then the conclusion of the theorem should hold as well. Let { K G = max k Z a G (k ) e G } { < = max k Z k < e G } + a G.

24 Then for the -vertex graphs, K G is the maximum number of disjoint G-designs that can be found using the techniques of Theorem.. The values of a G and K G for all -vertex graphs are summarized in Table.. Table.: Summary of -Vertex Graphs G a G a G e G K G

25 Chapter The Intersection Problem for Graphs with Vertices and Edges Throughout this chapter, we assume that all graphs have nontrivial components unless stated otherwise. Before we solve the intersection problem for some select vertex, edge graphs, we first need some definitions. Among these definitions is that of the intersection problem itself. Definition.. For a graph G, the intersection problem for G is the problem of determining all integer pairs n, k for which there exist G-designs B and B of order n such that B B = k. For a graph G, we need to know Spec(G) in order to solve the intersection problem. Then for each n Spec(G) we can define two sets that are closely related to the intersection problem. Definition.. For a graph G, we define the following two sets. Suppose there exists a G-design on a graph H. Let I G (H) be the set of all k such that there are G-designs B and B on H with B B = k. Clearly, if B is a G-design on H, then B I G (H). Let J G (H) be the set of all non-negative integers k such that ke G e H k = e H e G. except for If H = K n, then we write these sets as I G (n) and J G (n) respectively. The set I G (n) is the set of all realizable k for which there exist G-designs B and B of order n such that B B = k. Thus the solution to the intersection problem for a graph G

26 is to find I G (n) for each n Spec(G). Since the size of the intersection of two G-designs of order n must be nonnegative and cannot exceed the number of blocks in either design, the set J G (n) gives us the set of possible integers that two G-designs of order n could have in common. We can exclude the integer one less than the number of blocks in the G-design, for if two G-designs have that many blocks in common, the two designs are forced to have the remaining block in common as well. Thus, I G (n) J G (n) for any graph G and n Spec(G). The intersection problem was first solved for Steiner Triple Systems (K -designs) by C.C. Lindner and A. Rosa in []. Since then, the intersection problem has been solved for various other graphs including cycles of length less than 0 [, ], connected graphs with at most vertices or edges [,,9] (except there are a few undetermined values for K -designs of order n =, 8, ), star graphs with m edges [], a -cycle with a pendant edge [], and graphs with vertices, edges, and a -cycle subgraph []. The type of graphs for which we will be solving the intersection problem have at least vertices and edges. But first, several useful definitions and results must be introduced. Solving the intersection problem involves working with various sets of non-negative integers which necessitates defining some binary operations on these sets. Definition.. Let A and B be nonempty sets of non-negative integers, and let p be a positive integer. Then. A + B = {a + b a A and b B}. p A = A + A A p where A i = A for i p. There should be no ambiguity in this definition due to the associativity of addition.. For convenience, let 0 A = {0}. Example.. Let A = {0,, } and B = {0,, }. Then A + B = {0,,,,,,, 8} and A = {0,,,,, }.

27 Observation. Let A, A, B, and B be nonempty sets of non-negative integers such that A i B i for each i, and let p be a positive integer. Then. A + A B + B. p A i p B i for each i {, }. The following generalization of G-designs is another useful definition. Definition.. Let K be a collection of graphs. A K-decomposition of a graph H is an ordered pair (V, B) where V is the vertex set of H and B is a collection of subgraphs of H called blocks, each of which is isomorphic to some graph in K, such that each edge of H is in exactly one block of B. Notice that if K = {G}, then a K-decomposition of H is a G-design on H. Proposition.. If a K-decomposition of a graph H exists and there exists a G-design on K for each K K, then there exists a G-design on H. Proof. Suppose H is a graph with vertex set V and (V, B) is a K-decomposition of H for some collection of graphs K. Furthermore, suppose G is a graph such that for each graph K K there is a G-design on K. Construct a G-design on B with block set Γ B for each B B. Then Γ B is a G-design on H. B B Lemma.. Let n and t be positive integers with t. (a) There exists a {K n, K n,n }-decomposition of K nt. (b) There exists a {K n+, K n,n }-decomposition of K nt+. Proof. (a) There are t subgraphs of K nt induced by the sets {ni + r r n} where 0 i t that are isomorphic to K n. Note that the edges of K nt in these induced subgraphs must be of the form {u, v} where u = ni+r and v = ni+s where 0 i t and r < s n. There are ( t ) subgraphs of Knt isomorphic to K n,n formed by taking 8

