Planargraphs Planargraphs:graphsthatcanbedrawnwithoutcrossingedges TUe

(1)

Packing and covering: planar separator and shifting

S´ andor Kisfaludi-Bak

Computaional Geometry Summer semester 2020

(2)

Overview

• Planar separator theorem (slides by Mark de Berg)

(3)

Overview

• Planar separator theorem (slides by Mark de Berg)

• Indepedent set in planar graphs (slides by MdB)

(4)

Overview

• Planar separator theorem (slides by Mark de Berg)

• Indepedent set in planar graphs (slides by MdB)

• Exact algroithms for packing and covering

(5)

Overview

• Planar separator theorem (slides by Mark de Berg)

• Indepedent set in planar graphs (slides by MdB)

• Shifting strategy: approximation schemes

• Exact algroithms for packing and covering

(6)

TU e

Planar graphs

Planar graphs: graphs that can be drawn without crossing edges

slide by Mark de Berg

(7)

TU e

Planar graphs

Planar graphs: graphs that can be drawn without crossing edges

Planar Separator Theorem (Lipton,Tarjan 1979)

(8)

TU e

Planar graphs

Planar graphs: graphs that can be drawn without crossing edges

Planar Separator Theorem (Lipton,Tarjan 1979)

For any planar graph G = (V, E ) there is a separator S ⊂ V of size O( √

n) such that V \ S can be partitioned into subsets A and B , each of size at most

²₃

n and with no edges between them.

(9)

TU e

Planar graphs

Planar graphs: graphs that can be drawn without crossing edges

Planar Separator Theorem (Lipton,Tarjan 1979)

For any planar graph G = (V, E ) there is a separator S ⊂ V of size O( √

n) such that V \ S can be partitioned into subsets A and B , each of size at most

²₃

n and with no edges between them.

A B

S

(10)

TU e

Planar graphs

Planar graphs: graphs that can be drawn without crossing edges

Planar Separator Theorem (Lipton,Tarjan 1979)

For any planar graph G = (V, E ) there is a separator S ⊂ V of size O( √

n) such that V \ S can be partitioned into subsets A and B , each of size at most

²₃

n and with no edges between them.

Such a (2/3)-balanced separator can be computed in O (n) time.

A B

S

(11)

TU e

A geometric proof of the Planar Separator Theorem

Fact: Any planar graph is the contact graph of a set of interior-disjoint disks.

(12)

TU e

A geometric proof of the Planar Separator Theorem

Fact: Any planar graph is the contact graph of a set of interior-disjoint disks.

Proof idea: Find a square σ intersecting O( √

n) disks

that is a balanced separator.

(13)

TU e

A geometric proof of the Planar Separator Theorem

Theorem. For any contact graph of n interior-disjoint disks, there is an α-balanced separator of size O( √

n), where α = 36/37.

Proof.

(14)

TU e

A geometric proof of the Planar Separator Theorem

Theorem. For any contact graph of n interior-disjoint disks, there is an α-balanced separator of size O( √

n), where α = 36/37.

Proof.

smallest square

containing at least n/37 disks

3

(15)

TU e

A geometric proof of the Planar Separator Theorem

Theorem. For any contact graph of n interior-disjoint disks, there is an α-balanced separator of size O( √

n), where α = 36/37.

Proof.

smallest square

containing at least n/37 disks

3 √ n squares σ

₁

, . . . , σ

^√_n

at distance 1/ √

n from each other

(16)

TU e

A geometric proof of the Planar Separator Theorem

Theorem. For any contact graph of n interior-disjoint disks, there is an α-balanced separator of size O( √

n), where α = 36/37.

Proof.

smallest square

containing at least n/37 disks

3 √ n squares σ

₁

, . . . , σ

^√_n

at distance 1/ √

n from each other

(17)

TU e

A geometric proof of the Planar Separator Theorem

Theorem. For any contact graph of n interior-disjoint disks, there is an α-balanced separator of size O( √

n), where α = 36/37.

