Method of Evolving Junctions: Finding the Shortest Paths in - - PowerPoint PPT Presentation

Method of Evolving Junctions: Finding the Shortest Paths in Cluttered Environment Jun Lu

collaboration with Shui-Nee Chow, Magnus Egerstedt Yancy Diaz-Mercado, Haomin Zhou Georgia Institute of Technology SIAM Gators, 2014

Introduction

Shortest Path Problem Given a finite number of obstacles in a region M in R2 or R3, what is the shortest path connecting two given points X, Y in M while avoiding the obstacles. X Y

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Existing Methods

• Polygonal obstacles: optimal alogrithm with complexity

O(n log n). (Hershberger and Suri) X Y

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Existing Methods

• Polygonal obstacles: optimal alogrithm with complexity

O(n log n). (Hershberger and Suri) X Y

• Polyhedral obstacles, NP-hard in the framework configuration
• space. (Canny and Reif)

X Y

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• Approximation Methods. (Papadimitriou)

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• Approximation Methods. (Papadimitriou)
• Continuous Settings: Penalty Method

X Y

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• Approximation Methods. (Papadimitriou)
• Continuous Settings: Penalty Method

X Y

• Front Propagation (Dijkstra, A∗, HJB).

|∇d(x)| = 1, d(Y ) = 0.

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Shortest Path Problem

• Starting point X, ending point Y in domain M ⊂ Rn.

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Shortest Path Problem

• Starting point X, ending point Y in domain M ⊂ Rn.
• Obstacles {Pk}k≤1. The boundary ∂Pk of Pk are piecewise

C 2 and has level set representation φk(x) = 0.

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Shortest Path Problem

• Starting point X, ending point Y in domain M ⊂ Rn.
• Obstacles {Pk}k≤1. The boundary ∂Pk of Pk are piecewise

C 2 and has level set representation φk(x) = 0.

• Admissible paths connecting X and Y is

A(X, Y , M) = {γ : [0, 1] → M | γ(0) = X, γ(1) = Y , φk(γ(θ)) ≥ 0, 1 ≤ k ≤ N} .

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Shortest Path Problem

• Starting point X, ending point Y in domain M ⊂ Rn.
• Obstacles {Pk}k≤1. The boundary ∂Pk of Pk are piecewise

C 2 and has level set representation φk(x) = 0.

• Admissible paths connecting X and Y is

A(X, Y , M) = {γ : [0, 1] → M | γ(0) = X, γ(1) = Y , φk(γ(θ)) ≥ 0, 1 ≤ k ≤ N} .

• The shortest path is

γopt = argminγ∈A J(γ). where J(γ) is the Euclidean length of γ.

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Separable Path

• For any two points x, y ∈ M, define

γ0(x, y) = argminγ∈A(x,y,M) J(γ), = {z(λ) = (1 − λ)x + λy, λ ∈ [0, 1]} , J0(x, y) = J(γ0(x, y)) = x − y.

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Separable Path

• For any two points x, y ∈ M, define

γ0(x, y) = argminγ∈A(x,y,M) J(γ), = {z(λ) = (1 − λ)x + λy, λ ∈ [0, 1]} , J0(x, y) = J(γ0(x, y)) = x − y.

• For any two points x, y ∈ ∂Pk, k ≥ 1, define

γk(x, y) = argminγ∈A(x,y,∂Pk) J(γ), Jk(x, y) = J(γk(x, y)).

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Separable Path

A path γ ∈ A(X, Y , M) is SEPARABLE if there are finite number

• f points, called junctions

(x0, x1, · · · , xn+1) such that γ = γ0(x0, x1) · γk1(x1, x2) · · · · γ0(xn, xn+1). Here γ1 · γ2 is the concatenation defined by (γ1 · γ2)(θ) =

• γ1(2θ),

θ ∈ [0, 1

2];

γ2(2θ − 1), θ ∈ [ 1

2, 1].

xi xi+1 xi+2 xi+3 xi+4 xi+5

Figure : γ0(xi, xi+1) · γk1(xi+1, xi+2) · γ0(xi+2, xi+3)·

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Separable Path Theorem

γopt of the shortest path problem is separable.

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Separable Path Theorem

γopt of the shortest path problem is separable.

• γopt is completely determined by finitely many junctions.

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Separable Path Theorem

γopt of the shortest path problem is separable.

