Basic Notions on Graphs. Planar Graphs and Vertex Colourings. Joe Ryan. Presented by

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1 Basic Notions on Graphs Planar Graphs and Vertex Colourings Presented by Joe Ryan School of Electrical Engineering and Computer Science University of Newcastle, Australia

2 Planar graphs Graphs may be drawn in more than one way: A graph G is planar if it can be drawn in the plane in such a way that no two edges meet except at a vertex with which they are both incident. Any such drawing is a plane drawing of G. A graph G is non-planar if no plane drawing of G exists. 2

3 Planar graphs Three plane drawings of K 4 : The cube and the dodecahedron are also planar graphs: 3

4 Planar graphs The graph below is also planar: Problem Determine which of the following graphs are planar and, for those that are, give a plane drawing. 4

5 Chip Design Planarity is related to practical problems in constructing circuits where printed connectors may not cross. 5

6 Planar graphs K 3,3 is non-planar: every drawing must contain at least one crossing. Next we need to insert the edges ub, vc, wa: only one can be drawn inside, two must be outside: impossible without a crossing. 6

7 Planar graphs When studying planarity, we may restrict our attention to simple graphs. Problem Decide whether each of the following statements is true or false. (a) Every subgraph of a planar graph is planar. (b) Every subgraph of a non-planar graph is non-planar. (c) Every graph that contains a planar subgraph is planar. (d) Every graph that contains a non-planar subgraph is non-planar. 7

8 Euler s formula Euler s formula relates the number of vertices, edges and faces of a plane drawing of a planar graph. Every plane drawing of a planar graph divides the plane into a number of regions. K 4 divides the plane into 4 regions: 8

9 Euler s formula K 2,5 divides the plane into 5 regions: Let G be a planar graph. Then any plane drawing of G divides the set of points of the plane not lying on G into regions, called faces; one face is of infinite extent and is the infinite face. 9

10 Euler s formula Let G be a connected planar graph, and let f be any face of a plane drawing of G. Then the degree of f, denoted by deg f, is the number of edges encountered in a walk around the boundary of the face f. If all faces have the same degree g then G is face-regular of degree g. Handshaking Lemma for Planar Graphs In any plane drawing of a planar graph, the sum of the face degrees is equal to twice the number of edges. Proof. Each edge has two sides so it contributes exactly 2 to the sum of the face degrees. 10

11 Euler s formula Problem Verify the planar version of the handshaking lemma for each of the following plane drawings of planar graphs. Problem For each of the plane drawings above, count the number of vertices, edges and faces, and find the value of number of vertices number of edges + number of faces. 11

12 Euler s formula Euler s Formula for Planar Graphs Let G be a connected planar graph, and let n,m and f denote, respectively, the numbers of vertices, edges and faces in a plane drawing of G. Then n m + f = 2. Proof. A plane drawing of a connected planar graph G can be constructed by taking a spanning tree of G and adding edges to it, one at a time. We prove Euler s formula by showing that (a) For any spanning tree, n m + f = 2. (b) Adding an edge does not change the value of n m + f. 12

13 Euler s formula Problem Verify Euler s formula for each of the following graphs. (a) octahedron graph; (b) wheel with k spokes; (c) complete bipartite graph K 2,k ; (d) graph formed from the vertices and edges of a kxk square lattice. 13

14 Euler s Formula and Duality Theorem Let G be a plane drawing of a connected planar graph with n vertices, m edges and f faces. Then G* has f vertices, m edges and n faces. Proof. It follows directly from the construction of G* that G* has f vertices and m edges. If G* has f* faces then, by applying Euler s formula, to both G and G* we get for G: n m + f = 2 for G*: f m + f* = 2 and so f* = n. 14

15 Euler s formula Next we show how Euler s formula can be used to prove that some graphs are non-planar. Corollary 1 Let G be a simple connected planar graph with n ( 3) vertices and m edges. Then m 3n 6. Proof. Since a simple graph does not have multiple edges or loops then the degree of each face is at least 3. From the Handshaking Lemma for planar graphs we get 3f 2m. Substituting for f from Euler s formula we get 3m 3n + 6 2m and so m 3n 6 15

