Computational Movement Analysis Lecture 3: Curve simplification Joachim Gudmundsson

Computational Movement Analysis

Lecture 3: Curve simplification

Joachim Gudmundsson

Problem

Many algorithms handling movement data are slow, e.g. - similarity O(nm log nm)- approximate clustering O(n2+nml)- …

Observation:Many trajectories can be simplified without losing much information.

Problem

Many algorithms handling movement data are slow, e.g. - - similarity O(nm log nm)- approximate clustering O(n2+nml)- …


Example:

Problem

Many algorithms handling movement data are slow, e.g.

- similarity O(nm log nm)- approximate clustering O(n2+nml)- …

Observation:Many trajectories can be simplified without losing much information. Example:

Problem

Many algorithms handling movement data are slow, e.g. - similarity O(nm log nm)

- approximate clustering O(n2+nml)- …


Example:

Problem



Example:

Problem




Example:

Problem



Example:

Problem




Example:

Problem



Example:

Example

Generated trajectory:

Universe: 400x400Number of points: 10KError: 5 units 278 points

10 units 174 points Time: ~15ms

Car trajectory:

Universe: ~20km 20kmNumber of points: 85KError: 5 meter 2601 points

10 meter 1728 points Time: ~200ms

Outline

Simplifying polygonal curves

• Ramer–Douglas–Peucker 1973

• Driemel, Har-Peled and Wenk, 2010

• Imai-Iri 1988

• Agarwal, Har-Peled, Mustafa and Wang, 2005

• Take time into consideration

Ramer-Douglas-Peucker

1972 by Urs Ramer and 1973 by David Douglas and Thomas Peucker

The most successful simplification algorithm. Used in GIS, geography, computer vision, pattern recognition…

Very easy to implement and works well in practice.p

pjpi


Input polygonal path P = p1,…,pn and threshold

Initially i=1 and j=n

Algorithm DP(P,i,j) Find the vertex vf between pi and pj farthest from pipj. dist := the distance between vf and pipj.

if dist > thenDP(P, vi , vf)DP(P, vf , vj)

elseOutput(vivj)

pjpi



Initially i=1 and j=n

Algorithm DP(P,i,j) Find the vertex vf between pi and pj farthest from pipj. dist := the distance between vf and pipj.


elseOutput(vivj)

dist

>

>


>

>

>

>

p

q


p

q


Time complexity?

Testing a shortcut between pi and pj takes O(j-i) time.

Worst-case recursion?Algorithm DP(P,i,j) Find the vertex vf farthest from pipj. dist := the distance between vf and pipj.


elseOutput(vivj)

DP(P, vi , vi+1)DP(P, vi+1 , vj)

Time complexity T(n) = O(n) + T(n-1) = O(n2)

Summary: Ramer-Douglas-Peucker

Worst-case time complexity: O(n2)

In most realistic cases: T(n) = T(n1) + T(n2) + O(n) = O(n log n), where n1 and n2 are smaller than n/c for some constant c.

If the curve is in 2D and it does not self-intersect then the algorithm can be implemented in O(n log* n) time.

[Hershberger & Snoeiynk’98]

Does not give any bound on the complexity of the simplification!

Driemel et al.

Simple simplification(P = p1,…,pn, ) P’:= p1i:=1while i<n do q := pi

pi := first vertex pi in q,…,pn s.t. |q-pi|> if no such vertex then set i:=nadd pi to P’

endreturn P’

Driemel et al.

Simple simplification(P = p1,…,pn, ) Property 1: All edges (except the last one) have length at least .

Property 2: F(P,P’)

Driemel et al.

Simple simplification(P = p1,…,pn, ) Property 1: All edges (except the last one) have length at least .


Definition: A curve P is c-packed, if has finite length, and for any ball b(p,r) it holds |P b(p,r)| < cr.

Property 3: If P is c-packed then P’ is 6c-packed.

Driemel et al.

Aim: Prove Property 3 If P is c-packed then P’ is 6c-packed.

We need the following observation:Let P be a curve and let P’ be an -simplification of P |P B(p,r+ )| |P’ B(p,r)| for any ball B(p,r)

Proof:

B(p,r)

Pu

Driemel et al.

Observation:Let P be a curve and let P’ be an -simplification of P |P B(p,r+)| |P’ B(p,r)| for any ball B(p,r)

Proof:

B(p,r)

u

Driemel et al.


Proof:

B(p,r)

uPu

Hu

Pu must lie inside Hu

Driemel et al.


Proof:

B(p,r)

uPu

Hu

v

lv rv

Pu must intersect lv and rv |PuHv| |v|

|B(p,r) u| = |v|

Hv

Driemel et al.


