9.1. Vertex colouring. 9. Graph colouring. Vertex colouring.


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1 Vertex colouring. 9. Graph colouring kvertexcritical graphs Approximation algorithms Upper bounds for the vertex chromatic number Brooks Theorem and Hajós Conjecture Chromatic Polynomials Colouring of planar graphs Edge colouring. Vizing Theorem Edge chromatic minimal graphs Total colouring The Timetabling Problem Graph Theory Vertex colouring Problem : We wish to arrange the talks in a congress in such a way that no participant will be forced to miss a talk he would like to hear. Assuming a good supply of lecture rooms enabling us to hold as many parallel talks as we like, how long will the program have to last? The graphtheoretical model: : Let G be such a graph whose vertices are the talks and in which two talks are joined iff there is a particip ant wishing to attend both. What is the minimal value of k for which V(G) can be partitioned into k classes, say V 1,V 2,,V k, such that no edge joins two vertices of the same class? Graph Theory 7 2 G In a given colouring c of a graph G,, a set consisting of all vertices assigned the same colour is referred to as a colour class: V k = {v V(G) c(v) ) = k}. k We remark that we shall use real colours (red, blue,,.) only if there are few colours, otherwise the natural numbers will be our colours. Thus kcolouring of the vertices of G is a function c: V(G) {1,2, 1,2,,k}. We denote the minimal number of classes by (G) and call as the (vertex) chromatic number of G. If G is a graph for which (G) = k then G is kchromatic. A possible colouring of a graph G with 3 colours. Graph Theory 7 3 Graph Theory 7 4 1
2 An alternative definition: the chromatic number is the minimum number of independent subsets into which V(G) can be partitioned. Each independent set is then a colour class in the (G) G)colouring of G so defined. G: For some graphs, the chromatic number is quiet easy to determine: (C 2k ) = 2. (C 2k+1 ) = 3. (K n ) = n. (K n1, n 2,, n ) = k. k (G) 3. C 5 G and (C 5 ) = 3, it follows that (G) 3. So, (G) =3. Graph Theory 7 5 Graph Theory 7 6 Some easy statement: If a graph G does not have an odd cycle then (G) 2. If a graph G has an odd cycle then (G) 3. (G) ) = k iff G is kpartite but G is not lpartite for l < k. k If K k G then (G) k. If G does not contain h+1 independent edges then (G) G / h. Unfortunatelly,, for k 4 we do not have a similar characterization of graphs with chromatic number at least k, though there are some complicated characterization. Can we make a good characterization for kchromatic graphs? The 1chromatic graphs are the empty graphs. The 2chromatic graphs are the nonempty bipartite graphs. No value of k > 2 is such an applicable characterization known. The chromatic number of a disconnected graph is the maximum of the chromatic numbers of its components. The chromatic number of a connected graph with cutvertices is the maximum of the chromatic numbers of its blocks. So, we will concern with determining the chromatic number of non separable graphs; (i.e. graphs which do not have cutvertices). Graph Theory 7 7 Graph Theory 7 8 2
3 Vertexcritical Graphs A graph G is k(vertex) critical if (G) ) = k and (G v) ) = k 1 for all v V(G). (We also say that G is critically kchromatic.) Every kchromatic graph has a kcritical subgraph. if k = 2 then the only 2critical 2chromatic graph is K 2, and the odd cycles are the only 3critical 3chromatic graphs. For k 4 the kcritical graphs have not been characterized, although it is quite difficult, in general, to determine whether a given k chromatic graph is kcritical or not. Graph Theory 7 9 The Grötzsch graph a 4critical graph. Graph Theory 7 10 Theorem 9.1.: If G is kcritical then (G) k 1. By contrary: let us suppose that G is kcritical and (G) < k 1 and let v be a vertex of degree in G. Since G is kcritical, G v is (k( 1)colourable. Let V 1, V 2,,V k1 be the colourclasses classes of a (k 1)colouring( of G v. By definition, v is adjacent in G to < k 1 vertices, therefore there exists at least one j, 1 j k 1,, such that for any vertex of V j is not adjacent to v in G. If we colour v by the jth colour then we get a (k( 1)colouring for G. This is a contradiction since G is kcritical. Corollary 9.2.:.: Every kchromatic graph has at least k vertices of degree at least k 1. Let G be a kchromatic graph,, and let H be a kcritical subgraph of G. By Theorem 9.1, each vertex of H has degree at least k 1 in H,, and hence also in G. Since H,, being kchromatic, has at least k vertices. Can we say something generally about the kcolourable graphs? Graph Theory 7 11 Graph Theory
