λ-harmonious Graph Colouring Lauren DeDieu

Transcription

1 λ-haronious Graph Colouring Lauren DeDieu June 12, 2012

2 ABSTRACT In 198, Hopcroft and Krishnaoorthy defined a new type of graph colouring called haronious colouring. Haronious colouring is a proper vertex colouring such that no two edges share the sae colour pair. The least nuber of colours needed to haroniously colour a graph is called the haronious chroatic nuber. We will exaine the results found for the haronious chroatic nuber of paths, cycles, and trees. We will also extend the definition of haronious colouring and define λ-haronious colouring, which allows each edge colour pair to occur up to λ ties. We will explore λ-haronious colouring and will prove soe results for the λ-haronious chroatic nuber of coplete graphs, coplete bipartite graphs, paths, cycles, and wheels. 2

3 CONTENTS 1. Introduction Terinology Haronious Graph Colouring λ-haronious Graph Colouring Conclusion

4 LIST OF FIGURES.1 K 6 {v, w} has an eulerian path of length An eulerian subgraph of K 5 with 7 edges Eulerian subgraphs of K 6 with 9 and 8 edges An eulerian subgraph of K 6 with 11 edges An eulerian subgraph of K 7 with 1 edges Coplete bipartite graph, K,n, with n h 4 (K 8, ) = 5, h 4 (K 8,6 ) = H 2, h λ (K n 1,1 ) = q An eulerian subgraph of H λ,q h 4 (K 4,1 ) = 2, but h 4 (W 5 ) =

5 1. INTRODUCTION Graph colouring originated in the id-nineteenth century, when atheaticians began to as the question, Can a geographical ap always be coloured using 4 colours or less? Although this proble too alost a century to solve, the iniu nuber of colours needed to propely colour different types of graphs has been studied extensively in the literature [1]. The two ain types of graph colouring are vertex colouring and edge colouring. In this paper, we will explore vertex colouring. Vertex colouring is an assignent of colours to the vertices of a graph such that each vertex receives exactly one colour. A colouring is called proper if no two adjacent vertices share the sae colour. In 198, Hopcroft and Krishnaoorthy defined a new type of vertex colouring called haronious colouring []. Haronious colouring is a special type of graph colouring in which each edge is assigned a distinct colour pair, i.e. if one edge has the colours red and blue on its incident vertices, then no other edge can also have the colour pair {red,blue}. In this report, we will discuss several well-nown haronious colouring results for failies of siple graphs, and will extend these results to allow for each colour pair to occur up to λ ties. In Section, we will discuss haronious colouring. Here we will state the theores for paths, cycles, and trees, and will outline the proofs for paths and cycles. In Section 4, we will extend the haronious colouring results and discuss λ-haronious colouring, which allows each edge colour pair to occur up to λ ties. Here we will give the λ-haronious chroatic nuber for coplete graphs, coplete bipartite graphs, paths, cycles, and wheels. 5

6 2. TERMINOLOGY For additional terinology, see [2]. We will denote path - an alternating sequence of distinct vertices and edges that begins and ends with a vertex. For exaple, P : v 1 v 2 v would be a path of length 2, denoted P. cycle - a closed path where the first and last vertex are the sae. For exaple H = v 1 v 2 v v 1 would be a cycle of length, denoted C (also nown as a -cycle or a triangle). tree - a connected graph without cycles. V (G) - the nuber of vertices in a graph G. E(G) - the nuber of edges in a graph G. degree - the degree of a vertex v, or deg(v), refers to the nuber edges incident to v. vertex colouring - an assignent of colours to the vertices of a graph, such that each vertex receives exactly one colour. proper colouring - a vertex colouring such that no two adjacent vertices share the sae colour. edge colour pair - the pair of colours assigned to an edge s incident vertices. haronious colouring - a proper colouring such that no two edges share the sae colour pair. For exaple, if an edge s vertices are coloured red and blue, then there is no other edge with the colour pair {red, blue}. h(g) - the least nuber of colours needed to haroniously colour a graph G; called the haronious chroatic nuber. λ-haronious colouring - a proper colouring such that no λ+1 edges share the sae colour pair. Note that the traditional haronious colouring defined above is a 1-haronious colouring. h λ (G) - the least nuber of colours needed to λ-haroniously colour a graph G; called the λ- haronious chroatic nuber. 6

