1. Single Source Shortest-Path: The General Case (with negative edges) Bellman-Ford algorithm. Iteratively relax all edges |V|-1 times Running time? O(VE).

2. COSC 3101NJ. Elder All-Pairs Shortest Paths

3. Given a directed graph G = (V, E), weight function w : E → R, |V| = n. Assume no negative weight cycles. Goal: create an n × n matrix of shortest-path distances δ(u, v). Could run BELLMAN-FORD once from each vertex: O(V 2 E)—which is O(V 4 ) if the graph is dense (E =  (V2)). If no negative-weight edges, could run Dijkstra’s algorithm once from each vertex: O(V E lg V) with binary heap—O(V 3 lg V) if dense, We’ll see how to do in O(V 3 ) in all cases, with no fancy data structure.

4. All Pairs Shortest Path – Floyd-Warshall Algorithm Dynamic programming approach. Use optimal substructure of shortest paths: Any subpath of a shortest path is a shortest path. Create a 3-dimensional table: Let d ij (k) –shortest path weight of any path from i to j where all intermediate vertices are from the set {1,2, …, k}. Ultimately, we would like to know the values of d ij (n).

5. Computing d ij (k) Base condition: d ij (0) = ? d ij (0) = w ij. For k>0: Let p= be a shortest path from vertex i to vertex j with all intermediate vertices in {1,2, …, k}. If k is not an intermediate vertex, then all intermediate vertices are in {1,2, …, k-1}. If k is an intermediate vertex, then p is composed of 2 shortest subpaths drawn from {1,2, …, k-1}.

6. Recursive Formulation for d ij (k)

7. Algorithm Running time = ? O(n 3 ). Memory required = ? O(n 2 ) (if we drop the superscripts).

8. Example

9. Step 1

10. Step 2

11. Step 3

12. Step 4

13. Step 5

14. All-Pairs Shortest Path: Johnson’s Algorithm Idea: If the graph is sparse (|E|<<|V| 2 ), it pays to run Dijkstra’s algorithm once from each vertex. O(VE log V) using binary heap, O(V 2 log V + V E) using Fibonacci heap. But Dijkstra’s algorithm does not handle negative edges. Johnson’s Algorithm: reweight edges to form equivalent graph with non-negative edges. Floyd-Warshall still has advantages: very simple implementation no fancy data structures small constant.

15. COSC 3101NJ. Elder The Maximum Network Flow Problem

16. The Problem Use a graph to model material that flows through conduits. Each edge represents one conduit, and has a capacity, which is an upper bound on the flow rate = units/time. Can think of edges as pipes of different sizes. But flows don’t have to be of liquids. Want to compute max rate that we can ship material from a designated source to a designated sink.

17. What is a Flow Network? Each edge (u,v) has a nonnegative capacity c(u,v). If (u,v) is not in E, assume c(u,v)=0. We have a source s, and a sink t. Assume that every vertex v in V is on some path from s to t. c(s,v 1 )=16; c(v 1,s)=0; c(v 2,v 3 )=0

18. What is a Flow in a Network? For each edge (u,v), the flow f(u,v) is a real-valued function that must satisfy 3 conditions: Note that skew symmetry condition implies that f(u,u)=0.

19. Example of a Flow: f(v 2, v 1 ) = 1, c(v 2, v 1 ) = 4. f(v 1, v 2 ) = -1, c(v 1, v 2 ) = 10. f(v 3, s) + f(v 3, v 1 ) + f(v 3, v 2 ) + f(v 3, v 4 ) + f(v 3, t) = 0 + (-12) + 4 + (-7) + 15 = 0 flow capacity

20. The Value of a flow The value of a flow is given by This it the total flow leaving s = the total flow arriving in t.

21. Example: |f| = f(s, v 1 ) + f(s, v 2 ) + f(s, v 3 ) + f(s, v 4 ) + f(s, t) = 11 + 8 + 0 + 0 + 0 = 19 |f|= f(s, t) + f(v 1, t) + f(v 2, t) + f(v 3, t) + f(v 4, t) = 0 + 0 + 0 + 15 + 4 = 19

22. A flow in a network We assume that there is only flow in one direction at a time. Sending 7 trucks from Edmonton to Calgary and 3 trucks from Calgary to Edmonton has the same net effect as sending 4 trucks from Edmonton to Calgary.

