Time-Based Voronoi Diagram

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Time-Based Voronoi Diagram. D. T. Lee Institute of Information Science Academia Sinica, Taipei, Taiwan. dtlee@iis.sinica.edu.tw. Jointly with C. S. Liao, W. B. Wang, IIS. Outline . Introduction Preliminaries Good intersection condition General condition Conclusion. - PowerPoint PPT Presentation

Transcript of Time-Based Voronoi Diagram

Time-Based Voronoi DiagramTime-Based Voronoi Diagram

D. T. LeeInstitute of Information Science AcadeInstitute of Information Science Acade

mia Sinica, Taipei, Taiwanmia Sinica, Taipei, Taiwan

dtlee@iis.sinica.edu.twdtlee@iis.sinica.edu.tw

Jointly with C. S. Liao, W. B. Wang, IIS.

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Outline Introduction Preliminaries Good intersection condition General condition Conclusion

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Multiple Highways Model Input: A set S of points, S={p1, …, pn} in th

e plane and k highways L1, …, Lk, modeled as lines. Travelers can enter the highways at any point a

nd move along Li at speed vi in both directions. Off the highways travelers can move freely in a

ny direction at speed v0 << v1,…, vk. Output: A Voronoi diagram for the input ba

sed on traveling time, i.e. Time-based Voronoi Diagram

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Time Distance Given two points p, q in the plane, the short

est time path spt(p, q) is a path that takes the shortest time traveling between p to q.

The time distance dt(p, q) between p and q is the time required to follow any shortest time paths between p and q.

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One Highway Problem Abellanas, Hurtado, Sacristan, Icking, Ma,

Klein, Langetepe, Palop IPL, 2003 Assumption

L1 lies on the x-axis. sine = v0/v1 = 1/v1

L1+: the half-plane above L1

L1-: the half-plane below L1

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Where to Enter the Highway

α

p

q

sine = v0/v1 = 1/v1α

prpl L1

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Time Distance

L1

α

p

q

pL1

pr ql

1ˆ Lp

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Terminology : the symmetric point of p reflecting by L1. Given a site p, let be the half-ray with

endpoint p and of slope tan (-tan ) respectively.

1Lp

)ˆ( ˆ pp

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L1

p

q

1ˆ Lp1ˆ Lp

1L1Lp

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L1

p

q

p̂p̂

1L

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Approach Transform the 1-highway problem into

another problem in time distance. If q and p are on the same side, the time

distance between q to p must be one of the Euclidean distances from q to

Otherwise, the time distance between q to p must be one of the Euclidean distances from q to

11 ˆ,ˆ, LL ppp

ppp ˆ,ˆ,

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Vor() & Vort() Vor(x, X): the Euclidean Voronoi Region of

a site or a line x X with respect to the set X.

Vort(x, X): the time-based Voronoi Region of a site or a line x X with respect to the set X.

Xx

YxVorYXVor

),(),(

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Theorem [Abellanas, et al.] For p L+

For p L-

)),ˆ(),ˆ(),((

)),ˆ(),ˆ(),((),(

1

111

bbb

aLaLat

PpVorPpVorPpVorL

PpVorPpVorPpVorLSpVor

)),ˆ(),ˆ(),((

)),ˆ(),ˆ(),((),(11

1

1

bLbLb

aaat

PpVorPpVorPpVorL

PpVorPpVorPpVorLSpVor

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Envelope & Objects Involved

The envelope of the objects below L1

The Voronoi diagram above L1

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Two Highways Problem O is the intersection of L1 and L2

is the angle between L1 and L2

is the union of and for L1 is similarly defined for L2

Four “quadrants” Q0, Q1, Q2, Q3

1ˆ Lp p̂ p̂

2ˆ Lp

2010 /sin ,/sin21

vvvv LL

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L1

L2

O

Q3

Q1

Q2

Q0

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Two Highways Lemma 3.1

Suppose L1 + L2 = , for two points p, q on different highways.

The shortest time paths are not unique. One of the shortest time paths from p to q is to

walk along one highway then change to the other at the intersection.

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Two Highways

q

p

L1

L2

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Two Highways Lemma 3.2

Suppose L1 + L2 < , for two points p, q on different highways.

The shortest time path from p to q is to walk along one highway then change to the other at the intersection.

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Two Highways

A B

DC

q

p

L1

L2

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Two Highways Lemma 3.3

Suppose L1 + L2 > , for two points p, q on different highways.

The shortest time path from p to q is to walk along at most one highway. (shortcut)

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Two Highways

q1

p

L1

L2

q2

q3

L1

L2

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Good Condition for Highway Intersection

Highways L1, L2 are said to satisfy good intersection condition if and only if L1 + L2 .

Any shortest time path connecting two points on different highways that satisfy good intersection condition contains no shortcut.

