Some Results on Graph Representations and Closure Systems

Representations of graphs are a way of encoding the structure of a graph by using other discrete structures. The object of a representation of a graph is to encode its structure efficiently. Typically a graph can be encoded by an n× n adjacency matrix. It is possible, however, to encode graphs much more efficiently using other representation schemes. This work considers tree representations of graphs. A tree representation of a target graph G is an assignment of subtrees of a host tree to the vertices of G in such a way that if uv ∈ E(G), then the subtree assigned to the vertex u and the subtree assigned to the vertex v have at least t nodes in common. This study considers tree representations such that the host tree comes from the family of subdivided n-stars. The largest such representable asteroidal set is constructed, and a lower bound on the length of the longest cycle representable on this family of host tree is also discovered. Next we move to a different area of study. The study of closure systems and closure spaces is a relatively new direction in mathematics. Jamison writes in his new text that ‘the notion of closure is pervasive throughout mathematics’. Surely, closure and closed sets can be discussed in almost any mathematical setting. It has been shown by Pfaltz in 1995 that for any finite ground set S, with |S| = n such that n ≥ 10, there are n unique closure operators on S. In topology, the separation properties provide criteria for categorizing topological spaces. While not all closure spaces are topological spaces, we may still explore whether the separation properties hold under certain conditions. This work defines a class of closure operators on the integers and investigates the conditions for which the resulting closure space satisfies the different definitions of separability.

target graph G is an assignment of subtrees of a host tree to the vertices of G in such a way that if uv ∈ E(G), then the subtree assigned to the vertex u and the subtree assigned to the vertex v have at least t nodes in common. This study considers tree representations such that the host tree comes from the family of subdivided n-stars. The largest such representable asteroidal set is constructed, and a lower bound on the length of the longest cycle representable on this family of host tree is also discovered.
Next we move to a different area of study. The study of closure systems and closure spaces is a relatively new direction in mathematics. Jamison writes in his new text that 'the notion of closure is pervasive throughout mathematics'. Surely, closure and closed sets can be discussed in almost any mathematical setting. It has been shown by Pfaltz in 1995 that for any finite ground set S, with |S| = n such that n ≥ 10, there are n n unique closure operators on S.
In topology, the separation properties provide criteria for categorizing topological spaces. While not all closure spaces are topological spaces, we may still explore whether the separation properties hold under certain conditions. This work defines a class of closure operators on the integers and investigates the conditions for which the resulting closure space satisfies the different definitions of separability.

ACKNOWLEDGMENTS
I would like to thank my thesis advisor, Dr. Nancy Eaton, for her countless hours of collaboration and guidance.
A great debt of gratitude is owed to Dr. Robert Jamison (Clemson University) for his helpful discussions about graph representations and closure systems.  (1918 -2010) iv PREFACE This dissertation contains a mixture of background material, published work, and work that is in preparation for publication. In accordance with the guidelines set forth by the Graduate School at the University of Rhode Island, the dissertation is written in manuscript format.
The study of graph representations is an active area of research within graph theory. Multiple mathematicians, including Eaton, Erdős, Füredi, Golumbic, Grable, Jamison, Kostochka, Rödl, Trotter, and more have done work in this area. Much of the work done in this dissertation discusses problems in the theory of graph representations.
A representation of a graph consists of three objects, 1) a host set S, 2) an assignment function f , and 3) a conflict rule g. The host set may be any collection of objects, ranging from an arbitrary set of elements, to trees, to subsets of the plane, and anything in between. The assignment function assigns a subset of the host set to each vertex of a target graph (that is, a graph which we desire to represent).
Finally, the conflict rule compares these assigned subsets to determine whether or not two vertices should be adjacent. If, given a host set S and conflict rule g, there is a suitable assignment function such that the graph G is induced by the conflict rule, then we say that G is (S; g)-representable. Graph representation problems have interesting applications, for example, in the field of computing. In this dissertation, specific representation schemes are explored, and previously unknown limitations of these schemes are revealed.
Chapter 1 of this document will provide a short introduction into the basic graph theory and the notation commonly used throughout the dissertation. The next four chapters concern results on graph representations. Chapter 2 serves as an introduction to graph representations, providing motivating examples and a bit v of history. While the contents of this dissertation will focus specifically on tree representations of graphs, in the second chapter we will discuss the traditional set representations, tree representations, and one of the main differences between the two.
In the third chapter we discuss a result discovered by J.R. Walter in his 1972 dissertation. Here we verify that Walter's result holds in the discrete case. We also use this chapter as motivation for chapters 4 and 5 which can be viewed as extensions of his work.
In chapters 4 and 5 we discuss representations of asteroidal sets and of cycles, respectively, on a family of host trees called subdivided stars. Chapter 4 contains an original result which has been accepted for publication in the Journal of Combinatorial Mathematics and Combinatorial Computing; while Chapter 5 contains work towards generalizing a known result about tree representations of cycles.
Chapters 6 and 7 discuss work I have done in a different area of mathematics. The notion of closure and closed sets can be discussed in almost all areas of mathematics. For a ground set S, closure can be defined in many different ways. In fact, it has been shown by , that for a ground set S of size n, with n ≥ 10, there are n n unique closure operators on S.
Chapter 6 contains an introduction to closure systems and closure spaces. Four examples of closure operators on different ground sets are given and briefly explored.
Finally, Chapter 7 discusses original results about closure spaces resulting from defining a certain class of closure operators on the integers. We will examine these closure spaces specifically with respect to separability. The separation properties give mathematicians useful ways to classify spaces. Traditionally these properties have been used to classify topological spaces, but closure spaces are also a natural place to discuss separability. vi 1 CHAPTER 1 Some Background in Graph Theory

Introduction
Before we begin discussing the main topics of this dissertation, it will be necessary to have a bit of background in graph theory. This first chapter will provide the basic background necessary for reading the next four chapters of this work.

What Is A Graph?
In graph theory, a graph G is a pair of sets. That is, G = (V, E), where V is a set of elements called vertices and E is a set containing pairs of vertices, called edges.
Typically a graph can be drawn with each vertex denoted by a dot and an edge xy by an arc connecting the dots corresponding to x and y. While there are many ways to arrange the vertices and edges of a given graph, the particular drawing does not change the underlying structure of the graph. It can be beneficial to note, however, that some drawings are more desirable than others. In particular one may try to draw a given graph in such a way that the number of edge crossings is minimal. For example, consider the graph G = (V, E), where V = {x, y, z, w}, and E = {xy, xz, xw, yz, yw, zw}. The two drawings below both depict G, however, the first drawing has one edge crossing, while the second has no edge crossings. Again, both of the drawings depict the graph G, but the graph on the right shows that G is planar. That is, G can be drawn in the plane with no edge crossings.
A graph G = (V, E) is said to be simple if it contains no multiple edges, that is to say that the edge xy occurs at most once in E; there are no loops, meaning that there are no edges of the form xx in E; and edges do not have a direction associated with them, so the edge xy and the edge yx are no different. In graph theory it is also possible to consider multigraphs, where loops and multi-edges are permitted, as well as di-graphs, where edges are given direction. In this dissertation, when we refer to a graph we will always mean a simple graph.

More on Vertices and Edges
We say that two vertices x and y in a graph G are adjacent if xy is an edge in We also say that an edge is incident to a vertex x if x is one of the endpoints of that edge. The degree of a vertex x in a graph G, denoted deg G (x) (or deg(x) for short) is the number of edges incident to x in the graph G.

Subgraphs
We say that a graph H = (V , E ) is a subgraph of the graph G = (V, E) if V ⊂ V and E ⊂ E. In the event that H is a subgraph of G, we often write H ⊂ G.
Furthermore, a subgraph H of a graph G is said to be an induced subgraph of G if for any pair x and y of vertices in H such that xy is an edge in G, xy is also an edge in H. If G is a graph with vertex set V and U is a non-empty subset of V , then we denote by G[U ] the induced subgraph of G with vertex set U . See below for an example of a graph, one of it's subgraphs, and one of it's induced subgraphs.
A Graph A Subgraph An Induced Subgraph

Some Special Classes of Graphs
Throughout the dissertation we will refer to and make use of some special classes of graphs. A graph is said to be complete if that graph contains an edge between every pair of vertices. The complete graph on n vertices is denoted by K n and has n 2 edges. A path is a graph whose vertex set is V = {v 1 , v 2 , ..., v n } and whose edge set is E = {v 1 v 2 , v 2 v 3 , ..., v n−1 v n }. Note that the length of a path is typically measured by the number of edges it contains; and the path of length n is denoted by P n . A cycle is a graph whose vertex set is V = {v 1 , v 2 , ..., v n } and whose edge set is E = {v 1 v 2 , v 2 v 3 , ..., v n−1 v n , v n v 1 }. The cycle of length n is denoted by C n .
See below for examples of a complete graph, a path, and a cycle.
Another class of graphs, called trees, are connected graphs that contain no cycles.
Trees are graphs which will be important objects throughout chapters 2, 3, 4, and 5 of this dissertation. Any tree will have at least two vertices of degree one; and these degree one vertices are typically called the leaves of the tree. A star is a special type of tree with one central vertex connected to all of the other vertices.
Note that this forces all of the vertices except the central vertex to be leaves, since any additional edge would create a cycle. The star with n leaves is denoted by K 1,n .
Finally, we will be interested in looking at subdivisions of stars. A subdivision of a graph is created when a single edge is replaced with a path. See below for an example of a star and a subdivision of that star.
K 1,5 A Subdivision of K 1,5 Note that in the example above, our star has undergone multiple subdivisions, and that not all of the subdivisions replaced edges with paths of the same length.
In any case, we call such a graph a subdivision of K 1,5 . In general, within this dissertation, we label any graph which is a subdivision of K 1,n by K 1,n . These graphs in particular will be extremely important in chapters 3, 4, and 5 of this dissertation.
Note that this chapter is by no means a comprehensive overview of basic graph theory. It does, however, provide the reader with the basic graph theoretic knowledge required to proceed with reading this dissertation. New definitions will be presented throughout the dissertation as they are needed. Any common graph theoretic terms the reader may come across which have not been identified in this chapter may be found in [1]. Notice that in the previous example, a conflict existed whenever the class rosters had a non-empty intersection. However, a conflict may be defined in any way that one wishes. It may be the case that the dean of the school has determined that too many exam slots are used, and that it would be best to declare a conflict between two classes only if three or more students are enrolled in both. In this case, the tolerance for conflict has been increased from 1 to 3, and all of the exams from the previous example could be given concurrently.
One may also consider the reverse problem. That is, one can declare a set of elements, called a host set, and also define a conflict rule with the goal of discovering which graphs can be realized as a result of assigning subsets of the host set to vertices of a target graph and analyzing the conflicts that arise. This venture isn't completely superficial. Consider the problem of communicating the structure of the following graph to a computer.
The representation scheme commonly discussed in introductory courses in graph theory is the adjacency matrix. The adjacency matrix for such a graph is where a 1 in the (i, j)th entry of the matrix communicates that the edge v i v j exists in the graph. Such a representation is costly in terms of computing resources, though. In general, the adjacency matrix for a graph on n vertices is an n × n matrix. This means that communicating the graph structure of a graph on n vertices via this method requires defining an n × n array. Are there more efficient ways to communicate the graph structure to a computer? If so, what are their limitations? These are two questions which are interesting to explore. We can use graph representations, and the idea of the conflict graph to explore these questions. For example, consider the 6-vertex graph which was given as the previous example. Let S = {0, 1, 2, 3, 4, 5} and assign subsets of S to the vertices via the assignment function f (v i ) = {i mod 6, i + 1 mod 6}. Drawing the conflict graph with tolerance 1 for the this set assignment yields the graph from the previous example.
Notice that we only have to statically define the host set S, define a function which assigns the subsets to each of the vertices, and then to define a function which determines when a conflict exists. Also, once such a host set and conflict rule are communicated to the computer, we may modify the assignment function in order to construct new graphs. Consider the conflict graph defined by the host set and conflict rule above, but where the set assignments are given by the function The adjacency matrix for this graph is communicated by defining an entirely new 6 × 6 array. This shows that defining the graph by the graph representation is more versatile than the definition via the adjacency matrix.