4 H: G: Corollary 9.3.:.: For any graph G, (G) Let us suppose that the statement is not true. Then there exists a graph G with chromatic number By Theorem 9.2 then there exists at least + 2 vertices with vertex degree + 1. This is impossible. G is a 3chromatic graph. H is 3critical subgraph of G. For a complete graph K n, (K n ) = If G is an odd cycle, then (C n ) = 2 and (C n ) = 3 = Graph Theory 7 13 Graph Theory 7 14 Let S be a vertex cut of a connected graph G,, and let the components of G S have vertex sets V 1, V 2,..., V t. The subgraphs G i = G[V i S] are called the S components of G. We say that colourings of G 1, G 2,...,G, t agree on S if, for every v S, the same colour is assigned to v in each of the colourings. u v G u u u Graph Theory 7 15 v v v A grah G and its {u,v{ u,v}components Graph Theory
5 Theorem 9.4. In a k critical graph, no vertex cut is a clique. We prove by contradiction: suppose that G is a k critical graph, and suppose that G has a vertex cut S that is a clique. Denote the S componenents of G by G 1, G 2,..., G t. Since G is kcritical, each G i is (k( 1)colourable. Because S is a clique, the vertices in S must receive distinct colours in any (k( 1)colouring of G i. Then there are (k( 1)colourings of G 1, G 2,..., G t which agree on S. But these colourings together form a (k( 1)coloring of G,, which is a contradiction. Theorem 9.5. Every k critical graph is a block. If v is cutvertex, then {v}{ } is a vertex cut which is also a (trivially) clique. Then it follows from Theorem 9.4 that no critical graph has a cut vertex; equivalently, every kcritical graph is a block. An {u,v{ u,v}component G i of G is of type 1 if every (k( 1)colouring of G i assigns the same colour to u and v,, and of type 2 if every (k( 1) colouring of G i assigns different colours to u and v. Consequence: If a kcritical graph has a 2 vertex cut {u,v{ u,v}, then u and v cannot be adjacent. Graph Theory 7 17 Graph Theory 7 18 Theorem 9.6. (Dirac( Dirac,, 1953): Let G be a kcritical graph with a 2 vertex cut {u,v{ u,v}. Then (1) G = G 1 G 2, where G i is a {u,v{ u,v}component of type i (i i = 1,2), and (2) both G i + uv and G 2 uv are kcritical k (where G 2 uv denote the graph obtained from G 2 by identifying u and v). Graph Theory 7 19 Graph Theory
6 (1) Since G is critical, each {u,v{ u,v}component of G is (k( 1)colourable. Now there cannot exist (k( 1)colourings of these {u,v{ u,v}components all of which agree on {u,v{ u,v}, since such colourings would together yield a (k( 1)colouring of G. Therefore there are two {u,v{ u,v}components G 1 and G 2 such that no (k 1)colouring of G 1 agrees with any (k( 1)colouring of G 2. So, one, say G 1, must be of type 1 and the other, G 2, of type 2. Since G 1 and G 2 are of different types, the subgraph G 1 G 2 of G is not (k( 1)colourable. Therefore, because G is critical, we must have G = G 1 G 2. (2) Set H 1 = G 1 + uv.. Since G 1 is type 1, H 1 is kchromatic. We shall prove that H 1 is critical by showing that, for every edge e of H 1, H 1 e is (k( 1)colourable. This is so if e = uv,, since H 1 e = G 1. Let e be some other edge of H 1. In any (k( 1)colouring of G e,, the vertices u and v must receive different colours, since G 2 is a subgraph of G e. The restriction of such a colouring to the vertices of G 1 is a (k( 1) colouring of H 1 e. Thus G 1 + uv is kcritical. An analogous argument shows that G 2 uv is kcritical. Graph Theory 7 21 Graph Theory 7 22 Theorem 9.7.: Let G be a critical graph with 2vertex cut {u,v{ u,v}. Then d(u) ) + d(v) 3k 5. Let G 1 be the {u,v{ u,v}component of type 1 and G 2 the {u,v{ u,v}component of type 2. Set H 1 = G 1 + uv and H 2 = G 2 uv. By Theorem 9.1and Theorem 9.6 we get d (u) + d H1 H (v) 1 2k 2 and d (w) H2 k 1 where w is the new vertex obtained by identifying u and v. From the construction of G 1 and G 2 it follows that d (u) + d G1 G (v) 1 2k 4 which yields the desired result. and d (u) + d G2 G (v) 2 k 1 Graph Theory The Greedy Algorithm Order the vertices of the given graph G in any order. Colour x 1 by the colour 1, Colour x 2 by 1 if x 1 x 2 E(G) and by 2 otherwise. We colour each vertex the by the smallest colour it can have at the stage. The problem: this colouring may and usually does use many more colours than necessary. However,, it is also true that for every graph the vertices can be ordered in such a way that the greedy algorithm uses as few colours as possible. Graph Theory
7 x 1 x 3 x 5 x 7 x 1 x 8 x 5 x 6 x 2 x 4 x 6 x 8 x 7 x 2 x 3 x 4 In the order x 1,x 2,,x 8 the greedy algorithm needs four colours. Graph Theory 7 25 In the order x 1,x 2,,x 8 the greedy algorithm needs two colours. Graph Theory 7 26 What kind of other colouring algorithms can we give? Algorithm 1. (G)=max{ (G 1 ), (G 2 ), (G 3 )} = 4 The colouring of a graph reduces to the problem of colouring certain subgraphs of it. The process can be used if the graph is disconnected or has a cutvertex or, slightly more generally,, contains a complete subgraph whose vertex set disconnects the graph. Then we may colour each part separately since, at worst by a change of notation,, we can fit these colourings together to produce a colouring of the original graph, as shown in the next example. Graph Theory 7 27 (G 1 )=3 (G 2 )=3 (G 3 )=4 The vertex set of the thick triangle disconnect G Graph Theory
