Restricted Track Assignment with Applications 報告人：林添進

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

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?

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.

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

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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 ?

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.

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,

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.

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.

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

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

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

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

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

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

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 ?

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

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