11.4 Bipartite Multigraphs

Transcription

1 11.4 Bipartite Multigraphs

2 Introduction Definition A graph G is bipartite if we can partition the vertices into two disjoint subsets U and V such that every edge of G has one incident vertex in U and the other in V.

3 Examples Notice the colors - the property of being bipartite is equivalent to being 2-colorable. We will come back to this next chapter.

4 Example Suppose we had 5 workers and seven different tasks that could be performed. If each worker can only perform one task at a time, how can the tasks be assigned so that the most possible jobs can be completed? Worker Tasks 1 1,2,3,4,7 2 4,5 3 4,5 4 1,5,6 5 1,6,7

5 Example (cont.) The situation can be modeled as a bipartite graph. W 1 T 1 W 2 T 2 W 3 T 3 W 4 T 4 W 5 T 5 T 6 T 7

6 One Solution W 1 T 1 W 2 T 2 W 3 T 3 W 4 T 4 W 5 T 5 T 6 T 7 This solution is known as a matching.

7 Matchings Definition A nonempty set M of edges in G is called a matching of G if no two edges of M are incident with a common vertex.

8 Matchings Definition A nonempty set M of edges in G is called a matching of G if no two edges of M are incident with a common vertex. Definition A perfect matching has V 2 edges.

9 Example Three ladies x 1, x 2 and x 3 are courted by 5 men y 1... y 5. The bipartite graph with bipartition (X, Y) shows the situation such that two vertices are adjacent if and only if the two persons would consent to marry each other. X x 1 x 2 x 3 Y y 1 y 2 y 3 y 4 y 5 1 Is {x 1 y 1 } a matching? 2 Is {x 1 y 1, x 2 y 1 } a matching? 3 Is {x 1 y 5, x 2 y 1, x 3 y 2 } a matching? 4 Find the number of matchings with three edges.

10 When is a graph bipartite? Theorem A graph G is bipartite iff every circuit of G has even length.

11 When is a graph bipartite? Theorem A graph G is bipartite iff every circuit of G has even length. Proof: Note that it is sufficient to prove the theorem for connected bipartite graphs. We claim that if the theorem is true for each connected component of a disconnected bipartite graph G, then it is true for G.

12 When is a graph bipartite? Theorem A graph G is bipartite iff every circuit of G has even length. Proof: Note that it is sufficient to prove the theorem for connected bipartite graphs. We claim that if the theorem is true for each connected component of a disconnected bipartite graph G, then it is true for G. This claim follows from G being bipartite iff each of its components is bipartite, and any circuit in G having even length iff any circuit in each of its components has even length.

13 Proof (cont.) First we will show that if G is bipartite, then any circuit has even length. If G is bipartite so it can be drawn connecting a left vertex with a right vertex, then any circuit x 1 x 2 x 3... x n x 1 has alternately a left vertex, a right vertex, then a left vertex, and so on, assuming the first vertex x 1 is on the left. Odd subscripted numbers on the left and even subscripted numbers on the right. That is, there is an even number of vertices in any circuit. Any circuit has the same number of edges as vertices, and thus this circuit has even length. x 1 x 2 x 3 x 4 x 5 x 6.. x n 1 x n

14 Proof (cont.) Suppose next that any circuit in G (there may be no circuits) has even length. We show how to construct a bipartite arrangement of G by assuming a bipartition exists and using the properties of a bipartition to construct one.

15 Proof (cont.) Suppose next that any circuit in G (there may be no circuits) has even length. We show how to construct a bipartite arrangement of G by assuming a bipartition exists and using the properties of a bipartition to construct one. Take any vertex a and put it on the left. Put all vertices adjacent to a on the right. Next put all vertices that are two edges away from a, that is, at the end of some path of length 2 from a, on the left. In general, if there is a path of odd length between a and a vertex x, put it on the right. If there is a path of even length between a and x, put x on the left.

16 Proof (cont.) Suppose next that any circuit in G (there may be no circuits) has even length. We show how to construct a bipartite arrangement of G by assuming a bipartition exists and using the properties of a bipartition to construct one. Take any vertex a and put it on the left. Put all vertices adjacent to a on the right. Next put all vertices that are two edges away from a, that is, at the end of some path of length 2 from a, on the left. In general, if there is a path of odd length between a and a vertex x, put it on the right. If there is a path of even length between a and x, put x on the left. There cannot be distinct paths P and P between a and x of odd and even lengths, respectively, since taking P from a to x and then returning to a on P yields an odd length circuit. This is impossible since all circuits have even length.

