ON THE COMPLEXITY OF BARRIER RESILIENCE FOR FAT REGIONS Matias - - PowerPoint PPT Presentation

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ON THE COMPLEXITY OF BARRIER RESILIENCE FOR FAT REGIONS Matias - - PowerPoint PPT Presentation

Oct 07, 2022 239 likes •1.45k views

ON THE COMPLEXITY OF BARRIER RESILIENCE FOR FAT REGIONS Matias Korman Maarten Lffler Rodrigo Silveira Darren Strash INTRODUCTION Let t be a desirable point in the plane. Let t be a desirable point in the plane. t Let t be a desirable point

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SLIDE 1

Rodrigo Silveira Darren Strash Maarten Löffler ON THE COMPLEXITY Matias Korman OF BARRIER RESILIENCE FOR FAT REGIONS

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INTRODUCTION

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Let t be a desirable point in the plane.

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Let t be a desirable point in the plane.

t

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SLIDE 5

Let t be a desirable point in the plane. Let G be a set of guards.

t

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Let t be a desirable point in the plane. Let G be a set of guards.

G t

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Let t be a desirable point in the plane. Let G be a set of guards.

G

Each guard can detect movement in a given region.

t

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SLIDE 8

Let t be a desirable point in the plane. Let G be a set of guards.

G

Each guard can detect movement in a given region.

t

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SLIDE 9

Let t be a desirable point in the plane. Let G be a set of guards.

G

Each guard can detect movement in a given region.

t

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SLIDE 10

Let t be a desirable point in the plane. Let G be a set of guards.

G

Each guard can detect movement in a given region.

t

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SLIDE 11

Let t be a desirable point in the plane. Let G be a set of guards.

G

Each guard can detect movement in a given region.

!!! t

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Let t be a desirable point in the plane. Let G be a set of guards.

G

Each guard can detect movement in a given region.

t

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Let t be a desirable point in the plane. Let G be a set of guards.

G

Each guard can detect movement in a given region. Let R be the set of guarded regions.

t

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Let t be a desirable point in the plane. Let G be a set of guards.

G

Each guard can detect movement in a given region. Let R be the set of guarded regions.

R t

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Let t be a desirable point in the plane. Let G be a set of guards.

G

Each guard can detect movement in a given region. Let R be the set of guarded regions.

R

We say R guards t if every path to t passes through one of the regions.

t

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t

Sometimes, guards fail.

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t

Sometimes, guards fail.

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t

Sometimes, guards fail. When they do, t may no longer be guarded!

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t

Sometimes, guards fail. When they do, t may no longer be guarded! The resilience of R is the smallest number of guards that need to fail before R no longer guards t.

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t

Sometimes, guards fail. When they do, t may no longer be guarded! The resilience of R is the smallest number of guards that need to fail before R no longer guards t.

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SLIDE 21

t

Sometimes, guards fail. When they do, t may no longer be guarded! The resilience of R is the smallest number of guards that need to fail before R no longer guards t.

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SLIDE 22

t

Sometimes, guards fail. When they do, t may no longer be guarded! The resilience of R is the smallest number of guards that need to fail before R no longer guards t.

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SLIDE 23

t

Sometimes, guards fail. When they do, t may no longer be guarded! The resilience of R is the smallest number of guards that need to fail before R no longer guards t.

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SLIDE 24

t

Sometimes, guards fail. When they do, t may no longer be guarded! The resilience of R is the smallest number of guards that need to fail before R no longer guards t.

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SLIDE 25

t

Sometimes, guards fail. When they do, t may no longer be guarded! The resilience of R is the smallest number of guards that need to fail before R no longer guards t.

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SLIDE 26

t

Sometimes, guards fail. When they do, t may no longer be guarded! The resilience of R is the smallest number of guards that need to fail before R no longer guards t.

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SLIDE 27

t

Sometimes, guards fail. When they do, t may no longer be guarded! The resilience of R is the smallest number of guards that need to fail before R no longer guards t.

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SLIDE 28

t

Sometimes, guards fail. When they do, t may no longer be guarded! The resilience of R is the smallest number of guards that need to fail before R no longer guards t.

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SLIDE 29

t

Sometimes, guards fail. When they do, t may no longer be guarded! The resilience of R is the smallest number of guards that need to fail before R no longer guards t. Can we compute the resilience?

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SLIDE 30

KNOWN & NEW RESULTS

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When guarding a strip, the resilience can be computed in polynomial time.

[Kumar, Lai & Arora, 2005]

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When guarding a strip, the resilience can be computed in polynomial time.

