On independent set on B1

On independent set on B

-EPG graphs

Marin Bougeret (Lirmm, Montpellier, France) Joint Work with S. Bessy, D.Goncalves, C. Paul

WAOA 2015

Onlindependent set on B

-EPG graphs

Marin Bougeret (Lirmm, Montpellier, France) Joint Work with S. Bessy, D.Goncalves, C. Paul

WAOA 2015

On independent set on B

-EPG graphs

Marin Bougeret (Lirmm, Montpellier, France) Joint Work with S. Bessy, D.Goncalves, C. Paul

WAOA 2015

Outline

1 Introduction on EPG-Graphs

2 Approximability

3 Fixed parameter tractability

4 / 29

Definition of EPG-Graphs

EPG (for Edge intersection graphs of Paths on a Grid) graphs introduced in [GLS09]

In EPG-graphG = (V,E):

each vertexv corresponds to a pathPv

{u,v} ∈E iff Pu,Pv share a grid edge

3 4 1

2

an EPG-representation Gis a{xyq}-EPG graph

P1

P2

P3

P4

G

B_k-EPG: graphs having a representation where every path has at most k bends

X-EPG ⊆B₁-EPG (with X ⊆ {p,q,x,y}): paths can only have shapes inX

5 / 29

Definition of EPG-Graphs

EPG (for Edge intersection graphs of Paths on a Grid) graphs introduced in [GLS09]

In EPG-graphG = (V,E):

each vertexv corresponds to a pathPv

{u,v} ∈E iff Pu,Pv share a grid edge

3 4 1

2

an EPG-representation Gis a{xyq}-EPG graph

P1

P2

P3

P4

G

B_k-EPG: graphs having a representation where every path has at most k bends

X-EPG ⊆B₁-EPG (with X ⊆ {p,q,x,y}): paths can only have shapes inX

5 / 29

Properties of EPG-Graphs

Class inclusions

max deg ∆⊆B_∆-EPG [HKU14]

B0-EPG = interval graphs

B₁-EPG ⊂2-Track graphs⊂B₃-EPG

B₁-EPG K_3,3 induced free, K_3,3−e induced free, S_n≥4 induced free [GLS09]

2−Track representation Track A

Track B I₄^B I₂^B I₁^B

I₁^A I₃^A I₄^A

row 2 row 3 horizontal track

vertical track

horizontal track

vertical track

⇒

column 1

row 1

a

6 / 29

Properties of EPG-Graphs

Class inclusions

max deg ∆⊆B_∆-EPG [HKU14]

B0-EPG = interval graphs

B₁-EPG ⊂2-Track graphs⊂B₃-EPG

B₁-EPG K_3,3 induced free, K_3,3−e induced free, S_n≥4 induced free [GLS09]

2−Track representation Track A

Track B I₄^B I₂^B I₁^B

I₁^A I₃^A I₄^A

row 2 row 3 horizontal track

vertical track

horizontal track

vertical track

⇒

column 1

row 1

a

6 / 29

Properties of EPG-Graphs

Class inclusions

max deg ∆⊆B_∆-EPG [HKU14]

B0-EPG = interval graphs

B₁-EPG ⊂2-Track graphs⊂B₃-EPG

B₁-EPG K_3,3 induced free, K_3,3−e induced free, S_n≥4 induced free [GLS09]

2−Track representation Track A

Track B I₄^B I₂^B I₁^B

I₁^A I₃^A I₄^A

row 2 row 3 horizontal track

vertical track

horizontal track

vertical track

⇒

column 1

row 1

a

6 / 29

Properties of EPG-Graphs

Class inclusions

max deg ∆⊆B_∆-EPG [HKU14]

B0-EPG = interval graphs

B₁-EPG ⊂2-Track graphs⊂B₃-EPG

B₁-EPG K_3,3 induced free, K_3,3−e induced free, S_n≥4 induced free [GLS09]

2−Track representation Track A

Track B I₄^B I₂^B I₁^B

I₁^A I₃^A I₄^A

row 2 row 3 horizontal track

vertical track

horizontal track

vertical track

⇒

column 1

row 1

a

6 / 29

Properties of EPG-Graphs

Bad news aboutB1-EPG

B₁-EPG are not planar graphs (K_n∈ B₁-EPG)

