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KERNEL LOWER BOUNDS: A SURVEY

By

Ashutosh Rai

THE INSTITUTE OF MATHEMATICAL SCIENCES, CHENNAI.

A thesis submitted to theBoard of Studies in Theoretical Computer Science

In partial fulfillment of the requirements

For the Degree of

Master of Science

of

HOMI BHABHA NATIONAL INSTITUTE

April 2012

CERTIFICATE

I, hereby declare that the investigation presented in the

thesis has been carried out by me. The work is original and

the work has not been submitted earlier as a whole or in

part for a degree/diploma at this or any other Institution or

University.

Venkatesh Raman

ACKNOWLEDGEMENTS

First of all I thank my thesis advisor Professor Venkatesh Raman for his guid-

ance and his time. I feel very fortunate that I have got the opportunity to work

under him, and to have discussions with him, which invariably turn out to make

me learn something I would have struggled to know otherwise.

I thank Professor Saket Saurabh for all the pain he has gone through while

reading the earlier drafts of this thesis. He was available for all my questions

anytime. He has guided me in the past two years in more ways than I can count.

I am grateful to all the faculty at the Department of Theoretical Computer

Science at IMSc. Whatever little I know about Computer Science, I owe it to

them.

I also thank M S Ramanujan, Fahad Pnolan, Sudeshan Kolay and Esha Ghosh,

who took the pain of sitting through my presentations and tried to make sense

of what I said. Their questions during the course of presentations have led me

to explore things in a way which would not have been possible otherwise. Addi-

tional thanks to M S Ramanujan for the fonts and style files.

I also owe one to Gaurav, for helping me with his insights in mathematical as-

pects, whenever I needed them. I thank Anup, Joydeep, Raja, Nitin and Meesum

for all the help they provided me with in the matter of science, or otherwise.

Lastly, I thank my family, which took care of all other things, so that I could

concentrate on my work here.

Abstract

Data reduction techniques are widely applied to deal with computationally hard

problems in real world applications. It has been a long standing challenge to for-

mally express the efficiency and accuracy of these preprocessing procedures. The

framework of parameterized complexity is particularly suitable for analysis of

preprocessing heuristics. A kernelization algorithm is a preprocessing algorithm

which simplifies the instances in polynomial time and the extent of simplification

desired is quantified with the help of the additional parameter. The output of the

algorithm is called a kernel.While many problems were known to have polynomial kernels, no technique

was known to rule out polynomial sized kernels until the recent work of Bodlaen-

der et al. [2] and Fortnow and Santhanam [10]. These techniques, called com-

position and distillation, turned out to be very helpful in proving kernel lower

bounds on various problems like k-PATH, w-INDEPENDENT SET and many others.

The techniques and framework have been further strengthened and general-

ized recently. The main contribution came from Dell and Melkebeek [7], which

allowed nondeterminism in compositions and looked at the proof of Fortnow

and Santhanam which allowed them to give polynomial lower bounds. There

have been many new frameworks developed based on the results of Dell and

Melkebeek.

The aim of this thesis is to look at all the techniques used to get lower bounds

on kernelization. We look at each of the techniques one by one, and try to analyze

their similarities and differences. Finally, a unified view is presented at the end

of the thesis. This framework is easy to obtain from the earlier frameworks and

generalizes all of them.

4

Contents

1 Introduction 1

1.1 Organization of the thesis . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Notations, Definitions and Conventions . . . . . . . . . . . . . . . 3

1.2.1 Growth of Functions . . . . . . . . . . . . . . . . . . . . . 3

1.2.2 Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

I Distillation Based Lower Bounds 5

2 Composition and Distillation 6

2.1 Basics of Parameterized Complexity . . . . . . . . . . . . . . . . . 6

2.2 Ruling Out Polynomial Kernels . . . . . . . . . . . . . . . . . . . 7

2.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 Cross Composition 11

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.3 Machinery Based on Cross-composition . . . . . . . . . . . . . . 12

3.4 Results based on Cross-composition . . . . . . . . . . . . . . . . . 15

3.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

II Communication Protocol Based Lower Bounds 21

4 Communication Protocol based Lower Bounds 22

4.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

4.2 Communication Protocol . . . . . . . . . . . . . . . . . . . . . . . 23

4.3 Complementary Witness Lemma . . . . . . . . . . . . . . . . . . 24

4.4 Application to Lower Bounds: Vertex Cover . . . . . . . . . . . . 28

4.4.1 Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.4.2 Lower Bounds on Kernelization . . . . . . . . . . . . . . . 30

4.4.3 NP-completeness of MULTICOLORED BICLIQUE . . . . . . . 31

4.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

i

Contents

5 Weak Composition 33

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

5.2 Framework for Weak Composition . . . . . . . . . . . . . . . . . 34

5.3 An example of weak composition: RAMSEY . . . . . . . . . . . . . 36

5.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

6 Co-nondeterminism in Composition 42

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

6.2 coNP-Composition . . . . . . . . . . . . . . . . . . . . . . . . . . 43

6.3 Application of coNP-composition: RAMSEY . . . . . . . . . . . . . 45

6.4 coNP-composition for REFINEMENT RAMSEY . . . . . . . . . . . . 45

6.4.1 Embedding Construction . . . . . . . . . . . . . . . . . . . 46

6.4.2 Construction of Host Graph . . . . . . . . . . . . . . . . . 48

6.4.3 Composition . . . . . . . . . . . . . . . . . . . . . . . . . 49

6.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

III Summing Up 51

7 Unified View 52

7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

7.2 The Idea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

7.3 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

7.4 Relation to Earlier Techniques . . . . . . . . . . . . . . . . . . . . 56

7.4.1 Cross Composition . . . . . . . . . . . . . . . . . . . . . . 56

7.4.2 Weak Composition . . . . . . . . . . . . . . . . . . . . . . 56

7.4.3 coNP-composition . . . . . . . . . . . . . . . . . . . . . . 57

8 Conclusion 58

ii

1Introduction

When encountered with a hard problem, we try to simplify it by using some

preprocessing. Of course, the notion is vague and several questions arise. “How

do we go about quantifying the quality of the simplification procedure?”, “How

easy a rule should be to be called as a preprocessing rule?”It turns out that even

though the questions are natural, they did not have satisfactory answers for a

long time.

Single parameter approach of classical complexity makes it hard for us to an-

swer these questions. Going by the convention of easy in complexity theory, we

say that preprocessing is easy if it can be performed in polynomial time. But

we don’t expect them to solve the problem fully in polynomial time. So we say

that they are good if they reduce the size of the problem. But if we have a pre-

processing algorithm which can preprocess an instance, say I of an NP-complete

problem, and produces an instance I ′ of the same problem while preserving the

answer such that |I ′| < |I|, then we can solve the problem in polynomial time

by repeating the procedure and getting a trivial instant in the end. It implies

that P=NP, which is believed to be false. Also, in classical complexity we don’t

have anything else apart from the problem length to get our hands on, and this

makes it difficult to analyze preprocessing. The picture changed drastically af-

ter the arrival of parameterized complexity. Parameterized complexity considers

another parameter apart from the input length, and that helps us analyze the

preprocessing better.

In parameterized complexity, a parameter, say k is associated with the input.

Here, for the ease of preprocessing, like earlier, we call the preprocessing algo-

1

Chapter 1. Introduction

rithm good, if it can be carried out in polynomial time. For dealing with the

quality, now we have got another handle, namely the parameter k. We use it to

measure the ’quality’ of preprocessing rules. We require that the size of the out-

put instance of the preprocessing rules is f(k), independent of the initial input

length. In this case, the preprocessing algorithm is called a kernelization and the

output instance is called a kernel.Again, a few natural questions after defining the above notion is, “Does a

problem have a kernel?”and “Does a problem have a kernel of polynomial size?

”While the techniques to answer the first question have been known for last two

decades, the techniques to answer the second question have been developed only

recently.

In this survey, we will start with the concept of distillation and composition

and will see how these techniques are generalized and used to prove kernel

lower bounds for many problems. The main emphasis is on the techniques used

to arrive at lower bounds, but we will also give a few examples to illustrate the

techniques.