7 coplete graph - a graph in which each pair of vertices is connected by an edge; we denote a coplete graph on n vertices by K n. eulerian path - a trail in a graph that visits each edge exactly once. eulerian cycle - a closed trail in a graph that visits each edge exactly once. coplete bipartite graph - a graph whose vertex set can be decoposed into two disjoint sets such that no two vertices in the sae set are adjacent, and every pair of vertices in distinct sets are adjacent; we will denote a coplete bipartite graph by K,n where and n are the sizes of the disjoint sets, with n. star - a coplete bipartite graph such that n = 1. wheel - a graph on n vertices fored by connecting a single vertex to all vertices of a C n 1 ; we denote a wheel on n vertices as W n. 7

8 . HARMONIOUS GRAPH COLOURING The haronious chroatic nuber, h(g), is the iniu nuber of colours needed to properly colour a graph G such that no two edges share the sae colour pair. This paraeter was originally introduced by Hopcroft and Krishnaoorthy in 198 [], and since then, h(g) has been found for several different failies of graphs. We can gain a lower bound for h(g) by considering the requireent that each edge receives a unique colour pair. The binoial coefficient ( tells us how any ways colours can be arranged into pairs of 2. Therefore, we now that ( E(G). Let be the sallest integer such that this inequality holds. This gives us a lower bound for the haronious chroatic nuber: h(g). Note that our definition of the haronious chroatic nuber differs slightly fro how it was first introduced. Hopcroft and Krishnaoorthy s definition of h(g) did not require that G was properly coloured. The ajority of the literature, however, has since added the additional condition that G is properly coloured, so we will use this variation of h(g) as well. We will now briefly state the results found for the haronious chroatic nuber of paths, trees, and cycles. We will also outline the proofs for P n and C n, as Section 4 will use these proof techniques to find the λ-haronious chroatic nuber of paths and cycles. Theore 1: [6] The haronious chroatic nuber of a path with n vertices is as follows: Let r Z be deterined by the inequality ( ) ( 2r 1 2 < n 1 2r+1 ) 2. Then Proof: { ( 2r if n 1 2r ) 2 (r 1), h(p n ) = 2r + 1 otherwise. We can deterine h(p n ) by considering sallest such that the coplete graph on vertices, K, has a subgraph with an eulerian path of length n 1. An eulerian path of length n 1 corresponds to a haronious colouring of P n, since if we colour the vertices of K with different colours, this eans that our eulerian path of length n 1 is properly coloured such that each edge has a distinct colour pair. Since we are considering the sallest such that this is true, we now that the haronious chroatic nuber of P n will be. We now that a graph has an eulerian path if and only if it is connected and has at ost two vertices of odd degree [2]. Also note that coplete graphs K have ( 2 ) edges, so when we consider 8

