Computational Geometry · O (n log n) O (n) O (n) O (log n) O (n) O (1) Robots Create trapezoidal...

transcript

Computational Geometry

Thomas van Dijk Winter Semester 2018/19

Lecture #10

Motion Planning

Planning

current situation,desired situation

Planning

sequence of steps to reachthe one from the other

Path Planning

current location,desired location

Path Planning

path to reach theone from the other

Point-ShapedRobots

pstart

Point-ShapedRobots

• Create trapezoidal map of obstacle edges.

pstart

Point-ShapedRobots

• Create trapezoidal map of obstacle edges.• Remove vertical extensions inside obstacles.

pstart

Point-ShapedRobots

• Create trapezoidal map of obstacle edges.• Remove vertical extensions inside obstacles.• Vertices at centers of trapez. and vertical ext.

pstart

Point-ShapedRobots

• Connect “neighboring” vertices by line segm.• Vertices at centers of trapez. and vertical ext.

pstart

Point-ShapedRobots

• Connect “neighboring” vertices by line segm.• Locate pstart, pgoal in map

• Vertices at centers of trapez. and vertical ext.

pstart

Point-ShapedRobots

• Connect “neighboring” vertices by line segm.• Locate pstart, pgoal in map

∆start ∆goal

→ ∆start, ∆goal.

pstart

Point-ShapedRobots

• Connect “neighboring” vertices by line segm.• Locate pstart, pgoal in map• Do breadth-first search in the roadmap

to find a path π from ∆start to ∆goal.

∆start ∆goal

pstart

Point-ShapedRobots

∆start ∆goal

• Connect pstart, pgoal to π by line segments.

pstart

4 - 10

Point-ShapedRobots

∆start ∆goal

pstart

4 - 11

Point-Shaped

O(n log n)

Robots

∆start ∆goal

pstart

4 - 12

Point-Shaped

O(n log n)O(n)

Robots

∆start ∆goal

pstart

4 - 13

Point-Shaped

O(n log n)O(n)O(n)

Robots

∆start ∆goal

pstart

4 - 14

Point-Shaped

O(n log n)O(n)O(n)

Robots

∆start ∆goal

pstart

4 - 15

Point-Shaped

O(n log n)O(n)O(n)

O(log n)

Robots

∆start ∆goal

pstart

4 - 16

Point-Shaped

O(n log n)O(n)O(n)

O(log n)

Robots

∆start ∆goal

pstart

4 - 17

Point-Shaped

O(n log n)O(n)O(n)

O(log n)

Robots

∆start ∆goal

pstart

4 - 18

Point-Shaped

O(n log n)O(n)O(n)

O(log n)

Robots

∆start ∆goal

pstart

4 - 19

Point-Shaped

O(n log n)O(n)O(n)

O(log n)

Robots

∆start ∆goal

pstart

A First ResultTheorem: We can preprocess a set of polygonal obstacles

with a total of n edges in O(n log n) expectedtime such that, given a start and a goal position,we can find a collision-free path for a pointrobot in O(n) time if it exists.

A First ResultTheorem: We can preprocess a set of polygonal obstacles

with a total of n edges in O(n log n) expectedtime such that, given a start and a goal position,we can find a collision-free path for a pointrobot in O(n) time if it exists.

A First Result

What about, say, polygonal robots?

Theorem: We can preprocess a set of polygonal obstacleswith a total of n edges in O(n log n) expectedtime such that, given a start and a goal position,we can find a collision-free path for a pointrobot in O(n) time if it exists.

Degrees of FreedomEvery robot has some number d of degrees of freedom,meaning that its configuration with respect to the world canbe specified by d parameters.

2D translating robot

2D translating robot 2D translating, rotating robot

3D translating robot

Configuration Space

robotic arm

Configuration Space

The configuration space is the d-dimensional space ofall possible (i.e., obstacle avoiding) parameter valuecombinations.

robotic arm

Configuration Space

robotic arm

obstacle

work space

Configuration Space

robotic arm

obstacle

work spacereferencepoint

Configuration Space

robotic arm

obstacle

work space configuration spacereferencepoint

Configuration Space

robotic arm

obstacle

Configuration Space

robotic arm

obstacle

Configuration Space

robotic arm

obstacle

Configuration Space

The configuration space is the d-dimensional space ofall possible (i.e., obstacle avoiding) parameter valuecombinations.Path for a point through configuration space

robotic arm

obstacle

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Configuration Space

robotic arm

obstacle

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Configuration Space

path for the robot in the original space.

robotic arm

obstacle

Example: Translating 2D Polygonal Robots

R(0, 0)

R(x, y)

y′P1

R(x′, y′)

work space

(0, 0) xx′

(x, y)

configuration space

(x′, y′)

R(0, 0)

R(x, y)

y′P1

R(x′, y′)

work space

• Compute CP i = {(x, y) : R(x, y) ∩ Pi 6= ∅} for each Pi.