28 vertices u and v such that u {ni + r r n} and v {nj + s s n} where 0 i < j t and taking all edges {u, v} E(K nt ) such that u = ni + r and v = nj + s with 0 i < j t and r, s n. Clearly, none of the edges of K nt in the subgraphs isomorphic to K n overlap with those in the subgraphs isomorphic to K n,n. Counting the number of edges in these subgraphs gives ( ) n t + ( ) t n = nt(n ) + n t(t ) = nt(nt ) = ( ) nt = e Knt. Thus a {K n, K n,n }-decomposition of K nt exists. (For a visual representation of the proof, see Figure..) Figure.: {K n, K n,n }-Decomposition of K nt with t Copy of K.. n.. i Copy of K.. n. Copy of K n,n j t... Copy of K n.... Copy of K n (b) Let V (K nt+ ) = V (K nt ) { }. Extend a {K n, K n,n }-decomposition of K nt to a {K n+, K n,n }-decomposition of K nt+ by adding the vertex to each of the t copies of K n in the {K n, K n,n }-decomposition of K nt to get t copies of K n+. Hence a {K n+, K n,n }- decomposition of K nt+ with t copies of K n+ and ( t ) copies of Kn,n exists. (For a visual representation of the proof, see Figure..) 9

29 Figure.: {K n+, K n,n }-Decomposition of K nt+ with t.... i... Copies of K n+ Copy of K n,n j.... t... The following lemma will be used frequently for solving the intersection problem for several upcoming graphs. Lemma.. If there exists an {H, H,..., H m }-decomposition of K n with r i blocks of the decomposition isomorphic to H i for i m and there exists a G-design on H i for i m, then I G (n) r I G (H ) + r I G (H ) r m I G (H m ). Proof. Suppose k r I G (H ) + r I G (H ) r m I G (H m ). Then k = s + s s m where s i r i I G (H i ) for i m. Furthermore, each s i = t i, + t i, t i,ri where {t i,, t i,,..., t i,ri } I G (H i ) for i m. Thus there exist r i pairs of G-designs on the r i distinct blocks in the decomposition of K n isomorphic to H i, 0

30 say B ti,j and B t i,j, such that B ti,j B t i,j = t i,j for j r i. Let B Hi = r i j= B ti,j and B H i = r i j= B t i,j for i m. Then B Hi B H i = s i for i m. Let Then B B = m m B = B Hi and B = B H i. i= m s i = k. Since B and B are unions of G-designs on H i for i m i= which combined appropriately (i.e. taking the union of the blocks in each design) comprise {H, H,..., H m }-decompositions of K n, they must both be G-designs of order n. Therefore, k I G (n). Finally, a concept that will be used in solving the intersection problems for several upcoming graphs is that of a trade. Definition.. A pair of G-designs T and T on the same graph H are called mates if T T =. If T is a G-design with at least one mate and T = t, then T is said to be a trade of volume t. It should be noted that no trades of volume exist. As outlined in the following lemma from [], the intersection problem for a graph G is closely related to finding trades. Lemma.. Let B be a G-design of order n. Then there is a G-design B of order n with B B = k if and only if B contains a trade of volume ( n ) /eg k. Proof. Suppose B and B are G-designs of order n with B B = k. Let H be the subgraph of K n formed by removing each edge found in some block contained in B B from K n. Then T = B (B B ) is a G-design on H as is T = B (B B ). Furthermore, T T = meaning T and T are mates with T = T = ( n ) /eg k. Thus T is a trade of volume ( n ) /eg k contained in B. i=