Two Types of Graph Representations
While there are certainly many types of graph representation schemes (see [1] for a comprehensive list), we will limit ourselves to considering two types. We first consider what is commonly referred to as a set representation of a graph. We then look to a second representation scheme, and our main object of study, tree representations of graphs.

Set Representations
Graph representations have been a topic of interest to many graph theorists since Szpilrajn-Marczewski introduced the notion of a set representation of a graph [2] in 1945. Given a graph G = (V, E), a representation of G consists of the following collection of objects: (1) a set S, (2) a function f : V → P(S) (the power set of S), and (3) a function g : We call S the host set, f the assignment function, and g the conflict rule. A graph G is representable under a given host set S and conflict rule g if there exists a suitable assignment function f , in which case we say that G is (S; g)-representable.
The following theorem of Erdős,Goodman,and Pósa [3] is one of the most basic and important results in the theory of graph representations. Theorem 2.2.1 For any finite graph G = (V, E), there exists a host set S of sufficient size such that G is (S; 1)-representable.

Proof.
Given a finite graph G, let S = {1, 2, ..., |E|}, where |E| denotes the number of edges in G. Label each edge uv ∈ E with a distinct element from S, call it (uv). Now, assign to each vertex the set consisting of all of the elements assigned to the edges which that vertex is incident to. Note that this assigns each element of S uniquely to two vertices. That is, if uv ∈ E, then the sets assigned to vertex u and vertex v have the element (uv) in common. If uv ∈ E, then the sets assigned to the vertex u and the vertex v have an empty intersection. This, by definition, is an (S; 1)-representation of G. 2 Noting that the graph representation resulting from the above construction is unlikely to be the most efficient representation, those mathematicians pursuing the study of graph representations began the search for the most efficient graph representation scheme. That is mathematicians search for the answer to the question: among all possible conflict rules, which is the one which minimizes the size of the host set required to represent any graph on n vertices.
This new direction has resulted in the exploration of many different types of graph representations, including (but not limited to) interval representations, circular arc representations, box representations, and tree representations. In many cases the exploration of these representation schemes has led to new classes of graphs, such as interval graphs, circular arc graphs, visibility graphs, and more. For a more comprehensive description of the types of graph representations which have been explored, please consult [1].

Tree Representations
Tree representations of graphs are an interesting variation on the set representations that we have discussed so far. The host is a tree, giving more structure than merely a set of elements; objects assigned to vertices of a represented graph are subtrees of the host tree; and an edge exists between two vertices if and only if their assigned subtrees intersect in t or more nodes, where t is a prescribed conflict-tolerance.
It can be seen that tree representations and set representations have significant differences. For example, consider Theorem 2.2.1 about set representations. Could there possibly be a similar theorem about tree representations? After a few moments the reader may notice a problem. In 1974 Gavril proved that the class of graphs having tree representations with tolerance 1 consisted exactly of those graphs whose induced cycles are all 3-cycles [4].
Increasing the conflict tolerance allows for less restrictive classes of graphs to be representable: for example longer cycles can be represented if the conflict tolerance is raised to at least 3. Increasing this conflict tolerance does increase the difficulty of the tree representation problem, though. It has been both fruitful and interesting to study the classes of graphs representable with higher conflict tolerance, but on fixed families of host tree. Eaton and Faubert [5] studied graphs representable on caterpillars, where a caterpillar is defined as a tree whose vertices all lie on or adjacent to its longest path. Eaton and Barbato [6] studied cycles representable on subdivisions of the 3-star. Chapters 2, 3, and 4 explore further those graphs representable on subdivisions of stars.

Discretization of a Known Result
A Note to the Reader When Lekkerkerker and Boland discovered that the class of graphs representable on an interval are exactly the chordal, non-asteroidal graphs [1], they also proved that the path representable graphs were exactly the interval representable graphs. A path is the discrete version of an interval. In his dissertation, J.R. Walter classified the graphs representable on a union of three intervals joined at a common endpoint.
This result only appeared in Walter's dissertation and was never peer reviewed.
Unlike Lekkerkerker and Boland, Walter did not verify that his result would still apply on the discrete analog of his tree. Walter's tree can be discretized as a subdivision of the 3-star. The inclusion of this chapter serves two purposes. Firstly verification that Walter's result still holds in the discrete case is needed, and secondly, Walter's result provides motivation for the work done in chapters 4 and

5.
A Discretization of a Result of J. R. Walter by J.R. Walter, Communicated by A.J. Gilbert

Introduction
Walter's dissertation considered tree representations of graphs. Recall from Chapter 2 that tree representations are a variation on the traditional graph representation. The host is a tree, giving more structure than merely a set of elements; objects assigned to vertices of a represented graph are subtrees of the host tree; and an edge exists between two vertices if and only if their assigned subtrees have a non-empty intersection. An important distinction between set representations and tree representations is that, while all graphs are (S; 1)-representable for large enough S, there exist graphs which are not (T ; 1)-representable, for any host tree T . For example, graphs containing cycles of length four or greater are always forbidden in tree representations where the conflict-tolerance t is set at 1.
James Walter considered the fixed host tree consisting of three unit intervals joined at a common endpoint. We will discretize Walter's result by considering the host tree K 1,3 . That is, we consider the host tree to be a subdivision of the 3-star.
Throughout this chapter it will be useful to have specific notation for objects such as the target graph, the host trees, and collections of subtrees which induce a representation of the target graph. We will use the following conventions on notation to help us distinguish these objects.
1. host trees and host sets will be denoted by calligraphic letters (T , H, S, etc.).
2. the standard font will be used to denote the target graph, its subgraphs, and subtrees of the host tree (G, H, T , etc.).

3.
A representation of a graph (that is, the collection of assigned subtrees) will be denoted by script letters (T , S , etc.).
4. Under a representation T , of a target graph G, we use T v to denote the subtree from T assigned to the vertex v of the target graph.

The Main Result
The remainder of this chapter will be dedicated to proving and verifying Walter's result in the discrete case. We begin by giving a definition and follow with the result and its proof.
and ∀i, j, k ∈ {1, 2, 3} with i, j, and k distinct, there exists a path between v i and v j which does not intersect N (v k ). If |A| = m, then A is said to be an m-asteroid of G. Furthermore, if A is the maximum size asteroidal set contained in G, then G is said to be m- at most three asteroidal, and has the property that if u 1 , u 2 and v 1 , v 2 are vertices of two asteroidal triples of G, then any path from u 1 to u 2 must be adjacent to any path from v 1 to v 2 [2].
Observation 3.2.4 G is at most 3-asteroidal.
Proof. Let v i , v j , and v k be vertices contained in an asteroidal set of G. It is easy to see that v i , v j , and v k must be assigned subtrees contained on different branches of K 1,3 . Assume otherwise, then at least one of the assigned subtrees must contain the branch node (the unique node of degree 3) of K 1,3 , otherwise we would have an asteroidal triple represented on an interval, violating the 1962 result of Lekkerkerker and Boland [1].
Assume T v i contains the branch node of K 1,3 . If T v j and T v k are represented by subtrees contained on different branches of K 1,3 , then any path from v j to v k in G must contain a vertex whose assigned subtree contains the branch node of G. This vertex must be adjacent to v i , a contradiction that v i , v j , and v k are contained in an asteroidal set of G. So, T v j and T v k must be contained on the same branch of K 1,3 . This, however, means that one of T v j and T v k must be 'closer' to the branch node of K 1,3 . Say that T v j is closer. Any path from T v i to T v k must contain a vertex whose assigned subtree intersects T v j . That is, any path from v i to v k contains a vertex adjacent to v j , again a contradiction. Thus, T v i , T v j , and T v k must all lie on different branches of K 1,3 . Since there are only three branches, there can be only three such vertices.
Observation 3.2.5 If v 1 , v 2 and u 1 , u 2 are vertices contained in asteroidal triples of G, then any path from v 1 to v 2 in G, must be adjacent to any path from u 1 to u 2 in G.

Proof.
Note that T v 1 and T v 2 must lie on different branches of K 1,3 , as must T u 1 and T u 2 . Now, any path in G from v 1 to v 2 must contain a vertex v * , whose assigned subtree contains the branch node of K 1,3 . Similarly, any path in G from u 1 to u 2 must contain a vertex u * , whose assigned subtree also contains the branch node of K 1,3 . That is, v * u * is an edge in G, and the two paths are adjacent.
The three observations show that if G is (K 1,3 ; 1)-representable, then G is chordal, at most 3-asteroidal, and has property that if v 1 , v 2 and u 1 , u 2 are pairs of vertices from asteroidal triples of G, then any path from v 1 to v 2 in G must be adjacent to any path from u 1 to u 2 in G.
We now assume that G is chordal, at most 3-asteroidal, and has the property that if v 1 , v 2 and u 1 , u 2 are pairs of vertices from asteroidal triples on G, then any path in G from v 1 to v 2 is adjacent to any path in G from u 1 to u 2 . From now on we will refer to the third property as the 'adjacency of asteroidal paths property'. We proceed by cases, giving definitions at the point where they are first needed.
connected, and weakly simplicial otherwise.
Using subdivision if necessary, assign T u and T v to be disjoint subtrees contained in the intersection. Replacing T u with T u , and keeping all other subtrees the same, gives a representation of G on T .
Definition 3.2.9 A chordal graph G is called extremal if every simplicial vertex of G is strongly simplicial.
Case 1: Assume that G is extremal Observation 3.2.10 Let G be an extremal, reduced, chordal graph, then G is nasteroidal iff G has exactly n simplicial vertices.
Assume that G has n simplicial vertices v 1 , v 2 , ..., v n , all of which are strongly is connected for each p ∈ {i, j, k}, there exist a path in G − N [v p ] between the remaining pair of simplicial vertices. This path is also a path in G, and is clearly not adjacent to v p . Since this can be done for any choice of three simplicial vertices, we must have that {v 1 , v 2 , ..., v n } satisfies the definition of being an n-asteroid of

G.
Assume that G is asteroidal, with n-asteroid {v 1 , v 2 , ..., v n }. Assume that {v 1 , v 2 , ..., v n } is chosen to contain the largest number of simplicial vertices, but that at least one is not simplicial. Say v 1 is not a simplicial vertex. For each distinct pair (i, j) with i and j from the set {2, 3, ..., n}, let P 1 i,j be the collection of all paths in G from v i to v j , not adjacent to v 1 . It is obvious that P = ∪ i,j P 1 i,j is a connected subgraph of G, and by a result contained in the Lekkerkerker and Boland paper [1], each non-empty component of G − N [P ] contains a simplicial vertex of G. In particular, we look at the component containing v 1 , to find a simplicial vertex v * . Now, there is a path in this component from v * to v 1 , which is clearly not adjacent to any of the other vertices contained in our asteroidal set.
Appending this path to the existing asteroidal paths corresponding to v 1 gives the necessary paths to show that {v * , v 2 , ..., v n } is an n-asteroid consisting of more simplicial vertices than the original, contradicting our assumption that the original triple contained the most simplicial vertices.
Note that the extremal condition was not required to show that if our graph was n-asteroidal, then each of the asteroidal vertices can be assumed to be simplicial.
We will move forward with the extremal case, and then consider the remaining cases with asteroidal triples consisting of simplicial vertices.
We this observation in mind, we prove that such a chordal, extremal graph is representable on K 1,3 by induction.
The two 3-asteroidal, chordal graphs on six vertices are easily representable on We now assume G has n vertices, with n > 6, and that every 3-asteroidal, chordal, extremal graph on fewer vertices has a (K 1,3 ; 1) representation.
By what was shown in Observation 3.2.10, there is a unique asteroidal triple of G, consisting of simplicial vertices, v 1 , v 2 , and v 3 .
Let G = G−v 1 . If G is not asteroidal, then it is representable on a path P because of Lekkerkerker and Boland's result [1]. Furthermore, since N [v 1 ] is complete, then we let x be a node in this intersection. Let T = P ∪ xx * , where xx * denotes an edge between node x and a new node, x * , not originally on P.
Now, note that T is a subdivision of K 1,3 . We obtain a representation of G on T from the representation of G − v 1 on P by appending the edge xx * to the subtrees assigned to vertices in N (v 1 ), assigning T v 1 = {x * }, and leaving all other subtrees the same.
Thus we will assume that G − v 1 remains 3-asteroidal, with asteroidal triple {u 1 , u 2 , u 3 } consisting of simplicial vertices. Note that it must be the case that ] would imply that G has more than 3 simplicial vertices, contradicting that G is 3-asteroidal. By definition then, u 1 is a strongly simplicial vertex in G . If u is any other simplicial vertex of G , then Thus, G is extremal, and has a representation on K 1,3 .
Observe that this representation can be chosen in such a way that all of the subtrees assigned to members of N (v 1 ) contain a common leaf, p, of K 1,3 . In- Definition 3.2.11 A chordal graph, G, is said to be quasi-extremal if every asteroidal triple of simplicial vertices consists only of strongly simplicial vertices.

