algorithms for path-planning shuchi chawla (cmu/stanford/wisconsin) 10/06/05

Algorithms for Path-Planning

Shuchi Chawla(CMU/Stanford/Wisconsin)

10/06/05

Shuchi Chawla2 Algorithms for Path-Planning

A trick-or-treat problem• It’s Halloween…

Collect large amounts of candy between 6pm and 8pm

• Goal: Get as much candy as possible

• In what order should you visit houses?


A travelling repair-bot problem• Robot receives requests for repair• Requests come with a time-window for

servicing• Brownie points for each request serviced

• Goal: Maximize the total brownie points

• Cannot perform all of them – Takes time to service each request and

move from one to another

• The problem:– Which ones to accept?– How to schedule them?

SelectionOrdering


• Informally… planning and ordering of tasks

• Classic instance ― Traveling Salesman ProblemFind the shortest tour covering all given locations

• A natural extension ― Orienteering Cover as many locations as possible by a given deadline

Path-planning

• Many variants, applications:– Delivery & distribution problems– Production planning, Supply chain management– Robot navigation

• Studied in Operations Research for 2-3 decades• Mostly NP-hard; we look for approximation

algorithms


Approximation Results• A reward vs. time trade-off

Time taken

Rew

ard

obta

ined

A “quota” approximation

A “budget” approximation

OPT

(2 hrs, )

(2 hrs, )

(4 hrs, )


Approximation Results• A reward vs. time trade-off

• A budget on time; maximize reward– Orienteering– Deadline-TSP– TSP with Time-Windows

• A quota on reward; minimize time– TSP– k-TSP– Min-latency

• Optimize a combination of reward and time– Prize-Collecting TSP– Discounted-Reward TSP

single deadline on time different deadlines on different locations different time windows for diff. locations

visit all locations visit k locations visit all; minimize sum of times

minimize time plus reward foregone max. reward, reward decreases with time


1.5 [Christofides ’76] 2+ [Garg99] [AK00] [CGRT03] [Garg05] 3.59 [GK96] [CGRT03]

1.5 [Christofides ’76] 2+ [Garg99] [AK00] [CGRT03] [Garg05] 3.59 [GK96] [CGRT03]

• A reward vs. time trade-off

• A budget on time; maximize reward– Orienteering– Deadline-TSP– TSP with Time-Windows

• A quota on reward; minimize time– TSP– k-TSP– Min-latency

• Optimize a combination of reward and time– Prize-Collecting TSP– Discounted-Reward TSP

2 [Goemans Williamson ’92] ? 2 [Goemans Williamson ’92] 6.75+

Approximation Results

? ? ?

3 3 log n 3 log2 n

LP-based techniques

Use structural properties& Dynamic Programming

Joint work with Bansal, Blum, Karger, Meyerson, Minkoff & Lane


Back to Orienteering• The givens

A “map” G of locations, distances, start location sRewards on locationsDeadline D

• To findA path that collects as much reward as possible by deadline D


Back to Orienteering: why LPs don’t work• Naïve attempt: use (a variant of) k-TSP

– Fails when most reward is far from the start

• Algorithms for “quota” problems rely on the Goemans-Williamson primal-dual subroutine– Miss out on far-away reward

• Budget problems are ill-behaved w.r.t. small perturbations

• Solution: handle the “very-low slack” case separately

Bad case: low slack


Formalizing the “slack”• Excess of path = length of path – length of

shortest path

• An approx. for excess an approx. for Orienteering

Given approx for k-TSP: Good solution

to prefix of OPT

Given approx for excess: Good solution to suffix of OPT

• -approx to length : len(ALG) (shorest-len + excess)• -approx to excess : len(ALG) shortest-len + excess


– examine a slightly different problem

• Suppose OPT visits all nodes in order of increasing distance from start; Find OPT.– Can solve this exactly using dynamic programming!

