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Advanced Algorithms

Advanced Algorithms. Piyush Kumar (Lecture 5: Preflow Push). Welcome to COT5405. Today. Preflow Push. Review: Augmenting Path. Maintains a flow in each iteration Finds a path in the residual graph Saturates the path to make progress. 2. 5. 9. 10. 15. 15. 10. 4. 5.

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Advanced Algorithms

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  1. Advanced Algorithms Piyush Kumar (Lecture 5: Preflow Push) Welcome to COT5405

  2. Today • Preflow Push

  3. Review: Augmenting Path • Maintains a flow in each iteration • Finds a path in the residual graph • Saturates the path to make progress.

  4. 2 5 9 10 15 15 10 4 5 s 3 6 t 8 10 15 4 6 10 15 4 7 30 Flows • Def. An s-t flow is a function that satisfies: • For each e  E: (capacity) • For each v  V – {s, t}: (conservation) • Def. The value of a flow f is: 0 4 0 0 0 4 0 4 4 0 0 0 0 capacity flow 0 0 Value = 4

  5. A Preflow • Def. An preflow is a function that satisfies: • For each e  E: (capacity) • For each v  V – {s, t}: (weak conservation) • Def. The excess of a preflow f at node v : • Required to be non-negative except s,t • Algorithm maintains preflow (not flow) • Each vertex is associated with a height • Flow is “downhill” • Vertices with excess are sometimes “lifted/relabeled”.

  6. A Preflow Total excess = flow out of s - flow into t 1 7/10 Nodes with +ve Excess are called Active s 5/5 10/10 G: 2/2 3/3 2 t e=9

  7. Labelling/Height function • H: V -> N • Source and Sink condition • h(s) = n • h(t) = 0 • Steepness condition • For all (v,w) in Ef , h(v) <= h(w)+1 • Here Ef is the residual graph. Invariant Compatibility of preflows and labellings

  8. Lemma • If s-t preflow f is compatible with the labelling h, then there is no s-t path in the residual graph Gf. • Cor: If s-t flow f is compatible with a labelling h, then f is a flow of maximum value.

  9. Initial Preflow f • h(v) = 0 for all v <> s • h(s) = n • f(e) = ce for all e=(s,v) • f(e) = 0 for all other edges. The algorithm gradually transforms a preflow to a flow. Why?

  10. Lemma • An s-t preflow f is compatible with a labelling h, then there is no s-t path in the residual graph Gf. • Cor: An s-t flow f is maximum when its compatible with labelling h.

  11. Preflow-Push AlgorithmInitialization example s G: 10/10 10/10 2 3 1 2 t s e=10 e=10 Gf: 5 10 10 2 3 1 2 t 5

  12. Basic Operation 1Push forward edge • Suppose e(u)>0, cf(u,v)>0, (u,v) in Ef (u,v) is a forward edge h[u]= h[v]+1 • Push as much flow across (u,v) as possible r = min (e[u], c(u,v) – f(e)) Increase f(e) by r.

  13. Basic Operation 2Push backward edge • Suppose e(u)>0, cf(u,v)>0, (u,v) in Ef (u,v) is a backward edge h[u]= h[v]+1 • Push as much flow across (u,v) as possible r = min (e[u], f(e)) Decrease f(e) by r. The preflow push algorithm will try to push from Active nodes towards the sink (maintaining compatibility with the labelling/heights of the nodes)

  14. Basic Operation 3Lift/Relabel • When no push can be done from overflowing vertex (except s,t) • Suppose e[u]>0, and for all (u,v) in Ef : h[u] £ h[v], u¹s, u¹t • Set h[u] = h[u] + 1

  15. Preflow-Push • Initialize • While there is a node v <> t with ef(v) > 0 • If w such that push (f,h,v,w) is possible • Push(f,h,v,w) • Else • Relabel(f,h,v) E

  16. Preflow push demo

  17. Invariants • The labels are >= 0 and integers • f is a preflow, if the capacities are integral, the preflows are integral. • The preflow f and labelling h are compatible. • The height of a node never decreases.

  18. Lemma • Lef f be a preflow. If excessf(v) > 0 then there is a path in Gf from v to source s. • Pf: Let A denote all the nodes that have a path to s in Gf. B = V \ A. What happens to edges inside B? If e has only its head in B? If e has only its tail in B? –fout(B)

  19. Lemmas • The height of a node is upper bounded by 2n-1. • Cor: Each node is relabeled at most 2n-1 times, hence total relabels is upper bounded by 2n2

  20. Lemma: Saturating Pushes • Throughout the algorithm, the number of saturating push operations is at most 2mn. • Pf: Look at a saturating push along an edge. When can another saturating push happen thru the same edge? (If it needs lifting, how many lifts can one do on such a node in total?)

  21. Potential function arguments • Consider a joint bank account held by A, B, and C • The account starts with an initial deposit of at most n2. • A only makes withdrawals, each for 1 or more. • B makes fewer than n2 deposits. Each deposit is for at most 1. • C makes fewer than nm withdrawals and deposits. Each deposit is for at most n. • The account never goes below 0. • What is the maximum number of withdrawals for A?

  22. Lemma: Non-Saturating PushesPotential Functions Argument • Lemma: The number of non-saturating push operations is bounded by 4mn2 • Pf: Define: A non saturating push decreases it by at least 1. A saturating push increases it by at most 2n-1 A relabel increases it by exactly 1. If we pick the active node at maximum height For pushing, we can get O(n^3) non-sat. pushes.

  23. Modifications • A simple modification gives O(n3). • Theorem: The total running time of the algorithm is upper bounded by O(mn2)

  24. References • R.K. Ahuja, T.L. Magnanti, and J.B. Orlin. Network Flows. Prentice Hall, 1993. (Reserved in Dirac) • K. Mehlhorn and S. Naeher. The LEDA Platform for Combinatorial and Geometric Computing. Cambridge University Press, 1999. 1018 pages

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