Some Results on Graph Representations and Graph Colorings

A graph G(V,E) is a structure used to model pairwise relations between a set of objects. In this context, a graph is a collection of vertices (representing the objects) and a collection of edges (representing the relation) that connect pairs of vertices. It is possible to represent a graph using an adjacency matrix, but often this is not the most efficient representation of the relation. In studying graph representation, the object is to capture the structure of the graph more efficiently using a variety of other discrete structures. This work considers path representations of graphs. Consider a host graph, H. A path representation [H : r : q] of a target graph G is a labeling in which each vertex is assigned a unique path of length r found in H in such a way that if uv ∈ E(G), then the Pr assigned to u and the Pr assigned to v have at least a Pq in common. This study considers representations in which the host tree is the complete graph on n vertices, [Kn, r, q] which will be referred to as Pr,q-representations. This work also considers the area in graph theory known as vertex-coloring, specifically coloring planar graphs, and explores a special class of planar graphs called “coils”.


PREFACE
In mathematics, a graph is a data structure that consists of a finite set of of ordered pairs of vertices which represent edges. Graph theory is the area of mathematics which studies these data structures. An active area of research within graph theory is the study of graph representations.
Different data structures are used to capture the information contained in a graph: adjacency lists, incidence lists, adjacency matrices, incidence matrices.
Each has with it a time complexity cost associated with performing various operations on a graph, for example adding or removing a vertex or an edge.
Depending on the nature of the graph, certain structures are preferred over others. If the graph is sparse, it would be preferable to use a list. If the graph is dense (the number of edges |E| is close to the square of the number of vertices |V 2 |), it would be preferable to use a matrix. Many applications work with graphs with special structures. In studying graph representations we attempt to exploit this structure in order to obtain a simpler and more efficient representation which in turn will reduce the costs associated with computing.
A representation of a target graph G consists of three objects, 1) a host set H, 2) an assignment function f , and 3) a conflict rule g. The assignment function assigns a subset of the host set to each vertex of a target graph. The conflict rule compares these assigned subsets to determine whether or not two vertices should be adjacent in G. If, given a host set H and conflict rule g, there is a suitable assignment function such that the graph G is induced by the conflict rule, we say that G is [H : g]-representable.
vi The first part of this dissertation will look at a special type of graph representation, the [K n : P r , P q ]-representation where the host graph is the complete graph on n vertices, the vertices are labeled with paths of length r, and vertices are adjacent if they have a path of length q in common. We share our results to date and discuss our proposed work moving forward. This line of research provides ample room for future work. In my dissertation I look at some general cases of P (r, q)-representations, but spend considerable effort on P (3, 1). The immediate goal is to classify all graphs which are P (3, 1)-representable, and eventually look at P (4, 2)-representations.
For the second part of my dissertation I look at a special class of graphs called planar graphs. Planar graphs are graphs that may be drawn in the plane in such a way that no edges are crossing. Specifically, I look at a sub-class of planar graphs called coils, a planar graph whose depth-first search tree is a path. Here, we are not so concerned with representation but move toward producing a short proof that this special class of coils may be colored using four colors. For more than a century there has been interest in proving the Four-Color Theorem (4CT) by means of a short proof. The conjecture was first posed in 1852 by Francis Guthrie and remained open until 1976 when Appel and Haken of the University of Illinois discovered its first proof.
The proof was the subject of much controversy, as it relied on an assumed accuracy of computers to check almost two-thousand cases. This proof was the first of its kind. In the years since, many mathematicians have revisited the 4CT. In 1996, Robertson, Sanders, Seymour, and Thomas published a new computer-assisted proof by analyzing only 633 cases.
vii In 2004, Werner and Gonthier used the Coq proof assistant, a formal proof management system developed in France, to discover a proof based on the 1996 proof, but with some original content. Although the proofs mentioned are generally accepted within the mathematics community, a short self-contained proof is still desirable. Attempts to find such a proof have resulted in the exploration of interesting generalizations of graph colorings, including 'list colorings' and 'defective colorings'. These variations provide many interesting questions for further research opportunities. In this part of my dissertation I make a conjecture about a lower bound for the number of colorings that exist in coloring a coil using only four colors and outline a proof which is self-contained and used counting techniques.
Each part of my dissertation discusses open problems. There are many interesting questions which remain unanswered in both areas of graph representations and colorings, providing a lifetime worth of research opportunities. [?, ?, ?, ?, ?, ?, ?, ?, ?, ?]    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 H, (2) a function f : V → P(H), and (3) a function

Graphic models
A subgraph of a graph H is a graph G with V (G) ⊆ V (H) and E(G) ⊆ E(H).
