Chapter 1
Introduction
1.1 Background
In 1735, Swiss mathematician Leonhard Euler solved the “Ko¨nigsberg bridge problem”. This gave birth to Graph Theory and Leonhard Euler is regarded as the father of Graph Theory. The paper titled “Solutio problematis ad geometriam situs pertinentis” written by L. Euler [6] on the Seven Bridges of Ko¨nigsberg and published in 1741 is regarded as the first paper in the history of Graph theory.
A graph consists of objects and the relationship between the objects. A graph comprises of vertices (also called nodes or points) representing the objects and edges (also called arcs or lines) joining two vertices if they are related. A graph is denoted by G = (V, E) where V (G) is the vertex set and E(G) is the edge set. The order of a graph is the cardinality of its vertex set and size of a graph is the cardinality of its edge set. A loop is an edge having same end vertices [16]. Multiple edges are the edges which share same end points [16]. A graph without loops and multiple edges is called a “simple graph”[16]. Otherwise the graph is called a “pseudo graph”. For example, see Figure 1.1.
Figure 1.1: Simple and Pseudo graph
In the past few decades, Graph theory has been extensively applied in the field of Computer science, Chemistry, Sociology, Electrical engineering, Communication science etc. One of the most important graph theoretical concepts which has vast appication in day to day life is graph coloring. Some of the important applications of graph coloring could be seen in circuit design, astronomy, coding theory, data base management, communication network, channel assignment, etc.
Graph coloring is a function from the vertex set V (G) to the set of colors such that no two adjacent vertices receive the same color. In other words, it is a way of coloring the vertices of a graph such that any two adjacent vertices receive different colors. This is called a “proper vertex coloring”. While trying to color a map of the counties of England, Francis Guthrie postulated the “Four color conjecture” which stated that four colors were sufficient to color the map so that no regions sharing a common border receive the same color. The Four color conjceture was proved in 1976 by Kenneth Appel and Wolfgang Haken.
Over the course of time, different types of coloring have been developed. Some of them being T-coloring, L(2,1)-coloring, Radio coloring, Hamiltonian coloring, Total coloring etc. An L(2,1) coloring is an assignment of colors such that the color difference between the adjacent vertices is atleast two and the color difference between the vertices at distance two is atleast one.
1.2 Preliminaries
Every graph referred to in this project work is finite, undirected, simple, connected graph.
Definition 1.2.1. A graph G is an ordered pair (V, E) consisting of a set V of vertices and a set E of edges which are 2-element subsets of V . If the edge is an unordered pair then the graph is undirected.
Definition 1.2.2. A path of length k in a graph is an alternating sequence of distinct vertices and edges, v0, e0, v1, e1, v2, …, vk−1, ek−1, vk. We denote a path on n vertices by Pn.
Example:
Figure 1.2: A path on 5 vertices (P5)
Definition 1.2.3. A cycle graph is a path with the first and last vertices coinciding. We denote a cycle graph on n vertices by Cn.
Example:
Figure 1.3: A cycle on 5 vertices (C5)
Definition 1.2.4. A wheel graph on n vertices is a join of two graphs namely, complete graph
K1 and cycle Cn−1. We denote a wheel graph on n vertices by Wn. So, Wn = K1 + Cn−1
Example:
Figure 1.4: A wheel on 6 vertices (W6)
Definition 1.2.5. A fan graph on n vertices is a join of two graphs namely, complete graph K1
and a path Pn−1. We denote a fan graph on n vertices by Fn. So, Fn = K1 + Pn−1. Example:
Figure 1.5: A fan on 6 vertices (F6)
Definition 1.2.6. A star graph is a tree formed by joining a single vertex to pendant vertices(i.e. vertices of degree one). We denote a star graph on n vertices by K1,n−1.
Example:
Figure 1.6: A star on 5 vertices (K1,4)
Definition 1.2.7. A bistar Br,s on r + s + 2 vertices is a tree obtained by joining the central vertices of two stars K1,r and K1,s.
Example:
Figure 1.7: A bistar on 10 vertices (B3,5)
Definition 1.2.8. A prism graph GP (n, 1) is a planar graph obtained by the cartesian product of a cycle Cn and K2. So, GP (n, 1) = Cn K2.
Example:
Figure 1.8: A prsim graph on 16 vertices (GP (8, 1))
Definition 1.2.9. [16] Given a simple graph G = (V, E), the complement of G denoted by G is a simple graph with the vertex set V (G) and any two vertices are adjacent in G if and only if they are not adjacent in G.
