1. Restricted Track Assignment with Applications 報告人：林添進

2. Channel Assignment Problem Given a set of intervals S = {I i | 1 ≦ i ≦ n} 1. To find the degree of overlap S, k, where k = max {|S’| | S’  S and S’ is overlapping} 2. To partition S into k subsets such that the intervals in each subsets are pairwise non-overlapping

3. Algorithm for finding degree k 1.To sort 2n endpoints in the increasing order of their x- coordinates O(n log n) 2.To scan the 2n endpoints from left to right. O(n) k = 1 temp = 1 k = 2 temp = 2 k = 3 temp = 3 k = 3 temp = 2 k = 3 temp = 1 Why this algorithm works?

4. Algorithm for finding degree k is optimal Ω (n log n) lower bound on the time required to determine whether n intervals on the line are pairwise disjoint.

5. 1.To sort 2n endpoints in the increasing order of their x- coordinates 2.To find degree k 3.To scan the intervals form left to right 1)If a left endpoint is encountered, we assign the corresponding interval to the next available channel. 2)If a right endpoint is encountered, we release the channel occupied by the corresponding interval and make the released channel available again. Why the algorithm works ? Algorithm for partitioning S into k subsets

6. c1 c2 c4 c3 c5

7. k channel is necessary, but why k channels is also sufficient ? 當使得 degree 變成 k 的 interval I 進來 以前，若 k 個 channel 被用光了，則 在 I 進來以前必有某一個 interval 要 離開，否則 degree 就大於 k 了，這是 不可能的。 為何對 channel 的長度不固定的 Track Assignment Problem 或 RTAP ， k channels is not sufficient ?

8. Definitions Given a set of intervals I = {I i | 1 ≦ i ≦ n}, where I i = (l i, r i ). Two intervals I i = (l i, r i ) and I j = (l j, r j ) are 1.independent: if r i < l j or r j < l i ; 2.dependent: otherwise. Two dependent intervals are 1)crossing: if l i < l j < r i < r j or l j < l i < r j < r i. 2)I i contains I j : If l i < l j < r j < r i.

9. Definitions 1.Track Assignment Problem (TAP) is to assign the intervals into tracks 1 to t such that in each track, intervals are pairwise independent. The goal is to minimize the number of tracks t. Whether density of I is sufficient or not? 2.The density of a column c, denoted d c, is the number of interval in I that contain c. 3.The density of the set I is d = max c d c,

10. Definitions 1.Consider a set of points C = { c 1, c 2, …, c m }, called a restricted set. 2.The maximal subset of I that contains a point of C is denoted by I c ; intervals in I c are called restricted intervals.

11. Restricted Track Assignment Problem Given I and C, the restricted track assignment problem (RTAP) is the problem of assigning intervals into tracks 1 to t such that: p1) In each track, intervals are pairwise independent. p2) If a restricted interval I i contains another restricted interval I j then τ i > τ j, where τ a is the track to which I a is assigned. The goal is, as before, to minimize the number of tracks t. Note that when when C = { c 1 } then RTAP is equivalent to TAP. Here, we will show RTAP is NP-hard for |C|=2 (and |C|>2). Then we propose an approximation algorithm for solving an arbitrary instance (I, C) of RTAP.

12. c1c1 c2c2 Why density is not sufficient? d = 3 but It need at least 4 tracks

13. Consider an arbitrary instance (I, C) of TRAP, where I = { I 1, I 2, …, I n } and C = { c 1, c 2, …, c m }. Certainly, the number of tracks is lower-bounded by the density d. We will show that it is NP-complete to decide if d tracks are sufficient of assignment of intervals, when m=2. We shall transform the problem of coloring a circular-arc graph to RTAP. NP-hardness of RTAP

14. Consider a set A = {A 1, A 2,…, A s } of arcs, where = (a i, b i ), 1 ≦ a i, b i ≦ 2s. The intersection graph of A is called a circular-arc graph. Consider a point P on the circle. The set of arcs cut by a line segment starting at the center of the circle and passing through P is denoted by A P. The arcs in A P are said to be P-equivalent. The circular-arc coloring problem (CACP) is to assign the minimum number of colors to arcs of A such that every two P-equivalent arcs, for all P, are assigned distinct colors. The maximum of |A P |, over all P, is called the density of A and is denoted byδ. It is NP- complete to decide ifδcolors are sufficient to color an arbitrary instance A of CACP. NP-hardness of RTAP

16. We will show that CACP can be reduced to RTAP in polynomial-time. Lemma 1 An arbitrary instance A of CACP is δ-colorable if and only if (I A, C A ) can be assigned into δ tracks of RTAP. NP-hardness of RTAP

17. In this section, first we show there are instances of RTAP with t ≧ 2d-1, where d is the problem density. Then, we propose an approximation algorithm for an arbitrary instance of RTAP achieving t ≦ 2d-1. 為何與 C 無關？ An Approximation Algorithm

18. We define an instance (I, C) of TRAP that requires at least 2d- 1 tracks in a recursive manner. It consists of a collection of blocks. B 1 has a single unit-length interval. We obtain B i, i > 1, by combining two copies of B i-1 and two intervals, called top-intervals. The instance B d with C being the set of right points of all unit intervals is called a difficult example. Lemma 2 In an arbitrary instance of difficult example B d, t ≧ 2d-1. An Approximation Algorithm

19. 1.Consider an arbitrary instance (I, C) of the problem with density d. Let {k 1, k 2,…, k s } denote the set of max-density columns with k 1 < k 2 <…< k s. Among all intervals containing k 1 we select the one with the rightmost right point. This interval is denoted by I 1,1. Assume I 1,1 contains columns k 1,…, k j, does not contain column k j+1. Then among all intervals containing k j+1 we select the one with the rightmost right point. This interval is denoted by I 1,2. This task is repested until all columns {k 1, k 2,…, k s } is processed. The set of intervals selected is denoted by I 1 = {I 1,1, I 1,2, …} 2.The set I 1 is removed and the process is repeated for the rest of intervals ─ we obtain I 1, I 2, … respectively until no interval is remained. 3.We assign I i, 1 ≦ i ≦ d-1 to tracks 2d-2i+1 and 2d-2i. I d has density 1, and thus we assign it to track 1. Why the algorithm works ?

20. Prop 1. Density of each I i is at most two. Prop 2. After removing each I i the density of I is reduced by at least one. Prop 3. Consider any interval I i,a  I i and any interval I j,b  I j for i < j. Then I i,a is not contained in I j,b. Three Properties of the Algorithm

21. k1k1 k3k3 k2k2 k4k4 k1k1 k4k4 k3k3 k5k5 k2k2 k6k6 k1k1 k4k4 k3k3 k5k5 k2k2 k6k6 k1k1 k4k4 k3k3 k5k5 k2k2 k1k1 k3k3 k4k4 k2k2