B₁-EPG are not perfect graphs (C_n∈B₁-EPG) (but G[N(v)]

weakly chordal)

B₁-EPG do not benefit from the many results on "intersection graph class" as two paths can cross without creating an edge

N(v ) is a C

v

7 / 29

Properties of EPG-Graphs

Bad news aboutB1-EPG

B₁-EPG are not planar graphs (K_n∈ B₁-EPG)

B₁-EPG are not perfect graphs (C_n∈B₁-EPG) (but G[N(v)]

weakly chordal)

B₁-EPG do not benefit from the many results on "intersection graph class" as two paths can cross without creating an edge

N(v ) is a C

v

7 / 29

Properties of EPG-Graphs

Bad news aboutB1-EPG

B₁-EPG are not planar graphs (K_n∈ B₁-EPG)

B₁-EPG are not perfect graphs (C_n∈B₁-EPG) (but G[N(v)]

weakly chordal)

B₁-EPG do not benefit from the many results on "intersection graph class" as two paths can cross without creating an edge

N(v ) is a C

v

7 / 29

Motivation

Considered problem: MIS (Maximum Independent Set) on B₁-EPG graphs (supposing a representation is given)

In 2013, [EGM13] proved that MIS (and coloring) are NP-hard on B₁-EPG and admits a 4-approximation algorithm

Can we say more ? (approximability and fixed parameter tractability with standard parameterization)

8 / 29

Motivation

Considered problem: MIS (Maximum Independent Set) on B₁-EPG graphs (supposing a representation is given)

In 2013, [EGM13] proved that MIS (and coloring) are NP-hard on B₁-EPG and admits a 4-approximation algorithm

Can we say more ? (approximability and fixed parameter tractability with standard parameterization)

8 / 29

Motivation

Considered problem: MIS (Maximum Independent Set) on B₁-EPG graphs (supposing a representation is given)

In 2013, [EGM13] proved that MIS (and coloring) are NP-hard on B₁-EPG and admits a 4-approximation algorithm

Can we say more ? (approximability and fixed parameter tractability with standard parameterization)

8 / 29

Outline

1 Introduction on EPG-Graphs

2 Approximability

3 Fixed parameter tractability

9 / 29

Outline approximability

Related work

simple 4 approximation onB₁-EPG ([EGM13] or [BYHN⁺06]) no PTAS for MIS on 2-Track as max deg∆ =3⊆2-Track (recallB₁-EPG ⊂2-Track)

Our contributions

no PTAS for MIS on {p}-EPG, even if each path has its vertical part or its horizontal part of length at most 3 PTAS for MIS on B₁-EPG when each path has its horizontal part at mostc

10 / 29

Outline approximability

Related work

simple 4 approximation onB₁-EPG ([EGM13] or [BYHN⁺06]) no PTAS for MIS on 2-Track as max deg∆ =3⊆2-Track (recallB₁-EPG ⊂2-Track)

Our contributions

no PTAS for MIS on {p}-EPG, even if each path has its vertical part or its horizontal part of length at most 3 PTAS for MIS on B₁-EPG when each path has its horizontal part at mostc

10 / 29

Outline approximability

Related work

simple 4 approximation onB₁-EPG ([EGM13] or [BYHN⁺06]) no PTAS for MIS on 2-Track as max deg∆ =3⊆2-Track (recallB₁-EPG ⊂2-Track)

Our contributions

no PTAS for MIS on {p}-EPG, even if each path has its vertical part or its horizontal part of length at most 3 PTAS for MIS on B₁-EPG when each path has its horizontal part at mostc

10 / 29

Negative result

Lemma: (no PTAS for{p,q}-EPG)

There is a strict reduction from MAX-3-SAT(3) to{p,q}-EPG

Proof

Consider first the textbook reduction from MAX-3-SAT to MIS as : create one triangle for each clause

add an edge between any occurrence of literal and its negation assignment satisfying t clauses⇔ IS size t