8 Algorithm 2 (Reduction( Algorithm). Let a and b two nonadjacent vertices of G.. Let G' be obtained from G by joining a to b,, and let G'' be obtained from G by identifying a and b. (See the next slide!) a a ab Assertion 1.: The colourings of G in which a and b get distinct colours are in 11 correspondence with the colourings of G'.. Indeed c: V(G) {1,2, 1,2,,k} } is a colouring of G with c(a) c(b) iff c is a colouring of G'. G b G' b G'' Assertion 2.: The colourings of G in which a and b get the same colour are in 11 correspondence with the colourings of G'' ''.. Indeed c: V(G) {1,2, 1,2,,k} } is a colouring of G with c(a) ) = c(b) iff c is a colouring of G'' ''. Graph Theory 7 29 The graphs G, G and G'' Graph Theory 7 30 G 0 G i = 0 yes Let G i = k no Is G i complete? G = G G i+ 1 i i A kcolouring of G i can be lifted to a k colouring of G. (G) is the minimum order of a complete graph in which a sequence G 0,G 1, can terminate. G i=i+1 Graph Theory 7 31 Colouring a 3colourable graph Graph Theory
9 Beside these simple statements however,, it is not easy to see that there are trianglefree graphs of large chromatic number. It is difficault to give a characterization of such graphs which have large chromatic number. G We shall concentrate on finding efficient ways of colouring a graph. How should one try to colour the vertices of a graph with colours 1,2,,n using as few colours as possible? Colouring a 3colourable graph Graph Theory 7 33 Graph Theory 7 34 It is not to hard to improve the efficiency of the greedy algorithm: If we already know a subgraph H 0 which can be coloured with few colours, (H 0 ), then we may start our sequence with the vertices of H 0, colour H 0 in an efficient way and apply only then the algorithm to colour the remaining vertices. Theorem 9.8.: Let H 0 be a spanned subgraph of G and suppose every subgraph H satisfying H 0 H G, G V(H 0 ) V(H), contains a vertex x V(H) V(H 0 ) with d H (x) k. Then (G) max {k+1, (H 0 ) } Upper Bounds for the Chromatic Number. It is trivial. The only thing we need to do is to give the vertices in the appropriate order. Graph Theory 7 35 Theorem 9.9.: Let k = max H (H), where the maximum is taken over all spanned subgraphs of G.. Then (G) k+1. Let x n a vertex with d(x n ) k, and let H n1 = G {x n } By assumption H n1 has a vertex of degree at most k. Let x n1 be one of them and put H n2 = H n1 {x n1 }= G {x n,x n1 }. Continuing in this way we enumerate all vertices. The sequence x 1,x 2,,x n is such that each x j is joined to at most k vertices preceding it. Hence the greedy algorithm will never need colour k+2 to colour a vertex. Graph Theory
10 G: x 9 x 7 x 8 x 4 x 6 (G) k + 1 = 3. 3 k = max H (H) = 2 Graph Theory 7 37 x 5 x 1 x 2 x 3 Consequence 9.10.: (G) + 1, where = (G) is the maximum degree of G. This follows immediately from the fact that max Consequence 9.11.: If G is connected and not regular then clearly max (H) (G) 1,, and so (G). max H G max H G (H) (G). Theorem (Brooks, 1941): Let G be a connected graph with maximal degree. Suppose G is neither a complete graph nor an odd cycle. Then (G). (Lovász( Lovász,, 1973) We may assume that G is connected and regular. Furthermore,, we may assume that 3,, since a 2regular 3chromatic graph is an oddcycle cycle. Graph Theory 7 38 If G is 3connected, let x n be any vertex of G and let x 1, x 2 be two nonadjacent vertices in (x n ). If G is not 3connected, let x n be a vertex for which G x n is separable, has at least two blocks. Since G is 2connected, each endblock of G x n has a vertex adjacent to x n. Let x 1 and x 2 be such vertices belonging to different endblocks. In either case we have found vertices x 1,x 2 and x n such that G {x 1,x 2 } is connected, x 1 x 2 E(G) but x 1 x n E(G) and x 2 x n E(G). Let x n1 V V {x 1,x 2,x n } be a neighbour of x n, let x n2 be a neighbour of x n or x n1 etc. Then the order x 1,x 2,x 3,,x, n is such that each vertex other than x n is adjacent to at least one vertex following it. Thus the greedy algorithm will use at most colours,, since x 1 and x 2 get the same colour and x n is adjacent to both. Graph Theory 7 39 A blockcutvertex cutvertex tree and endblocks. Graph Theory
