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-partition3-partition

3. Tree splitting (edge partition) 2-split3-split

4. Objective functions min-maxmax-min minimize largest smalles t

5. Previous results tree vertex partition: (weighted) 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. For a path and the objective is to minimize the difference: polynomial time. The most uniform partition: 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. The tree k -splitting is NP -hard. For k  4, the existence of a k -splitting for any tree with ratio at most. 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. A simple 3-approximation algorithm for general k. –Experimental results included.

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

8. For k = 3 n/4 n/4  y  x  n/2 Y P0P0 X n/4  y  n/2 n/4  x  n/2

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

10. Case 1: n/4 Case 1: n/4  y  2n/5 Y P0P0 X T1T1 P1P1 P2P2 n/4  y  2n/5 P 1  2T 1 /3  n/2 P 2  T 1 /3  n/4

11. Case 2: Case 2: 2n/5 < y  x  n/2 Y P0P0 X X1X1 X2X2 n/5  X 1  2n/5 X1X2X1X2X1X2X1X2

12. Only need to consider n/5  x 1 < n/4 y/2  x 1 < y 2n/5 < y  n/2, y/2  x 1 < y n/4 < n-x 1 -y< 2n/5 n/4 < n-x 1 -y< 2n/5 (X 1, X 2  P 0, Y) is a desired splitting (X 1, X 2  P 0, Y) is a desired splitting Y P0P0 X1X1 X2X2 e(X2  P0)

13. For k=4 It can be prove in a similar way, but the cases are more complicated. 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. There is a simple algorithm to split a tree with ratio at most 3. Method: always split the maximum part of the previous splitting. Method: always split the maximum part of the previous splitting.

15.  2 22 2e e ee e3e Proof: By induction. By induction. e  3e  2 22 2e ratio  3

16. Experimental result

17. Thank you