As we reduce from MAX-3-SAT(3), the graph is{p,q}-EPG

C1 C2 C3

C1=not(x1)∨x2∨x3

C2=x1∨not(x2)∨x3

C3=x1∨x2∨not(x3)

x1 x1

not(x1) x3

x2 not(x2) x3 x2 not(x3)

11 / 29

Negative result

Lemma: (no PTAS for{p,q}-EPG)

There is a strict reduction from MAX-3-SAT(3) to{p,q}-EPG

Proof

Consider first the textbook reduction from MAX-3-SAT to MIS as : create one triangle for each clause

add an edge between any occurrence of literal and its negation assignment satisfying t clauses⇔ IS size t

As we reduce from MAX-3-SAT(3), the graph is{p,q}-EPG

C1 C2 C3

C1=not(x1)∨x2∨x3

C2=x1∨not(x2)∨x3

C3=x1∨x2∨not(x3)

x1 x1

not(x1) x3

x2 not(x2) x3 x2 not(x3)

11 / 29

Negative result

Lemma: (no PTAS for{p,q}-EPG)

There is a strict reduction from MAX-3-SAT(3) to{p,q}-EPG

Proof

Consider first the textbook reduction from MAX-3-SAT to MIS as : create one triangle for each clause

add an edge between any occurrence of literal and its negation assignment satisfying t clauses⇔ IS size t

As we reduce from MAX-3-SAT(3), the graph is{p,q}-EPG

C1 C2 C3

C1=not(x1)∨x2∨x3

C2=x1∨not(x2)∨x3

C3=x1∨x2∨not(x3)

x1 x1

not(x1) x3

x2 not(x2) x3 x2 not(x3)

11 / 29

Negative result

Lemma: (no PTAS for{p,q}-EPG)

There is a strict reduction from MAX-3-SAT(3) to{p,q}-EPG

Proof

Consider first the textbook reduction from MAX-3-SAT to MIS as : create one triangle for each clause

add an edge between any occurrence of literal and its negation assignment satisfying t clauses⇔ IS size t

As we reduce from MAX-3-SAT(3), the graph is {p,q}-EPG

C1 C2 C3

C1=not(x1)∨x2∨x3

C2=x1∨not(x2)∨x3

C3=x1∨x2∨not(x3)

x1 x1

not(x1) x3

x2 not(x2) x3 x2 not(x3)

C1 C2 C3

Line used forx1andnot(x1) Line used forx2andnot(x2) Line used forx3andnot(x3)

11 / 29

Negative result

How removing one type of shape ?

Given the previousG, subdivide 4 times each "literral" edge to get a new graph G⁰

G⁰ can now be drawn as{p}-EPG

∃ indep. set. S inG ⇔ ∃indep. set. S⁰ in G⁰ with

|S⁰|=|S|+2m

As ∃c such that m≤cOpt(G), we get an AP-reduction for MIS from{p,q}-EPG to{p}-EPG

C1 C2 C3

Line used forx3andnot(x3)

C1 C2 C3

x1 x1

not(x1) x3

x2 not(x2) x3 x2 not(x3)

12 / 29

Negative result

How removing one type of shape ?

Given the previousG, subdivide 4 times each "literral" edge to get a new graph G⁰

G⁰ can now be drawn as{p}-EPG

∃ indep. set. S inG ⇔ ∃indep. set. S⁰ in G⁰ with

|S⁰|=|S|+2m

As ∃c such that m≤cOpt(G), we get an AP-reduction for MIS from{p,q}-EPG to{p}-EPG

C1 C2 C3

Line used forx3andnot(x3)

C1 C2 C3

x1 x1

not(x1) x3

x2 not(x2) x3 x2 not(x3)

12 / 29

Negative result

How removing one type of shape ?

Given the previousG, subdivide 4 times each "literral" edge to get a new graph G⁰

G⁰ can now be drawn as{p}-EPG

∃ indep. set. S inG ⇔ ∃indep. set. S⁰ in G⁰ with

|S⁰|=|S|+2m

As ∃c such that m≤cOpt(G), we get an AP-reduction for MIS from{p,q}-EPG to{p}-EPG

C1 C2 C3

Line used forx3andnot(x3)

C1 C2 C3

x1 x1

not(x1) x3

x2 not(x2) x3 x2 not(x3)

12 / 29

Negative result

How removing one type of shape ?

