Approximation Algorithms for Path-Planning Problems

1
Approximation Algorithms for Path-Planning
Problems
Shuchi Chawla

• with
• Nikhil Bansal, Avrim Blum and Adam Meyerson

2
The Trick-o-Treaters Problem

• Collect as much candy as possible within 6pm and
  8pm
• More candy ? more popularity with the kids
• Some complicating constraints
• Limited amount of time
• Mr. X always gives twice as much candy as Mrs. Y,
  but his house is a long detour.
• Orienteering
• Given a metric and a starting point, cover as
  many high-reward locations as possible within
  a limited amount of time

3
Path-planning in the real world

• A robot-navigation problem
• Deliver packages to certain locations
• Faster delivery gt greater happiness
• Limited battery power
• Packages have different deadlines for delivery
• Assembly analysis
• Manufacturing
• Production planning

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A reward-time trade-off

• Given graph (metric) G, construct a path
  satisfying some constraints and optimizing some
  function.
• Classic formulation Traveling Salesman
• Find the shortest tour covering all locations
• Budget the path-length and maximize reward
• Orienteering Hard bound on path length
• Time Window Visit node v within Rv, Dv
• Impose a reward quota and minimize length
• k-Path Collect at least k reward

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A reward-time trade-off

• Given graph (metric) G, construct a path
  satisfying some constraints and optimizing some
  function.
• Classic formulation Traveling Salesman
• Find the shortest tour covering all locations
• Budget the path-length and maximize reward
• Orienteering 4 Blum C Karger03
• 3 Bansal Blum C Meyerson 04
• Time Window 3log2n Bansal Blum C Meyerson 04
• Impose a reward quota and minimize length
• k-Path 2 ? Chaudhury Godfrey Rao 03

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The rest of this talk

• A 3-approximation for Orienteering
• An O(log2n) approx for the Time-Window Problem
• Orienteering with deadlines
• Incorporating release-dates
• Extensions and Open Problems

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Orienteering and k-Path

• Orienteering length D maximize reward
• k-Path reward k minimize length
• Complementary problems
• Series of results on k-TSP (related to k-Path)
• BRV99 Garg99 AK00 CGRT03
• best approx (2?)
• None for Orienteering until recently!

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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!

OPT(d)
s
APPROX
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Why is Orienteering difficult?

• First attempt Use distance-based approximations
  to approximate reward
• Idea Modify the algorithm itself
• Doesnt help moat-growing always goes for
  shallow fruit
• Orienteering is inherently harder Perturbation
  of the input changes the output widely

OPT(d)
s
APPROX
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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

OPT
APPROX
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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, target reward k, find a path that
  collects reward at least k and minimizes ?(P)
  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
  Min-excess is strictly harder than
  K-path
• There is a (2?)-approximation for Min-Excess
• Blum, C, Karger, Meyerson, Minkoff, Lane,
  FOCS03
• Our algorithm returns a path with length
• d(s,t) (2?) ?(P)

excess
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A 3-approximation to Orienteering

• 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
• ? 3-approximation to orienteering

• 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)
Open Given an r-approx for min-excess (r 2 R ),
can we get r-approx to Orienteering?
v2
OPT
v1
APPROX
Excess of one path (?1?2?3)/3
Can afford an excess up to (?1?2?3)
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So far

• A 3-approximation for Orienteering
• An O(log2n) approx for the Time-Window Problem
• Orienteering with deadlines
• Incorporating release-dates
• Extensions and Open Problems

Coming up
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The Time-Window Problem

• Find a path visiting many nodes in their
  time-window
• school bus routing bank and postal
  deliveries
• industrial refuse collection newspaper
  delivery
• fuel oil delivery dial-a-ride
  service
• 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.

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The next step Deadline-TSP

• Every vertex has a deadline D(v) Find a path
  that maximizes nodes v visited before D(v)
• If the last node on the path has the min
  deadline, use Orienteering to approximate the
  reward
• Everything visited before the minimum deadline
• Dont 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!
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A segmentation of OPT
Deadline
Time
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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

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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

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

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The Rectangle Argument

1. 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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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
v
Require l(s,v) ? R(v)
? l(t,v) ? L-R(v)
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A Bicriteria Approximation

• Given any ? gt 0,
• Get O(log 1/?) fraction of reward
• Exceed deadlines by a (1?) factor
• O( log Dmax )-approximation
• 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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An overview of our results
Approximation
Problem
Orienteering
3
Deadline TSP
3 logn
Time-Window Problem
3 log2n
reward log 1/? deadlines 1?
Time-Window Problem - bicriteria
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Future Directions

• Better approximations
• can we get a constant factor for Time-Windows?
• special metrics such as trees or planar graphs
• hardness of approximation?
• Asymmetric Path-planning
• the graph is directed still obeys triangle
  inequality
• polylog-approximations and lower bounds for
  distance
• need entirely different ideas for
  asymmetric-Orienteering
• is it log-hard?

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Questions?
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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 ? d, ?(d?) 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 ? d, ?(d?) 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 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(vj1,t)
• consider P(u,t), where t t-l(u,vj1)
• pick the best path (best u) from the above

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Solving Min-Excess

• 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