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O-Domination Site pO is the O-domination site if

O is in the Voronoi region of O-domination site pO

),(min),( pOdpOd tSpOt

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-Distance-Line-from-O

L1

L2

O

2

Q3

213 )( O

Q2

Q1

Q0

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O-Domination Line The -distance-line-from-O, , is called

O-domination line in Qi, where = dt(O, pO).

)(Oi

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Trivial Site Any site which is not the O-domination site

is a trivial site

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Some Properties For a point qVort(p, S), if the shortest time

path from q to p passes through O, then the site p is the O-domination site.

For a point qVort(p, S), if the shortest time path from q to p enters both highways, the path must pass through O provided that the two highways satisfy good intersection condition.

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Some Properties (cont’d) For a point qVort(p, S), and p is a trivial si

te, then the q to p path never enters both highways.

For a trivial site p in Qi,Vort(p, S) Q(i+2) mod 4 =

We need not consider trivial sites in Qi when we compute the Voronoi diagram in Q(i+2) mod 4

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Notations Let Li+ be the line that borders quadrant Qi a

nd Q(i+1) mod 4, and Li- borders quadrant Qi and Q(i-1) mod 4

QiQ(i+1) mod 4

Q(i-1) mod 4

Li-

Li+

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Good Condition Case The time-based Voronoi diagram in Qi, is d

etermined by the set of objects Pi:

4 mod)1(4 mod)1(

ˆˆ

ˆˆ)(

ii

i

ii

Qpi

Qpi

Qp

Li

Lii

i

pp

pppOP

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Envelope & Objects Involved

Li+

Li-

O

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Time-Based Voronoi Diagram The time-based Voronoi diagram in a quadr

ant Qi is The time-based Voronoi diagram is

It is our general form.

ii QPVor )(

3

0

)(

i

ii QPVor

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Algorithm Find the O-domination site pO and let=dt(O, pO) Compute the O-domination line for Qi, i=0,1,2,3 Compute the set Pi of objects used for constructing the Vor

onoi diagram in each quadrant Qi for i=0,1,2,3. i.e, the envelope surrounding Qi, and all the sites in Qi

Compute the ordinary Voronoi diagram in Qi. i.e, Vor(Pi) Qi

For all sites p, collect all regions associated with ,and p

pp ˆ ,ˆ

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Theorem The Voronoi diagram for a set S of n sites i

n the presence of two highways L1 and L2 in the plane that satisfy the good intersection condition, can be computed in O(n log n) time.

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Multiple Highways Problem Idea

If good intersection condition holds, the problem is not hard.

Find domination site for each intersection. In each cell of the arrangement, only the sites in

the cell and neighboring cells determine the time-based Voronoi diagram in the cell.

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How to Find Domination Sites?

Insert highways one at a time in order of non-descending speeds.

Rewrite and update intersection domination sites.

Propagation subroutine.

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Propagation

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Time Complexity n sites, k highways To determine all intersection-domination sit

es with propagation costs O(kn + k3 log k) time

To compute all time-based Voronoi region costs O(n log n) time

The total time is O(kn + k3 log k + n log n)

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Two-Highway Model in General No good condition now. Lemma 5.1

Let p, q be any two points on the plane. If the number of shortest time path from p to q is finite, and the shortest time path walks along both highways, then the path must pass through the intersection of two highways.

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Two-Highway Model in General (cont’d) The time-based Voronoi diagram in Qi, is d

etermined by the set of objects Pi:

)(ˆˆˆˆ

ˆˆˆˆ

4 mod)1(4 mod)2(

4 mod)1(

Opppp

ppppSP

iQp

iLi

Qpii

Qp

Lii

Qp

Li

Li

i

i

i

i

i

i

i

ii

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Time-Based Voronoi Diagram The time-based Voronoi diagram in a quadr

ant Qi is The time-based Voronoi diagram is

The time-based Voronoi diagram for n points in the presence of two highways can be computed in O(n log n) time.

ii QPVor )(

3

0

)(

i

ii QPVor

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Special Cases Two parallel highways

12 LL

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Two Parallel highways Problem Idea

No origin-domination site No shortest time path along both highways Compute the envelope associated with a proper

set of hats

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Two Parallel Highways Problemp

q

qL1

qL2

L1

L2

21 vv

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L2 L1+ L1 nullifies L2

No shortest time path along both highways We solve the problem as in two parallel hig

hways case.

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L2 L1+

p

OL1

L2

1

ˆLO

2

ˆLO

21 vv

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General Multiple Highways Case Hard to determine the shortest time path Hard to determine the intersection dominati

on sites Propagation doesn’t work

OPEN?

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Conclusion n sites, k highways If good intersection condition holds, we can

solve the problem inO(k3 log k + kn + n log n) time

If good intersection condition doesn’t hold, we can solve two highways problem inO(n log n) time.