Definition 1.2.10. [4] Given a simple connected graph G = (V, E), an L(2, 1)-coloring is a function f from the vertex set V (G) to the set of non-negative integers such that:
(i) |f(u) − f(v)| ≥ 2 ∀ uv ∈ E(G)
(ii) |f(u) − f(v)| ≥ 1 if d(u, v) = 2
Example:
Figure 1.9: An L(2, 1)-coloring of P5
Definition 1.2.11. [12] Let f be an L(2, 1)-coloring of a graph G. The span of f, denoted by
λ(f) is the maximum color assigned to a vertex in G by f.
Definition 1.2.12. [4] The span of G, denoted by λ(G) is the smallest among all the f-spans.
λ(G) = min{λ(f) : f is an L(2,1)-coloring}
Definition 1.2.13. [12] A span coloring is an L(2, 1)-coloring in which the maximum color used is λ(G).
Figure 1.10: Span colorings of P5
Definition 1.2.14. [12] A span coloring f is said to be a full-coloring if f is onto.
Definition 1.2.15. [12] An L(2, 1)-coloring f is irreducible if there does not exist an L(2, 1)- coloring g such that g(u) ≤ f(u) for all u ∈ V (G) and g(v) < f(v) for some v ∈ V (G). Example:
Here, the first span coloring is reducible since the color 4 given to the right most vertex can be reduced to 2. The second span coloring is an irreduclible span coloring of P5.
Figure 1.11: Reducible and irreducible span colorings of P5
Definition 1.2.16. [12] Let f be an L(2, 1)-coloring of a graph G using the colors {0, …, k}. An integer h such that 0 < h < k is said to be a hole in f if f assigns the color h to none of the vertices in G. Let H denote the set of holes. Let Hλ(G) denote the maximum number of holes in an irreducible span coloring of G.
Example:
Figure 1.12: Irreducible span colorings of P5 with different set of holes
Here, the first irreducible span coloring has two holes namely, 1 and 3. The second irre- ducible span coloring has one hole namely, 2. The third irreducible span coloring has no holes. So, the maximum number of holes is 2.
Definition 1.2.17. [12] The index of G is the minimum number of unused colors in a span coloring of G. It is denoted by ρ(G).
Definition 1.2.18. [12] An L(2, 1)-coloring using colors from 0 to k is a no-hole coloring if it uses all the labels from 0 to k.
Definition 1.2.19. [12] An inh-coloring is defined as an irreducible no-hole coloring.
Definition 1.2.20. [12] Let f be an irreducible span coloring of a graph G. If f has the maximum number of holes among all the irreducible span colorings of G, then f is called a maximum-hole coloring.
1.3 Literature Review
The channel assignment problem was propsed by W.K. Hale [10] in 1980. Inspired by the chan- nel assignment problem, F. Roberts as a of his communication to J. Griggs in 1988, proposed the problem of optimally assigning frequencies to radio transmitters using nonnegative integers to represent channels, so that “close” stations are assigned different integers, and difference in the frequencies assigned to “very close” staions is atleast 2. This gave rise to the study of L(2, 1)-coloring of a graph which was first proposed by J. Griggs and R. Yeh [9] in 1992.
The concept of no hole 2 distance-coloring was introduced by F. Roberts [14] in 1993 for T -coloring of graphs. Here, the vertices of a graph are colored with positive integers so that the color difference between any two adjacent vertices is greater than 1 and so that the set of integers used as colors is a consecutive set (the no-hole requirement). F. Roberts [14] has studied what graphs have such colorings. He also studied what graphs have such colorings which are near-optimal, that is, separation between the largest and smallest colors used is no more than one larger than the minimum span in a non-adjacent coloring which may have holes.
D. Sakai and C. Wang [15] in 1993 studied a generalization of the no-hole 2-distant color- ings introduced by F. Roberts [14]. Here, the vertices of a graph are colored with consecutive positive integers so that the color difference between any two adjacent vertices get is greater than a fixed positive integer r > 1.
A.A. Bertossi and M.A. Bouccelli [2] have studied how to avoid hidden collision in the multi hop radio networks in 1995.
In the year 2000, P. Fishburn and F. Roberts [7] proved that all trees except stars are full- colorable. They have proved that C6 is the only connected graph with minimum number of vertices exceeding λ(G) which is not full-colorable. They have given a number of other con- nected graphs that are not full-colorable and have given details on full-colorability of graphs with maximum degree four or less than four.
P. Fishburn and F. Roberts [8] in 2003 introduced the parameter µ(G), which is defined to be the minimum integer k for which G has a no-hole coloring, if it exists. They have proved two main results in their paper. First, for every m ≥ 1 there is a graph G with ρ(G) = m and µ(G) = λ(G) + m. Second, for every m ≥ 2 there is a connected G with λ(G) = 2m, n = λ(G) + 2 and ρ(G) = m.