31 Suppose B is a G-design of order n containing a trade T of volume ( n ) /eg k. Then T is a G-design on some subgraph H of K n with a mate T. Consequently, T and T are G-designs on the same subgraph H with T T = and T = T = ( n ) /eg k. Moreover, there must be a G-design S on H (the complement of H in K n ) with S = k since T is contained in a G-design of order n; that is, B = T S with T S =. Let B = T S. Then B is a G-design of order n since T is a G-design on H and S is a G-design on H. Additionally, B B = (T S) (T S) = (T T ) S = S = k. Hence there is a G-design B of order n with B B = k For each graph G that will be discussed in the subsequent sections, there exists a G-design of orders n and n + as well as a G-design on K n,n for some positive integer n which essentially reduces the intersection problem for these graphs to a few small cases. (See Figures. and..) Figure.: G-Design of Order nt with t G-Design of Order n.... i G-Design of Order n... G-Design on K n,n j t... G-Design of Order n.... G-Design of Order n

32 Figure.: G-Design of Order nt + with t.... i... G-Designs of Order n + G-Design on K n,n j.... t.... A Pair of Dragons with Edges Definition.. A dragon D l (m) (l < m) is a graph with m edges consisting of a cycle of length l and an attached path called the tail. (See Figure. for a picture of the dragon D ().) Figure.: The Dragon D () The definition and notation describing dragons are a generalization of that given in []. In particular, a dragon D (m) containing a triangle with vertices {a, b, c} is denoted by (a, b, c; v, v,..., v m ) where the tail is attached to vertex c. Similarly, a dragon D (m) containing a -cycle with vertices {a, b, c, d} and edges {ab, bc, cd, da} is denoted by (a, b, c, d; v, v,..., v m ) where the tail is attached to vertex d. As for the intersection problem, several of the dragon graphs have been completely solved. D () (also known as

33 a -kite or just a kite) is solved in []. The dragons D () (also called a -kite) and D () are solved in [] and [] respectively. In solving the intersection problem for D (), the spectrum of D () must first be found. In order to find the spectrum, we first look at a few small D ()-designs. Example.. A D ()-design of order is given by B = {(,,, ;,, ), (,,, ;,, ), (,,, ;,, )}. See Figure. for a picture of the blocks as graphs. Figure.: B, a D ()-Design of Order Example.. A D ()-design of order 8 is given by B 8 = {(, 8,, ;,, ), (,, 8, ;,, ), (,, 8, ;,, ), (,,, ;,, 8)}. The blocks can be seen as graphs in Figure.. Figure.: B 8, a D ()-Design of Order

34 Example.. If V (K, ) = Z {, } with partitions A = Z {} and B = Z {}, then B, = {((i, ), (i +, ), (i +, ), (i, ); (i +, ), (i +, ), (i +, ) 0 i }, where addition in the first coordinates is done modulo, is a D ()-design on K,. To see that this is a design on K,, notice that for each j Z there is some edge {(x, ), (y, )} in each block of B, such that y x = j. Thus each edge of K, is in some block of B,, and B, = which is the number of blocks expected in an D ()-design on K,. Also, see Figure.8. Figure.8: B,, a Cyclic D ()-Design on K, (0, ) (, ) (, ) (, ) (, ) (, ) (, ) (0, ) (, ) (, ) (, ) (, ) (, ) (, ) For a view of each of the blocks, see Figure A. in the appendix. Now for the spectrum of D (). Theorem.. There exists a D ()-design of order n if and only if n 0 or (mod ). Proof. The necessity of n 0 or (mod ) is obvious since those are the only orders such that ( n ). The proof of sufficiency proceeds by checking two cases. (i) Suppose n 0 (mod ). For n =, a D ()-design of order n exists as shown in Example.. Also, a D ()-design on K, exists as shown in Example.. By Lemma., a {K, K, }-decomposition of K t exists for each t ; hence, a D ()-design of order t exists for each positive integer t according to Proposition..

