Parameterized Algorithms - PowerPoint Presentation

Slide1

Parameterized AlgorithmsAdvanced Kernelization Techniques

Bart M. P. Jansen

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August 20th 2014,

BędlewoSlide2

This lecture

2Slide3

Planar graphs

3Slide4

The definition

A planar graph is a graph that can be drawn in the real Euclidean plane

without crossing edgesThis is equivalent to drawing it on the surface of a sphereA plane graph is a planar graph with a chosen embeddingA face in an embedded plane graph is a maximal connected region that does not intersect the drawing 4Slide5

Planar problems are easier

5

While most NP-complete problems remain NP-complete when the input graph is restricted to be a planar, many problems do become

easier

on planar graphsSlide6

Why planar problems

are easier

The parameter is often large compared to the graph size

For several problems the only way a

big

planar graph can have a

small

solution, is simple to understand

In general graphs, it can be very complicated

Sparsity

Planar graphs are

sparse

); interactions are not as complex as in general graphs

You can exploit the drawing

In a planar graph, you can define a sense of “locality”

There is no direct interaction (no edges) between far away regions

Hardness proofs fail

Many gadgets for proving

-hardness require crossings

6Slide7

Properties of planar graphs

A

simple -vertex planar graph has at most edgesA simple -vertex bipartite graph has at most edgesEvery simple planar graph has a vertex of degree at most An -vertex planar graph has treewidth

7

Euler’s

formula

.

If

are the

number

of

vertices

,

edges

,

and

faces

of a

plane

graph

,

then

. Equality holds if is connected.

 Slide8

Edge counts in simple planar graphs (I)

Lemma. A simple

-vertex planar graph has at most edgesProof. Trivial for , so assume Let be the set of faces of a planar drawingFor a face , let

be the

edges

bounding

In

we

count

every

edge

at most

twice

8

 Slide9

Edge counts in simple planar graphs (II)

Since is a

simple graph and , the boundary of every face consists of at least edges

Apply Euler’s formula:

9Slide10

Edge counts in simple bipartite planar graphs

If is a

simple planar bipartite graph and , the boundary of every face consists of at least edges

Apply Euler’s formula:

10Slide11

Planar bipartite neighborhood

lemma

Lemma. Let be a simple planar bipartite graph with bipartition classes and . If each vertex in is of degree at least , then Proof.Let be the number of verticesThe number of edges in

is at least

Since

is simple, planar, and bipartite,

11Slide12

connected vertex cover

12Slide13

The Connected Vertex Cover

problem

Input: A graph and an integer Parameter: Question: Is there a vertex cover of at most vertices in , such that is connected?Such a set is a connected vertex cover of Connected Vertex Cover does not admit a polynomial kernel in general graphs

,

unless

In

sharp

contrast

to

the

-vertex kernel for

Vertex Cover

We

give

a

-

vertex kernel on

planar graphs

13Slide14

Reduction rules for

Connected Vertex Cover

(R1) If is an isolated vertex, then delete (R2) If is not connected, then answer no(R3) If or then decide the problem(R4) If vertex has multiple neighbors of degree , then delete all of them except one (exercise) 

14Slide15

Cut vertices of degree

15(R5) If there is a cutvertex of degree with neighbors and , then delete , add the edge , and decrease by oneLemma. If

is transformed into

by applying (

R5)

to vertex

,

then

is

yes

iff

is

yes

Proof.

Any

cvc

for

contains

If then , but then

is disconnectedSo is strictly smaller, and is connected by edge

Add to a cvc for , to get a cvc for By edge , one of

is in , connecting All edges of

are covered by

 Slide16

Solutions can avoid degree-2 non-cut vertices (I)

Lemma. If the connected graph

contains a non-cut vertex of degree , then there is a minimum connected vertex cover in that does not contain and therefore contains Proof. Let fix a minimum cvc

with

At least one of

is in

, say

If

then take

Connected because

has some other neighbor than

That neighbor is in

, since

is not

16Slide17

Solutions can avoid degree-2 non-cut vertices (II)

Lemma. If the connected graph

contains a non-cut vertex of degree , then there is a minimum connected vertex cover in that does not contain and therefore contains Proof. If , the set is a vertex cover but not connectedLook at the connected components

and

of

Since

is no

cutvertex

, there is a

path

in

At some point, the path crosses from

to

by visiting

one

vertex

The set

is a minimum

cvc

without

17Slide18

Non-cut vertices of degree

18(R6) If there is a vertex of degree that is not a cutvertex, then delete and add a degree- vertex to each former neighborLemma. If is transformed into

by applying (

R6)

to vertex

,

then

is

yes

iff

is

yes

Proof.