We start with a host graph H and two subgraphs, a prototype R, and a quota Q. Throughout we will assume that r is the order of R and q the order of Q. A subgraph of H that is isomorphic to a fixed graph C is a copy of C in H. An with adjacency determined as in ( * ).
Line graphs are a classical example. In 1968, Beineke ([2]) proved the following theorem, characterizing line graphs.
Theorem 1.2.2 The following statements are equivalent for a graph G.
1. G is the derived graph of some graph, that is, G is a line graph.
2. The edges of G can be partitioned into complete subgraphs in such a way that no vertex belongs to more than two of the subraphs.
3. The graph K 1,3 is not an induced subgraph of G; and if abc and bcd are distinct odd triangles, then a and d are adjacent.
4. None of the following nine graphs is an induced subgraph of G. Here the prototype is the path P 2 , the path on two nodes, and the quota is P 1 .
The class of all line graphs arising from graphs of order ≤ n is [K n , P 2 , P 1 ]. The class of all line graphs is [ * , P 2 , P 1 ].

Path Representation
We will now look at the structure of graphs in [ * , P r , P q ], where P k denotes a path on k nodes. For simplicity, we will use the notation P(r, q) to mean [ * , P r , P q ] and say that a graph has a P (r, q)-representation if the graph is in P(r, q).
Proposition 1.3.1 P(r, q) ⊆ P(s, q) for all s > r.
Proof. Let G be a graph in P(r, q). Since we are unrestricted on the host graph, attach to the end of each path P r a vertex not used anywhere in the P (r, q)- representation. This yields a P (r + 1, q)-representation. We repeat this process for a total of s − r times, creating a P (s, q)-representation.
Proof. Let G be arbitrary in P(r, q). Consider the set of all r-length paths used in a P (r, q)-representation of G. For each P r , form the line graph L(P r ), which is itself a P r−1 . The collection of these line graphs become the new label set for G, that is, if v 1 had label P 1 , its new label is L(P 1 ). For example, the path 1-2-3-4-5 is now 12-23-34-45=A-B-C-D. It is easily verifiable that vertices in the P r -labeling share a P q if and only if the vertices in the P r−1 -labeling share a P q−1 . Hence, G is in P(r − 1, q − 1).
The question arises: When is there strict containment? When is there equality?
We spent a great deal of time looking at this question. It lead us to consider the set of line graphs, L(G) and the set of line graphs of line graphs, L 2 (G).
There appears to be some nesting properties associated with L and P for small values of r and q. We are interested in exploring this idea further. We know that P(2, 1) = L(G) (based on their definitions.) We also know from Whitney ([11]) that with one exceptional case (the graphs K 3 and K 1,3 whose line graphs are both K 3 ) the structure of a graph G can be completely recovered from its line graph. In other words, G is known if its adjacencies are known. This might prove useful in further characterizing the relationships that exist between line graphs and P (r, q)-representations.

Proof.
Let H = K 7 with nodes labeled 1, 2, 3, 4, A, B, C. Notice that the induced subgraph G c = G \ C has a perfect matching of non-edges, as does G b = G \ B and G a = G \ A. Also notice that the non-edges in G form three non-adjacent cycles of length 6.
We found a P (4, 2) representation of a graph G 12 on 12 vertices that contains a perfect matching of non-edges ( Figure 3). Through exhaustion (see Appendix) we determined that this representation which uses exaclty four nodes from its host graph is the only way (up to isomorphism) to represent G 12 . Since there are only twelve unique paths using four labels it would be impossible to label three partitions on 18 vertices since the induced graph any two partitions G i ∪ G j is a copy of G 12 . Proposition 1.3.4 P(r, q) ⊆ P(r, q − 1) for all q > 1.
Proof. This follows from Propositions 1.3.2 and 1.3.1.
Proof. Let G have a P (n, 1)-representation. For each of the n nodes used in the P n to label the vertex in G, say 1-2-3-...-n, replace with the path 1 1 -...
It is easy to see that this gives to each vertex a P kn label and if two vertices shared a P 1 in the P (n, 1)-representation, they will share a P k . Hence, G is in P(kn, k).
The question arises: Is P(n, s) ⊆ P(kn, ks) for s = 1? We know that the technique used for proving Proposition 1.3.5 will not work for values of s > 1. This does not mean that these graphs are not P (kn, ks)representable, however. There may be a different method for labeling the vertices in such a way as to get a representation. Our initial thought, however, is that the answer to the question is negative.