(0, 0) xx′

(x, y)

configuration space

(x′, y′)

R(0, 0)

R(x, y)

y′P1

R(x′, y′)

work space

(0, 0) xx′

(x, y)

configuration space

(x′, y′)

R(0, 0)

R(x, y)

y′P1

R(x′, y′)

work space

• Compute their union Cforb =⋃

i CP i.• Compute CP i = {(x, y) : R(x, y) ∩ Pi 6= ∅} for each Pi.

(0, 0) xx′

(x, y)

configuration space

(x′, y′)

R(0, 0)

R(x, y)

y′P1

R(x′, y′)

work space

i CP i.• Compute CP i = {(x, y) : R(x, y) ∩ Pi 6= ∅} for each Pi.

(0, 0) xx′

(x, y)

configuration space

(x′, y′)

R(0, 0)

R(x, y)

y′P1

R(x′, y′)

work space

i CP i.• Find a path for a point in the complement Cfree of Cforb.

(0, 0) xx′

(x, y)

configuration space

(x′, y′)

CP2Cfree

R(0, 0)

R(x, y)

y′P1

R(x′, y′)

work space

i CP i.• Find a path for a point in the complement Cfree of Cforb.

(0, 0) xx′

(x, y)

configuration space

(x′, y′)

CP2Cfree

R(0, 0)

R(x, y)

y′P1

R(x′, y′)

work space

i CP i.• Find a path for a point in the complement Cfree of Cforb.⇒ collision-free path for the robot in work space

(0, 0) xx′

(x, y)

configuration space

(x′, y′)

CP2Cfree

R(0, 0)

R(x, y)

y′P1

R(x′, y′)

work space

i CP i.• Find a path for a point in the complement Cfree of Cforb.⇒ collision-free path for the robot in work space

(0, 0) xx′

(x, y)

configuration space

(x′, y′)

CP2Cfree

Some Linear AlgebraVector sums p

Some Linear AlgebraVector sumsAlgebra: (px, py) + (qx, qy) = (px + qx, py + qy)

Some Linear AlgebraVector sumsAlgebra:Geometry:

(px, py) + (qx, qy) = (px + qx, py + qy)place vectors head to tail

Some Linear AlgebraVector sums

Minkowski sums

Algebra:Geometry:

⊕ ==

Minkowski sums

Algebra:Geometry:

⊕ ==

Minkowski sums

Algebra:Geometry:

Algebra: S1 ⊕ S2 = {p + q | p ∈ S1, q ∈ S2}

⊕ ==

Minkowski sums

Algebra:Geometry:

S1 ⊕ S2 = {p + q | p ∈ S1, q ∈ S2}place copy of one shapeat every point of the other

⊕ ==

Minkowski sums

Algebra:Geometry:

Inversion

9 - 10

Minkowski sums

Algebra:Geometry:

Inversion

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Minkowski sums

Algebra:Geometry:

InversionAlgebra: −S = {−p | p ∈ S}

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Minkowski sums

Algebra:Geometry:

InversionAlgebra:Geometry:

−S = {−p | p ∈ S}rotate 180◦ (point-mirror)around reference point

10 - 1

Characterizing CP

R(0, 0)R(x, y)

Recall that CP = {(x, y) : R(x, y) ∩ P 6= ∅} for an obstacle P .

10 - 2

Characterizing CP

R(0, 0)R(x, y)

R(x, y) intersects P ⇔ (x, y) ∈ CP .In other words:

10 - 3

Characterizing CP

R(0, 0)R(x, y)

CP = P ⊕ (−R(0, 0))Theorem.

10 - 4

Characterizing CP

R(0, 0)R(x, y)

Proof.

10 - 5

Characterizing CP

R(0, 0)R(x, y)

Proof. Show: R(x, y) intersects P ⇔ (x, y) ∈ P ⊕ (−R(0, 0)).

