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### Minicourse on parameterized algorithms and complexity Part 7: Parameterized complexity

Dániel Marx

Jagiellonian University in Kraków April 21-23, 2015

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### Lower bounds

So far we have seen positive results: basic algorithmic techniques for fixed-parameter tractability.

What kind of negative results we have?

Can we show that a problem (e.g., Clique) isnotFPT?

Can we show that a problem (e.g., Vertex Cover) hasno algorithm with running time, say, 2o(k)·nO(1)?

This would require showing thatP6=NP: ifP=NP, then, e.g., k-Clique is polynomial-time solvable, hence FPT.

Can we give some evidence for negative results?

2

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### Lower bounds

So far we have seen positive results: basic algorithmic techniques for fixed-parameter tractability.

What kind of negative results we have?

Can we show that a problem (e.g., Clique) isnotFPT?

Can we show that a problem (e.g., Vertex Cover) hasno algorithm with running time, say, 2o(k)·nO(1)?

This would require showing thatP6=NP: ifP=NP, then, e.g., k-Cliqueis polynomial-time solvable, hence FPT.

Can we give some evidence for negative results?

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### Goals of this talk

Two goals:

1 Explain the theory behind parameterized intractability.

2 Show examples of parameterized reductions.

3

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### Classical complexity

Nondeterministic Turing Machine (NTM):single tape, finite alphabet, finite state, head can move left/right only one cell. In each step, the machine can branch into an arbitrary number of directions. Run is successful if at least one branch is successful.

NP:The class of all languages that can be recognized by a polynomial-time NTM.

Polynomial-time reductionfrom problemP to problemQ: a functionφwith the following properties:

φ(x) can be computed in time |x|O(1),

φ(x) is a yes-instance of Q if and only if x is a yes-instance of P.

Definition: Problem Q isNP-hard if any problem inNP can be reduced toQ.

If anNP-hard problem can be solved in polynomial time, then every

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### Parameterized complexity

To build a complexity theory for parameterized problems, we need two concepts:

An appropriate notion of reduction.

An appropriate hypothesis.

Polynomial-time reductions are not good for our purposes.

Example: Graph G has an independent setk if and only if it has a vertex cover of sizen−k.

⇒Transforming anIndependent Set instance(G,k) into a Vertex Coverinstance (G,n−k) is a correct polynomial-time reduction.

However,Vertex Coveris FPT, butIndependent Set is not known to be FPT.

5

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### Parameterized complexity

To build a complexity theory for parameterized problems, we need two concepts:

An appropriate notion of reduction.

An appropriate hypothesis.

Polynomial-time reductions are not good for our purposes.

Example: Graph G has an independent setk if and only if it has a vertex cover of sizen−k.

⇒Transforming anIndependent Set instance(G,k) into a Vertex Coverinstance (G,n−k) is a correct polynomial-time reduction.

However,Vertex Coveris FPT, butIndependent Set is not known to be FPT.

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### Parameterized reduction

Definition

Parameterized reductionfrom problemP to problem Q: a functionφwith the following properties:

φ(x) can be computed in time f(k)· |x|O(1), where k is the parameter of x,

φ(x) is a yes-instance of Q ⇐⇒ x is a yes-instance of P. If k is the parameter ofx andk0 is the parameter ofφ(x), thenk0 ≤g(k)for some functiong.

Fact: If there is a parameterized reduction from problemP to problemQ andQ is FPT, then P is also FPT.

Non-example: Transforming anIndependent Set instance (G,k) into a Vertex Cover instance(G,n−k)is nota parameterized reduction.

Example: Transforming anIndependent Set instance(G,k) into aClique instance(G,k) isa parameterized reduction.

6

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### Parameterized reduction

Definition

Parameterized reductionfrom problemP to problem Q: a functionφwith the following properties:

φ(x) can be computed in time f(k)· |x|O(1), where k is the parameter of x,

φ(x) is a yes-instance of Q ⇐⇒ x is a yes-instance of P. If k is the parameter ofx andk0 is the parameter ofφ(x), thenk0 ≤g(k)for some functiong.

Fact: If there is a parameterized reduction from problemP to problemQ andQ is FPT, then P is also FPT.

Non-example: Transforming anIndependent Set instance (G,k) into a Vertex Cover instance(G,n−k)is nota parameterized reduction.

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### Multicolored Clique

A useful variant ofClique:

Multicolored Clique: The vertices of the input graph G are colored withk colors and we have to find a clique containing one vertex from each color.

(orPartitioned Clique)

V1 V2 . . . Vk

Theorem

There is a parameterized reduction fromCliqueto Multicolored Clique.

CreateG0 by replacing each vertex v withk vertices, one in each color class. Ifu andv are adjacent in the original graph, connect all copies ofu with all copies ofv.

G G0

V1 V2 . . . Vk

v

u u1, . . . ,uk

v1, . . . ,vk

k-clique inG ⇐⇒ multicoloredk-clique in G0. Similarly: reduction toMulticolored Independent Set.

7

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### Multicolored Clique

Theorem

There is a parameterized reduction fromCliqueto Multicolored Clique.

CreateG0 by replacing each vertex v withk vertices, one in each color class. Ifu andv are adjacent in the original graph, connect all copies ofu with all copies ofv.

G G0

V1 V2 . . . Vk

v

u u1, . . . ,uk

v1, . . . ,vk

k-clique inG ⇐⇒ multicoloredk-clique inG0.

Similarly: reduction toMulticolored Independent Set.

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### Multicolored Clique

Theorem

There is a parameterized reduction fromCliqueto Multicolored Clique.

CreateG0 by replacing each vertex v withk vertices, one in each color class. Ifu andv are adjacent in the original graph, connect all copies ofu with all copies ofv.

G G0

V1 V2 . . . Vk

v

u u1, . . . ,uk

v1, . . . ,vk

k-clique inG ⇐⇒ multicoloredk-clique inG0. Similarly: reduction to Multicolored Independent Set.

7

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### Dominating Set

Theorem

There is a parameterized reduction fromMulticolored Independent Setto Dominating Set.