16 Euler s formula Example K 5 is non-planar. Proof is by contradiction. Suppose K 5 is planar. Since K 5 is a simple connected graph with 5 vertices and 10 edges, by Corollary 1, 10 (3x5) - 6 = 9 which is false. Therefore, K 5 is non-planar. Note that we cannot use Corollary 1 to prove that the complete bipartite graph K 3,3 is non-planar since 9 (3x6) - 6 = 12 does hold. We will need another argument. 16

17 Euler s formula Corollary 2 Let G be a simple connected planar graph with n ( 3) vertices, m edges and no triangles. Then m 2n 4. Proof. Consider a plane drawing of a simple connected planar graph G with f faces and no triangles. The degree of each face of such a graph is at least 4. It follows from the Handshaking Lemma for planar graphs that 4f 2m. Substituting for f from Euler s formula we get 2m 2n + 4 m and so m 2n 4. 17

18 Euler s formula Example K 3,3 is non-planar. Proof is by contradiction. Suppose K 3,3 is planar. Since K 3,3 is a simple connected graph with 6 vertices and 9 edges, and no triangles, by Corollary 2, 9 (2x6) - 4 = 8 which is false. Therefore, K 3,3 is non-planar. Corollary 3 Let G be a simple connected planar graph. Then G contains a vertex of degree 5 or less. Problem Prove Corollary 3. 18

19 Kuratowski s theorem A contraction of a graph is the result of a sequence of such edge contractions. 19

20 Kuratowski s theorem Kuratowski s Theorem A graph is planar if and only if it contains no subgraph that has K 5 or K 3,3 as a contraction. In practice, the Theorem is not used for testing planarity. Instead, several observations are used: A disconnected graph is planar if and only if each of its components is planar. 20

21 Kuratowski s theorem A graph that has a cut vertex (vertex whose removal disconnects the graph) is planar if and only if each of the subgraphs obtained when the graph is disconnected at the cut vertex is planar. A graph that has loops or multiple edges is planar if and only if the graph obtained by removing the loops and coalescing the multiple edges is planar. 21

22 Vertex colouring Example: Storing chemicals. Storing chemicals in a warehouse some chemicals react violently with each other so we need to separate pairs of chemicals. What is the smallest number of areas needed? 22

23 Vertex colouring Representing the situation by a graph: vertices are chemicals, edges between two chemicals indicate pairs of chemicals that must be kept apart. We can split the set of chemicals into 4 disjoint subsets Corresponding to the 4 areas: {a,e},{b,f},{c},{d,g}. 23

24 Vertex colouring Let G be a simple graph. A k-colouring of G is an assignment of at most k colours to the vertices of G in such a way that adjacent vertices are assigned different colours. If G has a k-colouring then G is k-colourable. The chromatic number of G, denoted by χ(g), is the smallest number k for which G is k-colourable. Note that the definition is given only for simple graphs. We usually show a k-colouring by writing the numbers 1,2,,k next to the vertices. Here χ(g) = 3. 24

25 Vertex colouring Problem Determine χ(g) for each of the following graphs G. Hint. For each graph, devise a suitable colouring and explain why there is no colouring with fewer colours. Problem What can you say about the graphs G for which χ(g) = 1? χ(g) = 2? 25

26 Vertex colouring Problem Write down the chromatic number for each of the following graphs. (a) the complete graph K n ; (b) the complete bipartite graph K r,s ; (c) the cycle graph C n (n 3); (d) a tree. Problem Decide whether each of the following statements about a graph G is true or false, and give a proof or counter-example, as appropriate. (a) If G contains the complete graph K r as a subgraph then χ(g) r? (b) If χ(g) r then G contains the complete graph K r as a subgraph. 26

27 Vertex colouring Given a graph G, how can we determine the chromatic number? Upper bound for χ(g) can be found by constructing an explicit colouring for the vertices of G. Lower bound for χ(g) can be found by finding the number of vertices in the largest complete subgraph of G. Note that if a graph has n vertices then χ(g) n and χ(g) = n only if G = K n. 27