Some optimal

cvc

for

contains

and

but not

(lemma)

This is also a cvc for Any cvc for contains both and

If one is missing, then the degree-1 neighbor covers the edgeBut then the cover is not connectedRemoving deg-1 vertices preserves connectivity, is connected and

cover all edges of

, so is a cvc in  Slide19

Boundary lemma

Lemma. If

is a yes-instance of Planar Connected Vertex Cover and (R1)-(R6) are not applicable, then Proof. Let be a connected vertex cover of size at most Consider the independent set By (R1), set

contains

no

vertices

of

degree

By

(R4), set

contains

at most

vertices

of

degree

By

(R5)

and

(R6), set

contains

no

vertices of degree  19

S

 Slide20

Boundary lemma

Lemma. If

is a yes-instance of Planar Connected Vertex Cover and (R1)-(R6) are not applicable, then Proof. Let be a connected vertex cover of size at most By the bipartite neighborhood lemma, set contains at most

vertices

of

degree

20

S

Theorem

.

Connected

Vertex Cover

has a

kernel

with

vertices when restricted to planar graphs

 Slide21

Turing kernelization

21Slide22

The limits of effective preprocessing

The composition framework presented yesterday shows that for some parameterized problems, we should not expect polynomial-size kernelsEven for problems that are FPTDoes that mean we cannot obtain useful and provably effective preprocessing routines for such problems?

No!We can slightly relax the requirements to circumvent the lower bounds from compositionality22Slide23

The Max Leaf Subtree problem

Input: A graph

and an integer Parameter: Question: Does have a tree with leaves as a subgraph?A leaf is a vertex with degree at most Max Leaf Subtree generalizes Max Leaf Spanning treeIn a connected graph, any -leaf tree can be extended to a spanning tree with at least leavesMax Leaf Subtree is NP-complete 23Slide24

Max Leaf Subtree is or

-compositional

Let be instances of Max Leaf SubtreeLet be the disjoint union of

is a

yes

-instance

iff

at least one input

is

yes

24

Theorem

.

-

Max

Leaf

Subtree

does not admit a polynomial kernel unless

 Slide25

Preprocessing for Max Leaf Subtree

We cannot efficiently reduce

to a single, equivalent instance of size However, we efficiently reduce to a list of instances each of size , such that is yes iff there is a yes-instance on the list

The instances on the list can be solved

in parallel

25Slide26

Reduction rule for Max Leaf Subtree

(R1) If there is a vertex

of degree , such that its neighbors also have degree , and , then remove and add the edge  26

Lemma.

If

is a

connected

graph to which (R1) cannot be applied, and

, then

contains a

-leaf

subtree

 Slide27

Preprocessing algorithm for

Max Leaf Subtree

Algorithm Preprocess(Graph , integer )while (R1) is applicable to vertex with neighbors remove and add the edge if a connected component of has

vertices

then

return

yes

for

each

connected

component

of

add

to the list of output

instances

,

size

is

By

the stated lemma,

algorithm is correct when saying yesSince a subtree is contained in 1 connected component, answer to is yes iff some

is a yes-instance 27Slide28

Formal definition?

What is the right definition for this type of preprocessing?The given procedure splits an instance

into a list of small instances, such that the answer to

is the logical

or

of the

However, the preprocessing would also be useful if there would be a

different

way of efficiently finding the answer to

from the answers to

Only important that

can efficiently be solved knowing just the answers to

-size instances

28Slide29

Turing kernelization

Let

be a parameterized problem and let A Turing kernelization for of size is an algorithm thatdecides whether a given instance

is in

in

time

polynomial

in

when

given access to an

oracle that

for any instance

with

,

decides whether

in a single step

29Slide30

Results on Turing kernelization

The preprocessing algorithm for Max Leaf Subtree shows:

Create the list of instances, query each small instance from the oracle, output yes if you get a yes-answer from the oracleThe Max Leaf Subtree algorithm is non-adaptiveIt formulates all oracle queries before making a queryThe definition of Turing kernelization also allows adaptivityFormulate next query based on previous answers30Theorem. -Max Leaf Subtree has a Turing kernel with

vertices and

bitsize

 Slide31

Turing kernelization for finding paths

[J, ESA 2014]The algorithm is crucially

adaptiveUnclear whether a non-adaptive Turing kernel existsTo show the idea, we consider the -Longest Cycle problemGiven , does have a simple cycle of length Behaves similarly as -Longest Path, but details are easier 31

Theorem

.