11 Although no necessary and sufficient condition for a graph to be k chromatic is known when k 3,, a plausible necessary condition has been proposed by Hajós (1961): A subdivision of a graph G is a graph that can be obtained from G by a sequence of edge divisions. Hajós' conjecture: If G is kchromatic, then G contains a subdivision of K k. For k = 1 and k = 2, the validity of Hajós conjecture is obvious. It is also easily verified k = 3, because a 3chromatic graph necessa ry contains an odd cycle, and every odd cycle is a subdivision of K 3. The k = 4 case was proved by Dirac in The conjecture has not yet been settled in general, and its resolution is known to be a very difficult problem. Graph Theory 7 41 A subdivision of K 4. Graph Theory 7 42 Theorem 9.13.: If G is 4chromatic, then G contains a subdividion of K 4. Let G be a 4chromatic graph. (Note that if some subgraph of G contains a subdivision of K 4, then so does G.) W.l.o.g.. we may assume that G is critical. Then G is a block with 3. If n = 4 then G is K 4, and the theorem is true. We will use induction on n. Assume the theorem is true for all four chromatic graphs with fewer than n vertices, and let n > 4. 4 First, we suppose that G has a vertex cut {u,v}. By Theorem 9.6, G has two {u,v{ u,v}components G 1 and G 2, where G 1 + uv is 4critical. Since V(G 1 + uv) < V(G), we can apply the induction hypothesis and deduce that G 1 + uv contains a subdivision of K 4. It follows that, if P is a (u,v( u,v)path in G 2, then G 1 P contains a subdivision of K 4. Hence, so does G, since G 1 P G. Graph Theory 7 43 Graph Theory
12 Now,, suppose that G is 3connected. Since length at least four. Let u and v be nonconsecutive vertices on C. 3, G has a cycle C of Since G {u,v}} is connected, there is a path P in G {u,v} connecting two components of C {u,v}. Assume that the two endvertices of P on C are x and y. Similarly, there is a path Q on G {x,y}. If P and Q have no vertex in common,, the C P Q is a subdivision of K 4. Otherwise, let w be the first vertex of P on Q,, and let P' denote (x,( w) section of P. Then C P' Q is a subdivision of K 4. Hence, in both cases, G contains a subdivision of K 4. G: C: x v u Q P' P w y Graph Theory 7 45 Graph Theory Chromatic Polynomials In the study of colourings, some insight can be gained by considering not only the existence of colourings but the number of such colourings. This approach was developed by Birkhoff (1912)) as a possible means of attacking the fourcolour conjecture. Let p k (G)) be the number of distinct colourings of G. p k (G) > 0 iff G is kcolourable. Two colourings are regarded as distinct if some vertex is assigned different colours in the two colourings. In case of colourings we do not speak about isomorphism: first we w label the vertices and two colourings are distinct if any labeled vertex has different colours in the two colourings: For example, a triangle (K( 3 ) has six distinct 3colourings. Even though there is exactly one vertex of each colour in each colouring, we still regard these six colourings as distinct. If G is empty, then each vertex can be independently assigned any one of the k available colours. Therefore p k (G) = k n. On the other hand, if G is complete, then there are k choices of colour for the first vertex, k 1 for the second, k 2 for the third, and so on. In this case, p k (G) ) = k(k 1)(k 2)...(k n + 1). Graph Theory 7 47 Graph Theory
13 Theorem 9.14.: If G is simple, then p k (G) ) = p k (G e) p k (G e) for any edge of G. Let e = (u,v( u,v).. To each kcolouring of G e that assigns the same colour to u and v,, there corresponds a kcolouring of G e in which the vertex of G e formed by identifying u and v is assigned the common colour of u and v. This correspondence is a bijection. Therefore p k (G e) is precisely the number of kcolourings of G e in which u and v are assigned the same colour. Since each kcolouring of G e that assigns different colours to u and v is a kcolouring of G,, and conversely, p k (G) is the number of k colourings of G e in which u and v are assigned different colours. So, the statement of the theorem follows. Graph Theory 7 49 Graph Theory 7 50 Consequence : p k (G e) = p k (G ) + p k (G e) Theorem 9.16.: Let G be a graph with n p n 1 i n i G ( k ) = ( 1 ) aik, i = 0 1 vertices, m edges. Then where a 0 =1, a 1 = m and a i > 0 for every i, 0 i n 1. We apply induction on n + m. If n + m =1 then the assertions hold so we pass to the induction step. If m > 0 then we pick two adjacent vertices of G, say a and b.. Let e= (ab). Then V(G e) = n and E(G e) = m 1. Similarly, V( V(G e) = n 1 and E( E(G e) = m 1. So, we can use the induction hypothesis for both of the graphs G e and G e. Therefore p (G e ) = x k where b i > 0 for each i,, and n ( m 1)k n 1 + n 1 i= 2 i n i i ( 1) b k If m = 0, (i.e( i.e.. no edge in G) ) then every map f: V(G) is a colouring of G,, we have p k (G) ) = k n. {1,2, 1,2,,k} where c i > 0 for each i. p ( G e ) = k n 1 k i = 2 n 1 i n i i, ( 1 ) c k Graph Theory 7 51 Graph Theory
