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Minimum Routing Cost Tree

Minimum Routing Cost Tree. Definition For two nodes u and v on a tree , there is a path between them. The sum of all edge weights on this path is called the routing cost of this pair of nodes. MRCT: Minimize the sum of routing costs between all pairs of nodes.

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Minimum Routing Cost Tree

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  1. Minimum Routing Cost Tree • Definition • For two nodes u and v on a tree, there is a path between them. • The sum of all edge weights on this path is called the routing cost of this pair of nodes. • MRCT: Minimize the sum of routing costs between all pairs of nodes. • cf. Minimum Spanning Tree

  2. Example 1 a 1 b 2 • Total sum of all pairs of routing costs of this tree: • RC(a,b)+RC(a,c)+RC(a,d)+RC(b,a)+RC(b,c)+RC(b,d)+RC(c,a)+RC(c,b)+RC(c,d)+RC(d,a)+RC(d,b)+RC(d,c) = 1+2+2+1+1+1+2+1+2+2+1+2 = 18 1 b 1 2 d c 1 1 1 1 a c d

  3. Example 2 a 1 b 2 • Total sum of all pairs of routing costs of this tree: • RC(a,b)+RC(a,c)+RC(a,d)+RC(b,a)+RC(b,c)+RC(b,d)+RC(c,a)+RC(c,b)+RC(c,d)+RC(d,a)+RC(d,b)+RC(d,c) = 30 1 a 1 2 d c 1 1 2 2 b c d

  4. Centroid of A Tree a i • Deleting a centroid of a tree will produce subgraphs such that each subgraph contains no more than n/2 nodes. h m b f g e c d

  5. Centroid (2) • Let T be rooted by a centroid m of T. • Every subtree of m contains no more than n/2 nodes. • Consider any node v in T. The subtree containing v contains no more than n/2 nodes. • At least n/2 paths between some node u (in other subtree) and v will pass through m. m

  6. b 1 1 1 a c d Centroid (3) • Note: we count every pair of nodes twice. • The path from u to v and the path from v to u will both be counted. • In the routing cost of this tree, the length of the path from any node u to m will be counted at least 2(n/2) = n times.

  7. 1-Star • A tree with only one internal node. • All other nodes are leaf nodes. …

  8. Approximation Algorithm for MRCT 1 For each node i 1-1 Form a 1-star Si rooted at i 1-2 Calculate the routing cost C(Si) 2 Return as the approximate solution

  9. Time Complexity • Step 1: n iterations • Step 1-1: O(n) • Step 1-2: O(n) • Step 2: O(n) • Total: O(n2)

  10. Performance Ratio … • Let C(S) denote the cost of 1-star S. • Assume T is a minimum routing cost spanning tree and m is a centroid of T. • Because of the triangular inequality, w(v,m) is less than RC(v,m) of T.

  11. Can we find a better ratio? • Yes. There is a PTAS.

  12. PTAS for MRCT • We shall use a k-star to approximate the optimal solution, and show • When k=1, this is exact the 1-star case we described in the previous section. • We shall use the case k=3 to illustrate the basic concept of this PTAS.

  13. k-Star • A k-star is a tree with exactly k internal nodes. d j b a c e i f a h

  14. -separator • For 0 <   ½, a -separator of a graph G is a minimum subgraph of G whose deletion will result in subgraphs, each of which contains no more than n nodes. • For =1/2, the -separator contains only one point, i.e. the controid. • We shall choose  = 2/(k+3) • For k=3,  = 1/3

  15. Analysis of Routing Cost r c e n a p m h b • Assume T is a minimum routing cost spanning tree. • Assume T is rooted at its centroid m. • Then at most two subtrees of m contain more than n/3 nodes. • Let a and b be the lowest nodes whose descendants have at least n/3 nodes. • P: the path from node a to node b. • P is a (1/3)-separator o d f j i g k q s

  16. Analysis of Routing Cost (2) • Let dt(v,P) denote the path length from v to P. • This path length must be counted at least 2n/3 times because P is a (1/3)-separator. • For each edge of P, the edge is counted at least (n/3)(2n/3) times in routing cost. • Each edge on P is counted (n/3 + )(2n/3 - ) times. • When   n/3, the above formula is always greater than or equal to 2n2/9. • Let w(P) denote the total path length of P, we have a b

  17. Partition The Nodes • Va consists of nodes whose lowest ancestors on P are a. • Vb consists of nodes whose lowest ancestors on P are b. • Vm consists of nodes whose lowest ancestors on P are m. • Vam consists of nodes whose lowest ancestors on P are between a and m. • Vbm consists of nodes whose lowest ancestors on P are between b and m. r c e n b a p m h o d f j i g k q s

  18. Transform to 3-Star (1) • For each node v in Va, Vb, and Vm, connect v to a, b, and m respectively. r c e n a p m h b o d f j i g k q s

  19. Transform to 3-Star (1) • For each node v in Va, Vb, and Vm, connect v to a, b, and m respectively. r c e n a p m h b o d f j i g k q s

  20. Transform to 3-Star (2) • Replace P by a path (a,m) and (b,m). • For each node v in Vam, connect v to a or m, depending on which one is nearer to v. • For each node v in Vbm, connect v to b or m, depending on which one is nearer to v. r c e n a p m h b o d f j i g k q s

  21. Transform to 3-Star (2) • Replace P by a path (a,m) and (b,m). • For each node v in Vam, connect v to a or m, depending on which one is nearer to v. • For each node v in Vbm, connect v to b or m, depending on which one is nearer to v. h r p c e n a m b o d f j i g k q s

  22. Comparison with 3-Star • For each edge (v,a), (v,b), (v,m) in the 3-star, it will be counted (n-1) times. • For edge (a,m) or (b,m), it is counted no more than (n/2)(n/2)=(n2/4).

  23. How to find the 3-Star a b c • Pick 3 nodes. • For i+j+k=n-3, connect i nodes to a, j nodes to b, and k nodes to c. • However, if we tries all combinations, it took exponential time. … … … i j k

  24. Minimum Perfect Bipartite Matching • In a bipartite graph, there are two sets of nodes, denoted as X and Y. • In our case, let X = V – {a,b,c}. Y contains i copies of a, j copies of node b, and k copies of c. • If a node v in X is matched with a node u in Y, v will be connected to u in the 3-star.

  25. PTAS • For all (a,b,c) where a, b and c are selected from V do • For all (i,j,k) where i+j+k = n-3 and I,j,j are all positive integers, do • Perform a perfect minimal bipartite matching to create a 3-star. • Compute the total routing cost C of this 3-star. • Choose the 3-star with minimum cost to be our approximate solution.

  26. Time Complexity • Total time complexity: O(n8) • O(n3) possible ways to select a,b and c. • O(n2) possible ways to select I,j and k. • The perfect minimum bipartite problem can be solved in O(n3) time. • For k-star, it takes O(n2k+2) time to obtain a solution which is less than or equal to (k+3)/(k+1) of the optimal solution. • Error ratio: 2/(k+1)

  27. Mid-Term • April 28th, Thursday • 9:00-12:00 • No class on April 26th • Closed Book

  28. Presentation • Each student chooses a journal paper about approximation algorithm and gives a 40-minute presentation.

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