Lecture 13: Polynomial-Time Algorithms for Min Cost Flows. (Reading: AM&O Chapter 10)

Transcription

1 Lecture 1: Polynomial-Time Algorithms for Min Cost Flows (Reading: AM&O Chapter 1)

2 Polynomial Algorithms for Min Cost Flows Improvements on the two algorithms for min cost flow: Successive Shortest Path scale costs scale capacities Cycle Cancelling augment along minimum mean cycles Assumption throughout: All data integer. C = largest cost of an arc U = largest capacity of an arc.

3 Capacity Scaling Idea: Augment along large capacity arcs. More specifically, let the current flow be x, the current shortest path values d i and threshold flow value of recall the -residual network G(x, ) of arcs of G(x) having residual capacity at least. Then the procedure is: Step 1: push flow along every arc of the residual graph G(x, ) having negative cost up to its residual capacity; Step : move flow from the set E( ) of nodes with excess at least to the set D( ) of nodes with deficit at least along shortest paths in G(x, ). Step is just the Shortest Successive Path Algorithm applied to the -reduced network, after an initialization stage where flow is pushed along large capacity cost-decreasing arcs.

4 The Capacity Scaling Algorithm for Min Cost Flows Initialize: Set x =, = U/ while 1 ( -scaling phase) Step 1: For each (i, j) in G(x, ) with negative reduced cost, push flow along (i, j) up to residual capacity. (All arcs with cost retain their original flow values.) Step : while E do Select node s E( ) and t D( ) and find a shortest (s, t)-path P in G(x, ). (Note that you may use the reduced costs from the previous shortest-path computations for this.) Set δ = min{r ij : (i, j) P } {e(k), e(l)}, and augment δ units of flow along path P. Update G(x, ), E( ), and D( ) where appropriate. end while Set = / end -scaling phase

5 u,c ij ij 45,$ 5,$5 b i ,$4 5,$7 64,$,$ 4 4 4,$1 5 s =,16 45,$ [] 4 1 π i = d i [ ] =8 [ ] 5,$5 (true costs) 5,$ 1,$ (reduced costs) [ 7] 4 4 t 64,$ 4,$1 5 [ 19],$ [ 17] [] 1 4,$ 4,$ 64,$14 15,$ 15 [ 19],$ [ 7] 4 4,$ 5 [ 17] push 15 units from 1 to t [5] ,$ 4,$ [] 1,$5 4,$ 5 64,$ [] =4,$ 15,$ [ ] [5] [ ] 5,$ 5,$ ,$4 49,$ 15,$ 4 4,$1 15,$ [],$ [ 1] s [] 15,$,$14 5 [ 1] no flow change, current flow optimal 4,$ Example of Capacity Scaling

6 Complexity of Capacity Scaling Algorithm potential function : Φ = sum of excesses 1. The number of scaling phases is O(logU). At the beginning of each -scaling phase, residual capacities in G(x, ) are between and, and Φ is at most n.. After Step 1 of the -scaling phase, Φ can be increased to at most n + m. 4. In Step of the -scaling phase, each augmentation decreases Φ by at least. Thus the number of augmentations is O(n+m) = O(m) Theorem 1.1 : The capacity scaling algorithm solves the min cost flow problem in O(m log U S(n, m, nc)), where S(n, m, nc) is the complexity of the best label-setting algorithm on an n-node, m-arc network with max cost nc.

7 Cost Scaling The Cost Scaling Algorithm is also a modification of the Shortest Augmenting Path Algorithm for finding max flows. The Cost Scaling Algorithm, however, augments flow along almost shortest paths. Specifically, let x be a pseudoflow, and π i a set of node-numbers representing almost shortest (i, t)-path lengths, and set reduced costs to c π ij = c ij π i + π j. For positive constant ϵ, we say that x and π satisfy the ϵ-optimality conditions if the associated reduced costs satisfy c π ij > ϵ (i, j) Gπ (x), If, in addition x is feasible, then we call x an ϵ-optimal flow.

8 Lemma 1.: For any ϵ < 1/n, an ϵ-optimal flow x is in fact optimal. Proof: Suppose not. Then by the Negative Cycle Optimality Conditions there must be a cycle W that has negative total cost, or what is the same, negative reduced costs c π (W ). But every arc in W has reduced cost greater than ϵ = 1/n. Thus c π (W ) > 1, and since all costs are integer, c π (W ). Thus there are no negative weight cycles in G(x), and so x is an optimal flow.

