Introduction to Parametrized Algorithms

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Introduction to Parametrized Algorithms

Marcin Pilipczuk most slides by Łukasz Kowalik

some slides by Marek Cygan and Michał Pilipczuk

Algorytmika, 18.05.2020

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Overview

1 A motivating example, definition of classes FPT, XP.

2 Technique: branching

3 Technique: kernelization

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The book

Everything can be found in

http://parameterized-algorithms.mimuw.edu.pl

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Motivation

We would like to solve NP-hard problems efficiently.

We are not interested in approximation, heuristics, etc.

We want exact solutions, with provable guarantees.

(Unless P=NP) we cannot solve all instances fast; let us just solve the simple ones fast.

Which istances are simple?

What do we mean by fast?

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Parameterized algorithms - example

There were 10⁴ measurements performed, each of them can report wrong output with probability p ≤ 0.1%.

Create a graph G = (V , E ):

I Vertices are measurements.

I u and v adjacent if measurements u and v contradict each other.

We want to find minimum sized subset X of measurements such that measurements in V \ X are non-contradictory.

We expect |X | ≤ 10.

The number 10 can be treated as a measure of hardness of the instance.

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Vertex cover

Vertex cover of a graph G = (V , E ) is a set S ⊆ V which touches all edges, i.e. for every uv ∈ E we have u ∈ S or v ∈ S.

Decision problem

Input: Graph G = (V , E ), a number k ∈ N.

Does G admit a vertex cover of size ≤ k?

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Vertex cover: algorithm in time O(2

^k

· (|V | + |E |))

Decision problem

Exhaustive search in time O(n^k · |E |) VertexCover(G = (V , E ),k)

1: for each X ⊆ V , |X | = k do

2: ifall edges incident to X return YES

3: return NO

Not very good for k = 10.

Greedy improvements (pick vertices of maximum degree) do not help.

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Vertex cover: algorithm in time O(2

^k

· (|V | + |E |))

Decision problem

Exhaustive search in time O(n^k · |E |) VertexCover(G = (V , E ),k)

1: for each X ⊆ V , |X | = k do

2: ifall edges incident to X return YES

3: return NO

Not very good for k = 10.

Greedy improvements (pick vertices of maximum degree) do not help.

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Vertex cover: algorithm in time O(2

^k

· (|V | + |E |))

Decision problem

Branching algorithm in time O(2^k· (|V | + |E |)) VertexCover(G = (V , E ),k)

1: if E = ∅ return YES

2: if k = 0 return NO

3: Pick any edge uv ∈ E

4: VertexCover(G − u,k − 1) or VertexCover(G − v ,k − 1) Reasonable running time for k = 10.

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Parameterized decision problems

Let Σ be a finite alphabet. Say, Σ = {0, 1}.

(Classical) decision problem

Decision problem is any subset L of words over Σ, i.e., L ⊆ Σ^∗. Intuition: L is a set of (binary encodings of) YES-instances.

Example: VertexCover is a set of binary encodings of pairs (G , k) such that G admits a vertex cover of size k.

Parameterized decision problem

Parameterized decision problem is any subset L ⊆ Σ^∗× N.

For an instance (x, k) ∈ Σ^∗× N, k is called a parameter.

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Parameterized decision problems: examples

Vertex cover parameterized by the solution size

L = {(e(G ), k) : G admits a vertex cover of size k}; e is an encoding

Clique parameterized by the maximum degree

L = {((G , k), ∆) : G contains a clique of size k and ∆ = ∆(G )} Independent Set parameterized by treewidth

L = {((G , k), t) :

G admits an independent set of size k and G has treewidth t} The central question of the area:

How does the complexity of a parameterized problem depend on the parameter?

