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Computational Movement Analysis Lecture 5: Segmentation, Popular Places and Regular Patterns Joachim Gudmundsson

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Computational Movement AnalysisLecture 5:

Segmentation, Popular Places and Regular PatternsJoachim Gudmundsson

Input:

x1, … , xn: moving entities in the plane

k: a positive integer

r: a positive real value

An axis aligned square σ of side length r is a popular place if at least k entities visit it.

σ

σ is a popular place for k 5

σ

Continuous model:

σ is a popular place if it is intersected by polylines from at least k different entities.

Discrete model:

σ is a popular place in the discrete model if it contains input points from at least k different entities.

T1

T2

Continuous model:

maximum number of visitorsO(2n2)

(2n2)

Discrete model:

maximum number of visitorsO(n log n)

(n log n)

- Detect if there exists a popular place?
- Report the popular place with the maximum
- number of visitors?
- Report “all” popular places?
- …

Problem:

Given a set of polygons, with holes, find a point that

stabs a maximum number of polygons.

Idea: Use a sweep-line algorithm

Complexity: O(2)

Straight forward approach (almost):

Running time:O(t2n2 log tn)

Space: O(t2n2)

Apply topological sweep by Edelsbrunner and Guibas:

Running time:O(t2n2)

Space: O(t2n)

Find the maximum “colour depth” using topological sweep by Edelsbrunner & Guibas.

Improvement (incl. topological sweep [Edelbsbrunner & Guibas’89])

Running time:O(t2n2)

Space: O(tn)

Non-trivial to implement due to the degenerate cases.

The problem has an (2n2) lower bound.

n=32

t=1000

Time: 5 seconds

Compression + finding all popular places

Discrete model:

maximum number of visitorsO(n log n)

(n log n)

Continuous model:

maximum number of visitorsO(2n2)

(2n2)

For analysis, it is often necessary to break a trajectory into pieces according to the behaviour of the entity (e.g., walking, flying, …).

Input: A trajectory T, where each point has a set of attribute values, and a set of criteria.

Attributes: speed, heading, curvature…

Criteria: bounded variance in speed, curvature, direction, distance…

Aim:Partition T into a minimum number of subtrajectories (so-called segments) such that each segment fulfils the criteria.

“Within each segment the points have similar attribute values”

Trajectory T sampled with equal time intervals.

Trajectory T sampled with equal time intervals.

Top right:speed cannot differ more than a factor 2

Trajectory T sampled with equal time intervals.

Bottom left:direction of motion differs by at most 90◦

Trajectory T sampled with equal time intervals.

Bottom right:both criteria are used in conjunction (speed and direction)

Observation:

Many trajectories span over several activities.

Goal:

Segment a trajectory into subtrajectories according to its behaviour.

We will only consider segmenting trajectories at vertices.

- Goal: Partition trajectory into a small number of segments
- such that a given criterion is fulfilled on each segment
- each segment are uniform
- e.g. heading angular range

Goal:Partition trajectory into a small number of segments

such that a given criterion is fulfilled on each segment

Criteria: heading, speed, location, curvature, sinuosity …

and combinations of these

e.g., describe movement characteristics

Definition: A criterion is decreasing monotone, if it holds on a segment, it holds on any subsegment.

- Examples: disk criterion (location), angular range (heading), speed…
- Theorem:A combination of conjunctions and disjunctions of decreasing monotone criteria is a decreasing monotone criterion.

Observation:

If criteria are decreasing monotone, a greedy strategy works.

Observation:

If criteria are decreasing monotone, a greedy strategy works.

For many decreasing monotone criteria Greedy requires O(n) time, e.g. for speed, heading…

- Observation:
For some criteria, iterative double & search is faster.

Double & search: An exponential search followed by a binary search.

- Boolean or linear combination of decreasing monotone criteria
- Greedy Algorithm

- incremental in O(n) time or constant-update criteriae.g. bounds on speed or heading
- double & search in O(n log n) time for non-constant update criteria e.g. staying within some radius

[Buchin et al.’11]

Observation:

For a combination of decreasing and increasing monotone criteria the

greedy strategy does not always work.