14 Hence, by Theorem 9.14 p ( G ) = p ( G e ) p ( G e ) = k = k = k k n n mk mk n 1 n 1 k + + n 1 i = 2 n 1 i = 2 ( 1 ) ( b i ( 1 ) a k i i i + c )k i n i where a i > 0 for each i. p k (G) will be called the chromatic polinomial of G., n i = Theorem 9.14 provides a means of calculating the chromatic polinomial of a graph recursively.it can be used in either of two ways: by repeatedly applying the recursion p k (G) ) = p k (G e) p k (G e), and thereby expressing p k (G) as a linear combination of chromatic polynomials of empty graphs. by repeatedly applying the recursion p k (G e) = p k (G )+ p k (G e) and thereby expressing p k (G) as a linear combination of chromatic polynomials of complete graphs. The first method is more suited to graphs with few edges, whereas the second one can be applied more effectively to graphs with many edges. Graph Theory 7 53 Graph Theory 7 54 = = = + = = = 3 +3 = = = k 4 3k 3 + 3k 2 k Graph Theory 7 55 = k(k 1)(k 2)(k 3)+2k(k 1)(k 2)+k(k 1) Graph Theory
15 The calculation of chromatic polynomials can sometimes be facilitated by the use of a number of formulae relating the chromatic polynomial of G to the chromatic polynomals of various subgraphs of G: if G is a cycle of length n,, then p k (G) ) = (k 1) n + (1)( n (k 1), if G is a tree then p k (G) = k(k 1) n1, if G has c components then p k (G) = p k (G 1 )... p k (G c ). Although many properties of chromatic polynomials are known, no one has yet discovered which polynomials are chromatic. Read (1968) conjectured that the sequence of coefficients of any chromatic polynomial must first rise in absolute value and then fall in other words, that no coefficient may be flanked by two coefficients having greater absolute value. The condition is not sufficient: k 4 3k 3 + 3k 2 is not chromatic polynomial of any graphs. No good algorithm is known for finding the chromatic polynomial of a graph. It would provide an efficient way to determine the chromatic number. Graph Theory 7 57 Graph Theory 7 58 One of the most famous problem in graph theory is the four colour problem: prove that every plane graph is 4colourable. The weaker assertion is almost immediate consequence of Euler's formula: Theorem 9.17.: Every plane graph is 5colourable Colouring of Planar Graphs. Let us suppose that the assertion is false,, and let G be 6colourable plane graph with minimal number of vertices. Then, by the Corollary 7.6, G has a vertex x of degree at most 5. Put H = G x. Then H is 5colourable, say with colours 1,2,3,4,5. Graph Theory 7 59 Each of these colours must be used to colour at least one neighbour hood of x, otherwise the missing colour could be used to colour x, and so G would be 5colourable. So, we may assume that d(x)=5,, and we denote them by x 1,x 2,,x 5, and we colour x i by i, i=1,2,,5. Denote by H(i,j) the subgraph of H spanned by the vertices of colour i and j: H(i,j) ) = max { H' H : v V(H') c(v) ) = i c(v) ) = j} j Suppose first that x 1 and x 3 belong to distinct components of H(1,3). Interchanging the colours 1 and 3 in the component of x 1 we obtain another colouring of H. However,, in this 5colouring both x 1 and x 3 get colour 3,, so 1 is not used to colour any of the vertices x 1,x 2,,x n. So, x 1 is colourable with the colour 1, which is a contradiction. Graph Theory
16 So, we can suppose that x 1 and x 3 belong to the same component.of H(1,3). Then, there is an x 1 x 3 path P 13 in H whose vertices are coloured 1 and 3. H: G: Similarly,, it is also true that x 2 and x 4 are in the same component and they can be coloured by 2 and 4,, and there exists an x 2 x 4 path P 24 in H. x 5 x 4 x x 3 Is this possible? The cycle x x 1 P 13 x 3 of G separates x 2 from x 4 but P 24 cannot meet this cycle. So, G is not a planar graph, which contradicts our assumption. x 1 x2 = 1 = 2 = 3 = 4 = 5 Graph Theory 7 61 Graph Theory 7 62 Is there any planar graph 3colorable? Not every planar graph is 3colourable. The simplest example is the complete graph K 4. Let 0 = max { (G): G is planar}. The consequence of the 5colour theorem: ? The original form of the four colour problem was posed by: Francis Guthrie in 1852: : show that every olane map can be coloured with four colours. His teacher,, de Morgan, circulated the problem among his colleagues, but the problem was first made popular in 1878 by Caley, who mentioned it before the Royal Society. The firstproofs proofs were given by Kempe (1879( 1879) ) and Tait (1880). Heawood's refutation of Kempe's proof was published in He modified the proof to obtain the five colour problem. Petersen in 1890 proved that Kempe's proof contained false assumptions: He proved that the four colour problem is equivalent to the conjecture that every cubic graph has edge chromatic number three. Graph Theory 7 63 Graph Theory