Proof:

B(p,r)

uPu

Hu

v

lv rv

v B(p,r)

B(p,r+)

|B(p,r) u| = |v|

Hv

Driemel et al.


Proof:

B(p,r)

uPu

Hu

v

lv rv

B(p,r+)

v B(p,r) Hv B(p,r+)

|B(p,r) u| = |v|

Hv

Driemel et al.


Proof:

B(p,r)

uPu

Hu

v

lv rv

B(p,r+)

v B(p,r) Hv B(p,r+) Pu Hv B(p,r+)

|B(p,r) u| = |v|

Hv

Driemel et al.


Proof:

B(p,r)

uPu

Hu

v

lv rv

B(p,r+)

v B(p,r) Hv B(p,r+) Pu Hv B(p,r+) |Pu B(p,r+)| |v|

|B(p,r) u| = |v|

Driemel et al.

Theorem: If P is c-packed then P’ is 6c-packed.

Observation:Let P be a curve and let P’ be an -simplification of P |P B(p,r+ )| |P’ B(p,r)| for any ball B(p,r)

Driemel et al.


Proof: Proof by contradiction.Assume |P’B(p,r)|>6cr for some B(p,r)

Two cases: r or r<

Case r : |P B(p,2r)|

Driemel et al.



Two cases: r or r<

Case r : |P B(p,2r)| |P B(p,r+)|

Driemel et al.



Two cases: r or r<


|P B(p,2r)| > 6cr

|P B(p,r’)| > 3cr’ which contradicts that P is c-packed


Driemel et al.



Two cases: r or r<


|P B(p,2r)| > 6cr

|P B(p,r’)| > 3cr’ which contradicts that P is c-packed

Case r < : Proof is similar.


Summary: Driemel et al.

Simple simplification: can be computed in O(n) time Property 1: All edges (except the last one) have length at least .


Definition: A curve P is c-packed, if has finite length, and for any ball b(p,r) it holds |P b(p,r)| < cr.

Property 3: If P is c-packed then P’ is 6c-packed.

Imai-Iri

Both previous algorithms are simple and fast but do not give a bound on the complexity of the simplification!

Imai-Iri 1988 gave an algorithm that produces a -simplification with the minimum number of links.

Imai-Iri


1. Build a graph G containing all valid shortcuts.2. Find a minimum link path from p1 to pn in G

Imai-Iri

Find all possible valid shortcuts

Imai-Iri


>

Imai-Iri


<<

Imai-Iri


>

Imai-Iri


Imai-Iri

All possible shortcuts!

Imai-Iri

1. Build a directed graph of valid shortcuts.2. Compute a shortest path from p1 to pn using breadth-first search.

Imai-Iri

Brute force running time: ? #possible shortcuts ?

Summary: Imai-Iri

Running time: O(n3) O(n2) possible shortcuts

O(n) per shortcut O(n3) to build graphO(n2) BFS in the graph

Output: A path with minimum number of edges

Improvements: Chan and Chin’92: O(n2)

Results so far

RDP: O(n2) time (simple and fast in practice) Output: A path with no bound on the size of the path

SS: O(n) time (simple and fast in practice)Output: A path with no bound on the size of the path

Imai-Iri: O(n2) time by Chan and Chin (complicated)Output: A path with minimum number of edges

RDP and II use the Hausdorff error measure, SS use Fréchet

Can we get something that is simple, fast and has a worst-case bound using the Fréchet error measure?

Agarwal et al.

Agarwal, Har-Peled, Mustafa and Wang, 2002

Time: Running time O(n log n)

Measure: Fréchet distance

Output: Path has at most the same complexity as a minimum link (/2)-simplification

Analysis - Agarwal et al.

Notation: Let (pipj) denote the Fréchet distance between (pi,pj) and the subpath (pi,pj) of P between pi and pj.

Let P = p1, p2, ... , pn be a polygonal curve pi

pm

pj

pl

Algorithm - Agarwal et al.

Algorithm(P = p1,…,pn, )

i := 1 P’= p1 while i < n do find any j>i such that ((pi,pj) and ((pi,pj+1) > ) or

((pi,pj) and j=n) P’= concat(P,pj) i := j end return P’


p1

pj

Pj+

1

Algorithm(P = p1,…,pn, )

i := 1 P’= p1 while i < n do find any j>i such that ((pi,pj) and ((pi,pj+1) > ) or

((pi,pj) and j=n) P’= concat(P,pj) i := j end return P’


Efficient algorithm?

