Backbone Colouring of Graphs

Transcription

1 Backbone Colouring of Graphs Arnoosh Golestanian Mathematics A thesis submitted in partial fulfilment of the requirements for the degree of Master of Science in Mathematics Facult of Mathematics and Science, Brock Universit St. Catharines, Ontario c 05

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3 Abstract Consider an undirected graph G and a subgraph of G, H. A q-backbone k-colouring of (G, H) is a mapping f : V (G) {,,..., k} such that G is properl coloured and for each edge of H, the colours of its endpoints differ b at least q. The minimum number k for which there is a backbone k-colouring of (G, H) is the backbone chromatic number, BBC q (G, H). It has been proved that the backbone k-colouring of (G, T ) is at most 4 if G is a connected C 4 -free planar graph or non-bipartite C 5 -free planar graph or C j -free, j {6, 7, 8} planar graph without adjacent triangles. In this thesis we improve the results mentioned above and prove that an connected planar graph without adjacent triangles has a -backbone k-colouring with at most 4 colours b using a discharging method. In the second part of this thesis we further improve these results b proving that for an graph G with χ(g) 4, BBC (G, T ) = χ(g). In fact, we prove the stronger result that a backbone tree T in G eists, such that uv T, f(u) f(v) = or f(u) f(v) k, k = χ(g). For the case that G is a planar graph, according to Four-Colour theorem, χ(g) = 4; so, BBC (G, T ) = 4. i

4 Acknowledgement Foremost, I would like to epress m sincere gratitude to m supervisors Dr. Babak Farad and Dr. Henrk Fukś for their continuous support of m Master of Science stud and research. I would like to thank them for their patience, motivation and for their brilliant guidance and assistance. M special thanks to Mrs. Susanna Ferreri for her great support. Last but not least, I would like to thank m famil: m parents and m brother, for supporting me spirituall throughout m life. ii

5 Contents Abstract i Introduction. Overview Preliminaries and Notions 3. Preliminaries Related Works Backbone Colouring 3. Upper bound on the backbone chromatic number of graphs Properties of a minimum countereample Properties of bad [4, 4, 4]-triangle Properties of a bad [5, 4, 4]-triangle The discharging method Proof of Theorem

6 4 Algorithm for Finding a Backbone Eistence of the backbone of ever graph G with χ(g) Eample of finding a backbone of graph Conclusion 4 Bibliograph 43 Table of Figures 46 iv

7 Chapter Introduction Graph Theor is the stud of graphs, which are discrete structures consisting of nodes and links between nodes. More specificall, nodes of a graph, called vertices, are connected b edges. One application of graph theor is the stud of social networks where vertices represent the users of the network and edges represent the connections between users. Also, a tpical application of graph colouring is the scheduling problem. In this problem we want to assign a set of jobs to time slots in which jobs with the same resources do not have conflict with each other.. Overview The purpose of the first section of Chapter is to familiarie the reader with commonl used notations in the stud of graph theor or specificall in this thesis. While the second section of Chapter is a summar of backbone colouring results

8 . Overview that have been previousl proved b earlier authors. In Chapter 3, we prove that ever connected planar graph without adjacent triangles has a backbone for which the backbone chromatic number is at most 4 b using discharging rules. In Chapter 4, we etend these results and prove that ever graph G with χ(g) 4 has a backbone for which the backbone chromatic number is equal to χ(g). Chapter 5 contains the concluding remarks and the discussions about possible future directions.

9 Chapter Preliminaries and Notions In this chapter we introduce graph theoretical definitions and terminolog that are used in this thesis. For undefined definitions and notations, we use [] as our reference.. Preliminaries Graph G consists of the verte set V (G) and the edge set E(G). Each edge has two endpoints which are adjacent to each other. Verte v is incident to edge e if v is an endpoint of e. The order of G is the number of vertices, V (G) and the sie of G is the number of edges, E(G). The graph obtained from G b deleting the edge e, e E(G) or the set of edges S are denoted b G e and G S, respectivel. Similarl, the graph obtained from G b deleting the verte v, v V (G) or the set of vertices S and all incident edges are denoted b G v and G S, respectivel. 3

10 . Preliminaries The degree of a verte v is the number of vertices adjacent to v and it is denoted b d(v); a k-verte is a verte of degree k. A leaf is a verte of graph G with degree. A k + -verte and k -verte is a verte of degree at least k and at most k, respectivel. Let N G(A) denotes the set of neighbours of vertices in A in graph G. Since each edge has two endpoints, we can sa that the sum of all degrees is eactl twice the number of edges, v V (G) d(v) = E(G). If all vertices of graph G have degree k, then G is said to be a k-regular. A connected -regular graph is called a ccle. A ccle graph of n vertices is denoted b C n. A complete graph is a graph whose vertices are all pairwise adjacent. The complete graph with n vertices is denoted b K n. Triangle is a common name for K 3. The minimum degree of all vertices of graph G is denoted b δ(g) and the maimum degree is denoted b (G). A path in a graph is a finite sequence of distinct edges which connect a sequence of vertices. A graph is connected if there is at least one path between an two vertices of graph; otherwise, it is a disconnected graph. A tree T is a connected graph in which there is eactl one path between an two vertices of it. A component is a largest connected subgraph of a graph. In other words, if G is disconnected, it has more than one component. The number of components of the graph G is denoted b comp(g). A cut verte or cut edge is a verte or edge of graph G, respectivel, whose deletion increases the number of components of G. 4

