Approximation Algorithms for Path-Planning Problems

1. Approximation Algorithms for Path-Planning Problems with Nikhil Bansal, Avrim Blum, David Karger, Adam Meyerson, Maria Minkoff Shuchi Chawla

2. The Lost-Wallet Problem • How should you go about finding a lost wallet? • Several possible locations; different likelihoods • Task: Find a good search strategy • visit a lot of places having high likelihood • faster search  greater likelihood of discovery • find it before someone else does ! • The trick-or-treaters problem • Collect as much candy as possibly between 6 and 8 • Mr. X always gives more candy than Mrs. Y Shuchi Chawla, Carnegie Mellon University

3. The Lost-Wallet Problem • Model as a graph problem • vertices are locations; labeled by likelihoods • edge lengths represent time taken to go from one location to another • Classic formulation – Traveling Salesman Find the shortest tour covering all locations • Some complicating constraints • The wallet could get stolen before you find it! • No candy after 8pm • Cover as many locations as possible • Give preference to the more likely ones Shuchi Chawla, Carnegie Mellon University

4. A probabilistic view Discounted-Reward TSP Orienteering • At every time step, there is a fixed probability (1-) that the wallet gets stolen • If a location with value (likelihood)  is visited at time t, the expected likelihood of discovery is t • Goal: Construct a path such that the total discounted reward collected is maximized “Discounted Reward” Alternately, you can only search for time D  Goal: Construct path of length at most D that collects maximum reward  Shuchi Chawla, Carnegie Mellon University

5. A time-reward trade-off Orienteering Dis. Rew. TSP reward • Given weighted graph G, root s, reward on nodes v • Construct a path P rooted at s • High level objective: Collect large reward in little time • Orienteering Maximize reward collected with path of length D • Discounted-Reward TSP Reward from node v, if reached at time t is vt time Shuchi Chawla, Carnegie Mellon University

6. A time-reward trade-off • Given weighted graph G, root s, reward on nodes v • Construct a path P rooted at s • High level objective: Collect large reward in little time • Orienteering Maximize reward collected with path of length D • Discounted-Reward TSP Reward from node v, if reached at time t is vt A related problem… • K-Traveling Salesperson Minimize length while collecting at least K in reward No approximation algorithm known previously for the rooted non-geometric version [GLV87] [Balas89] [AMN98] New problem Best: (2+)-approx [Garg99] [AK00] … Shuchi Chawla, Carnegie Mellon University

7. The FEDEX-guy Problem • The deliveryman has to deliver packages to various locations • Packages have time-windows for delivery • Some packages have higher priority than others • Deliver as many packages in their time-windows, as possible • Metric of success: total reward from packages delivered on time The Time-Window Problem Shuchi Chawla, Carnegie Mellon University

8. The Time-Window Problem • Find a path that visits many nodes in their time-window • Widely studied in scheduling and OR literature • Constant-approx known for points on a line, few different time-windows • No approximation known for the general case • A special case – The Deadline-TSP Problem • Vertices only have deadlines • All “release-times” are 0. Shuchi Chawla, Carnegie Mellon University

9. Our results Problem Approximation K-path [Chaudhuri et al ’03] 2+ min=kℓ(P) Min-Excess Path 2+ Orienteering “point-to-point” 3 maxℓ(P)=D(P) Discounted-Reward TSP 6.75+ max ∑t Deadline-TSP 3 log n max ∑t≤D(v)v Time-Window Problem 3 log2n max ∑R(v)≤t≤D(v)v Shuchi Chawla, Carnegie Mellon University

10. The rest of this talk • Point-to-point Orienteering and D-R TSP • Why is this difficult? • The Min-Excess problem and how to solve it • Using Min-Excess to solve Orienteering & D-R TSP • The Time-Window Problem • Orienteering with deadlines • Incorporating release-dates • Extensions and Open Problems Shuchi Chawla, Carnegie Mellon University

11. Why is Orienteering difficult? K-path min=kℓ(P) Orien. maxℓ(P)=D(P) DR-TSP max ∑t • First attempt – Use distance-based approximations to approximate reward • Let OPT(d) = max achievable reward with length d • A 2-approx for distance implies that ALG(d) ≥ OPT(d/2) • However, we may have OPT(d/2) << OPT(d) • Bad trade-off between distance and reward! OPT(d) s APPROX Shuchi Chawla, Carnegie Mellon University

12. Why is Orienteering difficult? K-path min=kℓ(P) Orien. maxℓ(P)=D(P) DR-TSP max ∑t • First attempt – Use distance-based approximations to approximate reward • Idea – Modify the algorithm itself • Doesn’t help – moat-growing always goes for shallow fruit • Orienteering is inherently harder; Perturbation of the input changes the output widely • Same problem with Discounted-Reward TSP • Multiplying the exponent with a constant gives a bad approximation OPT(d) s APPROX Shuchi Chawla, Carnegie Mellon University

13. Why is Orienteering difficult? K-path min=kℓ(P) Orien. maxℓ(P)=D(P) DR-TSP max ∑t t s • Second attempt – approximate subparts of the optimal path and shortcut other parts • If we stray away from the optimal path by a lot, we may not be able to cover reward that’s far away • Approximate the “extra” length taken by a path over the shortest path length OPT APPROX Shuchi Chawla, Carnegie Mellon University

14. Why is Orienteering difficult? K-path min=kℓ(P) Orien. maxℓ(P)=D(P) DR-TSP max ∑t Min-Excess Path Problem • Second attempt – approximate subparts of the optimal path and shortcut other parts • If we stray away from the optimal path by a lot, we may not be able to cover reward that’s far away • Approximate the “extra” length taken by a path over the shortest path length • If OPT obtains k reward with length d+, ALG should obtain the same reward with length d+ Shuchi Chawla, Carnegie Mellon University