Theorem 1.3.6 P(2, 1) = P(3, 2) Proof. P(3, 2) ⊆ P(2, 1) follows from Proposition 1.3.2 so we need only show P(2, 1) ⊆ P(3, 2). Assume G has a P (2, 1)-representation. Without loss of generality, assume that all labels are ordered chronologically, that is, label 5-4 would be 4-5. Consider a vertex v 1 labeled 1-2. v 1 is adjacent to all vertices labeled 1-x or 2-y where x = 2 and y = 1. v 1 is not adjacent to any vertex labeled w-z where w < z and w, z / ∈ {1, 2}. Re-label all vertices i-j with the new label i-A-j. v 1 is now labeled with the P 3 1-A-2. All vertices previously labeled 1-x or 2-y are now labeled 1-A-x or 2-A-y, respectively. All vertices previously labeled w-z where w < z and w, z / ∈ {1, 2} are re-labeled w-A-z. v i is still adjacent to 1-A-x or 2-A-y and not adjacent to vertices labeled w-A-z.

Characterization of P (3, 1)
Theorem 1.4.1 G has a P (3, 1)-representation if and only if there exists an edge covering of G into cliques such that each vertex belongs to exactly three cliques and there is no K 4 contained in the intersection of three cliques.
Proof. Assume that G has a P (3, 1)-labeling. Without loss of generality, assume the size of the host graph is minimal, that is, all its n nodes are used in the labeling of G. Each node in the host graph H represents a clique in G, that is, Since each vertex in G uses exactly three nodes from H in its label, create an edge covering of G where each vertex belongs to exactly three cliques. Since there is a P (3, 1)-labeling, there does not exist a K 4 contained in the intersection of three cliques in the covering. Now, assume you can edge cover G into cliques such that each vertex belongs to exactly three cliques and there is no K 4 contained in the intersection of three cliques. Label the cliques K 1 , K 2 , ..., K m and use the labels of the cliques to denote the P 3 in H with which to label each vertex. Since there are at most three vertices in the intersection of any of the the cliques, K a , K b , K c , there are three unique paths available to label each vertex: abc, acb, and bac. Hence, G has a P (3, 1)labeling.
• C is the set of all graphs not in K 1,4 that contain a K = K A in the intersection Table 1 for values of A, B, C, D, and note that ∅ 1 signifies that the vertex is adjacent to no vertices other than those in K A . See Figure 4.) • D is the set of all graphs not in K 1,4 that have an induced D 8 (a graph which contains two non-adjacent vertices, x and y that share a neighborhood that is itself a P 6 ) and are of the following form: there exists a vertex z which is not adjacent to x or y, yet it is adjacent to at least one of the vertices in the P 6 =w 1 , w 2 , w 3 , w 4 , w 5 , w 6 . Vertex z has the following characteristic: adjacent to the entire P 6 ; or adjacent to w 2 but not w 3 ; or adjacent to w 3 but not w 2 ; or adjacent to w 4 but not w 4 ; or adjacent to w 5 but not w 4 . (See Figure 5.) • W is the set of all graphs not in K 1,4 that are on 7 or 8 vertices and contain a vertex v of degree 6 or 7 respectively, such that for every clique K a in N (v), the induced subgraph N (v) \ K a yields either an ∅ 3 or a C 5 .
Theorem 1.4.3 If G is P (3, 1)-representable, no graph contained in the set F of forbidden graphs is an induced subgraph of G.
contains an induced ∅ 4 . In any edge covering of G K , v 1 is necessarily contained in four cliques, hence, by Theorem 1.4.1, G K has no P (3, 1)-representation.
. Without loss of generality, label v 1 with the path '1-2-3', v 2 with the path '4-5-6' and v 3 with the path '7-8-9'. Since each vertex in K is adjacent to each v i it must contain a label from the set A = {1, 2, 3}, a label from the set B = {4, 5, 6} and a label from the The P 3 's which use exactly one node from each set represent all possible labels for each vertex in K, assuming that each node in the P 3 is available. With no restrictions, there are 3 · 3 · 3 = 27 sets of size three which contain exactly one element from the set A, B, and C. Each of these sets of size three can be combined to represent three unique paths from the host graph H. For example, {1, 4, 7} can be used to form the labels 1-4-7, 1-7-4, and 4-1-7, so there are 27 · 3 = 81 paths available to label the vertices in K, so m = 82 is not P (3, 1)-representable.
Now assume that one of the vertices, v i , is also adjacent to another vertex, w i which is not in K. One of the nodes in the path used to label v i must also be contained in the path used to label w i , leaving only two nodes available for use in the labeling of the vertices in K. This leaves only 2 · 3 · 3 = 18 nodes to be used yielding 54 unique paths available to label the vertices in K, so m = 55 is not Now assume that the vertex v i is adjacent to not only w i but z i where w i and z i are not adjacent. One of the nodes in the path used to label v i must be contained in the path used to label w i , and a different node must be contained in the path used to label z i , leaving only 1 node available for use in the labeling of the vertices in K. This leaves only 1 · 3 · 3 = 9 nodes to be used yielding 27 unique paths available to label the vertices in K, so m = 28 is not P (3, 1)-representable.