Proof: Let G be a graph with color classes V1,. . .,Vk. We construct a graphH such that G has a multicoloredk-clique iffH has a dominating set of sizek.

V1

x1 y1 x2 y2 xk yk

u

v

V2 Vk

The dominating set has to contain one vertex from each of the k cliques V ,. . .,V to dominate every x andy.

For every edgee =uv, an additional vertex we ensures that these selections describe an independent set.

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### Dominating Set

Theorem

There is a parameterized reduction fromMulticolored Independent Setto Dominating Set.

Proof: Let G be a graph with color classes V1,. . .,Vk. We construct a graphH such that G has a multicoloredk-clique iffH has a dominating set of sizek.

V1

x1 y1 x2 y2 xk yk

u

v

we

V2 Vk

The dominating set has to contain one vertex from each of the k cliques V1,. . .,Vk to dominate every xi andyi.

For every edgee =uv, an additional vertex we ensures that

these selections describe an independent set. 8

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### Variants of Dominating Set

Dominating Set: Given a graph, findk vertices that dominate every vertex.

Red-Blue Dominating Set: Given a bipartite graph, find k vertices on the red side that dominate the blue side.

Set Cover: Given a set system, find k sets whose union covers the universe.

Hitting Set: Given a set system, findk elements that intersect every set in the system.

All of these problems are equivalent under parameterized reductions, hence at least as hard asClique.

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### Regular graphs

Theorem

There is a parameterized reduction fromCliqueto Cliqueon regular graphs.

Proof: Given a graph G and an integerk, letd be the maximum degree ofG. Taked copies ofG and for every v ∈V(G), fully connect every copy ofv with a set Vv ofd −d(v) vertices.

G G0

G1 G2 . . . Gd

v1

vn

Vv1

Vvn

v2

Vv2

Observe the edges incident toVv do not appear in any triangle, hence everyk-clique ofG0 is ak-clique ofG (assuming k ≥3).

10

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### Regular graphs

Theorem

There is a parameterized reduction fromCliqueto Cliqueon regular graphs.

Proof: Given a graph G and an integerk, letd be the maximum degree ofG. Taked copies ofG and for every v ∈V(G), fully connect every copy ofv with a set Vv ofd −d(v) vertices.

G G0

G1 G2 . . . Gd

v1

vn

Vv1

Vvn

Vv2

v2

Observe the edges incident toVv do not appear in any triangle,

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### Partial Vertex Cover

Partial Vertex Cover: Given a graph G, integers k and s, findk vertices that cover at least s edges.

Theorem

There is a parameterized reduction fromIndependent Set on regular graphs parameterized byk toPartial Vertex Cover parameterized byk.

Proof: If G isd-regular, then k vertices can cover s :=kd edges if and only if there is a independent set of sizek.

d =3,k=4,s =12

11

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### Partial Vertex Cover

Partial Vertex Cover: Given a graph G, integers k and s, findk vertices that cover at least s edges.

Theorem

There is a parameterized reduction fromIndependent Set on regular graphs parameterized byk toPartial Vertex Cover parameterized byk.

Proof: If G isd-regular, then k vertices can cover s :=kd edges if and only if there is a independent set of sizek.

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### Hard problems

Hundreds of parameterized problems are known to be at least as hard asClique:

Independent Set Set Cover

Hitting Set

Connected Dominating Set Independent Dominating Set

Partial Vertex Cover parameterized byk Dominating Setin bipartite graphs

. . .

We believe that none of these problems are FPT.

12

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### Basic hypotheses

It seems that parameterized complexity theory cannot be built on assumingP6=NP– we have to assume something stronger.

Let us choose a basic hypothesis:

Engineers’ Hypothesis

k-Cliquecannot be solved in time f(k)·nO(1).

Theorists’ Hypothesis

k-Step Halting Problem(is there a path of the given NTM that stops ink steps?) cannot be solved in timef(k)·nO(1).

Exponential Time Hypothesis (ETH)

n-variable 3SATcannot be solved in time2o(n). Which hypothesis is the most plausible?

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### Basic hypotheses

It seems that parameterized complexity theory cannot be built on assumingP6=NP– we have to assume something stronger.

Let us choose a basic hypothesis:

Engineers’ Hypothesis

k-Cliquecannot be solved in time f(k)·nO(1).

Theorists’ Hypothesis

k-Step Halting Problem(is there a path of the given NTM that stops ink steps?) cannot be solved in time f(k)·nO(1).

Exponential Time Hypothesis (ETH)

n-variable 3SATcannot be solved in time2o(n). Which hypothesis is the most plausible?

13

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### Basic hypotheses

It seems that parameterized complexity theory cannot be built on assumingP6=NP– we have to assume something stronger.

Let us choose a basic hypothesis:

Engineers’ Hypothesis

k-Cliquecannot be solved in time f(k)·nO(1).

Theorists’ Hypothesis

k-Step Halting Problem(is there a path of the given NTM that stops ink steps?) cannot be solved in time f(k)·nO(1).

Exponential Time Hypothesis (ETH)

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### Basic hypotheses

It seems that parameterized complexity theory cannot be built on assumingP6=NP– we have to assume something stronger.

Let us choose a basic hypothesis:

Engineers’ Hypothesis

k-Cliquecannot be solved in time f(k)·nO(1).

Theorists’ Hypothesis

k-Step Halting Problem(is there a path of the given NTM that stops ink steps?) cannot be solved in time f(k)·nO(1).

Exponential Time Hypothesis (ETH)

n-variable3SAT cannot be solved in time2o(n).

Which hypothesis is the most plausible? 13

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### Independent Set ⇒ Turing machines

Theorem

There is a parameterized reduction fromIndependent Set to thek-Step Halting Problem.

Proof: Given a graph G and an integerk, we construct a Turing machineM and an integerk0 =O(k2) such that M halts ink0 steps if and only ifG has an independent set of size k.

The alphabetΣof M is the set of vertices ofG.

In the first k steps, M nondeterministically writesk vertices to the first k cells.