15. The Min-Excess Problem K-path min=kℓ(P) Excess min=k(P) • Given graph G, start and end nodes s, t, reward on nodes v • Find a path from s to t collecting K reward and minimizing ℓ(P) – d(s,t) • At optimality, this is exactly the same as the K-path objective of minimizing ℓ(P) • However, approximation is different • -approx to K-path : ℓ(P) • -approx to min-excess : d + (ℓ(P) – d) = ℓ(P) – (-1)d • Min-excess is strictly harder than K-path Shuchi Chawla, Carnegie Mellon University

16. Solving Min-Excess K-path min=kℓ(P) Excess min=k(P) • OPT = d+; k-path gives us ALG = (d+) We want ALG = d +  • Note: When ≈d, (d+) ≈ d + O() • Idea: When  is large, approximate using k-path • What if  << d ? • Small   path is almost like a shortest path or “its distance from s mostly increases monotonically” Shuchi Chawla, Carnegie Mellon University

17. Solving Min-Excess K-path min=kℓ(P) Excess min=k(P) • OPT = d+; k-path gives us ALG = (d+) We want ALG = d +  • Note: When ≈d, (d+) ≈ d + O() • Idea: When  is large, approximate using k-path • What if  << d ? • Small   path is almost like a shortest path or “its distance from s mostly increases monotonically” • Idea: Completely monotone path  use dynamic programming to solve exactly! • Binary decision for each vertex – should it be in the path or not? • Compute P(vj,t) = the “best” path that has length t and ends at vj • P(vj+1,t) == • consider P(u,t’), where t’ = t-ℓ(u,vj+1) • pick the best path (best u) from the above Shuchi Chawla, Carnegie Mellon University

18. Solving Min-Excess K-path min=kℓ(P) Excess min=k(P) Dynamic Program Approximate • Idea: When  is large, approximate using k-path • Idea: Completely monotone path  use dynamic programming to solve exactly! Patch segments using dynamic programming t s OPT wiggly wiggly monotone monotone monotone Shuchi Chawla, Carnegie Mellon University

19. From Min-Excess to Orienteering Orien. maxℓ(P)=D(P) Excess min=k(P) t 3 s 1 2 • There exists a path from s to t, that • collects reward at least  • has length  D • Given a 3-approximation to min-excess: 1. Divide into 3 “equal-reward” parts (hypothetically) 2. Approximate the part with the smallest excess • Using an r-approx for Min-excess ( r Z+ ), we get an r-approximation for s-t Orienteering Excess of path P (P) = dP(u,v)– d(u,v) v2 OPT v1 APPROX Excess of one path · (1+2+3)/3 Can afford an excess up to (1+2+3) Shuchi Chawla, Carnegie Mellon University

20. Solving Discounted-Reward TSP DR-TSP max ∑t Excess min=k(P) t s v excess = 1 • WLOG,  = ½. Reward of v at time t = vt • An interesting observation: OPT collects half of its reward before the first node that has excess 1 • Therefore, approximate the min-excess from s to v • New path has excess 3. Reward  by factor of 23. • 16-approximation ’ = 2OPT(v,t) > OPT reward OPT/2 OPT length of entire remaining path decreases by 1 Shuchi Chawla, Carnegie Mellon University

21. So far… • (2+)-approximation for Min-excess • 3-approximation for Orienteering • (6.75+)-approximation for Discounted-Reward TSP • You learnt how to look for your lost wallet should’ve Shuchi Chawla, Carnegie Mellon University

22. Deadline-TSP • Every vertex has a deadline D(v); Find a path that maximizes nodes v visited before D(v) • If a path has length smaller than the minimum deadline, use Orienteering to approximate the reward in that path • Everything visited before the minimum deadline • Don’t need to bother about deadlines of other nodes • Does OPT always have a large subpath with the above property? • There are many subpaths of OPT with the above property that together contain all the reward NO! Shuchi Chawla, Carnegie Mellon University

23. A segmentation of OPT Deadline Time Shuchi Chawla, Carnegie Mellon University

24. Deadline-TSP • Segment graph into many parts, approximate each using Orienteering and patch them together • How do we find such a segmentation without knowing the optimal path? • In order to avoid double-counting of reward, segments should be node-disjoint • Our result – There exists a segmentation based only on deadlines, such that the resulting solution is a (3 log n)-approximation Shuchi Chawla, Carnegie Mellon University

25. From Deadlines to Time-Windows t s s t • Nodes have deadlines as well as release times • Note that release times are dual to deadlines – if we look at the path from the end to the start, release times become deadlines! • Log-approximation for deadlines  log-approximation for release dates • Algorithm for Time-Windows: • Run the approximation for Deadline-TSP • Replace Orienteering by Orienteering with release-dates • O(log2n)-approximation for the Time-Window problem ℓ(OPT) = L D(v) = L-R(v) OPT v Require ℓ(s,v)  R(v)  ℓ(t,v)  L-R(v) Shuchi Chawla, Carnegie Mellon University

26. Our results Problem Approximation Min-Excess Path 2+ Orienteering (point-to-point) 3 maxℓ(P)=D(P) Discounted-Reward TSP 6.75+ max ∑t Deadline-TSP 3 log n max ∑t≤D(v)v Time-Window Problem 3 log2n max ∑R(v)≤t≤D(v)v Shuchi Chawla, Carnegie Mellon University

27. Some extensions • Unrooted versions • Multiple tours • Max-reward Steiner tree of bounded size Shuchi Chawla, Carnegie Mellon University

28. Future work… • Improve the approximations • 2-approx for Orienteering? • Constant factor for Deadline-TSP • The Time-Window problem Can we get a constant or O(log n) in general graphs? Shuchi Chawla, Carnegie Mellon University