If we continue in this manner and consider all the different possible adjacencies for each v i , we see that Table 3.1 gives the various restrictions on m, and the set C of forbidden subgraphs are not P (3, 1)-representable.
D / ∈ P(3, 1): Assume G D is in D, that is, it contains an induced D 8 and a vertex z which is not adjacent to vertices x or y, both of which are adjacent to the same P 6 = (w 1 , w 2 , w 3 , w 4 , w 5 , w 6 ). Yet, it is adjacent to at least one of the vertices in the P 6 . Assume vertex z is adjacent to w 2 yet not adjacent to w 3 . If there were a P (3, 1)-labeling, there would exist an edge covering of G D in which neither x and y were in 4 cliques. In this case, the edge covering would necessarily contain the following six K 3 's: xw 1 w 2 , xw 3 w 4 , xw 5 w 6 , yw 1 w 2 , yw 3 w 4 , and yw 5 w 6 . In this case, w 2 would be in two cliques. Since z is adjacent to w 2 , but not w 3 , w 2 would be forced to be in a clique containing the edge (w 2 , z) and another containing the edge (w 2 , w 3 ). Hence, w 2 would necessarily be contained in four cliques. By Theorem 1.4.1, G D has no P (3, 1)-representation. The case is similar for z adjacent to w 3 and not w 2 ; z adjacent to w 4 and not w 5 ; and z adjacent to w 5 and not w 4 . Now assume that z is adjacent to the entire P 6 . Similar to the argument for covering the edges between x and y and the P 6 , if there were a P (3, 1)-labeling, in order to edge cover G D so that z is not in 4 cliques, the edge covering would necessarily contain the following three additional K 3 's: zw 1 w 2 , zw 3 w 4 , and zw 5 w 6 . Hence, each vertex in the P 6 would be contained in three cliques. However, the edges (w 2 , w 3 ) and (w 4 , w 5 ) would still need to be covered, forcing these vertices to be in four cliques. Therefore, by Theorem 1.4.1, G D has no P (3, 1)-representation.
Let K be an arbitrary clique in N (v 1 ) and assume the induced subgraph N (v 1 ) \ K yields an ∅ 3 . In any edge covering of G W , v 1 is contained in at least one clique which covers the edges connecting v 1 to K. In order to cover the edges connecting v 1 to each of the three vertices in the ∅ 3 , v 1 must be in three additional cliques, hence, v 1 is necessarily contained in four cliques and by Theorem 1.4.1 has no P (3, 1)-representation. Now assume that the induced subgraph N (v 1 ) \ K yields a C 5 . In this edge covering of G W , v 1 is contained in at least one clique which covers the edges connecting v 1 to K. In order to cover the edges connecting v 1 to the vertices in the C 5 , v 1 must be in three additional cliques, hence v 1 is necessarily contained in four cliques and by Theorem 1.4.1, G W has no P (3, 1)-representation.
Every induced subgraph of graph which is P (3, 1)-representable must also be P (3, 1)-representable. Hence, if one of the graphs in F is an induced subgraph of G, then G is not P (3, 1)-representable.
Theorem 1.4.4 If G is a tree and does not contain an induced K 1,4 , G has a P (3, 1)-representation.
Proof. Assume G is a tree and does not contain an induce K 1,4 . Edge cover G using each edge as a clique. Since there are no K 1,4 's, each vertex is in at most three cliques and by Theorem 1.4.1, G has a P (3, 1)-representation.
Theorem 1.4.5 If G is a graph on 6 vertices and does not contain an induced Proof. Assume G does not contain an induced K 1,4 . By Proposition 1.3.1, if G has a P (2, 1)-representation, then it has a P (3, 1)-representation, so by Theorem 1.2.2 we need only consider the nine Beineke graphs which do not have a P (2, 1)representation. Six of the nine graphs are on six vertices, and it is easy to show that these have a P (3, 1)-representation, so we will focus on the remaining three graphs.
graph on 6 vertices. There are 2 9 = 512 ways to form this union. Many of these have an induced K 1,4 or do not result in a connected graph. A large number of them are isomorphic to each other. We considered the cases where v a and v b are adjacent, and the cases where they are not adjacent. We noted that if either v a or v b (or both) is adjacent to v 0 , it must also be adjacent to at least one outer vertex in order to avoid a K 1,4 . We also noted that if v b is not adjacent to v 0 but is adjacent to v a , v b may not be adjacent to all three outer vertices or else there is a K 1,4 . In total, there were about 20 cases to check, and in doing so, we verified that if G does not contain a K 1,4 , it is P (3, 1)-representable.