1.1 Organization of the thesis

We introduce and motivate the concept of preprocessing in the first chapter. We

also give the basic notations and definitions will be used in the thesis.

In the second Chapter, we will briefly look at the basic concepts of param-

eterized complexity like FPT and kernelization. Then we define the notions of

composition and distillation and illustrate with help of one example how they can

be used to derive lower bounds.

In the third chapter, we see how the notion of composition can be generalized

so that it makes it easy to use for certain problems. The framework is called cross-composition and we will see with an example how it can be applied to yield lower

bounds.

In the fourth chapter, we will have a look at work by Dell and Melkebeek [7]

which generalizes the work of Fortnow and Santhanam [10] mentioned in the

second chapter. We give a proof of the main result in the paper, i.e. complemen-tary witness lemma which is mostly along the line of proof given by Fortnow and

2


Santhanam. Then we look at the example of VERTEX COVER and give a proof

that it does not have a kernel of size O(k2−ε).

Fifth and sixth chapters give the notion of weak d-composition and coNP-composition respectively. They give a framework which makes it easy to use

complementary witness lemma to arrive at lower bounds on kernels.

In the seventh chapter, we conclude by giving a unified view of all the frame-

works.

1.2 Notations, Definitions and Conventions

We assume that the reader is familiar with basic notions like sets, functions,

polynomials, relations, integers etc. In particular, for these notions we follow the

same notations as that in [17].

1.2.1 Growth of Functions

We employ mainly the big-Oh (O) notation (see [5]). Let f : N → N and g :

N → N be two functions from Natural numbers to Natural numbers. We say

that f(n) = O(g(n)) if there exist constants c, and n0 such that for all n ≥ n0,

f(n) ≤ c · g(n).

1.2.2 Graphs

An undirected graph G is a pair (V,E) where V and E are unordered sets. The

elements of V are called vertices of G. E consists of unordered pairs of vertices

and elements of E are called edges of G. A vertex u and a vertex v are said to

be adjacent if E contains the pair (u, v). The edge (u, v) is said to be incident on

the vertices u and v, while u and v are called the endpoints of the edge (u, v). An

undirected graph G is called a simple undirected graph if there is no edge in E

of the form (v, v) where v is a vertex of G. In this thesis, the graphs we consider

are all simple undirected graphs.

Let G = (V,E) be a graph. The graph G′ = (V ′, E ′) is subgraph of G if V ′ ⊆ V

and E ′ ⊆ E.

3


Let G = (V,E) be a graph and A ⊆ V then by G[A] we denote the subgraph

induced on A. Here G[A] = (A,E ′) where E ′ = E ∩ (A× A).

A simple graphG = (V,E) is called a clique if it contains all the edges possible,

i.e. (u, v) ∈ E for all u, v ∈ V, u 6= v . A graph G = (V,E) is called an independentset if it has no edges, i.e. E = φ.

We call a graph G = (U ] V,E) bipartite if G[U ] and G[V ] are independent

sets. Similarly, a graph G = (V1] ·s]Vk, E) is k-partite if G[Vi] is an independent

set for all i ∈ [k]. We call a bipartite graph G = (U ] V,E) a complete bipartitegraph if it has all the edges across, i.e. (u, v) ∈ E for all u ∈ U, v ∈ V . Similarly,

we call a k-partite graph a complete k-partite graph if it has all the edges possible

between different Vi’s, i.e. (u, v) ∈ E for all u ∈ Vi, v ∈ Vj for all i, j ∈ [k], i 6= j.

We will also need the notion of join of two graphs which we define as follows-

Definition 1. A graph H obtained by the join of two graphs G = (V,E) and G′ =

(V ′, E ′) is a graph on vertex set V ]V ′ and edge set E ]E ′ ]Eacross, where (u, v) ∈Eacross for all u ∈ V, v ∈ V ′.

4

Part I

Distillation Based Lower Bounds

5

2Composition and Distillation

In this chapter, we introduce the basic notions of parameterized complexity. We

also introduce the framework provided by Bodlaender et al. [2] and results by

Fortnow and Santhanam [10] and how these can be used to rule out polynomial

kernels with the help of an example.

2.1 Basics of Parameterized Complexity

Definition 2. [16] A parameterized problem is a language L ⊆ Σ∗ × N, where Σis a finite alphabet. The second component is called the parameter of the problem.

Definition 3. [16] A parameterized problem L is said to be Fixed Parameter Tractable(FPT) if it can be determined in time f(k). nO(1) whether or not (x, k) ∈ L, where fis a computable function depending only on k and n = |(x, k)|. The complexity classcontaining all fixed parameter tractable problems is called FPT. Unless specified oth-erwise, we will assume that the parameter k is a non negative integer encoded witha unary alphabet.

Definition 4. [2] We associate an instance (x, k) of a parameterized problemwith the unparameterized instance formed by the string x#1k , where # denotesa new character that we add to the alphabet and 1 is an arbitrary letter in Σ.The unparameterized version of a parameterized problem Q is the language Q =

x#1k|(x, k) ∈ Q.

Definition 5. Let Q be a parameterized problem over Σ∗ × N . A polynomial timecomputable function K : Σ∗ × N → Σ∗ × N is a kernelization of Q if there is a

6

Chapter 2. Composition and Distillation

computable function h : N→ N such that for all (x, k) ∈ Σ∗ × N we have

((x, k) ∈ Q⇔ K((x, k)) ∈ Q)

and|K((x, k))| ≤ h(k).

If K is a kernelization of Q, then for every instance (x, k) of Q the image K((x, k))

is called the kernel of (x, k) (under K).

K is called a polynomial kernelization of Q, if the function h is polynomial.

Observe that a kernelization is a polynomial time many-one reduction of a

problem to itself with the additional property that the image is bounded in terms

of the parameter of the argument.

Theorem 1. [9] For every parameterized problem (Q, k), the following are equiva-lent:

• (Q, k) ∈ FPT.

• (Q, k) has a kernelization.

2.2 Ruling Out Polynomial Kernels

In this section, we will look at the some techniques for proving lower bounds on

kernels. We first define the notion of polynomial parameter transformation, and

state a theorem which makes use of the definition to prove kernel lower bounds.

The definition is analogous to polynomial time many-one reductions.

Definition 6. [4] Let P,Q ⊆ Σ∗×N be parameterized problems. We say that f : Σ∗×N → Σ∗ × N is a polynomial parameter transformation (PPT) from P to Q if forall (x, k) ∈ Σ∗ ×N the following holds: (x, k) ∈ P if and only if (x′, k′) = f(x, k) ∈Q and k′ ≤ kO(1).

Theorem 2. [4, 8] Let P and Q be parameterized problems and P and Q be theunparameterized versions of P andQ respectively. Suppose that P is NP-hard and Q

7


is in NP. Assume that there is a polynomial parameter transformation from P to Q.Then if Q admits a polynomial kernel, so does P .

The technique is useful in deriving kernelization lower bound for one problem

from the other, but it does not give us a starting point. For that, we define

distillation and composition algorithms as follows-

Definition 7. [2] (Distillation). A distillation algorithm for a classical problemL ⊆ Σ∗ is an algorithm that

• receives as input a sequence (x1, . . . , xt), with xi ∈ Σ∗ for each 1 ≤ i ≤ t,

• uses time polynomial in∑t

i=1 |xi|,

• and outputs a string y ∈ Σ∗ with

1. y ∈ L⇔ xi ∈ L for some 1 ≤ i ≤ t,

2. |y| is polynomial in maxti=1 |xi|.

Definition 8. [2] (Composition). A composition algorithm for a parameterizedproblem L ⊆ Σ∗ × N is an algorithm that

• receives as input a sequence ((x1, k), . . . , (xt, k)), with (xi, k) ∈ Σ∗ × N foreach 1 ≤ i ≤ t,


i=1 |xi|+ k,

• and outputs a string (y, k′) ∈ Σ∗ × N with

1. (y, k′) ∈ L⇔ (xi, k) ∈ L for some 1 ≤ i ≤ t,

2. k′ is polynomial in k.

