Online Motion Planning MA-INF 1314 Summersemester 17 Escape Paths/Alternative Measure Elmar Langetepe

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Online Motion Planning MA-INF 1314

Summersemester 17 Escape Paths/Alternative Measure

Elmar Langetepe

University of Bonn

Juli 18th, 2017

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Rep.: Different performance measures

Set L_m of m line segmentss_i of unknown length |s_i| Dark corridors, escape, digging for oil

Test corridors successively

s_j₁ up to a certain distancex₁, thens_j₂ for another distance x₂ and so on

s₁ s2 s₃ s₄ s₅ s₆ s7 x₁ x₂

x₃ x₄

x₅ (i)

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Rep.: Partially informed

Assume distribution is known!

f₁ ≥f₂ ≥ · · · ≥f_m order of the length given Extreme cases! Good strategies!

s₁s₂s₃s₄s₅ s₆ s₇

(ii) (iii)s1 s2s₃s4 s5 s₆ s₇ x x x x x x x

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Rep.: Optimal strategy for this case

Theorem:For a set of sorted distancesFm (i.e.

f₁≥f₂ ≥ · · · ≥f_m) we have

maxTrav(Fm) := min

i i·f_i.

Proof:

Arbitrary strategy A

Less than min_ii·f_i means less than j·f_j for anyj Visiting depth d₁≥d₂ ≥ · · · ≥d_m

Not reached f1 byd1,

not reached f₂, since d₁+d₂<2f₂ and d₂ ≤d₁ and so on Not successful!

minii·fi always sufficient!

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Online Strategy

Fm with f1 ≥f2 ≥ · · · ≥fm not known Compete against maxTrav(Fm) := minii·fi

Dovetailing strategy: Rounds c = 1,2,3,4, . . . For any round c from left to right:

Path length of segment i is extendedup to distance_c

i

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Rep.: Online Strategy

Dovetailing strategy: Rounds c = 1,2,3,4, . . . For any round c from left to right:

Path length of segment i is extendedup to distance_c

i

1 1 2

2 3

3 4

4 5

5 6

6 7

7 8

8 9

9 10

10

10 11

2 3 4

4

5 6 6 6

7 8

8

9 f3= 3 9

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Rep.: Online Strategy!

Theorem:Hyperbolic traversal algorithm solves the multi-segment escape problem for any listFm with maximum traversal cost bounded by

D·(maxTrav(Fm) ln(min(m,maxTrav(Fm))) for some constantD.

Proof:(W.l.o.g.Fm integers)

Let minii·fi =j·fj for somej c with c =j·f_j exists (Round c) Overoll cost:

Xm t=1

jc t

k≤

min(m,c)X

t=1

c t ≤c+

Z min(m,c) 1

c

t dt =c(1+ln min(m,c)).

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Rep.: Matches Lower bound!

Theorem:For any deterministic online strategyAthat solves the multi-segment escape problem we can construct input sequences Fm(A,C) so thatA has cost at leastd·C ln min(C,m) and maxTrav(F_m(C,A))≤C holds for some constantd and arbitrarily large valuesC.

1 2 3 4 5 6 7 8

d1+/m=:f1(A, C)

d4< f₄⁰(C) =^C₄ f₁⁰(C) =^C₁

f2(A, C) :=d2+/m

f7(A, C) :=f₇⁰(C) f8(A, C) :=f₈⁰(C) f3(A, C) :=d3+/m

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Rep.: Different performance measure: Simple Polygon

Simple polygon, escape path unknown Searching for different cost measure

Polygonal extension of the list search problem Distance to the boundary x (estimation, given) Simple circular strategyx(1 +α_x)

x αx

P

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Rep.: Extreme cases, circular strategy

Circular escape path: Distribution of the length is known Extreme situations: x₁(1 + 2π), x₂(1 + 0)

xs α_x≈2π

(i)

s x αx= 0

ii)

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Circular strategy: Star shaped polygon

Optimal circular espape path for s ∈P: Πs(x) For any distance x a worst-caseα_s(x)

In total: minx x(1 +αs(x)) Π_s := min

x Π_s(x) = min

x x(1 +α_s(x)). Radial dist. function interpretation: Area plus height!