In the year 2007, S.S. Adam et al. [1] have shown that for each k ≥ 2, there exists an
r-regular graph G with ρ(G) = k so that ρ(G) and r are relatively prime integers.
R. Laskar et al. [13] in 2009 have shown that for any tree T that is not a star, inh-span of T is equal to its span. Here, the inh-span of a graph G, denoted by λinh(G), is the smallest number k such that there is an irreducible no-hole L(2, 1)-coloring of G using the integers 0, …, k.
In 2009, R. Laskar and G. Eyabi [12] have determined the exact values of maximum number of holes in the span colorings of certain classes of graphs such as paths, cycles, trees, complete bipartite graphs, complete multi-partite graphs. Also, they have characterized complete graphs in terms of their maximum number of holes.
1.4 Scope of Work
We review the paper “Holes in L(2, 1)-Coloring of Some Classes of Graphs” by R. Laskar and G. Eyabi [12]. We further extend their work by determining the exact values of maximum number of holes in the span colorings of complement of a cycle, wheel, fan, bistar. We also determine the bounds for maximum number of holes in span coloring for prism graph in this project. This work can be further extended by finding the exact values for maximum number of holes for prism graph and some other complex classes of graphs such as generalized Petersen graph, flower graph, different types of product of graphs etc.
Chapter 2
Holes in L(2,1)-Coloring of graphs
Now a days, the assignment of radio frequency to the radio stations is creating a lot of problems because day by day, the number of radio stations getting installed are increasing . As a result, there is an increase in the request for radio frequencies . Interference during communication is a difficulty in assigning channels to different radio stations. Two adjacent stations which have same frequency will have a direct interference or “direct collision” whereas station nearby the adjacent station perform hidden interference or “hidden collision”. Bertossi et al. [3, 2] studied the case of avoiding hidden collision in the multi hop radio networks. To avoid collisions from its neighbor and next to neighbour station we require distinct labeling.
The “channel assignment problem” is the task of optimally assigning frequencies to radio transmitters located at various places so that there is interference-free communication. The frequencies assigned to neighboring transmitters have to be significantly different and pairs of transmitters which are slightly farther away are given different frequencies so that there is no interference. Once, it was said that the number of usable frequencies was very much greater than the number of transmitters. However, this is no longer true.
Inspired by the channel assignment problem introduced by W.K. Hale [10] in 1980, F. Roberts introduced the L(2, 1)-coloring problem as a result of his communication to J. Griggs in 1988. As a result of this, J. Griggs and R. Yeh [9] in 1992 introduced the concept of L(2, 1)-
coloring. In this type of coloring, the colors (nonnegative integers) assigned to the vertices of a graph depend not only on whether two vertices are adjacent but also on whether two vertices are at distance 2.
The L(2, 1)-coloring problem is the problem of assigning frequencies (non-negative inte- gers) to radio transmitters so that “close” transmitters receive different frequencies and “very close” transmitters receive frequencies that are at least two frequencies apart. This problem can be modeled using a graph G. The vertices of the graph represent the transmitters. Based on the proximity of the transmitters and the power of the transmissions, edges are placed between vertices to represent possible interference. Two vertices are “very close” if they are adjacent in the graph and “close” if they are at distance two in the graph. The frequencies that are assigned to the radio stations are denoted by the nonnegative integers 0, …, λ.
The following are few preliminary results which are used to prove further results.
Proposition 2.0.1. [9] Let Pn be a path on n ≥ 2 vertices. Then,
2 if n = 2
λ(Pn) =
3 if n = 3, 4
4 if n ≥ 5.
Proposition 2.0.2. [9] Let Cn be a cycle on n ≥ 3 vertices. Then λ(Cn) = 4.
Theorem 2.0.3. [9] If T is a tree with maximum degree ∆ ≥ 1, then ∆ + 1 ≤ λ(T ) ≤ ∆ + 2. Proposition 2.0.4. [8] If G is not full colorable and n ≥ λ(G) + 2, then µ(G) ≤ λ(G) + ρ(G). Corollary 2.0.5. [8] G has a no-hole coloring if and only if n ≥ λ(G) + 1.
Theorem 2.0.6. [4] If G is a graph with maximum degree ∆ and λ(G) = ∆ + 1, then for any span coloring of G, a vertex of degree ∆ must be labeled 0(or ∆ + 1) and its neighbors must be labeled 2 + i (or i), i = 0, 1, …, ∆ − 1.