11 . Preliminaries Cut edges are commonl called bridges. A graph H(V, E) is a subgraph of G(V, E) if V (H) V (G) and E(H) E(G). A spanning subgraph of G is a subgraph that includes all the vertices of G. A spanning tree is a spanning subgraph that is a tree. A graph is called planar if it can be drawn in the plane in such a wa that no two edges intersect. An embedding of a planar graph divides the plane into a number of connected regions, called faces which are bounded b the edges of the graph. Verte v is incident to face f if v is the endpoint of an edge in the boundar of f. A k-face is a face f with k edges in the boundar. A k + -face and k -face is a face with at least k edges in its boundar and at most k edges in its boundar, respectivel. A k-colouring of graph G is a mapping f : V {,,..., k}. If f(u) f(v) for an two adjacent vertices u and v, f is a proper colouring. The smallest integer k such that G has a proper k-colouring is called a chromatic number of G and is denoted b χ(g). For a graph G and its subgraph H, a q-backbone k-colouring of (G, H) is a mapping f : V (G) {,,..., k} such that: q if uv E(H) f(u) f(v) if uv E(G) \ E(H) The minimum number k for which there is a q-backbone k-colouring of (G, H) is the backbone chromatic number of (G, H), denoted b BBC q (G, H). Let u and v be two endpoints of an edge e E(G) and f be a colouring of G. If f(u) f(v) 5

12 . Preliminaries 3 4 Figure.: Eample of backbone colouring. q, then the edge e is called an f-edge. For the case that H is spanning tree T we called T the backbone of G. A circular q-backbone k-colouring of (G, T ) is a special case of q-backbone k-colouring of (G, H) which is a proper colouring of vertices with k colours such that q f(u) f(v) k q, for each edge uv E(H). The circular q-backbone k-colouring chromatic number of (G, T ), denoted b CBC q (G, T ), is the minimum k such that for an spanning trees of G there is a circular q-backbone k-colouring. We can sa that if f is a circular q-backbone k-colouring of (G, T ), then f is also a q-backbone k-colouring of (G, T ). For the case that q = we simpl sa backbone k-colouring and use the notation BBC(G, T ). Similarl when q = we sa circular backbone k-colouring and use the notation CBC(G, T ). In Figure. edges of graph G are shown b black colour and edges of spanning tree T are shown b red colour. Graph G has a backbone k-colouring with 4 colours. Similarl, G has a circular k-colouring with 5 colours. 6

13 . Preliminaries u v 4 w p Figure.: Eample of (4,)-Kempe chain Breadth First Search is one of the simplest algorithms for searching a graph G with a distinguished source verte s. This algorithm eplores the neighbours of s first, before moving to the net level neighbours b computing a shortest path from s to vertices reachable from s. Consider a proper colouring of graph G with colour set S containing at least two distinct colours a and b. If v is a verte of graph G with colour a then the (a,b)- Kempe chain of G containing v is the maimal connected subset of vertices of G, which contains v and whose vertices are all coloured either a or b. We can use the (a,b)-kempe chain for switching the colour of vertices in this subset without changing the colour of all vertices in G. B switching the colour of vertices in the subset we mean that for a verte u in (a,b)-kempe chain of v, f (u) = b if f(u) = a and vice versa. We can observe that b recolouring vertices in this Kempe chain we did not change the proper colouring because adjacent vertices to an verte of Kempe chain of V have other colouring rather than a or b; otherwise, the would be in the (a, b)-kempe chain of v. In Figure., the (4, )-Kempe chain of v is set {,, v, w, p}. If we switch the colour of vertices in the (4, )-Kempe chain of v, the new colour of vertices will be 7

14 . Related Works f () = f (v) = f (p) = and f () = f (w) = 4.. Related Works Backbone colouring was firstl motivated b the colouring problems related to frequenc assignment and was introduced b Boersma et al. [] in 003. In addition to studing backbone colouring for (G, H), the studied BBC(G, P ) where P is a Hamiltonian Path, and for the special case of split graphs. Their main result was that for an connected graph G and spanning tree T, BBC(G, T ) χ(g). This result together, with Four-Colour Theorem [], implies that if G is a planar graph, then BBC(G, T ) 7. However, the conjecture that for an planar graph G, BBC(G, T ) 6 and b providing an eample in which BBC(G, T ) = 6, the showed that this upper bound, if true, is tight. In 03, Campos et al. [3] proved this conjecture when T has diameter at most 4. In [], the authors proved that if G is a split graph, BBC(G, T ) = χ(g) + and if ω(g) 3 and G contains a Hamiltonian path, then for ever Hamiltonian path P in G, BBC(G, T ) = χ(g) +. The general case of q-backbone k-colouring when q, was first defined b Broersma et al. [4] in 009. The consider the cases that backbone is either a collection of pairwise disjoint stars or a matching. According to their result, the minimum k for a star backbone S of G can differ b a multiplicative factor of at most q from χ(g). In [9], Havet et al. showed that BBC q (G, H) 3q + b Four-Colour Theorem [] when G is planar and H is forest. The improve this upper bound b showing that 8

15 . Related Works BBC q (G, H) q + 6 and for the special case of q = 3, BBC 3 (G, H) 8. Another wa of considering backbone colouring is to show that a spanning tree T of G eists, such that BBC q (G, T ) k, instead of showing that this is true for an spanning tree T of G. In 0, Bu and Zhang [5] proved that if G is a connected C 4 -free planar graph, then there eists a spanning tree T of G such that BBC(G, T ) 4. In 00, Zhang and Bu [6] proved that if G is a connected nonbipartite C 5 -free planar graph, then there eists a spanning tree T of G such that BBC(G, T ) = 4. Also, in 0, Bu and Li [7] proved that if G is connected C 6 -free or C 7 -free planar graphs without adjacent triangles, then there is a spanning tree T of G such that BBC(G, T ) 4. In 0, Wang [8] showed that if G is a connected planar graph without C 8 and adjacent triangles, then there is a spanning tree T of G such that BBC (G, T ) 4. In 05, Bu and Bao [] showed that if G is a connected C 8 -free or C 9 -free planar graph without adjacent 4-ccles, then there is a spanning tree T of G such that BBC(G, T ) 4. The purpose of part one of m thesis is to etend these results b allowing graph G to have C j where j = 3,..., n. Also, in the second part of m thesis, we improve these results for an graph with χ(g) 4. In 04, Havet et al. [9] defined the concept of circular backbone colouring for the first time. The showed that even though CBC q (G, H) q.χ(g), this is not the best possible upper bound and the proved that CBC q (G, H) q + 4. Also, the conjectured that CBC q (G, H) q + 3. In [0], Araujo et al. proved that if G is a planar graph without C 4 or C 5, then for an forest H that H G, CBC(G, H) 7. Also, the improve this upper bound 9

