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Approximation Algorithms for Path-Planning Problems

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### Approximation Algorithms for Path-Planning Problems

with

Nikhil Bansal, Avrim Blum and Adam Meyerson

Shuchi Chawla

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

Shuchi Chawla, Carnegie Mellon University

Path-planning in the real world

- A robot-navigation problem
- Deliver packages to certain locations
- Faster delivery => greater happiness
- Limited battery power
- Packages have different deadlines for delivery

- Assembly analysis
- Manufacturing
- Production planning

Shuchi Chawla, Carnegie Mellon University

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

Shuchi Chawla, Carnegie Mellon University

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
- Orienteering4 [Blum C Karger+03]
3 [Bansal Blum C Meyerson 04]

- Time Window3log2n [Bansal Blum C Meyerson 04]

- Orienteering4 [Blum C Karger+03]
- Impose a reward quota and minimize length
- k-Path 2 + [Chaudhury Godfrey Rao+ 03]

Shuchi Chawla, Carnegie Mellon University

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

Shuchi Chawla, Carnegie Mellon University

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!

Shuchi Chawla, Carnegie Mellon University

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) << OPT(d)
- Bad trade-off between distance and reward!

OPT(d)

s

APPROX

Shuchi Chawla, Carnegie Mellon University

Why is Orienteering difficult?

- 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

OPT(d)

s

APPROX

Shuchi Chawla, Carnegie Mellon University

Why is Orienteering difficult?

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

Why is Orienteering difficult?

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

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, FOCS’03]

- Our algorithm returns a path with length
d(s,t) + (2+) (P)

excess

Shuchi Chawla, Carnegie Mellon University

A 3-approximation to Orienteering

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

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 2R +),

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)

Shuchi Chawla, Carnegie Mellon University

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…

Shuchi Chawla, Carnegie Mellon University

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.

Shuchi Chawla, Carnegie Mellon University

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
- 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

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

A 2-dimensional view

minimal vertices

Deadline

“Disjoint Rectangles”

Time

Shuchi Chawla, Carnegie Mellon University

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

Shuchi Chawla, Carnegie Mellon University

The Rectangle Argument

Deadline

Time

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

Shuchi Chawla, Carnegie Mellon University

The Rectangle Argument

If there are between 2b and 2b+1 points in between,

then either the bth or a larger family contains exactly 1 point in the interval

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

Deadline

Time

Shuchi Chawla, Carnegie Mellon University

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

Shuchi Chawla, Carnegie Mellon University

The Rectangle Argument

Deadline

Time

2. We can approximate the best disjoint family

Shuchi Chawla, Carnegie Mellon University

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

Shuchi Chawla, Carnegie Mellon University

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

A Bicriteria Approximation

- Given any > 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

Shuchi Chawla, Carnegie Mellon University

An overview of our results

Problem

Approximation

Orienteering

3

Deadline TSP

3 logn

Time-Window Problem

3 log2n

reward: log 1/

deadlines: 1+

Time-Window Problem - bicriteria

Shuchi Chawla, Carnegie Mellon University

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?

Shuchi Chawla, Carnegie Mellon University

Questions?

Shuchi Chawla, Carnegie Mellon University

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