Proof.

smallest square

containing at least n/37 disks

3 √ n squares σ

₁

, . . . , σ

^√_n

at distance 1/ √

n from each other Constructing the separator:

Select a square σ

_i

that intersects O( √

n) disks and put these disks into the separator.

(18)

TU e

A geometric proof of the Planar Separator Theorem

Theorem. For any contact graph of n interior-disjoint disks, there is an α-balanced separator of size O( √

n), where α = 36/37.

Proof.

smallest square

containing at least n/37 disks

3 √ n squares σ

₁

, . . . , σ

^√_n

at distance 1/ √

n from each other Constructing the separator:

Select a square σ

_i

that intersects O ( √

n) disks and put these disks into the separator.

(19)

TU e

A geometric proof of the Planar Separator Theorem

Theorem. For any contact graph of n interior-disjoint disks, there is an α-balanced separator of size O( √

n), where α = 36/37.

Proof.

smallest square

containing at least n/37 disks

3 √ n squares σ

₁

, . . . , σ

^√_n

at distance 1/ √

n from each other Constructing the separator:

Select a square σ

_i

that intersects O ( √

n) disks and put these disks into the separator.

Things to check

• separator is (36/37)-balanced

• does square σ

_i

with the desired property actually exist ??

(20)

TU e

A geometric proof of the Planar Separator Theorem

Theorem. For any contact graph of n interior-disjoint disks, there is an α-balanced separator of size O( √

n), where α = 36/37.

Proof.

smallest square

containing at least n/37 disks

3 separator is (36/37)-balanced

(21)

TU e

A geometric proof of the Planar Separator Theorem

Theorem. For any contact graph of n interior-disjoint disks, there is an α-balanced separator of size O( √

n), where α = 36/37.

Proof.

smallest square

containing at least n/37 disks

3 separator is (36/37)-balanced

• at least n/37 disk inside

• at most 36n/37 disks inside

(22)

TU e

A geometric proof of the Planar Separator Theorem

Theorem. For any contact graph of n interior-disjoint disks, there is an α-balanced separator of size O( √

n), where α = 36/37.

Proof.

smallest square

containing at least n/37 disks

3 Does σ

_i

intersecting O ( √

n) disks exist?

total number of disk-square intersections

� �n_small

i=1 (1 + diam(Di) · √ n)

� nsmall + O(√

n) · �n_small i=1

�area(Di)

= O(n)

last step uses

• �n_small

i=1 area(Di) = O(1) (sort of . . . )

• �k i=1

√ai � �k i=1

�_�_k

i=1 a_i k

(23)

TU e

A geometric proof of the Planar Separator Theorem

Theorem. For any contact graph of n interior-disjoint disks, there is an α-balanced separator of size O( √

n), where α = 36/37.

Proof.

smallest square

containing at least n/37 disks

3 Does σ

_i

intersecting O ( √

n) disks exist?

� �n_small

i=1 (1 + diam(Di) · √ n)

� nsmall + O(√

�area(Di)

= O(n)

(24)

TU e

A geometric proof of the Planar Separator Theorem

Theorem. For any contact graph of n interior-disjoint disks, there is an α-balanced separator of size O( √

n), where α = 36/37.

Proof.

smallest square

containing at least n/37 disks

3 Does σ

_i

intersecting O ( √

n) disks exist?

� �n_small

i=1 (1 + diam(Di) · √ n)

� nsmall + O(√

�area(Di)

= O(n)

=⇒ one of the σi’s intersects O(√

n) disks

(25)

TU e

Subexponential algorithms on planar graphs

Theorem. Independent Set can be solved in 2

^O(^√ⁿ⁾

time in planar graphs.

(26)

TU e

Subexponential algorithms on planar graphs

Theorem. Independent Set can be solved in 2

^O(^√ⁿ⁾

time in planar graphs.

