8.2 Paths and Cycles

Transcription

1 8.2 Paths and Cycles

2 Degree a b c d e f Definition The degree of a vertex is the number of edges incident to it. A loop contributes 2 to the degree of the vertex. (G) is the maximum degree of G. δ(g) is the minimum degree of G.

3 Degree Sum Theorem Let G be an (n, m) graph with vertex set {v 1, v 2,... v n }. Then n deg(v i ) = 2m i=1

4 Degree Sum Theorem Let G be an (n, m) graph with vertex set {v 1, v 2,... v n }. Then Proof. n deg(v i ) = 2m i=1 Consider vertex v i. deg(v i ) gives the number of edges incident to v i. But, each of these edges is incident to some other vertex. Since every edge is incident to exactly two vertices, each of the vertices is counted when we count the m edges.

5 Parity of Vertices Theorem For any graph, there must be an even number of odd vertices.

6 Parity of Vertices Theorem For any graph, there must be an even number of odd vertices. Proof We have previously shown that the degree sum of any graph is even. We can partition the vertices of the graph into even and odd vertices, say V e and V o. Consider deg(vi ) = deg(v i ) + deg(v i ) v i V e v i V o

7 Proof (cont.) Proof. We know the left-hand side is even, so the right hand side must be even as well. Since v i V e deg(v i ) is the sum of even integers, it must be so that v i V o deg(v i ) must be even as well. But, since this is the sum of all odd integers, it can only be even if there is an even number of such integers.

8 Degree Sequence Definition The degree sequence of a graph is a list of degrees of each vertex. There will be n numbers listed if the order of G is n. a b c d e f Definition In a regular graph, all vertices have the same degree.

9 Graphical Sequence Is {6, 6, 5, 4, 3, 3, 2, 2, 1, 1, 1, 1} the degree sequence of a graph? Theorem A sequence S : d 1, d 2,..., d n of nonnegative integers with d 1 d 2... d n, n 2, d 1 1 is graphical iff the sequence S 1 : d 2 1, d 3 1,..., d d1 +1 1, d d1 +2,..., d n is graphical.

10 Examples Determine if the following sequences are graphical. 1 {8, 7, 7, 7, 7, 6, 5, 5} 2 {5, 5, 5, 5, 5, 5} 3 {7, 6, 5, 4, 3, 2, 1}

11 Examples Determine if the following sequences are graphical. 1 {8, 7, 7, 7, 7, 6, 5, 5} 2 {5, 5, 5, 5, 5, 5} 3 {7, 6, 5, 4, 3, 2, 1} Notice that the last sequence is all different integers. This is called an irregular graph.

12 Irregular Graph Theorem Irregular graphs do not exist.

13 Irregular Graph Theorem Irregular graphs do not exist. Proof. Suppose an irregular graph of order n exists with degree sequence {n 1, n 2,..., 0}. Then, the the vertex of degree n 1 would have to be adjacent to each of the other n 1 vertices of the graph, but the last vertex in the sequence already has degree 0.

14 Distance in Graphs Definition Let u and v be (not necessarily distinct) vertices of a graph G. A u v walk is an alternating finite sequence u = u 0, e 1, u 1, e 2, u 2,..., u k 1, e k, u k = v of vertices and edges, beginning at u and ending at v, such that e i = u i 1 u i for i = 1,..., k. The number k is the length of the walk. If u = v then the walk is closed; otherwise the walk is open. a b c d e f g h i j k

15 Distance in Graphs Definition A u v trail is a u v walk where no edge is repeated. A u v path is a u v walk where no vertex is repeated. a b c d e f g h i j k Definition A circuit is a nontrivial closed trail A cycle is circuit whose vertices are distinct, n 3.

16 Subgraphs Definition A subgraph H = (W, F) of a graph G = (V, E) is a graph such that W V and F E. Definition An induced subgraph H of G is a graph such that if u, v are vertices of H, then {u, v} is an edge of H iff {u, v} is an edge of G.

17 Subgraphs Definition A subgraph H = (W, F) of a graph G = (V, E) is a graph such that W V and F E. Definition An induced subgraph H of G is a graph such that if u, v are vertices of H, then {u, v} is an edge of H iff {u, v} is an edge of G. G H (induced) a b a f c f c e d e

18 Subgraphs G H (not induced) a b a b f c c e d

19 Spanning Subgraphs Definition A spanning subgraph of a graph G contains all vertices but not necessarily all edges. That is, if V(G) = V(H).

20 Spanning Subgraphs Definition A spanning subgraph of a graph G contains all vertices but not necessarily all edges. That is, if V(G) = V(H). G spanning subgraph a b a b f c f c e d e d

21 Connectivity and Complements Definition A graph is connected if for any two distinct vertices, there exists a path between the vertices. Each connected piece of a graph is called a component.

