Lecture 25

You may want to download the lecture slides that were used for these videos (PDF).

1. A Key Detail on Network Flow Problems

A network flow problem posed with integer capacities on edges always has a maximum flow in which the flow on every edge is an integer. The proof of this fact is an immediate consequence of the fact that the Ford-Fulkerson labelling algorithm uses only addition, subtraction and minimum as its three operations. (1:32)

2. Network Flow Problems with Capacities 1

This video discusses the exercise introduced at the end of the previous lecture, which was to carry out the Ford-Fulkerson labelling algorithm on a network flow that has capacities of 1. (3:47)

3. Disjoint Paths From x to y

There are two different notions of “disjoint”. We could simply require that two different paths share no edges. Or we could make the stronger requirement that they have no vertices in common other than x and y. Network flows will find the maximum number of disjoint paths in either case. (5:04)

4. Some Consequences

There are a few interesting graph theory results. This video discusses Menger’s Theorem, the edge version and the vertex version. (2:16)

5. Matchings in Graphs

A matching in a graph is a set of edges no two of which share an end point. Typically the problem is to find a maximum size matching. (3:24)

6. Matchings in Bipartite Graphs

We are particularly interested in finding a maximum matching in a bipartite graph. (9:12)

7. More on Matchings in Bipartite Graphs

A company has 9 open positions and 7 applicants. The graph has an edge from applicant x to position i when x is capable of performing i. A matching is then an employment plan, and it is natural to try to fill as many open positions as possible. Note that some applicants may not be capable of doing any job and there may be some jobs that no applicant can do. (2:15)

8. The Concept of Defect

Let G = (X, Y, E) be a bipartite graph. For each subset S of X, let N(S) denote the set of all elements y in
Y for which there is some x in X adjacent to y. We call N(S) the set of neighbors of S. (4:01)

9. Hall’s Theorem (Defect Form)

Let G = (X, Y, E) be a bipartite graph. Then the maximum size of a matching in G is |X| – d(G). (3:00)

10. Regular Balanced Bipartite Graphs

Let G = (X, Y, E) be a balanced regular bipartite graph. Then there is a matching of size |X| in G. (3:38)

11. Posets to Bipartite Graphs (1)

For each x in P, the bipartite graph contains two points labeled x’ and x’’ respectively. The edges in the graph have the form x’y’’ where x < y in P. (4:05)

12. Posets to Bipartite Graphs (2)

A matching in G determines a chain partition of P with x immediately below y in a chain when x’y’’ is one of the matching edges. (5:20)

13. Examples of Problems You Could Be Expected to Solve on a Test

In this video, Dr. Trotter explains some of the problems that students are expected to solve in a test. (2:13)

14. Questions

Is 4’6” an edge in the matching? Is 4 less than 6 in P? (2:21)

15. Chain Partitions from a Maximum Matching

We claim that if we form the chain partition from a maximum matching, the chain partition will be a Dilworth Partition. (5:34)

16. Is 19′ Labelled or Unlabelled?

Pick any chain in our partition. Suppose we choose the chain that has the vertex that is labelled 19. Is 19′ labelled or unlabelled? (4:50)

17. An Old Theorem

If we move down a chain, and the starting point has a labelled prime point, and the end point has an unlabelled double prime point, there is a point somewhere in the chain that has a particular property. (1:28)

18. Finding a Maximum Antichain

When the Ford-Fulkerson labelling algorithm halts, for each chain Ci in the partition, there is a point xi in Ci so that xi’ is labeled and xi’’ is not. These points form an antichain. In chain C1 = {e < c < f}, take f. In chain C2 = {d < a}, take a and C3 = {b}, take b (the only choice). Note that {f, a, b} is a 3- element antichain. (11:44)

19. Quiz

In the last few minutes of this lecture, Dr. Trotter assigned a quiz to his students. See if you can answer the question he assigned. (1:12)