We now look at the following theorems, which relate the above two concepts-

Theorem 3. [10] If any NP-complete problem has a distillation algorithm then NP⊆ coNP/poly.

Theorem 4. [2] Let L be a compositional parameterized problem whose unparam-eterized version L is NP-complete. If L has a polynomial kernel, then L is distillable(has a distillation algorithm).

8


Now, combining Theorems 3 and 4, we get the following theorem-

Theorem 5. Let L be a compositional parameterized problem whose unparameter-ized version L is NP-complete, then L does not have a polynomial kernel unless coNP⊆ NP/poly.

As an example to illustrate the technique, we will use Theorem 5 to prove that

k–PATH does not have a polynomial kernel. The proof here is due to Bodlaender

et al. [2]. We first define the problem as follows.

k–PATH

Instance: A graph G = (V,E), an integer k.

Parameter: k

Problem: Does G have a simple path of length k?

Unparameterized version of k–PATH is NP-complete by a simple reduction

from HAMILTONIAN PATH. We first give a composition algorithm for k–PATH .

We know that the problem is FPT due to Alon et al. [1] using the color coding

technique.

Given as input a sequence ((G1, k), . . . , (Gt, k)), with (Gi, k) for all 1 ≤ i ≤ t

being an instance of k–PATH , we output (G′, k), where G′ is disjoint union of the

t graphs in the input. We also observe that k′ = k. Now, we make the following

claim.

Claim 1. (G′, k′) ∈ k–PATH ⇔ (Gi, k) ∈ k–PATH for some 1 ≤ i ≤ t.

Proof. (⇐) If (Gi, k) is a YES instance, then Gi has a path a length k. Since G′

is disjoint union of Gi’s, the path of length k also exists in G′. Hence (G′, k′) ∈k–PATH .

(⇒) If (G′, k) is a YES instance then, G′ has a path of length k. This path

can not come from two dfferent Gi’s, since there is no edge between them. So,

the k-length path comes from some Gi, and hence (Gi, k) ∈ k–PATH for some

1 ≤ i ≤ t.

So, by definition 8, we have given a composition algorithm for k–PATH . Ap-

plying theorem 5, we get the following-

Theorem 6. [2] k–PATH does not have a polynomial kernel unless NP⊆ coNP/poly.

9


2.3 Summary

We started with defining the basic concepts of parameterized complexity, FPT

and kernelization. Then we defined the notion of polynomial parameter trans-

formations and how it can be used to rule out polynomial kernels. We also had

a look at composition and distillation algorithms defined by Bodlaender et al.

[2]. Then we saw how the result of Fortnow and Santhanam [10] combined

with these notions can be used to rule out polynomial kernels. To illustrate the

technique, we looked at the example of k–PATH . Polynomial kernels can be

ruled out for many other problems by just using the disjoint unions (k-CYCLE,

k-BOUNDED TREEWIDTH SUBGRAPH TEST) while some require a bit more work

(w-INDEPENDENT SET). For details, the reader is referred to [2, 8].

10

3Cross Composition

In this chapter, we look at one generalization of the notion of composition, that

makes it easier to derive lower bounds. We illustrate the framework by obtaining

a kerlen lower bound result on CLIQUE parameterized by the size of vertex cover.

3.1 Introduction

The standard OR-composition works by composing multiple instances of a pa-

rameterized problemQ into a single instance ofQwith parameter value bounded

by polynomial in the parameter value for the original instances. Bodleander et

al. [3] introduced a technique called Cross-composition, where it is sufficient to

compose the OR of any classical NP-hard problem into an instance of the param-

eterized problem Q for which we want to prove a lower-bound. The term “cross”

in the name comes from the fact that the source and target problems might not

be the same. Also, instead of requiring the source problem to be parameterized,

the new framework allows instances of some classical NP-complete problem as

a source. Furthermore, instead of the parameter being bounded to be a poly-

nomial in parameter of the original problem, it is required that the parameter

of the output is bounded by polynomial in largest input size. It also allows the

output parameter to depend polynomially on logarithm of the number of input

instances, which often simplifies the constructions and proofs. In the same pa-

per, the authors also introduce the concept of polynomial equivalence relation.

This is very useful in the cases where, in the composition algorithm, we need the

number of vertices or the parameters to be the same for all the instances.

11

Chapter 3. Cross Composition

3.2 Definitions

In this section, we define the notion of cross-composition, and give all the termi-

nology which is needed to apply the technique.

Definition 9. (Polynomial Equivalence Relation).[3] An equivalence relation Ron Σ∗ is called a polynomial equivalence relation if the following two conditionshold:

1. There is an algorithm that given two strings x, y ∈ Σ∗ decides whether x andy belong to the same equivalence class in (|x|+ |y|)O(1) time.

2. For any finite set S ⊆ Σ∗ the equivalence relation R partitions the elementsof S into at most (maxx∈S |x|)O(1) classes.

Definition 10. (Cross-composition).[3] Let L ⊆ Σ∗ be a set and let Q ⊆ Σ∗×Nbe a parameterized problem. We say that L cross-composes into Q if there is a poly-nomial equivalence relation R and an algorithm which, given t strings x1, x2, . . . , xtbelonging to the same equivalence class ofR, computes an instance (x∗, k∗) ∈ Σ∗×Nin time polynomial in

∑ti=1 |xi| such that:

1. (x∗, k∗) ∈ Q⇔ xi ∈ L for some 1 ≤ i ≤ t,

2. k∗ is bounded by a polynomial in maxti=1 |xi|+ log t.

3.3 Machinery Based on Cross-composition

The result of Fortnow and Santhanam [10] does not require the distilled lan-

guage to be same as the language we started with. In fact, it tells that an NP-

complete language can not have a distillation algorithm to any L ⊆ Σ∗. On the

other hand, the framework of Bodlaender et al [2] only makes use of the fact

that NP-complete languages are not self compressible. We recall the definition of

strong distillation from [2] and call it a weak distillation if the language output

by the algorithm is not same as the input language.

Definition 11. (Weak Distillation).[3] A weak distillation of a SAT into a lan-guage L ⊆ Σ∗ is an algorithm that:

12


• receives as input a sequence (x1, . . . .xt) of instance of SAT,


i=1 |xi|, and

• outputs a string y ∈ Σ∗ with

1. y ∈ L⇔ xi ∈ SAT for some 1 ≤ i ≤ t,

2. |y| is bounded by a polynomial in maxti=1 |xi|.

Theorem 7. [10] If there is a weak distillation of SAT into any set L ⊆ Σ∗ thenNP ⊆ coNP/poly and thus the polynomial-time hierarchy collapses to the third level.

Definition 12. [7] The OR of a language L ⊆ Σ∗ is the set OR(L) that consists ofall tuples (x1, . . . , xt) for which there is an index 1 ≤ i ≤ t with xi ∈ L.

Theorem 8. [3] Let L ⊆ Σ∗ be a set which is NP-hard under Karp reductions. IfL cross-composes into the parameterized problem Q and Q has a polynomial kernelthen there is a weak distillation of SAT into OR(Q) and NP ⊆ coNP/poly.

Proof. The proof is along the lines of proof by Bodlaender et al. [2] and gener-

alizes it. We assume that the premise of the theorem is true and then give an

algorithm which distills SAT into OR(Q).

The input to the distillation algorithm consists of a sequence (x1, . . . , xt) of

instances of SAT, which we may assume are elements of Σ∗. Define m :=

maxt1=1 |xi|. If, t > (|Σ|+ 1)m, then there must be duplicate inputs, since the

number of distinct inputs of length m′ ≤ m is |Σ|m′. We discard the duplicate

instances from the input and make sure that t ≤ (|Σ|+ 1)m, which gives us

log t ∈ O(m). Since L is NP-hard, we have a polynomial time reduction from

SAT to L. So, we first apply this reduction on each of the input instances, so

that we get a sequence of instances y1, . . . , yt of L. Clearly, ∃i : yi ∈ L if and

only if ∃i : xi ∈ SAT. Also, since the reductions take polynomial time in each of

the input instances, the size of yi is also polynomial in the size of largest input

instance, m.