9 Idea of Algorithm The Cost Scaling Algorithm starts with ϵ = C, and finds ϵ-optimal flows for successively smaller ϵ, until ϵ reaches 1/n, at which point the ϵ-optimal flow is optimal. It does this by starting with a pseudoflow x that is ϵ-optimal, and pushes flow from excess nodes to deficit nodes along ϵ/-shortest paths, that is, paths whose arcs satisfying c π ij < ϵ/ Clearly any pseudoflow obtained in this way will continue to satisfy the ϵ-optimality conditions. We search for these paths in much the same way as the Shortest Augmenting Path Algorithm for Max Flow (note that in the SAP version the paths must satisfy c π ij = ), updating the π values during retreats. When an ϵ/-optimal flow is found, we set ϵ = ϵ/ and start the process over again.

10 ϵ-scaling Phase: Start with a flow that is ϵ-optimal, and end with a flow that is ϵ/- optimal. Initialization of the phase: Use the old values of π, and set flow on G π (x) equal to residual capacity on arcs with c π ij ϵ/, adjusting residual capacities and excesses accordingly. This will make x an ϵ/-optimal pseudoflow. We call an arc (i, j) ϵ-admissible if c π ij < ϵ/. By only pushing flow along ϵ-admissible arcs, we create a true flow for which G π (x) satisfies the ϵ/-optimality conditions. To do this we perform the Shortest Augmenting Path Algorithm on G π (x), with the following changes: perform advances along ϵ-admissible arcs; relabel by adding ϵ/ to π i ;

11 The Cost Scaling Algorithm Initialize: Set π =, ϵ = C, x = Create G π (x) while ϵ 1/n do (ϵ-scaling phase) for each arc (i, j) with c π (i, j) ϵ/, push r(i, j) units of flow along arc (i, j), adjusting G π (x) accordingly. while E do (label/augment routine) Choose s E. Perform the Shortest Augmenting Path Algorithm on G π (x), doing advances along ϵ-admissible arcs and relabeling in a retreat from node i by adding ϵ/ to π i. When a node t D is reached, augment flow along ϵ/-shortest path P by min{r ij : (i, j) P } {e(s), e(t)}. If no such t is reached STOP, problem is infeasible. end do while Set ϵ = ϵ/. end ϵ-scaling phase

12 Example of the Cost Scaling Algorithm (all capacities are 15) Start of First Scaling Phase (ϵ = 16:) πi $9 5 c ij $9 $7 s b i 1 1 $7 $15 $7 t $16 $7 4 π-values Retreat Node (augment) Augment along path 1 6 7, δ = 1. End of first scaling phase. x, ij cπ ij $9 πi 8 $1 5 $7 5,$7 5,$7 5,$ , $ , $7 1, $7 $8 $7 4

13 Second Scaling Phase (ϵ = 8) Flow on arcs (1,),(,6),(6,7) set to. Starting network: x, ij π i 8 $1 5 cπ ij $9 $7 s b i 1 1 5,$7 5,$7 6 5,$ t 1 $8 $7 4 π-values Retreat Node (augment) Augment along path 1 5 7, δ = 1.

14 End of second scaling phase. x, ij cπ ij 5,$1 π i 1, $ ,$1 5 1, $1 1, $ 5,$ 1 $ $ 6 $ $ $ 4 4

15 Third Scaling Phase (ϵ = 4:) Flow on arc (5,7) set to. Starting flow: x, ij cπ ij 5,$1 π i 1, $ ,$1 5 1, $1 s 1 b i 5,$ 1 $ $ 6 $ t 1 $ $ 4 4 π-values Retreat Node (search ) (augment) Augment along path , δ = 1. End of third scaling phase. π c ij $ 1 πi 1 4 $1 5 $ 1 $ $1 6 $ ,$ 1,$ 5, $1 1,$1 4 6

16 Optimal Flow The current flow of 1 units on arcs (1,4) and (4,7) turns out to be optimal. The final residual graph, with the given π values has no nonnegative cycles. Solving the shortest path problem on G(x) with the final reduced costs (or the original costs) gives the following reduced costs satisfying the Reduced Cost Optimality Conditions: π c ij $ π 1 $ $ 6 $ ,$ i 14 1,$ $ 4 7 5,$ 5 5 $ 1,$