[Kumar, Lai & Arora, 2005]

The resilience of unit disks in the plane can be 5

6-approximated.

[Bereg & Kirkpatrick, 2009]

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SLIDE 33

When guarding a strip, the resilience can be computed in polynomial time.

[Kumar, Lai & Arora, 2005]

The resilience of unit disks in the plane can be 5

6-approximated.

[Bereg & Kirkpatrick, 2009]

Computing the resilience of line segments in the plane is NP-hard.

[Cabello, Giannopoulos & Knauer, 2011] [Tseng & Kirkpatrick, 2011]

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SLIDE 34

We can compute the resilience r of n unit disks in O(2f(r)n3) time.

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We can compute the resilience r of n unit disks in O(2f(r)n3) time. We can ε-approximate the resilience

  • f unit disks of ply δ in O(2f(δ,ε)n7) time.
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We can compute the resilience r of n unit disks in O(2f(r)n3) time. We can ε-approximate the resilience

  • f unit disks of ply δ in O(2f(δ,ε)n7) time.

These results extend to β-fat regions.

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SLIDE 37

We can compute the resilience r of n unit disks in O(2f(r)n3) time. We can ε-approximate the resilience

  • f unit disks of ply δ in O(2f(δ,ε)n7) time.

These results extend to β-fat regions. Computing the resilience of β-fat regions is NP-hard.

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SLIDE 38

FIXED PARAMETER ALGORITHM

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Consider two points s and t and a set

  • f unit disks R.
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Consider two points s and t and a set

  • f unit disks R.

s t

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Consider two points s and t and a set

  • f unit disks R.

s t

The thickness or R is half the shortest path from s to t in the arrangement.

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Consider two points s and t and a set

  • f unit disks R.

s t

The thickness or R is half the shortest path from s to t in the arrangement.

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Consider two points s and t and a set

  • f unit disks R.

This path gives an upper bound on the resilience.

s t

The thickness or R is half the shortest path from s to t in the arrangement.

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SLIDE 44

Consider two points s and t and a set

  • f unit disks R.

This path gives an upper bound on the resilience.

s t

The thickness or R is half the shortest path from s to t in the arrangement.

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SLIDE 45

Consider two points s and t and a set

  • f unit disks R.

This path gives an upper bound on the resilience.

s t

The thickness or R is half the shortest path from s to t in the arrangement.

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SLIDE 46

Consider two points s and t and a set

  • f unit disks R.

This path gives an upper bound on the resilience.

s t

The thickness or R is half the shortest path from s to t in the arrangement.

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SLIDE 47

Consider two points s and t and a set

  • f unit disks R.

This path gives an upper bound on the resilience.

s t

LEMMA: The thickness is at most twice the resilience.

[Bereg & Kirkpatrick, 2009]

The thickness or R is half the shortest path from s to t in the arrangement.

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SLIDE 48

Consider two points s and t and a set

  • f unit disks R.

This path gives an upper bound on the resilience.

s t

LEMMA: The thickness is at most twice the resilience.

[Bereg & Kirkpatrick, 2009]

The thickness or R is half the shortest path from s to t in the arrangement.

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SLIDE 49

Let r be our estimated resilience.

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Let r be our estimated resilience.

t s

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Let r be our estimated resilience. We only need to consider the part of R within distance O(r) from s and t.

t s

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Let r be our estimated resilience. We only need to consider the part of R within distance O(r) from s and t.

t s

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Let r be our estimated resilience. We only need to consider the part of R within distance O(r) from s and t.

W t s

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Let r be our estimated resilience. We only need to consider the part of R within distance O(r) from s and t. LEMMA: There exists a path π from s via t to the boundary of W that goes through at most O(r) disks.

W t s

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Let r be our estimated resilience. We only need to consider the part of R within distance O(r) from s and t. LEMMA: There exists a path π from s via t to the boundary of W that goes through at most O(r) disks.

W t π s

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Let r be our estimated resilience. We only need to consider the part of R within distance O(r) from s and t. LEMMA: There exists a path π from s via t to the boundary of W that goes through at most O(r) disks.

W

We guess which of these are in the solution: only 2O(r) possibilities.

t π s

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Let r be our estimated resilience. We only need to consider the part of R within distance O(r) from s and t. LEMMA: There exists a path π from s via t to the boundary of W that goes through at most O(r) disks. We guess which of these are in the solution: only 2O(r) possibilities.

W t π s

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IDEA: Cut open the domain along π.

W t π s

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IDEA: Cut open the domain along π.

W t s

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IDEA: Cut open the domain along π. Now the problem looks almost like computing the resilience of a strip!