10 - 6

Characterizing CP

R(0, 0)R(x, y)

Proof.

“⇒”

Show: R(x, y) intersects P ⇔ (x, y) ∈ P ⊕ (−R(0, 0)).

“⇐”

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Characterizing CP

R(0, 0)R(x, y)

Proof.

Suppose R(x, y) intersects P .“⇒”

“⇐”

10 - 8

Characterizing CP

R(0, 0)R(x, y)

Proof.

Suppose R(x, y) intersects P .Let q ∈ R(x, y) ∩ P . Then...

“⇒”

“⇐”

10 - 9

Characterizing CP

R(0, 0)R(x, y)

Proof.

“⇒”

“⇐”

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Characterizing CP

R(0, 0)R(x, y)

Proof.

“⇒”

“⇐”

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Characterizing CP

R(0, 0)R(x, y)

Proof.

“⇒”

“⇐” Let (x, y) ∈ P ⊕ (−R(0, 0)).

10 - 12

Characterizing CP

R(0, 0)R(x, y)

Proof.

“⇒”

“⇐” Let (x, y) ∈ P ⊕ (−R(0, 0)).Then there are pointsq ∈ P and r ∈ R(0, 0)such that . . .

10 - 13

Characterizing CP

R(0, 0)R(x, y)

Proof.

“⇒”

10 - 14

Characterizing CP

R(0, 0)R(x, y)

Proof.

“⇒”

10 - 15

Characterizing CP

R(0, 0)R(x, y)

Proof.

“⇒”

10 - 16

Characterizing CP

R(0, 0)R(x, y)

Proof.

“⇒”

11 - 1

Minkowski Sums: ComplexityIf P and R are convex polygons with n and medges, respectively, then P ⊕R is a convexpolygon with at most n + m edges.

P ⊕R

Theorem:

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P ⊕R

Theorem:

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P ⊕R

Theorem:

11 - 4

P ⊕R

Theorem:

11 - 5

P ⊕R

Theorem:

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P ⊕R

Theorem:

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P ⊕R

Theorem:

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P ⊕R

Theorem:

12 - 1

Minkowski Sums: Computation

P ⊕R

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P ⊕R

How would you compute P ⊕R given P and R?Task:

12 - 3

P ⊕R

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P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

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P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

12 - 6

P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

Problem:

12 - 7

P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

Problem:︸︷︷︸

complexity ∈ Θ( )

12 - 8

P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

complexity ∈ Θ( )|P| · |R| :-(

12 - 9

The Minkowski sum of two convex polygons Pand R can be computed in O(|P|+ |R|) time.

Theorem.

P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

12 - 10

Theorem.

P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

12 - 11

Theorem.

P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

12 - 12

Theorem.

P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

12 - 13

Theorem.

P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

12 - 14

Theorem.

P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

12 - 15

Theorem.

P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

12 - 16

Theorem.

P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

12 - 17

Theorem.

P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

12 - 18

Theorem.

P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

12 - 19

Theorem.

P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

12 - 20

Theorem.

P ⊕R

Idea: P ⊕R = CH({p + r | p ∈ P , r ∈ R}

)(Proof?)

13 - 1

Pseudodisks

A pair of planar objects o1 and o2 is a pair ofpseudodisks if:• ∂o1 ∩ int(o2) is connected, and• ∂o2 ∩ int(o1) is connected.

Definition:

13 - 2

Pseudodisks

Definition:

13 - 3

Pseudodisks

Definition:

p ∈ ∂o1 ∩ ∂o2 is a boundary crossing if ∂o1 crosses at p fromthe interior to the exterior of o2.

13 - 4

Pseudodisks

Definition:

p ∈ ∂o1 ∩ ∂o2 is a boundary crossing if ∂o1 crosses at p fromthe interior to the exterior of o2.

Observation: A pair of polygonal pseudodisks defines atmost two boundary crossings.

14 - 1

Extreme DirectionsObservation: Let P1, P2 be interior-disjoint convex polygons

14 - 2

Let d1 and d2 be directions in which P1 ismore extreme than P2.

14 - 3

Let d1 and d2 be directions in which P1 ismore extreme than P2.

14 - 4

Let d1 and d2 be directions in which P1 ismore extreme than P2.Then P1 is more extreme than P2 either in[d1, d2] or in [d2, d1].

