Theoretical Aspects of Intruder Search Course Wintersemester 2015/16 Geometric Firefighting Elmar Langetepe

Theoretical Aspects of Intruder Search

Course Wintersemester 2015/16 Geometric Firefighting

Elmar Langetepe

University of Bonn

December 8th, 2015

Connnected Search vs. non-connected search

Non-connected, other rules!

Differ in a factor of 2

1 Place a team of p guards on a vertex.

2 Move a team ofm guards along an edge.

3 Remove a team of r guards from a vertex.

Connnected Search vs. non-connected search

D_k denote a tree with rootr of degree three and three full binary trees,Bk−1, of depth k−1 connected to ther.

Lemma 31:For the graph Dk, we concludecs(Dk) =k+ 1.

ConsiderT₁,T₂ andT₃ atr! At most k+ 1

At least k+ 1

Connnected Search vs. non-connected search

D_k denote a tree with rootr of degree three and three full binary trees,Bk−1, of depth k−1 connected to ther.

Lemma 32:ForD2k−1 we concludes(D_2k−1)≤k+ 1.

k = 1 is trivial. So assume k >1

Place one agent at the root r and successively clean the copies of B2k−2 byk agents

This is shown by induction!

B_2k¹ ₋₂

v^k+1₂ v^k+1₁

B_2k² ₋₂ B_2k³ ₋₂ B_2k⁴ ₋₂ B_2(k+1)−2

Connnected Search vs. non-connected search

Corollary 33:There exists a treeT so that cs(T)≤2s(T)−2 holds.

T =D2k−1,s(D2k−1)≤k+ 1, cs(D2k−1) = 2k cs(T)

s(T) <2 for all treesT.

Geometric firefighting, Simple Polygon

Intruder/Contam. constant speed, exclude fire, fences First, inside a polygon, single fire source,

Build linear barriers with speedB, build barriers successively Instance:Simple polygon, fire spreads froms ∈P with speed 1,m line segmentbarriers,b_i successively constructed with speed B.

Output:Valid sequence of barriers constructed successively, area blocked from the fire is maximized.

Geometric firefigthing, simple polygon

Theorem 1:Computing an optimal-enclosement-sequence is NP-hard.

Approximation hard!

Our goal: Polynomial time constant approximation!

ai 2

Ai

Bi

Ci

Di

s r

Ai+1 ai+1

2

di s

r²−^a2ⁱ 2

=xi

C⁰_i h

h

Geometric firefigthing, simple polygon, approximation

General scheduling algorithm, working with profits 0.086-approximation of optimal profit (area).

Non-intersecting barriers, is an application!

Intersection is more difficult!

Framework: Set of jobs b₁,b₂, . . . ,b_m

Duration d_i, starting time s_i (start befores_i!) Algorithm: n steps scheduleJn= (bn1,bn2, . . . ,bn_ln) Size ln,n jobs considered, s_n⁰_k precise starting time Valid: Pj

k=1s_n⁰_k+d_nk ≤s_nj+1 for j = 1 tol_n−1 Job bi contribute with a profitAi to overall profitA

Geometric firefigthing, simple polygon, approximation

Profits might overlap!Ai ∪Aj 6=∅ Schedule:J_n= (b_n1,b_n2, . . . ,b_nln) b_j 6∈J_n, current profit! Can decrease!

Aj(Jn) :=Aj \



 [

b_nk∈Jn

Ank





Approximation scheme: GlobalGreedy

Empty schedule J₀, constantµ <1 Sort remaining jobs b_j by ^Aj(J_dⁿ⁾

j , process largest!

1 b_j can be scheduled somewhere in J_n. Insert b_j:J_n+1

2 b_j cannot be processed, overlaps with jobs in J_n. Find sequence in J_n that overlaps:

1. Profits of these jobs smaller thanµ times Aj(Jn).

2.b_j can be scheduled after deletion of the jobs.

Then build J_n+1 withb_j.

Deleted jobs will never be processed again.