9 coplete graphs, we now that our criterion ( E(G) is satisfied for =. If is odd, then all of the vertices of K have even degree, so n 1 ( 2 ), i.e. the length of P ust be less than or equal to the nuber of edges in K. If is even, then all vertices of K have odd degree, so we ust delete at least 2 1 edges to ensure we have at ost 2 vertices of odd degree. Therefore, if is even, then we ust have n 1 ( ) 2 ( 2 1). The theore then follows, letting r = 2. For exaple, in Figure.1 we can see that K 6 has 15 edges, and each vertex has odd degree. If we reove 2 edges, v and w, we are left with only 2 vertices of odd degree. Therefore, we can trace an eulerian path , and can see that h(p 11 ) = h(p 12 ) = h(p 1 ) = h(p 14 ) = v w Fig..1: K 6 {v, w} has an eulerian path of length 1. The haronious chroatic nuber of a path can also be stated as follows [7]: Let be the least integer such that n 1 (. Then, if odd, h(p n ) = or, if even and n 1 = ( ) 2 i, for i { 2 1, 2,..., 2}, + 1 otherwise. Since the haronious chroatic nuber of a path is either or + 1, one ight expect that the haronious chroatic nuber of a tree would also be close to. However, this is not the case. John Mitche [7] proved that the haronious chroatic nuber of a tree T with n vertices can fall anywhere between and n. We will state the theore forally: Theore 2: [7] Let be the least integer such that n 1 (. Then for each t such that t n, there is a tree T with n vertices such that h(t ) = t. Theore : [4] [5] The haronious chroatic nuber of a cycle with n vertices is as follows: Let be the least integer such that n (. Then, 9

10 h(c n ) = + 1 if odd and n ( i, for i {1, 2}, or, if even and n ( ) 2 i, for i {0, 1,..., 2 1}, otherwise. Proof: Let be the least integer such that n (. Then, as with paths, we can deterine the haronious chroatic nuber of a cycle, C n, by considering the sallest such that the coplete graph on vertices, K, has a subgraph with an eulerian cycle of length n. We now that a graph has an eulerian cycle if and only if it s connected and each of its vertices has even degree [2]. If is odd, then each vertex in K has even degree. Therefore, in order to create a subgraph J of K with an eulerian cycle of length n (i.e. a subgraph such that each vertex has even degree), we need to either delete zero edges, or delete the edges fro a cycle in K. Reoving the edge fro a cycle will eep the degree of each vertex even, since vertices in a cycle all have degree 2. We are able to do this unless n = ( i for i = 1 or 2, since no cycles of length 1 or 2 exist. So, if is odd and n ( i, for i {1, 2}, then h(cn ) =. For exaple, if n = 7, then = 5. So, if we delete the edges fro a cycle of length in K 5, then we are left with a subgraph of K 5 with an eulerian cycle of length 7 (see Figure.. Therefore, h(c 7 ) = = 5. K K 5 C Fig..2: An eulerian subgraph of K 5 with 7 edges. If n = ( i for i = 1 or 2, then we now that K has no eulerian subgraph of length n. However, we will show that such a subgraph does exist in K +1. Let s call this subgraph L. If i = 1, then we can create L by tracing a path of length 4 in K, deleting the edges fro P 4, and joining the end vertices of P 4 to a new vertex v not in K. This new graph L has ( 1 edges, +1 vertices, and all edges have even degree. Therefore, L is a subgraph of K +1 and has a eulerian cycle of length ( ) 2 1. So, h(cn ) = + 1 when n = ( 1. Siilarly, if i = 2, then we can create L by deleting the edges fro a P 5 in K, and joining the end vertices of this path to a new vertex v. Therefore, we also have h(c n ) = + 1 for n = ( 2. 10