Recall: Computing the Fréchet distance between (pi,pj) and (pi,pj) can be done in O(j-i) time (Lecture 1).

pi

pj

Pj+1


Efficient algorithm?

Recall: Computing the Fréchet distance between (pi,pj) and (pi,pj) can be done in O(j-i) time (Lecture 1).

Finding the first j such that (pi,pj) and (pi,pj+1) > takes O(n2) time.

pi

pj

Pj+1


How can we speed up the search for pj?

Note that it is enough to find any vertex pj such that

(pi,pj) and (pi,pj+1) >

Idea: Search for pj using exponential search followed by binary search!


Exponential search:Test pi+1, pi+2, pi+4, pi+8… until found pi+2k, such that (pi,pi+2k) > .Binary search:Search for pj between pi+2k-1 and pi+2k s.t.

(pi,pj) and (pi,pj+1) > .

i+2k-1 i+2k

>


> if then

search right

Binary search:Search for pj between pi+2k-1 and pi+2k s.t.


Exponential search:Test pi+1, pi+2, pi+4, pi+8… until found pi+2k, such that (pi,pi+2k) > .

i+2k-1 i+2k


>




i+2k-1 i+2k

if > thensearch left


>




i+2k-1 i+2kpj pj+1

>



(pi,pj) and (pi,pj+1) > . Iterations: O(log n)

Total time: O(n log n)

Exponential search:Test pi+1, pi+2, pi+4, pi+8… until found pi+2k, such that (pi,pi+2k) > . Iterations: O(log n)


Lemma 1: Let P = p1, p2, ... , pn be a polygonal curve. For l ≤ i ≤ j ≤ m, (pipj) ≤ 2·(plpm).

pi

pm

pj

pl



Proof:Fix matching that realises (plpm).Set = (plpm).

pi

pm

pj

pl

qj

qi




1. ((pipj),(qiqj)) ≤

pi

pm

pj

pl

qj

qi





pi

pj

qj

qi

(pipj)





2. ((qiqj),(pipj)) ≤

pi

pm

pj

pl

qj

qi






pi

pj

qj

qi

≤

≤






((pipj),pipj) ≤ ≤ ((pipj),(qiqj)) + ((qiqj),(pipj)) ≤ 2

pi

pm

pj

pl

qj

qi

□

(pipj)


Theorem: (P,P’) and |P’| Popt(/2)

Proof: (P,P’) follows by construction.


Theorem: |P’| Popt(/2)

Proof: Q = p1= pj1

,…, pjl=pn - optimal (/2)-

simplificationP’ = p1= pi1

,…, pik=pn

Prove (by induction) that im jm , m 1.

pim-1

pjm

pim



Proof: Q = p1= pj1



,…, pik=pn


Assume im-1 jm-1 and let i’ = im+1.

pim-1

pjm

pimpi’

pjm-1



Proof: Q = p1= pj1



,…, pik=pn


Assume im-1 jm-1 and let i’ = im+1. By construction we have: (pim-1

pi’) > and (pim-1pi’-1)

pim-1

pjm

pimpi’

pjm-1

If i’ > jm then we are done!

>



Proof: Q = p1= pj1



,…, pik=pn

(pim-1pi’) > and (pim-1

pi’-1)

Assume the opposite i.e. i’ jm

pim-1

pjm

pim

pi’

pjm-1



Proof: Q = p1= pj1



,…, pik=pn


pi’-1)


Q is an (/2)-simplification (pjm-1pjm

) /2

pim-1

pjm

pim

pi’

pjm-1



Proof: Q = p1= pj1



,…, pik=pn


pi’-1)


Q is an (/2)-simplification (pjm-1pjm

) /2

According to Lemma 1: (pim-1

pi’) 2 (pjm-1pjm

)

pim-1

pjm

pim

pi’

pjm-1



Proof: Q = p1= pj1



,…, pik=pn

According to Lemma 1: (pim-1

pi’) 2 (pjm-1pjm

)

This is a contradiction since (pjm-1pjm

) > by construction.

i’ > jm and we are done!

pim-1

pjm

pimpi’

pjm-1

□

Summary - Agarwal et al.

Theorem: Given a polygonal curve P in Rd and a parameter 0, an -simplification of P of size at most Popt(/2) can be constructed in O(n log n) time and O(n) space.

Summary

• Ramer–Douglas–Peucker 1973Used very often in practice. Easy to implement and

fast in practice. Hausdorff error

• Simple Simplification O(n) time - simple and fastSeveral nice properties. Fréchet error

• Imai-Iri 1988Gives an optimal solution but slow. Hausdorff error

• Agarwal, Har-Peled, Mustafa and Wang 2005Fast and easy to implement. Gives a worst-case

performance guarantee. Fréchet error

Handle time?