22 Connectivity and Complements Definition A graph is connected if for any two distinct vertices, there exists a path between the vertices. Each connected piece of a graph is called a component. Definition The complement G of G is that graph with vertex set V(G) such that two vertices are adjacent in G only when the vertices are not adjacent in G. E(G) + E(G) = ( ) n 2

23 Connectivity and Complements Theorem If G is disconnected, then G is connected.

24 Connectivity and Complements Theorem If G is disconnected, then G is connected. Proof. If G is disconnected then G has at least two components. Within any component of G, any two vertices that are not adjacent in G will be adjacent in G. Also, in G, the vertices in any one component are adjacent to no vertices in any other component of G. Therefore, each vertex in any component of G will be adjacent to every vertex of every other component in G. Also, for every two vertices adjacent in G, they will not be adjacent in G but they will be adjacent to a common vertex (i.e. all vertices from other components of G) and will be connected.

25 K.. onigsberg, 1736

26 Graph Representation A B C D

27 New Definitions Definition An Eulerian trail is an open trail of G containing all edges and vertices.

28 New Definitions Definition An Eulerian trail is an open trail of G containing all edges and vertices. Definition An Eulerian circuit is a closed trail containing all edges and vertices.

29 New Definitions Definition An Eulerian trail is an open trail of G containing all edges and vertices. Definition An Eulerian circuit is a closed trail containing all edges and vertices. Definition A graph containing an Eulerian circuit is called Eulerian.

30 Examples Which contain Eulerian trails? circuits? Do you see any patterns as to when one does/does not exist?

31 More Examples

32 Conclusions 1 If all vertices are even, an Eulerian circuit exists

33 Conclusions 1 If all vertices are even, an Eulerian circuit exists 2 If there are exactly 2 odd vertices, an Eulerian trail exists that begins at one odd vertex and ends at the other one

34 Conclusions 1 If all vertices are even, an Eulerian circuit exists 2 If there are exactly 2 odd vertices, an Eulerian trail exists that begins at one odd vertex and ends at the other one 3 If there are more than two odd vertices, no Eulerian trail or circuit exists

35 Necessary Condition Theorem A graph G contains an Eulerian circuit if and only if the degree of each vertex is even.

36 Necessary Condition Theorem A graph G contains an Eulerian circuit if and only if the degree of each vertex is even. Proof of necessity Suppose G contains an Eulerian circuit C. Then, for any choice of vertex v, C contains all the edges that are incident to v. Furthermore, as we traverse along C, we must enter and leave v the same number of times, and it follows that deg(v) must be even.

37 Example K 5 is 4-regular and so has all even vertices. start

38 Notes Euler presented proof of necessity in 1736

39 Notes Euler presented proof of necessity in 1736 His paper did not include the proof of the converse

40 Notes Euler presented proof of necessity in 1736 His paper did not include the proof of the converse Proof of sufficiency was presented in 1873 by a German mathematician named Carl Hierholzer

41 Proof of Sufficiency Proof. We prove by induction on the number of edges. For graphs with all vertices of even degree, the smallest possible number of edges is 3 in the case of simple graphs, and 2 in the case of multigraphs. In both cases, the graph trivially contains an Eulerian circuit.

42 Proof of Sufficiency Proof. We prove by induction on the number of edges. For graphs with all vertices of even degree, the smallest possible number of edges is 3 in the case of simple graphs, and 2 in the case of multigraphs. In both cases, the graph trivially contains an Eulerian circuit. The induction hypothesis then says: Let H be a connected graph with k edges. If every vertex of H has even degree, H contains an Eulerian circuit.

43 Proof of Sufficiency Proof. We prove by induction on the number of edges. For graphs with all vertices of even degree, the smallest possible number of edges is 3 in the case of simple graphs, and 2 in the case of multigraphs. In both cases, the graph trivially contains an Eulerian circuit. The induction hypothesis then says: Let H be a connected graph with k edges. If every vertex of H has even degree, H contains an Eulerian circuit. Now, let G be a graph with k + 1 edges, and every vertex has an even degree. Since there is no odd degree vertex, G cannot be a tree. Thus, G must contain a cycle C.

44 Proof of Sufficiency (cont.) Proof. Now, remove the edges of C from G, and consider the remaining graph G. Since removing C from G may disconnect the graph, G is a collection of connected components, namely G 1, G 2,....

45 Proof of Sufficiency (cont.) Proof. Now, remove the edges of C from G, and consider the remaining graph G. Since removing C from G may disconnect the graph, G is a collection of connected components, namely G 1, G 2,.... Furthermore, when the edges in C are removed from G, each vertex loses even number of adjacent edges. Thus, the parity of each vertex is unchanged in G. It follows that, for each connected component of G, every vertex has an even degree.