35 (ii) Suppose n (mod ). For n = 8, a D ()-design of order 8 exists as shown in Example.. Once again, a D ()-design on K, exists as shown in Example.. By Lemma., a {K 8, K, }-decomposition of K t+ exists for each t ; hence, a D ()- design of order t + exists for each positive integer t according to Proposition.. Therefore, a D ()-design of order n exists for each n 0 or (mod ). Directly solving the intersection problem for D () begins with looking at small cases. The techniques that will prove most useful in determining I D ()(n) will be permutation of vertices and finding trades. Example.. Starting with B from Example., consider the following three D ()-designs of order. B = {(,,, ;,, ), (,,, ;,, ), (,,, ;,, )} Transposing vertices and in B yields the design B 0 = {(,,, ;,, ), (,,, ;,, ), (,,, ;,, )} which is disjoint from B. Transposing vertices and in B yields a design B = {(,,, ;,, ), (,,, ;,, ), (,,, ;,, )} that shares exactly block with B. Thus I D ()() = {0,, } = J D ()(). Example.. Starting with B 8 from Example., consider the following three D ()-designs of order 8. B 8 = {(, 8,, ;,, ), (,, 8, ;,, ), (,, 8, ;,, ), (,,, ;,, 8)}

36 Transposing vertices and 8 in B 8 yields the following design that is disjoint from B 8. B 0 8 = {(,,, ; 8,, ), (, 8,, ;,, ), (,,, ;,, 8), (,, 8, ;,, )} Transposing vertices and in B 8 yields the following design that has exactly one block in common with B 8. B 8 = {(, 8,, ;,, ), (,, 8, ;,, ), (,, 8, ;,, ), (,,, ;,, 8)} The last two blocks in B 8 form a trade of volume in the design. The trade and a mate are illustrated in the subgraph of K 8 shown in Figure.9. Figure.9: A Trade of Volume and a Mate in a D ()-Design of Order 8 Trade Mate 8 8 Thus I D ()(8) = {0,,, } = J D ()(8). For the purposes of solving the intersection problem of D () for orders larger than 8, the existence of some elements of I D ()(K, ) also need to be shown. Example.. Looking at the D ()-design on K, of Example., B, = {((i, ), (i +, ), (i +, ), (i, ); (i +, ), (i +, ), (i +, ) 0 i },

37 a disjoint design is readily found by changing the second coordinates of each vertex in each block as shown below. B 0, = {((i, ), (i +, ), (i +, ), (i, ); (i +, ), (i +, ), (i +, ) 0 i } (Note that addition is done modulo in both designs.) Transposing vertices (0, ) and (, ) in B, yields a design B, on K, that has exactly one block in common with B,. (See Example A. in the appendix for a list of the blocks of B, which indicates the block that B, and B, have in common.) Hence I D ()(K, ) {0,, }. Example.8. From Lemma. and Theorem., a D ()-design of order can be constructed using D ()-designs of order, a D ()-design on K,, and a {K, K, }-decomposition of K. From Lemma. and Examples. and., I D ()() I D ()() + I D ()(K, ) {0,, } + {0,, } {0,,,,, } + {0,, } {0,,,,,,,, 8, 9, 0,, } J D ()(). Hence I D ()() = J D ()(). Finally the remaining orders n Spec(D ()) are solved in the following theorem. Theorem.. If n Spec(D ()), then I D ()(n) = J D ()(n). 8

38 Proof. As shown in Theorem., the values of n Spec(D ()) are n 0 or (mod ). Hence the proof proceeds by considering two cases. (i) Suppose n 0 (mod ). For n = and n =, I D ()(n) = J D ()(n) as shown in Examples. and.8 respectively. For n = t with t an integer, it was shown in Theorem. that a D ()-design of order n can be constructed using a {K, K, }-decomposition of K t with t blocks isomorphic to K and ( t ) blocks isomorphic to K,. From Lemma., I D ()(t) t I D ()() + ( t ) ID ()(K, ) t {0,, } + ( t ) {0, } {0,,,..., t, t, t} + { r 0 r ( t { 0,,,..., ( ( t ) + t, t ( ) + t, t ) } + t { 0,,,..., ( ( t ) + t, t ) + t, t t { 0,,,..., ( ( t ) + t, t ) + t, { ( 0,,,..., t ) (, t ) (, t )} J D ()(t). )} + t t(t ) } } (ii) Suppose n (mod ). For n = 8, I D ()(n) = J D ()(n) as shown in Example.. For n = t + with t an integer, it was shown in Theorem. that a D ()-design of order n can be constructed using a {K 8, K, }-decomposition of K t+ with t blocks isomorphic to K 8 and ( t ) blocks isomorphic to K,. Consequently I D ()(t + ) t I D ()(8) + ( t ) ID ()(K, ) t {0,,, } + { r 0 r ( t {0,,,..., t, t, t} + { r 0 r ( t 9 )} )}