For every 1≤i ≤k,M moves to thei-th cell, stores the vertex in the internal state, and goes through the tape to check that every other vertex is nonadjacent with the i-th vertex (otherwise M loops).

M does k checks and each check can be done in2k steps⇒ k0=O(k2).

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### Independent Set ⇒ Turing machines

Theorem

There is a parameterized reduction fromIndependent Set to thek-Step Halting Problem.

Proof: Given a graph G and an integerk, we construct a Turing machineM and an integerk0 =O(k2) such that M halts ink0 steps if and only ifG has an independent set of size k.

The alphabetΣof M is the set of vertices ofG.

In the first k steps, M nondeterministically writesk vertices to the first k cells.

For every 1≤i ≤k,M moves to thei-th cell, stores the vertex in the internal state, and goes through the tape to check that every other vertex is nonadjacent with the i-th vertex (otherwise M loops).

M does k checks and each check can be done in2k steps⇒ k0=O(k2).

14

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### Turing machines ⇒ Independent Set

Theorem

There is a parameterized reduction from thek-Step Halting ProblemtoIndependent Set.

Proof: Given a Turing machine M and an integerk, we construct a graphG that has an independent set of sizek0 := (k+1)2 if and only ifM halts ink steps.

cell0 cell1 cellk

before step1

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### Turing machines ⇒ Independent Set

Theorem

There is a parameterized reduction from thek-Step Halting ProblemtoIndependent Set.

Proof: Given a Turing machine M and an integerk, we construct a graphG that has an independent set of sizek0 := (k+1)2 if and only ifM halts ink steps.

G consists of(k+1)2 cliques, thus a k0-independent set has to contain one vertex from each.

The selected vertex from cliqueKi,j describes the situation before stepi at cell j: what is written there, is the head there, and if so, what the state is, and what the next transition is.

We add edges between the cliques to rule out inconsistencies:

head is at more than one location at the same time, wrong character is written, head moves in the wrong direction etc.

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### Summary

Independent Setandk-Step Halting Problemcan be reduced to each other ⇒ Engineers’ Hypothesis and Theorists’

Hypothesis are equivalent!

Independent Setandk-Step Halting Problemcan be reduced toDominating Set.

Is there a parameterized reduction from Dominating Setto Independent Set?

Probably not. Unlike inNP-completeness, where most problems are equivalent, here we have a hierarchy of hard problems.

Independent Setis W[1]-complete. Dominating SetisW[2]-complete.

Does not matter if we only care about whether a problem is FPT or not!

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### Summary

Independent Setandk-Step Halting Problemcan be reduced to each other ⇒ Engineers’ Hypothesis and Theorists’

Hypothesis are equivalent!

Independent Setandk-Step Halting Problemcan be reduced toDominating Set.

Is there a parameterized reduction from Dominating Setto Independent Set?

Probably not. Unlike inNP-completeness, where most problems are equivalent, here we have a hierarchy of hard problems.

Independent Setis W[1]-complete.

Dominating SetisW[2]-complete.

Does not matter if we only care about whether a problem is FPT or not!

16

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### Boolean circuit

ABoolean circuitconsists of input gates, negation gates, AND gates, OR gates, and a single output gate.

x1 x2 x3 x4 x6 x7

Circuit Satisfiability: Given a Boolean circuit C, decide if there is an assignment on the inputs ofC making the output true.

Weight of an assignment: number of true values.

Weighted Circuit Satisfiability: Given a Boolean circuit C and an integer k, decide if there is an assignment of weight k making the output true.

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### Boolean circuit

ABoolean circuitconsists of input gates, negation gates, AND gates, OR gates, and a single output gate.

x1 x2 x3 x4 x6 x7

Circuit Satisfiability: Given a Boolean circuit C, decide if there is an assignment on the inputs ofC making the output true.

Weight of an assignment: number of true values.

Weighted Circuit Satisfiability: Given a Boolean circuit C and an integer k, decide if there is an assignment of weight k

making the output true. 17

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### Weighted Circuit Satisfiability

Independent Setcan be reduced to Weighted Circuit Satisfiability:

x1 x2 x3 x4 x6 x7

Dominating Setcan be reduced toWeighted Circuit Satisfiability:

x1 x2 x3 x4 x6 x7

To expressDominating Set, we need more complicated circuits.

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### Weighted Circuit Satisfiability

Independent Setcan be reduced to Weighted Circuit Satisfiability:

x1 x2 x3 x4 x6 x7

Dominating Setcan be reduced toWeighted Circuit Satisfiability:

x1 x2 x3 x4 x6 x7

To expressDominating Set, we need more complicated circuits.

18

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### Depth and weft

Thedepth of a circuit is the maximum length of a path from an input to the output.

A gate islargeif it has more than2inputs. Theweft of a circuit is the maximum number of large gates on a path from an input to the output.

Independent Set: weft 1, depth3

x2 x3 x4 x6 x7

x1

Dominating Set: weft 2, depth 2

x1 x2 x3 x4 x6 x7

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### The W-hierarchy

LetC[t,d] be the set of all circuits having weft at mostt and depth at mostd.

Definition

A problemP is in the class W[t]if there is a constant d and a parameterized reduction from P toWeighted Circuit SatisfiabilityofC[t,d].

We have seen thatIndependent Set is inW[1]and Dominating Setis inW[2].

Fact: Independent Setis W[1]-complete.

Fact: Dominating SetisW[2]-complete.

If anyW[1]-complete problem is FPT, then FPT=W[1]andevery problem inW[1] is FPT.

If anyW[2]-complete problem is inW[1], thenW[1]=W[2].

⇒If there is a parameterized reduction from Dominating Setto Independent Set, thenW[1]=W[2].

20

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### The W-hierarchy

LetC[t,d] be the set of all circuits having weft at mostt and depth at mostd.

Definition

A problemP is in the class W[t]if there is a constant d and a parameterized reduction from P toWeighted Circuit SatisfiabilityofC[t,d].

We have seen thatIndependent Set is inW[1]and Dominating Setis inW[2].