46 Proof of Sufficiency (cont.) Proof. Now, remove the edges of C from G, and consider the remaining graph G. Since removing C from G may disconnect the graph, G is a collection of connected components, namely G 1, G 2,.... Furthermore, when the edges in C are removed from G, each vertex loses even number of adjacent edges. Thus, the parity of each vertex is unchanged in G. It follows that, for each connected component of G, every vertex has an even degree. Therefore, by the induction hypothesis, each of G 1, G 2,... has its own Eulerian circuit, namely C 1, C 2,....

47 Proof of Sufficiency (cont.) Proof. We can now build an Eulerian circuit for G. Pick an arbitrary vertex a from C. Traverse along C until we reach a vertex v i that belongs to one of the connected components G i.

48 Proof of Sufficiency (cont.) Proof. We can now build an Eulerian circuit for G. Pick an arbitrary vertex a from C. Traverse along C until we reach a vertex v i that belongs to one of the connected components G i. Then, traverse along its Eulerian circuit C i until we traverse all the edges of C i. We are now back at v i, and so we can continue on along C. In the end, we shall return back to the first starting vertex a, after visiting every edge exactly once.

49 Illustration of Theorem

50 Illustration of Theorem

51 Illistration of Theorem (cont..)

52 Illistration of Theorem (cont..)

53 Illistration of Theorem (cont..)

54 Illistration of Theorem (cont..)

55 Fleury s Algorithm 1 Check if the graph is connected, and every vertex is of even degree. Reject otherwise. 2 Pick any vertex v start to start. 3 While the graph contains at least one edge: 1 Pick an edge that is not a bridge. 2 Traverse that edge, and remove it from G.

56 Fleury s Algorithm 1 Check if the graph is connected, and every vertex is of even degree. Reject otherwise. 2 Pick any vertex v start to start. 3 While the graph contains at least one edge: 1 Pick an edge that is not a bridge. 2 Traverse that edge, and remove it from G. Definition A bridge is an edge whose removal makes a connected graph become disconnected.

57 Proof of Correctness To see that this algorithm terminates, suppose we are stuck at some vertex v before all the edges are removed. Then either v must have an odd degree, or G is not a connected graph to begin with. Both of these cases are impossible due to our first check. This is a contradiction. Note that, when the algorithm terminates, we must return to v start because every vertex has an even degree. Furthermore, the edge removal operation guarantees that each edge is visited exactly once. Therefore, the discovered tour is an Eulerian circuit.

58 Eulerian trail Theorem A graph contains an Eulerian path if and only if there are 0 or 2 odd degree vertices.

59 Eulerian trail Theorem A graph contains an Eulerian path if and only if there are 0 or 2 odd degree vertices. Proof. Suppose a graph G contains an Eulerian path P. Then, for every vertex v, P must enter and leave v the same number of times, except when it is either the starting vertex or the final vertex of P.

60 Eulerian trail Theorem A graph contains an Eulerian path if and only if there are 0 or 2 odd degree vertices. Proof. Suppose a graph G contains an Eulerian path P. Then, for every vertex v, P must enter and leave v the same number of times, except when it is either the starting vertex or the final vertex of P. When the starting and final vertices are distinct, there are precisely 2 odd degree vertices. When these two vertices coincide, there is no odd degree vertex.

61 Eulerian trail Proof. Conversely, suppose G contains 2 odd degree vertex u and v. (The case where G has no odd degree vertex is shown in the previous theorem.)

62 Eulerian trail Proof. Conversely, suppose G contains 2 odd degree vertex u and v. (The case where G has no odd degree vertex is shown in the previous theorem.) Then, temporarily add a dummy edge {u, v} to G. Now the modified graph contains no odd degree vertex.

63 Eulerian trail Proof. Conversely, suppose G contains 2 odd degree vertex u and v. (The case where G has no odd degree vertex is shown in the previous theorem.) Then, temporarily add a dummy edge {u, v} to G. Now the modified graph contains no odd degree vertex. By the last theorem, this graph contains an Eulerian circuit C that also contains {u, v}. Remove {u, v} from C, and now we have an Eulerian path where u and v serve as initial and final vertices.

64 Chinese Postman Problem Kwan, 1962 A mail carrier starting out at the post office must deliver letters to each block in a territory and return to the post office. What is the least number of repeated street necessary?

65 Chinese Postman Problem Kwan, 1962 A mail carrier starting out at the post office must deliver letters to each block in a territory and return to the post office. What is the least number of repeated street necessary?

66 Chinese Postman Problem (cont.) Edges must be added to create a multigraph because the simple graph contains no Eulerian circuit.