• γopt is completely determined by finitely many junctions.
• Dimension Reduction

Finding γopt = ⇒ Finding junctions (x0, · · · , xn+1) Infinite Dimensional = ⇒ Finite Dimensional

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Gradient Flow

• The length of the path connecting junction xi and xi+1 is

J(xi, xi+1) =

• J0(xi, xi+1),

i even; Jki(xi, xi+1), i odd.

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Gradient Flow

• The length of the path connecting junction xi and xi+1 is

J(xi, xi+1) =

• J0(xi, xi+1),

i even; Jki(xi, xi+1), i odd.

• The length of the shortest path is

J(x0, · · · , xn+1) =

n

• i=0

J(xi, xi+1), xi ∈ ∂Pki

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Gradient Flow

• The length of the path connecting junction xi and xi+1 is

J(xi, xi+1) =

• J0(xi, xi+1),

i even; Jki(xi, xi+1), i odd.

• The length of the shortest path is

J(x0, · · · , xn+1) =

n

• i=0

J(xi, xi+1), xi ∈ ∂Pki

• The gradient flow on the manifold ∂Pki is

dxi dθ = −∇J(x0, · · · , xn+1) = −∇xi

• J(xi−1, xi) + J(xi, xi+1)
• J(xi)
• .

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Gradient Flow

• For i odd, J(xi) = J0(xi−1, xi) + Jki(xi, xi+1).

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Gradient Flow

• For i odd, J(xi) = J0(xi−1, xi) + Jki(xi, xi+1).
• T(xi, xi+1) ∈ Txi∂Pki - unit tangent of γki(xi, xi+1) at xi.

xi xi+1 T(xi, xi+1) γki(xi, xi+1)

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Gradient Flow

• For i odd, J(xi) = J0(xi−1, xi) + Jki(xi, xi+1).
• T(xi, xi+1) ∈ Txi∂Pki - unit tangent of γki(xi, xi+1) at xi.

xi xi+1 T(xi, xi+1) γki(xi, xi+1)

• Gradient of Jki is

∇xiJki(xi, xi+1) = −T(xi, xi+1).

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Gradient Flow

• For i odd, J(xi) = J0(xi−1, xi) + Jki(xi, xi+1).
• T(xi, xi+1) ∈ Txi∂Pki - unit tangent of γki(xi, xi+1) at xi.

xi xi+1 T(xi, xi+1) γki(xi, xi+1)

• Gradient of Jki is

∇xiJki(xi, xi+1) = −T(xi, xi+1).

• Gradient of J0 is

∇xiJ0(xi−1, xi) = xi − xi−1 xi − xi−1 · T

• T

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Intermittent Diffusion

• The gradient flow on the manifold ∂Pki is

dxi dθ = − xi − xi−1 xi − xi−1 · T − 1

• · T.

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Intermittent Diffusion

• The gradient flow on the manifold ∂Pki is

dxi dθ = − xi − xi−1 xi − xi−1 · T − 1

• · T.
• Gradient descent only finds local minimizeers.

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Intermittent Diffusion

• The gradient flow on the manifold ∂Pki is

dxi dθ = − xi − xi−1 xi − xi−1 · T − 1

• · T.
• Gradient descent only finds local minimizeers.
• There could be 2N locally shortest paths in an environment

with N obstacles.

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Intermittent Diffusion

• Idea: add noise to gradient flow intermittently,

dxi = −∇xiJ(xi)dθ + σ(θ)dW(θ),

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Intermittent Diffusion

• Idea: add noise to gradient flow intermittently,

dxi = −∇xiJ(xi)dθ + σ(θ)dW(θ), where

• W(θ) is a standard Brownian motion in tangent space Txi∂Pki.

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Intermittent Diffusion

• Idea: add noise to gradient flow intermittently,

dxi = −∇xiJ(xi)dθ + σ(θ)dW(θ), where

• W(θ) is a standard Brownian motion in tangent space Txi∂Pki.
• For 0 = S1 < T1 < · · · < Sm,

σ(θ) =

m

• i=1

σiχ[Si,Ti](θ).

S1 T1 S2 T2 S3 T3 S4 σ1 σ2 σ3 x y 11 / 25

Intermittent Diffusion

• Diffusion on, get out of local trap.

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Intermittent Diffusion

• Diffusion on, get out of local trap.
• Diffusion off, converges to local minimizers.

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Intermittent Diffusion

• Diffusion on, get out of local trap.
• Diffusion off, converges to local minimizers.
• After m diffusions, obtain m local minimzers.

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Intermittent Diffusion

• Diffusion on, get out of local trap.
• Diffusion off, converges to local minimizers.
• After m diffusions, obtain m local minimzers.
• Set γopt to be the best minimizer.