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Extreme cases: Radial dist. function

Circular escape path: Distribution of the length is known Extreme situations: x₁(1 + 2π), x₂(1 + 0)

xs α_x≈2π

(i)

s x αx= 0

ii)

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Radial distance function of extreme cases

Optimal circular espape path

Hit the boundary by 90 degree wedge Area plus height! min_x x(1 +α_x)

0 2π

x α_x≈2π (i)

0 2π

x α_x= 0

ii)

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Different justifications

Simple, computation (polynomial), star-shaped vs. convex Natural extension of the discrete certificate (Kirkpatrik) Outperforms escape paths for known cases (diameter)

α

L β L

(i) (ii) (iii)

L

s

s L⁰

L⁰ L⁰

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Outperforms Zig-Zag path

For any position, better than the Zig-Zag path Formal arguments!

Zig-Zag cannot end in farthest vertex: Region R!

s x

d=ysinα=xsin 2α≈0.1214 x

α= arcsin



√1 28



≈10.9^◦ y= 2xcosα

x=√^√³ 28 D⁰

x 2α

C

B C β=⁵₄π B

A

0.125×(5π/4 + 1)<2x= 2√^√³ 28 R

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Interesting example

Distance distribution exactly resembles the polygon Analogy to discrete case! Sorting!

Log. spiral α_x for anyx is known:

x(φ)·(1 +α_x(φ)) withα_x(φ) = 2π−φandx(φ) =A·e^φ^cotβ

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Online Approximation!

Inside a polygon P at points, totally unknown

Leave the polygon, compare to certificate path for s ∈P Dovetailing strategy (discr. case)! Now spiral strategy (a, β)!

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Analysis of a spiral strategy!

Assume certificate: x(1 +αx) fors

Spiral reach distancex =a·e^(φ−α^x^{) cot(β)} at angle φ

Worst-case success at angleφ! (Increasing forα_x distances!) Ratio:

f(γ,a, β) =

a

cosβ ·e^φ^cot^β

a·e^{(φ−γ) cot}^β(1 +γ) = e^γ^cotβ

cosβ(1 +γ) for γ ∈[0,2π]

γ represents possibleαx!

(β,a) represents the spiral strategy!

Independent froma!

How to chooseβ?

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How to choose β ?

Ratio: f(γ, β) = _cos^e^γ_β(1+γ)^cot^β forγ ∈[0,2π]

Balance: Choose β s.th. extreme cases have the same ratio f(0, β) = _cos¹_β = _cos^e^2πcot_β(1+2π)^β =f(2π, β)

β = arccot

ln(2π+1) 2π

= 1.264714. . .

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Balance the extreme cases!

β := arccot

ln(2π+1) 2π

= 1.264714. . . Ratio: f(γ, β) = _cos^e^γ_β(1+γ)^cot^β forγ ∈[0,2π]

f(0, β) =f(2π, β) = 3.31864. . .

andf(γ, β)<3.31864. . .for γ ∈(0,2π)

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Spiral strategy for β = 1 . 264714 . . .

Theorem:There is a spiral strategy for any unknown starting points in any unknown environment P that approximates the certificate fors andP within a ratio of 3.31864.

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Online spiral strategy is almost optimal!

Lower bounds: Difficult to achieve!

Arbitrary strategy: Discretization, Reordering, Functionals!

2πn 1 2

3

4

5

6 7 8

s

Lower bound: Minimize

Pk−1 i=1

q

x_i²−2 cos(^2π_n)^xixi+1+x_i+1² xk(1+^2π_n) +

Pk−1 i=1

q

x_i²−2 cos(^2πn)^xixi+1+x_i+1² xk−n+1(1+2π)

for all n and k.

Gives at leastD = 6.62521. . .for the sum of Detours!

Detour at least D/2≥3.313126!