Fact: Independent Setis W[1]-complete.

Fact: Dominating SetisW[2]-complete.

If anyW[1]-complete problem is FPT, thenFPT=W[1]andevery problem inW[1] is FPT.

If anyW[2]-complete problem is inW[1], thenW[1]=W[2].

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### Weft

Weftis a term related to weaving cloth: it is the thread that runs from side to side in the fabric.

21

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### Parameterized reductions

TypicalNP-hardness proofs: reduction from e.g.,Clique or 3SAT, representing each vertex/edge/variable/clause with a gadget.

v1 v2 v3 v4 v5 v6

C1 C2 C3 C4

Usually does not work for parameterized reductions: cannot afford the parameter increase.

Types of parameterized reductions:

Reductions keeping the structure of the graph. CliqueIndependent Set

Independent Seton regular graphs Partial Vertex Cover

Reductions with vertex representations.

Multicolored Independent SetDominating Set Reductions with vertex and edge representations.

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### Parameterized reductions

TypicalNP-hardness proofs: reduction from e.g.,Clique or 3SAT, representing each vertex/edge/variable/clause with a gadget.

v1 v2 v3 v4 v5 v6

C1 C2 C3 C4

Usually does not work for parameterized reductions: cannot afford the parameter increase.

Types of parameterized reductions:

Reductions keeping the structure of the graph.

CliqueIndependent Set

Independent Seton regular graphs Partial Vertex Cover

Reductions with vertex representations.

Multicolored Independent SetDominating Set Reductions with vertex and edge representations.

22

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### Balanced Vertex Separator

Balanced Vertex Separator: Given a graphG and an inte- gerk, find a setS of at mostkvertices such that every component ofG −S has at most|V(G)|/2vertices.

Theorem

Balanced Vertex Separatorparameterized byk is W[1]-hard.

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### Balanced Vertex Separator

Theorem

Balanced Vertex Separatorparameterized byk is W[1]-hard.

Proof: By reduction from Clique.

G G0

K`

|V(G)|=11,|E(G)|=22, k=4,`=3

We formG0 by

Subdividing every edge ofG.

Making the original vertices of G a clique.

Adding an `-clique for `=|V(G)|+|E(G)| −2(k+ k2 ) (assuming the graph is sufficiently large, we have`≥1).

We have |V(G0)|= 2|V(G)|+2|E(G)| −2(k + k2

) and the “big component” ofG0 has size|V(G)|+|E(G)|. 23

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### Balanced Vertex Separator

Theorem

Balanced Vertex Separatorparameterized byk is W[1]-hard.

Proof: By reduction from Clique.

G G0

K`

|V(G)|=11,|E(G)|=22, k=4,`=3

We have |V(G0)|= 2|V(G)|+2|E(G)| −2(k + k2

) and the “big component” ofG0 has size|V(G)|+|E(G)|.

⇒: Ak-clique inG cuts away k2

vertices, reducing the size of the big component to|V(G)|+|E(G)| −(k+ k2

) =|V(G0)|/2.

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### Balanced Vertex Separator

Theorem

Balanced Vertex Separatorparameterized byk is W[1]-hard.

Proof: By reduction from Clique.

G G0

K`

|V(G)|=11,|E(G)|=22, k=4,`=3

We have |V(G0)|= 2|V(G)|+2|E(G)| −2(k + k2

) and the “big component” ofG0 has size|V(G)|+|E(G)|.

⇐: We need to reduce the size of the large component of G0 by k+ k2

by removingk vertices. This is only possible if thek vertices cut away k2

isolated vertices, i.e., the k-vertices form ak-clique in G.

23

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### List Coloring

List Coloringis a generalization of ordinary vertex coloring:

given a graph G,

a set of colors C, and

a list L(v)⊆C for each vertex v,

the task is to find a coloringc where c(v)∈L(v) for every v.

Theorem

Vertex Coloringis FPT parameterized by treewidth.

However, list coloring is more difficult:

Theorem

List Coloringis W[1]-hard parameterized by treewidth.

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### List Coloring

Theorem

List Coloringis W[1]-hard parameterized by treewidth.

Proof: By reduction from Multicolored Independent Set. Let G be a graph with color classes V1,. . .,Vk.

Set C of colors: the set of vertices ofG.

The colors appearing on vertices u1,. . .,uk correspond to the k vertices of the clique, hence we setL(ui) =Vi.

If x∈Vi andy ∈Vj are adjacent inG, then we need to ensure thatc(ui) =x andc(uj) =y are not true at the same time⇒ we add a vertex adjacent to ui anduj whose list is {x,y}.

u1:V1 u3:V3

u2:V2

u4:V4

u5:V5 25

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### List Coloring

Theorem

List Coloringis W[1]-hard parameterized by treewidth.

Proof: By reduction from Multicolored Independent Set. Let G be a graph with color classes V1,. . .,Vk.

Set C of colors: the set of vertices ofG.

The colors appearing on vertices u1,. . .,uk correspond to the k vertices of the clique, hence we setL(ui) =Vi.

If x∈Vi andy ∈Vj are adjacent in G, then we need to ensure thatc(ui) =x andc(uj) =y are not true at the same time⇒ we add a vertex adjacent to ui anduj whose list is {x,y}.

u1:V1 u3:V3

u2:V2

{x,y}

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### Vertex representation

Key idea

Represent the k vertices of the solution with k gadgets.

Connect the gadgets in a way that ensures that the represented values arecompatible.

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### Odd Set

Odd Set: Given a set systemF over a universeU and an integer k, find a setS of at mostk elements such that|S∩F|is odd for everyF ∈ F.

Theorem

Odd Setis W[1]-hard parameterized byk.

First try: Reduction fromMulticolored Independent Set. LetU =V1∪. . .Vk and introduce each setVi into F.

⇒The solution has to contain exactly one element from each Vi.

Ifxy ∈E(G), how can we express thatx∈Vi andy ∈Vj cannot be selected simultaneously? Seems difficult:

introducing {x,y} intoF forces that exactly oneofx andy appears in the solution,

introducing {x} ∪(Vj \ {y}) into F forces that eitherbothx andy or noneof x andy appear in the solution.

