Methoden der Oﬄine Bewegungsplanung Feuerbek¨ampfung Elmar Langetepe

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Elmar Langetepe

Universit¨at Bonn, Institut f¨ur Informatik

27.01.2014

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Bewegungsplanung: Feuerbek¨ ampfung

Wichtiges Problem:

Schlagen von Feuerschneisen Vorhandene Schneisen Neue Schneisen Einfaches Modell:

Gleichm¨aßige Ausbreitung Geschwindigkeit: s ∈[0,1)

s ≈0.38∈[0,1) s = cos(α), α∈(0, π/2] Feuer schnell umschließen (Zeitminimierung!) l0,l1,l2 gefahrlos minimieren!

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Bewegungsplanung: Feuerbek¨ ampfung

Wichtiges Problem:

α∈(0, π/2] Feuer schnell umschließen (Zeitminimierung!) l0,l1,l2 gefahrlos minimieren!

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Bewegungsplanung: Feuerbek¨ ampfung

Wichtiges Problem:

s ≈0.38∈[0,1) s = cos(α), α∈(0, π/2]

Feuer schnell umschließen (Zeitminimierung!) l0,l1,l2 gefahrlos minimieren!

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Bewegungsplanung: Feuerbek¨ ampfung

Wichtiges Problem:

1

s ≈0.38∈[0,1) s = cos(α), α∈(0, π/2]

l0,l1,l2 gefahrlos minimieren!

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Bewegungsplanung: Feuerbek¨ ampfung

Wichtiges Problem:

1

l

s ≈0.38∈[0,1) s = cos(α), α∈(0, π/2]

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Bewegungsplanung: Feuerbek¨ ampfung

Wichtiges Problem:

1 t1= 1 1 +s·t1

s ≈0.38∈[0,1) s = cos(α), α∈(0, π/2]

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Bewegungsplanung: Feuerbek¨ ampfung

Wichtiges Problem:

t2≈1.31 1

l 1 +s·t2

s ≈0.38∈[0,1) s = cos(α), α∈(0, π/2]

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Bewegungsplanung: Feuerbek¨ ampfung

Wichtiges Problem:

1

t₃= 4 1 +s·t₃

l₁

s ≈0.38∈[0,1) s = cos(α), α∈(0, π/2]

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Bewegungsplanung: Feuerbek¨ ampfung

Wichtiges Problem:

1

l

t₃= 4 1 +s·t₃

l₁

l

s ≈0.38∈[0,1) s = cos(α), α∈(0, π/2]

Feuer schnell umschließen (Zeitminimierung!)

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1

t₃= 4 1 +s·t₃

l₁

s ≈0.38∈[0,1) s = cos(α), α∈(0, π/2]

Feuer schnell umschließen (Zeitminimierung!) l ,l ,l gefahrlos

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Bewegungsplanung: Feuerbek¨ ampfung

s ≈0.38 = cos(1.18) Optimale L¨osung!

Minimieren der Werte Passieren des Feuers Zeit und Fl¨ache!

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Theorem: There exists a unique area and completion time optimal axis-parallel firebreak for allα∈(π/4, π/2]. Forα∈(0, π/4] no solution exists.

Beweis!

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Bewegungsplanung: Feuerbek¨ ampfung

Situation erster Teil!

A B= (0, b)

l (0,0) q= (qx,0)

p1

Ccosα(B, A+ cosα× |Π^p_q¹|)

A+ cosα× |Π^p_q¹|

Rp₁

Π1

ϕ

q_x (qx, b) p1

b

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Bewegungsplanung: Feuerbek¨ ampfung

F¨urα∈(0, π/4] keine L¨osung!

Geht das, wenn die Firefigther beliebig graben d¨urfen?

Idee: Starte nah am Feuer und bleibe dran!

Geschwindigkeit cos(α), α∈(0, π/2]

Kontinuierlich lokal in Richtung α!

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Bewegungsplanung: Feuerbek¨ ampfung

Weg: Logarithmische Spirale mit Exzentrizit¨atα!