As the definition of cross-composition guarantees a polynomial equivalence

relation R, we apply it on the t instances of L, namely y1, . . . , yt, by comparing

them pairwise and partitioning them into r groups. Let the groups be Y1, . . . , Yrsuch that the instances inside any Yi are equivalent under R. Note that r is

13


polynomial in m and this partitioning takes time polynomial in the input size by

the definition of polynomial equivalence relation.

We apply the cross-composition algorithm on each of the Yi’s. It is possible

to do so since all the instances in Yi are equivalent under R. Let the output

of the cross-composition algorithm be (zi, ki) for Yi, where (zi, ki) is an instance

of Q. By the definition of cross-composition, for all i ∈ [r], ki is bounded by a

polynomial in maxti=1 |yi|+ log t. But we know that log t ∈ O(m), and also for all

i ∈ [t], |yi| is polynomial in m. Hence, ki is bounded by polynomial in m for all

i ∈ [r].

In the next step of the algorithm, we kernelize each of the r instances of Q

to get instances whose sizes are polynomial in the parameter. Bus as we have

seen earlier that that parameters, ki’s are themselves polynomially bounded in

m, hence each of the kernelized instance is polynomially bounded in m. Also,

there are only polynomially many in m instances, since r is polynomial in m. Let

the instances after the kernelization be (z′i, k′i) for i ∈ [r] where |z′i| and k′i are

both polynomially bounded in m.

Then we convert all parameterized instances (z′i, k′i) of Q into unparameter-

ized instances of Q of the form zi := z′i#1k′i. This transformation takes polyno-

mial time and that the size each of the instances of Q is polynomially bounded

in m, since both |z′i| and k′i were polynomially bounded in m.

In the final step, we combine all the instances of Q to get one tuple x∗ =

(z1, z2, . . . , zr), which is an instance of OR(Q). Since the size of each of the in-

stances as well as the number of components (i.e. r) is polynomial in m, |x∗| ispolynomial in m. Now, by the definition of OR(Q), we know that x∗ ∈ OR(Q)

if and only if some component of the tuple is contained in Q. By tracking back

the series of equivalences, it we see that x∗ ∈ OR(Q) if and only if some input

xi is a YES instance of SAT. Since each of the steps take polynomial time, we

can construct the instance x∗ such that |x∗| is polynomial in m. Hence, we have

constructed a weak distillation of SAT into OR(Q). This, by theorem 7, implies

that NP ⊆ coNP/poly and polynomial time-hierarchy collapses, which concludes

the proof of the theorem.

Corollary 1. [3] If some set L is NP-hard under Karp reductions and L cross-composes into the parameterized problem Q then there is no polynomial kernel for

14


Q unless NP ⊆ coNP/poly.

A simple extension of theorem 8 allows us to rule out compression into small

instance of a different parameterized problem, the notion sometime referred as

bikernelization [11, 13]. In the proof of theorem, if we assume that Q polyno-

mially kernelizes into P , instead of Q itself, we get a weak distillation of SAT

into OR(P ), where P is the unparameterized version of P . So, by theorem 7, we

get NP ⊆ coNP/poly and polynomial time-hierarchy collapses. hence we get the

following corollary.

Corollary 2. If some set L is NP-hard under Karp reductions and L cross-composesinto the parameterized problem Q, then Q does not admit polynomial kernelizationor polynomial bikernelization unless NP ⊆ coNP/poly.

3.4 Results based on Cross-composition

In this section, we will see examples for how the cross-composition framework

allows us to rule out polynomial kernels. We take the problem to be CLIQUE

parameterized by vertex cover. It was already known [2] that CLIQUE does not

have polynomial kernel when parameterized by treewidth. Since we know that

the vertex cover of a graph is as large as treewidth, ruling out polynomial kernels

when parameterized by vertex cover is a stronger result. We first define the

problem as follows-

CLIQUE parameterized by Vertex Cover

Instance: A graph G on the vertex set V , integers k, l and a vertex cover

S ⊆ V such that |S| = k.

Parameter: k

Problem: Does G have a clique on l vertices?

Theorem 9. [3] CLIQUE parameterized by the size of vertex cover does not admita polynomial kernel unless NP ⊆ coNP/poly.

Proof. We prove the theorem by showing that CLIQUE cross-composes into CLIQUE

parameterized by vertex cover; by Corollary 1 this is sufficient to establish the

15


claim. An instance of the NP-complete CLIQUE problem is a tuple (G, l) and asks

whether the graph G contains a clique on l vertices. For the cross-composition,

we first need to define a polynomial equivalence relation R. We define R such

that all the pairs (Gi, li), (Gj, lj) go to the same equivalence class if |V (Gi)| =

|V (Gj)| and li = lj. We put all the instances which are not well formed in an-

other equivalence class. We say that the instance is not well formed, if either

it is not in the right format or if the size of the clique that is asked for exceeds

the number of vertices in the graph. Clearly, this relation satisfies both the prop-

erties to be a polynomial equivalence relation: Given a pair (Gi, li), (Gj, lj) of

well-formed instances of CLIQUE, we can check in polynomial time if they have

the same number of vertices and the same value of li. Also, since in a well formed

instance, li ≤ |V (Gi)|, hence the number of equivalence classes is at most m2 +1,

where m = maxti=1 |V (Gi)|. This is polynomial in largest input size, since |V (Gi)|is at most the largest input size.

Now, we have to give a cross-composition algorithm for instances belonging

to the same equivalence class. For the equivalence class containing the mal-

formed instances, we output a trivial NO instance. So, all that is left to give com-

position algorithm for an equivalence class where the number of the vertices and

the size of clique which is asked for are the same. So, we take (G1, l), . . . , (Gt, l)

to be the instances in the same equivalence class, such that |V (Gi)| = n for

all i ∈ [t]. Using these instances, we make an instance (G′, Z ′, k′, l′) of CLIQUE

parameterized by vertex cover, which consists of a graph G′ with vertex cover

Z ′ ⊆ V (G) of size k′ and an integer l′.

We number the vertices in each of theGi’s from 1 to n arbitrarily. We construct

G′ as follows. In the construction, -

1. Create ln vertices vi,j with i ∈ [l] and j ∈ [n]. Connect two vertices vi,j and

vi′,j′ if i 6= i′ and j 6= j′ . Let C denote the set of these vertices. It is easy to

see that any clique in G′ can contain only one vertex vi,m for all m ∈ [n] for

each choice of i ∈ [l]. Similarly, a clique in G′ can contain only one vertex

vm,j for all m ∈ [l] for each choice of j ∈ [n]. Thus any clique contains at

most l vertices from C.

2. For each pair (p, q) : 1 ≤ p < q ≤ n of distinct vertices from [n], (i.e.,

vertices of graphs Gi ), create three vertices: wp,q , wp,q, and wp,q. Let D

16


denote these 3(n2

)vertices. We make D adjacent to C as follows:

• wp,q is adjacent to all vertices from C,

• wp,q is adjacent to all vertices from C except for vi,q for all i ∈ [l], and

• wp,q is adjacent to all vertices from C except for vi,p for all i ∈ [l].

Furthermore, we add all edges between vertices in D that correspond to

distinct pairs from [n]. Which means that D is a complete(n2

)-partite graph,

with all the partitions having three vertices each. Any clique can contain at

most one out of the three vertices for each pair of distinct vertices from [n]

because the three vertices form an independent set.

3. For each graph Gi make a new vertex ui and connect it to all vertices in C.

Let B be the set of these t vertices. The adjacency of B to D is as follows:

• Make ui adjacent to wp,q if (p, q) is an edge in Gi.

• Otherwise make ui adjacent to wp,q and wp,q.

We put l′ := l + 1 +(n2

). Also, we put the set Z ′ = C ∪D as the vertex cover.

Clearly, Z ′ is a vertex cover forG′ since the remaining vertices areD, which forms

an independent set. Also, |Z ′| = nl+3(n2

). So, the parameter, k′ = |Z ′| = nl+3

(n2

)is bounded by a polynomial in n. That is it is polynomial in the largest input size.