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn 1 2

3

4

5

6 7 8

s

S i)

S

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn x1,1 1 2

3

4

5

6 7 8

s

S i)

S

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn x1,1 1 2

3

4

5

6 7 8

x2,2 s

S i)

S

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn x1,1 1 2

3

4

5

6 7 8

x2,2 x3,3

s

S i)

S

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn x1,1 1 2

3

4

5

6 7 8

x2,2 x3,3

x4,4 s

S i)

S

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn x1,1 1 2

3

4

5

6 7 8

x2,2 x3,3

x4,4

x5,5 s

S i)

S

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn x1,1 1 2

3

4

5

6 7 8

x2,2 x3,3

x4,4

x5,5 x6,6 x8,8

x7,7 s

S i)

S

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn x1,1 1 2

3

4

5

6 7 8

x2,2 x3,3

x4,4

x5,5 x6,6 x8,8

x7,7 s

x9,1

S i)

q

S

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn x1,1 1 2

3

4

5

6 7 8

x2,2 x3,3

x4,4

x5,5 x6,6 x8,8

x7,7 s

x9,1

S p

i)

q

S

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn x1,1 1 2

3

4

5

6 7 8

x2,2 x3,3

x4,4

x5,5 x6,6 x8,8

x7,7

x10,4 s

x9,1

S p

i)

q

S

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn x1,1 1 2

3

4

5

6 7 8

x2,2 x3,3

x4,4

x5,5 x6,6 x8,8

x7,7

x10,4 s

x9,1

x11,3

S p

i)

q

S

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn x1,1 1 2

3

4

5

6 7 8

x2,2 x3,3

x4,4

x5,5 x6,6 x8,8

x7,7

x10,4 x12,4

s x9,1

x11,3

x13,5 S p

i)

q

S

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn x1,1 1 2

3

4

5

6 7 8

x2,2 x3,3

x4,4

x5,5 x6,6 x8,8

x7,7

x10,4 x12,4

s x9,1

x11,3

x13,5

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn x1,1 1 2

3

4

5

6 7 8

x2,2 x3,3

x4,4

x5,5 x6,6 x8,8

x7,7

x10,4 x12,4

s x9,1

x11,3

x13,5 P1

P2

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn x1,1 1 2

3

4

5

6 7 8

x2,2 x3,3

x4,4

x5,5 x6,6 x8,8

x7,7

x10,4 x12,4

s x9,1

x11,3

x13,5 P1

P2

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn x1,1 1 2

3

4

5

6 7 8

x2,2 x3,3

x4,4

x5,5 x6,6 x8,8

x7,7

x10,4 x12,4

s x9,1

x11,3

x13,5 P1

P2

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn 1 2

3

4

5

6 7 8

s x8,8=x3

x2,2=x1 x1,1=x2

x11,3=x13 x5,5=x9

x10,4=x8 x12,4=x10

x13,5=x12 x9,1=x11

x3,3=x4 x4,4=x5

x6,6=x6 x7,7=x7

ii)

Pk−1 i=1

q

Pk−1 i=1

q

for all n and k.

n goes to infinity!

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Online spiral strategy is almost optimal!

2πn 1 2

3

4

5

6 7 8

s x8,8=x3

x2,2=x1 x1,1=x2

x11,3=x13 x5,5=x9

x10,4=x8 x12,4=x10

x13,5=x12 x9,1=x11

x3,3=x4 x4,4=x5

x6,6=x6 x7,7=x7

ii)

Pk−1 i=1

q

Pk−1 i=1

q

for all n andk.

n goes to infinity!

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Summary!

1

1 s

s 1 1 1

?

Adversary rotates/translates

2π n

x1,1 1 2

3

4

5

6 7 8

x2,2 x3,3

x4,4

x5,5 x6,6 x8,8

x7,7

x10,4 x12,4

s x9,1

x11,3

x13,5 P1

P2

S p q x⁰10,4

i)

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Online Motion Planning MA-INF 1314 Summersemester 17 Escape Paths/Alternative Measure Elmar Langetepe