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Parameterized decision problems: examples

Vertex cover parameterized by the solution size L = {(G , k) : G admits a vertex cover of size k}

L = {((G , k), ∆) : G contains a clique of size k and ∆ = ∆(G )} Independent Set parameterized by treewidth

L = {((G , k), t) :

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Parameterized decision problems: examples

L = {((G , k), ∆) : G contains a clique of size k and ∆ = ∆(G )}

Independent Set parameterized by treewidth L = {((G , k), t) :

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Parameterized decision problems: examples

G admits an independent set of size k and G has treewidth t}

The central question of the area:

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Parameterized decision problems: examples

G admits an independent set of size k and G has treewidth t}

The central question of the area:

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Parameterized algorithm: definition

Let L be a parameterized problem.

Parameterized algorithm

Algorithm A is a parameterized algorithm for L if there is a computable function f : N → N and a constant c such that

for every (x, k) ∈ Σ^∗× N, algorithm A correctly decides if (x, k) ∈ L in timef (k) · n^c

where n = |(x, k)| is the size of the instance.

Problem L is then called fixed parameter tractable (FPT)

A is also called FPT algorithm. Class of decision problems:

FPT = {L ⊆ Σ^∗× N : L admits an FPT algorithm}

Note: P ⊆ FPT, but FPT 6⊆ NP.

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Parameterized algorithm: definition

Algorithm A is a parameterized algorithm for L if there is a computable function f : N → N such that

for every (x, k) ∈ Σ^∗× N, algorithm A correctly decides if (x, k) ∈ L in timef (k) · n^O(1)

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Parameterized algorithm: definition

Examples: time O(2^kn), O(k! · n¹⁵), O(10^k+ n³), O(2²^kn³), O(2²^2kn⁵), O(2²²

···2

| {z }

k times

n²)

BAD examples: O(2ⁿk), O(n^k), O(n^{log log k})

FPT = {L ⊆ Σ^∗× N : L admits an FPT algorithm} Note: P ⊆ FPT, but FPT 6⊆ NP.

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Parameterized algorithm: definition

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Parameterized algorithm: definition

Problem L is then called fixed parameter tractable (FPT) A is also called FPT algorithm.

Class of decision problems:

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Parameterized algorithm: definition

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Parameterized algorithm: definition

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Parameterized algorithm: definition

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XP algorithm: definition

XP algorithm

Algorithm A is an XP algorithm for L if

there are two computable functions f , g : N → N such that

for every (x, k) ∈ Σ^∗× N, algorithm A correctly decides if (x, k) ∈ L in timef (k) · n^{g (k)}

think: polynomial for every fixed k. Examples: time O(2^kn), O(2²²^···2

| {z }

k times

n²), O(n^{log log k}), O(n^k), O(n²^2k). BAD examples: O(2ⁿk), O(kⁿ)

XP = {L ⊆ Σ^∗× N : L admits an XP algorithm} Note: P ⊆ FPT ⊆ XP.

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XP algorithm: definition

XP algorithm

think: polynomial for every fixed k.

Examples: time O(2^kn), O(2²²^···2

| {z }

k times

n²), O(n^{log log k}), O(n^k), O(n²^2k). BAD examples: O(2ⁿk), O(kⁿ)

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XP algorithm: definition

XP algorithm

| {z }

k times

n²), O(n^{log log k}), O(n^k), O(n²^2k).

BAD examples: O(2ⁿk), O(kⁿ) Class of decision problems:

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XP algorithm: definition

XP algorithm

| {z }

k times

BAD examples: O(2ⁿk), O(kⁿ)

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XP algorithm: definition

XP algorithm

| {z }

k times

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XP algorithm: definition

XP algorithm

| {z }

k times

XP = {L ⊆ Σ^∗× N : L admits an XP algorithm}

Note: P ⊆ FPT ⊆ XP.

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XP algorithm: definition

XP algorithm

| {z }

k times

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Technique: branching

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Vertex Cover

Branching algorithm in time O(2^k· (|V | + |E |)) VertexCover(G = (V , E ),k)

1: if E = ∅ return YES

3: Pick any edge uv ∈ E

4: VertexCover(G − u,k − 1) or VertexCover(G − v ,k − 1) Recursion tree of depth at most k.

Each node in the recursion tree has at most 2 children.