Given the previousG, subdivide 4 times each "literral" edge to get a new graph G⁰

G⁰ can now be drawn as{p}-EPG

∃ indep. set. S inG ⇔ ∃indep. set. S⁰ in G⁰ with

|S⁰|=|S|+2m

As ∃c such that m≤cOpt(G), we get an AP-reduction for MIS from{p,q}-EPG to{p}-EPG

C1 C2 C3

x1 x1

not(x1) x3

x2 not(x2) x3 x2 not(x3)

Lines used forx3andnot(x3)

12 / 29

Negative result

How removing one type of shape ?

Given the previousG, subdivide 4 times each "literral" edge to get a new graph G⁰

G⁰ can now be drawn as{p}-EPG

∃ indep. set. S inG ⇔ ∃indep. set. S⁰ in G⁰ with

|S⁰|=|S|+2m

As ∃c such that m≤cOpt(G), we get an AP-reduction for MIS from{p,q}-EPG to{p}-EPG

C1 C2 C3

x1 x1

not(x1) x3

x2 not(x2) x3 x2 not(x3)

Lines used forx3andnot(x3)

12 / 29

Negative result

How removing one type of shape ?

Given the previousG, subdivide 4 times each "literral" edge to get a new graph G⁰

G⁰ can now be drawn as{p}-EPG

∃ indep. set. S inG ⇔ ∃indep. set. S⁰ in G⁰ with

|S⁰|=|S|+2m

As ∃c such that m≤cOpt(G), we get an AP-reduction for MIS from{p,q}-EPG to{p}-EPG

C₁ C₂ C₃

x1 x1

not(x1) x3

x2 not(x2) x3 x2 not(x3)

Lines used forx3andnot(x3)

Theorem

There is no PTAS for MIS on{p}-EPG, even if each path has its vertical part or its horizontal part of length at most 3

12 / 29

Negative result

Can we improve this when both parts have constant size ? No.

Theorem

There is PTAS for MIS onB₁-EPG when each path has its horizontal part at mostc (which remains NP-hard)

Proof

We will prove this using classical Baker shifting technique

13 / 29

Positive result

Proof

Letk be a large integer (goal: sol of size|S| ≥ |OPT|(1−¹_k)) Givend ∈[kc−1], letR_d be the set of paths whose horizontal part crosses the vertical stripd,d +kc,d +2kc, ..

Let OPT_d =OPT ∩R_d

∈/R_d

kc kc

∈/R_d

kc

∈R_d

∈R_d

d+kc d+2kc

0 c d

14 / 29

Positive result

Proof

Letk be a large integer (goal: sol of size|S| ≥ |OPT|(1−¹_k)) Givend ∈[kc−1], letR_d be the set of paths whose horizontal part crosses the vertical stripd,d +kc,d +2kc, ..

Let OPT_d =OPT ∩R_d

∈/R_d

kc kc

∈/R_d

kc

∈R_d

∈R_d

d +kc d+2kc

0 c d

14 / 29

Positive result

Proof

Letk be a large integer (goal: sol of size|S| ≥ |OPT|(1−¹_k)) Givend ∈[kc−1], letR_d be the set of paths whose horizontal part crosses the vertical stripd,d +kc,d +2kc, ..

Let OPT_d =OPT ∩R_d

∈/R_d

kc kc

∈/R_d

kc

∈R_d

∈R_d

d +kc d+2kc

0 c d

14 / 29

Positive result

Lemma

∃d₀ such that |OPT_d0| ≤ ¹_k|OPT|

Proof Pkc−1

d=0 |OPT_d| ≤c|OPT|(as each vertex belongs to at most c different R_d)

as |OPT_d0|kc≤Pkc−1

d=0 |OPT_d|, we get the result

15 / 29

Positive result

Proof

Back to the PTAS proof:

thus, for each d we solve the problem optimally onG \R_d, and we output A: the best of these solutions

Previous Lemma ⇒ |A| ≥(1− ¹_k)|OPT|

16 / 29

Positive result

Proof

Back to the PTAS proof:

thus, for each d we solve the problem optimally onG \R_d, and we output A: the best of these solutions