17 Birkhoff's introduction of the chromatic polynomial at the beginning of the 20. century contributed to the solution. During the 20. century a lot of works were made by various authors giving lower bounds on the order of a possible counterexample. In 1943 Hadwiger made a deep conjecture containing the four colour problem as a special case: : if (G) = k then G is contractible to K k. The problem was at last solved by Appel and Haken in 1976 using fast electronic computer and considering more than 1900 reducible configurations to complete the proof. Graph Theory 7 65 Graph Theory Edge Colouring Problem 2.: Each of n businessmen wishes to hold confidental meetings with some of the others. Assuming that each meeting lasts a day and at each meeting exactly two businessmen are present,, in how many days can the meeting be over? The graphtheoretical model: : Let H be such a graph whose vertices correspond to the n businessmen and two vertices are adjacent iff the two businessmen wish to hold a meeting. What is the minimal value of k for which E(H) can be partitioned into k classes, say E 1,E 2,,E k, such that no two edges from the same class are adjecent? Graph Theory 7 67 Graph Theory
18 A kedge colouring C of a loopless graph G is an assignment of k colours, 1, 2,..., k, to the edges of G. The colouring C is proper if no two adjacent edges have the same colour. G: A proper kedge colouring is a kedge colouring (E( 1, E 2,..., E k ) in which each subset E i is a matching. A graph G is kedgecolourable if G has a proper kedgecolouring. Every loopless graph G is E edgecolourable, if G is kedgecolourable,, then G is also ledgecolourable for every k l E. The (edge) chromatic number '(G),, of a loopless graph G,, is the minimum k for which G is kedgecolourable. G is kedgechromatic if '(G) = k. Graph Theory 7 69 A proper 5edge colouring of a graph G. Graph Theory 7 70 The edge chromatic number of a graph G is at least as large as the maximum degree (G) max x d(x), so '(G) (G). d c e g f b a Since for the example we can not give a proper 3edge colouring,, so the inequality may be strict. The edge set E(G) of a bipartite graph can be partitioned into (G) classes of independent edges, so '(G) = (G). What about the upper bounds? E 1 = {a,g{ a,g} E 2 = {b,e{ b,e} E 3 = {c,f{ c,f} E 4 = {d}{ A graph which has a proper 4colouring. Graph Theory 7 71 Graph Theory
19 Theorem 9.18.: '(G) 2(G) 1. Each edge is adjacent to at most 2((G) (G) 1) edges. Apply the result of the Theorem 9.9 we will get the desired statement. Theorem 9.19.: If (G) 3, then '(G) 2(G) 2. The statement of the theorem follows immediately from the Brooks Theorem. Graph Theory 7 73 Theorem (Vizing( Vizing): (G) '(G) (G) + 1. Let = (G) ) and assume that we have used 1,2,,+1 to colour all but one of the edges. We are ready if we can show that this uncoloured edge can also be coloured one of the +1 used coloures. We say that a colour is missing at a vertex z if no edge incident with z gets that colour. If z is incident with d'(z) d(z) edges that have been coloured, then +1 d'(z) colours are missing at z. Since d'(z) > 0 then at each vertex at least one colour is missing. Our aim is to move around the colours and the uncoloured edge in such a way that a colour will be missing at both endvertices of the uncoloured edge, enabling us to complete the colouring. Graph Theory 7 74 G: x y 1 xy 1 is an uncoloured edge. Let s (red) and t 1 (blue) be missing colours at x and y 1 respectivelly. We shall construct a sequence of edges xy 1, xy 2,, and a sequence of colours t 1, t 2, such that t i is missing at y i and xy has colour t i. xy i+1 has Suppose we have constructed xy 1,,xy i and t 1,,t i. There is at most one edge xy of colour t i. If y {y 1,,y i }, we put y i+1 = y and pick a colour t i+1 missing at y i+1 otherwise we stop the sequence. These sequences have to terminate at most (G) terms.. Let xy 1,, xy h and t 1,,t,t h be the complete sequences. What kind of reasons may occur that a sequence terminates? No edge xy has colour t h. Then recolour the edges xy i, i < h, h giving xy i the colour t i. In the colouring we obtain every edge is coloured, except xy h. Since t h occurs neither at x nor at y h, we may complete the colouring by assigning t h to xy h. i+1, Graph Theory 7 75 Graph Theory