Also, it is easy to see that the construction can be done in polynomial time in the

size of input. So, all we need to show is that (G′, Z ′, k′, l′) is a YES instance of

CLIQUE parameterized by vertex cover, if and only if (Gi, l) was a YES instance

for some i ∈ [t].

(⇒) Let (G′, Z ′, k′, l′) is a YES instance of CLIQUE parameterized by vertex

cover. Let S ′ be the clique of size l + 1 +(n2

)in G′. As argued earlier, any clique

in G′ can take at most l vertices from C, at most(n2

)vertices from D and at most

one vertex from B. Let S = j ∈ [n]|vi,j ∈ S ′ for some i ∈ [l]. The size of the set

S is l since for each i ∈ [l], S ′ can contain at most one vi,· and no two vertices vi,jand vi′,j′ can come such that j = j′. Let the vertex in S ′ from B be ui∗. We argue

that (Gi∗ , l) is a YES instance of CLIQUE, and S is a clique in Gi∗.

Let p, q be any two vertices in S. We look at the vertices in D corresponding

to (p, q), namely wp,q, wp,q and wp,q. Since these three vertices form an indepen-

dent set, at most one of these is contained in S ′. Also, since we want(n2

)vertices

17


from D, so each such triple contributes exactly one vertex to the clique. By the

construction of S, the clique S ′ must contain two vertices vi,p and vi′,q for some

i and i′. Now, the clique can not contain any of the vertices wp,q and wp,q from

D, since they do not have edges to vi′,q and vi,p respectively, which are vertices in

the clique. So, wp,q is contained in the clique along with ui∗, which means (p, q)

is an edge in Gi∗. So, S is a clique of size l in Gi∗ and (Gi∗ , l) is a YES instance of

CLIQUE.

(⇐) Let (Gi∗ , l) be a YES instance of CLIQUE for some i∗. Let S = p1, . . . , pl ⊆[n] be the clique of size l in Gi∗. We show that (G′, Z ′, k′, l′) is a YES instance by

constructing a clique S ′ of size l + 1 +(n2

).

First we put the vertex ui∗ to S ′. Also, for each pi in S, we add vj,pj to S ′. All

these vertices are adjacent to ui∗ by construction. Also, they are adjacent among

themselves since for no two vertices vi,j and vi′,j′ added to S ′, i = i′ or j = j′.

This contributes l+1 vertices to the clique. To construct a clique of size l+1+(n2

),

we have to add exactly one vertex from wp,q, wp,q and wp,q for 1 ≤ p < q ≤ n.

For each pair (p, q), we look at the following two cases:

• If (p, q) is an edge in Gi∗ then the vertex ui∗ is adjacent to wp,q in G′ and

wp,q is adjacent to all vertices of C. We add wp,q to S ′.

• Otherwise, ui∗ is adjacent to both wp,q and wp,q. Since (p, q) is a non-edge in

Gi∗, S can not contain both p and q. If S contains p (and does not contain

q), then we add the vertex wp,q to S ′. It is adjacent to all the vertices in S ′

which are added already since the only vertices in C is is not adjacent to

are of the form v·,q, which are not added to S ′ by construction.

Hence we get a clique S ′ of size l′ = l + 1 +(n2

)in G′ and (G′, Z ′, k′, l′) is a

YES instance of CLIQUE parameterized by vertex cover. This concludes the proof

of the theorem.

Corollary 3. [3] IfF is a class of graphs containing all cliques, then VERTEX COVER

and INDEPENDENT SET parameterized by the minimum number of vertex deletionsto obtain a graph in F do not admit polynomial kernels unless NP ⊆ coNP/poly.In particular, VERTEX COVER and INDEPENDENT SET parameterized by co-cluster

18


deletion distance or cluster deletion distance do not admit polynomial kernels unlessNP ⊆ coNP/poly.

Proof. Here, we give polynomial parameter transformation from CLIQUE param-

eterized by vertex cover to INDEPENDENT SET and VERTEX COVER respectively

parameterized by the number of vertex deletions to obtain a clique.

Claim 2. There is a polynomial parameter transformation from CLIQUE parame-terized by vertex cover to INDEPENDENT SET parameterized by the number of vertexdeletions to obtain a clique.

Proof. Let (G,Z, k, l) be an instance of CLIQUE parameterized by vertex cover.

The polynomial parameter transformation outputs (G,Z, k, l) as an instance of

INDEPENDENT SET parameterized by the number of vertex deletions to obtain a

clique. Z being the vertex cover of G, deleting it would make the remaining

graph an independent set in G, hence deleting it would result in a clique in G.

Also, if there exists a set of vertices of size l inG such that by deleting it we obtain

a clique, in G, we obtain an independent set by deleting the same set of vertices

and vice versa. Hence (G,Z, k, l) is a YES instance of CLIQUE parameterized

by vertex cover if and only if (G,Z, k, l) is a YES instance of INDEPENDENT SET

parameterized by the number of vertex deletions to obtain a clique.

Similarly, we can prove that (G,Z, k, l) is a YES instance of CLIQUE parame-

terized by vertex cover if and only if (G,Z, k, |V (G)| − l) is a YES instance of

VERTEX COVER parameterized by the number of vertex deletions to obtain a

clique. Hence, There is a polynomial parameter transformation from CLIQUE

parameterized by vertex cover to VERTEX COVER parameterized by the number

of vertex deletions to obtain a clique, since the parameter remains to be k.

This, using Theorem 2 implies that INDEPENDENT SET and VERTEX COVER pa-

rameterized by the number of vertex deletions to obtain a clique do not have

polynomial kernel unless NP ⊆ coNP/poly.

Size of minimum deletion set to F is at most the number of vertex deletions to

obtain a clique. So, if a problem does not admit polynomial kernel in the number

of vertex deletions to obtain a clique, it will also not admit polynomial kernel

in size of minimum deletion set to F . Hence, we get that VERTEX COVER and

INDEPENDENT SET parameterized by the minimum number of vertex deletions to

19


obtain a graph in F do not admit polynomial kernels unless NP ⊆ coNP/poly,

where F contains all cliques.

3.5 Summary

In this chapter, we saw a way to define composition which allows more leeway

in choosing the input problem for the composition. The framework is called

cross-composition and was defined by Bodlaender et. al. [3]. Then we looked

at an example of CLIQUE parameterized by the size of vertex cover and saw that

it does not admit polynomial kernels. The framework helps us to rule out poly-

nomial kernels for many other problems, like CLIQUE and CHROMATIC NUMBER

parameterized by the size of the vertex cover. For details, the reader may refer

to [3].

20

Part II

Communication Protocol BasedLower Bounds

21

4Communication Protocol based Lower

Bounds

In the previous chapters we saw the complexity theoretic result by Fortnow and

Santhanam implying unlikeliness of distillation algorithms for NP-hard problems,

and its implication to kernel lower bounds by using the machinery given by Bod-

laender et al. [2]. We also saw one generalization of the technique by Bodlaen-

der et al. [3]. In this chapter, we will see some other results which generalize

and strengthen the existent techniques further.

4.1 Motivation

We recall the theorem of Fortnow and Santhanam which is used for obtaining

lower bounds for kernelization-

Theorem 10. [10] If there is a weak distillation of SAT into any set L ⊆ Σ∗ thenNP ⊆ coNP/poly and the polynomial-time hierarchy collapses to the third level.

Here weak distillation is defined as-

Definition 13. (Weak Distillation).[3] A weak distillation of a SAT into a lan-guage L ⊆ Σ∗ is an algorithm that:

• receives as input a sequence (x1, . . . .xt) of instance of SAT,


i=1 |xi|, and

22

Chapter 4. Communication Protocol based Lower Bounds

• outputs a string y ∈ Σ∗ with

1. y ∈ L⇔ xi ∈ SATforsome1 ≤ i ≤ t,

2. |y| is bounded by a polynomial in maxti=1 |xi|.