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### Odd Set

Theorem

Odd Setis W[1]-hard parameterized byk.

First try: Reduction fromMulticolored Independent Set. LetU =V1∪. . .Vk and introduce each setVi into F.

⇒The solution has to contain exactly one element from each Vi.

V1 V2 V3 V4 V5

Ifxy ∈E(G), how can we express thatx∈Vi andy ∈Vj cannot be selected simultaneously?

Seems difficult:

introducing {x,y} intoF forces that exactly oneofx andy appears in the solution,

introducing {x} ∪(Vj \ {y}) into F forces that eitherbothx andy or noneof x andy appear in the solution.

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### Odd Set

Theorem

Odd Setis W[1]-hard parameterized byk.

First try: Reduction fromMulticolored Independent Set. LetU =V1∪. . .Vk and introduce each setVi into F.

⇒The solution has to contain exactly one element from each Vi.

V1 V2 V3 V4 V5

x y

Ifxy ∈E(G), how can we express thatx∈Vi andy ∈Vj cannot be selected simultaneously? Seems difficult:

introducing {x,y} intoF forces that exactly oneofx andy appears in the solution,

introducing {x} ∪(Vj \ {y}) into F forces that eitherbothx andy or noneofx andy appear in the solution.

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### Odd Set

Theorem

Odd Setis W[1]-hard parameterized byk.

First try: Reduction fromMulticolored Independent Set. LetU =V1∪. . .Vk and introduce each setVi into F.

⇒The solution has to contain exactly one element from each Vi.

V1 V2 V3 V4 V5

x y

Ifxy ∈E(G), how can we express thatx∈Vi andy ∈Vj cannot be selected simultaneously? Seems difficult:

introducing {x,y} intoF forces that exactly oneofx andy appears in the solution,

introducing {x} ∪(Vj \ {y}) into F forces that eitherbothx andy or noneofx andy appear in the solution.

27

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### Odd Set

Theorem

Odd Setis W[1]-hard parameterized byk.

First try: Reduction fromMulticolored Independent Set. LetU =V1∪. . .Vk and introduce each setVi into F.

⇒The solution has to contain exactly one element from each Vi.

V1 V2 V3 V4 V5

x y

Ifxy ∈E(G), how can we express thatx∈Vi andy ∈Vj cannot be selected simultaneously? Seems difficult:

introducing {x,y} intoF forces that exactly oneofx andy appears in the solution,

(54)

### Odd Set

Theorem

Odd Setis W[1]-hard parameterized byk.

First try: Reduction fromMulticolored Independent Set. LetU =V1∪. . .Vk and introduce each setVi into F.

⇒The solution has to contain exactly one element from each Vi.

V1 V2 V3 V4 V5

x y

Ifxy ∈E(G), how can we express thatx∈Vi andy ∈Vj cannot be selected simultaneously? Seems difficult:

introducing {x,y} intoF forces that exactly oneofx andy appears in the solution,

introducing {x} ∪(Vj \ {y}) into F forces that eitherbothx andy or noneofx andy appear in the solution.

27

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### Odd Set

Reduction fromMulticolored Clique. U :=Sk

i=1Vi∪S

1≤i<j≤kEi,j. k0:=k+ k2

.

Let F containVi (1≤i ≤k) andEi,j (1≤i <j ≤k).

For every v ∈Vi andx6=i, we introduce the sets: (Vi\ {v})∪ {every edge from Ei,x with endpoint v} (Vi\ {v})∪ {every edge from Ex,i with endpoint v}

E1,2 E1,3 E1,4 E2,3 E2,4 E3,4

V1 V2 V3 V4

(56)

### Odd Set

Reduction fromMulticolored Clique. U :=Sk

i=1Vi∪S

1≤i<j≤kEi,j. k0:=k+ k2

.

Let F containVi (1≤i ≤k) andEi,j (1≤i <j ≤k).

For every v ∈Vi andx6=i, we introduce the sets:

(Vi\ {v})∪ {every edge from Ei,x with endpointv} (Vi\ {v})∪ {every edge from Ex,i with endpointv}

E1,2 E1,3 E1,4 E2,3 E2,4 E3,4

V1 V2 V3 V4

28

(57)

### Odd Set

Reduction fromMulticolored Clique. U :=Sk

i=1Vi∪S

1≤i<j≤kEi,j. k0:=k+ k2

.

Let F containVi (1≤i ≤k) andEi,j (1≤i <j ≤k).

For every v ∈Vi andx6=i, we introduce the sets:

(Vi\ {v})∪ {every edge from Ei,x with endpointv} (Vi\ {v})∪ {every edge from Ex,i with endpointv}

E1,2 E1,3 E1,4 E2,3 E2,4 E3,4

V1 V2 V3 V4

(58)

### Odd Set

Reduction fromMulticolored Clique. U :=Sk

i=1Vi∪S

1≤i<j≤kEi,j. k0:=k+ k2

.

Let F containVi (1≤i ≤k) andEi,j (1≤i <j ≤k).

For every v ∈Vi andx6=i, we introduce the sets:

(Vi\ {v})∪ {every edge from Ei,x with endpointv} (Vi\ {v})∪ {every edge from Ex,i with endpointv}

E1,2 E1,3 E1,4 E2,3 E2,4 E3,4

V1 V2 V3 V4

28

(59)

### Odd Set

Reduction fromMulticolored Clique.

For every v ∈Vi andx6=i, we introduce the sets:

(Vi\ {v})∪ {every edge from Ei,x with endpointv} (Vi\ {v})∪ {every edge from Ex,i with endpointv} v ∈Vi selected ⇐⇒ edges with endpointv are selected

fromEi,x andEx,i

vi ∈Vi selected

vj ∈Vj selected ⇐⇒ edgevivj is selected in Ei,x

E1,2 E1,3 E1,4 E2,3 E2,4 E3,4

V1 V2 V3 V4

(60)

### Odd Set

Reduction fromMulticolored Clique.