Case 2: G is quasi-extremal
Assume that G is reduced, and has a weakly simplicial vertex v.
be an asteroidal triple of simplicial vertices of G.
is an asteroidal triple, but v is weakly simplicial, contradicting quasi-extremality.
We also show that G 1 = G − K 1 can be represented on a path, P. Since v 1 , v 2 , and v 3 must all lie in different components of G − N [v], we know that v is not strongly simplicial in G − K 1 . Assuming that G − K 1 is not representable on a path, then G − K 1 must contain an asteroidal triple {u 1 , u 2 , u 3 }.
, as the common attachment vertices of K.
Note that u 1 / ∈ A(K 1 ), since otherwise {v 1 , u 2 , u 3 } would form an asteroidal triple of simplicial vertices of G, u 2 and u 3 would have to lie in different components , so we could have no path from u 2 to u 3 not adjacent to u 1 , contradicting that {u 1 , u 2 , u 3 } is an asteroidal triple. Since u 1 / ∈ A(K 1 ), we must have that u 1 is a simplicial vertex of G. By the quasiextremality assumption, each of u 1 , u 2 , and u 3 are strongly simplicial in G. Letting v 1 = u 4 gives {u 1 , u 2 , u 3 , u 4 }, a set of four strongly simplicial vertices of G, and thus is a 4-asteroid in G, contradicting that G was 3-asteroidal. So G must be representable on the path P 1 .
Also, we note that may be assigned subtrees containing a common leaf, p, of the path that G 2 is represented on. We verify this assertion with the following claim. Proof. We will consider cases: 1. Consider first, that v is strongly simplicial. Let G have a representation, T on the host path P. Now, consider X = P − ∪ u∈N [v] T u . Now, X must be a connected subset of P. This, more importantly, means that there is a leaf, p of P in X. Since all of the subtrees defined on X are part of the same clique, we may extend the subtree assignments of all of these vertices such that their assigned subtrees contain the leaf p.
2. We now consider that v is weakly simplicial. Then we must have that has more than one component. We call them K 1 , K 2 , ..., K m . Let  ii.
. Now, we must have a vertex u 1 ∈ K k such that u 1 is not adjacent to w 2 and u 2 ∈ K k+1 so that u 2 is not adjacent to w 1 . Now in K k ∪{w 1 }, we take a path from u 1 to w 1 that is not adjacent to w 2 , and add to this path the edge w 1 v 1 , giving us a path from II. Assume j = k. By the hypothesis on j we must have that Now that we have representations of two different graphs on two different host trees, it may be beneficial to provide some way to distinguish them. Let T 1 v denote the subtree of P 1 assigned to the vertex v of G 1 , and T 2 v denote the subtree of P 2 assigned to vertex v of G 2 .
We note that in our path representation of G 1 , it must be that Definition 3.2.14 Let v i be a simplicial vertex of G and let K be a component Case 3: G is neither extremal, nor quasi-extremal Also, since there is no full-neighbor of N (v 1 ), this non-empty intersection cannot intersect with a subtree assigned to a vertex outside of N (v 1 ).
If v is any vertex of K, then {v, v 2 , v 3 } forms an asteroidal triple of G. Suppose Note that δ ∩ H = ∅, and δ ∩ H can be assumed not to be exactly δ, by making use of subdivision if necessary. Let T v 1 = δ \ H, and we have a representation of G on K 1,3 .
Subcase 3: One of the components K i , with i = 0, contains a full-neighbor of N (v 1 ) as well as a vertex which is not a full neighbor of N (v 1 ).
Note that v 1 is strongly simplicial in G 2 , and that G 2 cannot be asteroidal since any path from v 2 to v 3 in K 0 would not be adjacent to the asteroidal paths between the vertices in G 2 , contradicting the adjacency of asteroidal paths condition. Since G 2 is not asteroidal, it has a representation, T 2 on a path, P. Furthermore, since each vertex of B(K 1 ) is adjacent to all of the other vertices in G 2 , we assume that By induction, we can also assume that G 1 is representable on K 1,3 , with representation T 1 . Since K 1 contains a full-neighbor of N (v 1 ) and K 1 also contains a simplicial vertex of G, call it v 4 , the set {v 2 , v 3 , v 4 } forms an asteroidal triple of G 1 , and as a result, the corresponding subtrees lie on different branches of K 1,3 .
Since there is a path from v 2 to v 3 missing N [v 1 ], this path also misses N [v] for all v ∈ K 1 , because there is no way to cross into K 1 from K 0 without going through Now consider γ, the connected portion of K 1,3 containing all of the subtrees assigned to vertices in K 1 , and containing no nodes of K 1,3 contained in subtrees assigned to vertices in the other components. Note that the branch node of K 1,3 is not in γ since that node must be contained in a subtree assigned to a vertex in This implies that γ is properly contained on the branch of K 1,3 , not containing T 1 v 2 or T 1 v 3 . So we have that δ and γ are properly contained on the same branch of K 1,3 , and that both γ and δ are paths. Note that δ ∩ γ = ∅, since K 1 contains a full neighbor of N (v 1 ). Now we can obtain a representation of G on K 1,3 by replacing γ by P (the path on which G 2 is represented), so that the leaf p, contained in This assignment of subtrees gives a representation of G on K 1,3 .
and v 3 is the only component of G containing a full neighbor of N (v 1 ).
Note that v 2 and v 3 cannot be full neighbors of N (v 1 ), so K 1 0 contains vertices that are full neighbors of N (v 1 ) as well as those that are not. We must consider four possibilities in proving Subcase 4.
Possibility 1: There is a choice of asteroidal triple of simplicial vertices, such that v 1 is weakly simplicial, but v 2 and v 3 are strongly simplicial.
. We will show that H is representable on a path, P in such a way that all subtrees assigned to vertices in N [v 1 ] contain a common leaf of P. Next we will construct a representation of G on K 1,3 .
If H is not as desired, then either H is asteroidal, or not all vertices of N [v 1 ] can be represented by subtrees containing a common leaf of P.
If the latter is true, then by Claim 3.2.13 there must exist vertices u 1 and u 2 of H such that there exists a path in H from u 1 to v 1 not adjacent to u 2 , and similarly a path from u 2 to v 1 not adjacent to u 1 . Note that since v 1 is weakly simplicial in H, all components, particularly the two containing u 1 and u 2 , contain simplicial vertices. Therefore, we replace u 1 and u 2 each with u 1 and u 2 , the simplicial vertex in the corresponding component. The existence of u 1 and u 2 gives a 4-asteroid, It must be the case, that H remains 3-asteroidal. Let {u 1 , u 2 , u 3 } be a 3-asteroid of H. Note that at most one of these vertices, say u 1 , is from N (v 1 ). If u 2 and u 3 are in the same component, then {v 1 , u 2 , u 3 } is an asteroidal triple of G, and we have contradicted the adjacency of asteroidal paths condition, since the path from v 2 to v 3 is completely contained in K 1 0 , and the path from u 2 to u 3 is completely contained within another component. Therefore, u 2 and u 3 are in , and, as a result, u 1 / ∈ N (v 1 ).
Since u 2 and u 3 are in different components of H − N [v 1 ], any path between u 2 and u 3 must go through N (v 1 ). Also, since containing a full-neighbor, we know that u 2 and u 3 are not full-neighbors of N (v 1 ).
Recall that v 2 and v 3 are part of an asteroidal triple with v 1 , so neither of these and v 3 are strongly simplicial in G, there is a path from u 2 to v 2 not adjacent to v 3 or to u 3 , and similarly there is a corresponding path avoiding v 2 , and there are corresponding paths for u 3 . Again, we have contradicted that G was at most 3-asteroidal. Now we know that H is representable on a path, P, such that all of the subtrees assigned to vertices of N [v 1 ] contain the common leaf, p.
Identifying the nodes p and p * gives a representation of G on K 1,3 .
Possibility 2: There is a choice of asteroidal triple of G such that v 1 and v 2 are weakly simplicial, but v 3 is strongly simplicial.
Assume {v 1 , v 2 , v 3 } is chosen such that |K 1 0 | is maximal, and so that, with v 1 already chosen, |K 2 0 | is also maximal. Under this assumption we can show that is an asteroidal triple of G, and there is a path from v 4 to z in K 1 1 , then appending the path from v 4 to v 1 (through z) to the asteroidal paths If v 4 were strongly simplicial, we could apply possibility 1 to obtain the desired representation of G It must be the case that cannot be asteroidal. Suppose that {u 1 , u 2 , u 3 } is an asteroidal triple of H (and therefore of G). Now, we must have, by Observation 3.2.16, that no two of these vertices lie in the same component of . Also, none of these vertices is in N (v 1 ), since any path between the other two would be adjacent to this vertex, As a result, all of these vertices lie in Since v 3 is strongly simplicial in G, there is a path from u i to u j not adjacent to v 3 . Also, since K 1 0 contains a full-neighbor z, of N (v 1 ), and any path from u i to u j not adjacent to u k contains a vertex of N (v 1 ), one can construct the necessary paths to show that {u 1 , u 2 , u 3 , v 3 } forms a 4-asteroid of G, contradicting that G was at most 3-asteroidal. Thus, H is non-asteroidal, and has a representation on a path, P.
Let H 1 consist of all the components, K 1 i , which are represented to the left of T v 1 in the representation on P. Similarly, let H 2 be those components represented to can be represented on another path, P 2 , so that all vertices of N [v 2 ] are assigned representatives containing a common leaf, p * 2 , of P 2 . Again, we have made use of Claim 3.2.13.
and v 3 are all strongly simplicial, we may assume that the representation is such that if p 1 , p 2 , and p 3 are the leaves of K 1,3 , then each vertex of N [v i ] is assigned a subtree which contains p i . Most importantly, the subtrees assigned to ] contain both p 1 and p 2 . Now, identifying p * 1 with p 1 as well as p * 2 with p 2 gives a representation of G on K 1,3 .
Possibility 3: G has the property that for each choice of asteroidal triple such that v 1 is weakly simplicial, then v 2 and v 3 must also be weakly simplicial.
Choose {v 1 , v 2 , v 3 } such that |K 1 0 | is maximum, and so that |K 2 0 | is maximum with the selection of v 1 already made, and finally so that |K 3 0 | is maximum with the selections of v 1 and v 2 already made.
Under these assumptions, we show that for each pair, (i, j), with i ∈ {1, 2, 3} Assuming the opposite, we let Note that w 1 and w 2 may not be distinct. Now, there is a path from v to w 1 in K 1 k ∪{w 1 }, and a path from w 1 to v 2 not adjacent to v 3 . The union of these paths is weakly simplicial, so by our assumptions v must be weakly simplicial. However, contradicting the maximality of |K 1 0 |. Therefore, we have either Similarly, the maximality of |K 2 0 | guarantees that for each Finally the maximality of |K 3 0 | induces the analogous implication that, for each k ∈ {1, 2, ..., n 3 }, then . This gives us that each K i j is also a K m n for j and m both at least 1.
We now define the following subgraphs of G , and so on.
Also, note that J contains all of the components of G − N [v i ] for i = 1, 2, 3.
We will show that we must have two simplicial vertices u 1 and u 2 so that there is a path in from u 1 to v 2 not adjacent to u 2 and another path from u 2 to v 2 not adjacent to u 1 . Since such paths exist, and since {v 1 , v 2 , v 3 } is an asteroidal triple of G, we must have that {v 1 , u 1 , u 2 } is an asteroidal triple of simplicial vertices in G. From Observation 3.2.16, we have that u 1 and u 2 must be in different components of J 3 .
Furthermore, since v 1 is weakly simplicial, u 1 and u 2 must be as well. Since we know that u 1 and u 2 are not full-neighbors of N (v 2 ). Therefore, the component such that all vertices of N [v 2 ] are assigned representatives containing a common leaf, p. Similar arguments work for Only two cases now remain.
To show the remaining cases, we let A( and that these graphs are reductions of Note that, as a consequence of this, no two vertices of any J i may be part of the same asteroidal triple of G, since otherwise we would contradict the adjacency of asteroidal paths condition on G. It can also be seen that if u i ∈ J i , then {u 1 , u 2 , u 3 } forms an asteroidal triple of G, regardless of the selection of u 1 , u 2 , or u 3 . To check this, let u 1 ∈ K 1 1 ∈ J 1 , u 2 ∈ K 3 1 ∈ J 2 , and u 3 ∈ K 2 1 ∈ J 3 . By the choices of the J i 's, we have the existence of There is a path from u 1 to z 1 to z 2 to u 2 , not adjacent to u 3 .  Note that J −K i cannot contain an asteroidal triple of G, otherwise we would have another asteroidal triple of strongly simplicial vertices of G. This gives us more than three non-adjacent strongly simplicial vertices, contradicting that G was at most 3-asteroidal.
Now, either all of the K i 's are components of J 0 , or at least one of them is a component of J j for j ∈ {1, 2, 3}, otherwise one can discover a larger asteroidal set in G.
Sub-Option 1: K 3 ∈ J 3 As before, v 1 , v 2 , and v 3 are strongly simplicial vertices of G − J, and G − J has a representation on K 1,3 so that subtrees assigned to vertices in N [v i ] are assigned subtrees containing p i .
] is representable on a path, and as in Option 1, we can form H 1 and H 2 , and attach their path representations onto K 1,3 at p 1 and p 2 respectively to represent G − J 3 on K 1,3 .
a common leaf, p * of P. We may attach the representation of by adding the edge p * p 2 , and altering the subtree assigned to T v 2 to contain this edge, obtaining a representation of G on K 1,3 .
Sub-Option 2: For each asteroidal triple of G, none of the vertices lie in J 1 , J 2 , or J 3 . Again, at least one of the J j 's must be empty. For simplicity, assume it is J 3 .
Imitating the construction in sub-option 1, with ( , respectively, gives the representation. That is, we have shown that a graph satisfying the hypotheses, however which is neither extremal, nor quasi-extremal, does have a (K 1,3 , 1)-representation.
We have provided an exhaustive proof that as long as a graph, G, satisfies the hypotheses, then we can construct a representation of G on K 1,3 . 2 After wading through the details in this proof, one can see how difficult completely classifying the graphs representable on a certain class of host tree can be. Walter's result remains the only case of the following conjecture which has been verified.
Conjecture 3.2.17 A graph, G, is (K 1,n ; 1)-representable iff it is chordal, at most n-asteroidal, and has the adjacency of asteroidal paths property.
Since it can be seen from Walter's result that two obstacles in representing graphs by subtrees of K 1,3 are asteroidal sets and cycles, it would be natural to consider these structures while trying to extend the result. This is exactly what we will do in chapters 4 and 5.
[3] F. Gavril, "The intersection graphs of subtrees of a tree are exactly the chordal graphs," Journal of Combinatorial Theory, Series B, vol. 16, pp. 47-56, 1974. abstract. Consider a simple undirected graph G = (V, E). A family of subtrees, {T v } v∈V , of a tree T is called a (T ; t)-representation of G provided uv ∈ E if and only if |T u ∩ T v | ≥ t. In this paper we consider (T ; t)-representations for graphs containing large asteroidal sets, where T is a subdivision of the n-star K 1,n . An asteroidal set in a graph G is a subset A of the vertex set such that for all 3-element subsets of A, there exists a path in G between any two of these vertices which avoids the neighborhood of the third vertex.
We construct a representation of an asteroidal set of size n + n k=2 n k t−2 k−1 and show that no graph containing a larger asteroidal set can be represented.