14 - 5

d′′

d′′′

14 - 6

d′′

d′′′

d′d′′

d′′′P2P2

14 - 7

d′′

d′′′

d′d′′

d′′′

P2 and P1equallyextreme

P2 moreextreme

P1 moreextremeP2P2

15 - 1

Polygonal PseudodisksIf P1 and P2 are convex polygons with disjointinteriors, and R is another convex polygon,then P1 ⊕R and P2 ⊕R is a pair of pseudodisks.

Theorem:Theorem:

15 - 2

Theorem:Theorem:

CP1 CP2

︸︷︷︸︸︷︷︸

15 - 3

Theorem:Theorem:

CP1 CP2

︸︷︷︸︸︷︷︸Proof. It suffices to show: CP1 \ CP2 is connected.

15 - 4

Theorem:Theorem:

CP1 CP2

Suppose CP1 \ CP2 is not connected...

15 - 5

Theorem:

CP1 CP2

15 - 6

Theorem:

CP1 CP2

15 - 7

Theorem:

CP1 CP2

15 - 8

Theorem:

CP1 CP2

CP2to previous observation!

15 - 9

Theorem:

CP1 CP2

since– dq and ds are also extreme for P1 and– dp and dr are also extreme for P2.

15 - 10

Theorem:

CP1 CP2

since– dq and ds are also extreme for P1 and– dp and dr are also extreme for P2.(and P1 and P2 are convex andinterior-disjoint).

16 - 1

Union ComplexityA collection S of convex polygonal pseudodiscswith n vtc in total has a union with ≤ 2n vtc.

Theorem:

16 - 2

Theorem:

Proof. Charge every vtx of the union to a polygon vtxs.t. every polygon vtx is charged at most twice.

16 - 3

Theorem:

16 - 4

Theorem:

boundary crossing

16 - 5

Theorem:

boundary crossing

adjacent vtxin the interiorof the union

16 - 6

Theorem:

boundary crossing

16 - 7

Theorem:

boundary crossing

16 - 8

Theorem:

boundary crossing

16 - 9

Theorem:

boundary crossing

16 - 10

Theorem:

boundary crossing

16 - 11

Theorem:

boundary crossing

16 - 12

Theorem:

boundary crossing

16 - 13

Theorem:

boundary crossing

16 - 14

Theorem:

boundary crossing

16 - 15

Theorem:

boundary crossing

17 - 1

Summary and Main ResultLet R be a constant-complexity convexrobot, translating among a set S of disjointpolygonal obstacles with n edges in total.

Theorem:

17 - 2

Theorem:

We can preprocess S in O(n log2 n) time suchthat, given any start and goal position, wecan compute in O(n) time a collision-freepath for R if it exists.

17 - 3

Proof.

Theorem:

17 - 4

Proof.

Theorem:

• Triangulate the obstacles if not convex.

17 - 5

Proof.

Theorem:

• Triangulate the obstacles if not convex. Ch.3

17 - 6

Proof.

Theorem:

• Triangulate the obstacles if not convex. Ch.3O(n log n)

17 - 7

Proof.

Theorem:

• Compute CP i for every convex obstacle Pi.

• Triangulate the obstacles if not convex. Ch.3O(n log n)

17 - 8

Proof.

Theorem:

• Triangulate the obstacles if not convex. Ch.3O(n log n)O(n)

17 - 9

Summary and Main Result

i CP iusing div. and conq. (merge by sweeping – Ch.2.3)

Let R be a constant-complexity convexrobot, translating among a set S of disjointpolygonal obstacles with n edges in total.

Proof.

Theorem:

. . . ?

Ch.3O(n log n)O(n)

17 - 10

Proof.

Theorem:

• Triangulate the obstacles if not convex. Ch.3O(n log n)O(n)

17 - 11

Proof.

Theorem:

• Triangulate the obstacles if not convex. Ch.3O(n log n)O(n)O(n log2 n)

17 - 12

Proof.

Theorem:

[Argue carefully about the number of intersection pts!]

Ch.3O(n log n)O(n)O(n log2 n)

17 - 13

• Find a path for a point in the complement Cfree.

Proof.

Theorem:

17 - 14

• Find a path for a point in the complement Cfree.

Proof.

Theorem:

Computational Geometry · O (n log n) O (n) O (n) O (log n) O (n) O (1) Robots Create trapezoidal...

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