1. Compute (2/3)-balanced separator S of size O(√ n).

2. For each independent set I_S ⊆ S (including empty set) do

(a) Recursively ﬁnd max independent set I_A for A \ {neighbors of I_S} (b) Recursively ﬁnd max independent set I_B for B \ {neighbors of I_S}

(c) I(S) := I_S ∪ I_A ∪ I_B

3. Return the largest of the sets I(S) found in Step 2.

(27)

TU e

Subexponential algorithms on planar graphs

Theorem. Independent Set can be solved in 2

^O(^√ⁿ⁾

time in planar graphs.

(c) I(S) := I_S ∪ I_A ∪ I_B

(28)

TU e

Subexponential algorithms on planar graphs

Theorem. Independent Set can be solved in 2

^O(^√ⁿ⁾

time in planar graphs.

(c) I(S) := I_S ∪ I_A ∪ I_B

(29)

TU e

Subexponential algorithms on planar graphs

Theorem. Independent Set can be solved in 2

^O(^√ⁿ⁾

time in planar graphs.

(c) I(S) := I_S ∪ I_A ∪ I_B

(30)

TU e

Subexponential algorithms on planar graphs

Theorem. Independent Set can be solved in 2

^O(^√ⁿ⁾

time in planar graphs.

(c) I(S) := I_S ∪ I_A ∪ I_B

(31)

TU e

Subexponential algorithms on planar graphs

Theorem. Independent Set can be solved in 2

^O(^√ⁿ⁾

time in planar graphs.

(c) I(S) := I_S ∪ I_A ∪ I_B

(32)

TU e

Subexponential algorithms on planar graphs

Theorem. Independent Set can be solved in 2

^O(^√ⁿ⁾

time in planar graphs.

(c) I(S) := I_S ∪ I_A ∪ I_B

(33)

TU e

Subexponential algorithms on planar graphs

Theorem. Independent Set can be solved in 2

^O(^√ⁿ⁾

time in planar graphs.

(c) I(S) := I_S ∪ I_A ∪ I_B

(34)

TU e

Subexponential algorithms on planar graphs

Theorem. Independent Set can be solved in 2

^O(^√ⁿ⁾

time in planar graphs.

(c) I(S) := I_S ∪ I_A ∪ I_B

(35)

TU e

Subexponential algorithms on planar graphs

Theorem. Independent Set can be solved in 2

^O(^√ⁿ⁾

time in planar graphs.

(c) I(S) := I_S ∪ I_A ∪ I_B

(36)

TU e

Subexponential algorithms on planar graphs

Theorem. Independent Set can be solved in 2

^O(^√ⁿ⁾

time in planar graphs.

(c) I(S) := I_S ∪ I_A ∪ I_B

(37)

TU e

Subexponential algorithms on planar graphs

Theorem. Independent Set can be solved in 2

^O(^√ⁿ⁾

time in planar graphs.

Running time

(38)

TU e

Subexponential algorithms on planar graphs

Theorem. Independent Set can be solved in 2

^O(^√ⁿ⁾

time in planar graphs.

T (n) � O(n) + 2

^O(^√ⁿ⁾

· T (2n/3) = ⇒ T (n) = 2

^O(^√ⁿ⁾

Running time

(39)

Overview

• Planar separator theorem (slides by Mark de Berg)

• Indepedent set in planar graphs (slides by MdB)

• Shifting strategy: approximation schemes

• Exact algroithms for packing and covering

(40)

Intersection graphs

Given a set S of n objects in R

^d

, their intersection graph has vertex set S and edge set

E [S ] := {ss

⁰

| s, s

⁰

∈ S and s ∩ s

⁰

6= ∅}

(41)

Intersection graphs

Given a set S of n objects in R

^d

, their intersection graph has vertex set S and edge set

E [S ] := {ss

⁰

| s, s

⁰

∈ S and s ∩ s

⁰

6= ∅}

arbitrary subset of R

^d

ball (disk) axis-parallel box

(42)

Intersection graphs

Given a set S of n objects in R

^d

, their intersection graph has vertex set S and edge set

E [S ] := {ss

⁰

| s, s

⁰

∈ S and s ∩ s

⁰

6= ∅}

arbitrary subset of R

^d

ball (disk) axis-parallel box

Planar graphs ⊂ Disk graphs (object: disks in R

²

)

(43)

Packing: discrete vs continuous

Continuous:

Given n objects, do they fit

in some other object without

overlap?