11 For exaple, in Figure. we can see that if n = 9, then we can delete the edges fro a P 4 in K 5 and join the end vertices of this path to v. If n = 8, we delete the edges fro a P 5 in K 5, and join the end vertices to v. 6 v 6 C 9 v 1 1 C Fig..: Eulerian subgraphs of K 6 with 9 and 8 edges. When is even, we now that all vertices in K have odd degree. Therefore, we ust delete at least 2 edges fro K in order to create a subgraph of K such that all vertices have even degree. So, let s consider n = ( ) 2 i for i = 2,..., 2. Note that we only consider i up to 2, because ( ) ( 2 ( 1) = 1 ) 2, and so if i = 1, then would not be the iniu integer that satisfies n ( ) 2. Let i = 2 + t. We can create an eulerian subgraph of K with ( i edges as follows. If t = 0, then reove 2 non-adjacent vertices fro K. This will ae the degree all vertices even. If t 0, then reove fro K the edges of a star K 2t+1,1 as well as 2 t 1 edges that are not adjacent to each other, or to any of the edges of K 2t+1,1. Doing so will reove 2t+1+ 2 t 1 = 2 +t = i edges fro K, and will ae the degree all vertices even. Therefore, h(c n ) = for n = ( ) 2 i when i = 2,..., 2, since there exists an eulerian subgraph of K of size n. For exaple, when n = 11, we have = 6, i = 4 and t = 1. Therefore, we can create our eulerian subgraph by deleting the edges fro a K,1 in K 6 and deleting one other edge which is not adjacent to any of the edges of K,1 (see Figure.4). When i < 2, it s not possible to create an eulerian subgraph of K, but we can create an eulerian subgraph of K +1. Since is even, we now that + 1 is odd, so all of K +1 s vertices have even degree. Let t = ( ) +1 2 n. If t + 1, then we can create an eulerian subgraph of K+1, by deleting the edges fro a cycle of length t in K +1. If t > + 1, then we will not be able to trace a cycle of length t in K +1 without repeating an edge. Therefore, we will instead create our eulerian subgraph of size n by deleting t edges fro two edge disjoint cycles in K +1, such that the two cycles use all + 1 vertices. Therefore, h(c n ) = + 1 for n = ( ) 2 i when i = 0,...,

12 Fig..4: An eulerian subgraph of K 6 with 11 edges. For exaple, when n = 1, we have = 6 and t = 8. So, we can create our eulerian subgraph of K 7 by deleting the edges fro two C 4 s in K 7 (see Figure.5) Fig..5: An eulerian subgraph of K 7 with 1 edges. 12

13 4. λ-harmonious GRAPH COLOURING In the previous chapter we explored the haronious chroatic nuber: the least nuber of colours needed to properly colour a graph such that no two edges share the sae colour pair. We will now extend these results to allow for a proper colouring such that no λ + 1 edges ay share the sae colour pair. We call this colouring λ-haronious colouring. The least nuber of colours needed to λ-haroniously colour a graph G is called the λ-haronious chroatic nuber, and will be denoted by h λ (G). As with haronious colouring, we can gain a lower bound for the λ-haronious chroatic nuber by considering the requireent that no λ + 1 edges ay share the sae colour pair. The binoial coefficient ( tells us how any ways colours can be arranged into pairs of 2. Therefore, since we are only allowed to use each colour pair at ost λ ties, we ust have E(G) λ (. Let be the sallest integer such that this inequality holds. Then, h λ (G). We will now give our results for the λ-haronious chroatic nuber of coplete graphs, coplete bipartite graphs, paths, cycles, and wheels. Theore 4: The λ-haronious chroatic nuber of a coplete graph on n vertices equals n, i.e. h λ (K n ) = n. Proof: By definition, a λ-haronious colouring ust be a proper colouring. Since each vertex in K n is adjacent to all n 1 reaining vertices, we need n colours in order to properly colour K n. Therefore, h λ (K n ) = n. Theore 5: The λ-haronious chroatic nuber of a coplete bipartite graph K,n with n is: Proof: h λ (K,n ) = in λ + q n ; 1 q λ q, q Z. The proof is quite straightforward for λ = 1, 2, and, so we will explore these cases before proving the theore in general. We will denote the disjoint vertex set of size in K,n by A, and will denote the the disjoint vertex set of size n by B (see Figure 4.1). 1