20 For some j < h the edge xy j has colour t h. Recolour the edges xy i, i < j, j giving xy i the colour t i. In this colouring the uncoloured edge is xy j. Let H(s,t h ) be the subgraph of G formed by the edges of colour s and t h. Each vertex of H(s,t h ) is incident with at most 2 edges in H(s,t h ) (one of colour s and the other of colour t h ) so the components of H(s,t h ) are paths and cycles. Each of the vertices x, y j and y h has degree at most 1 in H(s,t h ), so they cannot belong to the same component of H(s,t h ). So, at least one of the following two cases has to hold: The vertices x and y j belong to distinct components of H(s,t h ). Interchange the colours s and t h in the component containing y j. Then the colour s is missing at both x and y j, so we may complete the colouring by giving xy j the colour s. The vertices x and y h belong to distinct components of H(s,t h ). Continue the recolouring of the edges incident with x by giving xy i the colour t i for each i < h, thereby making xy h the uncolour ed edge. So, xy h will this recolouring does not involve edges of colours s and t h, so H(s,t h ) has not been altered. Switch around the colours in the component containing y h. This switch makes sure that the colour s is missing at both x and y h, so we can use s to colour the so far uncoloured edge xy h. Graph Theory 7 77 Graph Theory 7 78 The maximum number of edges joining two vertices in G is called the multiplicity of G,, and it is denoted by (G). Vizig proved a more general theorem: If G is loopless then (G) '(G) (G) + (G). This theorem is best possible in the sense that, for any, there exists a graph G such that '(G) = (G) + (G). It leaves open one interesting question: which simple graphs satisfy '(G) = (G) (G)? A nonempty graph is said to be of class one if '(G) = (G) and of class two if '(G) = (G) + 1. C n (n 3) ) is of class one if n is even and of class two if n is odd. K n is class one if n is even and of class two if n is odd. Every regular graph of odd order is of class two. It is not true that every regular graph of even order is of class s one: the Petersen graph is of class two. There are considerable more class one graphs than class two graphs: Erds and Wilson (1977( 1977) ) proved that the probability that a graph of order n is of class one approches 1 as n approaches infinity. An independent set of edges in a graph G is a set of edges, each two of which are nonadjacent. The edge independent number 1(G) is the maximum cardinality among the independent sets of edges of G. Graph Theory 7 79 Graph Theory
21 Theorem 9.23.: Let G be a graph of size m.. If m > (G) is of class two. Assume that G is of class one. Then '(G) = (G) (G).. Let a '(G)colouring of G be given. Each edgecolour class of G has at most Therefore m(g) 1(G) (G). A graph G is overfull if m > (G) n/2 1(G) edges. Corollary 9.24.: Every overfull graph is of class two. 1(G) then G A graph G is minimal with respect to edge chromatic number if (G e) ) = (G) 1 for all e E(G). Theorem 9.25.(Vizing, 1965): Let G be a connected graph of class two that is minimal with respect to edge chromatic number. Then every vertex of G is adjacent to at least two vertices of degree (G) (G). In particular, G contains at least three vertices of degree (G) (G). Theorem 9.26.(Vizing, 1965): Let G be a connected graph of class two that is minimal with respect to edge chromatic number. If u and v are adjacent vertices with deg u = k, then v is adjacent to at least (G) k + 1 vertices of degree (G) (G). Graph Theory 7 81 Graph Theory 7 82 What is the situation with planar graphs? It is easy to find planar graphs G of class one for which (G) = d for each d 2 since all star graphs are planar and of class one. There exist planar graphs G of class two with (G) = d for d = 2, 3, 4, 5. It is not known whether there exist planar graphs of class two having maximum degree 6 or 7. Vizing proved that if G is planar and (G) (G) 8, then G must be of class one. (G) = 2 (G) = 3 (G) = 4 (G) = 5 Graph Theory 7 83 Graph Theory 7 Planar graphs of class two
22 There is a colouring that assigns colours to both the vertices and a edges of a graph: a total colouring of a graph G is an assignment of colours to the elements (vertices and edges) of G so that adjacent elements and incident elements of G are coloured differently. A ktotal colouring is a total colouring that uses k colours. The minimum k for which a graph G admits ktotal colouring is called the total chromatic number of G and denoted by T (G). It is trivial that T (G) (G) ) + 1. Total Colouring Conjecture: for every graph G, T (G) (G) ) + 2. The Timetabling Problem The problem: : In a school,, there are m teachers X 1, X 2,, X m, and n classes Y 1, Y 2,, Y n. Given that teacher X i is required to teach class Y for p ij periods.. Schedule a complete timetable in the minimum possible number of periods. We have, at least, the assumption that in any one period each teacher can teach at most one class,, and each class can be thaght by at most one teacher. The model: : we represent the teaching requirements by a bipartite graph G with bipartition (X,Y), where X = {x 1, x 2,, x m } Y = {y{ 1, y 2,, y n } and vertices are joined by p ij edges. Graph Theory 7 85 Graph Theory 7 86 In this model a teaching schedule for one period corresponds to a matching in the bipartite graph and, conversely,, each matching corresponds to a possible assignement of teachers to classes for one period. Therefore, our problem is to partition the edges of G into as few matching as possible or, equivalently,, to properly colour the edges of G with as few colurs as possible. Since G is bipartite,, we know,, that ' =. Case A.: if no teacher teaches for more than p periods, and if no class is thaught for more than p periods, the teaching requirements can be scheduled in a pperiod period timetable. Since we need to solve a matching problem, there is a good algorithm for constructing such a timetable. Case B : if only a limited number of classrooms are available then how many periods are needed to schedule a complete timetable? Suppose that altogether there are l lessons to be given,, and that they have been scheduled in a pperiodperiod timetable. Since this timetable requires an average of l/p lessons to be given in each period,, so at least l/p rooms are occupied in any one period. This is the consequence of the following theorem. Graph Theory 7 87 Graph Theory
23 Lemma 9.27.: Let M and N be disjoint matchings of G with M > N. Then there are disjoint matchings M' and N' of G such that M' = M 1,, N' = N + 1 and M' N' = M N. Consider the graph H = G[M N]. Then each component of H is either an even cycle with edges alternately in M and N, or else a path with edges alternately in M and N. Since M > N,, some path component P of H must start and end with edges of M. Let P = v 0 e 1 v 1 e 2 e 2n+1 v 2n+1, and set M' = (M \ {e 1,e 3,, e 2n+1 }) {e{ 2, e 4,, e 2n } M' = (M \ {e 2,e 4,, e 2n }) {e{ 1, e 3,, e 2n+1 } Then M' and N' are matchings of G that satisfy the conditions of the lemma. Graph Theory 7 89 Theorem 9.28.: If G is bipartite,, and if p,, then there exist p disjoint matchings M 1, M 2,, M p of G such that E(G) = M 1 M 2 M p (*) and for 1 i p [ E(G) / p] M i E(G) / p. (**) First we note that the last inequalities say that any two matchings M i and M j differ in size by at most one. Let G be a bipartite graph. Then the edges of G can be partitioned into matchings M 1,M 2,...,M. Therefore,, for any p,, there exist p disjoint matchings M 1,M2,..., M p (with M i = for i > ) ) such that E ( G ) = M M 1 2 M p Graph Theory 7 90 By repeatedly applying Lemma 9.27 to pairs of these matchings that differ in size by more than one, we eventually obtain p disjoint matchings M 1, M 2,,M, p of G satisfying (*) and (**), as required. Let us consider the following example: : suppose there are four teachers and five classes,, and the teaching requirement matrix is given. One possible 4period timetable is given on the next page. We can represent this timetable by a decomposition into matchings of the edge set of the bipartite graph G coresponding to P. From the timetable we see that four classes are taught in period 1, and so, four rooms are needed. Graph Theory 7 91 However, E(G) = 11 and so, by Theorem 9.28,, a 4period time table can be arranged so that in each period either 2 ( = [11[ / 4] ) or 3 ( = {11{ / 3} ) classes are thaught. Let M 1 be the matching with the red lines and M 4 the matching with the green lines. ( M 4 = 4 and M 1 = 2.) We can now find a 4period 3room timetable by considering G[ M 1 M 4 ]. G[ M 1 M 4 ] has two components, each consisting of a path of length three. Both paths start and end with red edges and so, by interchanging the matchings on one of the two paths, we shall reduce the red matching to one of three edges, and the same time increase the green matching to one of three edges. This gives the revised timetable where only three rooms are needed at any time. Graph Theory
24 Suppose now that there are just two rooms available. Theorem 9.28 tells us that there must be a 6period timetable that satisfies our requierements since {11/6{ 11/6} } = 2.. Such a timetable is given on the next figure X 1 X 2 X 3 X 4 Y 4 Y 3 Y 1 Y 1 Y 2 Y 4 Y 4 Y 3 Y 2 Y 4 Y 5 P = Y 1 Y 2 Y 3 Y 4 Y X 1 X 2 X 3 X 4 X 1 X 1 Y 1 Y 1 Y 2 X 2 X 2 Y 4 Y 4 X 3 X 4 X 3 Y 3 Y 4 Y 4 Y 2 X Y 5 Y 3 In practice, most problems on timetabling are complicated by preassignements (that is, conditions specifying the periods during which certain teachers and classes must meet). Such problem has been studied by Dempster and de Werra. Graph Theory 7 93 Y 1 Y 2 Y 3 Y 4 Y 5 Graph Theory 7 94 X 1 X 2 X 3 X 4 G[ M 1 M 4 ]. Y 1 Y 2 Y 3 Y 4 Y 5 X 1 X 2 X 3 X 4 Y 1 Y 2 Y 3 Y 4 Y 5 X 1 X 2 X 3 X Y 4 Y 2 Y 4 Y 1 Y 3 Y 1 Y 4 Y 5 Y 3 Y 2 Y 4 Graph Theory
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