(44)

Packing: discrete vs continuous

Continuous:

Given n objects, do they fit

in some other object without

overlap?

(45)

Packing: discrete vs continuous

Continuous:

Given n objects, do they fit

in some other object without

overlap?

(46)

Packing: discrete vs continuous

Continuous: Discrete:

Given n objects, do they fit in some other object without overlap?

Given n objects, find maximum subset of

non-overlapping objects

(47)

Packing: discrete vs continuous

Continuous: Discrete:

Given n objects, do they fit in some other object without overlap?

Given n objects, find maximum subset of

non-overlapping objects

(48)

Packing: discrete vs continuous

Continuous: Discrete:

Given n objects, do they fit in some other object without overlap?

Given n objects, find maximum subset of

non-overlapping objects

Same as max. independent

set in intersection graph

(49)

Exact algorithm for discrete packing

Theorem. Independent set in intersection graphs of disks can be solved in n

^O⁽

√k)

time, where k = size of max indep. set.

(50)

Exact algorithm for discrete packing

Theorem. Independent set in intersection graphs of disks can be solved in n

^O⁽

√k)

time, where k = size of max indep. set.

Proof.

Solution I has k interior-disjoint disks.

There is a balanced separator square σ intersecting O( √

k)

disks from I .

(51)

Exact algorithm for discrete packing

Theorem. Independent set in intersection graphs of disks can be solved in n

^O⁽

√k)

time, where k = size of max indep. set.

Proof.

Solution I has k interior-disjoint disks.

There is a balanced separator square σ intersecting O( √ k) disks from I .

Claim. Given S , we can compute a

family Y of poly(n) squares containing

all attainable square separators of all

subsets of S .

(52)

Exact algorithm for discrete packing II

for each separator σ ∈ Y do

for each intersecting I

_σ

⊂ S of size O( √

k) do Remove disks in S intersecting σ

Remove neighbors of I

_σ

Recurse on disks inside σ

Recurse on disks outside σ

return largest indep. set found

(53)

Exact algorithm for discrete packing II

for each separator σ ∈ Y do

for each intersecting I

_σ

⊂ S of size O( √

k) do Remove disks in S intersecting σ

Remove neighbors of I

_σ

Recurse on disks inside σ Recurse on disks outside σ return largest indep. set found T (n, k) = poly(n) · n

^O⁽

√k)

· 2T

n, 36 37 k

(54)

Exact algorithm for discrete packing II

for each separator σ ∈ Y do

for each intersecting I

_σ

⊂ S of size O( √

k) do Remove disks in S intersecting σ

Remove neighbors of I

_σ

Recurse on disks inside σ Recurse on disks outside σ return largest indep. set found T (n, k) = poly(n) · n

^O⁽

√k)

· 2T

n, 36 37 k

T (n, k) = n

^c

√k+c

√

(36/37)k+c

√

(36/37)²k+...

= n

^O(

√k)

(55)

Is n

^O(

√k)

good?

General graphs: Independent set is NP-hard, has O(n

^k

k

²

) algo

(56)

Is n

^O(

√k)

good?

General graphs: Independent set is NP-hard, has O(n

^k

k

²

) algo Exponential-Time Hypothesis (ETH). There is γ > 0 such that there is no 2

^γn

algorithm for 3-SAT on n varaibles.

ETH ⇒ P6= NP

(57)

Is n

^O(

√k)

good?