The remaining two graphs, B 4 and B 9 , are on five vertices, so we need only check that the addition of a single vertex, v, which does not result in a K 1,4 is P (3, 1)representable. We give each graph a relaxed -P (2, 1) representation, that is, one in which duplicate labels are allowed. Hence, each vertex has two labels. It is possible to partition the vertices into three cliques, K a , K b and K c where |K i | ≤ 3. Label vertex v 'a-b-c,and give to any vertx in K i which is adjacent to v the additional label i and if it is not, give to it an arbitrary unused label. Since there are no more than three vertices in a partition, it is possible to re-order the labels to form up to three unique paths, giving each graph a P (3, 1)-representation.

Future Work
It remains to write out the details of the proof of the following statement: If G is a graph on 7 vertices and does not contain an induced K 1,4 or an induced subgraph from the forbidden set W, then G is P (3, 1)-representable.
Outline of proof. Six of the nine graphs are on six vertices. Five of these have a relaxed -P (2, 1)-representation as well as a vertex set that may be partitioned into three cliques of size less than or equal to three. Hence, these graphs are P (3, 1)-representable. The other graph on six vertices, B 7 = W 6 a wheel on six vertices does not have a relaxed -P (2, 1)-representation. The addition a single vertex, v, adjacent only to the vertex of degree 5 is in W and does not have a P (3, 1)-representation. In every other case, B 7 ∪ v is P (3, 1)-representable.
Two of the three remaining graphs are on five vertices, B 4 and B 9 . We checked all the ways in which two additional vertices could be added. If the newly formed graph does not have a K 1,4 , then it is P (3, 1)-representable.
The final graph B 1 = K 1,3 is on four vertices. We checked all the ways to add three vertices to B 1 . If the newly formed graph does not have a K 1.4 and is not in W, then it is P (3, 1)-representable.
These cases, however, are important in that we may now focus on graphs that have size at least 8.
Conjecture 1.5.1 A graph G is P (3, 1)-representable if and only if no graph contained in the set F ∪ F of forbidden graphs is an induced subgraph of G.
We are working on proving that a graph which contains no forbidden subgraphs is P (3, 1)-representable. We have developed an ordered list of claims that build upon each other. In proving these claims, we hope to identify what the set F includes or determine that it is empty and out current list of forbidden graphs is complete.
However, there are many questions still open regarding P (r, q)-representations.
Once we fully classify P (3, 1), we will begin to look at P (4, 2) and P (5, 3) and look for possible generalizations.

CHAPTER 2
Coloring Planar Graphs

Four-Coloring a Coil
A coil is an inner-triangulated graph whose depth-first search tree T (G) = (v 1 , v 2 , . . . , v n ) is a path with the property that for all i, the up-neighborhood We conjecture that a coil G is 4-colorable with at least 4 · 3 n−1 2 3 m 3 4 β−1 distinct colorings, where m is the number of edges other than path edges, and β is the number of nonempty up-neighborhoods in G.

Introduction and Definitions
In this paper, we assume that G is a simple, inner-triangulated, and 2-connected plane graph and all 3-cycles, except possibly the outer-circuit, are empty.
For C = (w 1 , w 2 , . . . , w k ), a cycle in G, we denote the vertices embedded in the interior by I(C) and define G(C ) to be the induced subgraph on I(C ) ∪ C .
Let C = (v 1 , v 2 , . . . , v * i ) be the outer-circuit. Induced subgraphs on the neighborhoods of vertices in I(C) form cycles and those in C form paths. Whenever we refer to the neighbors of a vertex as a cycle or a path, we'll assume they are listed in a clockwise direction. Starting with v 1 , we produce an ordering of the vertices, . . , v n ), as they are encountered when forming a depth-first-search tree of G. Thus, the second vertex v 2 is the neighbor of v 1 on the outer-circuit clockwise of v 1 and we proceed in a clockwise direction, using the rule: take the next most clockwise neighbor that hasn't been taken yet. The outer-circuit is the initial subsequence of T (G).
The labeling of the vertices of G suggests an orientation of its edges -(v s , v r ) is a directed edge if and only r < s. Also the children of a vertex T (G) are given an order that indicates the order in which they were chosen in T (G).
Definition 2.2.1 We consider an interior vertex v r and its neighborhood (w 1 , w 2 , . . . , w k ), where w 1 is its parent in T (G) and w j is the first child. It is clear that j > 2. The neighbors w 2 , w 3 , . . . , w j−1 of v r are referred to as its upneighbors and U r = (w 2 , w 3 , . . . , w r−1 ) as its up-neighborhood. We define edges of the form v s v r , where r < s − 1, as crossing edges. If T (G) is a path and all upneighborhoods are intervals, we say that T (G) is a coil. Thus, each up-neighborhood is an up-interval.
