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On the uniform edge-partition of a tree

On the uniform edge-partition of a tree. 吳邦一 樹德科大 資工系 王弘倫 台大 資工系 管世達 樹德科大 資工系 趙坤茂 台大 資工系. vertex partition of a tree. 2-partition. 3-partition. Tree splitting (edge partition). 2-split. 3-split. largest. min-max. max-min. minimize. smallest. Objective functions.

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On the uniform edge-partition of a tree

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  1. On the uniform edge-partition of a tree 吳邦一 樹德科大 資工系 王弘倫 台大 資工系 管世達 樹德科大 資工系 趙坤茂 台大 資工系

  2. vertex partition of a tree 2-partition 3-partition

  3. Tree splitting (edge partition) 2-split 3-split

  4. largest min-max max-min minimize smallest Objective functions

  5. Previous results • tree vertex partition: (weighted) • min-max or max-min: polynomial time • most-uniform: unknown • For a path and the objective is to minimize the difference: polynomial time. • The most uniform partition: • No report (to our best knowledge) even for set partition. • tree splitting: apparently NP-hard (3-partition) even for unweighted edges.

  6. Our results • The tree k-splitting is NP-hard. • For k 4, the existence of a k -splitting for any tree with ratio at most. • a 2-approximation algorithm • A simple 3-approximation algorithm for general k. • Experimental results included.

  7. Y Y Y A simple property • For any 1    e(T), we can split T into (T1, T2) at a vertex v in linear time such that   e(T1)  2. Corollary: A tree can be spit into T1 and T2, n/3  e(T1) , e(T2)  2n/3 each y  

  8. For k = 3 • n/4  y  x  n/2 P0 X Y n/4  y  n/2 n/4  x  n/2

  9. Two cases • y  2n/5 • 2n/5 < y  x  n/2

  10. Case 1: n/4  y  2n/5 T1 P2  T1/3  n/4 P0 P1 P2 X Y n/4 y  2n/5 P1 2T1/3  n/2

  11. Case 2: 2n/5 < y  x  n/2 X1X2 X1 P0 X X2 Y n/5  X1 2n/5

  12. Only need to consider n/5  x1 < n/4 • 2n/5 < y  n/2, y/2  x1 < y • n/4 < n-x1-y< 2n/5 • (X1, X2P0, Y) is a desired splitting X1 P0 X2 Y e(X2P0)

  13. For k=4 • It can be prove in a similar way, but the cases are more complicated.

  14. A simple algorithm • There is a simple algorithm to split a tree with ratio at most 3. • Method: always split the maximum part of the previous splitting.

  15. Proof: • By induction. ratio3  3e 2e e 3e e ratio3 2e

  16. Experimental result

  17. Thank you

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