General graphs: Independent set is NP-hard, has O(n

^k

k

²

) algo Exponential-Time Hypothesis (ETH). There is γ > 0 such that there is no 2

^γn

algorithm for 3-SAT on n varaibles.

ETH ⇒ P6= NP

Theorem. There is no f (k)n

^o(k)

algorithm for Independent

Set for any computable f , unless ETH fails.

(58)

Is n

^O(

√k)

good?

General graphs: Independent set is NP-hard, has O(n

^k

k

²

) algo Exponential-Time Hypothesis (ETH). There is γ > 0 such that there is no 2

^γn

algorithm for 3-SAT on n varaibles.

ETH ⇒ P6= NP

Theorem. There is no f (k)n

^o(k)

algorithm for Independent Set for any computable f , unless ETH fails.

Theorem. There is no f (k)n

^o(

√k)

algorithm for Independent

Set in planar graphs for any computable f , unless ETH fails.

(59)

Is n

^O(

√k)

good?

General graphs: Independent set is NP-hard, has O(n

^k

k

²

) algo Exponential-Time Hypothesis (ETH). There is γ > 0 such that there is no 2

^γn

algorithm for 3-SAT on n varaibles.

ETH ⇒ P6= NP

Theorem. There is no f (k)n

^o(k)

algorithm for Independent Set for any computable f , unless ETH fails.

Theorem. There is no f (k)n

^o(

√k)

algorithm for Independent Set in planar graphs for any computable ⇒ f , unless ETH fails.

Theorem. There is no f (k)n

^o(

√k)

algorithm for Independent

Set in disk graphs for any computable f , unless ETH fails.

(60)

Geometric set cover: discrete vs continuous

Set cover : given m subsets of {1, . . . , n}, are there k among them whose union is {1, . . . , n}

very hard, can’t be approximated efficiently

(61)

Geometric set cover: discrete vs continuous

Set cover : given m subsets of {1, . . . , n}, are there k among them whose union is {1, . . . , n}

very hard, can’t be approximated efficiently

Continuous:

Geometric set cover:

Given P ⊂ R

²

, can we cover

P with k unit disks?

(62)

Geometric set cover: discrete vs continuous

Set cover : given m subsets of {1, . . . , n}, are there k among them whose union is {1, . . . , n}

very hard, can’t be approximated efficiently

Continuous:

Geometric set cover:

Given P ⊂ R

²

, can we cover P with k unit disks?

similar to cont. k-center!

(63)

Geometric set cover: discrete vs continuous

Set cover : given m subsets of {1, . . . , n}, are there k among them whose union is {1, . . . , n}

very hard, can’t be approximated efficiently

Continuous:

Geometric set cover:

Given P ⊂ R

²

, can we cover P with k unit disks?

similar to cont. k-center!

Discrete:

Given P ⊂ R

²

and m unit

disks D , can we cover P with

k disks from D ?

(64)

Exact algroithms for covering

Theorem (Marx–Pilipczuk, 2015) Discrete geometric set cover with disks can be solved in m

^O⁽

√k)

poly(n) time, where

k = size of min cover.

(65)

Exact algroithms for covering

Theorem (Marx–Pilipczuk, 2015) Discrete geometric set cover with disks can be solved in m

^O⁽

√k)

poly(n) time, where k = size of min cover.

Proof based on guessing separator in solution’s Voronoi diagram.

Theorem (Marx–Pilipczuk, 2015). There is no f (k)(m + n)

^o(

√k)

algorithm for covering points with disks for

any computable f , unless ETH fails.

(66)

Shifting grids

Approximation schemes

Hochbaum–Maass 1985

(67)

PTASes

Definition. A polynomial time approximation scheme (PTAS) for a minimization problem is an algorithm, which given ε > 0 and the input instance, outputs a feasible solution of value at most (1 + ε)OP T in poly

_ε

(n) time.

E.g. possible running time: O (n

^1/ε

) or n

^O(2^1/ε⁾

(68)

PTASes

Definition. A polynomial time approximation scheme (PTAS) for a minimization problem is an algorithm, which given ε > 0 and the input instance, outputs a feasible solution of value at most (1 + ε)OP T in poly

_ε

(n) time.