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Faster Vertex Cover

Faster branching algorithm VertexCover(G = (V , E ),k)

1: ifmaximum degree ≤ 2, solve in linear time

3: Pick v of maximum degree

4: VertexCover(G − v ,k − 1) or VertexCover(G − N(v ),k − |N(v )|)

T (k) ≤ T (k − 1) + T (k − 3)

T (k) ≤ 1.4656^k.

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Faster Vertex Cover

T (k) ≤ T (k − 1) + T (k − 3)

T (k) ≤ 1.4656^k.

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Faster Vertex Cover

T (k) ≤ T (k − 1) + T (k − 3)

T (k) ≤ 1.4656^k.

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General Idea

A recursive algorithm

At every recursive call, we branch into f (k) cases Depth of the recursion is g (k).

Example:

Vertex Cover, first algorithm: f (k) = 2, g (k) = k.

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FVS in tournaments

Feedback Vertex Set (FVS) in Tournaments Input: a tournament (oriented clique) T , integer k Question: is there a subset X ⊆ V (T ) of size at most k,

such that T \ X is acyclic

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FVS in tournaments

Feedback Vertex Set (FVS) in Tournaments Input: a tournament (oriented clique) T , integer k Question: is there a subset X ⊆ V (T ) of size at most k,

such that T \ X is acyclic

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FVS in tournaments

Lemma

If a tournament contains a cycle, then it contains a 3-cycle.

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FVS in tournaments

Lemma

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FVS in tournaments

Lemma

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FVS in tournaments

Lemma

Branching algorithm in time O^∗(3^kn³) FVS(G = (V , E ),k)

1: if k < 0 return NO

2: if G has no directed 3-cycle return YES

3: Let abc be a directed 3-cycle

4: return FVS(G − a,k − 1) or FVS(G − b,k − 1) or FVS(G − c,k − 1)

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Technique: kernelization

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Kernelization: FPT by preprocessing

NP-hard problem instance

k k

poly time

size ≤ f (k) want: f (k) = poly(k) Instead of saying ‘L has a kernelization algorithm which returns an instance of size f (k)’ we say shortly ‘L has a kernel of size f (k)’. Kernel of size f (k) for a problem in NP implies FPT algorithm in time O(exp(poly(f (k))))

One can show that an FPT algorithm implies a kernel of size f (k). Usually, we are interested in polynomial kernels (of size k^O(1)).

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Kernelization: FPT by preprocessing

NP-hard problem instance k

k

poly time

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Kernelization: FPT by preprocessing

k

poly time

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Kernelization: FPT by preprocessing

k

poly time

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Kernelization: FPT by preprocessing

k

poly time

size ≤ f (k)

want: f (k) = poly(k) Instead of saying ‘L has a kernelization algorithm which returns an instance of size f (k)’ we say shortly ‘L has a kernel of size f (k)’. Kernel of size f (k) for a problem in NP implies FPT algorithm in time O(exp(poly(f (k))))

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Kernelization: FPT by preprocessing

k

poly time

size ≤ f (k) want: f (k) = poly(k)

Instead of saying ‘L has a kernelization algorithm which returns an instance of size f (k)’ we say shortly ‘L has a kernel of size f (k)’. Kernel of size f (k) for a problem in NP implies FPT algorithm in time O(exp(poly(f (k))))

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Kernelization: FPT by preprocessing

k

poly time

size ≤ f (k) want: f (k) = poly(k) Kernelization algorithm for a parameterized problem L is any polynomial-time algorithm which, given an instance (x, k) returns an instance (x⁰, k⁰) such that:

Instead of saying ‘L has a kernelization algorithm which returns an instance of size f (k)’ we say shortly ‘L has a kernel of size f (k)’. Kernel of size f (k) for a problem in NP implies FPT algorithm in time O(exp(poly(f (k))))

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Kernelization: FPT by preprocessing

k

poly time

size ≤ f (k) want: f (k) = poly(k) Instead of saying ‘L has a kernelization algorithm which returns an instance of size f (k)’ we say shortly ‘L has a kernel of size f (k)’.