The following are few results based on holes in L(2, 1)-coloring.
2.1 Maximum Number of Holes in Paths
Lemma 2.1.1. [8] If G is a graph with span λ, then
(i) 0, λ 6∈ H and
(ii) h ∈ H implies that h − 1, h + 1 6∈ H.
Theorem 2.1.2. [12] Let Pn be a path on n vertices, n > 4. Then,
(i) ρ(Pn) = 0
(ii) Hλ(Pn) = 2
Proof. (i) Since all paths Pn (n > 4), are full colorable, we have ρ(Pn) = 0.
(ii) We first order the vertices of path Pn as v1, v2, …, vn. Here v1 is the first vertex, vn is the last vertex and for every m, m = 1, 2, …, n − 1, vmvm+1 ∈ E(Pn). We define an L(2, 1)-coloring f of Pn as follows:
f(vk) =
0 if k 1(mod 3)
2 if k ≡ 2(mod 3)
4 if k ≡ 0(mod 3)
Here, f is an irreducible span coloring on Pn with two holes, namely, 1 and 3. So, we have,
Hλ(Pn) ≥ 2.
Now, we prove that Hλ(G) ≤ 2. By Lemma 2.1.1 (i), 0, 4 6∈ H since λ(Pn) = 4 for n > 4. So, the remaining labels which can be used in a span coloring of Pn are 1, 2 and 3. By Lemma
2.1.1 (ii), all these labels cannot be holes. That is, at least one of these labels must be in f(Pn). Thus, we have Hλ(G) ≤ 2. Therefore, Hλ(Pn) = 2.
2.2 Maximum Number of Holes in Cycles
Lemma 2.2.1. [12] Let Cn be a cycle on n vertices and let f be an irreducible span coloring on Cn. Then there exists v ∈ V (Cn) such that f(v) = 1 if and only if there exists u ∈ V (Cn) such that f(u) = 3.
Proof. We know that every vertex in a cycle has degree two. Let v ∈ V (Cn) such that f(v) = 1. Then, the labels 0, 1, and 2 cannot be used for neighbors of v. So one of the neighbors of v must be labeled 3 since λ(Cn) = 4. So, the forward part is proved.
Conversely, suppose there exists a vertex u ∈ V (Cn) such that f(u) = 3. We see that the labels 2, 3 and 4 cannot be used for neighbors of u. So, one of the neighbors of u must be labeled 1 which proves the converse part. Hence the lemma.
Theorem 2.2.2. [12] Let Cn be a cycle on n vertices, n > 4. Then,
(i) ρ(Cn) = 0 except for C6
(ii) hi(C6) = 0
(iii) if n ≡ 0(mod 3), then Hλ(Cn) = 2
(iv) if n 6≡ 0(mod 3) and n ≡ 0(mod 4), then Hλ(Cn) = 1
(v) if n 6≡ 0(mod 3) and n 6≡ 0(mod 4), then Hλ(Cn) = 0
Here, hi(G) denotes the minimum number of holes over all irreducible L(2, 1)-colorings of G. Proof. (i) Since all cycles Cn (n > 4) are full colorable except for C6, we have ρ(Cn) = 0 for
all n > 4. Also, ρ(C6) = 2.
(ii) C6 is inh-colorable as shown in Figure 2.1. Hence, hi(C6) = 0.
Figure 2.1: inh-coloring of C6
(iii) Let n ≡ 0(mod 3). We first irder the vertices of cycle Cn as v1, v2, …, vn where vnv1 ∈ E(Cn) and for every m, m = 1, 2, …, n − 1, vmvm+1 ∈ E(Cn). We define an L(2, 1)- coloring f of Cn as follows:
f(vk) =
2 if k 1(mod 3)
0 if k ≡ 2(mod 3)
4 if k ≡ 0(mod 3)
f is an irreducible span coloring on Cn with two holes, namely, 1 and 3. So, Hλ(Cn) ≥ 2.
To prove the reverse inequality, we use the same arguments as in the case of proving the reverse inequality of Theorem 2.1.2 and see that Hλ(Cn) ≤ 2. Thus, Hλ(Cn) = 2 which proves part (iii) of the theorem.
(iv) Let n 6≡ 0(mod 3) but n ≡ 0(mod4). We have two cases here.
Case 1. n ≡ 1(mod 3) and n ≡ 0(mod 4).
Using Lemma 2.1.1 and Lemma 2.2.1, there are only 3 possible sets of labels available for color- ing any cycle on n vertices, n > 4. The sets are: S1 = {0, 2, 4} with two holes, S2 = {0, 1, 3, 4} with one hole and S3 = {0, 1, 2, 3, 4} with no holes.
