Approximation Algorithms for PathPlanning Problems

1
Approximation Algorithms for Path-Planning
Problems
Shuchi Chawla

• with
• Nikhil Bansal, Avrim Blum,
• David Karger, Adam Meyerson,
• Maria Minkoff, Terran Lane

2
The Lost-Wallet Problem

• How should you go about finding a lost wallet?
• Several possible locations
• Different likelihoods of finding the wallet at
  different locations
• Some chance of getting stolen before you find it
• Task Find a good search strategy
• visit a lot of places having high likelihood
• visit them quickly
• faster search ? greater likelihood of discovery

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!
• The mall is only open until 5pm
• Cover as many locations as possible
• Give preference to the more likely ones

4
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
Alternately, you can only search for time D
?
Goal Construct path of length at most D that
collects maximum reward
?
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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

time
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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
7
Our results
Approximation
Problem
2?
K-path Chaudhuri et al FOCS,03
2?
Min-Excess Path
3
Orienteering
point-to-point
6.75?
Discounted-Reward TSP
8
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
• Orienteering with deadlines Deadline-TSP
• Extensions and Open Problems

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Why is Orienteering difficult?

• 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) ltlt OPT(d)
• Bad trade-off between distance and reward!
• Same problem with Discounted-Reward TSP
• Multiplying the exponent with a constant gives a
  bad approximation

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Why is Orienteering difficult?

• 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 thats far
  away
• Approximate the extra length taken by a path
  over the shortest path length

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Why is Orienteering difficult?

• 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 thats 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??

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The Min-Excess Problem

• 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

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Solving Min-Excess

• OPT d? k-path gives us ALG ?(d?)
• We want ALG d ??
• Note When ? ¼ d, ?(d?) ¼ d O(?) ?
• Idea When ? is large, approximate using k-path
• What if ? ltlt d ?
• Small ? ? path is almost like a shortest path
• or its distance from s mostly increases
  monotonically

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Solving Min-Excess

• OPT d? k-path gives us ALG ?(d?)
• We want ALG d ??
• Note When ? ¼ d, ?(d?) ¼ d O(?) ?
• Idea When ? is large, approximate using k-path
• What if ? ltlt d ?
• Small ? ? path is almost like a shortest path
• or its distance from s mostly increases
  monotonically
• Idea Completely monotone path ? use dynamic
  programming!

Patch segments using dynamic programming
t
s
wiggly
wiggly
monotone
monotone
monotone
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From Min-Excess to Orienteering

• There exists a path from s to t, that
• collects reward at least ?
• has length D
• Given an r-approximation to min-excess ( r 2
  Z )
• 1. Divide into r equal-reward parts
• 2. Approximate the part with the smallest excess

• Using an r-approx for Min-excess,
  we get an r-approximation for
  s-t Orienteering

Excess of path P extra time taken over
the shortest path length dP(u,v) d(u,v)
Excess of one path (?1?2?3)/3
Can afford an excess up to (?1?2?3)
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Solving Discounted-Reward TSP

• 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) gt ?OPT
reward ?OPT/2
length of entire remaining path decreases by 1
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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
shouldve
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The FEDEX-guy Problem

• The deliveryman has to deliver packages to
  various locations
• Packages have deadlines
• Some packages have higher priority than others
• Deliver as many packages before their deadlines,
  as possible
• Metric of success total reward from packages
  delivered on time

Deadline-TSP
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The Deadline-TSP

• Find a path that visits many nodes before their
  deadlines
• A generalization The Time-Window Problem
• Each vertex can be visited within a particular
  time-window
• Release times as well as deadlines
• 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

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The Deadline-TSP Using Orienteering

• If the last node visited by a path has the
  minimum deadline, use Orienteering to approximate
  the reward in that path
• Everything visited by the minimum deadline
• Dont need to bother about deadlines of other
  nodes
• Does OPT always have a large subpath with the
  above property?
• Approximate many such subpaths of OPT and patch
  them together
• How do we segment nodes?
• How do we ensure there is no double counting?