Kernel of size f (k) for a problem in NP implies FPT algorithm in time O(exp(poly(f (k))))

One can show that an FPT algorithm implies a kernel of size f (k).

Usually, we are interested in polynomial kernels (of size k^O(1)).

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A simple kernel for vertex cover

Kernelization algorithm:

1 While there is v ∈ V with deg(v ) > k, remove v and decrease k by 1.

2 While there is v ∈ V with deg(v ) = 0, remove v .

3 If now |E | > k² return NO.

4 return (G , k).

Running time O(|E | + |V |). Theorem

Vertex Cover admits a kernel with at most k² edges. Theorem

Vertex Cover admits a kernel with at most 2k² vertices. Proof:

|V | ≤² X

v ∈V

deg(v ) = 2|E | ≤ 2k²

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A simple kernel for vertex cover

Why correct?

(G , k) ∈ VertexCover ⇔ (G − v , k − 1) ∈ VertexCover

(⇐) Let X⁰ be a vertex cover in G − v .

Then X⁰∪ {v } is a vertex cover of size k in G . (⇒) Let X be a vertex cover in G , |X | ≤ k.

ThenX contains v (why?).

So X \ {v } is a vertex cover of size ≤ k − 1 in G − v .

4 return (G , k).

|V | ≤² X

v ∈V

deg(v ) = 2|E | ≤ 2k²

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A simple kernel for vertex cover

4 return (G , k).

|V | ≤² X

v ∈V

deg(v ) = 2|E | ≤ 2k²

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A simple kernel for vertex cover

2 While there is v ∈ V with deg(v ) = 0, remove v . Why correct?

(G , k) ∈ VertexCover ⇔ (G − v , k) ∈ VertexCover

(⇐) Let X⁰ be a vertex cover of size k in G − v . Then X⁰ is a vertex cover of size k in G .

(⇒) Let X be a minimum vertex cover in G , |X | ≤ k.

Thenv 6∈ X (why?).

So X is a vertex cover of size ≤ k in G − v .

4 return (G , k).

|V | ≤² X

v ∈V

deg(v ) = 2|E | ≤ 2k²

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A simple kernel for vertex cover

4 return (G , k).

|V | ≤² X

v ∈V

deg(v ) = 2|E | ≤ 2k²

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A simple kernel for vertex cover

3 If now |E | > k² return NO. Why correct?

If YES, i.e., exists a vertex cover X , |X | ≤ k then

|E | ≤X

x ∈X

deg(x ) ≤ X

x ∈X

k ≤ k².

4 return (G , k).

|V | ≤² X

v ∈V

deg(v ) = 2|E | ≤ 2k²

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A simple kernel for vertex cover

4 return (G , k).

|V | ≤² X

v ∈V

deg(v ) = 2|E | ≤ 2k²

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A simple kernel for vertex cover

4 return (G , k).

Running time O(|E | + |V |).

Theorem

|V | ≤² X

v ∈V

deg(v ) = 2|E | ≤ 2k²

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A simple kernel for vertex cover

4 return (G , k).

Theorem

Vertex Cover admits a kernel with at most k² edges.

Theorem

|V | ≤² X

v ∈V

deg(v ) = 2|E | ≤ 2k²

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A simple kernel for vertex cover

4 return (G , k).

Theorem

Vertex Cover admits a kernel with at most k² edges.

Theorem

Vertex Cover admits a kernel with at most 2k² vertices.

Proof:

|V | ≤² X

v ∈V

deg(v ) = 2|E | ≤ 2k²

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FPT algorithms from the kernel

Theorem

Vertex Cover admits a kernel with at most 2k² vertices, and it can be computed in time O(|E | + |V |).

Exhaustive search: ^2k_k² = k^O(k).

Whole FPT algorithm: O(|E | + |V | + k^O(k))

Branching: 2^k · k². Whole FPT algorithm: O(|E | + |V | + 2^k· k²).

Compare with O(2^k· (|V | + |E |))

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Kernel of size 2k vertices for VertexCover

Consider the standard integer linear program for finding minimum vertex cover.

min P

v ∈V xv

x_u+ x_v ≥ 1 for each uv ∈ E xv ≥ 0 for each v ∈ V . xv ∈ Z for each v ∈ V .