W t s

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IDEA: Cut open the domain along π. Now the problem looks almost like computing the resilience of a strip!

W

. . . but, the optimal solution may still cross π . . .

t s

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IDEA: Cut open the domain along π. Now the problem looks almost like computing the resilience of a strip!

W

. . . but, the optimal solution may still cross π . . .

t s

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IDEA: Cut open the domain along π. Now the problem looks almost like computing the resilience of a strip!

W

. . . but, the optimal solution may still cross π . . . LEMMA: There are only 2O(r) possible crossings patterns.

t s

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IDEA: Cut open the domain along π. Now the problem looks almost like computing the resilience of a strip!

W

. . . but, the optimal solution may still cross π . . . LEMMA: There are only 2O(r) possible crossings patterns.

  • So. . . we just try them all.

t s

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IDEA: Cut open the domain along π. Now the problem looks almost like computing the resilience of a strip!

W

. . . but, the optimal solution may still cross π . . . LEMMA: There are only 2O(r) possible crossings patterns.

  • So. . . we just try them all.

t s

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Now we have a nice, clean problem.

W t s

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Now we have a nice, clean problem. Given a simply connected region with pairs of points on the boundary, remove the smallest number of disks such that all pairs are connected.

W t s

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Now we have a nice, clean problem. Given a simply connected region with pairs of points on the boundary, remove the smallest number of disks such that all pairs are connected.

W t s

OBSERVATION: The geometry no longer matters.

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Now we have a nice, clean problem. Given a simply connected region with pairs of points on the boundary, remove the smallest number of disks such that all pairs are connected. OBSERVATION: The geometry no longer matters.

t s W

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t s W

Consider the intersection graph.

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t s

Consider the intersection graph.

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t s

Add special terminal vertices for pieces of boundary (at most O(r)). Consider the intersection graph.

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t s

Add special terminal vertices for pieces of boundary (at most O(r)). Consider the intersection graph.

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t s

Add special terminal vertices for pieces of boundary (at most O(r)). We want to cut this graph “along the dotted lines”. Consider the intersection graph.

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t s

Add special terminal vertices for pieces of boundary (at most O(r)). We want to cut this graph “along the dotted lines”. OBSERVATION: This is the same as cutting between all pairs of terminal vertices that are separated by at least one dotted line. Consider the intersection graph.

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t s

Add special terminal vertices for pieces of boundary (at most O(r)). We want to cut this graph “along the dotted lines”. OBSERVATION: This is the same as cutting between all pairs of terminal vertices that are separated by at least one dotted line. Consider the intersection graph.

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t s

Add special terminal vertices for pieces of boundary (at most O(r)). We want to cut this graph “along the dotted lines”. OBSERVATION: This is the same as cutting between all pairs of terminal vertices that are separated by at least one dotted line. Consider the intersection graph.

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SLIDE 78

t s

Add special terminal vertices for pieces of boundary (at most O(r)). We want to cut this graph “along the dotted lines”. OBSERVATION: This is the same as cutting between all pairs of terminal vertices that are separated by at least one dotted line. Consider the intersection graph.

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This is known as vertex multicut.

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This is known as vertex multicut. PROBLEM: Given a graph G = (V, E) and a set of forbidden pairs of vertices L, compute the smallest subset D ⊂ V such that all pairs in L are disconnected in G \ D.

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This is known as vertex multicut. PROBLEM: Given a graph G = (V, E) and a set of forbidden pairs of vertices L, compute the smallest subset D ⊂ V such that all pairs in L are disconnected in G \ D.

[Xiao, 2010]

An optimal vertex multicut can be computed in O(2f(|D|,|L|)n3) time.

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This is known as vertex multicut. PROBLEM: Given a graph G = (V, E) and a set of forbidden pairs of vertices L, compute the smallest subset D ⊂ V such that all pairs in L are disconnected in G \ D.

[Xiao, 2010]

An optimal vertex multicut can be computed in O(2f(|D|,|L|)n3) time.

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This is known as vertex multicut. PROBLEM: Given a graph G = (V, E) and a set of forbidden pairs of vertices L, compute the smallest subset D ⊂ V such that all pairs in L are disconnected in G \ D.

[Xiao, 2010]

An optimal vertex multicut can be computed in O(2f(|D|,|L|)n3) time.

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SLIDE 84

This is known as vertex multicut. PROBLEM: Given a graph G = (V, E) and a set of forbidden pairs of vertices L, compute the smallest subset D ⊂ V such that all pairs in L are disconnected in G \ D.