39 { 0,,,..., ( ( t ) + t, t ( ) + t, t ) } + t { 0,,,..., ( ( t ) + t, t ) } + t, t t + 8t { 0,,,..., ( ( t ) + t, t ) } + t, t(t+) { ( 0,,,..., t+ ) (, t+ ) (, t+ )} J D ()(t + ). Therefore, I D ()(n) = J D ()(n) for each n Spec(D ()). Using the similar techniques, the intersection problem for D () can be solved as well. Also, similar notation will be used when denoting D () as a block in a design. Example.9. If the graph D () has vertex set V (D ()) = {a, b, c, d, e, f, g}, then its edge set is E(D ()) = {ab, bc, cd, de, ef, af, fg}. For brevity, this graph will be denoted by the vector (a, b, c, d, e, f; g) henceforth. Figure.0 gives an illustration of this graph. Figure.0: The Dragon D () c b d a e f g Once again, the spectrum of the graph needs to be found before solving the intersection problem which will be done through several examples and Lemma.. Example.0. The following set forms a D ()-design of order : B = {(,,,,, ; ), (,,,,, ; ), (,,,,, ; )}. See Figure. to see how the blocks look as graphs. 0

40 Figure.: B, a D ()-Design of Order Example.. The following set forms a D ()-design of order 8: B 8 = {(, 8,,,, ; ), (,,, 8,, ; ), (,, 8,,, ; ), (,,,,, ; 8)}. To see the blocks as graphs, view Figure.. Figure.: B 8, a D ()-Design of Order Example.. Let V (K, ) = Z {, } with partitions A = Z {} and B = Z {}. Then B, = {((i, ), (i +, ), (i +, ), (i +, ), (i +, ), (i, ); (i +, )) 0 i } where addition in the first coordinates is done modulo is a D ()-design on K,. To see that this is a design on K,, notice that for each j Z there is some edge {(x, ), (y, )} in each block of B, such that y x = j. Thus each edge of K, is in some block of B,, and B, = which is the number of blocks expected in a D ()-design on K,. Also, see Figure..

41 Figure.: B,, a Cyclic D ()-Design on K, (0, ) (, ) (, ) (, ) (, ) (, ) (, ) (0, ) (, ) (, ) (, ) (, ) (, ) (, ) For an illustration of the individual blocks of B,, see Figure A. in the appendix. The spectrum of D () is outlined in the next theorem. Theorem.. There exists a D ()-design of order n if and only if n 0 or (mod ). Proof. The necessity of the orders n 0 or (mod ) is clear given that these are the only orders such that ( n ). Showing their sufficiency proceeds by checking two cases. (i) Suppose n 0 (mod ). For n =, a D ()-design of order n has been exhibited in Example.0. For n = t (t ), a D ()-design of order n exists according to Proposition. and Lemma. because there exists both a D ()-design of order and a D ()-design on K, (Example.). Hence there is a D ()-design of order t for each positive integer t. (ii) Suppose n (mod ). For n = 8, a D ()-design of order n has been exhibited in Example.. For n = t + (t ), a D ()-design of order n exists according to Proposition. and Lemma. because there exists both a D ()-design of order 8 and a D ()-design on K,. Thus there is an D ()-design of order t + for each positive integer t. Therefore a D ()-design of order n exists if and only if n 0 or (mod ). Similarly to D (), solving the intersection problem for D () proceeds by considering a few small cases.