17 Lemma 1: Let x and π be the current ϵ/- optimal pseudoflow and node numbers, respectively, at some point of the ϵ/-phase of the Cost Scaling Algorithm, and let x and π be the flow and node numbers, respectively, at the end of the previous ϵ-phase. Let P be an (s, t)-path in G(x) such that the (t, s)-path P of reverse arcs of P existed in G(x ). Then (π s π s ) (π t π t ) < nϵ. Proof: Let l(p ) and l( P ) be the length of P and P, respectively, and let p be the number of arcs in each. Then But c π ij l(p ) = (i,j) P l( P ) = (i,j) P c π ij π s + π t c π ij π t + π s ϵ/, since x is ϵ/-optimal; c π ij ϵ, since x was ϵ-optimal; l( P ) = l(p ), since their edges are oppositely directed. Thus pϵ π s + π t l( P ) = l(p ) pϵ/ + π t π s. and so (π s π s ) (π t π t ) < pϵ nϵ.

18 Lemma 1.4 No node will be relabeled more than n times during an ϵ-scaling phase. Proof: Let x and π be the ϵ-optimal flow and node numbers at the end of the ϵ-scaling phase of the Cost Scaling Algorithm, and let x and π be the ϵ/-optimal pseudoflow and node numbers at some point in the label/augment routine from node s during the ϵ/-scaling phase. Now consider the flow x x. Since x is a feasible flow on G, then x x is then a flow on G(x) with supplies and demands being the imbalances e. We can therefore decompose this flow into a set of E-to-D paths and cycles on G(x), and further, the reverse of these paths and cycles also appeared in G(x ). Thus the conditions of Lemma 1 are satisfied on any of these paths. From the description of the Cost Scaling Algorithm it follows that a deficit node t is never relabeled until its deficit is satisfied, so that π t = π t. Thus for any excess node s, Lemma 1.4 insures that π s will never increase more than nϵ during the label/augment part of the algorithm. But since each relabeling of a node increases its node number by ϵ/, then there can be no more than n relabels of any node.

19 Complexity of the Algorithm Facts: 1. After each ϵ-scaling phase, the current x is ϵ/-optimal; when ϵ < 1/n the current x is optimal (Lemma 1.).. ϵ starts at value C, the algorithm ends when ϵ 1/n. Thus the total number of ϵ-scaling phases is log nc.. No node will be relabeled more than n times during an ϵ-scaling phase (Lemma 1.4). 4. During an ϵ-scaling phase: for each s E chosen there will be at most nm total augment steps, with total time complexity O(n m). (This follows exactly the same argument as for the Shortest Augmenting Path Algorithm, independent of how many t nodes are reached.) 5. The total time for an ϵ-scaling phase is therefore O(n m). Theorem: The cost scaling algorithm runs in O(n m log nc) time.

20 The Minimum Mean Cycle-Canceling Algorithm Recall the Cycle Canceling Algorithm: Initialize: Find some feasible flow x in G and form the residual graph G(x). while G(x) contains a negative-cost cycle Find some negative cost cycle W in G(x) Set δ = min{r ij : (i, j) W } Augment δ units of flow around W and update G(x) end while Minimum mean cycle-canceling algorithm: In the cycle-cancelling algorithm, choose the cycle W with minimum mean cost c(w )/ W ( W = the number of arcs in W ). Theorem 1.18: The minimum mean cycle-canceling algorithm performs O(nm log n) iterations and runs in O(n m log n) time. Proof: (Two sections and seven pages. Pass.)

21 Preflow Push Algorithms Most of the most theoretically efficient algorithms for min-cost flow involve pushing flow along arcs rather than paths. We outline some of them here. Algorithm Description Complexity Preflow Push Cap. Scaling Wave Implementation Double Scaling Repeated Capacity Scaling Enhanced Capacity Scaling Capacity scaling using the preflow push technique. Modification of cost scaling algorithm where nodes of the (acyclic) admissible network are examinined in topological order. Capacity scaling algorithm is embedded in the ϵ-scaling phase of the cost scaling algorithm (applied to the transportation transformation of the problem). Capacity scaling algorithm is used to find the optimal solution to the dual problem, and the min flow is constructed from this. Capacity scaling algorithm, but using the previous flow in the succeeding phase rather than starting a new flow. O(n m log(nc)) O(n log(nc)) O(nm log U log(nc)) O(n log n S(n, m)) O(m log n S(n, m)) where S(n, m) is the best strongly polynomial complexity of a shortest path algorithm using nonnegative arc weights (currently O(m+n log m)). The Enhanced Scaling Algorithm is the best known strongly polynomial algorithm.