# 1 Primal-dual Approximation Algorithms for Integral Flow and Multicut in Trees Garg, N., Vazirani, V. V. and Yannakakis, M., Algorithmica, Vol. 18, 1997,

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1 Primal-dual Approximation Algorithms for Integral Flow and Multicut in Trees Garg, N., Vazirani, V. V. and Yannakakis, M., Algorithmica, Vol. 18, 1997, pp. 3-20. Advisor: R. C. T. Lee Speaker: F. L. Lin National Chi Nan University

2 Minimum Multicut Problem Given a graph G=(V, E) with a positive capacity c e on every edge, and a list of vertex pairs, {(s 1, t 1 ), …, (s k, t k )}, find a minimum weight set of edges separating each pair of vertices in the list. –e st denotes the edge st. Vertex pairs ={(a, f), (c, f), (h, d)} Multicut = {e ah, e bf, e ef } a h g c d f e b 5 7 1 1 2 3 1 2 1 a h g c d f e b 5 7 1 1 2 3 1 2 1

3 The minimum multicut problem is NP-hard even if it is restricted to trees of height 1 and unit capacity edges. This paper deals with undirected trees. – denotes the unique path between s i and t i in the tree. vertex pairs ={(b, e), (c, d), (a, f)} p be = bghj i e p cd = cghj i d p af = ahjr f r jf h i agde bc 4 4 33 4 2 1 1 1 1

4 Since the minimum multicut problem is NP-hard, we have to find an approximation algorithm for it. This paper used the primal-dual approach. But we found that we could give the algorithm without mentioning the primal-dual approach which is quite hard to understand and confusing.

5 For this problem, we can compute a lower bound of our solutions immediately. This lower bound is obtained by reducing the cost of each edge in the path and there will be at least one zero cost in every path. Let. We observe that is a lower bound of cutting the path. –For example, f ac = min{4, 2, 1} =1 which is a lower bound of cutting p ac. c e = c e -,. Let Q be the set of edges whose cost, c e, becomes zero in this step. h ag bc 4 2 1 1

6 For example –f ac =min{4, 2, 1}=1, the lower bound for cutting the path p ac is 1. –Q={e gc } –f ab =min{3,1,1}=1, the lower bound for cutting the path p ab is 1. –f ac +f ab =2, the lower bound for cutting the path p ac and p ab is 2. –Q={e gc, e hg, e gb } h ag bc 4 2 1 1 vertex pairs ={(a, c), (a, b)} p ac = ahgc p ab = ahgb h ag bc 3 1 1 0 h ag bc 4-1 2-1 1 1-1 h ag bc 3 1 1 0 h ag bc 3-1 1-1 0 h ag bc 2 0 0 0

7 Because the graph is a tree, we may omit redundant edges from the multicut. When some pairs have the same least common ancestor, v, and the edges are the ancestor and the child in Q, the path passes through the child and the path must pass through the ancestor. We only retain the ancestor. Let frontier (v) = Q – the edge that is the descendant of other edges in Q. Claim The union of all frontiers is a multicut.

8 For example, pairs (a, c) and (a, b) have the same least common ancestor, h, and e hg is the ancestor of e gb and e gc. The edge e gb and e gc are omitted. frontier(h)={e hg }. 0 0 h ag bc 2 0 vertex pairs = {(a, c), (a, b)} p ac = ahgc p ab = ahgb Q={e gc, e hg, e gb } frontier(h)={e hg }

9 We move down the tree one level at a time and join frontiers to build the multicut. Considering vertex v, we include an edge only if no edge along the path from e to v is already included in the multicut. Let M be the set of edges picked. For example, suppose that M={e fg } and consider frontier(f), e fg is in the path from e gd to f, e gd is not picked to M. Lemma M is a multicut. vertex pairs = {(b, d),(a, c)} p bd = bfgd p ac = aefgc frontier(e) = {e fg } M = {e fg } frontier(f) = {e gd } e af bg cd 0 0 2 1 2 2

10 Algorithm multicut_integral-flow(r) Input : a tree T = (V, E) with a positive capacity c e on every edge, and a list of vertex pairs, {(s 1, t 1 ), …, (s k, t k )}. 1.for current_level = max_level downto 0 do for all current_level do 1.1 for each pair (s i, t i ) such that its least common ancestor is at v do denotes the unique path between s i and t i in the tree. = min{ c e },. c e = c e -,. 1.2 Let Q be the set of edges such that c i =0 in this step. frontier (v) = Q – the edge that is the descendant of other edges in Q. 2. 2.1 2.2 for current_level = 0 to max_level do for all current_level do for all do If no edge is on the path from e to v then 3. return (Multicut, F)

11 1 33 2 1 Example vertex pairs ={(b, e), (c, d), (a, f)} p be = bghj i e p cd = cghj i d p af = ahjr f r jf h i agde bc 4 4 4 1 1 Pass 1.1 The first node which is a least common ancestor of some vertex pair is j. j = lca(b, e) f be = min{1, 2, 3, 3, 1}=1 Subtract 1 from all edges in p be.