1 + cos(α)·t

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Bewegungsplanung: Feuerbek¨ ampfung

1 + cos(α)·t

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Bewegungsplanung: Feuerbek¨ ampfung

1 + cos(α)·t

α

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Bewegungsplanung: Feuerbek¨ ampfung

1 + cos(α)·t

α x

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Bewegungsplanung: Feuerbek¨ ampfung

1 + cos(α)·t

α x

xcos(α)

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Bewegungsplanung: Feuerbek¨ ampfung

1 + cos(α)·t

α x

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Bewegungsplanung: Feuerbek¨ ampfung

1 + cos(α)·t

α x

xcos(α) α

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Bewegungsplanung: Feuerbek¨ ampfung

1 + cos(α)·t

α x

α

Z A

B SAB

ZA+ cos(α)·SAB=ZB

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Bewegungsplanung: Feuerbek¨ ampfung

1 + cos(α)·t

α x

xcos(α) α

Z A

B SAB

ZA+ cos(α)·SAB=ZB 1 + cos(α)·(t+SAB)

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1 + cos(α)·t

α x

α

Z A

B SAB

ZA+ cos(α)·SAB=ZB 1 + cos(α)·(t+SAB)

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Bewegungsplanung: Feuerbek¨ ampfung

Mit Spiralen eine bessere/optimale L¨osung erstellen!

Verschiedene Kandidaten!

Optimale L¨osung: Zeitminimierung f¨ur jedes α!!!

Begr¨undung! (Beweis!) Gilt f¨ur alle α∈(0, π/2]

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Bewegungsplanung: Feuerbek¨ ampfung

1 x

α

O

a

B Sab

Sab=x 1 + cos(α)·x

A

b

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1 x

α

O

a Sab

A

b

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Bewegungsplanung: Feuerbek¨ ampfung

1 x

α

O

a

B Sab

A

b

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1 x

α

O

a Sab

A

b

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Bewegungsplanung: Feuerbek¨ ampfung

1 x

α

O

a

B Sab

A

b

Achtung: Nicht optimal f¨ur Fl¨achenminimierung!

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Waldbr¨ande, Grunds¨atze

Ausbreitungsgeschwindigkeit klein: 0.5 km/h Graben mit Winkel

Ver¨anderte Ausbreitungskurven Modellerweiterungen

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Diskrete Problemstellung: Strategic Deployment

Starting form a basis camp with a set of agents Take over and control some settlements

Resistance in the outback

Enough agents for the movements between the settlements and forcontrolling the settlements

Task: Move efficientlyaround and occupy the settlements Historic examples!

Gaius Julius Ceasar: Conquer of the Gauls (58 to 51 B.C.) Alexander the Great (356 B.C. to 323 B.C.): Alexander’s campaign

Modeled by an edge an vertex-weigthed graph

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Diskrete Problemstellung: Strategic Deployment

Starting form a basis camp with a set of agents

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Diskrete Problemstellung: Strategic Deployment

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Diskrete Problemstellung: Strategic Deployment

Historic examples!

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Diskrete Problemstellung: Strategic Deployment

Enough agents for themovements between the settlements and forcontrolling the settlements

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Diskrete Problemstellung: Strategic Deployment

Task: Move efficientlyaround and occupy the settlements

campaign

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Diskrete Problemstellung: Strategic Deployment

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Model of the Problem

Edge- and vertex-weigthed graphG, Graph is fully known

Some start vertexv_s Rules for the movement:

1 Edgee, weightwe:Lower boundon the number of agents required for traversal ofe.

2 Vertexv, weightw_v: Number of agents that have to be placed at the vertex.

3 First visit ofv: Full amountw_v have to be placed, these agents cannot be removed any more.

Visit and occupy the vertices accordingly Interesting computational questions:

How many agents are required in total? How long does this take?

k agents given: How many settlements can we get?

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Model of the Problem

Edge- and vertex-weigthed graphG, Graph is fully known Some start vertexv_s

Rules for the movement:

at the vertex.

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Model of the Problem

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Model of the Problem

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Model of the Problem

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Model of the Problem

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Model of the Problem

Visit and occupy the vertices accordingly

Interesting computational questions:

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Model of the Problem

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Model of the Problem

How many agents are required in total?

How long does this take?

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Model of the Problem

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Model of the Problem

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Example: Minimal number of agents required is 23!

1 At least w_e for traversing edgee are required

2 At the first visit of v exactlywv have to be placed and will never by removed!

1

1 1

15 1 v3

v5 v1=vs

v2

v4 25

1 1

20 22

23

7

4 agents remain unsettled! No other strategy is better!! Is the problem clear?