... and its relaxation (a linear program):

min P

v ∈V x_v

xu+ xv ≥ 1 for each uv ∈ E xv ≥ 0 for each v ∈ V .

Theorem (Khachiyan 1979)

Linear programming admits a polynomial time algorithm.

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Kernel of size 2k vertices for VertexCover

Consider the standard integer linear program for finding minimum vertex cover.

min P

v ∈V xv

x_u+ x_v ≥ 1 for each uv ∈ E xv ≥ 0 for each v ∈ V . xv ∈ Z for each v ∈ V . ... and its relaxation (a linear program):

min P

v ∈V x_v

xu+ xv ≥ 1 for each uv ∈ E x_v ≥ 0 for each v ∈ V .

Theorem (Khachiyan 1979)

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Vertex Cover LP

cost: n − 1

cost: n/2

Vertex Cover LP

min X

v ∈V (G )

xv

s.t. x_u+ x_v ≥ 1 ∀_{uv ∈E (G )}

xv ≥ 0 ∀_{v ∈V (G )}

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Vertex Cover LP

cost: n − 1

cost: n/2

Vertex Cover LP

min X

v ∈V (G )

xv

s.t. x_u+ x_v ≥ 1 ∀_{uv ∈E (G )}

xv ≥ 0 ∀_{v ∈V (G )}

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Vertex Cover LP

cost: n − 1

cost: n/2

Vertex Cover LP

min X

v ∈V (G )

xv

s.t. x_u+ x_v ≥ 1 ∀_{uv ∈E (G )}

xv ≥ 0 ∀_{v ∈V (G )}

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Vertex Cover LP: vertex classes

xv = 0 ⇐⇒ v ∈ V₀

0 < xv < ¹₂ ⇐⇒ v ∈ Vε

xv =¹₂ ⇐⇒ v ∈ V¹

2

1

2< xv < 1 ⇐⇒ v ∈ V1−ε

x_v = 1 ⇐⇒ v ∈ V1

V_<¹

2 = V0∪ Vε

V_>1

2 = V₁∪ V1−ε

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Vertex Cover LP: analysis

+ε −ε

1 V_<¹

2

independent, E (V_<1 2, V¹

2) = ∅,

⇒ ∀_{v ∈V}

< 12

N(v ) ⊆ V_>¹

2.

2 |V_<1 2

| ≥ |V_>1 2

|, otherwise +ε on V_<¹

2, −ε on V_>¹

2.

3 ∀_{X ⊆V}

> 12

|X | ≤ |N(X ) ∩ V_<1 2

|, otherwise +ε on N(X ) ∩ V_<¹

2, −ε on X .

4 Hall’s theorem

⇒ matching saturating V_>1 2.

5 Greedily take V_>1 2

and discard V_<1 2

.

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Vertex Cover LP: analysis

+ε −ε

1 V_<¹

2

2) = ∅,

⇒ ∀_{v ∈V}

< 12

N(v ) ⊆ V_>¹

2.

2 |V_<1 2

| ≥ |V_>1 2

2, −ε on V_>¹

2.

3 ∀_{X ⊆V}

> 12

|X | ≤ |N(X ) ∩ V_<1 2

2, −ε on X .

4 Hall’s theorem

and discard V_<1 2

.

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Vertex Cover LP: analysis

+ε −ε

1 V_<¹

2

2) = ∅,

⇒ ∀_{v ∈V}

< 12

N(v ) ⊆ V_>¹

2.

2 |V_<1 2

| ≥ |V_>1 2

2, −ε on V_>¹

2.

3 ∀_{X ⊆V}

> 12

|X | ≤ |N(X ) ∩ V_<1 2

2, −ε on X .

4 Hall’s theorem

and discard V_<1 2

.

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Vertex Cover LP: analysis

+ε −ε

1 V_<¹

2

2) = ∅,

⇒ ∀_{v ∈V}

< 12

N(v ) ⊆ V_>¹

2.