Introduction
The study of graph representations is an active research area in graph theory.
Given a graph G = (V, E), a representation of G is the following collection of objects: (1) a set S, (2) a function f : V → P(S) (the power set of S), and (3) a We call S the host set, f the assignment function, and g the conflict rule. A graph G is representable under a given host set S and conflict rule g if there exists a suitable assignment function f , in which case we say that G is (S; g)-representable.
Much is known about graph representations when the conflict rule depends on the size of the intersection between assigned subsets. Such intersection representations have been studied extensively. A comprehensive list of authors can be found in [1].
Given conflict-tolerance t = 1, all graphs are (S; 1)-representable for large enough S. A central objective in the theory of graph representations is to find the smallest host set on which certain classes of graphs may be represented with respect to some given conflict rule. For example, the cycle C n can be represented on a host set of size n. Indeed, assigning the set {i, i + 1 mod (n)} to vertex v i gives a set representation of C n , however there is no possible assignment of subsets from a smaller host set that will induce C n with t = 1.
In this paper we consider tree representations of graphs. Tree representations are a variation on the traditional graph representation. The host is a tree, giving more structure than merely a set of elements; objects assigned to vertices of a represented graph are subtrees of the host tree; and an edge exists between two vertices if and only if their assigned subtrees intersect in t or more nodes, where t is a prescribed conflict-tolerance. An important distinction between set representations and tree representations is that given a host tree, not all graphs have a representation where the conflict-tolerance is t = 1, even if we allow subdivision of the tree. For example, cycles of length four or greater are always forbidden.
The classes of graphs representable using different host trees differ significantly (see [2,3,4,5,6,7]), and are thus interesting to study. A well known and interesting example is the result of Lekkerkerker and Boland [8]. We first give a definition and then state their result.
and ∀i, j, k ∈ {1, 2, 3} with i, j, and k distinct, there exists a path between v i and v j which does not intersect N (v k ). If |A| = m, then A is said to be an m-asteroid of G. Furthermore, if A is the maximum size asteroidal set contained in G, then G is said to be m- and non-asteroidal.
For convenience of notation, consider K 1,n to be a subdivision of the tree K 1,n .
In 1972 James Walter wrote his dissertation [9] on graphs representable on K 1,3 with tolerance t = 1. This can be thought of as a generalization of the result of Lekkerkerker and Boland, since K 1,3 is a logical extension from studying paths.
Walter discovered that graphs containing certain induced cycles and certain asteroidal configurations were not representable under this host tree and conflict rule.
Walter's result provides motivation for the work done in this paper, and is stated as follows. Given this result, it seems that cycles and asteroidal sets are interesting structures to study while considering tree representations of graphs. It was Jamison who stated that it would be interesting to explore what happens if the conflict-tolerance is greater than one. In 2001, Eaton and Barbato [2] studied representations of cycles on K 1,3 with arbitrary tolerance t. They described all cycles representable on K 1,3 with conflict tolerance t, showing that arbitrarily large cycles can be represented on K 1,3 at the cost of increasing the tolerance. Their theorem is restated below.
Theorem 4.1.4 For t = 3, 4, and 5 the maximum n such that C n is (K 1,3 ; t)representable is 3t−3. For t ≥ 6 the largest such n satisfies the following inequality A related result on cycles is due to Eaton and Faubert [4]. The two considered tree representations where the host tree is a caterpillar. That is, a tree in which every node is either on, or adjacent to, its longest path. Again, we provide a definition and then state their result. The two went on to show that C n ∈ cat [4,3] for all values of n. They also completely classified the graphs which are in cat [2, t] and cat [3,1] [3].
Going back to the result of Eaton and Barbato, note that for k ≥ 6, the cycle C k contains an asteroidal set of size k 2 . That is, their result on cycles implies that graphs containing arbitrarily large asteroidal sets are representable on K 1,3 . We explore this result further, and construct the largest asteroidal set representable on K 1,n with arbitrary conflict tolerance t > 1 and n ≥ 3. The remainder of this paper will be devoted to proving the following main theorem and discussing a few open problems. k−1 which is (K 1,n ; t)-representable exists. Furthermore, any graph containing a larger asteroidal set is not (K 1,n ; t)-representable.
Note that any terms not defined in this introduction, but used throughout the paper, can be found in [10].

Construction of the Graph Containing the Asteroidal Set 4.2.1 Vertices and their Assigned Subtrees
For convenience, we will refer to the graph we are constructing as the target graph.
Recall that our host tree is K 1,n . It shall be sufficient to assume that each branch of the host tree contains t nodes. For clarity, the word node will be used to indicate a vertex of the host tree, distinguishing these from vertices of the target graph.
We call the unique node of degree n the branching node of the host tree, and the branches of the host tree are labeled with the integers 0, 1, ..., n − 1. Furthermore, within this paper, any reference to the size of a subtree or of an intersection should be interpreted in terms of number of nodes. It should be noted that each subtree defined below will correspond to a vertex in the target graph. The vertex set of the target graph will be described in terms of four disjoint subsets V, W, P, and Q. Note that the subscripts of the vertices in V and P will give information as to which branches their corresponding subtrees exist on, and should be interpreted modulo n. The desired asteroidal set will be a subset of the vertex set of the target graph. The description of each set of vertices below is accompanied by an image, giving an example of the subtree representative of one of the vertices from that set.
The sample host tree K 1,3 is shown; the tolerance used in the examples is t = 5; and the selected subtrees are denoted by thick edges and filled nodes.
1. Let V be a collection of vertices represented by distinct subtrees, each of size exactly t, and contain the leaf of a branch of K 1,n . These subtrees are paths on the exterior portion of each branch of K 1,n . Since there are n branches of K 1,n , we will have n vertices in V. We will refer to these vertices later as v 0 , ..., v n−1 , where the subscripts denote the branch on which the corresponding assigned subtree exists.
The subtree assigned to one of the vertices in V 2. For 2 ≤ k ≤ n, we define W k to be the set of vertices represented by distinct subtrees which are of size exactly t and exist non-trivially on exactly k branches of K 1,n . There are n k t−2 k−1 such subtrees, and therefore the same number of corresponding vertices. We take W to be the union, over all values of k, of the sets W k . Clearly {W k } n k=2 forms a partition of W, and therefore |W| = n k=2 n k t−2 k−1 . Note that we assume the convention that a b = 0 for b > a.
The subtree assigned to one of the vertices in W 2 The subtree assigned to one of the vertices in W 3 The subtree assigned to one of the vertices in P 3. Construct P, the collection of all vertices representable by subtrees existing non-trivially on exactly two consecutive branches of K 1,n , and which extend out to the leaf node on each of these branches. We define vertex p i , whose assigned subtree we denote by T p i and contains the entirety of branches i and i + 1 for 0 ≤ i ≤ n − 1.
4. Let Q be a collection of vertices, each assigned a subtree which is an extension of a subtree assigned to a vertex in W. That is, for each vertex in W we create k(w) (recall that k is the number of legs of K 1,n on which T w , the subtree assigned to w, lives non-trivially) new vertices in Q. Each of these vertices is assigned a subtree which has been created by elongating, out to the second to last node, exactly one of the pre-existing legs of the subtree assigned to the corresponding vertex w. Consider the following example where we show T w , the subtree assigned to one of the vertices from W and one of its extensions, a subtree assigned to a vertex in Q.
T w , the subtree assigned to the vertex w ∈ W One of the three extensions of T w , giving the subtree assigned to one of vertices in Q

Basic Claims About Adjacencies
Consider the graph G = (V ∪ W ∪ P ∪ Q , E), where E is the edge set defined by the conflict tolerance relationship. Given the assignment of subtrees of K 1,n from the previous section, we have the following claims and observations about E. Note that we will use the notation T x to refer to the subtree assigned to vertex x of G.   Proof. Note that for each w ∈ W, we have that |T w | = t. Let w i , w j be distinct vertices in W. Then their assigned subtrees are distinct. That is, there exists at least one node of K 1,n which is a node of T w i ∪ T w j , but which is not a node of T w i ∩ T w j . This gives that |T w i ∩ T w j | < |T w i | = t. Therefore, given any two vertices from W, their assigned subtrees have an intersection of fewer than t nodes, and thus the vertices cannot be adjacent. In short, W is an independent set.  Proof. If j = i − 1 or i + 1 then p i and p j are assigned subtrees which exist on distinct pairs of branches of K 1,n . That is, their assigned subtrees intersect in exactly a single node (the branching node). Since t > 1, we have that p i and p j are not adjacent. Proof. Consider p ∈ P and w ∈ W with pw ∈ E. Then |T p ∩ T w | ≥ t. However, since |T w | = t, this implies that T p ∩ T w = T w . Now, since T w exists on at least two branches of K 1,n , and we know that T w ⊂ T p , then T w exists non-trivially on the same two branches of K 1,n as T p (which are indeed consecutive).
If T w exists non-trivially only on two consecutive branches of K 1,n (call them branch i and i + 1), then T w ⊂ T p i . Therefore, |T w ∩ T p i | = t, implying that p i ∈ N (w).

