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

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. Discounted-Reward TSP
A probabilistic view
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”
Discounted-Reward TSP
Alternately, you can only search for time D  
Orienteering
Goal: Construct path of length at most D that collects maximum reward
Shuchi Chawla, Carnegie Mellon University

5. 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
reward
Orienteering
Dis. Rew. TSP
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 Time-Window Problem
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=k ℓ(P)
2+
Min-Excess Path
2+
Orienteering
“point-to-point”
maxℓ(P)=D (P) 3
Discounted-Reward TSP
max ∑t
6.75+
Deadline-TSP
max ∑t≤D(v)v
3 log n
Time-Window Problem
max ∑R(v)≤t≤D(v)v
3 log2n
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?
Orien. maxℓ(P)=D (P)
DR-TSP max ∑t
K-path min=k ℓ(P)
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?
Orien. maxℓ(P)=D (P)
DR-TSP max ∑t
K-path min=k ℓ(P)
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?
Orien. maxℓ(P)=D (P)
DR-TSP max ∑t
K-path min=k ℓ(P)
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 s t
Shuchi Chawla, Carnegie Mellon University

14. Why is Orienteering difficult?
Orien. maxℓ(P)=D (P)
DR-TSP max ∑t
K-path min=k ℓ(P)
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+
Min-Excess Path Problem
Shuchi Chawla, Carnegie Mellon University

15. The Min-Excess Problem
K-path min=k ℓ(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 OPT = d+; k-path gives us ALG = (d+)
K-path min=k ℓ(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 OPT = d+; k-path gives us ALG = (d+)
K-path min=k ℓ(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. Patch segments using dynamic programming
Solving Min-Excess
K-path min=k ℓ(P)
Excess min=k (P)
Idea: When  is large, approximate using k-path
Idea: Completely monotone path  use dynamic programming to solve exactly!
Patch segments using dynamic programming
Dynamic Program
Approximate t s
OPT
wiggly
wiggly
monotone
monotone
monotone
Shuchi Chawla, Carnegie Mellon University

19. From Min-Excess to Orienteering
Excess min=k (P)
Orien. maxℓ(P)=D (P)
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 s t 1 2 3 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
Excess min=k (P)
DR-TSP max ∑t
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 s t
OPT
excess = 1 v
length of entire remaining path decreases by 1
Shuchi Chawla, Carnegie Mellon University

21. learnt how to look for your lost wallet
So far…
(2+)-approximation for Min-excess
3-approximation for Orienteering
(6.75+)-approximation for Discounted-Reward TSP
You
should’ve
learnt how to look for your lost wallet
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
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 s t s t 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)
maxℓ(P)=D (P) 3
Discounted-Reward TSP
max ∑t
6.75+
Deadline-TSP
max ∑t≤D(v)v
3 log n
Time-Window Problem
max ∑R(v)≤t≤D(v)v
3 log2n
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