11.2 Eulerian Trails

Transcription

1 11.2 Eulerian Trails

2 K.. onigsberg, 1736

3 Graph Representation A B C D

4 Do You Remember... Definition A u v trail is a u v walk where no edge is repeated.

5 Do You Remember... Definition A u v trail is a u v walk where no edge is repeated. Definition A circuit is a nontrivial closed trail.

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

7 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.

8 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.

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

10 More Examples

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

12 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

13 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

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

15 Necessary Condition Theorem A graph G contains an Eulerian circuit if and 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.

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

17 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

18 Proof of Sufficiency 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.

19 Proof of Sufficiency 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.

20 Proof of Sufficiency 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.

21 Proof of Sufficiency (cont.) 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,....

22 Proof of Sufficiency (cont.) 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.

23 Proof of Sufficiency (cont.) 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,....

24 Proof of Sufficiency (cont.) 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.

25 Proof of Sufficiency (cont.) 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.

26 Illustration of Theorem

27 Illustration of Theorem

28 Illistration of Theorem (cont..)

29 Illistration of Theorem (cont..)

30 Illistration of Theorem (cont..)

31 Illistration of Theorem (cont..)

32 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.

33 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.

34 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.

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

36 Eulerian trail Theorem A graph contains an Eulerian trail if and only if there are 0 or 2 odd degree vertices. Suppose a graph G contains a Eulerian trail 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.

37 Eulerian trail Theorem A graph contains an Eulerian trail if and only if there are 0 or 2 odd degree vertices. Suppose a graph G contains a Eulerian trail 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. 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 a Eulerian circuit C that also contains {u, v}. Remove {u, v} from C, and now we have a Eulerian trail where u and v serve as initial and final vertices.

38 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?

39 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?

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