First we show that S1 cannot be used to color Cn in this case. Consider the ordering of the vertices of Cn described in the proof of part (iii) above. Trying to color Cn by alternating the labels 0, 2 and 4, we observe that f(vn) = f(v1) which contradicts the definition of an L(2, 1)- coloring. So, S1 cannot be used to color Cn. Thus, Hλ(Cn) < 2 or Hλ(Cn) ≤ 1.
We now investigate S2. Again, considering the ordering of the vertices of Cn described in the
proof of part (iii) above, we define an L(2, 1)-coloring f of Cn as follows:
f(vk) =
0 if k 1(mod 4)
3 if k ≡ 2(mod 4)
1 if k 3(mod 4)
4 if k ≡ 0(mod 4)
Here, f is an irreducible span coloring of Cn with one hole which is 2. So, we have Hλ(Cn) ≥ 1. Therefore, Hλ(Cn) = 1 in this case.
Case 2. n ≡ 2(mod 3) and n ≡ 0(mod 4).
Consider the ordering of the vertices of Cn as above. Trying to color Cn by alternating 0, 2 and 4, we observe that f(vn−1) = f(v1) and f(vn) = f(v2) which contradicts the definition of an L(2, 1)-coloring. So, S1 cannot be used to color Cn in this case. The same argument as in Case 1 above proves that S2 can be used to color Cn in this case. Thus, Hλ(Cn) = 1 in this case.
(v) Let n 6≡ 0(mod 3) and n 6≡ 0(mod 4). In part (iv), we proved that if n 6≡ 0(mod 3), then S1 cannot be used to color Cn. Here, it remains to prove that, if n 6≡ 0(mod 4), then S2 cannot be used to color Cn. Consider the ordering of the vertices of Cn described in the proof part (iii). Color consecutive groups of 4 vertices by alternating the labels 0, 3, 1 and 4 starting with any of the labels. We observe that:
(a) for n ≡ 1(mod 4), f(vn) = f(v1) which contradicts the definition of an L(2, 1)-coloring, (b) for n ≡ 2(mod 4), f(vn) = f(v2) which contradicts the definition of an L(2, 1)-coloring
and
(c) for n ≡ 3(mod 4), |f(vn) − f(v1)| = 1 which contradicts the definition of an L(2, 1)- coloring.
Thus, S2 cannot be used to label Cn in this case. We know that Cn is full colorable for all n > 4 except for C6. Hence, we conclude that only S3 can be used to color Cn in this case. Therefore, Hλ(Cn) = 0.
2.3 Maximum Number of Holes in Trees
Lemma 2.3.1. [12] Let T be a star on n vertices, with ∆ ≥ 3. Then Hλ(T ) = 1 = ρ(T ). Proof. We know that for any star T , λ(T ) = 1 + ∆. Let v denote the central vertex of the star
with f(v) = m. Since the vertex v has maximum degree, by Theorem 2.0.6, we have m = 0 or m = ∆+1. So, the neighbors of v can be labeled 2, 3, …, 1+∆ or 0, 1, 2, …, ∆−1 respectively. Since, there is only one hole in both the cases, we have Hλ(T ) = ρ(T ) = 1.
Theorem 2.3.2. [12] Let T be a tree that is not a star with ∆ ≥ 2. Then,
(i) λ(T ) = 1 + ∆ ⇒ Hλ(T ) ≤ 1
(ii) λ(T ) = 2 + ∆ ⇒ Hλ(T ) ≤ 2
Proof. By Theorem 2.0.3, we have λ(T ) = 1 + ∆ or λ(T ) = 2 + ∆. We investigate both the cases.
(i) Let λ(T ) = 1+∆. By Theorem 2.0.6, the maximum degree vertices must be 0 or 1+∆. Let v be a vertex of maximum degree in T . Consider the star T 0, a subgraph of T , which is centered at v. By Lemma 2.3.1, Hλ(T 0) = 1. Since λ(T 0) = λ(T ) and T 0 is a subgraph of T , we observe that each hole in T should be a hole in T 0. So, Hλ(T ) ≤ Hλ(T 0) = 1. Therefore, Hλ(T ) ≤ 1.
(ii) Let λ(T ) = 2 + ∆. In this case, there are no restrictions for the labels of the maximum degree vertices. Let a star T 0 be a subgraph of T , whose center is some maximum degree vertex v. Set f(v) = k where k = 0, k = 1 + ∆ and k = 2 + ∆. While assigning labels to the neighbors of T 0, we must avoid only the labels k − 1, k and k + 1. Here, we have two holes namely, k −1 and k +1 where the maximum label in T 0 is 2+∆. This is an irreducible coloring. While labeling the remaining vertices of T , at he most we maintain these two holes since the
maximum label in T 0 is 2 + ∆ = λ(T ). Thus, Hλ(T ) ≤ 2.