2 |V_<1 2

| ≥ |V_>1 2

2, −ε on V_>¹

2.

3 ∀_{X ⊆V}

> 12

|X | ≤ |N(X ) ∩ V_<1 2

2, −ε on X .

4 Hall’s theorem

and discard V_<1 2

.

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Vertex Cover LP: analysis

+ε −ε

1 V_<¹

2

2) = ∅,

⇒ ∀_{v ∈V}

< 12

N(v ) ⊆ V_>¹

2.

2 |V_<1 2

| ≥ |V_>1 2

2, −ε on V_>¹

2.

3 ∀_{X ⊆V}

> 12

|X | ≤ |N(X ) ∩ V_<1 2

2, −ε on X .

4 Hall’s theorem

and discard V_<1 2

.

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Vertex Cover LP: analysis

+ε −ε

1 V_<¹

2

2) = ∅,

⇒ ∀_{v ∈V}

< 12

N(v ) ⊆ V_>¹

2.

2 |V_<1 2

| ≥ |V_>1 2

2, −ε on V_>¹

2.

3 ∀_{X ⊆V}

> 12

|X | ≤ |N(X ) ∩ V_<1 2

2, −ε on X .

4 Hall’s theorem

and discard V_<1 2

.

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Vertex Cover LP: analysis

−ε +ε

+ε −ε

1 E (V0, V1−ε) = ∅,

⇒ N(V_1−ε) ∩ V_<¹

2

⊆ V_ε.

2 if |V_1−ε| ≤ |V_ε|, then

we can +ε on V_1−ε, −ε on V_ε.

3 if |V_1−ε| ≥ |V_ε|, then

we can +ε on V_ε, −ε on V_1−ε.

4 conclusion: there exists LP optimum with V = V0∪ V1

2 ∪ V₁.

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Vertex Cover LP: analysis

−ε +ε

+ε −ε

1 E (V0, V1−ε) = ∅,

⇒ N(V_1−ε) ∩ V_<¹

2

⊆ V_ε.

2 if |V_1−ε| ≤ |V_ε|, then

3 if |V_1−ε| ≥ |V_ε|, then

2 ∪ V₁.

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Vertex Cover LP: analysis

−ε +ε

+ε −ε

1 E (V0, V1−ε) = ∅,

⇒ N(V_1−ε) ∩ V_<¹

2

⊆ V_ε.

2 if |V1−ε| ≤ |V_ε|, then

3 if |V_1−ε| ≥ |V_ε|, then

2 ∪ V₁.

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Vertex Cover LP: analysis

−ε +ε

+ε −ε

1 E (V0, V1−ε) = ∅,

⇒ N(V_1−ε) ∩ V_<¹

2

⊆ V_ε.

3 if |V_1−ε| ≥ |V_ε|, then

2 ∪ V₁.

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Vertex Cover LP: analysis

−ε +ε

+ε −ε

1 E (V0, V1−ε) = ∅,

⇒ N(V_1−ε) ∩ V_<¹

2

⊆ V_ε.

3 if |V_1−ε| ≥ |V_ε|, then

2 ∪ V₁.

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Vertex Cover LP: summary of analysis

1 There exists LP optimum with V = V0∪ V¹

2 ∪ V₁.

I Half-integrality

2 If V¹

2 6= V , then we can reduce something.

I Take V1 into solution, and discard V0.

unique LP optimum

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Vertex Cover LP: summary of analysis

1 There exists LP optimum with V = V0∪ V¹

2 ∪ V₁.

I Half-integrality

2 If V¹

2 6= V , then we can reduce something.

I Take V1 into solution, and discard V0.

unique LP optimum

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Kernel of size 2k vertices for VertexCover

Theorem (Nemhauser, Trotter 1974) Let {x_v}_{v ∈V} be any optimal solution of the vertex cover LP. Partition V :

V0 = {v ∈ V (G ) : xv < ¹₂}, V_1/2= {v ∈ V (G ) : xv = ¹₂}, V1 = {v ∈ V (G ) : x_v > ¹₂}.

