ESA On approximating a geometric prize-collecting traveling salesman problem with time windows Reuven Bar-Yehuda – Technion IIT Guy Even – Tel Aviv Univ. Shimon (Moni) Shahar –Tel Aviv Univ.

ESA :00-5:00 7:00-6:00 12:00-8:00 6:00-7:00 leave office at 5:00 get back at 20:00 10:00-11:00 16:00-17:00 17:00-18: hour rest 1 hour 2 hours 1.5 hours 2 hours 4 hours 2 hours Motivation – postman distributing packages

ESA Prize-collecting TSP with time windows A scheduling problem with locations. Definition: –Sites in a metric space (e.g. the plane). –A time-interval for each site (release-time, deadline). –Moving agent with speed in [0,1]. –Goal: max #sites the agent visits on-time. –Extension: service-time per site.

ESA Known results: scheduling with locations Feasibility is NPC for points on a line [Tsitsiklis92]. Polynomial algorithm for the case where all intervals are [0,t i ] (using dynamic programming) [Tsitsiklis92 Khanna02] Min makespan (completion time of last job): –1.5-approx for points on a line with release times, processing times, and no deadlines [KNI98]. –2-approx for points on a line, no deadlines, multiple agents (vehicles) [KN01]. –PTAS for trees with O(1) leaves, single & multiple agents [AS02].

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ESA speed = 1/slope  Slope in [45 0, ] Chop intervals outside of visibility cone

ESA Now we rotate the view by 45 0 ….

ESA After we rotate the view by 45 0, Slope of tour  [0, 90]

ESA Longest monotone path

ESA Special case: zero length Longest monotone subsequence

ESA Approach: Longest path on a DAG

ESA Approach: Longest path on a DAG

ESA Approach: Longest path on a DAG

ESA Approach: Longest path on a DAG

ESA x y t Approach works for any dimension: Longest path on a DAG

ESA x y t Longest path on a DAG

ESA Polynomial time algorithms:  If all interval times have zero length  If Max length ≤ k * Min Length Grid path: Opt in time Poly(n,2 k ) General: 2-approx in time Poly(n,2 k ) General: O(log k /loglogn) - approx  No assumptions: O(log(n)) - approximation

ESA Construct a DAG. V={(x,y): (x,y)  R 2 } Direct the grid up & right. Assign edge weights (#intersecting intervals). Find a longest path on the obtained DAG.  k-apx Grid path: |interval  grid|  k

ESA Grid path: |interval  grid|  k Construct a DAG V = {(x,y),b 1 …b k : (x,y)  R 2 and b i  {0,1}} Directed “right” edges (x,y)0b 2 …b k  (x+1,y)b 2 …b k 0 (x,y)0b 2 …b k  (x+1,y)b 2 …b k 1 Directed “up” edges (x,y)1b2…bk  (x,y+1)b2…bk0 (x,y)1b2…bk  (x,y+1)b2…bk1 Assign edge weights and find longest path in the DAG (2,2) (2,3) 0 (2,2)1001  (2,3)0010

ESA A 2-approx for length  [1,2) Construct a  2/4 square grid. Each interval intersects at most 8 grid lines. Find optimal grid path (k=8). Time complexity: Poly(n) Claim:  (optimal) path P:  grid paths P1 P2, s.t P1 and P2 cover all intervals intersected by P

ESA  (optimal) path P:  grid paths P1, P2, s.t P1 and P2 cover all intervals intersected by P P: an optimal path P1: Upper grid path P P1 P2 P2: Lower grid path

ESA A 2-approx for length  [1,2)  2log(I max /I min )- apx for the general case

ESA A 2-approx for length  [1,k)  2log(I max /I min )/logk- apx for the general case k=O(logn)  time is Poly(n)

ESA Recursive bisection Claim:  separating vertical line (at most half the intervals lie strictly on each side).

ESA Recursive bisection (cont.) bisect recursively  log(n) “combs” Level 1 Level 2 2nd comb A comb defines subset of intervals that intersect exactly one comb-tooth.  comb C i such that: C i  OPT contains at least OPT/log n intervals.

ESA O(log(n)) Approximation Partition the intervals into log n combs. For each comb 2-apx.  2log(n)- approximation.

ESA Approximation for comb Form a grid. Construct a DAG. V = set of horz segments Set of edges: (i, j)  (i+1, k) If j  k. Edge weights is the number Of new intersected intervals ii+1 j k weight((i, j), (i+1, k))=4

ESA Approx ratio = 2 Decompose OPT into alternating sub-tours: –horizontal sub-tours inside a slice –vertical sub-tours between two comb teeth –Each “covered segment” must cross P1 or P2 P2 P1

ESA Zigzags: source of hardness special case: no zigzags between intervals Dynamic programming finds optimal tour (even if distances are asymmetric). Extension: density = bound on number of zigzag between intervals. apx ratio=density (same dynamic programming)

ESA Further research Improve approximation ratio in 1-D. Nothing known for 2-D.