42 Example.. Starting with the D ()-design of order given in Example.0, permutation of select vertices will yield a disjoint design and a design that shares one block in common with the original design. As a reminder, the original design is B = {(,,,,, ; ), (,,,,, ; ), (,,,,, ; )}. Transposing vertices and gives the following D ()-design of order which is disjoint from the original design. B 0 = {(,,,,, ; ), (,,,,, ; ), (,,,,, ; )}. Transposing vertices and and transposing vertices and gives a design shown below that shares exactly one block in common with B. B = {(,,,,, ; ), (,,,,, ; ), (,,,,, ; )}. Consequently I D ()() = J D ()(). Example.. For the D ()-design of order 8 of Example. B 8 = {(, 8,,,, ; ), (,,, 8,, ; ), (,, 8,,, ; ), (,,,,, ; 8)}, a disjoint design can be found by transposing vertices and as shown in the design below. B 0 8 = {(, 8,,,, ; ), (,,, 8,, ; ), (,, 8,,, ; ), (,,,,, ; 8)}. To obtain a design that shares exactly one block with B 8, transpose vertices and, and transpose vertices and 8 as seen in the design below. B 8 = {(, 8,,,, ; ), (,,,, 8, ; ), (,, 8,,, ; ), (,,,,, ; 8)}

43 Notice that the last two blocks of B 8 form a trade of volume in the design as shown in Figure.. Figure.: A Trade of Volume and a Mate in a D ()-Design of Order 8 Trade Mate 8 8 Thus I D ()(8) = J D ()(8). To solve the intersection problem for larger orders in Spec(D ()), the existence of some of the elements in I D ()(K, ) needs to be shown. Example.. Looking at the D ()-design on K, of Example., B, = {((i, ), (i +, ), (i +, ), (i +, ), (i +, ), (i, ); (i +, )) 0 i }, a disjoint design is readily found by changing the second coordinates of each vertex in each block as shown below. B 0, = {((i, ), (i +, ), (i +, ), (i +, ), (i +, ), (i, ); (i +, )) 0 i } (Note that addition is done modulo in both designs.) Transposing vertices (, ) and (, ) in B, yields a design B, on K, that has exactly two blocks in common with B,. (See Example A..) Hence I D ()(K, ) {0,, }.

44 The intersection problem for order is more easily handled separately from the other higher orders. The solution is shown in the next example. Example.. From Lemma. and Theorem., a D ()-design of order can be constructed by combining two D ()-designs of order and one D ()-design on K, in a {K, K, }-decomposition of K. From Lemma. and Examples. and., I D ()() I D ()() + I D ()(K, ) {0,, } + {0,, } {0,,,,, } + {0,, } {0,,,,,,,, 8, 9, 0,, } J D ()(). Thus I D ()() = J D ()(). The solution to the intersection problem for D () is completed in the following theorem. Theorem.8. If n Spec(D ()), then I D ()(n) = J D ()(n). Proof. For n =, n = 8, and n =, I D ()(n) = J D ()(n) as shown in Examples.,., and. respectively. For the remaining orders in Spec(D ()), two cases are considered. (i) Suppose n 0 (mod ) and n ; that is, n = t for some integer t. From Theorem., a D ()-design of order t (t ) can be constructed containing t D ()- designs of order and ( t ) D ()-designs on K,. According to Lemma., if t, I D ()(t) t I D ()() + ( t ) ID ()(K, ) t {0,, } + ( t ) {0, }

45 { 0,,,..., ( t ), ( t ), ( t )} (See proof of Theorem..) J D ()(t). (ii) Suppose n (mod ) and n ; that is, n = t + for some integer t. From Theorem., a D ())-design of order t + (t ) can be constructed containing t D ()-designs of order 8 and ( t ) D ()-designs on K,. By Lemma., if t, I D ()(t + ) t I D ()(8) + ( t ) ID ()(K, ) t {0,,, } + { r 0 r ( t { 0,,,..., J D ()(t + ). ( t+ ), ( t+ )} ), ( t+ )} (See Theorem..) Therefore I D ()(n) = J D ()(n) for each n Spec(D ()).. A Graph Containing a -Cycle, Pendant Edges on Vertex in the Cycle, and a Single Pendant Edge on an Adjacent Vertex in the Cycle Since there is no well known convention for naming the graph of interest, one is given in the following definition. Definition.. The graph R k (l, m) (l m) is a graph containing a cycle of length k, l > 0 pendant edges on one vertex in the k-cycle, and m > 0 pendant edges on an adjacent vertex in the k-cycle. Example.. The graph R (, ) has vertex set V (R (, )) = {a, b, c, d, e, f, g} and edge set E(R (, )) = {ab, bc, cd, de, be, ef, eg}. For convenience, this graph will be denoted by the vector R(a; b, c, d, e; f, g) from now on. This graph is illustrated in Figure..