14 v A w n B Fig. 4.1: Coplete bipartite graph, K,n, with n. For λ = 1 (i.e. regular haronious colouring), we now that no two edges can share the sae colour pair. If we give a vertex v in A the colour 1, then all vertices in B ust receive distinct colours, since v is adjacent to each vertex in B. Let s suppose w is a vertex in B that received the colour 2. Since w is adjacent to every vertex in A, we ust give the vertices in A distinct colours as well. Therefore, h 1 (K,n ) = + n. For λ = 2, we are allowed to use each colour pair twice. Therefore, the ost efficient way to colour K,n would be to group the vertices of A into groups of 2, giving each group a distinct colour. This forces us to use n distinct colours in B. Since n, it will always be better to repeat colours in A, rather than B. Therefore, h 2 (K,n ) = 2 + n. Siilarly, h (K,n ) = + n. When λ = 4, things get a little ore interesting. Depending on the graph, it ay be better to use 4 colours in A, and use n colours in B, or soeties it ay be better to use 2 colours in A, and n 2 colours in B. For exaple, in Figure 4.2, if we loo at K 8, we see that 4 + n = 2 + = = = 6, but for K 8,6 we have 4 + n = = = 4 + = Fig. 4.2: h 4 (K 8, ) = 5, h 4 (K 8,6 ) = 7. 14

15 Therefore, h 4 (K,n ) = { in ( 4 + n, 2 + n )} 2. Siilarly, when λ = 9, depending on the graph, it ay be best to use 9 + n colours, 4 + n 2 colours, or + n colours, as in K9,1, K 4,2, and K,, respectively. In general, for each λ there will be several different colouring options for K,n, and the ost appropriate colouring choice will depend on the size of and n. Each colouring choice will allow colours to repeat t ties in A, and s ties in B. Since n, we will always have t s. For exaple, for λ = 9, we could choose to use each colour twice in A (t =, and 4 ties in B (s = 4), but choosing t = 4, s = 2 would always be better, since A is never saller than B. We also now that each colour pair can only be used up to λ ties, so t s λ. Therefore, our choice of s always depends on our choice of t, and so, we ay choose {t = λ, s = 1}, {t = λ 2, s = λ λ 2},..., or {t = λ q, s = q}, for 1 q, q Z. We do not need to consider q >, because { ( )} this would give us s > t. Therefore, h λ (K,n ) = in λ + n λ q ; 1 q, q Z. q λ Note that for graphs where A is significantly larger than B, such as stars, we have h λ (K,n ) = + n, i.e. q = 1. However, when the sizes of A and B are closer together, the value of q becoes less clear. Theore 6: The λ-haronious chroatic nuber of a path, P n, is as follows: Let r Z be deterined by the inequality λ ( ) ( 2r 1 2 < n 1 λ 2r+1 ) 2. Then 2r if λ is even and n 1 λ ( ) 2r 2, h λ (P n ) = or, if λ is odd and n 1 λ ( ) 2r 2 (r 1), 2r + 1 otherwise. Proof: We gave the proof for the case where λ = 1 in Section. The proof for a general λ is quite siiliar, but we ust tae a few ore factors into account. To deterine h 1 (P n ), we considered sallest such that the coplete graph on vertices, K, had a subgraph with an eulerian path of length n 1. However, for a general λ, we are allowed to use each edge pair λ ties. Therefore, instead of considering K, we will instead consider a graph on vertices in which each pair of vertices is connected by λ edges. Let s call this new graph H λ,. For exaple, when λ = 2, H 2,5 would be K 5 with double edges (see Figure 4.). In order to deterine h λ (P n ), we will consider the sallest such that H λ, has a subgraph with an eulerian path of length n 1. When λ is even, the degree of each vertex in H λ, will be even, since the degree of each vertex in H λ, is a ultiple of λ. We now that K has ( ) 2 edges, so Hλ, will have λ ( edges. Therefore, 15