16 . Related Works when H is a path forest. In other words, for the same class of graphs when the connected components of forest H are paths, CBC q (G, H) 6. However, b using discharging method, recentl we proved that if G is a connected C j -free planar graph, j 4, 5, 6, 7, then for ever spanning tree T of G, CBC(G, T ) 7. 0

17 Chapter 3 Backbone Colouring 3. Upper bound on the backbone chromatic number of graphs Observation. Suppose that f and f are two verte colourings of G and v V (G), f (v) + f(v) = k +. Then f is a backbone k-colouring of (G, T ), if and onl if, f is a backbone k-colouring of (G, T ). We call f the smmetric colouring of f and vice versa. Proof. Assume that f is a backbone k-colouring of (G, T ). According to the definition of backbone colouring: If uv E(T ), f(u) f(v), f (u) = k + f(u) and f (v) = k + f(v). Thus, f (u) f (v) = f(u) f(v). If uv E(G) E(T ), f(u) f(v), then f (u) f (v).

18 3. Properties of a minimum countereample 3 Theorem. If G is a connected planar graph without adjacent triangles, then there eists a spanning tree T of G such that BBC(G, T ) 4. We will use proof b contradiction in the following two parts. First we assume for contradiction that there eists a planar graph G without adjacent triangles such that for a spanning tree T of G, BBC(G, T ) 5. Then we determine some properties for the graph G. In the second part, we show that graph G does not eist using the discharging method. In other words, b assigning charges to each verte and to each face of G, we carefull redistribute those charges such that the total charge of G is nonnegative. However, b using Euler s formula we can show that initiall the total charge of G was negative. This obvious contradiction states that the assumption of the eistence of graph G was incorrect. 3. Properties of a minimum countereample Let G(V, E) be a countereample to Theorem with minimum V. So, G is a connected planar graph without adjacent triangles and for ever spanning tree T of G, BBC(G, T ) 5. Also, for ever G (V, E ) without adjacent triangles if V < V, then there is a spanning tree T such that BBC(G, T ) 4. We will use the color set S = {,, 3, 4}. Lemma. δ(g) 4. Proof. We will show that G does not contain a leaf, -verte and 3-verte. These three properties are demonstrated separatel below.

19 3. Properties of a minimum countereample 3. G has no leaf. Assume for a contradiction that G contains a leaf u adjacent to a verte. Let G = G u with V number of vertices and E number of edges. Since V = V, b the minimalit of G there is a spanning tree T of G such that (G, T ) has a backbone 4-colouring. For an colouring of we can find a proper colour for u. Without loss of generalit, assume that f() {, } (set {3, 4} is a smmetric colouring for {, }). In this case, colour u with 4; so, u is an f-edge. Let spanning tree T = T {u}. Hence there is a spanning tree T of G such that (G, T ) has a backbone 4-colouring; this is a contradiction.. G has no -verte. Suppose to the contrar that G contains a -verte u with two neighbours,. Let G = G u. Graph G remains connected if u is not a cut verte. Consider the cases of u being a cut verte and u not being a cut verte separatel. (a) u is a cut verte. In this case G has two components C and C. Without loss of generalit, assume that C and C. B the minimalit of G, there is a spanning tree T i of C i such that (C i, T i ) has a backbone 4-colouring for i =,. Without loss of generalit, assume that f() {, } and f() f(). If {f(), f()} = {, }, then colour u with 4. For the other cases, if f() f() consider the smmetric colouring of C and 3

20 3. Properties of a minimum countereample 3 colour u with. If f() = f(), colour u with 4; thus, u and u are f-edges. Suppose that T = T T {u, u}. So, T is a backbone of G such that BBC(G, T ) 4, contradiction. (b) u is not a cut verte. Here G is connected and, G. b the minimalit of G, there is a spanning tree T of G such that (G, T ) has a backbone 4-colouring. Without loss of generalit, assume that f() {, }. For an colouring of and, let f(u) = ma({,, 3, 4} {f(), f()}). One can easil observe that ma( f(u) f(), f(u) f() ). Without loss of generalit, assume that f(u) f() ; so, u is an f-edge. Suppose that T = T {u}. Thus, T is a backbone of G such that BBC(G, T ) 4, contradiction. 3. G has no 3-verte. Assume for a contradiction that G contains a 3-verte u with neighbours,,. Let set A = {,, }. If u is a cut verte, graph G u has more than one component which lead to the following three cases. (a) u is not a cut verte. Here G is connected and,, G. b the minimalit of G, there is a spanning tree T of G such that BBC(G, T ) 4. Without loss of generalit, assume that f() {, }. For an colouring of, and, let f(u) = ma({,, 3, 4} {f(), f(), f()}). One can show that ma( f(u) f(), f(u) f(), f(u) f() ). Without 4

21 3. Properties of a minimum countereample 3 loss of generalit, assume that f(u) f(). Hence, u is an f-edge and for a spanning tree T = T {u}, BBC(G, T ) 4, contradiction. (b) u is a cut verte and comp(g ) =. Assume that C and C are two components of G u and let, C and C. In graph G u we identif with and we obtain graph G ; so, in G, f() = f(). B the minimalit of G, there is a spanning tree T of G such that (G, T ) has a backbone 4-colouring. Let T i be C i T for i =,. Without loss of generalit, we assume that f() = f() = {, }. If u is an f-edge in G then clearl there eists a spanning tree T = T T {u} such that BBC(G, T )(G, T ) 4, contradiction. If u is not an f-edge, then regarding to the colouring of,, u we have the following three cases: If f() = f() = and f(u) =. Consider the smmetric colouring of C and f () = 4. Hence, u is an f-edge in G and T = T T {u}, So for the spanning tree T in G, BBC(G, T ) 4, contradiction. If f() = f() = and f(u) =. Consider the smmetric colouring of C with f () = 3. Hence, u is an f-edge in G and T = T T {u}, So for the spanning tree T in G, BBC(G, T ) 4, contradiction. If f() = f() = and f(u) = 3. 5

