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Unformatted text preview: 1 15.082 and 6.855J March 11, 2003 Max Flows 3 PreflowPush Algorithms 2 Review of Augmenting Paths At each iteration: maintain a flow x Let G(x) be the residual network At each iteration, find a path from s to t in G(x). In the shortest augmenting path algorithm, we kept distance labels d( ), and we sent flow along the shortest path in G(x). 3 Preflows At each intermediate stages we permit more flow arriving at nodes than leaving (except for s) A preflow is a function x: A → R s.t. 0 ≤ x ≤ u and such that e(i) = ∑ j ∈ N x ji ∑ j ∈ N x ij ≥ 0, for all i ∈ N – {s, t}. i.e., e(i) = excess at i = net excess flow into node i. The excess is required to be nonnegative. 4 A Feasible Preflow The excess e(j) at each node j ≠ s, t is the flow in minus the flow out. s 3 4 2 5 t 3 3 3 2 2 2 2 1 2 1 Note: total excess = flow out of s minus flow into t. 5 Active nodes Nodes with positive excess are called active . s 3 4 2 5 t 3 3 3 2 2 2 2 1 2 1 The preflow push algorithm will try to push flow from active nodes towards the sink, relying on d( ). 3 2 6 Review of Distance Labels Distance labels d( ) are valid for G(x) if i. d(t) = 0 ii. d(i) ≤ d(j) + 1 for each (i,j) ∈ G(x) Defn. An arc (i,j) is admissible if r ij > 0 and d(i) = d(j) + 1. Lemma. Let d( ) be a valid distance label. Then d(i) is a lower bound on the distance from i to t in the residual network. 7 Push/Relabel, the fundamental subroutine Suppose we have selected an active node i. Procedure Push/Relabel(i) begin if the network contains an admissible arc (i,j) then push δ : = min{ e(i), r ij } units of flow from i to j; else replace d(i) by min{d(j) + 1 : (i,j) ∈ A(i) and r ij > 0} end ; 8 Pushing using current arcs Tail Head Res. Cap Admissible ? 4 1 0 No 4 2 1 No 4 3 4 Yes 4 5 0 No 4 6 2 Yes Suppose that node 4 is active, and has excess. 3 4 1 2 3 5 6 1 3 2 2 2 e(4) = 2 Scan arcs in A(4) one at a time using “Current Arc” till an admissible arc is found. Push on (4,3) 9 Pushing on (4,3) Tail Head Res. Cap Admissible ? 4 1 0 No 4 2 1 No 4 3 4 Yes 4 5 0 No 4 6 2 Yes 3 4 1 2 3 5 6 1 3 2 2 2 e(4) = 2 Push on (4,3) e(3) = 1 Send min (e(4), r 43 ) = 2 units of flow. 2 e(4) = 0 e(3) = 3 For the next push from node 4, start with arc (4,3)....
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This note was uploaded on 05/05/2010 for the course EE 15.082 taught by Professor Orlin during the Spring '10 term at Visayas State University.
 Spring '10
 Orlin

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