For every v ∈Vi andx6=i, we introduce the sets:

(Vi\ {v})∪ {every edge from Ei,x with endpointv} (Vi\ {v})∪ {every edge from Ex,i with endpointv} v ∈Vi selected ⇐⇒ edges with endpointv are selected

fromEi,x andEx,i

vi ∈Vi selected

vj ∈Vj selected ⇐⇒ edgevivj is selected in Ei,x

E1,2 E1,3 E1,4 E2,3 E2,4 E3,4

V1 V2 V3 V4

28

(61)

### Vertex and edge representation

Key idea

Represent the vertices of the clique by k gadgets.

Represent the edges of the clique by k2

Connect edge gadgetEi,j to vertex gadgetsVi andVj such that if Ei,j represents the edge betweenx ∈Vi andy ∈Vj, then it forcesVi tox andVj toy.

(62)

### Variants of Odd Set

The following problems areW[1]-hard:

Odd Set

Exact Odd Set(find a set of size exactly k . . . ) Exact Even Set

Unique Hitting Set

(at mostk elements that hit each set exactly once) Exact Unique Hitting Set

(exactly k elements that hit each set exactly once)

Open question:

## ?

Even Set: Given a set systemF and an integerk, find a nonempty setS of at mostk elements such|F∩S|is even for every F ∈ F.

30

(63)

### Variants of Odd Set

The following problems areW[1]-hard:

Odd Set

Exact Odd Set(find a set of size exactly k . . . ) Exact Even Set

Unique Hitting Set

(at mostk elements that hit each set exactly once) Exact Unique Hitting Set

(exactly k elements that hit each set exactly once) Open question:

## ?

Even Set: Given a set systemF and an integerk, find a nonempty setS of at mostk elements such|F∩S|is even for every F ∈ F.

(64)

### Summary

By parameterized reductions, we can show that lots of

parameterized problems are at least as hard asClique, hence unlikely to be fixed-parameter tractable.

Connection with Turing machines gives some supporting evidence for hardness (only of theoretical interest).

TheW-hierarchy classifies the problems according to hardness (only of theoretical interest).

Important trick inW[1]-hardness proofs: vertex and edge representations.

31

(65)

### Exponential Time Hypothesis (ETH)

Hypothesis introduced by Impagliazzo, Paturi, and Zane:

Exponential Time Hypothesis (ETH)

There is no2o(n)-time algorithm for n-variable3SAT. Note: current best algorithm is 1.30704n [Hertli 2011]. Note: an n-variable3SATformula can have Ω(n3) clauses.

Sparsification Lemma[Impagliazzo, Paturi, Zane 2001] There is a 2o(n)-time algorithm for n-variable 3SAT.

m

There is a 2o(m)-time algorithm for m-clause3SAT.

(66)

### Exponential Time Hypothesis (ETH)

Hypothesis introduced by Impagliazzo, Paturi, and Zane:

Exponential Time Hypothesis (ETH)

There is no2o(n)-time algorithm for n-variable3SAT. Note: current best algorithm is 1.30704n [Hertli 2011]. Note: an n-variable3SATformula can have Ω(n3) clauses.

Sparsification Lemma[Impagliazzo, Paturi, Zane 2001]

There is a2o(n)-time algorithm for n-variable 3SAT. m

There is a 2o(m)-time algorithm for m-clause3SAT.

32

(67)

### Lower bounds based on ETH

Exponential Time Hypothesis (ETH)

There is no2o(m)-time algorithm for m-clause3SAT. The textbook reduction from3SAT to3-Coloring:

3SATformula φ n variables

m clauses

GraphG O(n+m) vertices

O(n+m) edges

Corollary

Assuming ETH, there is no2o(n) algorithm for3-Coloringon an n-vertex graphG.

(68)

### Lower bounds based on ETH

Exponential Time Hypothesis (ETH)

There is no2o(m)-time algorithm for m-clause3SAT. The textbook reduction from3SAT to3-Coloring:

3SATformula φ n variables

m clauses

GraphG O(m) vertices

O(m) edges

Corollary

Assuming ETH, there is no2o(n) algorithm for3-Coloringon an n-vertex graphG.

33

(69)

### Transfering bounds

There are polynomial-time reductions from, say,3-Coloring to many other problems such that the reduction increases the number of vertices by at most a constant factor.

Consequence: Assuming ETH, there is no2o(n) time algorithm on n-vertex graphs for

Independent Set Clique

Dominating Set Vertex Cover Hamiltonian Path Feedback Vertex Set . . .

(70)

### Transfering bounds

There are polynomial-time reductions from, say,3-Coloring to many other problems such that the reduction increases the number of vertices by at most a constant factor.

Consequence: Assuming ETH, there is no 2o(k)·nO(1) time algo- rithm for

k-Independent Set k-Clique

k-Dominating Set k-Vertex Cover k-Path

k-Feedback Vertex Set . . .

34

(71)

### Transfering bounds

There are polynomial-time reductions from, say,3-Coloring to many other problems such that the reduction increases the number of vertices by at most a constant factor.

Consequence: Assuming ETH, there is no 2o(k)·nO(1) time algo- rithm for

k-Independent Set k-Clique

k-Dominating Set k-Vertex Cover k-Path

k-Feedback Vertex Set . . .

(72)

### Lower bounds based on ETH

The textbook reduction from3-Coloringto Planar

3-Coloringuses a “crossover gadget” with4 external connectors:

In every 3-coloring of the gadget, opposite external connectors have the same color.

Every coloring of the external connectors where the opposite vertices have the same color can be extended to the whole gadget.

If two edges cross, replace them with a crossover gadget. 35

(73)

### Lower bounds based on ETH

The textbook reduction from3-Coloringto Planar

3-Coloringuses a “crossover gadget” with4 external connectors:

In every 3-coloring of the gadget, opposite external connectors have the same color.