2.4 Maximum Number of Holes in Some Other Classes of
Graphs
Theorem 2.4.1. [12] Let G be a simple graph on n vertices, n > 1. Then Hλ(G) = n − 1 if and only if G = Kn.
Proof. Assume that G = Kn. Kn has a unique L(2, 1) optimum labeling f with f(Kn) =
0, 2, 4, …, 2n − 2 and hence λ(Kn) = 2n − 2. The holes are the first n − 1 odd numbers. This proves the forward part.
We prove the converse part by induction on n. We see that the result is true when n = 2. So, assume that the result is true for graphs of order n−1. Let G be a graph with n vertices such that Hλ(G) = n − 1. Let v be any vertex of G. Now no graph on n − 1 vertices can have more than n − 2 holes in a λ-labeling (and in this case the labels must be 0, 2, …, 2n − 4). So, G − v must be complete and this forces G also to be complete since otherwise Hλ(G) = Hλ(G−v) = n−2 which is a contradiction. This proves the converse part. Hence the theorem.
Corollary 2.4.2. [2] Let G be a graph on n vertices. Then 0 ≤ Hλ(G) ≤ n − 1.
Proof. Suppose G = Cn where n 6≡ 0(mod 3) and n 6≡ 0(mod 4). By Theorem 2.2.2, Hλ(G) =
0, showing it is tight, since Hλ(G) ≥ 0 for any graph G. By Theorem 2.4.1, Hλ(G) ≤ n − 1
since any graph G is a subgraph of a complete graph Kn.
Proposition 2.4.3. [2] Let Kn,m be a complete bipartite graph. Then Hλ(Kn,m) = 1.
Proof. It is easy to see that λ(Kn,m) = n + m. There are n + m + 1 labels in [0, n + m] and n + m vertices in Kn,m. Since diameter of any complete bipartite graph is two, all the vertices must be given distinct labels. We see that every span coloring of Kn,m must have exactly (n + m + 1) − (n + m) = 1 hole. Therefore, Hλ(Kn,m) = 1.
2.5 Conclusion
In this chapter, we had a brief introduction to the concept of L(2, 1)-coloring and holes in L(2, 1)-coloring. We have determined the maximum number of holes in irreducible span color- ing of a path, cycle, star, complete bipartite graph. Also, we have characterized complete graphs in terms of their maximum number of holes. Further, we extend this study by determining the maximum number of holes in irreducible span coloring for complement of cycle graph, wheel, fan, bistar and bounds for prism graph.
Chapter 3
New Results, Observations and Conclusion
To determine the maximum number of holes in L(2, 1)-coloring of a graph, we have to define an irreducible span coloring for that graph. For that, we first find the span of the graph λ(G). Then, an irreducible span coloring can be found for the graph using span of the graph. In this chapter, we determine the maximum number of holes in irreducible span coloring for complement of cycle graph, wheel, fan, bistar and bounds for prism graph.
3.1 Maximum Number of Holes in L(2,1)-Coloring for Com- plement of a Cycle, Wheel, Fan, Bistar and Prism
In addition to the above mentioned results, we use the following results for further discussion.
Theorem 3.1.1. [9] For every positive integer t, λ(K1,t) = t + 1.
5 if n ≡ 0(mod 3)
Theorem 3.1.2. [11] λ(Cn P2) =
6 if n 6≡ 0(mod 3)
We begin with the problem of determining the maximum number of holes in L(2, 1)-coloring for complement of a cycle as follows:
Theorem 3.1.3. Let Cn be the complement of a cycle Cn, n ≥ 5. Then, Hλ(Cn) = 0.
Proof. Order the vertices of Cn as v1, v2, …, vn. Since diameter of Cn(n ≥ 5) is 2, all the vertices have to be given distinct colors. Define a coloring f as follows:
f(vi) = i − 1 ∀ i = 1, 2, …, n
Note that a vertex vi is at a distance 2 from the vertices vi−1 and vi+1 and adjacent to rest of the vertices where i is under addition modulo n.
We now prove that the coloring f is an L(2, 1)-coloring.