Then there is a minimum vertex cover S of G such that V₁ ⊆ S, V₀∩ S = ∅.

Vertex Cover LP:

min P

v ∈Vxv

∀ uv ∈ E x_u+ x_v ≥ 1

∀ v ∈ V x_v ≥ 0.

(G , k) ∈ VertexCover ⇔ (G [V1/2], k − |V₁|) ∈ VertexCover (⇒) SinceV1 ⊆ S and|S| ≤ k we have |S \ X1|=|S| − |V₁|≤k − |V₁|.

Also, S \ V₁ is a vertex cover in G [V_1/2]. (check it!)

0

1 Solve the vertex cover LP and get an optimal solution {x_v}_{v ∈V}.

2 If |V_1/2| > 2k return NO.

3 return (G [V_1/2], k − |V₁|). Theorem

Vertex cover admits a kernel of vertex size at most 2k.

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Kernel of size 2k vertices for VertexCover

Vertex Cover LP:

min P

v ∈Vxv

∀ uv ∈ E x_u+ x_v ≥ 1

∀ v ∈ V x_v ≥ 0.

3 return (G [V_1/2], k − |V1|). Theorem

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Kernel of size 2k vertices for VertexCover

Vertex Cover LP:

min P

v ∈Vxv

∀ uv ∈ E x_u+ x_v ≥ 1

∀ v ∈ V x_v ≥ 0.

2 If |V_1/2| > 2k return NO. Why correct?

Then,P

x_v > k, so the integral solution has also value > k.

3 return (G [V_1/2], k − |V1|). Theorem

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Kernel of size 2k vertices for VertexCover

Vertex Cover LP:

min P

v ∈Vxv

∀ uv ∈ E x_u+ x_v ≥ 1

∀ v ∈ V x_v ≥ 0.

3 return (G [V_1/2], k − |V1|).

Theorem

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Kernel of size 2k vertices for VertexCover

Vertex Cover LP:

min P

v ∈Vxv

∀ uv ∈ E x_u+ x_v ≥ 1

∀ v ∈ V x_v ≥ 0.

3 return (G [V_1/2], k − |V1|).

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Vertex Cover LP branching

Assumption: V1

2 = V is the unique LP optimum.

Pick any vertex v .

Branch 1:

Take v into solution. Force x_v = 1 in the LP. LP value goes up by ¹₂. (k − LP) goes down by ¹₂.

Branch 2:

Forbid v to be in the solution. Force x_v = 0 in the LP.

LP value goes up by at least ¹₂. (k − LP) goes down by at least ¹₂.

Theorem

Vertex Cover can be solved in 4^k−LP · (|V | + |E |)^O(1) time.

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Vertex Cover LP branching

Assumption: V1

Pick any vertex v . Branch 1:

Take v into solution.

Force x_v = 1 in the LP.

LP value goes up by ¹₂. (k − LP) goes down by ¹₂.

Branch 2:

Forbid v to be in the solution. Force x_v = 0 in the LP.

Theorem

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Vertex Cover LP branching

Assumption: V1

Branch 2:

Forbid v to be in the solution.

Theorem

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Vertex Cover LP branching

Assumption: V1

Branch 2:

Forbid v to be in the solution.

Theorem

k−LP O(1)

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A few examples of kernels

The problems below ask for a set vertices of size k and k is the parameter.

Size of the kernel = number of vertices.

VertexCover: 2k

Feedback Vertex Set: O(k²) Odd Cycle Transversal: k^O(1) For planar graphs:

Dominating Set 67k,

Feedback Vertex Set 13k, Induced Matching 28k,

Connected Vertex Cover ¹¹₃k,

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Thanks and credits

Thanks!

Credits:

Slides mostly compiled from previous lectures of:

I Łukasz Kowalik

I Marek Cygan

I Michał Pilipczuk

Some nice figures by Felix Reidl.

Tikz faces based on a code by Raoul Kessels,

http://www.texample.net/tikz/examples/emoticons/, under Creative Commons Attribution 2.5 license (CC BY 2.5)