1. 1 Maximum flow sender receiver Capacity constraint Lecture 6: Jan 25

2. 2 Network transmission Given a directed graph G A source node s A sink node t Goal: To send as much information from s to t

3. 3 Flows An s-t flow is a function f which satisfies: (capacity constraint) (conservation of flows) An s-t flow is a function f which satisfies: (capacity constraint) (conservation of flows ( at intermediate vertices )

4. 4 Value of the flow st 10 9 8 4 6 2 3 9 9 9 7 0 G: 6 Value = 19 Maximum flow problem: maximize this value

5. 5 Flow decomposition Any flow can be decomposed into at most m flow paths. The same idea applies to the Chinese postman problem

6. 6 An upper bound sender receiver

7. 7 Cuts An s-t cut is a set of edges whose removal disconnect s and t The capacity of a cut is defined as the sum of the capacity of the edges in the cut Minimum s-t cut problem: minimize this capacity of a s-t cut

8. 8 Flows ≤ cuts Let C be a cut and S be the connected component of G-C containing s. Then:

9. 9 Main result Value of max s-t flow ≤ capacity of min s-t cut (Ford Fulkerson 1956) Max flow = Min cut A polynomial time algorithm

10. 10 Greedy method? Find an s-t path where every edge has f(e) < c(e) Add this path to the flow Repeat until no such path can be found. Does it work?

11. 11 A counterexample Hint: Find an augmenting path

12. 12 Residual graph Key idea: allow flows to push back c(e) = 10 f(e) = 2 c(e) = 8 c(e) = 2 Advantage of this representation is not to distinguish send forward or push back (which are irrelevant) Can send 8 units forward or push 2 units back.

13. 13 Finding an augmenting path Find an s-t path in the residual graph Add it to the current flow to obtain a larger flow. Why? 1.Flow conservations 2.More flow going out from s Key: don’t think about flow paths!

14. 14 Ford-Fulkerson Algorithm 1. Start from an empty flow f 2. While there is an s-t path P in G update f along P 3. Return f

15. 15 Max-flow min-cut theorem Consider the set S of all vertices reachable from s So, s is in S, but t is not in S No incoming flow coming in S (otherwise push back) Achieve full capacity from S to T Min cut!

16. 16 Integrality theorem If every edge has integer capacity, then there is a flow of integer value.

17. 17 Complexity Assume edge capacity between 1 to C At most nC iterations Finding an s-t path can be done in O(m) time Total running time O(nmC)

18. 18 Speeding up Capacity scaling (find paths with large capacity) Find a shortest s-t path  time Preflow-push

19. 19 Faster Algorithms

20. 20 Even Faster Algorithms

21. 21 Applications of the algorithm of the min-max theorem of the integrality theorem

22. 22 Multi-source multi-sink A set of sources S = {s1,…,sk} A set of sinks T = {t1,…,tm} Maximum flow from S to T

23. 23 Bipartite matching Bipartite matching <= Maximum flow

24. 24 Disjoint paths Find the maximum number of disjoint s-t paths directed edge => directed vertex (vertex splitting) directed vertex => undirected vertex (bidirecting) undirected vertex => undirected edge (line graph)

25. 25 Minimum Path Cover Given a directed graph, find a minimum number of paths to cover all vertices Directed graph => Bipartite graph

26. 26 Matrix Rounding Round the entries to “keep” the row sums and column sums

27. 27 League winner See if your favorite team can still win the leaque

28. 28 Bonus Question 3 (25%) In a soccer tournament of n teams, every pair of teams plays one match. The winner gets 3 points, the loser gets 0, while both teams receive 1 point in a draw. Is there a polynomial algorithm to decide whether a given score sequence (a score for each team) can be the score sequence at the end of a valid championship? League winner version is NP-hard.