12 Pass 1.1 Consider the pair (c, d) whose lca is j. j = lca(c, d) f cd = min{1, 1, 2, 2, 1}=1 Subtract 1 from all edges in p cd. The lower bound of cutting the paths p be and p cd is f be + f cd =2. 1 0 22 1 0 Example r jf h i agde bc 4 4 41 f be = 1 f cd = 1 vertex pairs ={(b, e), (c, d), (a, f)} p be = bghj i e p cd = cghj i d p af = ahjr f

13 0 0 0 11 0 0 Example r jf h i agde bc 4 4 4 vertex pairs ={(b, e), (c, d), (a, f)} p be = bghj i e p cd = cghj i d p af = ahjr f Pass 1.2 Let Q be the set of edges such that c i =0. Q = {e gb, e ie, e hg, e gc, e id }. Pairs (b,e) and (c,d) have the same least common ancestor, j, and e hg is the ancestor of e gb and e gc in Q. e gb and e gc are not added to frontier(j). frontier(j) = Q - {e gb, e gc } = {e ie, e hg, e id }. f be = 1 f cd = 1

14 4 4 4 0 0 0 11 0 0 Example r jf h i agde bc vertex pairs ={(b, e), (c, d), (a, f)} p be = bghj i e p cd = cghj i d p af = ahjr f Pass 1.1 Consider the pair (a, f) whose lca is r. r = lca(a, f) f af = min {4, 1, 4, 4}=1 Minus 1 from all edges in p af. The lower bound of cutting the path p be, p cd and p af is f be + f cd + f af =3. f be = 1 f cd = 1 f af = 1 frontier(j)={e ie, e hg, e id }

15 3 3 3 0 0 0 10 0 0 Example r jf h i agde bc vertex pairs ={(b, e), (c, d), (a, f)} p be = bghj i e p cd = cghj i d p af = ahjr f Pass 1.2 In this step, Q = {e jh } and frontier(j) = Q - { } = {e jh }. frontier(j)={e ie, e hg, e id } frontier(r)={e jh }.

16 3 3 3 0 0 0 10 0 0 Example r jf h i agde bc vertex pairs ={(b, e), (c, d), (a, f)} p be = bghj i e p cd = cghj i d p af = ahjr f Pass 2. Consider the vertex up to down to build the multicut. Multicut = { }. The first node of least common ancestor is r. frontier(r) = {e jh }. We check e jh. No edge in Multicut is on the path from e jh to r, e jh is included in Multicut. Multicut = {e jh }. frontier(j)={e ie, e hg, e id } frontier(r)={e jh }

17 3 3 3 0 0 0 10 0 0 Example r jf h i agde bc vertex pairs ={(b, e), (c, d), (a, f)} p be = bghj i e p cd = cghj i d p af = ahjr f Pass 2. Consider the node j. frontier(j) = {e ie, e hg, e id }. Check e ie. No edge in Multicut is on the path from e ie to j, e ie is included in Multicut. Multicut = {e jh, e ie }. Check e hg. e jh along the path from e hg to j is already included in Multicut, e hg is not selected. Multicut = {e jh, e ie }. x

18 3 3 3 0 0 0 10 0 0 Example r jf h i agde bc vertex pairs ={(b, e), (c, d), (a, f)} p be = bghj i e p cd = cghj i d p af = ahjr f Pass 2. Continue to consider the node j. frontier(j) = {e ie, e hg, e id }. Check e id. No edge in Multicut is on the path from e id to j, e id is included in Multicut. Multicut = {e jh, e ie, e id }. Cost of Multicut = 3+1+1 = 5. The optimal solution is actually e jh where cost is 3.

19 Lemma Let s i -t i be a path and let v be the least common ancestor of s i, t i. Then Multucut contains at most one edge from the path s i -v and one edge from the path t i - v. Whenever an edge e is chosen, its cost must be a summation of s. That is, for every c e in Multicut. v sisi titi

20 The cost of Multicut is, then, where l=|Multicut|. Thus,. From the algorithm, every corresponds to a path. From Lemma 1, we know that for every path, only two edges of this path can be chosen.

21 Thus, the cost of every edge of path which is chosen contains only one. Finally, only two s appear in. Thus,. Therefore the cost of Multicut is at most.

22 The algorithm achieves approximation guarantees of factor 2 for the minimum multicut problem on tree.

23 This paper interpreted the dual program as specifying a multicommodity flow in T, with a separate commodity corresponding to each vertex pair (s i, t i ). will denote the amount of this commodity routed along the unique path from s i to t i. For example: vertex pairs ={(b, e), (c, d), (a, f)} p be = bghj i e p cd = cghj i d p af = ahjr f 2 1 1 1 4 4 4 3 r jf h i agde bc 3 1

24 Let the value of which gets from the algorithm be the solution of the maximum integer multicommodity flow problem. The algorithm achieves approximation guarantees of factor 1/2 for the maximum integer multicommodity flow problem on trees.

25 Theorem The algorithm achieves approximation guarantees of factor 2 for the minimum multicut problem and factor 1/2 for the maximum integer multicommodity flow problem on trees.

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28 Thank you.

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