NO!
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A 2-dimensional view
Deadline
Disjoint Rectangles
Time
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The Rectangle Argument

• Approximate reward contained in a disjoint
  family of rectangles
• Every pair of rectangles is non-overlapping in
  BOTH dimensions
• We construct O(log n) families of disjoint
  rectangles
• 1. These cover ALL the reward in OPT
• 2. We can approximate the best of them
• We get an O(log n)-approximation

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The Rectangle Argument

• There are O(log n) families of disjoint
  rectangles that cover all the reward in OPT

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The Rectangle Argument

• There are O(log n) families of disjoint
  rectangles that cover all the reward in OPT

Deadline
Time
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The Rectangle Argument

• 2. We can approximate the best disjoint family
• Suppose we know the minimal vertices
• Just try out all the log n families
• Problem - Minimal vertices depend on the optimal
  tour!
• Solution
• Try all possibilities.
• They are ordered by deadlines, so use a simple
  dynamic program

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The Rectangle Argument

• 2. We can approximate the best disjoint family

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The O(log n)-approximation

• Approximate reward contained in a disjoint
  family of rectangles
• Every pair of rectangles is non-overlapping in
  BOTH dimensions
• We construct O(log n) families of disjoint
  rectangles
• 1. These cover ALL the reward in OPT
• 2. We can approximate the best of them
• Obtain an O(log n)-approximation

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A Bicriteria Approximation

• Given any ? gt 0,
• Get O(log 1/?) fraction of reward
• Exceed deadlines by a (1?) factor
• Constant factor approximation if we can exceed
  deadlines by a small constant factor
• Nice trade-off
• Halving the extra time taken, increases the
  approximation factor by only an additive 1

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A summary of our main results
Approximation
Problem
2?
Min-Excess Path
3
Orienteering (point-to-point)
6.75?
Discounted-Reward TSP
3 log n
Deadline-TSP
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Some extensions

• Unrooted versions
• Multiple tours
• Max-reward Steiner tree of bounded size

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Future work

• Improve the approximations
• 2-approx for Orienteering?
• Constant factor for different deadlines
• The time-window problem
• Can we get a constant or log factor in general
  graphs?

32
Some related problems

• Basic task reach as many locations as possible,
  as fast as possible
• Deliver a fraction of the packages as fast as
  possible
• Minimize time taken plus value of the packages
  not delivered

k-Traveling Salesperson
Garg AroraKarpinski
Best 2? approx
Prize-collecting TSP
GoemansWilliamson
Best 2-approx
Note All these variants are NP-complete
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Approximating the length traveled

• Distance-based approximations are not good enough
  for Robot Navigation
• Consider a 2-approx for distance
• - Suppose OPT has a path of length t
• - Our solution has length 2t
• All rewards of OPT are discounted by at most ?t
  Our rewards may be discounted
  by ?2t
• We may incur a factor of ?t extra !
• Bad if the tour is long

34
A fixed-length formulation

• Impose a hard constraint on the length of the
  tour
• Goal Given a parameter D, construct path of
  length at most D that collects maximum reward
• Discounted-reward is a soft-constraint version
• As we get farther, reward diminishes steadily
  and gradually

• No approximation algorithm known previously for
  the rooted non-geometric version

time
35
Approximating Orienteering

• Using a distance-based approx
• Divide the optimal path into many segments
• Approximate the max reward segment using distance
  saved by short-cutting other segments
• If min-distance between s and v is d, we spend at
  least d in going to v, regardless of the path

t
v
s
36
The Min-Excess Problem

• Using distance-based approx for orienteering
• Short-cut a part of the optimal path
• Used the saved length to approximate another part
• If min-distance between s and t is d, we spend at
  least d in going to t, regardless of the path
• 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??

37
Our contributions
Approximation
Source/Reduction
Problem
2?
Chaudhuri et al03
K-path (?CP)
2.5?
1.5 ?CP 0.5
Min-Excess Path (?EP)
2?
4
1?EP
8.12?
(1?EP)(11/?EP)?EP
6.75?
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A Robot Navigation Problem

• Task deliver packages to locations in
  a building
• Faster delivery gt greater happiness
• Classic formulation Traveling Salesperson
  Problem
• Find the shortest tour covering all locations
• Uncertainty in robots lifetime/behavior
• battery failure sensor error
• Robot may fail before delivering all packages
• Deliver as many packages as possible
• Some packages have higher priority than others

39
Robot Navigation A probabilistic view

• At every time step, the robot has a fixed
  probability (1-?) of failing
• If a package with value ? is delivered at time t,
  the expected reward is ??t
• Goal Construct a path such that the total
  discounted reward collected is maximized

Discounted Reward
Alternately, robot has a fixed battery life D
?
Goal Construct path of length at most D that
collects maximum reward
?
40
Approximating Orienteering

• Using a distance-based approx
• Divide the optimal path into many segments
• Approximate the max reward segment using distance
  saved by short-cutting other segments
• If min-distance between s and v is d, we spend at
  least d in going to v, regardless of the path

41
Approximating Orienteering

• Using a distance-based approx
• Divide the optimal path into many segments
• Approximate the max reward segment using distance
  saved by short-cutting other segments
• If min-distance between s and v is d, we spend at
  least d in going to v, regardless of the path
• 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??