Claim 4.2.8
If q is a vertex from Q, then q and p j are adjacent iff T q was obtained from some T w by elongating branch j or branch j + 1.
Proof. If T q exists non-trivially on exactly two branches of K 1,n , then the result follows directly from the previous claim. We therefore assume that T q exists nontrivially on at least 3 branches of K 1,n . Let qp j ∈ E. Assume that T q was obtained from T w by an elongation of branch i, where i / ∈ {j, j + 1}. Now, since T p j exists non-trivially only on branches j and j + 1 of K 1,n , we have a j + a j+1 ≥ t − 1, where a j and a j+1 are the lengths of the legs of T q on branch j and j + 1 of K 1,n .
Furthermore, since i = j and i = j + 1, we must have that the previous statement is also true about T w . This, however, gives a contradiction since |T w | = t and T w lives non-trivially on at least 3 branches of K 1,n .
Proof. Let w 1 , w 2 ∈ W with w 1 = w 2 . Assume that N (w 1 ) ⊂ N (w 2 ). Now, consider Q w 1 the set of all q ∈ Q so that T q can be obtained via an elongation of one of the legs of T w 1 . Then, since N ( ∀q ∈ Q w 1 . This gives us that T w 2 ⊂ ∩ q∈Q w 1 T q . However, we know that ∩ q∈Q w 1 T q is exactly T w 1 . Therefore, we have that T w 2 ⊂ T w 1 ; but, since |T w 1 | = |T w 2 |, we know T w 2 = T w 1 . Thus w 1 = w 2 , a contradiction. This gives that there must exist a q 1 ∈ N (w 1 ) \ N (w 2 ). Proof. We recall that an asteroidal set A is an independent set which has the property that for any selection of 3 vertices, a 1 , a 2 , a 3 ∈ A there is a path connecting any two of them, which avoids the neighborhood of the third.

Verification of the Asteroidal Properties of V ∪ W
Firstly, notice that C = v 0 p 0 v 1 ..., v n−1 p n−1 v 0 is a cycle in G. This cycle will be used extensively in the verification of the properties of the asteroidal set. Also worth noting is that, for convenience of notation we will use P (u, v) to denote a path in G connecting the vertices u and v instead of the more commonly used (uv)-path.
The following claims 2.11 through 2.16 verify that this definition is satisfied on V ∪ W.

Claim 4.2.11
For v 1 , v 2 , v 3 ∈ V, ∀i, j, k ∈ {1, 2, 3}, with i, j, and k distinct, there exists a path, Proof. Let v 1 , v 2 , v 3 ∈ V. Now, we construct a path between v i and v j which Then the cycle C provides two paths from v i to v j , one of which must avoid the sequence p k v k p k+1 .

Claim 4.2.12
For v 1 , v 2 ∈ V, w ∈ W, there exists a path, P (v 1 , v 2 ), so that Proof. The cycle C connects the vertices v 1 and v 2 via two paths. Now, recall that N (w) contains at most one of the vertices on this cycle. If N (w) ∩ P = ∅, then we take either portion of the cycle as our path. If N (w) ∩ P is non-empty, then there must be exactly one vertex p ∈ N (w) ∩ P, so we traverse the portion of the cycle from v 1 to v 2 in the direction which avoids p. This gives us the desired path.
Proof. We begin from w. Let Q w denote the collection of all q ∈ Q, so that T q was obtained from T w by elongating one of its legs. Now, notice that |N (w) ∩ Q w | ≥ 2, so there exists q 1 , q 2 ∈ N (w) ∩ Q w . Then, recall ∀q ∈ Q, there exists two vertices p a , p b ∈ N (q) ∩ P. Since q 1 and q 2 are distinct, there are at least three distinct vertices p a , p b , p c ∈ N (q 1 )∪N (q 2 ). Also, recall that N (v j )∩P = {p j−1 , p j }.
Therefore, there exists p ∈ {p a , p b , p c } with p / ∈ {p j−1 , p j }. We take the path from w to this vertex p. We again traverse the path from vertex p to v i which uses the portion of the cycle C avoiding the sequence p j−1 v j p j .
Proof. Recall the existence of q i ∈ N (w i ) so that q i / ∈ N (w k ), and q j ∈ N (w j ) so that q j / ∈ N (w k ). We traverse from w i to q i and from w j to q j . If q i = q j , then we have the desired path already, so assume they are not equal. Now, q i and q j may be adjacent, however, we are unsure, so we travel on. Recall that N (w k ) may contain at most one vertex from P. Also, recall that each vertex from Q is adjacent to exactly two vertices from P. Therefore we can extend from q i to at least one of its neighbors p i and from q j to at least one of its neighbors p j without crossing into N (w k ). Again, if p i = p j we have the desired path, so we assume this is not the case. Now, we are on the cycle C. If N (w k ) ∩ P = ∅, then we connect p i to p j via either portion of the cycle. Otherwise N (w) ∩ P is a single vertex, p, and we connect p i to p j by a path along the cycle in the direction which avoids p.
In either case we have completed a path connecting the vertices w i and w j which avoids N (w k ).

Claim 4.2.15
For w 1 , w 2 ∈ W, v ∈ V, there exists a path, P (w 1 , w 2 ), so that Proof. If N (w 1 ) ∩ N (w 2 ) = ∅ then we can draw a path from w 1 to w 2 via their common neighbor. Recall that there are no edges between the partitions V and Q, so this path satisfies the requirements. Otherwise, N (w 1 ) ∩ N (w 2 ) = ∅. Then, similar to the proof of claim 2.13, there exist paths from w 1 to p a and from w 2 to . Now, we can use the cycle C to connect the vertices p a and p b with a path that does not intersect N (v).
Proof. We begin from w i . Recall that there exists q ∈ N (w i ), so that q / ∈ N (w j ).
We move from w i to q. Also, recall that N (w j ) contains at most one vertex in the partition P, and that q is adjacent to exactly two vertices from P. That is, we can extend from q to one of its neighbors in P without crossing into N (w j ). Now we find ourselves again on the cycle C. If N (w j ) ∩ P = ∅, we travel either half of the cycle from p to v. Otherwise, N (w j ) ∩ P consists of a single vertex, and we travel the half of the cycle connecting p and v which avoids this vertex. The union of the two selected paths provides a single path from w i to v with the desired property.
We have shown that V ∪ W satisfies the definition of an asteroidal set. Recall that |V| = n and |W| = n k=2 n k t−2 k−1 , and that V ∩ W = ∅. Therefore we have exhibited an asteroidal set of size n + n k=2 n k t−2 k−1 .

No Larger Asteroid is Representable Under the Restrictions of n and t
In the current section we show that the construction of a larger asteroidal set under the current restrictions of n and t is impossible. Note that showing this will prove the following theorem: Proof. Firstly, observe that an asteroidal set must be an independent set. Furthermore, it is easily seen that at most n k=1 n k t−2 k−1 vertices may be assigned subtrees which contain the branching node of K 1,n such that no two subtrees intersect in t or more nodes. Now, note that we have almost 'saturated' the branching node of the host tree with subtrees while constructing the asteroidal set of size n + n k=2 n k t−2 k−1 . Let M be the set of vertices which are assigned subtrees containing the branching node and extending non-trivially only on a single branch of K 1,n . We need only to consider whether the vertices of M and V can coexist in the same asteroidal set. We consider any path beginning at v i and ending at a. Along this path there must be a vertex u , the last vertex along the path whose assigned subtree does not contain the branching node of K 1,n . Now, we must have that T u +1 contains the branching node of the host tree, and also has an intersection of size at least t with inducing either the edge u +1 m i or u m i . That is, any path connecting v i with another asteroidal vertex must be adjacent to the vertex m i .
The previous claim directly implies that the vertices m i and v i cannot be part of the same asteroidal set. A similar agument can be used to show the same result about v i and a vertex whose subtree is a proper sub-path of the i th branch of the host tree. This implies that per branch we may only have one vertex whose assigned subtree exists non-trivially only on that branch, no matter the configuration of subtrees. Note that this property is satisfied in the configuration constructed in section 2.
Combining the result from section 2 with this result, we have shown the main result from the introduction, and have rediscovered the surprising corollary originally seen by Eaton and Barbato: Corollary 4.3.3 Graphs containing arbitrarily large asteroidal sets are repre-sentable on K 1,3 [2].
Since the size of the largest cycle which is (K 1,3 ; t)-representable is eventually quadratic in t, and the size of the largest representable asteroidal set grows exponentially, the following observation can be made directly from combining the main result here with Eaton and Barbato's result on cycles. Observation 4.3.4 If an asteroidal configuration of size m is (K 1,3 ; t)representable, it is not neccessary that every asteroidal configuration of size m or smaller has such a representation. In fact, the size gap can be made arbitrarily large.
The following fairly obvious observation can be made, however. Observation 4.3.5 If an asteroidal configuration of size m is (K 1,n ; t)representable, then there exists an asteroidal configuration of size m − k for each k ≤ m − 3 which is also (K 1,n ; t)-representable.

Open Problems
There are still many interesting open problems in the theory of tree representations, as well as some questions stemming from the main result in this paper. Answers to the following questions would be interesting. We can already see that the answer is non-trivial. The cycle gives an example showing that even though some large asteroid may be representable, not all smaller asteroidal configurations can be represented. The number of m-asteroidal configurations grows quickly, so given the current tools, it seems that analyzing relatively small cases is in order.

Conjecture 4.4.2
A graph G is (K 1,n ; 1)-representable iff G is chordal, at most n asteroidal, and G satisfies the condition that given any two pairs of vertices from asteroidal sets in G, any path connecting the first pair must be adjacent to any path connecting the second pair (Walter [9]).
The argument for necessity in Walter's conjecture is straight forward. The argument towards sufficiency, however, seems to require surgical detail, and has not yet been resolved. Problem 4.4.3 Exactly which graphs are (K 1,n ; t)-representable?
A complete characterization of the class of (K 1,n ; t)-representable graphs is the ultimate goal here. The fact that Walter's 1972 conjecture still remains undecided may be a good indicator that a full solution is still out of reach, however. [5] N. Eaton, Z. Füredi, A. Kostochka, and J. Skokan, "Tree representations of graphs," European Journal of Combinatorics, vol. 28(4), pp. 1087-1098, 2007.

List of References
[6] F. Gavril, "The intersection graphs of subtrees of a tree are exactly the chordal graphs," Journal of Combinatorial Theory, Series B, vol. 16, pp. 47-56, 1974. [7] R. Jamison and H. Mulder, "Constant tolerance representations of graphs in trees," in Graph Theory and Computing (Boca Raton, Florida), 2000, pp. 175-192. abstract. Consider a simple undirected graph G = (V, E). A family of subtrees, {T v } v∈V , of a tree T is called a (T , t)-representation of G provided uv ∈ E if and only if |T u ∩ T v | ≥ t. In this paper we consider (T , t)-representations for cycles, where T is a K 1,n -subdivision, denoted by K 1,n . We show that the maximum length cycle which is (K 1,4 ; t)representable is at least on the order of t 3 . Furthermore, we conjecture that the length of the longest cycle which is (K 1,n ; t)-representable is O(t n−1 ).