[Xiao, 2010]

An optimal vertex multicut can be computed in O(2f(|D|,|L|)n3) time.

t s

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SLIDE 85

This is known as vertex multicut. PROBLEM: Given a graph G = (V, E) and a set of forbidden pairs of vertices L, compute the smallest subset D ⊂ V such that all pairs in L are disconnected in G \ D.

[Xiao, 2010]

An optimal vertex multicut can be computed in O(2f(|D|,|L|)n3) time.

s

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SLIDE 86

This is known as vertex multicut. PROBLEM: Given a graph G = (V, E) and a set of forbidden pairs of vertices L, compute the smallest subset D ⊂ V such that all pairs in L are disconnected in G \ D.

[Xiao, 2010]

An optimal vertex multicut can be computed in O(2f(|D|,|L|)n3) time.

s

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SLIDE 87

This is known as vertex multicut. PROBLEM: Given a graph G = (V, E) and a set of forbidden pairs of vertices L, compute the smallest subset D ⊂ V such that all pairs in L are disconnected in G \ D.

[Xiao, 2010]

An optimal vertex multicut can be computed in O(2f(|D|,|L|)n3) time.

W t s

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SLIDE 88

This is known as vertex multicut. PROBLEM: Given a graph G = (V, E) and a set of forbidden pairs of vertices L, compute the smallest subset D ⊂ V such that all pairs in L are disconnected in G \ D.

[Xiao, 2010]

An optimal vertex multicut can be computed in O(2f(|D|,|L|)n3) time.

s t

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SLIDE 89

This is known as vertex multicut. PROBLEM: Given a graph G = (V, E) and a set of forbidden pairs of vertices L, compute the smallest subset D ⊂ V such that all pairs in L are disconnected in G \ D.

[Xiao, 2010]

An optimal vertex multicut can be computed in O(2f(|D|,|L|)n3) time.

s t

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SLIDE 90

This is known as vertex multicut. PROBLEM: Given a graph G = (V, E) and a set of forbidden pairs of vertices L, compute the smallest subset D ⊂ V such that all pairs in L are disconnected in G \ D.

[Xiao, 2010]

An optimal vertex multicut can be computed in O(2f(|D|,|L|)n3) time.

s t

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ε-APPROXIMATION ALGORITHM

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The ply δ of a point is the total number of regions that contain it.

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The ply δ of a point is the total number of regions that contain it.

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The ply δ of a point is the total number of regions that contain it.

δ = 1 δ = 3

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The ply δ of a point is the total number of regions that contain it.

δ = 1 δ = 3

The ply ∆ of R is the maximum ply

  • ver all points in the plane.
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The ply δ of a point is the total number of regions that contain it.

δ = 1 δ = 3 ∆ = 4

The ply ∆ of R is the maximum ply

  • ver all points in the plane.
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Let k = ⌈4∆/ε2⌉.

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Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm.

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SLIDE 99

Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm.

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SLIDE 100

s t

Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm.

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SLIDE 101

s t r = 1

Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm.

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SLIDE 102

t s

Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm.

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SLIDE 103

t s r = 1

Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm.

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SLIDE 104

s r = 2 t

Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm.

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SLIDE 105

s t r = 1

Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm.

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SLIDE 106

s t

Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm.

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s t r =?

Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm.

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Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm. Augment the dual graph of R with extra edges.

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Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm. Augment the dual graph of R with extra edges.

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SLIDE 110

Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm. Augment the dual graph of R with extra edges.

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SLIDE 111

Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm. Augment the dual graph of R with extra edges.

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SLIDE 112

Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm. Augment the dual graph of R with extra edges.

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SLIDE 113

Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm. Augment the dual graph of R with extra edges. Compute the shortest path in the resulting graph.

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Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm. Augment the dual graph of R with extra edges. Compute the shortest path in the resulting graph.

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CLAIM: This is a 1 + ε approximation. Let k = ⌈4∆/ε2⌉. Compute all pairs of resilience at most k exactly, using the FPT algorithm. Augment the dual graph of R with extra edges. Compute the shortest path in the resulting graph.

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CONCLUSIONS

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We show that resilience for unit disks is FPT, and leads this to an EPTAS.

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We show that resilience for unit disks is FPT, and leads this to an EPTAS. On the other hand, resilience of many classes of regions is NP-hard.

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We show that resilience for unit disks is FPT, and leads this to an EPTAS. On the other hand, resilience of many classes of regions is NP-hard. It seems all hardness reductions require regions to cross each other.

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We show that resilience for unit disks is FPT, and leads this to an EPTAS. On the other hand, resilience of many classes of regions is NP-hard. It seems all hardness reductions require regions to cross each other. OPEN QUESTION: Can the resilience

  • f unit disks be computed in

polynomial time?

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THANKS! ;)