-

Longest

Path

has a polynomial Turing kernel

when restricted to planar graphs

 Slide32

Long cycles through

-separators

 Claim. Let such that , , and there are no edges between

and

Let

be the vertices on a longest

path in

If

has a cycle of length

, then:

The graph

has a cycle of length

, or

The graph

has a cycle of length

32Slide33

Turing reduction rule for

-Longest Cycle

 If there are such that , is a minimal separator,

and

:

If

has

a cycle of length at least

,

output

yes

If

does not have

a cycle of length at least

:

Query the oracle for the vertices

of a longest

path in

If

, then conclude that the answer is

yes

Else, remove the vertices of

from the graph

33

This info can be obtained from the decision oracle for

-

Longest Cycle

by self-reduction on the

-size

subgraph

Query the oracle for the instance

with at most vertices

 Slide34

Splitting rule for

-Longest Cycle

 If there is a connected component of that is not biconnected, then split it into its biconnected components 34Slide35

Turing kernelization

for -Longest Cycle

 If neither of the reduction rules can be applied to a planar graph , and there is a connected component with vertices, then the answer is yesCombination of nontrivial off-the-shelf graph-theoretic results proves the existence of a -cycleIf all connected components have at most vertices, query the oracle for each one to determine if it contains a -cycleAfterward, we know the answer to the instanceIn Turing kernelization, we can use the solutions to small instances of difficult problems to help us reduce!

35Slide36

Advanced discussion

36Slide37

ProtrusionsFor planar graphs, there is a single algorithmic idea that simultaneously gives polynomial-size (often even linear-size) kernels for a

wide variety of graph problemsBased on the idea of protrusion replacement

37Slide38

How to replace

Goal is to replace a large but structurally simple part of the graph by a smaller gadget that enforces the same constraintsA list of possible gadgets is hardcoded into the algorithm

To determine which gadget we should use, we have to be able to analyze the behavior of the problem on the protrusionWe demand that it has constant treewidthIf a graph problem satisfies certain simple conditions and is expressible in a general type of logic, this approach yields polynomial kernels for problems on planar graphsMeta-kernelization38Slide39

Meta-kernelization results

39Slide40

Beyond planar graphs

Planar graphs have nice algorithmic propertiesSimilar algorithmic properties can be derived for generalizations of planar graphs

Graphs of bounded genusGraphs that can be drawn without crossings onto a sphere to which a constant number of “handles” have been attachedThe handles allow some edges between distant regions, but the graph cannot be too wildMinor-free graphsA graph is planar iff it contains neither nor as a minorIf a family of graphs does not contain a fixed graph as a minor, then the family contains only sparse graphsMany algorithms first developed for planar graphs were later generalized to bounded-genus and minor-free graphs 40Slide41

Beyond planar graphs for Connected Vertex Cover

The reduction rules for Connected Vertex Cover

did not rely on planarityThey are also correct in general graphsThe boundary lemma (exhaustively reduced instances with more than vertices have answer no) did rely on planarityThe analysis can be adapted for graphs of genus Graphs that can be drawn onto a sphere with handlesYou can prove an upper bound of vertices for exhaustively reduced yes-instances 41Slide42

outlook

42Slide43

Open problem: Turing kernel for

-Longest Path

 The -Longest Path problem has a polynomial Turing kernel when restricted to planar graphsIs there a polynomial Turing kernel in general graphs?This is wide open! 43Slide44

Open problem: Lower bounds for Turing kernels

The composition framework allows us to prove that some problems do not have polynomial-size (standard) kernelsHow can we prove that a problem does not have a polynomial Turing kernel?Hermelin et al. [IPEC 2013] suggest that two specific problems do not admit polynomial-size Turing kernels:

Hitting Set parameterized by # of elementsHitting Set parameterized by # of setsHowever, no complexity-theoretic evidence is known!Does the polynomial-time hierarchy collapse if these problems have polynomial Turing kernels?44Slide45

Exercises

45Slide46

ConclusionPlanar graphs have many nice properties that can be exploited for

kernelizationTuring kernelization is a relaxed form of preprocessing that can sometimes circumvent lower bounds for normal

kernelizationMany interesting open problems remain in this area46Slide47

The end