We see that he premise of the theorem asks for a weak distillation algorithm

for unbounded number of instances, t. This motivates us to ask the question

if such a strong premise is really necessary. In other words, can we bound the

number of instances and still arrive at a similar result? Other ways to weaken the

premise and therefore to strengthen the theorem are to allow non-determinism

and oracle access in distillation. Dell and Melkebeek [7] modeled the problem

as a communication protocol.

4.2 Communication Protocol

A two-player communication protocol is described by strategies that tell each of

the players when and what to communicate to the other player. It also allows the

information being communicated to be a function of the input and the commu-

nication history. As a specific case of Communication Protocol, we define Oracle

communication protocol as follows-

Definition 14. [7] (Oracle Communication Protocol). An oracle communica-tion protocol for a language L is a communication protocol between two players.The first player is given the input x and has to execute in time polynomial in thelength of the input; the second player is computationally unbounded (an oracle)but is not given any part of x. At the end of the protocol the first player shouldbe able to decide whether x ∈ L. The cost of the protocol is the number of bits ofcommunication from the first player to the second player.

Remark: We note that the bits communicated from the second player to the

first player do not count in the cost.

We model the first player as a polynomial time Turing machine M and the

second player as a function f . The machine M has a special oracle query tape,

oracle query symbol and oracle answer tape. Whenever M writes the special

oracle query symbol on the oracle query tape, in a single computation step the

23


contents of the answer tape is replaced by f(q), where q represents the contents

of the oracle query tape at that time. In the oracle communication protocol, the

second player has no access to the input. To model that, the function f is inde-

pendent of the input given to M . Also, we assume that the oracle query tape is

write-only and one way, so that the previous queries are available to the func-

tion. This is used to model that the strategy used by the second player can be

adaptive. We can also assume that f(q) is not more than polynomial number of

bits in the input size for any q. The assumption is valid because if it is so, then

the machine M can not read it all being a polynomial time Turing machine.

We say that the oracle communication protocol decides a parameterized prob-

lem (L, k) if M with oracle f accepts an input x if and only if x ∈ L. The cost

c(k) of the protocol is the maximal number of bits written on the oracle query

tape over all inputs x with parameter k(x) = k.

Also, here we have described the model where the Turing machine M is de-

terministic. We get other variants of oracle communication protocol in following

ways-

• Allowing M to be randomized

• Allowing M to be nondeterministic.

In particular, we will be interested in the variant, where the machine is conon-

deterministic. Whenever there are multiple possible valid executions (as in the

case of conondeterministic protocols), we define the cost as the maximum cost

over all of them, i.e., we consider the worst case.

4.3 Complementary Witness Lemma

Now, we prove the following lemma, which is the main result of Dell and Melke-

beek [7] and try to mimic the proof.

Lemma 1. (Complementary Witness Lemma).[7] Let L be a language andt : N → N \ 0 be polynomially bounded such that the problem of decidingwhether at least one out of t(s) inputs of length at most s belongs to L has an

24


oracle communication protocol of cost O(t(s) log t(s)), where the first player can beconondeterministic. Then L ∈ coNP/poly.

Proof. First we consider the case where the first player is deterministic. As men-

tioned earlier, we model the deterministic oracle communication protocol P for

OR(L) by a deterministic polynomial-time Turing machine M and a function f .

If we carefully observe the proof of Fortnow and Santhanam, then we find out

that we never needed the number of input instances to be more than polynomial

in largest input size. So, we fix a polynomial number of instances, t(s) of L and

try to see how big a cost for OR(L) we can get to imply L ∈ coNP/poly.

So, we want to design a nondeterministic polynomial time machine, with

polynomial advice which, given an input string x, outputs YES on at least one

path if x ∈ L and outputs NO on all paths if x /∈ L. In short, we want to

design an NP machine for L with polynomial advice which will prove L is in

NP/poly or L is in coNP/poly. In doing so, we use the fact that OR(L) has a

small communication protocol. We also make use of notion of communication

transcript on a given input x. Such a transcript consists of the sequence of all

queries P makes on input x (i.e., the contents of M ’s oracle query tape at the

end of the protocol) as well as the answers f(q) to each of the oracle queries q.

We first make the following observation-

Observation 1. An instance x of bitlength s is in L if and only if there exists asequence x2, . . . , xt(s) of instances of bitlength s such that P (x, x2, . . . , xt(s)) rejects.

Proof. (⇐) Clearly, if P , being a protocol for OR(L) rejects (x, x2, . . . , xt(s)), then

@xi : xi ∈ L. Hence x ∈ L.

(⇒) To prove this, we take xi = x for i = 1, 2, . . . , t(s). Clearly, since x ∈ L,

(x, x2, . . . , xt(s)) /∈ OR(L) and hence P (x, x2, . . . , xt(s)) rejects.

So, we device the proof system for L as below, including a large enough set

As of communication transcripts and the value of t(s) as advice-

1. Guess a sequence x2, . . . , xt(s) where each xi has bitlength s.

2. Check whether there is a communication transcript τ inAs that is consistent

with P on input (x, x2, . . . , xt(s)) and that P (x, x2, . . . , xt(s)) rejects. If so,

accept; otherwise, reject.

25


The check for a given transcript τ involves running the first player on input

(x, x2, . . . , xt(s)). Whenever the first player sends a bit to the second player (by

writing on the oracle query tape), verify that it agrees with the corresponding bit

in τ . Whenever the first player expects a bit from the second player (by reading

from the oracle answer tape), use the corresponding bit in τ . This process

continues until a discrepancy is detected or the first player halts.

x /∈ L: In this case, x ∈ L. So, for all choices of x2, . . . , xt(s), P will accept.

If all the the communication transcripts in As are consistent with P then above

proof system will reject (since none of the transcripts in As can reject), conclud-

ing x /∈ L. So, if x /∈ L, the above proof system says NO on all computational

paths. The running time will depend on the size of As which we will talk about

later.

x ∈ L : In this case, we have to show a set As exists, such that for every

x ∈ L, there is a choice of (x, x2, . . . , xt(s)) such that the transcript τ resulting

from (x, x2, . . . , xt(s)) is in As, and P rejects on τ . Clearly, if we take As to be

the set of all transcripts, then it is true, by taking the values of xi’s same as x, as

we did in the observation above. But the problem is that the set having all the

transcripts can be huge, and we ae allowed only polynomial size advice.

Consider instances x1, . . . , xt(s) of L of bitlength s, and let T (x1, . . . , xt(s)) de-

note the communication transcript of P on input (x1, . . . , xt(s)). Since the second

player is not given the input (x1, . . . , xt(s)), the transcript T (x1, . . . , xt(s)) is deter-

mined solely by the bits sent from the first player to the second player. Therefore,

the number of distinct such transcripts is less than 2c(s)+1 , where c(s) denotes

the cost of the protocol on inputs consisting of t(s) instances of bitlength s each.

We say that a rejecting transcript τ covers an instance x ∈ L of bitlength s if

there exists a sequence x2, . . . , xt(s) of instances of bitlength s each such that

T (x1, . . . , xt(s)) = τ . From the discussion above, it suffices to show that a small

As exists which covers all x ∈ L. We construct As in the following way.

We start with As empty and successively pick a rejecting communication tran-

script τ that covers the largest number of instances x ∈ L of length s that are

not covered thus far, and add τ to As . We keep doing so until there are no more

instances x ∈ L of bitlength s left to cover.

Consider one step in the construction of As and let F denote the set of un-

covered instances x ∈ L of bitlength s at the beginning of the step. Since every

26


tuple in F t(s) is mapped by T to one of the rejecting transcripts above and there

are less than 2c(s)+1 distinct such transcripts, there exists a rejecting transcript τ

such that at least a fraction 1/2c(s)+1 of the tuples in F t(s) are mapped by T to

this particular τ , i.e., |T−1(τ ) ∩ F t(s)| ≥ |F |t(s)/2c(s)+1 . Now, each component of

a tuple in T−1(τ )∩ F t(s) is covered by τ since we can regard the input of T as an

unordered sequence. Thus, if we let G denote the subset of F that is covered by

τ , we have that (T−1(τ ) ∩ F t(s)) ⊆ Gt(s). From this, we get that

|G|t(s) ≥ |T−1(τ ) ∪ F t(s)| ≥ |F |t(s)/2c(s)+1

From which, we get |G| ≥ ϕ(s)· |F | where ϕ(s) = 1/2(c(s)+1)/t(s).