Let v α be the vertex with the smallest subscript such that U α = ∅. Note α ≥ 3. Let β = n−(α −1) be the number of (not necessarily distinct) non-empty up-intervals.
We see that m, the number of crossing edges, is at most 2n − 5.
We denote the rooted full-ternary tree with n levels and root r by T . The coloring φ assigns colors {1, 2, 3, 4} to the vertices of T according to the following rule: r gets 1 and the three children of each parent are given distinct colors that are different from the parent's color. We denote by T = (T, φ) the tree T that is colored by φ. If a proper coloring of G exists, it is represented by some path in T from the root to level n. In this paper, we derive a minimum positive bound on the number of such paths that depends on the number of vertices, crossing edges, and nonempty up-intervals in G.
Definition 2.2.2 Let U α = (v 1 , v 2 , . . . , v α ). We let T (U α ) denote the subtree of T that consists of all of the intervals from levels 1 through α that are colored with at most three colors. We refer to the paths in T * (U α ) that extend from the root to its leaves as interval-paths.
Define X n+1 to be the P 3 : v n , v n−1 , v n−2 , and G j to be the union n+1 i=j X i .

Definition 2.2.4
The up-degree of a vertex v i is the size of its up-interval; updeg(v i ) = |U i |.

Preliminary Lemmas
Lemma 2.3.1 A coil G on n vertices, with k crossing edges and one up-interval has at least C = 3 n−1 2 3 k proper colorings, such that the color of v 1 is 1.
Proof. Note that n = k + 2 so that 3 n−1 2 3 k = 3 k+1 2 3 k = 3 · 2 k . G is the wheel W n on k + 2 vertices, and it is clear that W n can be colored as desired.
Proof. We are concerned only with the portion of the tree which represents the up-interval U n , so we consider the subtree T * (U n ) of height k. For k = 2, there are 3 = 3(2 2−1 − 1) nodes at level k, so the base case holds. By induction, assume the Lemma is true for n − 1 = k + 1 and the tree T * (U n−1 ) of height k − 1 has 3(2 k−2 − 1) nodes at level k − 1.
In adding one more vertex to the coil, the up-interval U n has k vertices. We see T * (U n ) as an extension of T * (U n−1 ) to the k th level so that each interval-path through the first k levels uses no more than three colors. Each node at level k − 1 has at least two children, since the k − 1 interval path from level 1 to level k − 1 is missing at least one of the 4 colors. Also, there are exactly three nodes that have a third child, since exactly three of the paths are missing two colors (the root node is fixed at color 1). This gives 2(3(2 k−2 − 1)) + 3 = 3(2 k−1 − 1) nodes at level k as desired.
Lemma 2.3.3 Let G be a coil on n vertices, with k crossing edges and one upinterval. There exists a corresponding tree T (G) that contains C = 3 · 2 k full paths from the root to level n, which represent proper colorings of G, such that the color of v 1 is 1.
Proof. Assume the root node of T * (U n ) is colored 1. Consider the rooted subtrees of T * (U n ) consisting of the bottom three levels of the tree, that is, the collections of subtrees whose roots are at level k and contains nine leaves at level k + 2. In each of these subtrees, three of the nine leaves are colored the same color as the root node and the remaining six leaves are colored with the remaining three colors -two each. Note that no matter how large k is, exactly three paths will be missing two colors.
Note that the full paths through T * (U n ) represent colorings of G in which T (G) is properly colored and the vertices contained in U n use at most three colors. We have left only to remove those leaves (corresponding to v n ) whose color is used in the interval-path above it (corresponding to U n .) Consider an interval-path through the first k levels. This path uses either 1, 2 or 3 colors.
Clearly only one color is used as the interval is represented by a single node colored 1, and we keep the leaves which are colored by one of the remaining three unused colors: 2,3 or 4. From the distribution of the nine leaves, we keep 2/3 of the colorings, and we see that 9(2/3) = 3 · 2 1 .
Clearly each of the interval-paths use only two colors, and we keep the leaves which are colored by one of the remaining two unused colors. From the distribution of the nine leaves of the sub-tree beneath a given interval-path, we keep 4/9 of the colorings. So, there are 27 full paths, none of which used more than three colors in the first k levels, and 4/9 of the paths corresponded to proper colorings, and we see that 27(4/9) = 3 · 2 2 .
We consider the subtrees whose roots are on level k and whose leaves are the leaves are on level n. Exactly three of the 3(2 k−1 − 1) interval-paths from level 1 to level k (see Lemma 2.3.2) use only two colors, and we keep the leaves of the subtrees beneath those intervals which are colored by one of the remaining two unused colors. Similar to case 2, we keep 4/9 of the colorings in these three subtrees. The rest of the interval-paths use three colors, and we keep the leaves which are colored the same as the remaining unused color.