Previous Lemma ⇒ |A| ≥(1− ¹_k)|OPT|

16 / 29

Positive result

Proof

It remains to solve the problem optimally onG\R_d:

G \R_d has several connected componentCC_l, where each component is drawn on a grid of constant widthkc

we solve optimally on each CC_l using a dyn. prog. algorithm

kc kc

kc

∈R_d

d+kc d+2kc

0 d

CC₁ CC₂ CC₃

c

17 / 29

Positive result

Proof

It remains to solve the problem optimally onG\R_d:

G \R_d has several connected componentCC_l, where each component is drawn on a grid of constant widthkc

we solve optimally on each CC_l using a dyn. prog. algorithm

kc kc

kc

∈R_d

d +kc d+2kc

0 d

CC₁ CC₂ CC₃

c

17 / 29

Positive result

Proof

It remains to solve the problem optimally onG\R_d:

G \R_d has several connected componentCC_l, where each component is drawn on a grid of constant widthkc

we solve optimally on each CC_l using a dyn. prog. algorithm

kc kc

kc

∈R_d

d +kc d+2kc

0 d

CC₁ CC₂ CC₃

c

17 / 29

Outline

1 Introduction on EPG-Graphs

2 Approximability

3 Fixed parameter tractability

18 / 29

Outline fixed parameter tractability

We consider the problemOPT ≤k?parameterized by k.

Question: is this problem FPT ? (i.e. can be solved inf(k)poly(n))

Related work

B₁-EPG ⊂2-Track ⊂B₃-EPG+MIS is W₁-hard on unit 2-Track [Jia10]⇒ MIS isW₁-hard onB₃-EPG

Our contributions

MIS is FPT on {x,y,p}-EPG MIS is W₁-hard on B₂-EPG

We will prove that MIS is FPT on{x}-EPG using a branching algorithm.

Definition

Given a pathP, let cor(P) = (x,y) be the coordinates of the corner ofP in the grid (orienting as usual ↑_y →_x)

19 / 29

Outline fixed parameter tractability

We consider the problemOPT ≤k?parameterized by k.

Question: is this problem FPT ? (i.e. can be solved inf(k)poly(n))

Related work

B₁-EPG ⊂2-Track⊂ B₃-EPG+MIS is W₁-hard on unit 2-Track [Jia10]⇒ MIS isW₁-hard onB₃-EPG

Our contributions

MIS is FPT on {x,y,p}-EPG MIS is W₁-hard on B₂-EPG

We will prove that MIS is FPT on{x}-EPG using a branching algorithm.

Definition

Given a pathP, let cor(P) = (x,y) be the coordinates of the corner ofP in the grid (orienting as usual ↑_y →_x)

19 / 29

Outline fixed parameter tractability

We consider the problemOPT ≤k?parameterized by k.

Question: is this problem FPT ? (i.e. can be solved inf(k)poly(n))

Related work

B₁-EPG ⊂2-Track⊂ B₃-EPG+MIS is W₁-hard on unit 2-Track [Jia10]⇒ MIS isW₁-hard onB₃-EPG

Our contributions

MIS is FPT on {x,y,p}-EPG MIS is W₁-hard onB₂-EPG

We will prove that MIS is FPT on{x}-EPG using a branching algorithm.

Definition

Given a pathP, let cor(P) = (x,y) be the coordinates of the corner ofP in the grid (orienting as usual ↑_y →_x)

19 / 29

Outline fixed parameter tractability

We consider the problemOPT ≤k?parameterized by k.

Question: is this problem FPT ? (i.e. can be solved inf(k)poly(n))

Related work

B₁-EPG ⊂2-Track⊂ B₃-EPG+MIS is W₁-hard on unit 2-Track [Jia10]⇒ MIS isW₁-hard onB₃-EPG

Our contributions

MIS is FPT on {x,y,p}-EPG MIS is W₁-hard onB₂-EPG

We will prove that MIS is FPT on{x}-EPG using a branching algorithm.