Approximating the excess• Excess of path = length of path – length of

shortest path

• -approx to length : len(ALG) (shortest-len + excess)

• -approx to excess : len(ALG) shortest-len + excess

• The simple case : excess > shortest-len– 2-approx to length gives 3-approx to excess– Solution: use k-TSP

• The harder case : excess << shortest-len


Approximating the excess• The large excess

case:excess > shortest-len

• Use k-TSP

• The small excess case:excess ≈ 0

(OPT is “monotone”)• Use Dynamic

Programming

Use k-TSPDynamic Program

monotone

monotone

monotone

wiggly wiggly

Patch segments using dynamic programming

OPT

Gives a (2+)-approximation for Min-Excess

In order of increasing distance from start

What about the intermediate case?


• Construct a path from s to t, that has length D and collects maximum possible reward

• Given a 3-approximation to min-excess:1. Divide OPT into 3 “equal-reward” parts (hypothetically)2. Approximate the part with the smallest excess

3-approximation to Orienteering

An algorithm for Orienteering

Excess of ALG 1+2+3 ; Reward of ALG = ⅓ reward of OPT

1 2

3

Excess of path from u to v

(u,v) = ℓ(u,v)–d(u,v)

An r-approximation for min-excess gives

a r-approximation for Orienteering

v1

v2

OPT

ALG


So far…• A new problem: min-excess• An approximation to min-excess• A 3-approximation to Orienteering

• Coming up next:Planning with deadlines – a (3 log n)-

approximation


Planning with deadlines: Deadline-TSP• The travelling repair-bot problem:

– Every node (request) v has a deadline D(v)– Reward obtained at v, if v is visited before D(v)

– In Orienteering, D(v) = D for all nodes v

• Constant approx known for special cases:– points on a line– Few different deadlines (based on Orienteering)

• Nothing known in general


Approximating Deadline-TSP• Our approach:

– Divide the problem into “Orienteering sub-problems”

– Solve these using algorithm developed previously

• When can we use Orienteering?If the last vertex visited by OPT has the smallest

deadline– Remove nodes of smaller deadline– Reduce all other deadlines to the smallest one– OPT remains unchanged

• Does OPT have a large subpath with this property?


A pictorial representation of OPT

Time

Dead

line “Good” rectangle

Can be approximated

via Orienteering


The high-level idea• A single “good” rectangle may not contain

enough reward

• Idea: Approximate many rectangles; stitch together the approximations

• Main bottleneck: avoiding double-counting of reward– Divide graph into disjoint subsets Vi – Approximate the i-th rectangle over Vi


Avoiding double-counting

• Divide nodes into sets by deadlines

• Guess last node on OPT in each group

• Approximate Orienteering on each group

Time

Dead

line

V1

V2

V3

V4

length of seg. 2

length of seg. 3


Avoiding double-counting

• To visit nodes before deadlines, rectangles must be disjoint along the time-axis

• Still consider the same sets Vi of vertices

• Obtain reward contained in axis-disjoint rectangles

Time

Dead

line

V1

V2

V3

V4


The disjoint-rectangle argument• Approximate reward contained in a family of

“disjoint” rectangles

• We construct (log n) families of disjoint rectangles– The families together contain all the nodes of OPT– We can approximate the best family via

Orienteering and dynamic programming

– Reward obtained ⅓ 1/(log n)

– Gives a (3 log n)-approximation


Deadlines to Time-Windows• Nodes have release-times as well as deadlines• Must visit a node within its time-window to

obtain reward

• Straightforward extension of Deadline-TSP algorithm (3 log2 n)-approximation to Time-Windows-TSP

• A bicriteria approximation:Given any k > 0, get O(1/k) fraction of reward

exceed deadlines by a (1+2-k) factor O(log Dmax)-approximation


A recap of our results• First approximations for Orienteering, Deadline-

TSP, Time-Windows-TSP, Discounted-Reward-TSP, …

• Introducing the new min-excess objective and approximating it


Open Problems• Hardness of planning problems

• Approximations for directed path-planning– Chekuri, Pal give quasi-polytime polylog-

approximations

• Dealing with capacity & precedence constraints

• Techniques for approximating planning problems on MDPs

Questions?

algorithms for path-planning shuchi chawla (cmu/stanford/wisconsin) 10/06/05

Documents