3 No such sequence exists in J_n. Rejectb_j!

Color scheme: Green profit/jobs (inserted), grey profit/jobs (deleted afterwards)

All profits (universe) red in the beginning!

Approximation scheme: GlobalGreedy Example

b₁= (a,b), b₂ = (c,d), b₃ = (e,f),|b₁|= 3, |b₂|= 0.3,

|b3|= 0.5, speed 2,µ= 0.2

Approximation scheme: GlobalGreedy Example

p1= 702 = ^2A₃¹,p2 = ^2A_0.3² = 1083,p3 = ^2A_0.5³ = 755 J₂ = (b₂,b₃), (0.3 + 0.5 + 3)/2 = 1.9>d_P(s,a) µ·A₁>A₃, (0.3 +d(a,b))/2<d(s,a),J₃ = (b₂,b₁).

GlobalGreedy: Green and Grey

J_n(grey) andJ_n(green) colored green/grey during the construction of J_n.

J_n⁰: All jobs that where inserted, green/grey Lemma 52:J_m(grey)≤ 1−µ^µ J_m(green).

By induction on the jobs processed during GlobalGreedy Base: Holds forJ₀

Assume that the lemma holds after n steps forJ_n. Consider step n+ 1.

GlobalGreedy: Green and Grey

Lemma 52:J_m(grey)≤ 1−µ^µ J_m(green).

Inductive step:n →n+ 1, considerb_j withA_j(J_n) No job deleted (Rules 1.,3.): Only green can increases!

J_n(grey) =J_n+1(grey)≤ µ

1−µJ_n(green)≤ µ

1−µJ_n+1(grey). Rule 2., some jobs deleted: smallerµ timesA_j(J_n)

µ

1−µJn+1(green) ≥ µ

1−µ(Jn(green) + (1−µ)A_j(Jn))

≥ µ

1−µJ_n(green) +µA_j(J_n)

≥ J_n(grey) +µA_j(J_n)≥J_n+1(grey),

Relationsship to optimal sequence, J opt

Green and grey profits/jobs

Jopt: Red profits finally not colored green or grey, colored blue!

Example: Job b₃ will be scheduled, no blue color!

Assign, blue profit to the first job in Jopt, that covers profit!

|Jopt| ≤Jm(blue) +Jm(green) +Jm(grey).

Relationsship to optimal sequence, J opt

Green, grey, blue profits/jobs are disjoint!

Expresse blue profit in terms of grey and green profit Payment scheme! Green/grey (J_m⁰ ) pay to blue jobs!

bi ∈J_m⁰ gets unique execution time! Pays to somebj ∈Jopt!

1 If the execution interval of bj ∈Jopt is fully included in the execution interval of b_i ∈J_m⁰ , the jobb_i pays its green or grey profit times ^d_d^j

i <1 tobj.

2 If the execution interval of bj ∈Jopt overlaps with the execution interval of b_i ∈J_m⁰ , the jobb_i pays its green or grey profit times ¹_µ tob_j.

Relationsship to optimal sequence, J opt

1 If the execution interval of bj ∈Jopt is fully included in the execution interval of b_i ∈J_m⁰ , the jobb_i pays its green or grey profit times ^d_d^j

i <1 tobj.

2 If the execution interval of b_j ∈Jopt overlaps with the execution interval of bi ∈J_m⁰ , the jobbi pays its green or grey profit times ¹_µ tob_j.

Lemma 53:Any single green or grey job fromJ_m⁰ pays in total at most 1 +_µ² times its profit to the blue jobs.

Lemma 54:Any single blue job fromJopt achieves at least a payment in the size of its blue profit from the green and grey jobs.