22 3. Properties of a minimum countereample 3 Since u is an f-edge, f() = and we can recolour u with 4. Hence, u is an f-edge in G and there is a spanning tree T = T T {u} in (G, T ) such that BBC(G, T ) 4, contradiction. (c) u is a cut verte and comp(g ) = 3. Assume that C, C and C 3 are three components of G and let C, C and C 3. B the minimalit of G, there is a spanning tree T i of C i such that (C i, T i ) has a backbone 4-colouring for i =,, 3. Without loss of generalit, assume that f() f() f(). If an two vertices in set A have different colours, then f() {3, 4}. In this case consider the smmetric colouring of C 3 and let f(u) = 4. If there eists two vertices v and v in A such that f(v ) = f(v ), then similar to the case a we can find a proper colouring for u. Since u, u and u are f-edges, then T = T T T 3 { u, u, u } is a spanning tree of G such that (G, T ) has a backbone 4-colouring, contradiction. Lemma. Graph G has at most one bridge. Proof. Assume otherwise and suppose that graph G has at least two bridges, e, e. We remove e and G is the remaining graph with two components, C and C. Assume that e C with endpoints v and v. According to Lemma, δ(g) 4 which states that δ(g ) 3. Let {v, v 3, v 4 } N G(v ). Spanning tree T of graph G must contain e because it does not lie on an ccle. So, assume that in G, 6

23 3. Properties of a minimum countereample 3 w 3 w w 3 3 w w w (a) bad [4, 4, 4]-triangle (b) bad [5, 4, 4]-triangle Figure 3.: Eamples of Lemma 3, bad triangles. f(v ) =, f(v 3 ) = and f(v 4 ) = 4. Now f(v ) {,, 3, 4} and BBC(G, T ) 5, which is a contradiction. A 4-face is weak if it is adjacent to four triangles. A [4, 4, 4]-triangle is bad if it is adjacent to three weak 4-faces. Similarl, a [5, 4, 4]-triangle is bad if it is adjacent to two weak 4-faces. Lemma 3. If, and are three mutuall adjacent 4-vertices in G, then,, form a bad triangle, see Figure 3.a. Proof. Suppose that G contains three mutuall adjacent 4-vertices,, and and let G = G {,, }. B the minimalit of G, there is a spanning tree T of G such that (G, T ) has a backbone 4-colouring. For now assume that G is connected. At the end we show that if G is not connected, we have the choice of epanding the spanning subgraph of G to the connected one. Let v and v be two neighbours of v, v A = {,, } which are in B = N(A) A, 7

24 3. Properties of a minimum countereample 3 w 3 w 0 0 w Figure 3.: Eample of graph G. i.e., B = {,,,,, }. Without loss of generalit, assume that f( ) f( ) and f( ) f( ) and f( ) f( ). For ever v A and v, v B, list L(v) is a set of available colours for verte v which is S {f(v ), f(v )}. For finding the spanning tree T we have the following two cases:. L() = L() = L() = {c, c }. In this case, we construct a graph G (V, E ) b adding two new edges to G {,, } which do not create an adjacent triangles in G, see Figure 3.. Here b adding and to G {,, }, G does not contain adjacent triangles. Since f( ) is the same as f( ), for ever T, BBC(G, T ) 5. B the minimalit of G and the fact that V < V, there is a spanning tree T of G, BBC(G, T ) 4, which is a contradiction. The onl challenging case is when b adding an two edges to G, where BBC(G, T ) 5, adjacent triangles are created, as in Figure 3.a. Accord- 8

25 3. Properties of a minimum countereample Figure 3.3: Three mutuall adjacent 4-vertices. ing to the definition this case is a bad [4, 4, 4]-triangle.. u, v A, L(u) L(v) or u A, L(u) = 3. If u, v A, L(u) L(v). Without loss of generalit, assume that L() L(). In this case we provide a formula to assign proper colours to, and in order to epand T to T, spanning tree of graph G. f() = min ( L() L() ) min ( L() {f()} ) f() = ma ( L() {f()} ) f() = ma ( L() {f()} ) : L() : L() One can easil observe that if {, 4} {f(), f(), f()}, alwas we can find a subset of edges for epanding T to spanning tree T. Without loss of generalit, assume that f() = and f() = 4. According to our assumption L() = L() = ; so, for an colouring of,,,, 9

26 3. Properties of a minimum countereample 3 edges and are alwas f-edges. Also, for an colouring of, one of the edges or is alwas an f-edge; if f() =, is an f-edge and if f() = 3, is an f-edge. If {, 4} {f(), f(), f()}, the set of colours for,, is either {, 3, 4} or its smmetric colouring {,, 3}. Without loss of generalit, assume that f() =, f() = 4 and f() = 3. Edges and are f-edges. If f( ) =, then is an f-edge. However, if f( ), it means that L(). Since f() = 3, min ( L() {f()} ) which is true onl if f() = ; this contradicts the fact that f() = 4. So, in this case f( ) =. If u, v A, L(u) L(v) and u A, L(u) = 3. In this case we provide a formula to assign proper colours to, and in order to epand T to T. f() = min(l()) f() = ma(l()) f() = ( L() {f()}) {f()} ) Without loss of generalit, assume that u A, L(u) = {, 3, 4} or L(u) = {, 3, 4}. If u A, L(u) = {, 3, 4} then according to above formula colour with, with 4 and with 3. So, edges, and are f- edges. If u A, L(u) = {, 3, 4} then according to above formula colour with, with 4 and with 3. So, edges, and are f- edges. 0

27 3. Properties of a minimum countereample 3 As a result, alwas we can find a subset of edges for epanding T to spanning tree T such that BBC(G, T ) 4. This contradicts the minimalit of G. Now assume that G is disconnected. Since G has at most one bridge, G cannot have more that three components, C, C and C 3 (three components for the case that, and are cut vertices of G). For finding a subset of edges to epand T, whether we can use the current colouring of vertices or the smmetric colouring of components with recolouring vertices. B the minimalit of G, backbone colouring of each component is at most 4. If two edges in the set {,, } are f-edges, then easil we can etend the spanning tree of each component to T. However, if just one of the edges in the set {,, } is an f-edge and b considering the smmetric colouring of one component, we cannot find two f-edges in that set, then we have the following case: u, v A, L(u) = L(v) = {, 3}. Without loss of generalit, assume that L() = L() = {, 3} and is an f- edge. In this case we consider a smaller graph G = G. Since is not an f-edge, for an spanning tree T of G, BBC(G, T ) 5. This contradicts the minimalit of G. So, this case does not eist. For the case that u A, L(u) = {, 3} or u A, L(u) = {, 4}, vertices, and form a bad [4, 4, 4]-triangle. The proof can also be given with a similar argument to Case.

