1 / 36

This class

CS 575 Design and Analysis of Computer Algorithms Professor Michal Cutler Lecture 19 November 10, 2005. This class. Dijkstra The max-flow min-cut problem The flow network What is a flow? The residual network Augmenting paths Cuts. Shortest Path Algorithms.

paulguzman
Télécharger la présentation

This class

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CS 575Design and Analysis ofComputer AlgorithmsProfessor Michal CutlerLecture 19November 10, 2005

  2. This class • Dijkstra • The max-flow min-cut problem • The flow network • What is a flow? • The residual network • Augmenting paths • Cuts

  3. Shortest Path Algorithms • Dijkstra’s algorithm does NOT allow negative edges. • Uses a greedy heuristic. • undirected/directed graph • Bellman-Ford’s and Floyd’s algorithms work correctly for any graph and can detect negative cycles.

  4. Main idea • Assume that the shortest distances from the starting node s to the rest of the nodes are d(s, s)  d(s, s1)  d(s, s2)  …  d(s, sn-1) • In this case a shortest path from s to si may include any of the vertices {s1, s2… si-1} but cannot include any sj where j > i. • Dijkstra’s main idea is to select the nodes and compute the shortest distances in the order s, s1, s2 ,…, sn-1

  5. Dijkstra’s greedy selection rule • Assume s1, s2… si-1 have been selected, and their shortest distances have been stored in Solution • Select node si and save d(s, si) if si has the shortest distance from s on a path that may include only s1, s2… si-1 as intermediate nodes. We call such paths special • To apply this selection rule efficiently, we need to maintain for each unselected node v the distance of the shortest special path from s to v, D[v].

  6. Updating D[ ] • After adding near to Solution, D[v] of all nodes v Ï Solution are updated if there is a shorter special path from s to v that contains node near, i.e., if (D[near] + w(near, v ) < D[v]) then D[v]=D[near] + w(near, v ) D[near ] = 5 2 Solution after adding near 2 3 s D[ v ] = 9 is updated to D[v]=5+2=7 3 6

  7. Dijkstra’s Algorithm for Finding the Shortest Distance from a Single Source Dijkstra(G,s)1. for eachv V2. doD [ v ]   3. D [ s ]  04. PQ make-PQ(D,V)5. whilePQ   6. donear  PQ.extractMin () 7. for eachv  Adj(near )8 ifv  PQ &&D [ v ] > D [ near ] +w( near ,v )9. then D [ v ]  D [ near ] +w( near, v )10. PQ.decreasePriorityValue (D[v], v )11. return the label D[u] of each vertex u

  8. 1. for eachv V2. doD [ v ]   3. D [ s ]  04. PQ make-PQ(D,V)5. whilePQ   6. donear  PQ.extractMin () 7. for eachv  Adj(near )8 ifD [ v ] > D [ near ] + w( near ,v )9. then D [ v ]  D[near] + w(near,v)10. PQ.decreasePriorityValue (D[v ], v )11. return the label D[u] of each vertex u Assume a node in PQ can be accessed in O(1) ** Decrease key for v requires O(lgV ) provided the node in heap with v’s data can be accessed in O(1) Using Heap implementation Lines 1 -4 run in O (V ) Max Size of PQ is | V | (5) Loop = O (V ) - Only decreases(6+(5)) O (V )  O( lg V )(7+(5)) Loop = O(deg(near) ) =O( E ) (8+(7+(5))) O(1)O( E )(9) O(1)(10+(7+(5))) Decrease- Key operation on the heap can be implemented in O( lg V)  O( E ). So total time for Dijkstra's Algorithm is O ( V lg V + E lg V ) What is O(E ) ?For Sparse Graph = O(V lg V ) For Dense Graph = O(V2 lg V ) Time Analysis

  9. Example a 4 4 2 b c 2 1 10 5 d e

  10. Solution for example

  11. Maximum flow and minimum cut Cornerstone problem in combinatorial optimization Many applications: Network connectivity, bipartite matching, project selection, network reliability, data mining, image processing, job scheduling for real time applications, baseball elimination, etc.

  12. A flow network v1 5 10 v3 t 7 15 12 s v2 6 • A directed graphG=(V, E) in which each edge (u, v) E has a nonnegative capacityc(u, v)  0. • If (u, v) E , c(u, v) = 0 • Contains a source node s, and a sink node t • Every node u V is on a directed path from s to t.

  13. Networks

  14. 11,15 Flow • An s-t flow is a function f: VV-> R that satisfies: • Capacity Constraint: (the flow cannot exceed the capacity) for all u, v, f(u, v) c(u, v) • Skew symmetry: (if f is flowing from u to v, –f is flowing from v to u) for all u, v, f(u, v) = -f(v, u)

  15. Flow • Flow conservation: • sum of flows from a node is 0, or • the sum of the flows that leave a node is equal to the sum of the flows that enter it • for all uV –{s, t}

  16. Flow conservation example v1 5, 5 v3 v4 11,15 6,12 v2 • f(v3, v1) +f(v3, v2)+f(v3, v4)=0 • f(v3, v1)= -f(v1, v3)= -5 • f(v3, v2)= -f(v2, v3)= -6 • -5 - 6 + 11= 0 • Flow entering v3 =f(v1, v3) + f(v2, v3)= 5+6 =11, • Flow leaving v3 = f(v3, v4) = 11

  17. An example of a flow v1 5, 5 5,10 v3 t 0, 7 11,15 6,12 s v2 6, 6 • The capacity constraints and skewed symmetry are satisfied for all edges • Conservation is satisfied.