Relationsship to optimal sequence, J opt

Lemmate 53/54:

J_m(blue)≤

1 + 2 µ

(J_m(green) +J_m(grey)). Lemma 52:

J_m(grey)≤ µ

1−µJ_m(green).

|Jopt| ≤ Jm(blue) +Jm(green) +Jm(grey) (1)

≤

2 + 2 µ

(Jm(green) +Jm(grey)) (2)

≤ 2(µ+ 1)

µ (J_m(green) + µ

1−µJ_m(green)) (3)

≤ 2(µ+ 1) µ

1

1−µJ_m(green) (4)

≤ 2 µ+ 1

µ(1−µ)J_m(green)≤2 µ+ 1

µ(1−µ)|J_m|. (5)

Relationsship to optimal sequence, J opt

|Jopt| ≤ 2 µ+ 1 µ(1−µ)|J_m|. Minimize:f(µ) := 2_µ(1−µ)^µ+1

Byµ=√

2−1 this givesf(µ) = 6 + 4√

2≈11.657

Theorem 55:For the geometric firefighter problem inside a simple polygon with non-intersecting barriers there is an approximation algorithms that saves at least ¹

6+4√

2 = ³₂−√

2≈0.086 times the area of the optimal barrier solution.

Applicable to the barrier construction problem!

Relationsship to optimal sequence, J opt

1 If the execution interval of bj ∈Jopt is fully included in the execution interval of b_i ∈J_m⁰ , the jobb_i pays its green or grey profit times ^d_d^j

i <1 tobj.

2 If the execution interval of b_j ∈Jopt overlaps with the execution interval of bi ∈J_m⁰ , the jobbi pays its green or grey profit times ¹_µ tob_j.

Lemma 53:Any single green or grey job fromJ_m⁰ pays in total at most 1 +_µ² times its profit to the blue jobs.

Lemma 54:Any single blue job fromJopt achieves at least a payment in the size of its blue profit from the green and grey jobs.

J_m(blue)≤

1 + 2 µ

(J_m(green) +J_m(grey)).

Relationsship to optimal sequence, J opt

1 If the execution interval of b_j ∈Jopt is fully included in the execution interval of bi ∈J_m⁰ , the jobbi pays its green or grey profit times ^d_d^j

i <1 tob_j.

2 If the execution interval of b_j ∈Jopt overlaps with the execution interval of b_i ∈J_m⁰ , the jobb_i pays its green or grey profit times ¹_µ tobj.

Lemma 53:Any single green or grey job fromJ_m⁰ pays in total at most 1 +_µ² times its profit to the blue jobs.

b_i ∈J_m⁰ has fixed execution intervalI_i with start- and endtime Interval ofb_j ∈Jopt fully insideI_i: ^d_d^j

i, sums up to at most 1 for all b_j ∈Jopt

Relationsship to optimal sequence, J opt

1 If the execution interval of b_j ∈Jopt is fully included in the execution interval of bi ∈J_m⁰ , the jobbi pays its green or grey profit times ^d_d^j

i <1 tob_j.

2 If the execution interval of bj ∈Jopt overlaps with the execution interval of bi ∈J_m⁰ , the jobbi pays its green or grey profit times ¹_µ tob_j.

Lemma 54:Any single blue job fromJopt achieves at least a payment in the size of its blue profit from the green and grey jobs.

Blue job bj ∈Jopt, job has to be rejected in stepk+ 1, Consider execution time interval of b_j ∈Jopt

Subset J_k of J_k = (b_k1,b_k2, . . . ,b_klk) thatminimallyoverlaps with execution interval for b_j

Total profit Jk larger thanµtimes curr. red profit Aj(Jk) ofbj

Larger µtimes the final blue part ofb_j b_j less priority: ^Ai(J_d^k)

i ≥ ^Aj(J_dj^k)

Relationsship to optimal sequence, J opt

Lemma 54:Any single blue job fromJopt achieves at least a payment in the size of its blue profit from the green and grey jobs.

1 Total profitJ_k larger than µtimes final blue profit of b_j (≤A_j(J_k))

2 bj less priority: ^Ai(J_d^k)

i ≥ ^Aj(J_dj^k)

|J_k|= 1 for single job, sayb_i

b_j ∈Jopt might be fully inside the execution time ofb_i: Pay: A_i(J_k)d_j

di ≥A_j(J_k)d_i di

=A_j(J_k) For |J_k| ≥1, execution interval of b_j overlaps with all execution intervals in Jk: