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Published byBilly Nunley Modified over 2 years ago

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Min Cost Flow: Polynomial Algorithms

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Overview Recap: Min Cost Flow, Residual Network Potential and Reduced Cost Polynomial Algorithms Approach Capacity Scaling Successive Shortest Path Algorithm Recap Incorporating Scaling Cost Scaling Preflow/Push Algorithm Recap Incorporating Scaling Double Scaling Algorithm - Idea

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Min Cost Flow - Recap v1v1 v2v2 v3v3 v4v4 v5v5 5 -2 -3 4,14,1 3,43,4 5,15,1 1,11,1 3,33,3

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Min Cost Flow - Recap fdsfds Compute feasible flow with min cost

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Residual Network - Recap

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Reduced Cost - Recap

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Min Cost Flow: Polynomial Algorithms

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Approach We have seen several algorithm for the MCF problem, but none polynomial – in logU, logC. Idea: Scaling! Capacity/Flow values Costs both Next Week: Algorithms with running time independent of logU, logC Strongly Polynomial Will solve problems with irrational data

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Capacity Scaling

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Successive Shortest Path - Recap

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Algorithm Complexity: Assuming integrality, at most nU iterations. In each iteration, compute shortest paths, Using Dijkstra, bounded by O(m+nlogn) per iteration

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Capacity Scaling - Scheme Successive Shortest Path Algorithm may push very little in each iteration. Fix idea: use scaling Modify algorithm to push units of flow Ignore edges with residual capacity < until there is no node with excess or no node with deficit Decrease by factor 2, and repeat Until < 1.

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Definitions 34 33 3 4 2 1 G(x)

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Definitions 34 33 3 4 G(x, 3)

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Main Observation in Algorithm Observation: Augmentation of units must start at a node in S( ), along a path in G(x, ), and end at a node in T( ). In the -phase, we find shortest paths in G(x, ), and augment over them. Thus, edges in G(x, ), will satisfy the reduced optimality conditions. We will consider edges with less residual capacity later.

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Initializing phases i j

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Capacity Scaling Algorithm

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Initial values. 0 pseudoflow and potential (optimal!) Large enough

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Capacity Scaling Algorithm In beginning of -phase, fix optimality condition on new edges with resid. Cap. r ij < 2 by saturation

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Capacity Scaling Algorithm augment path in G(x, ) from node in S( ) to node in T( )

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Capacity Scaling Algorithm - Correctness

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Capacity Scaling Algorithm - Assumption We assume path from k to l in G(x, ) exists. And we assume we can compute shortest distances from k to rest of nodes. Quick fix: initially, add dummy node D with artificial edges (1,D) and (D,1) with infinite capacity and very large cost.

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Capacity Scaling Algorithm – Complexity The algorithm has O(log U) phases. We analyze each phase separately.

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Capacity Scaling Algorithm – Phase Complexity D E

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Capacity Scaling Algorithm – Phase Complexity – Cont.

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Capacity Scaling Algorithm – Complexity

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Cost Scaling

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Approximate Optimality

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Approximate Optimality Properties a b c d 4 -5 3 -2

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Algorithm Strategy

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Preflow Push Recap

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Distance Labels Distance Labels Satisfy: d(t) = 0, d(s) = n d(v) d(w) + 1 if r(v,w) > 0 d(v) is at most the distance from v to t in the residual network. s must be disconnected from t …

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Terms Nodes with positive excess are called active. Admissible arc in the residual graph: w v d(v) = d(w) + 1

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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) }

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Running Time The # of relabelings is (2n-1)(n-2) < 2n 2 The # of saturating pushes is at most 2nm The # of nonsaturating pushes is at most 4n 2 m – using potential Φ = Σ v active d(v)

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Back to Min Cost Flow…

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Applying Preflow Pushs technique j i

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Initialization v w -10

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Push/relabel until no active nodes exist

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Correctness Lemma 1: Let x be pseudo-flow, and x a feasible flow. Then, for every node v with excess in x, there exists a path P in G(x) ending at a node w with deficit, and its reversal is a path in G(x). Proof: Look at the difference x-x, and observe the underlying graph (edges with negative difference are reversed).

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Lemma 1: Proof Cont. Proof: Look at the difference x-x, and observe the underlying graph (edges with negative difference are reversed). 34 33 3 4 2 1 32 34 2 5 2 0 -

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Lemma 1: Proof Cont. Proof: Look at the difference x-x, and observe the underlying graph (edges with negative difference are reversed). v w S There must be a node with deficit reachable, otherwise x isnt feasible

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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

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Correctness – Cont.

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Correctness (cont)

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Complexity Lemma : a node is relabeled at most 3n times.

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Lemma 2 – Cont.

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Complexity Analysis (Cont.) Lemma: The # of saturating pushes is at most O(nm). Proof: same as in Preflow Push.

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Complexity Analysis – non Saturating Pushes Def: The admissible network is the graph of admissible edges. 4 -2 2 -4 2

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Complexity Analysis – non Saturating Pushes Def: The admissible network is the graph of admissible edges. -2 -4

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Complexity Analysis – non Saturating Pushes Def: The admissible network is the graph of admissible edges. Lemma: The admissible network is acyclic throughout the algorithm. Proof: induction.

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Complexity Analysis – non Saturating Pushes – Cont. Lemma: The # of nonsaturating pushes is O(n 2 m). Proof: Let g(i) be the # of nodes reachable from i in admissible network Let Φ = Σ i active g(i)

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Complexity Analysis – non Saturating Pushes – Cont. Φ = Σ i active g(i) By acyclicity, decreases (by at least one) by every nonsaturating push ij

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Complexity Analysis – non Saturating Pushes – Cont. Φ = Σ i active g(i) Initially g(i) = 1. Increases by at most n by a saturating push : total increase O(n 2 m) Increases by each relabeling by at most n (no incoming edges become admissible): total increase < O(n 3 ) O(n 2 m) non saturating pushes

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Cost Scaling Summary Total complexity O(n 2 mlog(nC)). Can be improved using ideas used to improve preflow push

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Double Scaling (If Time Permits)

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Double Scaling Idea

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Network Transformation i b(i) j b(j) C ij, u ij x ij i b(i) (i,j) C ij, x ij j b(j) -u ij 0, r ij

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Improve approximation Initialization N1N1 N2N2 +

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Capacity Scaling - Scheme

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Double Scaling - Correctness Assuming algorithm ends, immediate, since we augment along admissible path. (Residual path from excess to node to deficit node always exists – see cost scaling algorithm) We relabel when indeed no outgoing admissible edges.

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Double Scaling - Complexity O(log U) phases. In each phase, each augmentation clears a node from S( ) and doesnt introduce a new one. so O(m) augmentations per phase.

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Double Scaling Complexity – Cont. In each augmentation, We find a path of length at most 2n (admissible network is acyclic and bipartite) Need to account retreats.

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Double Scaling Complexity – Cont. In each retreat, we relabel. Using above lemma, potential cannot grow more than O(l), where l is a length of path to a deficit node. Since graph is bipartite, l = O(n). So in all augmentations, O(n (m+n)) = O(mn) retreats. N1N1 N2N2

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Double Scaling Complexity – Cont. To sum up: Implemented improve-approximation using capacity scaling O(log U) phases. O(m) augmentations per phase. O(n) nodes in path O(mn) node retreats in total. Total complexity: O(log(nC) log(U) mn)

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Thank You

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