Introduction
The study of graph representations is an active research area in graph theory. Given a graph G = (V, E), a representation of G is the following collection of objects: (1) a set S, (2) a function f : V → P(S), and (3) a function g : We call S the host set, f the assignment function, and g the conflict rule. We say that a graph G is representable under a given host set S and conflict rule g if there exists a suitable assignment function f .
Much is known about graph representations when the conflict rule depends on intersection between assigned subsets. Intersection Representations have been wellstudied by many authors, see [1] for a comprehensive list.
Certain substructures within a graph can make that graph difficult or impossible represent with certain host sets and conflict rules. One common structure which has been given as a forbidden subgraph for certain conflict rules is the cycle. That is, giving restrictions for n and t prohibits representability of graphs containing induced cycles with length above a certain threshold dependent on n and t.

Preliminaries
In this paper we consider tree representations of graphs. A tree representation of a graph is a representation where the host set is a given tree T , subsets assigned to vertices of a representable graph G must be subtrees of T , and the conflict rule is a threshold intersection rule, where two vertices of G are adjacent if and only if their assigned subtrees intersect in at least some prescribed tolerance t. We will study the cycles which are (K 1,n ; t)-representable.
At this point we recall the 2001 result by Eaton and Barbato [2] which describes all cycles which are (K 1,3 ; t)-representable. Their paper shows that arbitrarily large cycles can be represented on K 1,3 at the cost of increasing the tolerance. We explore the same structure, but generalize the host tree to K 1,n .
We begin by taking generalizations of two preliminary lemmas from the paper of Eaton and Barbato.
Lemma 5.2.1 In a (K 1,n ; t)-representation of C p for p > 3, each vertex must be assigned a subtree such that the degree of the branching node, x, is at least 2.
Proof. The proof is by induction on p. We first show the case for p = 4. Assume that for at least one of the assigned subtrees, T v 1 , T v 2 , T v 3 , T v 4 , the degree of x (the branching node of T ) is at most one. Without loss of generality we shall assume that it is T v 1 . Let P 1 , ..., P n denote the branches of T . Again, without loss of generality, assume that T v 1 ⊂ P 1 . Now, we also know that T v 2 and T v 4 must each intersect T v 1 in at least t nodes, but may not intersect one another in t or more nodes. This again forces at least one of T v 2 or T v 4 to be a subset of P 1 . Now, T v 3 must intersect T v 2 and T v 4 in at least t nodes each, but may not intersect T v 1 in t or more nodes. However, since K 1,n is a tree, there is a unique path between T v 2 and T v 4 , and T v 1 contains this path. Similarly, for T v 3 to intersect T v 1 and T v 4 in at least t nodes each, it must contain this path. Now, it is easily seen that Thus, such a constructions of a C 4 is impossible.
We now consider p > 4. Assume that C p has a representation on T such that in at least one of the assigned subtrees, T v 1 , deg Tv 1 (x) < 2. Assume T v 1 ⊂ P 1 and again, T vp and T v 2 must each intersect T v 1 in at least t nodes, but may not intersect eachother in t or more nodes. This forces T v 2 ⊂ P 1 as well. Let Note that T v * is also a subset of P 1 . Furthermore, note that no new adjacencies we know this cannot be the case because we would have v k adjacent to both v 2 and v 1 .
Thus, the merging of the two subtrees creates no new adjacencies. Therefore, we have a representation of C p−1 with a subtree in which the branching node has degree less than 2. This gives that if C p is representable, then C p−1 is also representable. It must be the case then that for no p > 4 does C p have such a representation. Otherwise, we would ultimately arrive at a contradiction stating that C 4 has such a representation, which we proved was impossible. 2 Lemma 5.2.2 If C p has a (K 1,n ; t)-representation, then there is a (K 1,n ; t)representation of C p so that the following conditions hold: 1. |T u ∩ T v | = t for all subtrees assigned to adjacent vertices u and v 2. |T v | = t + 1 for all assigned subtrees Proof. Let p be the maximum value such that C p is (K 1,n ; t)-representable. To prove 1), suppose that |T v 1 ∩ T v 2 | ≥ t + 1 and show that we can reduce |T v 1 ∩ T v 2 | and still have a (K 1,n ; t)-representation of C p . We first remove any nodes of T v 2 which are not contained in T v 1 or T v 3 since they are extraneous. Note that this process may not disconnect T v 2 since the branching node, x, is contained in every assigned subtree. Let x a b denote the node on branch b, and which is of distance a from x, then for any k ∈ [n] and i ∈ {1, 2, ..} if for some v ∈ C p , x i k / ∈ T v , then for all j > i, we have x j k / ∈ T v . Thus, any nodes removed from T v 2 must form disjoint subpaths of T v 2 , each containing a leaf of T v 2 and not containing x. Now, observe that there must be at least two nodes of loss of generality, assume that at least one of these nodes is on branch one, and x is a node in all of the representatives, we know that x j 1 / ∈ T v 3 for any j > i. This means that there is a by one, and we still have a (K 1,n ; t)-representation for C p as was claimed. To prove 2), suppose |T v 1 | ≥ t + 2 and we show that we can replace T v 1 with two subtrees, yielding a representation of a larger cycle. Firstly, we again remove the nodes of T v 1 which are not contained in T v 2 or T vp since they are extraneous. Once again, the process exists two vertices 1 and 2 , such that 1 is a leaf of which intersect eachother in at least t nodes, since |T v 1 | ≥ t + 2. This gives a (K 1,n ; t)-representation of C p+1 , a contradiction to the assumption that C p was the longest cycle with such a representation. 2

Developing Notation
We take T v = (x 1 , x 2 , ..., x n ) to describe the subtree extending out to the x th i node along branch i of the host tree. From this observation we can see that each subtree can be identified as a point in a finite subset of N n . With respect to the two lemmas cited in the previous section, we can see that we are interested in the subset consisting of points whose coordinate sum is exactly t. Note that there are exactly n k=1 n k t−2 k−1 such points. As a direct result of Lemmas 5.2.1 and 5.2.2, we obtain the following useful corollary.
Corollary 5.3.1 Let X be the subset of N n consisting only of points whose coordinate sum is t. All of the assigned subtrees of K 1,n used to represent any cycle of length at least four correspond to points in X.
Since X is a subset of a vector space we will make use of the operations of scalar multiplication and vector addition where it makes notation convenient. We will use e i to denote the i th standard basis vector for R n .
We define edges on the pointset X as follows: ( x, y) ∈ E iff x = y + e i − e j for i, j ∈ [n] with i = j. We now consider the graph H = (X, E), in which we will search for the largest induced cycle. Proof. This follows immediately by the construction of H. 2

The Structure of H
We can now categorize the points of X according to certain common characteristics It is obvious that no simplicial point will be used as an assigned subtree in a representation of a cycle of length greater than three, since any two points adjacent to a simplicial point are also adjacent. Interior points have 'full neighborhoods'. That is, the neighborhood of an interior point contains 2 n 2 points.

Isolation Tubes
Note that there is a radius, r, from each originating point, t e i , such that, given two distinct originating points, t e i and t e j , and distinct points x and y within a distance of r from t e i and t e j , respectively, then x and y are not adjacent in H.
Definition 5.3.7 The isolation tube corresponding to the point t e i consists of all the nodes which are of distance no more than t 2 − 1 from t e i .
Note 5.3.8 We will refer to points of the form (t − k) e i + Ω k as being on the k th level of the isolation tube corresponding to e i , where Ω k denotes some linear combination of the remaining basis vectors, the sum of whose coefficients is k.
Theorem 5.3.9 If x and y are points from distinct isolation tubes, then they are not adjacent in H. That is, points from different isolation tubes are isolated from one another.
Proof. Consider, t is even. Let x = ( t 2 + 1) e i + Ω t 2 −1 and y = ( t 2 + 1) e j + Ω t 2 −1 . Recalling the rules for adjacency in H, we see that if there is any chance for x and y to be adjacent, then Ω t 2 −1 = ( t 2 − 1) e j and Ω t 2 −1 = ( t 2 − 1) e i . Thus, we have Now, it is easily seen that x and y differ by 2 in two coordinates. Since adjacency requires exactly two coordinates which differ by 1, and all other coordinates equal, we have shown that there can be no edge connecting x and y.
We now consider the case when t is odd: and, Again, if we hope for x and y to be adjacent we must have that Ω t 2 −1 = t 2 − 1 e j and Ω t 2 −1 = t 2 − 1 e i . Thus, Again, the difference in two of the non-zero coordinates is too large, and adjacency is not achieved. 2 Thus, we have shown that the isolation tubes truly do isolate the points of H.
Furthermore, it is easily seen that increasing the distance from the originating point beyond this threshold will invite adjacencies. These observations result in the fact that the radius of each isolation tube truly is t 2 − 1.

Finding the Largest Cycle in H
Using what we have previously discussed about the isolation tubes, a good place to begin is by constructing the longest induced path within an isolation tube. Note that since all isolation tubes are isomorphic, if we are successful, we can obtain n such longest paths which can be joined to make a cycle, giving us a non-trivial lower bound on the number we search for.

More About the Isolation Tubes
Before we can find a longest path through an isolation tube, we must gain some more information about its 'geography'. Noting that the radius of the isolation tubes is t 2 − 1 will allow us to have our discussion without using cases for t even and t odd. Also, since the isolation tubes are symmetric, we will consider the tube corresponding to t e 1 as the representative case, and refer to t e 1 as the originating point. i + n − 2 n − 2 points total contained within any isolation tube.
Proof. Note that points at distance k from t e 1 must be of the form x = (t − k) e 1 + Ω k . Since the sum of the coefficients must be exactly t, we know that there are k+n−2 n−2 ways to assign coefficients to the remaining n − 1 basis vectors, since the first coefficient is fixed at (t − k). Thus, there are k+n−2 n−2 points at level k of each isolation tube.
Clearly the levels partition the isolation tube, so summing over all the levels gives the total number of points in the isolation tube, as desired. 2 Now that we know the total number of points within each isolation tube, it would be beneficial to know more about the neighborhood of each point. That is, how many neighbors does each point have, and exactly where are these neighbors?
Definition 5.4.2 For a point x at level k of an isolation tube, we define the back neighbors of x to be neighbors of x at level (k − 1) of the isolation tube, the cross neighbors to be neighbors at level k, and the forward neighbors to be neighbors at level k + 1.
Claim 5.4.3 Consider a point, x, at level k, (k ≥ 1), of an isolation tube. The following must be true of x, 1.
x has (j − 1) back neighbors, all occurring at the (k − 1) st level.
4. x has n − 1 forward neighbors, all occuring at the (k + 1) st level.
1. Since x is at level k of the isolation tube, we must have x = (t − k) e 1 + Ω k .
Again, since the coefficients must sum to t, we know that the sum of the remaining coefficients is k. Since k is at least 1, we must have that at least one of the other basis vectors has a non-zero coefficient, implying that x is at most a type-(n − 2) boundary point. We may have up to min{k, (n − 1)} of the remaining coefficients being non-zero. Thus, x must be as claimed.
2. By the way the edges of H are defined, it is clear that any back neighbors of x must be at the (k − 1) st level of the isolation tube. By 1 we may assume that x is a type-(n−j) boundary point. To move from the k th level of the isolation tube to the (k − 1) st level we must increase the coefficient of e 1 by one. Thus, we may choose to decrease the coefficient of any of the (j − 1) other vectors with non-zero coefficients in order to make increasing the coefficient of e 1 possible. That is, we have (j − 1) choices for back neighbor.
3. Similar to 2, in order to move to a neighbor while remaining on level k, we must leave the coefficient of e 1 the same, but decrease the coefficient of one of the vectors with a non-zero coefficient by one and increase the coefficient on any of the other n − 2 vectors by one. Thus, we have (j − 1) other vectors with non-zero coefficients to choose from, and then (n − 2) viable vectors whose coefficient we wish to increase by one. By the fundamental principle of counting, this gives (j − 1)(n − 2) cross neighbors.