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Example: Minimal number of agents required is 23!

v3

v5 v1=vs

v2

v4 25

1 1

20 22 22←23

1

15 7

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Example: Minimal number of agents required is 23!

v3

v5 v1=vs

v2

v4 25

1 1

20 22

7 21←23

1

15

1

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Example: Minimal number of agents required is 23!

v3

v5 v1=vs

v2

v4 25

1 1

20 22

7

20←23 1

1

15

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Example: Minimal number of agents required is 23!

v3

v5 v1=vs

v2

v4 25

1 1

20 22

7

1

15 1

20←23 1

1

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Example: Minimal number of agents required is 23!

v3

v5 v1=vs

v2

v4 25

1 1

20 22

7 5←23

1

15

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Example: Minimal number of agents required is 23!

v3

v5 v1=vs

v2

v4 25

1 1

20 22

7

1

15 1

20←23 1

1

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Example: Minimal number of agents required is 23!

v3

v5 v1=vs

v2

v4 25

1 1

20 22

7 20←23

1

15

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Example: Minimal number of agents required is 23!

v3

v5 v1=vs

v2

v4 25

1 1

20 22

7 1

1

15

1 19←23

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Example: Minimal number of agents required is 23!

v3

v5 v1=vs

v2

v4 25

1 1

20 22

7 19←23

1

15

1

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Example: Minimal number of agents required is 23!

v3

v5 v1=vs

v2

v4 25

1 1

20 22

7 1

1

15 19←23 1

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Example: Minimal number of agents required is 23!

v3

v5 v1=vs

v2

v4 25

1 1

20 22

7 4←23

1 1

1

15 Done!!

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Example: Minimal number of agents required is 23!

v3

v5 v1=vs

v2

v4 25

1 1

20 22

7 4←23

1 1

1

15 Done!!

4 agents remain unsettled! No other strategy is better!!

Is the problem clear?

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e

v3

v5 v1=vs

v2

v4 25

1 1

20 22

7 4←23

1 1

1

15 Done!!

4 agents remain unsettled! No other strategy is better!!

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Number of agents required: Simple bounds!

G = (V,E),N :=P

v∈V w_v wmax := max{we|e ∈E}

N+wmax onG is sufficient! Strategy S:

wS := max{we|e was visited byS} S requires at most N+w_S

S requires at least max{N,w_S}

Minimum Spanning Tree (MST), wMST := max{w_e|e ∈ MST} N+wMST on MST is sufficient

Any Strategy S onG requires at least max{N,wMST}

Lemma: Optimal Strategy for MST gives 2-Approximation for G

1

v₃

v₅ v₁=v_s

v₂

v₄ 25

1 1

20

22 1

1

15 7

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Number of agents required: Simple bounds!

G = (V,E),N :=P

v∈V w_v wmax := max{we|e ∈E}

N+wmax onG is sufficient! Strategy S:

1

v₃

v₅ v₁=v_s

v₂

v₄ 25

1 1

20

22 1

1

15 N= 19

7 w_max= 25

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Number of agents required: Simple bounds!

G = (V,E),N :=P

v∈V w_v wmax := max{we|e ∈E} N+wmax onG is sufficient!

Strategy S:

1

v₃

v₅ v₁=v_s

v₂

v₄ 25

1 1

20

22 1

1

15 N= 19

7 w_max= 25

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Number of agents required: Simple bounds!

G = (V,E),N :=P

Strategy S:

S requires at least max{N,w_S} Minimum Spanning Tree (MST), wMST := max{w_e|e ∈ MST} N+wMST on MST is sufficient

1

v₃

v₅ v₁=v_s

v₂

v₄ 25

1 1

20

22 1

1

15 N= 19

7 w_max= 25

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Number of agents required: Simple bounds!

G = (V,E),N :=P

Strategy S:

wS := max{we|e was visited byS}

S requires at most N+w_S S requires at least max{N,w_S} Minimum Spanning Tree (MST), wMST := max{w_e|e ∈ MST} N+wMST on MST is sufficient

1

v₃

v₅ v₁=v_s

v₂

v₄ 25

1 1

20

22 1

1

15 N= 19

7 w_max= 25

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Number of agents required: Simple bounds!

G = (V,E),N :=P

Strategy S:

1

v₃

v₅ v₁=v_s

v₂

v₄ 25

1 1

20

22 1

1

15 N= 19

7 w_max= 25

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Number of agents required: Simple bounds!

G = (V,E),N :=P

Strategy S:

S requires at least max{N,w_S} Minimum Spanning Tree (MST), wMST := max{w_e|e ∈ MST}

N+wMST on MST is sufficient

1

v₃

v₅ v₁=v_s

v₂

v₄ 25

1 1

20

22 1

1

15 N= 19

7 w_max= 25

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Number of agents required: Simple bounds!

G = (V,E),N :=P

Strategy S:

S requires at least max{N,w_S} Minimum Spanning Tree (MST), wMST := max{w_e|e ∈ MST}

N+wMST on MST is sufficient

1

v₃

v₅ v₁=v_s

v₂

v₄ 25

1 1

20

22 1

1

15 N= 19

7 w_max= 25 wMST= 20

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Number of agents required: Simple bounds!