28 3. Properties of a minimum countereample 3 w 3 3 w 0 0 w Figure 3.4: Eample of graph G. Lemma 4. If, and are three mutuall adjacent vertices with degrees 5, 4, 4 in G, then,, form a bad triangle, as shown in Figure 3.b. Proof. Suppose that G contains three mutuall adjacent vertices with degrees 4, 4, 5 and without loss of generalit, assume that d() = 5 and d() = d() = 4. Let G = G {,, } and for now assume that G is connected. B the minimalit of G, there is a spanning tree T of G such that BBC(G, T ) 4. Assume that A = {,, } and set B = N(A) A = {,, 3,,,, }. For finding the spanning tree T, we consider the following two cases:. L() = L() and L() L() and L() =. Without loss of generalit, assume that f( ) = f( ) and f( ) = f( ). In this case, we construct a graph G b adding two new edges to G {,, } which do not create an adjacent triangles in G, Figure 3.4. Here b adding and to G {,, }, G does not contain adjacent triangles and for ever spanning tree T, BBC(G, T ) 5. Since, graph G is smaller

29 3. Properties of a minimum countereample 3 3 Figure 3.5: Three mutuall adjacent vertices with degrees 5,4,4. than G, b the minimalit of G there eists a spanning tree T of G such that BBC(G, T ) 4, which is a contradiction. The onl challenging case is when b adding an two edges to G {,, }, where BBC(G, T ) 5, adjacent triangles in G are created, Figure 3.. According to the definition, this case is a bad [5, 4, 4]-triangle.. u, v A, L(u) L(v), L(u) L(v) or u A, L(u) = 3. (a) If u, v A, L(u) L(v), L(u) L(v) and u A, L(u). If u A, L(u) =, we can use the same formula as the Case of Lemma 3 to assign proper colours to, and. The proof can also be given with a similar argument to the Case of Lemma 3. Now, if L() =, we can assign proper colours to, and b using the following formula and epand T to T. f() = L() f() = min ( L() {f()} ) f() = ma ( (L() {f()}) {f()} ) 3

30 3. Properties of a minimum countereample 3 One can easil observe that if {, 4} {f(), f(), f()}, alwas we can find a subset of edges for epanding T to T. Without loss of generalit, assume that f() = and f() = 4. As a result, 3 and are f- edges. If f() =, will be an f-edge and if f() = 3, will be an f-edge. If {, 4} {f(), f(), f()}, also we can find a subset of edges for epanding T to T. Without loss of generalit, assume that f() =, f() = and f() = 3. So, 3 and are f-edges. If at least one of the edges in the set {,,, } is an f-edge easil we can epand spanning tree T to T. In this case, none of the edges in the set {,,, } are f-edges, switch the name of and in the formula. For eample, if f( ) =, f( ) = 3, f( ) = and f( ) = 4, b switching and in the formula, f() = 4 and f() =. Hence, and are f-edges and we can epand T to T. (b) If u A, L(u) = 3. Without loss of generalit, assume that L() = 3. f() = min ( L() ) f() = ma ( L() ) f() = ma ( (L() {f()}) {f()} ) The proof can be given with a similar argument that is discussed in the previous case. Now assume that G is disconnected. Since G has at most one bridge, G cannot 4

31 3. Properties of a minimum countereample 3 have more that four components, C, C, C 3 and C 4. Suppose that i C, 3 C, i C 3 and i C 4, i =, (four components for the case that, and are cut vertices of G and one of the edges in set {,, 3 } is a bridge). For finding a subset of edges to epand T, whether we can use the current colouring of vertices or the smmetric colouring of components with recolouring vertices. B the minimalit of G, backbone colouring of each component is at most 4. If two edges in the set {,, } are f-edges, then easil we can etend the spanning tree of each component to T. However, if just one of the edges in the set {,, } is an f-edge and b considering the smmetric colouring of components, we cannot find two f-edges in that set, then we have the following case: u, v A, L(u) L(v) {, 3}. Without loss of generalit, assume that L() L() {, 3} and is an f-edge. In this case we consider a smaller graph G = G. Since is not an f-edge, for an spanning tree T of G, BBC(G, T ) 5, but this contradicts our assumption that G is a minimum countereample. So, this case does not eist. For the case that u A, L(u) {, 3} or u A, L(u) {, 4}, vertices, and form a bad triangle. The proof can also be given with a similar argument to Case. Corollar. If a [4, 4, 4]-triangle is bad then in an colouring of G {,, }, L() = L() = {p, q} and L() L(), {p, q} {,, 3, 4}. 5

32 3.3 Properties of bad [4, 4, 4]-triangle 3 w w 3 w Figure 3.6: Eample of sponsor vertices of bad [4, 4, 4]-triangle with degrees 4. Corollar. If a [5, 4, 4]-triangle is bad then in an colouring of G {,, }, L() = L() = {p, q} and L() L(), {p, q} {,, 3, 4} and p q. 3.3 Properties of bad [4, 4, 4]-triangle We call vertices in set B = {,,,,, } the sponsor vertices of the bad triangle. Vertices w i, i =,, 3 in Figure 3.6 are the common neighbours of adjacent sponsor vertices. Let w be the common neighbour of and. According to Corollar, if S {p, q} = {r, s}, {f( ), f( )} = {f( ), f( )} = {f( ), f( )} = {r, s}. Following lemmas will be frequentl applied. Lemma 5. The degree of sponsor vertices is at least 5. Proof. First we prove it for, d( ) 5. Assume otherwise, that is, is a 4- verte and N( ) = {,,, w }, see Figure 3.6. Consider a colouring of G 6