28 Vertex colouring Theorem: Let G be a simple graph whose maximum vertex degree is d. Then χ(g) d + 1. Proof is by mathematical induction on n, the number of vertices of G. The statement is true for K 1 since χ(k 1 ) = 1 and d = 0. Next we assume that χ(h) d + 1 holds for all simple graphs H with fewer than n vertices. We will show that χ(g) d + 1 holds for all simple graphs G with n vertices. Let H be any graph obtained from G by removing a vertex v and the edges incident with it. 28

29 Vertex colouring Proof (cont.) By assumption, χ(h) d + 1 so the graph H is (d+1)-colourable. We can now obtain a (d+1)-colouring of G by colouring v with any not assigned to the (at most d) vertices adjacent to v Since these vertices can be coloured with at most d colours, it follows that χ(g) d

30 Vertex colouring Brooks Theorem Let G be a connected simple graph whose maximum vertex degree is d. If G is neither a cycle graph with an odd number of vertices, nor a complete graph, then χ(g) d. Proof is omitted. To illustrate the use of Brooks theorem, consider the graph G on the right. We already know that χ(g) 4 since G contains the complete graph K 4. By Brooks theorem, χ(g) 4. Therefore, χ(g) = 4. 30

31 Vertex colouring This approach does not always work. For example, if G is a complete bipartite graph K 1,12 then Brooks theorem gives χ(g) 12 but the actual value is χ(g) = 2. 31

32 Vertex colouring Problem For each of the following graphs G, write down (a) the lower bound for χ(g) given by the size of the largest complete subgraph in G; (b) the upper bound for χ(g) given by Brooks theorem; (c) the actual value of χ(g) and a colouring using χ(g) colours. 32

33 Vertex colouring To find the chromatic number χ(g) of a simple graph G. Try to find an upper bound and a lower bound that are the same; then χ(g) is equal to this common value. Possible upper bounds for χ(g): the number of colours used in an explicit vertex colouring of G; the number n of vertices in G; d+1, where d is the maximum vertex degree in G (Theorem); d, where d is the maximum vertex degree in G, provided that G is not C n (for odd n) or K n (Brooks theorem). Possible lower bounds for χ(g): the number of vertices in the largest complete subgraph in G. 33

34 Vertex colouring of planar graphs Six Colour Theorem for Planar Graphs The vertices of any simple connected planar graph G can be coloured with 6 (or fewer) colours in such a way that adjacent vertices are coloured differently. Proof is by mathematical induction on n, the number of vertices of G. The statement is trivially true when n=1. Assuming that the vertices of all simple connected planar graphs with fewer that n vertices can be coloured with 6 or fewer colours, we will show that the vertices of all simple connected planar graphs with n vertices can be coloured with 6 or fewer colours. Let G be a simple connected planar graphs with n vertices. 34

35 Vertex colouring of planar graphs Proof (cont.) By Corollary 3, G contains a vertex v of degree 5 or less. We remove v and its incident edges. The resulting planar graph H has fewer than n vertices and so, By our assumption, H (or each component of H if H is disconnected) can be vertex coloured with 6 colours In such a way that adjacent vertices are coloured differently. 35

36 Vertex colouring of planar graphs Proof (cont.) We now reinstate the vertex v. Since v has at most 5 neighbours and 6 colours are available, there is a spare colour that can be used for v. 36

37 Vertex colouring of planar graphs Five Colour Theorem for Planar Graphs The vertices of any simple connected planar graph G can be coloured with 5 (or fewer) colours in such a way that adjacent vertices are coloured differently. Proof is omitted Four Colour Theorem for Planar Graphs The vertices of any simple connected planar graph G can be coloured with 4 (or fewer) colours in such a way that adjacent vertices are coloured differently. Proof is omitted. 37

38 Map Colouring Example We wish to determine the smallest number of colours needed to colour the countries of a map in such a way that any two countries with a common boundary are coloured differently. How many colours are needed to colour this map? 38