Every coloring of the external connectors where the opposite vertices have the same color can be extended to the whole

(74)

### Lower bounds based on ETH

The textbook reduction from3-Coloringto Planar

3-Coloringuses a “crossover gadget” with4 external connectors:

In every 3-coloring of the gadget, opposite external connectors have the same color.

Every coloring of the external connectors where the opposite vertices have the same color can be extended to the whole gadget.

If two edges cross, replace them with a crossover gadget. 35

(75)

### Lower bounds based on ETH

The reduction from 3-ColoringtoPlanar 3-Coloring introducesO(1) new edges/vertices for each crossing.

A graph with medges can be drawn with O(m2) crossings.

3SATformula φ n variables

m clauses

GraphG O(m) vertices

O(m) edges

Planar graph G0 O(m2) vertices O(m2)edges

Corollary

Assuming ETH, there is no2o(

n) algorithm for3-Coloringon ann-vertex planar graph G.

(76)

### Lower bounds for planar problems

Consequence: Assuming ETH, there is no 2o(

n) time algorithm onn-vertex planar graphsfor

Independent Set Dominating Set Vertex Cover Hamiltonian Path Feedback Vertex Set . . .

Note: Reduction to planar graphs does not work for Clique (why?).

37

(77)

### Lower bounds for planar problems

Consequence: Assuming ETH, there is no2o(

k)·nO(1) time algo- rithm onplanar graphs for

k-Independent Set k-Dominating Set k-Vertex Cover k-Path

k-Feedback Vertex Set . . .

Note: Reduction to planar graphs does not work for Clique (why?).

(78)

### Lower bounds for planar problems

Consequence: Assuming ETH, there is no2o(

k)·nO(1) time algo- rithm onplanar graphs for

k-Independent Set k-Dominating Set k-Vertex Cover k-Path

k-Feedback Vertex Set . . .

Note: Reduction to planar graphs does not work for Clique (why?).

37

(79)

### Exponential Time Hypothesis

Engineers’ Hypothesis

k-Cliquecannot be solved in time f(k)·nO(1). m

Theorists’ Hypothesis

k-Step Halting Problem(is there a path of the given NTM that stops ink steps?) cannot be solved in time f(k)·nO(1).

⇑ Exponential Time Hypothesis (ETH)

n-variable3SAT cannot be solved in time2o(n).

What do we have to show to prove that ETH implies Engineers’

Hypothesis?

(80)

### Exponential Time Hypothesis

Engineers’ Hypothesis

k-Cliquecannot be solved in time f(k)·nO(1). m

Theorists’ Hypothesis

k-Step Halting Problem(is there a path of the given NTM that stops ink steps?) cannot be solved in time f(k)·nO(1).

⇑ Exponential Time Hypothesis (ETH)

n-variable3SAT cannot be solved in time2o(n).

What do we have to show to prove that ETH implies Engineers’

Hypothesis?

We have to show that anf(k)·nO(1) algorithm implies that there is a2o(n) time algorithm forn-variable3SAT.

38

(81)

### Exponential Time Hypothesis

Engineers’ Hypothesis

k-Cliquecannot be solved in time f(k)·nO(1). m

Theorists’ Hypothesis

k-Step Halting Problem(is there a path of the given NTM that stops ink steps?) cannot be solved in time f(k)·nO(1).

⇑ Exponential Time Hypothesis (ETH)

n-variable3SAT cannot be solved in time2o(n).

We actually show something much stronger and more interesting:

Theorem

(82)

### Lower bound on the exponent

Theorem

Assuming ETH, there is nof(k)·no(k) algorithm fork-Cliquefor any computable functionf.

Suppose thatk-Cliquecan be solved in time f(k)·nk/s(k), where s(k) is a monotone increasing unbounded function. We use this algorithm to solve3-Coloringon ann-vertex graphG in time 2o(n).

Letk be the largest integer such thatf(k)≤n andkk/s(k) ≤n. Functionk :=k(n) is monotone increasing and unbounded.

Split the vertices ofG intok groups. Let us build a graph H where each vertex corresponds to a proper3-coloring of one of the groups. Connect two vertices if they are not conflicting.

Everyk-clique ofH corresponds to a proper 3-coloring ofG.

⇒A 3-coloring of G can be found in time

f(k)· |V(H)|k/s(k)≤n·(k3n/k)k/s(k)=n·kk/s(k)·3n/s(k)=2o(n).

39

(83)

### Lower bound on the exponent

Theorem

Assuming ETH, there is nof(k)·no(k) algorithm fork-Cliquefor any computable functionf.

Suppose thatk-Cliquecan be solved in time f(k)·nk/s(k), where s(k) is a monotone increasing unbounded function. We use this algorithm to solve3-Coloringon ann-vertex graphG in time 2o(n).

Letk be the largest integer such thatf(k)≤n andkk/s(k) ≤n.

Functionk :=k(n) is monotone increasing and unbounded.

Split the vertices ofG intok groups. Let us build a graph H where each vertex corresponds to a proper3-coloring of one of the groups.

Connect two vertices if they are not conflicting.

Everyk-clique ofH corresponds to a proper 3-coloring ofG.

⇒A 3-coloring of G can be found in time

f(k)· |V(H)|k/s(k)≤n·(k3n/k)k/s(k)=n·kk/s(k)·3n/s(k)=2o(n).

(84)

### Lower bound on the exponent

Theorem

Assuming ETH, there is nof(k)·no(k) algorithm fork-Cliquefor any computable functionf.

Suppose thatk-Cliquecan be solved in time f(k)·nk/s(k), where s(k) is a monotone increasing unbounded function. We use this algorithm to solve3-Coloringon ann-vertex graphG in time 2o(n).

Letk be the largest integer such thatf(k)≤n andkk/s(k) ≤n.

Functionk :=k(n) is monotone increasing and unbounded.

Split the vertices ofG intok groups. Let us build a graph H where each vertex corresponds to a proper3-coloring of one of the groups.

Connect two vertices if they are not conflicting.

Everyk-clique ofH corresponds to a proper 3-coloring ofG.