G = (V,E),N :=P

Strategy S:

S requires at least max{N,w_S} Minimum Spanning Tree (MST), wMST := max{w_e|e ∈ MST} N+wMST on MST is sufficient

Any StrategyS onG requires at least max{N,wMST}

1

v₃

v₅ v₁=v_s

v₂

v₄ 25

1 1

20

22 1

1

15 N= 19

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Number of agents required: Simple bounds!

G = (V,E),N :=P

Strategy S:

S requires at least max{N,w_S} Minimum Spanning Tree (MST), wMST := max{w_e|e ∈ MST} N+wMST on MST is sufficient Any StrategyS on G requires at least max{N,wMST}

1

v₃

v₅ v₁=v_s

v₂

v₄ 25

1 1

20

22 1

1

15 N= 19

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Number of agents required: Simple bounds!

G = (V,E),N :=P

Strategy S:

S requires at least max{N,w_S} Minimum Spanning Tree (MST), wMST := max{w_e|e ∈ MST} N+wMST on MST is sufficient Any StrategyS on G requires at least max{N,wMST}

1

v₃

v₅ v₁=v_s

v₂

v₄ 25

1 1

20

22 1

1

15 N= 19

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Variants: Return or No-Return!

(No-return) It suffices to fill the vertices as required, no agents have to return to the start vertex.

Comparable toroutes(round-trips) andtours (open paths) in TSP Reporting the success formally means:

Set,M, of agents return to v_s, the union of allvertices visited by the members ofM equals V.

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Variants: Return or No-Return!

(Return) Finally some agents have to return to the start vertex and report the success of the whole operation.

Set,M, of agents return to v_s, the union of allvertices visited by the members ofM equals V.

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Variants: Return or No-Return!

Comparable toroutes(round-trips) andtours (open paths) in TSP

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Variants: Return or No-Return!

Set,M, of agents return to v_s, the union ofall vertices visited by the members ofM equals V.

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Optimal Algorithm for Trees: Return Variant

Computational lower bound and algorithmic idea! Example!

vs

n

1 1

0

n−1 2

1 1

n−2

Example: Visitn−2 before n

Lemma:Computational lower bound Ω(nlogn) by sorting (both variants, but real weights)!

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Optimal Algorithm for Trees: Return Variant

vs

n

1 1

0

n−1 2

1 1

n−2

Optimal strategy:n+ 1 agents, visit vertices in order of decreasing edge weights:n,n−1,n−2,. . . ,2,1

Any other order will increase the number! Example: Visitn−2 before n

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Optimal Algorithm for Trees: Return Variant

vs

n

1 1

0

n−1 2

1 1

n−2

Any other order will increase the number!

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Optimal Algorithm for Trees: Return Variant

vs

n

1 1

0

n−1 2

1 1

n−2

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Optimal Algorithm for Trees: Return Variant

vs

n

1 1

0

n−1 2

1 1

n−2 n+ 1

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Optimal Algorithm for Trees: Return Variant

vs

n

1 0

n−1 2

1 1

n−2 1

n←n+ 1

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Optimal Algorithm for Trees: Return Variant

vs

n

1 0

n−1 2

1 1

n−2 1 n←n+ 1

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Optimal Algorithm for Trees: Return Variant

vs

n

1 0

n−1 2

1 1

n−2 1 n←n+ 1 currentnnot enough,

at leastn+ 2 in total!

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vs

n

1 0

n−1 2

1 1

n−2 1 n←n+ 1 currentnnot enough,

at leastn+ 2 in total!

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Datenstruktur: Trapezzerlegung nach Seidel

Lokalisation Black Box Polygon: Dreieckszerlegung

Anfrage: F¨urp ∈P finde DreieckT mit p∈T K¨urzeste Wege in Polygonen/auf Polyeder Aufbau:O(nlog^∗n), Query:O(logn)

Zerlegung in Trapeze, monotone Polygone, danach in Dreiecke Hier Trapezzerlegung!

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s₁ t₂

t₃ t₁

t₄ t₅

t₆

s₂ t₇

s₃ t₈ t₉

s₄ t₁₀

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Datenstruktur: Trapezzerlegung nach Seidel

s₁

r(s₁)

l(s₁)

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s₁

r(s₁)

l(s₁)

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Datenstruktur: Trapezzerlegung nach Seidel

s₁

l(s₁) t₂

t₁

r(s₁)

l(s₁)

t₂ t₁

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s₁

l(s₁) t₂

t₁

r(s₁)

l(s₁)

t₂ t₁

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Datenstruktur: Trapezzerlegung nach Seidel

s₁

l(s₁) t₂

t₃ t₁

r(s₁)

l(s₁)

t₁ r(s₁) t₂ t₃