If vi and vj are any two adjacent vertices in Cn(i < j), then
|f(vj − f(vi)| = |(j − 1) − (i − 1)|
= j − i
≥ 2
If vi and vj are any two non-adjacent vertices in Cn(i < j), then they are at distance two. Hence,
|f(vj − f(vi)| = |(j − 1) − (i − 1)|
= j − i
≥ 1
Hence, f is an L(2, 1)-coloring that uses the colors {0, 1, …, n − 1}. Since f(vn) = n − 1 is the largest color that has been used, λ(Cn) ≤ n − 1.
We claim that λ(Cn) = n − 1. Suppose that λ(Cn) < n − 1. Let f be an L(2, 1)-coloring of Cn using the colors {0, 1, 2, …, n−2}. Since every two distinct vertices of Cn are either adjacent or at distance 2, f must assign distinct colors to all vertices of Cn. Therefore, exactly n number of distinct colors are required to color Cn, but the number of colors in {0, 1, 2, …, n − 2} is n − 1. Hence, λ(Cn) = n − 1.
f is an irreducible span coloring and all the colors in {0, 1, 2, …, n−1} are used for coloring.
So, there are no holes, i.e. Hλ(Cn) ≥ 0. But, Hλ(Cn) ≯ 0 since exactly n colors are required
for coloring. Hence, Hλ(Cn) = 0.
Figure 3.1: Irreducible span coloring for C6 and C7
Remark 3.1.4. Since cycle is a 2 regular graph, the degree of each vertex in Cn is (n−1)−2 =
n − 3. So, complement of a cycle is an n-3 regular graph.
Remark 3.1.5. The diameter of complement of cycle is 2 since each vertex in Cn is adjacent to all other vertices except its succeeding and preceeding vertices. So, any two vertices in Cn are either adjacent or at distance two.
Remark 3.1.6. Here, n ≥ 5 is considered because for n = 3 and n = 4, we get C3 and C4 as disconnected graphs which we are not interested.
Theorem 3.1.7. Let Wn be Wheel graph on n vertices(n ≥ 5). Then,
2 if n=5
Hλ(Wn) =
1 if n ≥ 6
Proof. Order the vertices of Wn as v1, v2, …, vn where vn is the central vertex. Since diameter of Wn is 2, all the vertices have to be given distinct colors. Define an L(2,1)-coloring f as
follows:
Case 1. n = 5
Irreducible span colorings for W5 is shown in Figure 3.2.
Figure 3.2: Irreducible span colorings for W5
Thus, λ(W5) ≤ 6. We claim that λ(W5) = 6. Suppose that λ(W5) < 6. Let f be an L(2, 1)-coloring of W5 using the colors {0, 1, 2, 3, 4, 5} such that λ(f) = λ(W5) = 5. Since W5 has diameter two, all the vertices have to be given distinct colors.
If f(v5) = 5, the color 4 cannot be used. The colors remaining are {0, 1, 2, 3} which how- ever cannot be used to color the remaining vertices since the definition of L(2, 1)-coloring is violated.
Further, if f(v5) = 0, the color 1 cannot be used. The colors remaining are {2, 3, 4, 5} which however cannot be used to color the remaining vertices since the definition of L(2, 1)-coloring is violated.
Thus, λ(W5) = 6. The above given colorings are the only two colorings possible with λ(W5) =
6. For the first span coloring, 1 and 4 are holes. For the second span coloring, 2 and 5 are holes. We see that Hλ(W5) = 2.
Case 2a. n is even (n ≥ 6)
f(vn) = n
f(vi) = 2i − 2 1 ≤ i ≤ n
f(vj) = f(vj−( n−1 ) + 1
2 + 1 ≤ j ≤ n − 1
Case 2b. n is odd (n ≥ 7)
f(vn) = n
f(vi) = 2i − 2 1 ≤ i ≤ n−1
n+1
f(vj) = c(vj−( n−1 ) + 1
2 ≤ j ≤ n − 1
The maximum color that is used is f(vn) = n. Thus, λ(Wn) ≤ n (n ≥ 6). To prove the reverse inequality, consider the star K1,n−1 on n vertices which is a spanning subgraph of Wn. By Theorem 3.1.1, λ(K1,n−1) = (n − 1) + 1 = n. Since K1,n−1 is a spanning subgraph of Wn, we have, λ(K1,n−1) ≤ λ(Wn). That is, λ(Wn) ≥ n. Thus, λ(Wn) = n.
f is an irreducible span coloring with one hole, namely, n − 1 for all wheels (n ≥ 6) i.e. Hλ(Wn) ≥ 1 (n ≥ 6). Consider the star K1,n−1, which is a spanning subgraph of Wn. By Lemma 2.3.1, Hλ(K1,n−1) = 1. Since λ(K1,n−1) = λ(Wn) and K1,n−1 is a spanning subgraph of Wn, we see that every hole in Wn must be a hole in K1,n−1. That is, Hλ(Wn) ≤ Hλ(K1,n−1) = 1 and so Hλ(Wn) ≤ 1. Hence, Hλ(Wn) = 1.