So, in every step, we are coveting ϕ fraction of the remaining instances in L.

Initially there are at most 2s instances to start with, so after l steps, we are left

with

≤ (1− ϕ(s))l· 2s

≤ exp(−ϕ(s)l)· 2s

instances, so there will be no more instances left to cover after O(s/ϕ(s)) steps.

Clearly, (s/ϕ(s)) is polynomially bounded in s if 1/ϕ(s) is polynomially bounded

in s. 1/ϕ(s) = 2(c(s)+1)/t(s) is polynomially bounded in t(s) and hence in s (for

polynomially bounded t(s)) as long as c(s) = O(t(s) log t(s)). This As covers all

instances in L and the number of transcripts, as well as length of each transcript

is polynomially bounded. So, the advise length is polynomially bounded in s.

Also, the running time of the proof system is also polynomially bounded in s and

hence the resulting algorithm for L runs in NP/poly. This finishes the proof for

the deterministic communication protocol case.

For cononderministic protocols, we define T (x1, . . . , xt(s)) to be arbitrary tran-

script of an execution on which P produces the correct output. The first step of

the proof system remains the same. In the second step, instead of determin-

istically simulating the first player, now we simulate nondeterministically. The

advise remains the same. Clearly, if x /∈ L, then being a conondeterministic pro-

tocol, none of the transcripts of P will reject and hence the proof system will say

NO. If x ∈ L then we argue same as above that a polynomially big advise will

exist which will cover all the instances in L. Clearly, if (x, x2, . . . , xt(s)) /∈ OR(L)

27


then there will be a valid path in P which will lead to reject. Here, he proof

system will guess that path, validate it and will finally accept the string. The

counting arguments carries over the same as in the deterministic case.

4.4 Application to Lower Bounds: Vertex Cover

Yap [18] showed that if coNP⊆ NP/poly, then the polynomial hierarchy collapses

to the third level. We somehow try to exploit this result along with Comple-

mentary Witness Lemma to get some lower bounds for kernels. We see that if

we have a communication protocol P for input (x1, . . . , xt(s)) for OR(L) of cost

O(t(s) log t(s)), then L ∈ coNP/Poly or L ∈ NP/poly . Now, if we take L to be

an NP -hard language, L is coNP-hard, but we know that L ∈ NP/poly, hence we

get coNP ⊆ NP/poly and hence the collapse.

So, in short, we can not get a protocol of cost O(t(s) log t(s)) for OR(L),

where L is an NP -hard language, unless polynomial hierarchy collapses. We can

use this fact to disallow kernels of certain size, while earlier, in the framework

given by Fortnow and Santhanam, we could rule out only polynomial kernels.

We will give one example of VERTEX COVER and show that it does not admit a

kernel of size O(k2−ε). The proof here is due to Dell and Marx [6].

4.4.1 Reduction

We know that vertex cover admits a linear kernel in number of vertices, but

no kernel is known which is linear in number of edges. We show, using Com-

plementary Witness Lemma, that it is unlikely to happen. For that, we define

MULTICOLORED BICLIQUE as follows-

MULTICOLORED BICLIQUE

Instance: A bipartite graph B on the vertex set U ∪W , an integer k and

partitions U = (U1, . . . Uk) and W = (W1, . . .Wk).

Problem: Does B have an biclique Kk,k that has one vertex from each Ui and

Wi, 1 ≤ i ≤ k?

28


We take L to be MULTICOLORED BICLIQUE. Given an instance (B1, B2, . . . , Bt)

of OR(L), we can assume that every instance Bi has the same number k of

groups in the partitions and every group in every instance Bi has the same size

n. If not, we achieve this property by doing the following. Let k and n denote the

maximum number of groups and the maximum number of vertices in a group

respectively in any of the instances. Firstly, if any group has size less than n,

we add some independent vertices in the group to make the size n. Secondly, if

the number of groups is less than k in some instance, then we add new groups

on n vertices each to U and W and make them adjacent to all the vertices in W

and U respectively. It is easy to see that it preserves equivalence and increases

the size of OR(L) instance by at most a polynomial factor. Furthermore, we can

assume that√t is an integer, if not, we repeat a few instances, not increasing

the number of instances by more than a factor of 2. In the following, we refer

to the t instances of MULTICOLORED BICLIQUE in the OR(L) instance as B(i,j) for

1 ≤ i, j ≤√t; let U(i,j) and W(i,j) be the two bipartite classes of B(i,j).

First we modify each of the B(i,j) such that all U(i,j) and W(i,j) become com-

plete k-partite graphs. We do so by adding edges between any two vertices of

U(i,j) and W(i,j) if they are in different groups. We call the new graphs B′(i,j).

Clearly, the original graph B(i,j) had a multicolored biclique Kk,k if and only if

B′(i,j) has a clique on 2k vertices.

Now, we construct a graphG by introducing 2√t sets U1, . . . , U

√t,W 1, . . . ,W

√t

of kn vertices each. For every 1 ≤ i, j ≤√t, we copy the graph B′(i, j) to the

vertex set U i ∪W j by mapping U(i,j) to U i and W(i,j) to W j. We see that U(i,j)

and W(i,j) induce the same graph for every B′(i,j) for every i and j. So, while

copying, it can be made sure that G[U i] receives the same set of edges as B′(i,j)while copying B′(i,j) to it. Similarly we can argue about G[W j]. Hence, we can

say that G[U i ∪W j] is isomorphic to B′(i,j) for every 1 ≤ i, j ≤√t.

Claim 3. G has a 2k clique if and only if at least one B′(i,j) has a 2k clique.

Proof. If one of the B′(i,j) has a 2k clique, then G has a 2k clique, because a graph

isomorphic to every B′(i,j) is a subgraph of G. Now, for the forward direction,

suppose G has a 2k sized clique. We observer that there are no edges between

U i and U i′ in G for i 6= i′. Similarly, there are no edges between W j and W j′

in G for j 6= j′. Thus, the 2k clique of G is contained in G[U i ∪W j] for some

29


1 ≤ i, j ≤√t. But we know that each G[U i∪W j] is isomorphic to B′(i,j). So, B′(i,j)

has a 2k sized clique.

As argued earlier, the original graph B(i,j) had a multicolored biclique Kk,k if

and only if B′(i,j) has a clique on 2k vertices. So, from the claim we get that the

original graph B(i,j) had a multicolored biclique Kk,k if and only if G has a 2k

clique. We also know that G has a 2k sized clique if and only if G has a N − 2k

sized vertex cover, where the N is number of vertices in the graph. Hence, we

output (G,N − 2k) as an instance of vertex cover, which is YES instance if and

only if at least one of the t instances of MULTICOLORED BICLIQUE was a YES

instance. We note that here the target size is N − 2k = 2√t · kn − 2k is at most

t1/2 · s2, where s is the size of largest input instance, and also that the algorithm

is deterministic and runs in time polynomial in input size.

4.4.2 Lower Bounds on Kernelization

Let us assume that VERTEX COVER has a kernel of size O(k2−ε). We take the in-

stance output by the previous algorithm and kernelize it. Since we have assumed

that VERTEX COVER has a kernel of size O(k2−ε), we get an instance of size

O((t1/2 · s2)2−ε)

which is t1−(ε/2)s4−2ε. Now, we choose a polynomial t : N→ N such that t(s)ε/2 =

Ω(s4−2ε) and hence the final instance becomes of size O(t(s)).

Now, for getting the protocol of cost O(t(s) log t(s)) for OR(L), where L is

MULTICOLORED BICLIQUE, the first player, given t(s) instances of L first generates

the instance of size O(t(s)1/2 · s2) by using the the algorithm given above, then

kernelizes is to get an instance of size O(t(s)) and writes it on the oracle query

tape. Then the second player decides whether the instance is YES or NO and

conveys it to the first player using one bit. The correctness of the protocol follows

from correctness of the reduction and kernelization.