39 Map Colouring Example (cont.) It can be done with 4 colours. 39

40 Map Colouring Example (cont.) We represent the situation as a vertex colouring problem. 40

41 Algorithm for vertex colouring No efficient algorithm is known for vertex colouring. Here we present a heuristic algorithm that in practice often works well. Greedy Algorithm for Vertex Colouring START with a graph G and a list of colours 1,2, Step 1 label the vertices a,b,c, in any manner. Step 2 identify the uncoloured vertex labelled with the earliest letter in the alphabet; colour it with the first colour in the list not used for any adjacent coloured vertex. Repeat Step 2 until all vertices are coloured, then STOP. A vertex colouring of G is obtained. The number of colours used depends on the labelling chosen for the vertices (Step 1) 41

42 Algorithm for vertex colouring Illustration A Find a vertex colouring of the following graph: Step 1. We label the vertices a,b,c,d,e,f. 42

43 Algorithm for vertex colouring Illustration A (cont.) Step 2. We successively colour vertex a with colour 1, vertex b with colour 2, vertex c with colour 1, vertex d with colour 2, vertex e with colour 3, vertex f with colour 4. All the vertices are now coloured so we STOP. 43

44 Algorithm for vertex colouring Illustration B Find a vertex colouring of the following graph: Step 1. We label the vertices a,b,c,d,e,f. 44

45 Algorithm for vertex colouring Illustration B (cont.) Step 2. We successively colour vertex a with colour 1, vertex b with colour 1, vertex c with colour 1, vertex d with colour 2, vertex e with colour 3, vertex f with colour 2. All the vertices are now coloured so we STOP. Using the same graph and the same greedy algorithm, we have obtained two different vertex colourings. 45

46 Algorithm for vertex colouring Exercise: Use the greedy algorithm to colour the vertices of the following graph G, using each of the given labellings. What is the actual value of χ(g)? 46

47 Vertex decomposition Some of the most interesting problems in graph theory involve the decomposition of a graph G into subgraphs of a particular type. In some problems we split the set of vertices into disjoint subsets; this is called a vertex decomposition of G. Example. A vertex decomposition of the graph below: {a,b,c}, {d,e,f,g}, {h} disjoint subsets that correspond to the components of G. 47

48 Storing Chemicals (colouring) Example: A chemical manufacturer wishes to store chemicals a,b,c,d,e,f,g in a warehouse. It is necessary to keep some chemicals separate from others. To determine the smallest number of areas needed, we draw a graph: 2 vertices are joined by an edge whenever the corresponding chemicals must be kept separate. The assignment of chemicals to areas is a vertex colouring problem: colours correspond to areas. The colouring gives rise to a vertex decomposition: {a,e}, {b,f}, {c}, {d,g}. The minimum number of subsets is the chromatic number of the corresponding graph. 48

49 Map Colouring Problem Consider the following map. (a) Find a colouring of this map by trial and error. (b) Draw the corresponding graph, and show how the 4-colouring in part (a) leads to a vertex decomposition of this graph in which no two vertices in the same subset are adjacent. 49

50 Communication Links (domination) Suppose that communication links are to be set up between a number of cities, and transmitting stations are to be built in some of these cities so that each city can receive messages from at least one transmitting station. How can this be done using the smallest possible number of transmitting stations? This can be represented by a graph whose vertices correspond to the cities, and edges correspond to pairs of cities that can communicate directly with each other. Each city must either contain a transmitting station or be adjacent to a city that contains a transmitting station: we wish to find a set of vertices that (between them) are adjacent to all other vertices of the graph. 50

51 Communication Links (domination) Example: Location of Transmitting Stations Suppose that this graph corresponds the communication links between 6 cities. The transmitting stations can be located at A,C,E since each of the other vertices is adjacent to at least one of A,C,E. The vertices A,C,E form a dominating set. We obtain a vertex decomposition into subset of cities served by the same transmitting station: {A,B,F}, {C,D}, {E}. A minimum dominating set contains the smallest number (called dominating number) of vertices. 51

52 Communication Links (domination) Problem Find a minimum dominating set in each of the following graphs. In each case, write down a vertex decomposition in which each subset contains a vertex adjacent to all the other vertices in that subset. 52

53 Revision (and terms to know) Planar Graphs Euler s Formula Kuratowski s Theorem Vertex colouring Vertex chromatic number Greedy algorithm Vertex decomposition 53

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