Compress trajectories s.t. standard database queries are “sound” after compression.

The five standard database operations on a trajectory T are:

• where-at(T,t): location of e at time t.

• when-at(T,(x,y),L): time at which e is within distance L of (x,y)

• intersect: • nearest-neighbour: • spatial-join:

• where-at(T,t): location of e at time t.

time t

point (x,y)

• when-at(T,(x,y),L): time at which e is within distance L of (x,y).

Handle time?

Soundness

Define distance between simplification and original trajectory.

Definition: A distance function is sound for a query q, if for each >0 there exists a >0, s.t. for every trajectory T and its -simplification T’ we have

||q(T)-q(T’)||< .

[Cao, Wolfson & Trajcevski’06]

Path simplification

1

45

32

7

6

1

2

7

1

2

7

6

6?

Input path Traditional path simplification

Aim

Soundness

Trajectory = Polygonal path in 3D (x,y,time)

Transform time units into space units (x,y,z), where z = time and is the maximum speed along the trajectory.

Use the Euclidean distance function.

Theorem: Where-at, spatial-join, NN, intersect are sound When-at is NOT sound

Where-at(T,t): Spatial error = |pq|

= |pr|/sin(rqp) 2

pq

rt

Soundness

Theorem: Where-at, spatial-join, NN, intersect are sound When-at is approximately sound

Apx-When-at(T,(x,y),L): When within distance L from (x,y)?

Set-Apx-when-at(T,(x,y),L) Set of time points when T is within distance L from (x, y)

Given a query point q=(x, y), let t1 be the time reported by apx-when-at(T’,(x, y), L+). There exists a time point t2 in set-apx-when-at (T, (x, y), L+2) such that |t1 − t2| / and |T(t1) − T’(t2)| .

Summary

Simplifying polygonal curves• Ramer–Douglas–Peucker 1973

• Driemel, Har-Peled and Wenk 2010

• Imai-Iri 1988

• Agarwal, Har-Peled, Mustafa and Wang 2005

• Take time into consideration

Other interesting simplification algorithms• Abam, de Berg, Hachenberger, Zarei 2007

Constraint FSD

Recall the free space diagram (fsd)Consider the fsd of two curves P and Q with respect to a distance r. (qQ,pP) in the fsd is in the free space iff dist(q,p) r.

(q1,p1)

(qm,pn)

P

Q

Can we use it for other measures?

qp

Constraint FSD

Example: (qQ,pP) in the fsd is in the free space iff p can “see” q.

q

p

Can the owner and the dog walk along P and Q such that they always see each other?

Constraint FSD

Example: (qQ,pP) in the fsd is in the free space iff p can “see” q.

q

p

Can the owner and the dog walk along P and Q such that they always see each other? Yes!

Constraint FSD

Example: The entities must walk with similar speed (similar ~ within a factor of 2?).

Can the entities walk along P and Q with similar speed while keeping a distance of at most r?

P

Q

Constraint FSD

Example: The entities must walk with similar speed (similar ~ within a factor of 2?).

Can the entities walk along P and Q with similar speed while keeping a distance of at most r? No!

P

Q

Open Problem

1. Can we find a subquadtratic algorithm that computed a simplification with a minimal number of edges, or within a constant factor?

References

• A. Driemel, S. Har-Peled and C. Wenk. Approximating the Fréchet Distance for Realistic Curves in Near Linear Time. Discrete & Computational Geometry , 2012.

• H. Cao, O. Wolfson, and G. Trajcevski. Spatio-temporal data reduction with deterministic error bounds. The VLDB Journal, 2006.

• W. S. Chan and F. Chin. Approximation of polygonal curves with minimum number of line segments. In Proceedings of the 3rd International Symposium on Algorithms and Computation, 1992.

• D. H. Douglas and T. K. Peucker. Algorithms for the reduction of the number of points required to represent a digitized line or its caricature. The Canadian Cartographer, 1973.

• J. Gudmundsson, J. Katajainen, D. Merrick, C. Ong and T. Wolle. Compressing spatio-temporal trajectories. Computational Geometry - Theory and Applications, 2009.

• P. K. Agarwal, S. Har-Peled, N. Mustafa, and Y. Wang. Near-linear time approximation algorithms for curve simplication in two and three dimensions. Algorithmica, 2005.

Computational Movement Analysis Lecture 3: Curve simplification Joachim Gudmundsson

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Transcript of Computational Movement Analysis Lecture 3: Curve simplification Joachim Gudmundsson