  18. Max flow problem v1 5, 5 9,11 v3 t 4, 7 15,15 10,12 s v2 6, 6 • Find s-t flow f* that maximizes net flow out of the source • The current value of the flow: • For our example |f| = f(s, v1) + f(s, v2) = 9 +6 = 15 is a max flow

  19. t1 s1 s2 S t2 s3 What if there is more than one source? Or sink?    s t   • Add a super sources and a super sinkt, and an edge from s to each source and from each sink to t. The capacity of these edge is 

  20. Working with flows • If X and Y are sets of vertices L1 • For all XV, f(X, X) = 0 • For all X, YV, f(X, Y) = -f(Y, X) • For all X, Y, ZV, with XY = , f(XY, Z)=f(X, Z)+f(Y, Z) and f(Z, XY)= f(Z, X) + f(Z, Y)

  21. Working with flows v1 5, 5 5,10 v3 t 0, 7 11,15 6,12 s v2 6, 6 • Let X={s, v1} Y={v2, v3}, Z={t} • f(X, X)=f(s, v1)+f(v1, s)=5–5= 0 • f(X, Y)= f(s, v2) +f(v1, v2)+f(v1, v3) = 11, f(Y, X) = -11 • f(XY, Z)=f(v3, t) = 11, f(X, Z)=0 and f(Y,Z)=f(v3, t) =11

  22. |f| = total flow out of source = total flow into the sink |f| = f(s,V) (By def.) = f(V,V) - f(V - s, V) (L.1-3) = - f(V - s, V) (L.1-1) = f(V, V - s) (L.1-2) = f(V, t) + f(V, V - s - t) (L.1-3) = f(V, t) (flow conservation) • For next slideU is the maximum capacity, V the number of nodes and E the number of edges in the network.

  23. Some History

  24. The Ford and Fulkerson method • Initialize flow f to 0 • while there exists an augmenting path p in residual network • augment flow f along p • returnf • Need to explain what are augmenting paths (and residual networks)

  25. Residual Networks • The amount of additional flow we can push from node u to v before exceeding the capacity c(u, v) is the residual capacity of (u, v) cf(u, v) = c(u, v) –f(u, v) Examples c(u, v)=16 and f(u, v)=11-> cf(u, v) =5 c(u, v)=16 and f(u, v)=-4 -> cf(u, v) =20

  26. Residual networks • Given a flow network G=(V, E) and a flow f, the residual network of Ginduced by f is Gf = (V, Ef) where Ef = {(u, v)VV| cf(u, v) > 0} • The edges in Efare either edges in E or their reversals

  27. Residual networks – how many edges? • If f(u, v) < c(u, v) for (u, v) E -> cf(u, v) > 0 and (u, v)  Ef • If f(u, v)<0, for (v, u) E, and f(v, u)>0 -> cf(u, v) > 0 and (u, v)  Ef • If neither (u, v) E or (v, u) E then c(u, v)=c(v, u)=0, f(u, v)=f(v, u)=0, and cf(u, v)=cf(v, u)=0 ((u, v)  Ef) • So |Ef| 2|E|

  28. v1 5, 5 G 5,11 v3 t 0, 7 11,15 6,12 s v2 6, 6 v1 5 11 Gf 5 6 v3 t 7 4 6 s v2 6 6 The residual network

  29. Lemma L.2 • Given a flow network G with flow f, and a residual network induced by f with flow f’. There is a flow in G with value |f + f’|=|f| + |f’| • Proof outline: • Verify that skew symmetry, capacity constraint and flow conservation are obeyed, and the total flow from s is |f| + |f’|

  30. Augmenting paths • An augmenting pathp is a path from s to t in the residual network Gf. • The maximum amount by which we can increase the flow on an augmenting path p is the residual capacity of p, cf(p) = min{cf(u, v)| (u, v) is on p}

  31. v1 5 11 5 6 v3 t 7 4 6 s v2 6 6 An augmenting path • cf(p) = min{cf(u, v)| (u, v) is on p} = min{6, 7, 6, 4}=4 • Adding 4 to the edges in G results in the maximum flow (for this example) Gf

  32. The updated flow v1 5, 5 9,11 v3 t 4, 7 15,15 10,12 s v2 6, 6

  33. Cuts • A cut (S,T), is a partition of V into S and T=V-S, such that sS, and t  T • The net flow across a cut (S, T) is f(S, T) • The capacity of the cut (S, T) is c(S,T) • Only the capacity of edgesfromStoT are included in c(S, T) .

  34. Examples of Cuts T S v3 v1 6, 6 5, 5 1, 6 11, 13 v4 t 5, 5 17, 20 12, 12 s v2 6, 6 • c(S, T) = 6 + 12=18 and f(S, T) = 6 + 12 – 1= 17 = f  c(S,T). • Only the capacity of edges from S to T are included in c(S, T) .

  35. Examples of Cuts v3 v1 6, 6 5, 5 1, 6 11, 13 v4 t 5, 5 17, 20 12, 12 s v2 6, 6 • f(S, T) = c(S, T) = f = 5+ 12=17

  36. Demos • Relevant Algorithm Animations/Visualizations (in Java) • http://www.hig.no/~algmet/animate.html • http://www-b2.is.tokushima-u.ac.jp/~ikeda/suuri/maxflow/Maxflow.shtml • (Alt-tab)

More Related