Maximum Flow. Reading: CLRS Sections (skipping Lemma 26.8, Theorem 26.9 (2 nd edtion) or Lemma 26.7, Theorem 26.

Transcription

1 Maximum Flow Reading: CLRS Sections (skipping Lemma 26.8, Theorem 26.9 (2 nd edtion) or Lemma 26.7, Theorem 26.8 (3rd edition) CSE 2331 Steve Lai

2 Flow Network A low network G = ( V, E) is a directed graph with a source node s V, a sink node t V, a capacity unction c. Each edge ( uv, ) Ehas a nonnegative capacity cuv (, ) 0. I ( uv, ) E, assume cuv (, ) = 0. Also, assume that every node v is on some path rom s to t. 2

3 Flows G = ( V, E), capacity unction c, source s, sink t. A low is a real-valued unction : V V R, satisying Capacity constraint: uv, V, ( uv, ) cuv (, ). Skew symmetry: uv, V, ( uv, ) = ( v, u). Flow conservation: u V { s, t}, v V ( uv, ) = 0 The value o a low is = ( sv, ). v V The maxlow problem is to ind a maximum low. 3

4 Some Properties o Flows I no edge between uand v, then ( uv, ) = ( vu, ) = 0. Flow conservation implies: u V { s, t}, Total positive low into u= Total positive low out o u. For XY, V, deine ( XY, ) = ( xy, ). x X y Y ( X, X) = 0. ( XY, ) = Y (, X). ( X Y, Z) = ( X, Z) + ( Y, Z), i X Y =. ( Z, X Y) = ( Z, X) + ( Z, Y), i X Y =. 4

5 Residual networks and augmenting paths Let G = ( V, E) be a low network and a low. Residual capacity o ( uv, ) is c ( uv, ) = cuv (, ) ( uv, ). Residual network induced by is G = ( V, E ), where { } E = ( uv, ) V V: c ( uv, ) > 0. Augmenting path: a simple path p rom s to t in G. Residual capacity o p : { } c ( p) = min c ( uv, ) : ( uv, ) is in p. 5

6 Ford-Fulkerson Method Given a low network G = ( V, E) with source s and sink t, Initialize low to 0 while there exists an augmenting path p do do augment low along p return 6

7 Ford-Fulkerson Gst,, ( ) or each edge ( uv, ) EG ( ) do ( uv, ) 0 (,) vu 0 while there exists an augmenting path p in residual network do { } c ( p) = min c ( uv, ) : ( uv, ) is on p or each edge ( uv, ) is in p do ( uv, ) = ( uv, ) + c ( p) (,) vu = (,) uv G 7

8 Analysis The running time depends on how the augmenting path p is determined. ( ) I capacities are integers, the running time is O E, where is the value o the maxlow. Each iteration can be done in O( E) time. At most iterations. Integrality Theorem. I all capacities are integers, the low produced by the Ford-Fulkerson method has the property that ( uv, ) is an integer or all u, v V. 8

9 Example s a 10 1/4 c 12/12 11/14 b 7/7 d t s a 11 3 c b 7 d t (a) Flow network and low (b) Residual network and augmenting path p with c ( p ) = 4 s a 10 1/4 c 12/12 11/14 b 7/7 d t s a 11 3 c b 7 d t (c) Augmented low (d) No augmenting path 9

10 Lemma 1. Let G be the residual network induced by low. Let be a low in G. Then + is a low in G with + = +. Proo. Skew symmetry: ( + )( uv, ) = ( uv, ) + ( uv, ) = + = + Capacity constraint: ( ) ( ( vu, ) ( vu, )) ( )( vu, ) + ( uv, ) ( uv, ) + c ( uv, ) = ( uv, ) + cuv (, ) ( uv, ) = cuv (, ) ( ) 10

11 ( ) ( ) { } Flow conservation: or all u V st,, Finally, + ( uv, ) = ( uv, ) + ( uv, ) v V v V v V = 0+ 0= 0 + = + (,) s v v V = (,) sv + (,) sv v V = + v V 11

12 Lemma 2. I p is an augmenting path in G, then is a low in G with = c ( p) > 0, where p p c ( p) i ( uv, ) p p( uv, ) = c ( p) i ( vu, ) p 0 otherwise Corollary 3. I p is an augmenting path in G, then augmenting along p yields a low in G with value + > p. This gives part o the basis o Ford-Fulkerson. Next we show that when the algorithm terminates it returns a maxlow. 12

13 Cuts ( ) A cut ST, o a low network G= ( V, E) is a partition o o V into S and T = V S such that s S and t T. I is a low, ( ST, ) denotes the net low across the cut ( ST, ). Capacity o ( ST, ) is cs (, T) = cuv (, ). u Sv T Example: CST (, ) = cab (, ) + ccd (, ) = = 26. ( S, T) = ( a, b) + ( c, d) + ( c, b) = = 19. a 12/12 b s S 10 1/4 c 11/14 7/7 d t T 13

14 Lemma 4. For any cut ( ST, ), ( S, T) =. Proo. Note that ( uv, ) = 0 u st,. ( S, T) = ( S, V) ( S, S) = ( SV, ) = ( s, V) + ( S sv, ) = (, s V ) = Corollary 5. = ( ST, ) c( S, T ). 14

15 The Max-low Min-cut Theorem. Theorem. The ollowing conditions are equivalent: 1. is a maxlow in G. 2. The residual network G contains no augmenting paths. 3. = cst (, ) or some cut ( ST, ) in G. Proo. (3) (1): Immediately ollows rom Corollary 5. (1) (2): Immediately ollows rom Corollary 3. (I G contains an augmenting path p, augmenting along p increase the low.) will 15