17 Proof (cont.) Similarly, we argue that there cannot be an edge between vertices, say b and c, both on the left. There must exist even length paths Q and Q joining a with b and c, respectively, since b and c are on the left. a Q b c Q

18 Proof(cont.) a Q b c Observe that Q followed by the edge {c, b} yields an odd length path from a to b. This is impossible, since we just proved that there cannot be both an even length path (Q) and an odd length path (Q plus {a, b}) from a to any other vertex in G. By similar reasoning, two vertices on the right cannot be adjacent. This, we have a bipartite arrangement if G. Q

19 Another Theorem Theorem Let G be a bipartite graph with bipartition (X, Y). 1 If X Y, then G does not have a Hamiltonian cycle.

20 Another Theorem Theorem Let G be a bipartite graph with bipartition (X, Y). 1 If X Y, then G does not have a Hamiltonian cycle. 2 If X = Y, then G does not have a Hamiltonian path that begins at a vertex in X and ends at a vertex in X.

21 Another Theorem Theorem Let G be a bipartite graph with bipartition (X, Y). 1 If X Y, then G does not have a Hamiltonian cycle. 2 If X = Y, then G does not have a Hamiltonian path that begins at a vertex in X and ends at a vertex in X. 3 If X and Y differ by at least 2, then G does not have a Hamiltonian path.

22 Another Theorem Theorem Let G be a bipartite graph with bipartition (X, Y). 1 If X Y, then G does not have a Hamiltonian cycle. 2 If X = Y, then G does not have a Hamiltonian path that begins at a vertex in X and ends at a vertex in X. 3 If X and Y differ by at least 2, then G does not have a Hamiltonian path. 4 If X = Y + 1, then G does not have a Hamiltonian path that begins at a vertex in X and ends at a vertex of Y, and vice versa.

23 Complete Bipartite Graphs Definition A bipartite graph is complete if each vertex in X is adjacent to every vertex in Y.

24 Complete Bipartite Graphs Definition A bipartite graph is complete if each vertex in X is adjacent to every vertex in Y. Example K 2,3

25 Complete Bipartite Graph Example K 1,4 Complete bipartite graphs of the form K 1,n are called stars.

26 The Knight s Tour Problem The knight s tour problem involves the knight graph, which is a graph on vertices in which each vertex represents a square in an chessboard, and each edge corresponds to a legal move by a knight. The process is this: begin with a square on the grid and and travel around the board using only L shaped moves. K K K

27 The Knight s Tour Problem The knight s tour problem involves the knight graph, which is a graph on vertices in which each vertex represents a square in an chessboard, and each edge corresponds to a legal move by a knight. The process is this: begin with a square on the grid and and travel around the board using only L shaped moves. K K K Definition A knight s tour is a sequence of moves by a knight such that each square of the board is visited exactly once. It is therefore a Hamiltonian path on the corresponding knight graph.

28 When do knight s tours exist? Example Does a knight s tour exist for a 3 3 board?

29 When do knight s tours exist? Example Does a knight s tour exist for a 3 3 board? Consider the following board: K Where do we go from here?

30 When do knight s tours exist? Example Does a knight s tour exist for a 3 3 board? Consider the following board: K Where do we go from here? What if we started here? K

31 Knight s Tours and Hamiltonian Graphs Knight s Tour

32 Knight s Tours and Hamiltonian Graphs Knight s Tour This corresponds to the graph:

33 Knight s Tours and Hamiltonian Graphs Knight s Tour This corresponds to the graph: We could invoke the last theorem...

34 Justification for 3 3 board Theorem Let G be a bipartite graph with bipartition (X, Y). 1 If X Y, then G does not have a Hamiltonian cycle. 2 If X = Y, then G does not have a Hamiltonian path that begins at a vertex in X and ends at a vertex in X. 3 If X and Y differ by at least 2, then G does not have a Hamiltonian path. 4 If X = Y + 1, then G does not have a Hamiltonian path that begins at a vertex in X and ends at a vertex of Y, and vice versa.

35 Results Conrad et al. (1994) shows that a knight s path exists on an n n board iff n 5. And for homework, you will provide an example for n = 5, 6, 7.