Maximum flow: The preflow/push method of Goldberg and Tarjan (87)

1. Maximum flow: The preflow/push method of Goldberg and Tarjan (87)

2. Definitions • G=(V,E) is a directed graph • capacity u(v,w) for every v,wV: If (v,w)  E then u(v,w) = 0 • Two distinguished vertices s and t. s 3 4 1 a b 3 3 2 c d 4 3 t

3. Definitions (cont) • A flow is a function on the edges which satisfies the following requirements • f(v,w) = -f(w,v) skew symmetry • f(v,w)  u(v,w) • For every v except s and t wf(v,w) = 0 The value of the flow |f| = wf(s,w) The maxflow problem is to find f with maximum value

4. Flows and s-t cuts Let (X,X’) be a cut such that s X, t X’. s t Flow is the same across any cut: f(X,X’) = f(v,w) = f(v,w) - f(v,w) = |f| - 0 = |f| v X, w  X v X, w  X’ v X, w  V so |f|  cap(X,X’) = u(v,w) The value of the maximum flow is smaller than the minimum capacity of a cut.

5. More definitions The residual capacity of a flow is a function r on the edges such that r(v,w) = u(v,w) - f(v,w) a 2, 1 d Interpretation: We can push r(v,w) more flow from v to w by increasing f(v,w) and decreasing f(w,v)

6. More definitions (cont) We define the residual graph R on V such that there is an arc from v to w with capacity r(v,w) for every v and w such that r(v,w) > 0 An augmenting path p  R is a path from s to t in R r(p) = min r(v,w) (v,w)  p We can increase the flow by r(p)

7. Example 1 4 1 3 3 3 3 1 1 1 3 1 3 1 2 2 1 2 2 2 1 3 3 3 3 4 1 The residual network A flow

8. Basic theorem (1) f is max flow <==> (2) There is no augmenting path in R <==> (3) |f| = cap(X,X’) for some X Proof. (3) ==> (1), (1) ==> (2) obvious To prove (2) ==>(3) let X be all vertices reachable from s in R. By assumption t  X. So (X,X’) is an s-t cut. Since there is no edge from X to X’ in R |f| = f(X,X’) = f(v,w) = u(v,w) = cap(X,X’)

9. Augmenting path methods Repeat the following step: Find an augmenting path in R, increase the flow, update R Stop when s and t are disconnected in R. Need to be careful about how you choose those augmenting paths ! The best algorithm in this family is Dinic’s algorithm, that can be implemented in O(nmlog(n)) time

10. But we’ll go for the preflow/push method

11. Distance labels • Defined with respect to residual capacities • d(t) = 0, d(s) = n • d(v) ≤ d(w) + 1 if r(v,w) > 0

12. Example (distance labels) 1 3 3 3 4 1 3 1 1 1 3 1 3 1 2 2 1 2 2 2 1 3 3 3 3 4 1 The residual network A flow

13. Example (distance labels) 1 6 3 3 3 4 1 3 1 1 2 2 1 3 1 3 1 2 2 1 2 2 2 1 1 1 3 3 3 3 4 1 0 The residual network A flow Is this a valid distance labeling ?

14. Distance labels – basic lemma Lemma: d(v) is a lower bound on the length of the shortest path from v to the sink Proof: Let the s.p. to the sink be: v v2 v1 t d(v) ≤ d(v1) + 1 ≤ d(v2) + 2 ..... ≤ d(t) + k = k

15. Preflow (definition) • A preflow is a function on the edges which satisfies the following requirements • f(v,w) = -f(w,v) skew symmetry • f(v,w)  u(v,w) • For every w, except s and t, vf(v,w) ≥ 0 Let e(w) = vf(v,w) be the excess at the node v (we’ll also have e(t) ≥ 0, and e(s) ≤ 0)

16. Example (preflow) 2 0 2 2 3 1 3 2 s t 0 3 2 1 Nodes with positive excess are called active. The preflow push algorithm will try to push flow from active nodes towards the sink, relying on d( ).

17. 3 4 4 3 1 3 3 2 3 4 Initialization (preflow) 3 3 4 4 0 1 3 0 3 0 2 0 3 0 4 0

18. 4 3 Initialization (distance labels) 6 3 3 4 4 3 4 0 1 1 0 0 3 0 3 0 3 3 2 2 0 0 0 3 0 3 4 0 4 0 Recall: s must be disconnected from t when d(s) = n, and the labeling is valid…

19. Admissible arc in the residual graph v w d(v) = d(w) + 1

20. The preflow push algorithm While there is an active node { pick an active node v and push/relabel(v) } Push/relabel(v) { If there is an admissible arc (v,w) then { push = min {e(v) , r(v,w)} flow from v to w } else { d(v) := min{d(w) + 1 | r(v,w) > 0} (relabel) }

34. s v S Correctness Lemma 1: The source is reachable from every active vertex in the residual network Proof: Assume that’s not the case: Which means that no flow enters S -- A contradiction

35. Correctness (cont) Corollary: There is an outgoing residual arc incident with every active vertex Corollary: So we can push-relabel as long as there is an active vertex

36. Correctness (cont) Lemma 2:Distance labels only increase and remain valid at all times Proof: By induction on the number of push and relabel operations. For relabel this is clear by the definition of relabel For push: w v d(v) = d(w) + 1 so even if we add (w,v) to the residual network then it is still a valid labeling

37. Correctness (cont) Lemma 3:When (and if) the algorithm stops the preflow is a maximum flow Proof: It is a flow since there is no active vertex. It is maximum since the sink is not reachable from the source in the residual network. (d(s) = n, and the labeling is valid)

38. Complexity analysis

39. Another example