E.g. possible running time: O (n

^1/ε

) or n

^O(2^1/ε⁾

related complexity classes: PTAS versus APX-hardness

P is APX-hard ⇒ P has no PTAS unless P=NP

(69)

PTASes

Definition. A polynomial time approximation scheme (PTAS) for a minimization problem is an algorithm, which given ε > 0 and the input instance, outputs a feasible solution of value at most (1 + ε)OP T in poly

_ε

(n) time.

E.g. possible running time: O (n

^1/ε

) or n

^O(2^1/ε⁾

related complexity classes: PTAS versus APX-hardness P is APX-hard ⇒ P has no PTAS unless P=NP

Example: Independent set is APX-hard on general graphs.

But! Independent set in planar graphs has a PTAS. (Baker ’83)

(70)

Packing unit disks via shifting

Theorem. The discrete packing of unit disks has a PTAS:

given n unit disks, we can compute an independent set of size

(1 − ε)OP T in n

^O(1/ε)

time.

(71)

Packing unit disks via shifting

Theorem. The discrete packing of unit disks has a PTAS:

given n unit disks, we can compute an independent set of size (1 − ε)OP T in n

^O(1/ε)

time.

Proof.

Grid of distance 2

⇒ each (open) disk intersects ≤ 1

horizontal and ≤ 1 vertical grid line

(72)

Packing unit disks via shifting

Theorem. The discrete packing of unit disks has a PTAS:

given n unit disks, we can compute an independent set of size (1 − ε)OP T in n

^O(1/ε)

time.

Proof.

Grid of distance 2

⇒ each (open) disk intersects ≤ 1 horizontal and ≤ 1 vertical grid line Let t = d2/εe.

For a shift (a, b) (a, b ∈ {0, . . . , t − 1}), select horizontal lines a, a +t, a +2t, . . . select vertical lines b, b + t, b+2t, . . .

a

a + t

a + 2t

b b + t b + 2t

(73)

Packing unit disks via shifting

Theorem. The discrete packing of unit disks has a PTAS:

given n unit disks, we can compute an independent set of size (1 − ε)OP T in n

^O(1/ε)

time.

Proof.

Grid of distance 2

⇒ each (open) disk intersects ≤ 1 horizontal and ≤ 1 vertical grid line Let t = d2/εe.

For a shift (a, b) (a, b ∈ {0, . . . , t − 1}), select horizontal lines a, a +t, a +2t, . . . select vertical lines b, b + t, b+2t, . . .

a

a + t

a + 2t

b b + t b + 2t

Remove disks intersecting selected lines

(74)

a

a + t

a + 2t

b b + t b + 2t

Shifting startegy: solving cells

(75)

a

a + t

a + 2t

b b + t b + 2t

Shifting startegy: solving cells

Large cells have area O(1/ε

²

)

⇒ max indep. set has size k = O(1/ε

²

)

⇒ max indep. set found in n

^O(

√k)

= n

^O(1/ε)

time.

(76)

a

a + t

a + 2t

b b + t b + 2t

Shifting startegy: solving cells

Large cells have area O(1/ε

²

)

⇒ max indep. set has size k = O(1/ε

²

)

⇒ max indep. set found in n

^O(

√k)

= n

^O(1/ε)

time.

Claim. The union of large cell solutions has size at least

(1 − ε)OP T for some shift (a, b).

(77)

a

a + t

a + 2t

b b + t b + 2t

Shifting startegy: solving cells

Large cells have area O(1/ε

²

)

⇒ max indep. set has size k = O(1/ε

²

)

⇒ max indep. set found in n

^O(

√k)

= n

^O(1/ε)

time.

Proof. Of the t = d2/εe shifts for horizontals, there is some a ∈ {0, . . . , t − 1} intersecting ≤

^ε₂

OP T solution disks.