23. Multiple Sources Network We have several sources and several targets. Want to maximize the total flow from all sources to all targets. Reduce to max-flow by creating a supersource and a supersink:

24. The Ford-Fulkerson Method Try to improve the flow, until we reach the maximum. The residual capacity of the network with a flow f is given by: Always nonnegative (why?)

25. Example of residual capacities Network: Residual Network: Augmenting path

26. The residual network The edges of the residual network are the edges on which the residual capacity is positive.

27. Augmenting Paths An augmenting path p is a simple path from s to t on the residual network. We can put more flow from s to t through p. We call the maximum capacity by which we can increase the flow on p the residual capacity of p.

28. Augmenting Paths - example The residual capacity of p = 4. Can improve the flow along p by 4.

29. Ford-Fulkerson Method

30. Example Flow(1) Residual(1) Flow(2) Residual(2) No more augmenting paths  max flow attained.

31. Augmenting Paths – example The maximum possible flow through the cut = 12 + 7 + 4 = 23 The network has a capacity of at most 23. This is called a minimum cut. Flow(2)

32. Cuts of Flow Networks A cut in a network is a partition of V into S and T=V-S so that s is in S and t is in T.

33. The net flow through a cut (S,T) f(S,T) = 12 – 4 + 11 = 19

34. The capacity of (S,T) c(S,T)= 12+ 0 + 14 = 26

35. Lemma 26.5 For any cut (S,T), the net flow across (S,T) is f(S,T)=|f|.

36. Corollary 26.6 The value of any flow f in a flow network G is bounded from above by the capacity of any cut of G.

37. Theorem 26.7 (Max-flow min-cut theorem) If f is a flow in a flow network G=(V,E), with source s and sink t, then the following conditions are equivalent: 1. f is a maximum flow in G. 2. The residual network G f contains no augmented paths. 3. |f| = c(S,T) for some cut (S,T) (a min-cut).

38. The Basic Ford-Fulkerson Algorithm

39. Example Resulting Flow = Original Network augmenting path 4

40. Example Resulting Flow = 4 Residual Network augmenting path

41. Example Residual Network Resulting Flow = 11

42. Example Resulting Flow = 11 Residual Network augmenting path

43. Example Residual Network Resulting Flow = 19

44. Example Resulting Flow = 19 Residual Network augmenting path

45. Example Residual Network Resulting Flow = 23

46. Residual Network Example Resulting Flow = 23 No augmenting path: Maxflow=23

47. Analysis O(E)

48. Analysis If capacities are all integer, then each augmenting path raises |f| by ≥ 1. If max flow is f*, then need ≤ |f*| iterations ⇒ time is O(E|f*|). Note that this running time is not polynomial in input size. It depends on |f*|, which is not a function of |V| or |E|. If capacities are rational, can scale them to integers. If irrational, FORD-FULKERSON might never terminate!

49. The basic Ford-Fulkerson Algorithm With time O ( E |f*|), the algorithm is not polynomial. This problem is real: Ford-Fulkerson may perform very badly if we are unlucky: |f*|=2,000,000

50. Run Ford-Fulkerson on this example Augmenting Path Residual Network

51. Run Ford-Fulkerson on this example Augmenting Path Residual Network

52. Run Ford-Fulkerson on this example Repeat 999,999 more times…

53. The Edmonds-Karp Algorithm A small fix to the Ford-Fulkerson algorithm makes it work in polynomial time. Specify how to compute the path in line 4.

54. The Edmonds-Karp Algorithm Compute the path in line 4 using breadth-first search on residual network. The augmenting path p is the shortest path from s to t in the residual network (treating all edge weights as 1). Runs in time O(V E 2 ).

55. The Edmonds-Karp Algorithm - example Edmonds-Karp’s algorithm runs only 2 iterations on this graph.

56. Further Improvements Push-relabel algorithm ([CLRS, 26.4]) – O(V 2 E). The relabel-to-front algorithm ([CLRS, 26.5) – O(V 3 ).