33 3.3 Properties of bad [4, 4, 4]-triangle 3 w w w 3 q q Figure 3.7: Eample of vertices w i, i =,, 3 with degrees 4. {,, }. Assume that f( ) = r. Since and are connected, f( ) f( ). Let f( ) = s (r s). B Corollar, f( ) f( ) and f( ) = f( ) = s. So, could be coloured with f(w ). As a result, all neighbours of are in two different colours, s and f(), and there are two colours for, S {s, f()}. This contradicts our assumption that for ever colouring of G, the colour of sponsor vertices is r or s. Second with the same argument we can prove that d(v) 5, v B. Lemma 6. If two adjacent vertices in the set B = {,,,,, } are 5- vertices, then their common verte in the set {w, w, w 3 } (if there eists an) has degree at least 5. Proof. Assume for a contradiction that d( ) = 5 and d( ) 5 and d(w ) = 4. Consider a proper colouring of graph G = G {,, }. According to Corollar 7

34 3.3 Properties of bad [4, 4, 4]-triangle 3, {f( ), f( )} = {f( ), f( )} = {r, s}. Without loss of generalit assume that f( ) = f( ) = r and f( ) = f( ) = s. If we can recolour with colour p or q, then vertices, and can be coloured with the colour set S for which BBC(G, T ) 4. According to figure 3.7, N G( ) = {,, 3, w } and { 3, 4 } N G( ). Since, f( ) = f( ) = s and the onl available colour for is r, then {f( 3 ), f(w )} = {p, q}. Similarl, since f( ) = f( ) = r and the onl available colour for is s, then {f(w ), f( 3 )} = {p, q}. If {f(q ), f(q )} = {r, s}, then we can recolour with {p, q} {f( 3 )} because graph G does not have adjacent triangles which means that w is not connected to 3, contradiction. Now if r {f(q ), f(q )} or s {f(q ), f(q )}, then we can recolour w with r or s, f (w ) {r, s}. Hence, or can be coloured with f(w ) which is p or q, contradiction. Lemma 7. If the sponsor verte sponsored more than one bad triangle, then it should be a 6 + -verte. Proof. Suppose for a contradiction that is a 5-verte and it sponsors at least two bad triangle. Then it should be adjacent to more than three triangles. In this case two of these triangles are adjacent, the this contradicts our choice of G. 8

35 3.4 Properties of a bad [5, 4, 4]-triangle 3 w 3 w 3 w Figure 3.8: Eample of sponsor Vertices of bad [5, 4, 4]-triangle with degree Properties of a bad [5, 4, 4]-triangle According to Corollar, if S {p, q} = {r, s}, {f( ), f( )} = {f( ), f( )} = {r, s} and {r, s} {f( ), f( ), f( 3 )}. Without loss of generalit, assume that f( ) = r and f( ) = s. Lemma 8. The degree of vertices,, and is at least 5. Proof. First we prove it for, d( ) 5. Assume otherwise, that is, is a 4- verte where N( ) = {,,, w }, as in Figure 3.8. Assume that f( ) = s. Since and are connected, f( ) f( ) and f( ) = r; also, f( ) f( ) and f( ) = f( ) = r and f() could be the same colour as f(w ). As a result, all neighbours of are in two different colours, r and f(), and there are two available colours for in the set S {f()}. This contradicts our assumption that for ever colouring of G, the colour of sponsor vertices is r or s. Second with the same argument we can prove that d( ) 5, d( ) 5 and d( ) 5. 9

36 3.5 The discharging method 3 Lemma 9. The degree of w is at least 5. The proof is similar to the proof of Lemma. 3.5 The discharging method In this section we will complete our proof b using discharging rules. Since G is a planar graph, then; according to Euler s formula, G has the characteristic that V (G) E(G) + F (G) =. Hence, b using the fact that v V (G) d(v) = f F (G) For ever connected planar graph, d(f) = E(G), we have the following result: (d(v) 4) + (d(f) 4) v V (G) f F (G) = E(G) 4 V (G) + E(G) 4 F (G) = 4( V (G) E(G) + F (G) ) = 8. We define an initial charge function of ω() = d() 4 where V (G) F (G). Then we design appropriate discharging rules and redistribute charges accordingl. After charges are distributed according to discharging rules, a new charge function ω is produced. When discharging is in progress, the sum of all charges is fied. On 30

37 3.6 Proof of Theorem 3 the other hand, (d(v) 4) + (d(f) 4) = ω (). v V (G) f F (G) V (G) E(G) We denote the amount of charges transformed from to b τ( ) where, V (G) F (G). Our discharging rules are as follows: Discharging rules: R Ever 5 + -verte gives 3 charge to each [5+, 5 +, 5 + ]-triangle. R Ever 5 + -verte gives charge to each [4, 5+, 5 + ]-triangle. R3 A sponsor verte gives its residual charges to a bad triangle. 3.6 Proof of Theorem For each V (G) F (G) we will calculate ω () and we will show that ω () 0. Proof. If V (G), according to Lemma, δ(g) 4, ω () = d() 4 0. If f is a [4,4,4]-triangle, f F (G). According to Lemma 3, [4, 4, 4]-triangles in G are alwas bad, Figure 3.6. Let f = []. 3