Example:

Min duration 2ANDMax speed range 4

5

speed

1

Goal: Delineate stopover sites of migratory geese

- Two behavioural types
- stopover
- migration flight

- Input:
- GPS tracks
- expert description of behaviour

Data

Spring migration tracks

- White-fronted geese
- 4-5 positions per day
- March – June
Up to 10 stopovers during spring migration

- Stopover: 48 h within radius 30 km
- Flight: change in heading <120°

Kees

Adri

manual

computed

Few local differences:

Shorter stops, extra cuts in computed segmentation

A combination of decreasing and increasing monotone criteria

stopover

migration flight

Within radius 30km

At least 48h

- Change in heading
- <120°

AND

OR

- Many Criteria are not (decreasing) monotone:
- Minimum time
- Standard deviation
- Fixed percentage of outliers
- Example: Geese Migration
- For these Aronov et al. introduced the start-stop diagram

- Algorithmic approach:

input trajectory compute start-stop diagram compute segmentation

- Given a trajectory T over time interval I = {t0,…,t} and criterion C
- The start-stop diagram D is (the upper diagonal half of) the n x n grid,
- where each point (i,j) is associated to segment [ti,tj] with
- (i,j) is in free spaceif C holds on [ti,tj]
- (i,j) is in forbidden space if C does not hold on [ti,tj]
- A (minimal) segmentation of T corresponds to a (min-link) staircasein D

forbidden space

free space

staircase

A (minimal) segmentation of T corresponds to a (min-link) staircasein D.

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A (minimal) segmentation of T corresponds to a (min-link) staircasein D.

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- Discrete case:
- A non-monotone segmentation can be computed in O(n2) time.

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- Discrete case:
- A non-monotone segmentation can be computed in O(n2) time.

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[Aronov et al.13]

- Definition:
- A criterion is stableif and only if where
- = number of changes of validity on segments [0,i], [1,i],…, [i-1,i]

- Definition:
- A criterion is stableif and only if where
- = number of changes of validity on segments [0,i], [1,i],…, [i-1,i]
- Observations:
- Decreasing and increasing monotone criteria are stable.
- A conjunction or disjunction of stable criterion are stable.

For stable criteria the start-stop diagram can be compressed by applying run-length encoding.

Examples:

decreasing monotone

increasing monotone

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For a decreasing criterion consider the algorithm:

ComputeLongestValid(critC, trajT)

Algorithm:

Move two pointersi,j from n to 0 over the trajectory. For every trajectory index j the smallest index i for which [I,j] satisfies the criterion C is stored.

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j

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For a decreasing criterion ComputeLongestValid(crit C, traj T):

Move two pointers i,j from n to 0

over the trajectory

i

j

9

Requires a data structure for segment [i,j] allowing the operations isValid, extend, and shorten, e.g., a balanced binary search tree on attribute values for range or bound criteria.

Runs in O(nc(n)) time where c(n) is the time to update & query.

Analogously for increasing criteria.

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The start-stop diagram of a conjunction (or disjunction) of two stable criteria is their intersection (or union).

The start-stop diagram of a

negated criteria is its inverse.

The corresponding compressed

start-stop diagrams can be

computed in O(n) time.

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- Examples of stable criteria
- Lower bound/Upper bound on attribute
- Angular range criterion
- Disk criterion
- Allow a fraction of outliers
- …

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- Observation: The optimal segmentation for [0,i] is either one segment, or an optimal sequence of segments for [0,j<i] appended with a segment [j,i], where j is an index such [j,i] is valid.
- Dynamic programming algorithm
- for each row from 0 to n
- find white cell with min link
- That is, iteratively compute a table
- S[0,n] where entry S[i] for row i
- stores
- last: index of last link
- count: number of links so far
- Runs in O(n2) time

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S

0

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- More efficient dynamic programming algorithm for compressed diagrams
- Process blocks of white cellsusing a
- range query in a binary search tree T
- (instead of table S) storing
- index: row index
- last: index of last link
- count: number of links so far
- augmented by minimal count in
- subtree
- Runs in O(nlog n)time

T

[Alewijnse et al.’14]

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- Greedy algorithm for decreasing monotone criteria
- O(n) or O(n log n) time
- [Buchin, Driemel, van Kreveld, Sacristan, 2010]
- Case Study: Geese Migration
- [Buchin, Kruckenberg, Kölzsch, 2012]
- Start-stop diagram for arbitrary criteria
- O(n2) time
- [Aronov, Driemel, van Kreveld, Löffler, Staals, 2012]
- Compressed start-stop diagram for stable criteria
- O(n log n) time
- [Alewijnse, Buchin, Buchin, Sijben, Westenberg,2014]

Given a query region, does an object (animal, vehicle, …) regularly visit the region?

Regular behaviour?

Every day?

Every month?

Annual migration?

Regular?

75% of all Saturdays I play

football. Regular?

Time

start

end

outside

inside

Time

start

end

Length = 5

offset

period

length

Time

start

end

Length = 9

Visit 78% of the times

offset

period

length

Time

start

end

period

offset

period

length

Given period offset and period length.

period

length

Time

0 1 1 0 1 1 1 0 1 1 0

period

offset

period

offset

Problem:

Given a bitstring S of length n and a constant 0<c<1, find the longest subsequence S’ of S such that there are at least c|S’| 1s in S’.

