1. Interval packing problem Multicommodity demand flow in a line Jian Li Sep. 2005

2. Interval packing problem  Input:  1) A set I of n intervals, I 1, I 2,..., I n, with each I i = [l i, r i ] and having a THICKNESS (or HEIGHT) h i, such that l i and r i are all integers in {1,2,..., n} and h i is an integer > 0  2) n-1 CAPACITY constraints c i, with each c i located at the point p i = (i + 1/2, 0) on the x-axis, for every i = 1,2,..., n-1, such that each c i is an integer > 0. We may call each p i a thickness constrained point  Output:  The MAXIMUM NUMBER of intervals in I such that at each thickness constrained point p i = (i + 1/2, 0), the sum of thickness values of all output intervals that contain p i is less then c i.

3.  In fact, it is a special case of the multicommodity demand flow problem

4. Multicommodity demand flow problem  T=(V,E,u) be a capacitated network, where each edge capacity u e is an integer.  A number of demand vertices pairs (s i,t i ) with demand value d i and profit w i  GOAL: simultaneously route the demand flow without violating the capacity constraints and maximize the profit.  NOTE: di must be fully satisfied to obtain a profit wi

5. Interval packing problem can be modeled as multicommodity flow problem in a line.

6. LP formulation max  i w i x i  i:e2 path(i) d i x i · u e x i =0,1

7. Some special case  All capacities are one.  All demand are one.  The problem is exactly to compute a independent set in a interval graph, which is in P.

8.  Algorithm:  Sort intervals according their left endpoints.  Use the reverse order to greedy construct the independent set. (reverse perfect elimination order)

9. Some special case.  the problem can be formulated as the following min-cost flow problems.  There are n vertice, namely v1,v2,…..,vn, and arcs from vi to v(i+1) with capacity u i and cost zero and arc from vn to v1 with capacity +infinity and cost zero.  For every interval I i, there is a arc from v(r i ) to v(l i ) with capacity d i and cost -1/d i (we call it back- arc).

10. Capacity=u i, cost=-1/d i

11. Some special case  However, the min cost circulation may not give the correct answer, because in the original problem only in case that the back-arc are saturated(flow=capacity) can incur a cost 1. So we can treat it as the relax of the original problem. In fact it solve the relaxed LP of the original problem.  Moreover, if all d i is one, then this special case can be solved optimally in polytime.

12. Some special case  if the number of intervals passing p i is bounded( less than a constant C), then we can solve the problem optimally by standard dynamic programming technique.

13. Small and Large demand  Bottleneck capacity b(f): the smallest capacity edge on Path(f)   -large demand: d(f)¸  b(f)  Otherwise,  -small

14. Bounding large demand  Lemma:  If d max · u min (a wild assumption)in a feasible solution, the number of  -large demands that cross any edge is at most 2(1/  2 )  So, we can use DP to solve the case with only large task optimally.

15. Bounding large demand  Proof of the lemma:  Fix a feasible solution S, consider an edge e.  Let S e be the set of all  -large demand cross e.  S e =S l [ S r, where S l is the set where its bottleneck is to the left of e(including e), and S r otherwise. SlSl e SrSr

16. Bounding large demand  Proof Cont.  We can show S l <=1/  2 (similar for S r )  Let A be the set of bottleneck edge for S l. Let e’ be the rightmost one in A. Suppose e’ is the bottleneck edge for demand I j, thus d j >  c(e’). But all demands in S l pass through e’(since e’ is rightmost one). But each demand I k is  -large. d k ¸  b k ¸  c min.   c min S l · c(e’), thus  S l · |c(e’)/  c min | · |d i /  2 c min | · 1/  2 \QED

17.  Suppose we have a  -approximation algorithm for the case where only small demands exist.  Then we choose the larger one from the optimal solution for large demand and approximation solution for small demand  We get 2  approximation for original problem OPT<=OPT large +OPT small <=SOL large +  SOL small <=2max{SOL large +  SOL small } <= 2  max{SOL large + SOL small }

18. Dealing with Small demands  Randomized rounding: G.Calineascu et IPCO2002 for uniform capacity. (1+e)-approx. A.Chakrabarti et APPROX2002 for general case. O(1)-approx.  Packing integer program C.Chekuri et ICALP2003 for general case. (1+e)-approx.