38 3.6 Proof of Theorem 3 All 3-faces of graph G have an initial charge of. B Lemma 5 and Lemma, d(v) 5, v {w,,, }. So, b using Rules R and R we have the following charge distributions: τ(v f) = 3, v {w,, }, f = w τ(v f) =, v {, }, f = Thus, ω ( ) =. With the same argument we can show that the new charge 6 of all sponsor vertices of [] is. Since bad triangle has si sponsor ver- 6 tices, b using Rule R3 we have the following result: ω [] = = 0 If sponsor vertices are 6 + -vertices, the are adjacent to at most three triangles. Let be a 6-verte, then ω ( ) = = 7 6. So, can give at least 6 charge to f. If f is a [5,4,4]-triangle, f F (G). According to Lemma 4, [5, 4, 4]-triangle in G are alwas bad, Figure 3.6. Let f = []. B Lemma 8 and Lemma 9, d(v) 5, v {w,,,, }. Also, is a 5-verte and ω() =. So, b using Rules R and R, we have the following charge distributions: τ(v f) = 3, v {w,, }, f = w τ(v f) =, where v {, }, f = or v {, }, f = τ( f ) = 3

39 3.6 Proof of Theorem Figure 3.9: Eample of [4,5,5]-triangle. B Rule R3, ω ( ) = ω ( ) = ω ( ) = 6, that is ω (f ) = If verte v, v {,,,, w } is 6 + -verte, it is adjacent to at most three triangles. Then ω (v) = = 7 6 and v can give at least 6 charge to f. If f is a [4 +, 5 +, 5 + ]-triangle, f F (G). Let f = [], Figure 3.9. One can easil observe that, for a k-verte v in graph G, v is adjacent to at most k triangles. Let and be two 5+ - vertices of []. So, ω(v) = k 4, v {, }. B Rule R4, τ(v f ) = k 4 k, v {, }. If k = 5, v is adjacent to at most two triangle. So, τ(v f ). If k 6, τ(v f ) = k 4 k 8 k 4 6. As a result, τ(v f ), where v {, }; ω (f )

40 3.6 Proof of Theorem 3 If f is a 4 + -face, f F (G), ω (f ) = d(f ) 4 0. Thus, we proved that ω () 0 for V (G) F (G) which leads the following contradiction, 0 ω () = 8. V (G) E(G) This contradiction shows that the assumption of eistence of a minimum countereample G was incorrect. As a result, we have succeeded in proving that backbone k-colouring of all connected planar graphs without adjacent triangles is at most 4. 34

41 Chapter 4 Algorithm for Finding a Backbone In this chapter we will etend the result obtained in Chapter 3 to all classes of graphs. In Chapter 3, as we coloured all vertices of graph G properl, we made the backbone of a graph. However, in this chapter, after colouring all vertices of the graph properl, we will find a backbone tree with the algorithm to be described below. 4. Eistence of the backbone of ever graph G with χ(g) 4 Theorem. For ever graph G with the chromatic number χ(g) = k, there is a spanning tree T and a proper k-colouring f such that uv E(T ), f(u) f(v) =. Proof. Consider a proper colouring of G with a colour function f : V (G) S, S = χ(g). An edge uv of the graph G is called good if f(u) f(v) =. We also define 35

42 4. Eistence of the backbone of ever graph G with χ(g) 4 4 an edge uv of the graph G as being bad if f(u) f(v). Let be an arbitrar verte in G. Let T be the Breadth First Search tree with source. If all edges of T are good, simpl we consider T as a backbone of G. Otherwise assume that T has q number of bad edges. Assume that set B is the set of all bad edges of T, B = q. Let e be the first bad edge in T with endpoints u and v where f(u) = r and f(v) = s, such that r s. Let C and C v be two components of T B where, u C and v C v. In other words, C is the maimal subgraph of T that contains and no other bad edges. Our goal is to epand C step b step in a wa that no other bad edges are left in C. If there eists one good edge in G between C and C v, simpl we add that edge to T e. So, it can be seen that the number of vertices in C is increased b at least. But if there does not eist an good edge between C and C v in G, we have the following two cases: Case I. r s =. Without loss of generalit, assume that s r =. In this case we consider the (s, s )-Kempe chain of v and switch the colour of vertices in the Kempe chain. So f (v) = s = r + and uv is a good edge; as a result, we add uv to T B and it is clear that the number of vertices in C is increased b at least. It is not possible that the (s, s )-Kempe chain of v has an intersection with C. Assume otherwise and let be an intersection of (s, s )-Kempe chain of v and C. Let be a neighbour of in C v and, also, in (s, s )- 36

43 4. Eistence of the backbone of ever graph G with χ(g) 4 4 Kempe chain of v. Since both and are in (s, s )-Kempe chain of v and the are adjacent to each other, f() f() = and will be a good edge. But we assumed that there are not an good edges between C and C v, contradiction. Case II. r s 3. Let r s = t and without loss of generalit, assume that r = s + t. We consider the following algorithm for recolouring vertices. Let i be a counter from to t. Consider (s + i, s + i)-kempe chain of v and switch the colour of vertices in the Kempe chain; then add to the counter i. Do this step until i = t. So, f (v) = s + t = r and uv is a good edge; then, we add uv to T B. Also, it is clear that the number of vertices in C is increased b at least. Until here we replace the first bad edge b another good edge and we will repeat this step until there are not an more bad edges in T. As a result, uv E(T ), f(u) f(v) =. Note: The reason that we did not consider (s, r ) directl is that it might have an intersection with C. For eample, consider Figure 4.. In this figure, black edges are edges of graph G and edges of the spanning tree T are shown b red colour. Let uv be a first bad edge in T. Since there are no other good edges between C and C v, we should 37

44 4. Eistence of the backbone of ever graph G with χ(g) 4 4 C 4 w p 3 4 u v 4 Figure 4.: Eample of (4, )-Kempe chain. consider the Kempe chain of v. If we consider (4, )-Kempe chain of v directl, the subset of vertices in the Kempe chain will be {,, v, w, p}. These vertices are shown b blue colour in Figure 4.. B switching colours of (4, )-Kempe chain, f (v) = and f () =. But we were not allowed to change the colour of an vertices in C. Corollar 3. For ever graph G if χ(g) 4, then there is a spanning tree T such that BBC(G, T ) = χ(g). Proof. Assign vertices of G into k, k = χ(g) colour classes and put these sets in an increasing order. Net, we add edges of T in between the sets. According to Theorem, there are edges just between two consecutive pair of colour classes. Net step is to change the permutation of colour classes in the following order:,, 3,..., k, 4,...,, 3,... It is clear that the colour class k will be in the middle of permutation or at the end, corresponding to k being even or odd respectivel. 38