⇒A 3-coloring of G can be found in time

f(k)· |V(H)|k/s(k)≤n·(k3n/k)k/s(k)=n·kk/s(k)·3n/s(k)=2o(n). 39

(85)

### Tight bounds

Theorem

Assuming ETH, there is nof(k)·no(k) algorithm fork-Cliquefor any computable functionf.

Transfering to other problems:

k-Clique

(x,k) ⇒ Problem A

(x0,O(k))

f(k)·no(k)

algorithm ⇐ f(k)·no(k)) algorithm

Bottom line:

To rule out f(k)·no(k) algorithms, we need a parameterized reduction that blows up the parameter at mostlinearly. To rule out f(k)·no(

k) algorithms, we need a parameterized reduction that blows up the parameter at mostquadratically.

(86)

### Tight bounds

Theorem

Assuming ETH, there is nof(k)·no(k) algorithm fork-Cliquefor any computable functionf.

Transfering to other problems:

k-Clique

(x,k) ⇒ Problem A

(x0,g(k))

f(k)·no(k)

algorithm ⇐ f(k)·no(g−1(k)) algorithm

Bottom line:

To rule out f(k)·no(k) algorithms, we need a parameterized reduction that blows up the parameter at mostlinearly. To rule out f(k)·no(

k) algorithms, we need a parameterized reduction that blows up the parameter at mostquadratically.

40

(87)

### Tight bounds

Theorem

Assuming ETH, there is nof(k)·no(k) algorithm fork-Cliquefor any computable functionf.

Transfering to other problems:

k-Clique

(x,k) ⇒ Problem A

(x0,k2)

f(k)·no(k)

algorithm ⇐ f(k)·no(

k)

algorithm

Bottom line:

To rule out f(k)·no(k) algorithms, we need a parameterized reduction that blows up the parameter at mostlinearly. To rule out f(k)·no(

k) algorithms, we need a parameterized reduction that blows up the parameter at mostquadratically.

(88)

### Tight bounds

Theorem

Assuming ETH, there is nof(k)·no(k) algorithm fork-Cliquefor any computable functionf.

Transfering to other problems:

k-Clique

(x,k) ⇒ Problem A

(x0,k2)

f(k)·no(k)

algorithm ⇐ f(k)·no(

k)

algorithm Bottom line:

To rule out f(k)·no(k) algorithms, we need a parameterized reduction that blows up the parameter at mostlinearly.

To rule out f(k)·no(

k) algorithms, we need a parameterized reduction that blows up the parameter at mostquadratically.

40

(89)

### Tight bounds

Assuming ETH, there is nof(k)no(k) time algorithms for Set Cover

Hitting Set

Connected Dominating Set Independent Dominating Set Partial Vertex Cover

Dominating Setin bipartite graphs . . .

More problems are FPT, more difficult to proveW[1]-hardness. The problem Grid Tilingis the key to many of these results.

(90)

### Tight bounds

Assuming ETH, there is nof(k)no(k) time algorithms for Set Cover

Hitting Set

Connected Dominating Set Independent Dominating Set Partial Vertex Cover

Dominating Setin bipartite graphs . . .

More problems are FPT, more difficult to proveW[1]-hardness.

The problem Grid Tilingis the key to many of these results.

41

(91)

### Grid Tiling

Grid Tiling

Input: A k ×k matrix and a set of pairs Si,j ⊆ [D]×[D] for each cell.

Find:

A pairsi,j ∈Si,j for each cell such that

Vertical neighbors agree in the 1st coordinate.

Horizontal neighbors agree in the 2nd coordinate.

(1,1) (3,1) (2,4)

(5,1) (1,4) (5,3)

(1,1) (2,4) (3,3) (2,2)

(1,4)

(3,1) (1,2)

(2,2) (2,3) (1,3)

(2,3) (1,1) (1,3)

(2,3) (5,3)

(92)

### Grid Tiling

Grid Tiling

Input: A k ×k matrix and a set of pairs Si,j ⊆ [D]×[D] for each cell.

Find:

A pairsi,j ∈Si,j for each cell such that

Vertical neighbors agree in the 1st coordinate.

Horizontal neighbors agree in the 2nd coordinate.

(1,1) (3,1) (2,4)

(5,1) (1,4) (5,3)

(1,1) (2,4) (3,3) (2,2)

(1,4)

(3,1) (1,2)

(2,2) (2,3) (1,3)

(2,3) (3,3)

(1,1) (1,3)

(2,3) (5,3) k =3,D =5

42

(93)

### Grid Tiling

Grid Tiling

Input: A k ×k matrix and a set of pairs Si,j ⊆ [D]×[D] for each cell.

Find:

A pairsi,j ∈Si,j for each cell such that

Vertical neighbors agree in the 1st coordinate.

Horizontal neighbors agree in the 2nd coordinate.

Fact

There is a parameterized reduction fromk-Clique tok×k Grid Tiling.

(94)

### Grid Tiling is W[1]-hard

Reduction fromk-Clique Definition of the sets:

For i =j: (x,y)∈Si,j ⇐⇒ x =y

For i 6=j: (x,y)∈Si,j ⇐⇒ x andy are adjacent.

(vi,vi)

Each diagonal cell defines a valuevi. . .

43

(95)

### Grid Tiling is W[1]-hard

Reduction fromk-Clique Definition of the sets:

For i =j: (x,y)∈Si,j ⇐⇒ x =y

For i 6=j: (x,y)∈Si,j ⇐⇒ x andy are adjacent.

(vi, .)

(.,vi) (vi,vi) (.,vi) (.,vi) (.,vi) (vi, .)

(vi.,) (vi, .)

(96)

### Grid Tiling is W[1]-hard

Reduction fromk-Clique Definition of the sets:

For i =j: (x,y)∈Si,j ⇐⇒ x =y

For i 6=j: (x,y)∈Si,j ⇐⇒ x andy are adjacent.

(vi, .)

(.,vi) (vi,vi) (.,vi) (.,vi) (.,vi) (vi, .)

(vi, .) (vj,vj) (vi, .)

vi andvj are adjacent for every 1≤i <j ≤k.

43

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