Definition

Given a pathP, let cor(P) = (x,y) be the coordinates of the corner ofP in the grid (orienting as usual ↑_y →_x)

19 / 29

MIS is FPT on { x }-EPG

Proof

Let P_v0 (cor(P_v0) = (x₀,y₀)): a path whose corner is

"top-right most" (highest line of the grid, and then the right most)

Suppose |OPT|=k

We know that∃P_v^∗ ∈N[P_v0]∩OPT (cor(P_v^∗) = (x^∗,y^∗)) Let us find this P_v^∗ (or aP⁰ ∈OPT⁰) using at most f(k) branches

y₀

x₀ P_v0

20 / 29

MIS is FPT on { x }-EPG

Proof

Let P_v0 (cor(P_v0) = (x₀,y₀)): a path whose corner is

"top-right most" (highest line of the grid, and then the right most)

Suppose |OPT|=k

We know that∃P_v^∗ ∈N[P_v0]∩OPT (cor(P_v^∗) = (x^∗,y^∗)) Let us find this P_v^∗ (or aP⁰ ∈OPT⁰) using at most f(k) branches

y₀

x₀ P_v0

20 / 29

MIS is FPT on { x }-EPG

Proof

Let P_v0 (cor(P_v0) = (x₀,y₀)): a path whose corner is

"top-right most" (highest line of the grid, and then the right most)

Suppose |OPT|=k

We know that∃P_v^∗ ∈N[P_v0]∩OPT (cor(P_v^∗) = (x^∗,y^∗)) Let us find this P_v^∗ (or aP⁰ ∈OPT⁰) using at most f(k) branches

y₀

x₀ P_v0

20 / 29

MIS is FPT on { x }-EPG

Proof

Let P_v0 (cor(P_v0) = (x₀,y₀)): a path whose corner is

"top-right most" (highest line of the grid, and then the right most)

Suppose |OPT|=k

We know that∃P_v^∗ ∈N[P_v0]∩OPT (cor(P_v^∗) = (x^∗,y^∗)) Let us find this P_v^∗ (or aP⁰ ∈OPT⁰) using at most f(k) branches

y₀

x₀ P_v0

20 / 29

MIS is FPT on { x }-EPG

Proof

Let P_v0 (cor(P_v0) = (x₀,y₀)): a path whose corner is

"top-right most" (highest line of the grid, and then the right most)

Suppose |OPT|=k

We know that∃P_v^∗ ∈N[P_v0]∩OPT (cor(P_v^∗) = (x^∗,y^∗)) Let us find this P_v^∗ (or aP⁰ ∈OPT⁰) using at most f(k) branches

y₀

x₀ P_v0

20 / 29

MIS is FPT on { x }-EPG

Proof

We branch on the following cases:

1 x^∗ =x₀ andy^∗ =y₀: easy, takeP_v0

2 x^∗ =x₀ andy^∗ <y₀

3 x^∗ <x₀ andy^∗ =y₀

Let us only treat case 2 here (case 3 works similarly)

y₀

x₀ P_v0

21 / 29

MIS is FPT on { x }-EPG

Proof

We branch on the following cases:

1 x^∗ =x₀ andy^∗ =y₀: easy, takeP_v0

2 x^∗ =x₀ andy^∗ <y₀

3 x^∗ <x₀ andy^∗ =y₀

Let us only treat case 2 here (case 3 works similarly)

y₀

x₀ P_v0

P_v^∗

21 / 29

MIS is FPT on { x }-EPG

Proof

We branch on the following cases:

1 x^∗ =x₀ andy^∗ =y₀: easy, takeP_v0

2 x^∗ =x₀ andy^∗ <y₀

3 x^∗ <x₀ andy^∗ =y₀

Let us only treat case 2 here (case 3 works similarly)

y₀

x₀ P_v0

P_v^∗

21 / 29

MIS is FPT on { x }-EPG

Proof

We branch on the following cases:

1 x^∗ =x₀ andy^∗ =y₀: easy, takeP_v0

2 x^∗ =x₀ andy^∗ <y₀

3 x^∗ <x₀ andy^∗ =y₀

Let us only treat case 2 here (case 3 works similarly)

y₀

x₀ P_v0 P_v^∗

21 / 29

MIS is FPT on { x }-EPG

Proof

We branch on the following cases:

1 x^∗ =x₀ andy^∗ =y₀: easy, takeP_v0

2 x^∗ =x₀ andy^∗ <y₀

3 x^∗ <x₀ andy^∗ =y₀

Let us only treat case 2 here (case 3 works similarly)

y₀

x₀ P_v0

P_v^∗

21 / 29

MIS is FPT on { x }-EPG

Proof

Let X be the "vertical" neighbors of P_v0 (P_v^∗ ∈X).