45 4. Eample of finding a backbone of graph u v m 3 w Figure 4.: Eample of finding a backbone of graph G with an arbitrar spanning tree T. p 4 Hence, uv T, f(u) f(v) and all edges of T are f-edges. So, we find a spanning tree T of G such that BBC(G, T ) = χ(g). Also, we prove the stronger result that uv T, f(u) f(v) = or f(u) f(v) k. 4. Eample of finding a backbone of graph Here is an eample of using this algorithm for finding the backbone tree of graph G, Figure 4.. In this figure black edges are edges of graph G. Let be an arbitrar verte in G and T be the breadth-first tree with source. Edges of the spanning tree T are shown b red colour. Also, green numbers are the verte colouring of G. Let u is the first bad edge in T. Set of all bad edges in T is B = {u, v, mw}. Graph T u has two components, C, C v and there are not an good edges between C and C v. So, we consider the Kempe(, ) of u in G and switch the colour of vertices. According to Figure 4., Kempe(, ) = {u, v,, w} and after 39

46 4. Eample of finding a backbone of graph u v m 3 4 w p Figure 4.3: Switching the colour of vertices in the Kempe(, ) of u in G. 3 4 u v m 3 w p 4 Figure 4.4: Replacing the first bad edge of T b another good edge from G. switching colours, f (u) =, f () = f (w) = f (v) = as we can see in Figure 4.3. Now u is a good edge and we keep it in T. Net step is to find a first bad edge of T in Figure 4.3 which is v. So, B = {v}. Components C and C v are two components of T B. In graph G, there eists a good edge vw between C and C v and simpl we add vw to T B. B using the Breadth First Search algorithm we can see that there are not an more bad edges in T. So, we find the spanning tree T such that all edges are good edges, as in Figure 4.4. After that, we assign vertices of G into the 4 colour classes and put these sets in 40

47 4. Eample of finding a backbone of graph 4 v w u m p Figure 4.5: Recolouring each colour class of G u v m w p 3 4 Figure 4.6: The spanning tree T is the backbone of graph G. an increasing order, as in Figure 4.5. Now we should recolour each colour class of G in the following order:, 4, 3, 4 3. Hence, according to the proper colouring of G in Figure 4.6, all red edges of graph are f-edges. As a result, spanning tree T is the backbone of G and BBC(G, T ) = 4. 4

48 Chapter 5 Conclusion In this research it has been proved that for ever connected planar graph without adjacent triangles, backbone k-colouring is at most 4. Also, we proved the stronger result that for ever graph G with χ(g) 4, a spanning tree T of G eists, such that BBC(G, T ) = χ(g). In future work, the following questions could be answered: Is it possible to prove that for ever connected planar graph and ever spanning tree T of G, the circular backbone k-colouring of (G, T ) is at most 7? Can we find an tight upper bound for the circular q-backbone k-colouring chromatic number of planar graphs? Is it possible to find an tight upper bound for the q-backbone k-colouring chromatic number of all classes of graphs? 4

49 Bibliograph [] Appel, K. and W. Haken, Ever planar map is four colourable, Bulletin of American mathematical societ 8 (976). [] H. Broersma, F.V. Fomin, P.A. Golovach, G.J. Woeginger, Backbone colourings for networks, in: Proceedings of the 9th International Workshop on Graph-Theoretic Concepts in Computer Science, WG 003, in: LNCS, vol. 880, 003, pp The full paper version: Backbone colourings for graphs: tree and path backbones, Journal of Graph Theor 55 () (007) [3] V. Campos, F. Havet, R. Sampaio, A. Silva, Backbone colouring: Tree backbones with small diameter in planar graphs, Theoretical Computer Science 487 (03) [4] H.J. Broersma, J. Fujisawa, L. Marchal, D. Paulusma, A.N.M. Salman, K. Yoshimoto, λ-backbone colourings along pairwise disjoint stars and matchings, Discrete Mathematics 309 (009), pp

50 BIBLIOGRAPHY 5 [5] Y. Bu, S. Zhang, Backbone Colouring for C 4 free-planar graphs, Science China Mathematics 4 () (0) [6] S. Zhang, Y. Bu, Backbone colouring for C 5 -free planar graphs, Journal of Mathematical Stud (00) [7] Y. Bu, Y. Li, Backbone colouring of planar graphs without special circles, Theoretical Computer Science 4 (0) [8] X. Wang, Backbone colouring of connected planar graphs without C 8 and adjacent triangles, Algebra Colloq., () (0). [9] F. Havet, A. D. King, M. Liedloff and I. Todinca, (Circular) backbone colouring: Forest backbones in planar graphs, Discrete Applied Mathematics 69 (04), pp [0] J. Araujo, F. Havet, M. Schmitt. Steinberg-like theorems for backbone colouring. RR-864, INRIA Sophia Antipolis. 04. <hal > [] Y. Bu, X. Bao, Backbone colouring of planar graphs for C 8 -free or C 9 -free, Theoretical Computer Science 580 (05) [] D. B. West, Introduction to Graph Theor - second edition, Prentice Hall (00). 44

51 List of Figures. Eample of backbone colouring Eample of (4,)-Kempe chain Eamples of Lemma 3, bad triangles Eample of graph G Three mutuall adjacent 4-vertices Eample of graph G Three mutuall adjacent vertices with degrees 5,4, Eample of sponsor vertices of bad [4, 4, 4]-triangle with degrees Eample of vertices w i, i =,, 3 with degrees Eample of sponsor Vertices of bad [5, 4, 4]-triangle with degree Eample of [4,5,5]-triangle Eample of (4, )-Kempe chain

52 LIST OF FIGURES 5 4. Eample of finding a backbone of graph G with an arbitrar spanning tree T Switching the colour of vertices in the Kempe(, ) of u in G Replacing the first bad edge of T b another good edge from G Recolouring each colour class of G The spanning tree T is the backbone of graph G