Restricted Track Assignment with Applications

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

2. Channel Assignment Problem • Given a set of intervals S = {Ii| 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. k = 1 temp = 1 k = 2 temp = 2 k = 3 temp = 3 k = 3 temp = 2 k = 3 temp = 1 Algorithm for finding degree k • To sort 2n endpoints in the increasing order of their x-coordinates O(n log n) • To scan the 2n endpoints from left to right. O(n) • 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. Algorithm for partitioning S into k subsets • To sort 2n endpoints in the increasing order of their x-coordinates • To find degree k • To scan the intervals form left to right • If a left endpoint is encountered, we assign the corresponding interval to the next available channel. • 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 ?

6. c1 c2 c3 c4 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 = {Ii| 1≦i ≦ n}, where Ii = (li, ri). • Two intervals Ii = (li, ri) and Ij = (lj, rj) are • independent: if ri <lj or rj <li ; • dependent: otherwise. Two dependent intervals are • crossing: if li <lj< ri <rj or lj <li< rj<ri . • Ii contains Ij: If li <lj< rj <ri.

9. Definitions • 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? • Thedensity of a column c, denoted dc, is the number of interval in I that contain c. • The density of the set I is d = maxc dc,

10. Definitions • Consider a set of points C = { c1, c2, …, cm}, called a restricted set. • The maximal subset of I that contains a point of C is denoted by Ic; intervals in Ic 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 Ii contains another restricted interval Ij then τi > τj , where τa is the track to which Ia is assigned. The goal is, as before, to minimize the number of tracks t. Note that when when C = { c1} 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. Why density is not sufficient? c1 c2 d = 3 but It need at least 4 tracks

13. NP-hardness of RTAP Consider an arbitrary instance (I, C) of TRAP , where I = { I1, I2, …, In} and C = { c1, c2, …, cm}. 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.

14. NP-hardness of RTAP Consider a set A = {A1, A2,…, As } of arcs, where = (ai, bi), 1≦ ai, bi ≦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 AP. The arcs in AP 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 |AP|, 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.

16. 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 (IA, CA) can be assigned into δ tracks of RTAP.

17. An Approximation Algorithm 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？

18. 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. B1 has a single unit-length interval. We obtain Bi, i > 1 , by combining two copies of Bi-1 and two intervals, called top-intervals. The instance Bd 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 Bd, t≧ 2d-1.

19. An Approximation Algorithm • Consider an arbitrary instance (I, C) of the problem with density d. Let {k1, k2,…, ks} denote the set of max-density columns with k1< k2<…< ks. Among all intervals containing k1 we select the one with the rightmost right point. This interval is denoted by I1,1. Assume I1,1 contains columns k1,…, kj , does not contain column kj+1. Then among all intervals containing kj+1 we select the one with the rightmost right point. This interval is denoted by I1,2. This task is repested until all columns {k1, k2,…, ks} is processed. The set of intervals selected is denoted by I1 = {I1,1, I1,2, …} • The set I1 is removed and the process is repeated for the rest of intervals ─ we obtain I1, I2, … respectively until no interval is remained. • We assign Ii, 1≦ i ≦d-1 to tracks 2d-2i+1 and 2d-2i. Id has density 1, and thus we assign it to track 1. • Why the algorithm works ?

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

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