1. Performances of List Scheduling for set partition problems 吳邦一歐秀慧 樹德科技大學 資訊工程系 報告人：歐秀慧

2. Set partition problems Given: a set of positive numbers (not necessary distinct) Divide them into m parts Objective functions: Minimize the maximum part Minimize the sum of squares (L 2 metric) others

3. S= 5 3211 3 5 LS Method: M 1 = M 2 = M 3 = The incoming number is put into the currently smallest subset.

4. S= 5 3 21 1 35 LPT Method: M1= M2= M3= It first sorts the numbers in decreasing order. The incoming number is put into the currently smallest subset. 55 3 32 1 1

5. Previous results Min-max LS: at most 2-1/m time of the optimal. LPT: at most 4/3-1/(3m) time of the optimal. L 2 metric: LPT: a 25/24 worse case upper bound Tight bound unknown.

6. Our results The worse-case performance ratios of LS for the following problems: Maximizing the minimum part (max-min problem):the tight bound is m Maximizing the sum of the smallest K parts for any 1  k  m (max-k-min problem) :the tight bound is m/k Minimizing the sum of the largest k parts for any 1  k  m (min-k-max problem) :the tight bound is 2-m/k Minimizing the ratio of the largest to the smallest part (min-ratio problem) :the tight bound is m+1

7. For any Q i and Q j in the LS partition, w ( Q i )- x i  w ( Q j ), LS Property: Q jQ j x i Q i x i Q i Q jQ j X i : the last element in Q i

8. The max-min problem : the tight bound is m LS 1 、 1 2 1 、 1 、 1 3 、 3 1 、 1 、 1 、 1 4 、 4 、 4 … 1 、 1 、 1 、 1 、........ 1 、 1 m 、 m 、 m 、 ……..m 、 m m 234...m C* min 234 … m C Ls min 111...1 C* min  m C Ls min The worst cases:

9. xixi x1x1 m S Q1Q1 QiQi QmQm

10. If x i > In the optimal partition, x i must be in a part of singleton The min part will not be reduced if we replacing x i with We can assume that No element is larger than

11. The min part is always less than or equal to the mean Theorem: The tight bound of LS for the max-min partition is m

12. Lemma : 在任意的 partition 中, 最大的 k 塊必不小 於最大的 k 個 elements 之和 The Max-k-Min problem Corollary : 對任意的 m-partition, 若 s 1 是 s 中 任意 m – k 個 elements, 則 最小的 k 塊之和 w* k  w (s) – w (s 1 ) Proof: by induction. ( 否則最大的 k 個數要放到哪裡 ?)

13. max-k-min problem : the tight bound is m/k X i Q i Q jQ j

14. The tight bound is m / k

15. Future Research The tight bound of LS/LPT for L 2 metric The Minimum sum of squares problem Good approximation algorithm for the min-ratio problem

16. Thank you