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Intermittent Diffusion

• Diffusion on, get out of local trap.
• Diffusion off, converges to local minimizers.
• After m diffusions, obtain m local minimzers.
• Set γopt to be the best minimizer.

Theorem

For any given δ > 0, the probability that γopt is the globally shortest path is larger that 1 − δ for appropriate selected σ.

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Solving SDEs

• The full SDE is

dxi = − xi − xi−1 xi − xi−1 · T − 1

• dθ · T + σ(θ)dW(θ)

= f (xi)dθ · T + σ(θ)dW(θ)· (1)

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Solving SDEs

• The full SDE is

dxi = − xi − xi−1 xi − xi−1 · T − 1

• dθ · T + σ(θ)dW(θ)

= f (xi)dθ · T + σ(θ)dW(θ)· (1)

• SDE (1) can be discretized by

x∗

i − xn i = f (xn i )∆θT + σn

√ ∆θξ, ξ ∼ N(0, I).

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Solving SDEs

• The full SDE is

dxi = − xi − xi−1 xi − xi−1 · T − 1

• dθ · T + σ(θ)dW(θ)

= f (xi)dθ · T + σ(θ)dW(θ)· (1)

• SDE (1) can be discretized by

x∗

i − xn i = f (xn i )∆θT + σn

√ ∆θξ, ξ ∼ N(0, I).

• Projection. x∗

i is then projected onto ∂Pki( φki = 0)

xn+1

i

= x∗

i − φki(x∗ i )N(x∗ i ).

xn

i

x∗

i

xn+1

i

T N

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Handling Dimension Changes

• Add junctions when γ0(xi, xi+1) intersect with one obstacle.

xi−2 xi−1 xi xi+4 xi+5 xi+1 xi+2 xi+3

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Handling Dimension Changes

• Add junctions when γ0(xi, xi+1) intersect with one obstacle.

xi−2 xi−1 xi xi+4 xi+5 xi+1 xi+2 xi+3

• Remove junctions when xi = xi+1.

xi−3 xi−2 xi−1 xi = xi+1 xi+2 xi+3 xi+4

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One Realization of the Algorithm

One Realization

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Complexity

A comparison with different algorithms Algorithm Complexity A∗ O(( 1

ǫ)3 log 1 ǫ)

Papadimitriou O( 1

ǫ)

Choi, et. al O( 1

ǫ)

Choi, et. al (unique shortest path) O(log 1

ǫ)

Evolving Junctions O(log 1

ǫ)

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Example: 4 convex obstacles

(a) L = 1.0264, 26 times (b) L = 1.0966, 6 times (c) L = 1.1320, 9 times (d) L = 1.1501, 9 times

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Example: 4 convex obstacles in R3

−2 2 −2 2 1 2 North (Meters) Minimizer 1, Length = 5.8660 meters East (Meters) Up (Meters) −2 2 −2 2 1 2 North (Meters) Minimizer 2, Length = 5.9527 meters East (Meters) Up (Meters) −2 2 −2 2 1 2 North (Meters) Minimizer 3, Length = 6.0403 meters East (Meters) Up (Meters) −2 2 −2 2 1 2 North (Meters) Minimizer 7, Length = 6.2305 meters East (Meters) Up (Meters)

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Example: 4 convex obstacles in R3

Minimizer Length (m) Times Obtained Out of 100 200 300 1 5.8660 48 103 159 2 5.9527 50 91 128 3 6.0403 1 4 6 4 6.0594 1 1 5 6.0919 1 6 6.2286 3 7 6.2305 1 1 2

Table : Improvement on Probability of Obtaining Shortest Path

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Example: One Non-convex Obstacle

(e) L = 1.1874, 46 times (f) L = 1.3393, 4 times

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Example: Polyhedral Obstacles

(g) Occurs 6 times, L=2.600 (h) Occurs 3 times, L=2.623 (i) Occurs once, L=3.000

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Example: Polyhedral Obstacles

(j) N=13, L=7.0803 (k) N=2, L=7.0956 (l) N=2, L=7.3404

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Example: Polyhedral Obstacles

(m) N=1, L=7.3436 (n) N=1, L=7.4314 (o) N=1, L=7.5253

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Example: Disk Mover

(p) L = 0.6266, r = 0 (q) L = 0.6347, r = 0.02 (r) L = 0.6952, r = 0.05

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Example: Shortest Path between two sets

(s) L = 0.2840 (t) L = 0.2979

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Questions?

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