S = 100011100110

i

f(i) = si

Maximize b - a

s.t. f(b) - f(a) c(b-a)

g(i)= f(i) – ci

Maximize b - a

s.t. g(b) - g(a) 0

j=1

1

6

g(i)

f(i)

0

3

-1

0

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S = 100011100110

Maximize b - a

s.t. g(b) g(a)

g(b) must lie on the URE

g(a) must lie on the LLE

g(i)= f(i) – ci

Maximize b - a

s.t. g(b) - g(a) 0

1

URE

0

1

g(i)

0

-1

LLE

-1

0

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Longest Dense Substring can be computed in O(n) time

Compressed case:

11100010100000011100000 3,3,1,1,1,6,3,5

size n size k

Compressed LDS can be computed in O(k) time

Given period length, but no period offset

Time

Time

Time

Find the longest vertical segment A such that at least a fraction c of the horizontal segments it intersects are “inside”.

A

offset

Theorem:

Given a set of intervals I, a constant 0<c1 and a period length p the longest c-dense repetitive pattern over all offsets can be computed in O(n3/2log2 n) time using O(n log n) space.

Time

What if neither period offset nor period length is known?

(no two consecutive visits in the “same region”)

Observation:

The longest pattern can be translated and scaled such that one of its “visits” coincides with the start point of an interval and one “visit” coincides the start/end point of an interval.

Theorem:

There are O(n2) pairs of distances. Each distance may generate n/c period lengths, thus O(n3/c) possible period lengths in total. This also fixes the offset!

The longest c-dense repetitive pattern for each period length can be found in O(n) time, thus O(n4/c) time in total.

Can be improved to O(n7/2log3 n).

[Djordjevic et al.’10]

Theorem:

Offset & period lengthO(n)

Period lengthO(n3/2log2 n)

Period length O(n log n) - approximate

NothingO(n7/2log3 n)

Approximate version of the general version?

Improve running times?

Lower bounds?

Input:A Euclidean graph G in the plane

PreprocessG into a data structure such that given a polygonal query

curve Q and a positive constant

report all subcurves Gfor which

(,Q) ≤

G

Q

( does not need to start and end at a vertex of G)

Assume is given as part of the input

Input: polygonal path P of size n and a constant >0

Query: segment Q (length >6)

Q

P

de Berg et al.’11: (Counting)

6-distance approximation

Preprocessing O(n2 + s polylog n)

Space O(s polylog n)

Query time O(n/s polylog n), for n<s<n2

A geometric graph G is c-packed if

the total length of all the edges of

G inside any ball is at most c times

the radius of the ball.

Input: O(1)-packed tree T of size n and a constant >0

Query: polygonal path Q of size m (each segment >2)

T

Gudmundsson and Smid:

(3+)-distance approximation

Preprocessing O(n polylog n)

Space O(n polylog n)

Query time O(m polylog n)

A geometric graph G is c-packed if

the total length of all the edges of

G inside any ball is at most c times

the radius of the ball.

Input: O(1)-packed tree T of size n and a constant >0

Query: polygonal path Q of size m (each segment >2)

T

Q

Gudmundsson and Smid:

(3+)-distance approximation

Preprocessing O(n polylog n)

Space O(n polylog n)

Query time O(m polylog n)

Pros: Extends to trees (instead of paths)

Query path instead of query segment

Approximation improved from 6 to

(1+)for query segments (3+) for query paths

Improved preprocessing time

Cons: Input must be O(1)-packed

If G is a tree the edges must be “long”

Q

G

References

- M. Benkert, B. Djordjevic, J. Gudmundsson and T. Wolle. Finding popular places. IJCGA, 2010.
- S. Alewijnse, T. Bagautdinov, M. de Berg, Q. Bouts, A. ten Brink, K. Buchin and M. Westenberg. Progressive Geometric Algorithms. SoCG, 2014.
- M. Buchin, A. Driemel, M. J. van Kreveld and V. Sacristan. Segmenting trajectories: A framework and algorithms using spatiotemporal criteria. Journal of Spatial Information Science, 2011.
- B. Aronov, A. Driemel, M. J. Kreveld, M. Loffler and F. Staals. Segmentation of Trajectories for Non-Monotone Criteria. SODA, 2013.
- M. Buchin, H. Kruckenberg and A. Kölzsch. Segmenting Trajectories based on Movement States. SDH, 2012.
- B. Djordjevic, J. Gudmundsson, A. Pham and T. Wolle. Detecting Regular Visit Patterns. Algorithmica, 2011.

References

- B. A. Burton. Searching a bitstream in linear time for the longest substring of any given density, Algorithmica, 2011.
- B. Djordjevic and J. Gudmundsson. Detecting areas visited regularly. COCOON 2010.
- M. de Berg, A. F. Cook IV and J. Gudmundsson. Fast Frechet Queries. CGTA, 2013.
- J. Gudmundsson and M. Smid. Frechetqueries in geometric trees. ESA, 2013.