Definition 2.3.4 The color of the root node of T * (U n ) is called the primary color and the remaining three colors are the secondary colors. A node that has the primary color is called a primary node and likewise, a node that has the secondary color is called a secondary node.
Lemma 2.3.5 Let G be a coil on n vertices, with k crossing edges and one upinterval. If k > 1, T * (U n ) has the following distribution of nodes at level k.
Furthermore, the secondary nodes are equally distributed among the three secondary colors.
Proof. Again, we are concerned only with the portion of the tree which represents the up-interval, so we consider the subtree T * (U n ) of height k. It is clear the base case holds for k = 2 and k = 3.
This total can be broken into primary nodes, P k−1 and secondary nodes S k−1with the same number of each secondary node. Then t k−1 = P k−1 + S k−1 . From Lemma 2.3.2, t k = 2t k−1 +3 = 2(P k +S k )+3. Since the root node is primary, every path through t * (U n ) contains primary nodes. At the level k − 1, any secondary will be extended to primary nodes at the k level, and no primary node at level k − 1 will be extended to the primary color. So, P k = S k−1 . Since nodes are either primary or secondary, we know that S k = t k − P k = 2(P k−1 + S k−1 ) + 3 − P k = 2P k−1 + 2S k−1 + 3 − S k−1 = 2P k−1 + S k−1 + 3. Using the initial values P 1 = 1, P 2 = 0, P 3 = 3, S 1 = 0, S 2 = 3, S 3 = 6, solving the difference equation yields the desired results.
By the symmetry of the tree, the distribution of the secondary nodes must be equal.
Lemma 2.3.6 Of all the colorings in T (G), which result from Lemma 2.3.3, there exists a collection S of size at least 3/4 of these colorings where each color class at level n − 1 other than the primary color is of equal size and holds at least 2/9 of the colorings in this collection.
Proof. Again, for the sake of simplicity, we assume the root node is colored 1, that is, 1 is the primary color of T * (U n ) and 2,3 and 4 are the secondary colors.
Let the notation s indicate the form of an interval that does not use the secondary color s. Note that an interval may be of the form s i , s i s j , or s i s j s k .
Let the notation (C 1 , C 2 , C 3 , C 4 ) indicate the distribution of the colorings which survive after Lemma 2.3.3. C i gives the number of colorings that have the color i at the n − 1 level. Given an interval-path, let φ denote the color of the bottom node.
Case 3.1 k is odd.
There are 2 k−1 −1 interval-paths beginning with the primary color 1 and also ending in the primary color, that is, φ = 1. Three of these are two-colored intervals.
From the trees hanging beneath these interval-paths ending in the secondary colors 2,3 and 4, we keep a distribution of (3z,z,z,z) where z = 2 k −2 3 . This gives a total of 6z + 3x colorings with the distribution (3z, z + x, z + x, z + x) at the n − 1 level. The sizes of the secondary color classes at the n − 1 level are equal. Since x > z, z + x > 2z and therefore, z + x is clearly at least 2/9 of the colorings.
Case 3.2 k is even.
From the trees hanging beneath these interval paths ending is the primary color 1, we keep a weighted distribution of (0,x,x,x) where x = 2 k −4 3 . There are 2 k − 1 interval-paths beginning with the primary color 1, ending in a secondary color 2, 3 or 4, that is, φ = 2, 3, or 4. Three of these are two-colored intervals.
From the trees hanging beneath these interval paths ending in a secondary number, we keep a combined weighted distribution of (3z,z,z,z) where z = 2 k +2 3 . This gives a total of 6z + 3x colorings with the distribution (3z, z + x, z + x, z + x) at the n − 1 level.
Leaving out 3 colorings that use the primary color in level n − 1, we obtain a collection of size 6z + 3x − 3 with distribution (3z − 3, z + x, z + x, z + x). We see that the size of this subcollection is larger than 3/4 of 6z + 3x and that the sizes of the secondary color classes at the n − 1 level are equal and hold 2/9 of the colorings in this collection.
Lemma 2.3.7 Let A and B be non-negative integers, R be the residue class modulo 4 of 3(A + 3B), and r the residue class modulo 2 of B. If A + R 3 + 2r 3 ≤ 3B, then there exists an integer C of size at least 3 4 (A + 3B) such that B is at least 2 9 C.
we are done.

Main Conjecture
Conjecture 2.4.1 Let G be a coil on n vertices with m crossing edges and β nonempty up-intervals. Denote by α the least subscript of a vertex that has a nonempty up-interval. There is a corresponding colored tree T (G) that contains at least C = 3 n−1 2 3 m 3 4 β−1 full-paths, each of which represents a proper coloring of G. Also, there exists a sub-collection S of these paths of size at least 3/4 · C, where each color class at level α − 1 other than the primary color P α holds an equal number s of colorings, where s ≥ 2/9 · |S|.