Similarly there is b s.t. verticals intersect ≤

₂^ε

OP T .

⇒ (a, b) works.

Claim. The union of large cell solutions has size at least

(1 − ε)OP T for some shift (a, b).

(78)

Discrete packing outlook

• Extends to unit balls in higher dimensions: n

^O(1/ε^d−1⁾

• n

^O(1/ε)

is essentially tight in R

²

(Marx 2007)

• Local search: slower PTAS for “pseudodisks” (last lecture?)

(79)

Discrete packing outlook

• Extends to unit balls in higher dimensions: n

^O(1/ε^d−1⁾

• n

^O(1/ε)

is essentially tight in R

²

(Marx 2007)

• Local search: slower PTAS for “pseudodisks” (last lecture?)

But! major open problem:

Is there a PTAS for Independent set of axis-parallel rectangles?

or for axis parallel segments?

(80)

Discrete packing outlook

• Extends to unit balls in higher dimensions: n

^O(1/ε^d−1⁾

• n

^O(1/ε)

is essentially tight in R

²

(Marx 2007)

• Local search: slower PTAS for “pseudodisks” (last lecture?)

But! major open problem:

Is there a PTAS for Independent set of axis-parallel rectangles?

or for axis parallel segments?

Best known: n

^{O((log log} ^n/ε)⁴⁾

(Chuzhoy–Ene 2016)

(81)

Continous covering: canonizing

Theorem. There is a PTAS for the coninuous covering of

points with unit disks with running time n

^O(1/ε)

.

(82)

Continous covering: canonizing

Theorem. There is a PTAS for the coninuous covering of points with unit disks with running time n

^O(1/ε)

.

Proof. A unit disk is canonical if it has 2 input points on its

boundary, or its topmost point is an input point.

(83)

Continous covering: canonizing

Theorem. There is a PTAS for the coninuous covering of points with unit disks with running time n

^O(1/ε)

.

Proof. A unit disk is canonical if it has 2 input points on its boundary, or its topmost point is an input point.

There is a cover of size k ⇔ there is a canonical cover of size k.

2 disks per point pair p, p

⁰

∈ P , one disk for each p ∈ P

2

ⁿ₂

+ n ≤ n

²

canonical disks

(84)

Shifting for set cover with unit disks

a

a + t

a + 2t

b b + t b + 2t

Grid of side length 2, set t = d6/εe Cell disks: canonical disks inside

and those intersecting the boundary

(85)

Shifting for set cover with unit disks

a

a + t

a + 2t

b b + t b + 2t

Grid of side length 2, set t = d6/εe Cell disks: canonical disks inside

and those intersecting the boundary

(86)

Shifting for set cover with unit disks

a

a + t

a + 2t

b b + t b + 2t

Grid of side length 2, set t = d6/εe Cell disks: canonical disks inside

and those intersecting the boundary Whole cell can be covered by

O(1/ε

²

) (non-canoncical) disks.

⇒ Min cover in a cell solved in (n

²

)

^O⁽

√

1/ε²)

= n

^O(1/ε)

In C, solution |S (C )| ≤ |OP T (C )|. Return U := S

C

S (C )

(87)

Shifting for set cover with unit disks

a

a + t

a + 2t

b b + t b + 2t

Grid of side length 2, set t = d6/εe Cell disks: canonical disks inside

and those intersecting the boundary Whole cell can be covered by

O(1/ε

²

) (non-canoncical) disks.

⇒ Min cover in a cell solved in (n

²

)

^O⁽

√

1/ε²)

= n

^O(1/ε)

In C, solution |S (C )| ≤ |OP T (C )|. Return U := S

C

S (C ) For some shift a blue intersects ≤ |OP T |/t disks.

⇒ ∃(a, b) intersecting 2|OP T |/t ≤ ε|OP T |/3 disks.

Each disk of OPT counted in ≤ 4 cells.

|U | ≤ X

C