4.
A forward neighbor must occur at level (k + 1) since it is impossible for a point to have a neighbor further than one level away. To move from level k to level (k + 1) we must decrease the coefficient of e 1 by one and increase the coefficient of any other vector by one. There are certainly (n−1) possibilities for this, as claimed.
5. Summing over all of the back, cross, and forward neighbors gives jn − j total neighbors, as desired.
2 5.5 Construction of a Cycle for n = 4 In this section we construct a cycle in the case where n = 4, that is, we construct a long cycle having a (K 1,4 ; t)-representation. As discussed in the previous section, our goal will be to construct long paths through the isolation tubes and then to connect the paths together in such a way that a single cycle is formed.
Consider the isolation tube corresponding to e 1 . We begin by selecting two points from level 1 of the isolation tube, namely (t − 1) e 1 + e 2 and (t − 1) e 1 + e 3 . Proceed by selecting all points of the form (t − k) e 1 + k e 2 , for k ≤ t 2 − 1. Note that these points are along an induced path extending through all levels of the isolation tube. Furthermore, the selection of these points eliminates only points of the form (t − k) e 1 + (k − 1) e 2 + e j for j ∈ {3, 4} from consideration as points along our long path. This is certainly a minimal elimination, as can be seen from the discussion about neighborhoods in the previous section.
We must now decide how to extend the path from (t − 1) e 1 + e 3 through the isolation tube, to level t 2 − 1. Certainly we could take the same approach as we did with the other end of the path, but it is clear that for large enough t, the resulting path would not be of maximum length.

Methods for Choosing Points
Before we begin constructing the desired cycle, we define a few new terms and processes.
Definition 5.5.1 Recall that a point of the form (t−k) e 1 +Ω k is said to be at level k of the isolation tube. We will say that a point of the form (t − k) e 1 + j e 2 + Ω k−j is on the j th stage of the k th level of the isolation tube.
We define the following 'moves' as strategies for choosing points at certain levels throughout the isolation tube. It is best to think of these moves in terms of how they affect the coefficient of e 2 .
Definition 5.5.2 An up-one move will be a move from the point (t − k) e 1 + Ω k to (t − (k + 1)) e 1 + e 2 + Ω k . That is, such a move will jump down a level in the isolation tube to a point where the coefficient of e 2 has been increased by one.
Up-one moves will occur at odd levels of an isolation tube.
Definition 5.5.3 An up-all move will be series of moves contained on a single level, collecting a maximal number of points of the form (t − k) e 1 + i e 2 + j e 3 + (k − i − j) e 4 , for i ∈ {2, ..., k − 2}, such that these points are along an induced path of H. An up-all move will only occur on an even level of an isolation tube.
Definition 5.5.4 A down-all move will be a series of moves contained on a single level, collecting a maximal number of points of the form (t − k) e 1 + i e 2 + j e 3 + (k − i − j) e 4 , for i ∈ {0, 1, ..., k − 2}, such that these points are along an induced path of H. A down-all move will only occur at an even level of an isolation tube.
Definition 5.5.5 A down-one move will be a move from the point (t − k) e 1 + j e 2 + Ω k−j to (t − k) e 1 + (j − 1) e 2 + Ω k−j−1 . That is, such a move remains on the same level, but will jump down a stage in the isolation tube. Down-one moves will occur at odd stages within a down-all move.

How Many Points are Chosen?
At each level, other than levels one and two of the isolation tubes, we will perform exactly one of the moves defined in the previous section. We start by statically defining the points taken at levels one and two. These points are (t − 1) e 1 + e 3 at level one, and (t − 2) e 1 + e 3 + e 4 as well as (t − 2) e 1 + 2 e 4 at level two. We continue by repeating the four step process: up-one, down-all, up-one, up-all. We provide the following claims and observations about the number of points taken when performing each move.
Observation 5.5.6 An up-one move takes a single point at the level which the move was performed.
Definition 5.5.7 We call a point heavy in a coordinate corresponding to e p if that point is the leaf of a path, the coefficient of e p does not determine a level or stage of an isolation tube, and the coefficient of e p is the only non-zero coefficient other than those on the basis vectors determining level and stage.
Claim 5.5.8 An up-all move at level k takes k 2 4 − 1 points. Furthermore, if an up-all move were to begin from (t − k) e 1 + 2 e 2 + (k − 2) e p will end at the point (t − k) e 1 + (k − 2) e 2 + 2 e p , for p ∈ {3, 4}. That is, an up-all move will maintain 'heavy' coordinates.
Proof. Let an up-all move occur at level k. Any up-all move must occur after an up-one move at the previous level. That is, the first point taken at level k will be of the form (t − k) e 1 + 2 e 2 + (k − 2) e p , for p ∈ {3, 4}.
Note that all odd stages will contribute only one point, think of this as a miniature up-one move. All even stages (let j = 2 ), except the final stage will contribute The final stage will contribute 2+1 2−1 = 3 points. Note that it is impossible to take k − 2 + 1 points at all of the even levels since we would have two points adjacent to the point obtained from the miniature up-one move to the next stage. Indeed, the points (t − k) e 1 + 2 e 2 + e q + (k − 2 − 1) e p and (t − k) e 1 + 2 e 2 + (k − 2 ) e p for p, q ∈ {3, 4} with p and q distinct, are both adjacent to (t − k) e 1 + (2 + 1) e 2 + (k − 2 − 1) e p in H.
Note also that the selected points constitute an induced path in H. Certainly there are only two points at each even stage which have a neighbor in the previous or next stage. Also, no selected point at an even stage has three or more neighbors which were also selected.
Consider neighbors of the points selected at an odd stage of level k. Clearly there can only be one neighbor, since the coefficient of e 1 must remain constant, and the coefficient of e 2 must be increased by one. The unique neighbor is (t − k) e 1 + j e 2 + (k − j) e p .
Consider two points at stage j of level k x = (t − k) e 1 + j e 2 + a e 3 + (k − (j + a)) e 4 y = (t − k) e 1 + j e 2 + b e 3 + (k − (j + b)) e 4 such that a, b ∈ {1, 2, ..., k − j − 1}. By the rules for adjacency in H, x and y are adjacent iff a = b + 1 or a = b − 1. Note that both of these points were selected as part of our path, and the degree of each such point is two, as desired.
It is clear that the point (t − k) e 1 + j e 2 + (k − j) e p can only have a single neighbor on stage j, that point is (t − k) e 1 + j e 2 + (k − (j − 1)) e p + e q . Thus, the degree of (t − k) e 1 + j e 2 + (k − j) e p is 2 in the graph induced on the selected points. Also, since the point (t − k) e 1 + j e 2 + (k − j) e q is not selected in stage j, the degree of (t − k) e 1 + j e 2 + e p + (k − j − 1) e q in the graph induced on the selected points must also be two.
Now that we have verified that an up-all move selects points forming an induced path, calculating the total points taken by summing the number of points taken at each stage, we obtain the desired result.
There is exactly one point taken at each odd stage other than 1 and k − 1. Thus, contributing k 2 − 2 points. The even stages account for a total of points.
Adding the number of points from even stages and odd stages gives − 1 total points, as claimed.
Furthermore, note that an up-all move will only occur at a level k which is equivalent to 2 mod 4. This means that k − 2 ≡ 0 mod 4. From this observation we can see that if an up-all move were to begin from (t − k) e 1 + 2 e 2 + (k − 2) e 3 , then such a move would end at the point (t − k) e 1 + (k − 2) e 2 + 2 e 3 . Note that this must be, since if j ≡ 2 mod 4, we move from a 'heavy' e 3 to a 'heavy' e 4 through stage j, whereas if j ≡ 0 mod 4, we move from a 'heavy' e 4 back to a 'heavy' e 3 .
That is, an up-all move maintains the 'heavy' coordinate.
Claim 5.5.9 A down-all move will take k 2 4 + k total points. Furthermore, if a down-all move were to begin from (t − k) e 1 + (k − 2) e 2 + 2 e p , it will end at the point (t − k) e 1 + k e q , for p, q ∈ {3, 4}. That is, a down-all move will maintain the 'heavy' coordinate.
Proof. Let a down-all move occur at level k. Any down-all move must occur after an up-one move at the previous level. That is, the first point taken at level k will be of the form (t − k) e 1 + (k − 2) e 2 + 2 e p , for p ∈ {3, 4}.
Similarly to the up-all move, any even stages, except the first, will contribute k−2 +1 2−1 − 1 = k − 2 points. The first stage will contribute 3 points. All of the odd stages, except for k − 1 will contribute a single point.
Again, it is impossible to take k − 2 + 1 points at all of the even levels since we would have two points adjacent to the point obtained from the miniature down-one move from the previous stage. Indeed, the points (t−k) e 1 +2 e 2 + e p +(k −2 −1) e q and (t − k) e 1 + 2 e 2 + (k − 2 ) e q for i, j ∈ {3, 4} with p and q distinct, are both adjacent to (t − k) e 1 + (2 + 1) e 2 + (k − 2 − 1) e q in H.
Note also that the selected points constitute an induced path in H. The verification of this fact is made in the same manner as it was in the case of the up-all move and will be omitted. Now, calculating the total number of points taken, by summing the number of points taken at each stage, we obtain the desired result.
Since there is exactly one point taken at each odd stage other than k − 1, the odd stages contribute k 2 − 1 points. The even stages account for a total of points.
Adding the number of points from even stages and odd stages gives k 2 4 + k 2 + 1 + k 2 − 1 = k 2 4 + k total points, as claimed.
Furthermore, note that a down-all move will only occur at a level k ≡ 0 mod 4.
This means that k − 2 ≡ 2 mod 4. From this observation we can see that if a down-all move were to begin from (t − k) e 1 + (k − 2) e 2 + 2 e 3 , then such a move would end at the point (t − k) e 1 + k e 3 . Note that this must be, since if j ≡ 2 mod 4, we move from a 'heavy' e 3 to a 'heavy' e 4 through stage j, whereas if j ≡ 0 mod 4, we move from a 'heavy' e 4 back to a 'heavy' e 3 . Since 0 ≡ 0 mod 4, we must have that we end at (t − k) e 1 + k e 3 . That is, a down-all move maintains the 'heavy' coordinate.

Connecting the Paths Through An Entire Tube
We have identified a sequence of paths of varying lengths at each level of the isolation tube. Here we show that the collection of all of the corresponding points induces a single path through all of the levels of the isolation tube.
Recall that we began from level one by selecting the single point (t−1) e 1 + e 3 . Recall that, on the second level we chose the points (t − 2) e 1 + e 3 + e 4 and (t − 2) e 1 + 2 e 4 ; at level three, we perform an up-one move; at level four, a down-all move; at level five another up-one move; and at level six an up-all move. We repeat this pattern, performing an up-one move at level k if k is odd, a down-all move if k ≡ 0 mod 4, and an up-all move if k ≡ 2 mod 4.
We can count the number of points selected by considering cases based on the value of t 2 − 1 mod 4.
This gives that there are t 2 −3 2 +1 total up-one moves, and therefore the same

Paths Through Other Tubes
We have constructed a path through the isolation tube corresponding to e 1 . Now we wish to obtain similar paths through the other three isolation tubes. We can construct such paths by permuting the roles of the basis vectors e 1 , e 2 , e 3 , and e 4 .
For the isolation tube corresponding to e i , we transpose the roles of e 1 and e i and also transpose the roles of the remaining two basis vectors. This gives us a path through each of the isolation tubes.

Joining Paths
We now wish to join the constructed paths in order to complete a cycle. We proceed by cases, depending on the leaves of the paths constructed within the tubes. Note that we will have four situations to explore, each depending on the value of t 2 − 1 mod 4. Recalling the lengths of the paths through the isolation tubes from section four, which we labeled by Q i , we have the following main theorem.

Proof.
We proceed by selecting points which will join the paths through the isolation tubes in such a way that a single cycle is formed.
i) The leaves of the paths through the isolation tubes are as follows: We connect the remaining leaf of tube 3 to that of tube 1 by selecting all Similarly to Case iii), one can verify that no unwanted edges exist. Again noting that some double-counting has occurred in the case where t is even gives us the cycle of the claimed length. 2 Recall that for all of the results in this chapter, we assumed t ≥ 14. Cycles for smaller values of t can be constructed using similar methods to those discussed here. Note, however, that we must have t > 5 for our isolation tubes to exist. For t ≤ 5, it is conjectured that the length of the longest (K 1,n ; t)-representable cycle is linear in t. In all cases for t ≥ 14 the length of the cycle we constructed is on the order of t 3 , which can describe by saying the length of the cycle is O(t 3 ). We conjecture that the length of the longest (K 1,4 ; t)-representable cycle truly is O(t 3 ) for large enough t.