16 Fig. 4.: H 2,5. when λ is even, our will be the sallest such that n 1 λ (, i.e. the length of Pn ust be less than or equal to the nuber of edges in H λ,. When λ is odd and is odd, the degree of each vertex in H λ, will also be even, since the degree of each vertex in H λ, is a ultiple of 1. Therefore, in this case, will also be the sallest such that n 1 λ (. When λ is odd and is even, the degree of each vertex in H λ, will be odd. Therefore, in order to create an eulerian path, we ust delete at least 2 1 edges, in order to be left with at ost two vertices of odd degree. Therefore, our will be the sallest such that n 1 λ ( ) 2 ( 2 1). The theore follows, with r = 2. Theore 7: The λ-haronious chroatic nuber of a cycle, C n, is as follows: Let be the least integer such that n λ (. Then h λ (C n ) = + 1 if one of the following four conditions hold: i) λ is even and n λ ( 1, ii) λ 1, λ is odd, is odd, and n λ ( 1, ii) λ = 1, is odd, and n λ ( i for i = 1, 2, iv) λ is odd, is even, and n λ ( ) 2 i for i = , otherwise., Proof: We will deterine the λ-haronious chroatic nuber for C n by extending the proof for λ = 1 given by Lee and Mitche [4], which we outlined in Section. We will do this by considering H λ,, which is defined the sae as in Theore 6. Let be the sallest integer such that H λ, has a subgraph with an eulerian cycle of length n. When λ is even, or when λ is odd and is odd, then each vertex in H λ, will have even degree. 16

17 We now that a connected graph G has an eulerian cycle if and only if each vertex in G has even degree. Therefore, in these two cases, we can create a subgraph of H λ, with an eulerian cycle of length n by either deleting zero edges, or by deleting the edges fro a cycle in H λ,. If n = λ ( 1 we won t be able to do this, since no cycles of length 1 exist. If n = λ ( 2, we can delete a cycle of length 2 in our graph if λ 1 (i.e. H λ, always has ultiple edges between vertices when λ > 1). Therefore, in these two cases, if λ 1, then h λ (C n ) =, where is the least integer such that n λ (, i.e. the length of Cn ust be less than or equal to the nuber of edges in H λ,. For the case where λ = 1 and is odd, see Theore. If n = λ ( 1, we cannot create this eulerian subgraph, but we can create it for Hλ,+1. As in Theore, we can create an eulerian subgraph of H λ,+1 of length n by deleting the edges fro a P 4 in H λ, that uses four distinct vertices, and then joining the end vertices of P 4 to a new edge v not in H λ,. Therefore, in these two cases, if n = λ ( 1, then we will have hλ (C n ) = + 1. When λ is odd and is even, then all vertices in H λ, will have odd degree. Therefore, if ) i for i = 0... n λ ( 2 2 1, then we will have h λ(c n ) =, since we can create a subgraph of H λ, with an eulerian cycle of length n in the sae way as the even case of Theore. If ) i for i = 0... n = λ ( 2 2 1, then we will have h λ(c n ) = + 1, since we can t create a subgraph of H λ, with an eulerian cycle of length n, but, we can create it for H λ,+1 as in the even case of Theore. Theore 8: Let q = h λ (K n 1,1 ). If q > 4, then h λ (W n ) = q. Proof: A wheel on n vertices consists of a single vertex, called the hub, and a cycle of length n 1, such that the hub is adjacent to all vertices of C n 1. Therefore, each wheel contains a star K n 1,1 and a cycle C n 1. Let h λ (K n 1,1 ) = q. Since W n contains K n 1,1 as a subgraph, we now that h λ (W n ) q. When q > 4 we will show that, in fact, h λ (W n ) = q. We will do this by considering H λ,q 1, which is defined the sae as in Theore 6. We will show that H λ,q 1 has a subgraph with an eulerian cycle of length n 1 such that each vertex in the cycle is only used at ost λ ties. Therefore, the C n 1 of the wheel can be haroniously coloured with q 1 colours, and this leaves one colour left over for the hub, which will be adjacent to each colour at ost λ ties. Hence, h λ (W n ) = q. Fro Theore 5, we now that h λ (K n 1,1 ) = n 1 λ + 1. Since we require q colours to colour our star, this eans that the n 1 end vertices of K n 1,1 (i.e. the vertices of degree one) ust partition into q 2 groups of size λ and one group of size less than or equal to λ, λ i (see Figure 4.4). Now, we can create a subgraph of H λ,q 1 which corresponds to the partitions of the n 1 end 17