46 Figure.: The Graph R (, ) a b f c e d g The intersection problem for R (, ) is solved in []. In the following discussion, the intersection problem for R (, ) is solved, but first the spectrum of R (, ) must be established. As usual, some small designs are given first. Example.8. The following block set is an R (, )-design of order. B = {R(;,,, ;, ), R(;,,, ;, ), R(;,,, ;, )} The blocks are illustrated as graphs in Figure.. Figure.: B, an R (, )-Design of Order Example.9. The following block set is an R (, )-design of order 8. B 8 = {R(8;,,, ;, ), R(;,, 8, ;, ), R(; 8,,, ;, ), R(;,,, ;, 8)} The blocks are shown as graphs in Figure..

47 Figure.: B 8, an R (, )-Design of Order Example.0. Let V (K, ) = Z {, } with partitions A = Z {} and B = Z {}. Then the following set is an R (, )-design on K,. B, = {R((i +, ); (i, ), (i +, ), (i +, ), (i, ); (i +, ), (i +, )) 0 i } The addition in the first coordinates is done modulo. To see that this is a design on K,, notice that for each j Z there is some edge {(x, ), (y, )} in each block of B, such that y x = j. Thus each edge of K, is in some block of B,, and B, = which is the number of blocks expected in an R (, )-design on K,. Also, see Figure.8. Figure.8: B,, a Cyclic R (, )-Design on K, (0, ) (, ) (, ) (, ) (, ) (, ) (, ) (0, ) (, ) (, ) (, ) (, ) (, ) (, ) Look at Figure A. in the appendix to see each of the blocks in B,. Theorem.9. There exists an R (, )-design of order n if and only if n 0 or (mod ). 8

48 Proof. The necessity of n 0 or (mod ) is obvious since those are the only orders such that ( n ). The proof of sufficiency proceeds by checking two cases. (i) Suppose n 0 (mod ). For n =, an R (, )-design of order n exists as shown in Example.8. Also, an R (, )-design on K, exists as shown in Example.0. By Lemma., a {K, K, }-decomposition of K t exists for each t ; consequently, an R (, )-design of order t exists for each positive integer t according to Proposition.. (ii) Suppose n (mod ). For n = 8, an R (, )-design of order 8 exists as shown in Example.9. Once again, an R (, )-design on K, exists as shown in Example.0. By Lemma., a {K 8, K, }-decomposition of K t+ exists for each t ; thus, an R (, )-design of order t + exists for each positive integer t according to Proposition.. Therefore, an R (, )-design of order n exists for each n 0 or (mod ). With the spectrum of R (, ) established, the intersection problem is now solved by considering a few small designs. Example.. Starting with B from Example.8, consider the following three R (, )-designs of order. B = {R(;,,, ;, ), R(;,,, ;, ), R(;,,, ;, )} Transposing vertices and in B yields the design B 0 = {R(;,,, ;, ), R(;,,, ;, ), R(;,,, ;, )} which is disjoint from B. Transposing vertices and in B yields a design B = {R(;,,, ;, ), R(;,,, ;, ), R(;,,, ;, )} 9

49 that shares exactly block with B. Thus I R (,)() = J R (,)(). Example.. Starting with B 8 from Example.9, consider the following three R (, )-designs of order 8. B 8 = {R(8;,,, ;, ), R(;,, 8, ;, ), R(; 8,,, ;, ), R(;,,, ;, 8)} Transposing vertices and 8 in B 8 yields the following design that is disjoint from B 8. B 0 8 = {R(;,,, ;, 8), R(;, 8,, ;, ), R(;,,, ;, 8), R(; 8,,, ;, )} Transposing vertices and in B 8 yields the following design that has exactly one block in common with B 8. B 8 = {R(8;,,, ;, ), R(;,, 8, ;, ), R(; 8,,, ;, ), R(;,,, ;, 8)} The last two blocks in B 8 form a trade of volume in the design. The trade and a mate are illustrated in the subgraph of K 8 shown in Figure.9. Figure.9: A Trade of Volume and a Mate in an R (, )-Design of Order 8 Trade Mate 8 8 Thus I R (,)(8) = J R (,)(8). 0