Let partition X =G¹∪G²..∪G^t

Sufficient to keep the left most pathPⁱ in each Gⁱ

X y₀

x₀ P_v0

G¹ G²

G^t

22 / 29

MIS is FPT on { x }-EPG

Proof

Let X be the "vertical" neighbors of P_v0 (P_v^∗ ∈X).

Let partition X =G¹∪G²..∪G^t

Sufficient to keep the left most pathPⁱ in each Gⁱ

y₀

x₀ P_v0

P¹ P²

P^t

22 / 29

MIS is FPT on { x }-EPG

Proof

Now, two cases are possible

1 P_v^∗ ∈ {P¹, . . . ,P^k}: easy, guess which Pⁱ to take

2 P_v^∗ ∈ {P^k+1, . . . ,P^t}:

as|OPT|=k, there existsPⁱ∈ {P¹, . . . ,P^k}such that there is noP⁰ ∈OPT in the "horizontal line" ofPⁱ

guess thisPⁱ and take it

y₀

x₀ P_v0

P¹ P²

P^t

23 / 29

MIS is FPT on { x }-EPG

Proof

Now, two cases are possible

1 P_v^∗ ∈ {P¹, . . . ,P^k}: easy, guess which Pⁱ to take

2 P_v^∗ ∈ {P^k+1, . . . ,P^t}:

as|OPT|=k, there existsPⁱ∈ {P¹, . . . ,P^k}such that there is noP⁰ ∈OPT in the "horizontal line" ofPⁱ

guess thisPⁱ and take it

y₀

x₀ P_v0

P¹ P²

P^t P^k

P_v^∗

23 / 29

MIS is FPT on { x }-EPG

Proof

Now, two cases are possible

1 P_v^∗ ∈ {P¹, . . . ,P^k}: easy, guess which Pⁱ to take

2 P_v^∗ ∈ {P^k+1, . . . ,P^t}:

as|OPT|=k, there existsPⁱ∈ {P¹, . . . ,P^k}such that there is noP⁰ ∈OPT in the "horizontal line" ofPⁱ

guess thisPⁱ and take it

y₀

x₀ P_v0

P¹ P²

P^t P^k

P_v^∗

∈OPT ∈OPT

23 / 29

MIS is FPT on { x }-EPG

Proof

Now, two cases are possible

1 P_v^∗ ∈ {P¹, . . . ,P^k}: easy, guess which Pⁱ to take

2 P_v^∗ ∈ {P^k+1, . . . ,P^t}:

as|OPT|=k, there existsPⁱ∈ {P¹, . . . ,P^k}such that there is noP⁰ ∈OPT in the "horizontal line" ofPⁱ

guess thisPⁱ and take it

y₀

x₀ P_v0

P¹ P²

P^t P^k

P_v^∗

∈OPT ∈OPT

Pi

23 / 29

Conclusion on FPT

Positive results

MIS is FPT on{x}-EPG as we can find aP_v^∗∈OPT using at most f(k) branches

The previous appproach doesn’t work (even for 2 shapes)..

but we use the same kind of arguments to prove that MIS is FPT with 3 shapes

Negative results

MIS is W₁-hard on B₂-EPG

24 / 29

Conclusion on FPT

Positive results

MIS is FPT on{x}-EPG as we can find aP_v^∗∈OPT using at most f(k) branches

The previous appproach doesn’t work (even for 2 shapes)..

but we use the same kind of arguments to prove that MIS is FPT with 3 shapes

Negative results

MIS is W₁-hard on B₂-EPG

24 / 29

Conclusion on FPT

Positive results

MIS is FPT on{x}-EPG as we can find aP_v^∗∈OPT using at most f(k) branches

The previous appproach doesn’t work (even for 2 shapes)..