16 2. The residual network G contains no augmenting paths. 3. = cst (, ) or some cut ( ST, ) in G. (2) (3): Suppose G contains no augmenting path. Deine { } S = v: there is a path rom s to v in G, T = V S. ( ST, ) is a cut since s S and t T (no path rom s to t in G ). For all u S, v T, we have ( uv, ) = cuv (, ), since otherwise ( uv, ) E and vwould be in S. So, By Lemma 4, = ( ST, ) = c( S, T). ( S, T) = cs (, T ). 16

17 Edmonds-Karp Algorithm In the while loop o Ford-Fulkerson, ind the augmenting path p with a breadth-irst search, that is, the augmenting path is a shortest path rom s to t in the residual network, where "shortest" is in terms o number o edges. ( 2 ) Running time: O VE 17

18 Flow: an alternative deinition (CLRS, 3rd ed.) Let G = ( V, E) be a low network with a capacity unction c, source s, and i.e., sink t. Assume G i ( uv, ) Ethen ( vu, ) E. has no parallel edges, A low is a real-valued unction : V V R, satisying Capacity constraint: uv, V, 0 ( uv, ) cuv (, ). Flow conservation: u V { s, t}, v V ( ) ( vu, ) = ( uv, ) i.e. V, u = u, v V ( V ) ( ) ( ) ( V s) The value o a low is = s, V,. Note: when ( uv, ) E, ( uv, ) = 0. 18

19 Some o these properties do not hold any more I no edge between uand v, then ( uv, ) = ( vu, ) = 0. Flow conservation implies: u V { s, t}, Total positive low into u= Total positive low out o u. For XY, V, deine ( XY, ) = ( xy, ). x X y Y ( X, X) = 0. ( XY, ) = Y (, X). ( X Y, Z) = ( X, Z) + ( Y, Z), i X Y =. ( Z, X Y) = ( Z, X) + ( Z, Y), i X Y =. 19

20 Residual networks and augmenting paths Let G = ( V, E) be a low network and a low. Residual capacity o ( uv, ) is cuv (, ) ( uv, ) i ( uv, ) E c (,) uv == (,) vu i (,) vu E 0 otherwise Residual network induced by is G = ( V, E ), where { > } E = ( uv, ) V V: c ( uv, ) 0. Augmenting path: a simple path p rom s to t in G. 20

21 Networks with multiple sources and sinks G = ( V E) { 1 2 m} { }, : low network with m sources: s, s,, s n sinks: t, t,, t 1 2 m s s1 s2 s1 G t1 t2 t3 t 21

22 Maximum Bipartite Matching G = ( V E), : undirected graph Bipartite graph: i V can be partitioned into L and R such that all edges in E go between L and R. Theorem: G is bipartite i it has no cycles o odd length. Matching: a set o edges M E such that every vertex in V is an endpoint o at most one edge in M. Maximum matching: a matching with the max cardinality. A maximum matching o a bipartite graph can be ound using the Ford-Fulkerson method. 22

23 Maximum Bipartite Matching G = ( V E), : undirected graph Bipartite graph: i V can be partitioned into L and R such that all edges in E go between L and R. Theorem: G is bipartite i it has no cycles o odd length. Matching: a set o edges M E such that every vertex in V is an endpoint o at most one edge in M. Maximum matching: a matching with the max cardinality. A maximum matching o a bipartite graph can be ound using the Ford-Fulkerson method. 23

24 Edge-Disjoint Paths G = ( V E), : a graph Edge-disjoint paths: two paths are edge-disjoint i they do not share any edges. ( V E) Problem: Given a directed graph G =, and two nodes s, t, ind a maximum number o edge-disjoint paths rom s to t. 24

25 Node-Disjoint Paths G = ( V E), : a graph Node-disjoint paths: two paths rom s to t are node-disjoint i they do not share any intermediate nodes. ( V E) Problem: Given a di rect ed graph G =, and two nodes s, t, ind a maximum number o node-disjoint paths rom s to t. 25

26 Image Segmentation A undamental problem in computer vision. Given a digital image (a set o pixels), we want to partition it into multiple segments. In a simple case, we just want to divide the image into two segments: the oreground and the background. Represent the image by an undirected graph G = ( V, E), where V is the set o pixels and there is an edge between two pixels i there are neighbors. Each pixel i has a likelihood/goodness ai > 0 to be in the oreground and a likelihood/goodness bi > 0 to be in to the background. 26

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28 Each edge ( i, j) E is associated with a separation penalty p = p > 0, which is incurred i pixels i and j ij ji are placed in dierent segments. Proble m: Given a pixel graph G = ( V, E), likelyhood unctions + + a, b: E R and penalty unction p: E R, we want to partition V into two sets A and B that maximize { } i i ij q( A, B) = a + b p : ( i, j) E, i, j in di segments i A i B Or, equivalently, minimiz q ( AB, ) = a+ b qab (, ) i V i V e i i a { : (, ),, in dierent segments} i b i pij i j E i j = + + i B i A 28

29 We can solve the image segmentation problem by converting it to a low network. Let G = ( V, E) be the pixel graph. Introduce two new vertices: a source s and a sink t. Connect s to each pixel i V with capacity ai. Connect t to each pixel i V with capacity bi. Replace each edge ( i, j) E with two directed edges ( i, j) and ( ji, ) with capacities pij and pji. What's the relationship between the pixel graph G = ( V, E) and ( V E ) the constructed low network G =,? 29

30 a i p ij Pixel graph G = ( V, E) b k 30