Figure 3.3: Irreducible span coloring for W6
Remark 3.1.8. Here, n = 4 is not considered because for n = 4, we get a complete graph on 4 vertices for which the maximum number of holes is already determined.
Corollary 3.1.9. Let Fn be Fan graph on n vertices(n ≥ 5). Then, Hλ(Fn) = 1. Proof. Irreducible span coloring for F5 is shown in Figure 3.5.
Figure 3.4: Irreducible span coloring for W7
Figure 3.5: Irreducible span coloring for F5
The maximum color used is 5. So, λ(F5) ≤ 5. Now, we prove that λ(F5) ≥ 5. Consider K1,4, which is a spanning subgraph of F5. Since, λ(K1,4) = 5, we see that λ(K1,4) ≤ λ(F5), i.e. λ(F5) ≥ 5. Thus, λ(F5) = 5. The above given coloring is the only coloring possible with λ(F5) = 5. We see that there is one hole, namely 4. Therefore, we have Hλ(F5) = 1. The coloring f defined for wheel itself is an irreducible span coloring for Fan graphs for n ≥ 6. Hence, λ(Fn) = n and Hλ(Fn) = 1 for n ≥ 5.
Figure 3.6: Irreducible span coloring for F6
Figure 3.7: Irreducible span coloring for F7
Theorem 3.1.10. Let Br,s be a bistar on r + s + 2 vertices (r, s ≥ 1). Then,
0 if r = s
Hλ(Br,s) =
1 if r = s
Proof. Let {u0, u1, u2, …, ur} S{v0, v1, v2, …, vs} be the vertex set of Br,s where u0 and v0 are
the central vertices of Br,s. The diameter of a bistar is 3. So, the colors could be repeated. We define an L(2, 1)-coloring of Br,s as follows:
Case 1. r = s f(u0) = 0
f(ui) = i + 1 1 ≤ i ≤ r
f(v0) = ∆ + 1
f(vj) = j 1 ≤ j ≤ s
Case 2. r = s
Without loss of generality, assume that r < s.
f(u0) = ∆ − 1
f(ui) = i − 1 1 ≤ i ≤ r f(v0) = ∆ + 1
f(vj) = j − 1 1 ≤ j ≤ s
From Cases 1 and 2, λ(Br,s) ≤ ∆ + 1. Since Br,s is a tree with ∆ ≥ 1, by Theorem 2.0.3,
∆ + 1 ≤ λ(Br,s) ≤ ∆ + 2. Hence, λ(Br,s) = ∆ + 1.
The coloring f defined above is an irreducible span coloring of Br,s. When r = s, all the colors in {0, 1, 2, …, ∆ + 1} are used and hence there are no holes. Thus, Hλ(Br,s) ≥ 0. Since λ(Br,s) = ∆ + 1, by Theorem 2.0.6, u0 of degree ∆ must be labeled 0 and its neighbors must be labeled 2 + i, i = 0, 1, …, ∆ − 1. Then, v0 of degree ∆ is labeled ∆ + 1 and its neighbors must be labeled i, i = 0, 1, …, ∆ − 1. Thus, all the colors in {0, 1, 2, …, ∆ + 1} are used and hence there are no holes. Thus, Hλ(Br,s) = 0 when r = s.
When r = s, there is one hole namely ∆. That is, Hλ(Br,s) ≥ 1. To prove the reverse inequality, by Theorem 2.3.2(i), λ(Br,s) = ∆ + 1 ⇒ Hλ(Br,s) ≤ 1. Thus, Hλ(Br,s) = 1 when r = s.
Figure 3.8: Irreducible span coloring for B3,3 and B3,5
Remark 3.1.11. The diameter of a bistar is three. So, the vertices at distance three can be assigned same colors since no conditions are put upon for coloring vertices at distance three or more.
Theorem 3.1.12. Let GP (n, 1) be a Prism graph (n ≥ 3) on 2n vertices. Then,
(i) Hλ(GP (n, 1)) =
3 if n = 4, 8
1 if n = 5
(ii) 0 ≤ Hλ(GP (n, 1)) ≤ 2 if n ≡ 0(mod 3)
(iii) 0 ≤ Hλ(GP (n, 1)) ≤ 3 if n 6≡ 0(mod 3) where n = 4, 5, 8
Proof. A prism graph GP (n, 1) is the Cartesian product of cycle Cn and path P2. By Theorem