Constructing a Cycle for Arbitrary n
Consider the construction in Section 5. The same technique used in this construction can be generalized to arbitrary n. One can construct a long path through the first isolation tube by generalizing the up-all and down-all moves to account for the extra dimensions. It is for this reason that we conjecture.
Conjecture 5.6.1 The length of the longest (K 1,n ; t)-representable cycle is Our conjecture is supported by the work in this paper and the theorem of Eaton and Barbato which was cited in Chapter 3 and is restated below.
Theorem 5.6.2 For t = 3, 4, and 5 the maximum n such that C n is (K 1,3 ; t)representable is 3t−3. For t ≥ 6 the largest such n satisfies the following inequality

Future Work
There are many questions left unanswered here. Again, we have only found a lower bound on the length of the longest (K 1,n ; t)-representable cycle. Improvements to this lower bound could be made by attempting to maximize the lengths of the connecting paths between the isolation tubes. Also of interest would be an upper bound on the length of the longest cycle having a (K 1,n ; t)-representation. The goal here would be to find the exact length of such a longest cycle. I am hopeful that the work done in chapters 4 and 5 will be useful stepping stones in the pursuit to completely classify those graphs which are (K 1,n ; t)-representable. [3,4] List of References the real line is closed iff it contains all of its endpoints. We are all very familiar with this notion of closure. Let's look at closure in some more abstract spaces.

Poker Closure
Consider a game of poker in which the cards 5, 6,7,9 are (1) any initial hand containing three or fewer cards is closed, (2) the poker closure of any initial hand containing four consecutive cards is the entire space, (3) the poker closure of any hand having a dependency extends to the whole space, Note that the resistance levels of the remaining uninfected vertices keep them from becoming infected. Observe, in general, that any vertex satisfying |N (v)| < r(v) is immune to the virus, unless it is one of the initial infected vertices. The idea of outbreak closure can be applied to study problems involving efficient vaccination strategies, as well as those problems centered around biological warfare.

Closure Spaces
A closure system consists of a ground set, X, a closure operator, φ, and the collection of all of the φ-closed subsets of X. If the family of subsets is closed under intersections, then the collection of all φ-closed subsets of X is called a closure space.

Introduction
Gap closures are natural extensions of poker closure. In gap closure, the ground set is the integers. An ordered pair (g, h) of non-negative integers will be called a gap type. For S ⊂ Z, a point p ∈ Z is a (g, h)-gap point of S provided the g integers to the left of p are in S and the h integers to the right of p also belong to S. Thus for S := {1, 2,3,5,6,8,10,11,12,16, 17}, 4 is a (3, 2)-gap point, 7 is a (2, 1)-gap point, and 9 is a (1, 3)-gap point. Moreover, 13 is a (3, 0)-gap point and 15 is a (0, 2)-gap point. Note that 14 is not a gap point for any non-trivial gap type. Note that even though 2 ∈ S, we still say that 2 is a (1, 1)-gap point of S. If S contains all of its (g, h)-gap points, then we say that S is (g, h)-closed. If G is a set of gap types, then S is G-closed if and only if S is (g, h)-closed for all (g, h) ∈ G.
If g ≤ j and h ≤ k, then every (j, k)-gap point is also a (g, h)-gap point. In particular, every (g, h)-closed set is also (j, k)-closed. This suggests ordering the gap types by (g, h) ≤ (j, k) iff g ≤ j and h ≤ k, making the gap types into a partially ordered set.

Preliminary Notation and a Result
In order to study the separation of disjoint G-closed sets, some tools and notation will be useful. A coloring of Z is a map f : Z → {1, 2, o}. The set of integers colored 1 will be denoted by A and those colored 2 will be denoted by B. The o's represent uncolored integers. Note that this means that any partial G-convex coloring, f , can be extended to a maximal G-convex coloring. These partial G-convex colorings will be our objects of study. In particular, we will be interested in partial convex colorings which extend to all of Z. In general, though, it is not always possible to extend a partial convex coloring of Z to a complete convex coloring of Z.
(A) Let g = h = 2 : 1121o2122 Let g = 1 and h = 3 : 1211o1222 The two color patterns in (A), above, are obstructions to completing any coloring in which they occur. More generally, given positive integers g, h, j, and k, we call the standard [g, h, j, k]-obstruction a string colored in the following manner: 1 g 21 h−1 o2 j−1 12 k , where 1 p denotes a string of p consecutive integers colored by 1, and 2 q denotes a string of q consecutive integers colored by 2. The space in this pattern cannot be filled without violating either (g, h)-convexity (if o → 1) or (j, k)-convexity (if o → 2).

Separation in Gap Closures
We are now ready to discuss the separation properties in the closure spaces resulting from gap closures.

Separation Properties
The separation properties are used as a way to categorize topological spaces. Although a closure space is not necessarily a topological space, the separation proper-Z itself. In the case of (1, 0)-closure, if p and q are distinct integers contained in (1, 0)-closed subsets of Z, then both subsets must contain {x|x ≥ max{p, q}}.
That is, there is no way to have two non-empty, disjoint, (1, 0)-closed subsets of Z.
Let G be a non-degenerate set of gap types, and let (g, h) be a minimal gap type in G. Suppose first that g = 0 and h > 1. Consider the partial coloring 12 h−1 o1 h−1 .
All of the gaps in this coloring have types smaller than (0, h), so since (0, h) is minimal we have that this coloring is G-convex. However, if the hole is filled with a 1, we have 12 h−1 1 h , making the rightmost original 2 into a (0, h) gap point, destroying G-convexity. Similarly if the hole is filled with a 2, we have 12 h 1 h−1 , making the original first 1 into a (0, h) gap point, which also destroys G-convexity.
The case that h = 0 and g > 1 is symmetric.
It remains to consider that g > 0 and h > 0. In this case, the standard [g, h, g, h]obstruction establishes the result. That is, the partial coloring 1 g 21 h−1 o2 g−1 12 h cannot be completed without destroying G-convexity. 2 This is not the end of the story for S 4 , however. Separation by hemispaces is not possible, but how close can we get? This question will be explored further in a dissertation by A. Mia Heissan.

The Other Separation Axioms
In this section we will provide examples of gap closures which do and do not satisfy the remaining separation properties.
Proof. Every point is trivially a gap point in (0, 0)-closure. Thus, the only (0, 0)closed set is Z. The only set which is open, then, is the empty set, which is not a neighborhood of any points of Z. Therefore, separation, even in the S 0 sense, is impossible. 2 Theorem 7.3.4 (0, 1)-closure is S 0 , but is not S 1 .
Proof. Consider two points, a and b, with a < b. Let A be the downset at a. It is clear that A is G-closed, so A is G-open and contains b but not a. Note that there is no G-closed set containing b which does not contain a, since a is in the downset at b. Thus, (0, 1)-closure is S 0 , but is not S 1 . 2 Note that in order to prove that a closure space is S 2 , for any two integers we need two disjoint open sets, each containing one of the integers. Since we are finding open sets by looking at the compliments of closed sets, we must find two closed sets whose union is all of Z, such that their compliments will be disjoint. That is, if G the set of gap types inducing our closure space, we will be looking for a complete G-convex coloring of Z. Suppose now that we have a complete convex coloring, f of the integers, that separates a and b. Let a < b and a be colored by 1, while b is colored by 2. We know there are no consecutive integers colored by 2 above a by (0, 2)-convexity, and likewise, no consecutive integers colored by 1 above b. Hence, f must alternate 1's and 2's above b, but this is impossible since f is G-convex. That is, no such complete G-convex coloring exists. 2 Theorem 7.3.6 (1, 2)-closure is S 2 , but is not S 3 .
Proof. Consider two integers, a and b, colored by 1 and 2, respectively. Let a < b, color all of the integers to the left of b by 1, and all of the integers to the right of b by 2. Note that the resulting coloring is a complete (2, 2)-convex coloring of Z, which separates a and b. Thus (1, 2)-closure is S 2 .
Consider the partial coloring 121oo1, placing a 1 in the first open position gives 1211o1, making the 2 a (1, 2)-gap point of the 1's. Thus, the first position must be filled with a 2, giving 1212o1. Note that no matter how the remaining hole is filled, we have a resulting (1, 2)-gap point. 2 Theorem 7.3.7 (2, 2)-closure is S 3 .
Proof. Consider trying to separate a (2, 2)-closed set of 1's from a single given 2.
We denote the original 2 by 2 * .
If the two positions to the right of 2 * are 1's, fill that entire side by 1's. Otherwise, fill all positions below 2 * by 1. Note that this is admissible since otherwise the original coloring would be of the form ...112 * 11..., for which the set of 1's is not (2, 2)-closed.
Using symmetry, assume that all positions below 2 * are 1. Note that we must and h such that (g, h) ∈ G. The weight of G is the minimum over all g + h such that (g, h) ∈ G.
Theorem 7.3.9 Given a set of gap types G, the weight of G is the largest n such that every set with fewer than n elements is closed.
Proof. Any set having a (g, h)-gap point must certainly have g + h elements.
Therefore, every set having fewer than g + h elements is certainly (g, h)-closed.
The set of elements corresponding to the coloring ...oo1 g o1 h oo... is not (g, h)-closed.
Thus, the width of a set G of gap types is the largest n such that any set consisting of fewer than n elements is G-closed. Proof. To show S 3 we must separate a single integer colored by 1 from a (0, h)closed set of integers color by 2. We must fill in any holes with 1's and 2's in such a way that convexity of the coloring is preserved. Let 1 * denote the initial 1, and fill all of the positions below 1 * with 2's. Consider a hole H that remains. This hole must be above 1 * . a) If H has even width and 1 * bounds H, then we fill H by alternating 2's and 1's respectively. b) If H has even width, but 1 * does not bound H, then fill H with 1's in the first two positions, followed by alternating 2's and 1's. c) If H has odd width, fill it with alternating 1's and 2's respectively.
These rules may generate pairs of consecutive 1's, but since h is at least 3, we have created no dependencies. The rules generate no strings of 2's. Therefore, convexity is preserved.
The pattern 1 * 2oo2 shows that the 1 * cannot be separated via (0, 2)-convex coloring from the two 2's. To prove that S 2 is satisfied, consider a pattern 1 * oo...oo2 * . Again, color all of the integers below the 1 * by 1. Now, if the width of the hole between 1 * and 2 * is even, then fill the hole by alternating 2's and 1's.
If the hole has odd width, then fill it by alternating 1's and 2's. In either case, fill all positions above 2 * by alternating 1's and 2's. 2 Theorem 7.3.11 Let G be a set of gap types, a) if G has mesh at least 1, then the resulting closure space is S 2 .
b) If G has mesh at least 3, then the resulting closure space is S 3 . Proof.
a): Let G be a set of gap types with mesh at least 1. Consider p and q, with p < q, to be two points that we wish two separate. Take p and its downset to be H. Since G has mesh at least 1, then both H and H are G-closed. That is, H is a separating hemispace.
b): Let G be a set of gap types with mesh at least 3, and let p be a point disjoint from a G-closed set F . Let C denote the complement of F , and consider the component K ⊂ C which contains p. Color K with 2's and F with 1's. Now, if |K| > 1, fill all remaining holes with 2's. Otherwise |K| = 1, and we proceed more carefully. If all of the remaining holes to the right of p are of width one, then color all of these holes by 1. Otherwise there are holes of width at least two to the right of p. Locate the hole of width at least two which is closest to p, and color it by 1's. Color all of the holes of width 1 between p and this newly colored sting of 1's by 1 as well. Color all of the holes above this string of 1's by 2's. Analogously fill the holes to the left of p. Note that we have extended the partial coloring to a complete coloring of Z while preserving G-convexity of the coloring.