18 λ vertices in each λ i vertices 1 2 q 1 q Fig. 4.4: h λ (K n 1,1 ) = q. vertices of our star, i.e. q 2 vertices of H λ,q 1 will have degree 2λ and one vertex will have degree 2(λ i). Such a subgraph will have an eulerian cycle, since each vertex will have even degree. We can create this subgraph in the following way. Tae q 1 vertices and give the all distinct colours {1, 2,..., q 1}. Now, put λ edges between the vertices coloured 1 and 2, λ edges between the vertices coloured 2 and, and so on until you put λ edges between the vertices coloured q and q 2. Now, put λ i edges between the vertices coloured q 2 and q 1, λ i edges between the vertices coloured q 1 and 1, and i edges between the vertices coloured 1 and q 2. We are left with a subgraph of H λ,q 1 with n 1 edges such that q 2 vertices have degree 2λ, and one has degree 2(λ i) (see Figure 4.5). λ i edges q 1 7 q i edges λ edges 2 4 Fig. 4.5: An eulerian subgraph of H λ,q 1. Therefore, we can colour our wheel W n by using the colouring of this eulerian cycle to colour 18

19 C n 1 and using a distinct colour for the hub. Our colouring ensured that each colour on our cycle was only used at ost λ ties, so our q-colouring is λ-haronious, and it is the best we can possibly do, since h λ (K n 1,1 ) = q. Therefore, h λ (W n ) = q. Note that when q = 4, q =, or q = 2 we ay not have h λ (W n ) = q. For exaple, h 4 (K 4,1 ) = 2, but we cannot properly colour W 5 with 2 colours (see Figure 4.6). Also, h (K 8,1 ) = 4, but Theore 7 tells us that h (C 8 ) = 4, so we need at least 5 colours to colour W 9. Siilarly, h 2 (K 5,1 ) = 4, but h 2 (C 5 ) = 4, so we need at least 5 colours to colour W Fig. 4.6: h 4 (K 4,1 ) = 2, but h 4 (W 5 ) =. 19

20 5. CONCLUSION Although the haronious chroatic nuber has been found for several failies of graphs, deterining h(g) is, in general, NP-coplete. In the appendix of [], David S. Johnson shows that the independent set proble can be transfored in polynoial tie into the haronious colouring proble. Since we now that the independent set proble is NP-coplete, this tells us that the haronious colouring proble is also NP-coplete. We can forally state the λ-haronious colouring proble as follows: Given a graph G and a positive integer V (G), can G be λ-haroniously coloured with colours? In other words, does there exist a proper vertex colouring of G with colours such that each edge colour pair occurs at ost λ ties? We suspect that the λ-haronious colouring proble is also NP-coplete, but we haven t been able to prove it yet. 20

21 BIBLIOGRAPHY [1] G. Agnarsson, R. Greenlaw, Graph Theory: Modeling, Applications, and Algoriths, Pearson Education Inc., [2] J.A. Bondy, U.S.R. Murty, Graph Theory, Springer, [] J.E. Hopcroft, M.S. Krishnaoorthy, On the haronious coloring of graphs, SIAM. J. Algebraic and Discrete Methods 4 (198), [4] S. Lee, J. Mitche, An upper bound for the haronious chroatic nuber of a graph, J. Graph Theory 11 (1987), [5] Y. Ielan, On the (upper) line-distinguishing and (upper) haronious chroatic nuber of a graph, Masters Thesis, University of Johannesburg, [6] Z. Miller, D. Pritiin, The haronious coloring nuber of a graph, Discrete Matheatics 9 (1991), [7] J. Mitche, On the haronious chroatic nuber of a graph, Discrete Matheatics 74 (1989),