Preflow-Push Algorithm

Transcription

1 Preflow-Push Algorithm An s-t preflow obeys the following constraints: Capacity constraints: 0 f(e) c(e) e f (v) = f (u,v) f (v,w) 0 u V w V A labeling h assigns nonnegative integers to vertices. h(t) = 0, h(s) = n, h(u) 0 for all other vertices u in V. A labeling h is compatible with an s-t preflow f if: h(t) = 0, h(s) = n. for all edges (v,w) in E f, h(v) h(w) + 1.

2 Initial preflow f assigns f(e) = c(e) for each edge leaving s, and f(e) = 0 for all other edges. Initial labeling h assigns h(t) = 0, h(s) = n. Initial preflow f and labeling h are compatible. Lemma 1. There is no s-t path in G f if s-t preflow f is compatible with labeling h. Lemma 2. If s-t flow f is compatible with labeling h, then f is a flow of maximum value. 2

3 Copyright 2005 Pearson Addison-Wesley. All rights reserved

4 Algorithm 7.4, page 360 Copyright 2005 Pearson Addison-Wesley. All rights reserved

5 Algorithm 7.5, page 360 Copyright 2005 Pearson Addison-Wesley. All rights reserved

6 Lemma 3. Throughout the preflow-push algorithm, the preflow f and the labeling h are compatible. Pf. When push operation is performed from vertex u to vertex v, the edge (v,u) can appear in G f. h(v) h(u) + 1. Why? When relabel operation is performed on vertex u, h(u) < h(v) for all edges (u,v) in E f. Why? For any edge (v,u) in E f, h(v) Why? h(u) + 1 after the relabel operation. 6

7 Algorithm 7.6, page 360 Copyright 2005 Pearson Addison-Wesley. All rights reserved

8 Preflow-Push Algorithm Analysis Lemma 4. If vertex v has excess preflow e f (v) then there is a path from v to s in G f. Pf. Let all vertices w such that there is a path in G f from w to s be in set A. The remaining vertices are put in set B. There can be no positive flow f(e) on any edge (u,v) in G f going out of set A. Why? Consider the sum of all excess in vertices in set B. (v) 0 e in out f = f (v) f (v). v B v B There can be no positive flow into B. This implies that each excess in B is 0. Why? 8

9 Preflow-Push Algorithm Analysis Number of relabel operations: Any vertex v in V can have label h(v) at most 2n 1. Why? Consider a vertex v with excess. v s can be changed only by a relabel operation. This there is a path in G f from v to s. This implies that the label of v can be at most 2n 1. Thus the total number of relabel operations is at most 2n 2. 9

10 Number of saturating pushes: (A push is saturating is it saturates the edge) When a push occurs along edge (u,v), the edge dissappears from G f. It can reappear only after a push along edge (v,u). But the label of vertex v must have increased by at least 2 for that to happen. Why? Thus vertex v can participate at most (2n 1)/2 times in push along edge (u,v). Therefore there are at most 2mn saturating pushes. 10

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12 Number of nonsaturating pushes: Let φ( f,h) = h(v) v:e f >0 be the sum of the heights of all vertices with excess. A nonsaturating push decreases φ( f,h) by at least 1. Why? A relabel operation increases φ( f,h) by at most 1. There are at most 2n 2 relabel operations. Each saturating push increases φ( f,h) There are at most 2nm saturating pushes. by at most 2n 1. Why? Thus φ( f,h) increases by at most 4mn 2. Since φ( f,h) is nonnegative, and decreases by at least for each nonsaturating push, there are at most 4mn 2 nonsaturating pushes. 12

13 Implementation Data Structure: All vertices with excess can be placed in a list. Choosing a vertex on this list, as well as updating the list (when a vertex with excess no longer has excess, or a new vertex with excess has to be added) can be done in O(1) time. Why? In the Ford-Fulkerson Algorithm, choosing the right augmenting path lead to a better algorithm. Similarly, in the Preflow-Push Algorithm, can choosing the right vertex with excess lead to a better algorithm? Choose the vertex with maximum height. 13

14 Choosing vertex with maximum height Algorithm: At each stage, a vertex with excess at maximum height is chosen for a push operation. What is an upper bound on the number of nonsaturating push operations? The maximum number of relabel operations is 2n 2. Therefore the total increase in height is at most 2n 2. Therefore the number of times the maximum height H changes is at most 4n 2. Each vertex can have at most one nonsaturating push operation at a given H. Why? Therefore the maximum number of nonsaturating push operations at a given H is at most n. Therefore the maximum number of nonsaturating push operations in total is at most 4n 3. 14

15 Data Structure: Maintain a (doubly-linked) list of all heights. For a given height, maintain a list of all vertices with excess. Maintain an adjacency list representation of the graph, with both forward edges (as in the original graph G) and backward edges (which can arise in the residual graph). For each forward (backward) edge, maintain a pointer to the backward (forward) edge. Also keep the flow value f, as well as the capacity of the edge. The above data structure and the initial flow and labeling can be constructed in O(m) time. 15

16 A pointer to the list with maximum height can be maintained. This can be used to choose a vertex with excess at maximum height. Once a vertex v is chosen, how is an edge for push chosen? For each vertex, all the edges adjacent on it are maintained in some arbitrary order in a list. For each such list, a pointer current(v) points to the current edge considered for push. When a vertex with excess (at maximum height) is chosen for push, the edge pointed to by current(v) is chosen for push. After nonsaturating push, current(v) points to the same edge. After saturating push, current(v) points to the next edge in the list. If we come to the end of the list, then current(v) is reset to the start of the list. 16

17 Algorithm: Choose a vertex v at maximum height. Choose edge pointed to by current(v) to push flow. Update current(v) by advancing if push saturating. Continue with vertex v if push saturating. If push nonsaturating, choose another vertex at maximum height H (if possible) or go to height H-1 to choose another vertex. If no other vertex at H, then there is a vertex with excess flow at H-1. Why? Relabel v if push saturating, and current(v) advances to end of list. After relabeling v, reset current(v) to start of list. 17

18 Lemma 5. After current(v) is advanced from edge (v,w) to next edge in list, push cannot be applied to edge (v,w) until v gets relabeled. Pf. If a push occurred on edge (v,w), then current(v) stays with edge (v,w) if push is nonsaturating. It will be used again if v gets relabeled. If the push was saturating, then the edge dissappears, and reappears only after v gets relabeled. Why? If no push occurred, and current(v) advances to the next edge in list, then edge (v,w) not present in G f. Either (v,w) is saturated (and v has to be relabeled to reappear), or h(v) relabeled for edge to reappear in G f. h(w). Againv has to get 18

19 Running time Analysis: Each time a vertex gets relabeled, current(v) reset to start of list. Total number of times a vertex can get relabeled is at most 2n. Total number of times current(v) gets advanced is 2nd v (where d v is the number edges adjacent to v). Total number of times current gets advanced (for all vertices) is at most 2n = O(mn). d v v V Total number of nonsaturating push operations is O(n 3 ). Therefore total running time is O(n 3 ). 19

20 Copyright 2005 Pearson Addison-Wesley. All rights reserved

21 Copyright 2005 Pearson Addison-Wesley. All rights reserved