Corollary 2.4.2 Based on this conjecture, all coils are 4-colorable.
Idea for Proof. We make great strides for proving this conjecture with a number of important lemmas and observations.
We will prove the case for P α = 1. The proof is by induction on β. proper coloring of G α+1 such that the color of v 1 is 1 and S α+1 is the sub-collection of these paths of size at least 3/4 of C α+1 , where each color class at level α other than P α+1 holds an equal number of colorings, the size of which is at least 2/9 of the colorings in S α+1 .
STEP 1 T (U α ) is the tree with height is u and is colored (as previously described) so that the color of the children are distinct and do not equal that of the parent.
For simplicity, assume the color of the root node is 1. Then 1 is the primary color of the interval, while {2, 3, 4} are the set of secondary colors.
There are 3 u−1 full paths in T (U α ).
STEP 2 T * (U α ) is formed by removing (if necessary) from T (U α ) the intervalpaths that use more than three colors. As was shown in Lemma 2.3.2, T * (U α ) has 3(2 u−1 − 1) unique interval-paths for u > 1. For u = 1, T * (U α ) is the single node colored by the primary color 1.
There are X = 1 for u = 1 3(2 u−1 − 1) for u > 1 full paths in T * (U α ). We calculate the number C of full paths in T (G α ), each representing a proper coloring of G α = G after STEP 4 by considering three cases.
Case 1 u = 1 P α = P α+1 = 1 and U α is an interval of type 234. We see that 3(2/9)=2/3 are Case 2 u = 2 Notice that after STEP 4, a node x at level α − 1 may have 0,1,2, or 3 children remaining in T (G α ) depending on the number of its children that were trimmed.
Consider all of the leaves of color s in T * (U α ). In STEP 3, the trees T (G α+1 ) that are hung from those leaves, are exactly the same. So each interval-path of type s in T (U α ) accounts for the same number of each of the colors in level α − 1 of T (G α ). Thus, the distribution of colors at level α − 1 depends upon φ, the color of the leaf node at level u and the type of interval-path that extends from the root to that node.
Consider the following tables. Table 2. Distribution of next level upon trimming.
The sequences (C 1 , C 2 , C 3 , C 4 ) in Table 2 show the distribution of the colors in row α − 1 of T (G α ), where φ is the color of the leaf of T * (U α ) and s is the type of interval-path that extends from the root to the given leaf. So that C s is the number of surviving colorings with color s at level α − 1 -counting multiple children of a node at level α − 1.
Case 1 u = 2 There are three interval paths, all starting with the primary color 1 ending in one of each φ = 2, 3, and 4. These paths are colored 34, 24, and 23 respectively.
Case 2 u ≥ 3 Case 2.1 u is odd.
There are 2 u−1 − 1 interval paths from level 1 through level u beginning with the primary color 1 and also ending in the primary color, that is, φ = 1. Three of these are two-colored intervals. Considering the subtrees T (G α+1 ) hanging from these vertices at level u, we use φ = 1 and Table 1.
Case 2.2 u is even.
There are 2 u − 1 interval paths beginning with the primary color 1, ending in a secondary color 2, 3 or 4, that is, φ = 2, 3, or 4. Three of these are two-colored intervals.
Applying Lemma 2.3.7, we need only show that P which is easily verified.
There are 2 p−1 − 1 interval paths from level 1 through level p beginning with the primary color 1 and also ending in the primary color, that is, φ = 1. Three of these are two-colored intervals: 34, 24, and 23. Of the remaining 2 p−1 − 4 interval paths, they are equally distributed 2, 3, and 4. Since we are keeping 2's, 3's, and 4's, at level α, we look at what we are leaving at level α − 1 for each of the 6 specific situations as described in Table 3 , φ = 1.
There are 2 p −2 3 interval paths from level 1 through level p beginning with the primary color 1 and ending in the secondary color 2, that is, φ = 2.
Hence, we need to show that y ≤ 3x + v.
Case 3.1.2 p is even.
The argument is similar, yielding the same inequality.
We use the results from the previous case.

Future Work
The depth-first search tree of a planar graph breaks the graph into coils. There is no record of any attempt to four-color a coil, so this is a new unsolved problem.
In our attempt to solve this problem, we discovered many propositions which have led us to our current state. We have a computer program that has validated the conjecture for workable values of n. We are looking to see the pattern of behavior for our last case. If we can isolate a pattern, we might understand better the inequality. So far, the computer generated patterns show us that the inequality that we need does hold, that is, our conjecture has not been disproven.  (N-K)))))) if (isempty(intersect(P3(1,:),P3(a,:))))