but we use the same kind of arguments to prove that MIS is FPT with 3 shapes

Negative results

MIS is W₁-hard on B₂-EPG

24 / 29

Conclusion on FPT

Positive results

MIS is FPT on{x}-EPG as we can find aP_v^∗∈OPT using at most f(k) branches

The previous appproach doesn’t work (even for 2 shapes)..

but we use the same kind of arguments to prove that MIS is FPT with 3 shapes

Negative results

MIS is W₁-hard onB₂-EPG

24 / 29

Conclusion

Approximability

No PTAS unlessP =NP

PTAS when one side is always small Simple 4 approximation

Open: c <4 approximation ?

Fixed parameter tractability FPT with 3 shapes W₁-hard onB₂-EPG Open: FPT onB₁-EPG ?

25 / 29

Thank you for your attention ! ...

26 / 29

MIS isW₁-hard on B₂-EPG as:

0000 0000 0000 0000 0000 0000 0000 0000 00

1111 1111 1111 1111 1111 1111 1111 1111 11

0000 0000 0000 0000 0000 0000 00

1111 1111 1111 1111 1111 1111 11

0000 0000 0000 0000 0000 0000 00

1111 1111 1111 1111 1111 1111 11

0000 0000 0000 0000 0000 0000 00

1111 1111 1111 1111 1111 1111 11

0000 0000 0000 0000 00

1111 1111 1111 1111 11

000000 000000 000000 000000 000

111111 111111 111111 111111 111

0000 0000 0000 0000 0000 0000 0000 0000 00

1111 1111 1111 1111 1111 1111 1111 1111 11

000000 000000 000000 000000 000

111111 111111 111111 111111 111

0000 0000 0000 0000 0000 0000 0000 0000 00

1111 1111 1111 1111 1111 1111 1111 1111 11

0000 0000 0000 0000 0000 0000 0000 0000 00

1111 1111 1111 1111 1111 1111 1111 1111 11

0000 0000 0000 0000 0000 0000 00

1111 1111 1111 1111 1111 1111 11

0000 0000 0000 0000 00

1111 1111 1111 1111 11 000000

000000 000000 000000 000000 000000

111111 111111 111111 111111 111111 111111

000000 000000 000000 111111 111111 111111

000000 000000 000000 000000 000000 000000

111111 111111 111111 111111 111111 111111

000000 000000 000000 111111 111111 111111

linesL¹_3,x linesL²_3,x

linesL³_2,x T₃

T₂ T1

T2,3

T_1,3 T_1,2

pathsa⁰<sub>`</sub>anda`fromT1,2

pathsb<sub>`</sub><sup>0</sup>andb`fromT1,2

27 / 29

Bibliography

[BYHN⁺06] Reuven Bar-Yehuda, Magnús M Halldórsson, Joseph Seffi Naor, Hadas Shachnai, and Irina Shapira.

Scheduling split intervals.

SIAM J. Comput., 36(1):1–15, 2006.

[EGM13] Dror Epstein, MartinCharles Golumbic, and Gila Morgenstern.

Approximation algorithms forb₁-epg graphs.

In WADS 2013: Algorithms and Data Structures, volume 8037 of Lecture Notes in Computer Science, pages 328–340. Springer Berlin Heidelberg, 2013.

[GLS09] Martin Charles Golumbic, Marina Lipshteyn, and Michal Stern.

Edge intersection graphs of single bend paths on a grid.

Networks, 54(3):130–138, 2009.

[HKU14] Daniel Heldt, Kolja Knauer, and Torsten Ueckerdt.

Edge-intersection graphs of grid paths: The bend-number.

Discrete Applied Mathematics, 167(0):144 – 162, 2014.

[Jia10] Minghui Jiang.

On the parameterized complexity of some optim ization problems related to multiple-interval graphs.

Theoretical Computer Science, 411:4253–4262, 2010.

28 / 29

Why does it fail for 2 shapes ?

Pv^∗∈ {P_k+1, . . . ,Pt}

but we cannot restructure OPT to chose one of the{P₁, . . . ,Pk} P4

P3

P_v0

P1

Pk=P2

P_v^∗

29 / 29