By complementary witness lemma, we get that MULTICOLORED BICLIQUE ∈coNP/poly. Now, if we can prove that MULTICOLORED BICLIQUE is NP-hard, then

assuming vertex cover has kernel of size O(k2−ε), we have shown that if coNP

⊆ NP/poly and hence PH collapses to the third level. We delay the proof of NP-

30


completeness of MULTICOLORED BICLIQUE till the next subsection, and state the

following theorem.

Theorem 11. VERTEX COVER does not have a kernel of size O(k2−ε) unless thepolynomial hierarchy collapses.

4.4.3 NP-completeness of MULTICOLORED BICLIQUE

Lemma 2. [6] MULTICOLORED BICLIQUE is NP-complete.

Proof. We prove by reducing from CLIQUE. Let (G, k be an instance of CLIQUE.

Let vi | 1 ≤ i ≤ n be the vertex set ofG. We make an instance of MULTICOLORED BICLIQUE

by constructing a bipartite graphB on vertex set ui,j, wi,j | 1 ≤ i ≤ k, 1 ≤ j ≤ n.Basically, we make k copies of the vertex set of G in both the partitions. The first

index of a vertex in B represents which copy it is coming from (1 to k), and the

second index tells about which vertex it is image of in G (1 to n). Now, to make

the adjacencies, we make two vertices in U and W adjacent if they are copies

are of same vertex in G and have the same first index. Also, we make copy of

a vertex adjacent to copies of all it’s neighbors in G if they have different first

index. Formally, we make vertices ui,j and wi′,j′ adjacent if and only if

• either i = i′ and j = j′, or

• i 6= i′ and vj and vj′ are adjacent in G.

Let us consider the partition U = U1 ∪ . . .∪Uk and W = W1 ∪ . . .∪Wk where

Ui = ui,j|1 ≤ j ≤ n and Wi = wi,j|1 ≤ j ≤ n. We claim that B contains a

biclique Kk,k respecting these partitions if and only if G contains a k-clique.

(⇐) Let va1 , . . . , vak be a clique inG, then we look at u1,a1 , . . . , uk,ak , w1,a1 , . . . , wk,ak.Clearly, u1,a1 , . . . , uk,ak and w1,a1 , . . . , wk,ak are independent set. We look at a

pair of the form (ui,ai , wj,aj). Clearly, if i = j, then there is an edge between the

two vertices by the first rule. If i 6= j, then we know that vai and vaj are adjacent

in G and hence there is an edge between these two vertices by the second rule.

So, B has a biclique Kk,k respecting the partitions.

(⇒) Let u1,a1 , . . . , uk,ak , w1,b1 , . . . , wk,bk be the biclique respecting the parti-

tions in B. We claim that ai = bi for all 1 ≤ i ≤ k, otherwise ui,ai and ui,bi will not

be adjacent. Now, we claim that va1 , . . . , vak is a clique in G. If not so, then vai

31


and vai′ are not adjacent in G, implying ui,ai and wi′,ai′ = wi′,bi′ are not adjacent

in B, which is a contradiction to the assumption made. Hence, va1 , . . . , vak is a

clique in G.

4.5 Summary

We started this chapter with the motivation behind looking at the existing lower

bound machinery more closely and the possibilities of extending it by restricting

the number of instances required for “distillation” to be only polynomially many

in the largest instance size, allowing nondeterminism and oracle access. We saw

how Complementary Witness Lemma takes all this into account and also allows

the cost of protocol to grow slowly with t(s), i.e. the number of input instances.

Then we saw one example how this setting can be used to prove unlikeliness of

polynomial kernels of certain size, instead of proving unlikeliness of existence of

any polynomial kernel. So, the technique can be useful for proving tight bounds

for problems for which polynomial kernels are already known. In particular,

we saw that VERTEX COVER can not have a kernel of size O(k2−ε) unless the

polynomial hierarchy collapses, while the kernels of size O(k2) have been known

for long time.

32

5Weak Composition

In the previous chapter, we saw how the tools given my Dell and Melkebeek [7]

allow us to get finer lower bounds for problems which are already shown to have

polynomial kernels. In this chapter we will formalize the notion, and see how

we can rule out kernels of size kd−ε for a fixed d, assuming coNP * NP/poly.

5.1 Introduction

Dell and Melkebeek proved that VERTEX COVER does not admit a kernel of size

k2−ε. The result is important in the sense that it rules out kernels of polynomial

size, which was not possible using the frameworks given earlier by Bodlaender et

al. [2, 3]. In parameterized complexity, the result gave rise to a new framework,

which abstracts out the details of Communication Protocol, and allows us to work

with problems in the domain of parameterized complexity and rule out kernels

of polynomial size.

The notion of weak composition introduced by Hermelin and Wu [12] pro-

vides a framework based on Complementary Witness Lemma for ruling out poly-

nomial kernels of certain size. It does so by allowing the parameter of the output

problem of the composition algorithm to depend on t, instead of requiring it to

be completely independent on t, and hence arriving at weaker consequences,

which are useful in many cases. In particular, they gave a negative answer to the

question posed by Dom et al., proving that d-SET COVER does not have kernels

of size O(kd−3−ε) for any ε > 0 assuming coNP * NP/poly.

33

Chapter 5. Weak Composition

5.2 Framework for Weak Composition

We recall the definition of Oracle Communication Protocol, the statement of

Complementary Witness Lemma and other notation we are going to use from

Chapter 4. Also, we observe the following.

Observation 2. If L ⊆ 0, 1∗ × N has a kernel of size f(k), then L has an oraclecommunication protocol of cost f(k).

Proof. The first player will convert the input of L to an input of L with the

parameter being k, and kernelize it to an instance of size f(k) and sends it to

the second player. The second player answers accordingly, and the equivalence

is maintained, resulting in a oracle communication protocol of cost f(k).

Definition 15. [12] (Weak d-composition.) Let d ≥ 2 be a constant and letL1, L2 ⊆ 0, 1∗ × N be two parameterized problems. A weak d-composition fromL1 to L2 is an algorithm A that on input (x1, k), . . . (xt, k) ∈ 0, 1∗ × N, outputsan instance (y, k′) ∈ 0, 1∗ × N such that

• A runs in polynomial time with respect to∑t

i=1(|xi|+ k)

• (y, k′) ∈ L2 ⇔ (xi, k) ∈ L1 for some i, and

• k′ ≤ t1/dkO(1)

We note that in the regular compositions the output parameter is required to

be polynomially bounded by the input parameter, while in weak d-compositions

it is also allowed to depend on the number of inputs t.

Lemma 3. [12] Let d ≥ 2 be a constant and let L1, L2 ⊆ 0, 1∗ × N be twoparameterized problems such that L1 is NP-hard. A weak d-composition from L1 toL2 implies that L2 has no kernel of size O(kd−ε) for all ε > 0 unless NP ⊆ coNP/poly.

Proof. We try to mimic the proof of Hermelin and Wu [12].

We assume that L2 has a kernel of sizeO(kd−ε) for some ε > 0. By Observation

2 L2 has a communication protocol of cost O(kd−ε).

We have to show that OR(L1) (see Definition 12 in Chapter 3 for definition of

OR(L)) has a communication protocol of cost O(t(s) log t(s)) for some polynomial

34


t(s), where the OR(L1) has t(s) instances and the largest instance size is s, and

hence we get the collapse by Complementary Witness Lemma 1.

Now, given an instance (x1, . . . , xt(s)) of OR(L1), we first convert it to t in-

stances (x1, k1), . . . , (xt(s), kt(s)) of L1. Now, we do the following sequence of

operations-

1. We divide the instances of into groups, such that problem instances in each

group has the same parameter value. Let Y1, . . . , Yr be the groups. Clearly,

r ≤ maxi ki ≤ s.

2. For each of the groups, we apply the d-composition from L1 to L2. This

will result in at most s instances of L2, with parameter of each bounded by

k′ ≤ t(s)1/dkO(1), where k is the parameter for that particular group. So,

parameter of each of the output instances is at most t1/dsO(1), since k ≤ s.

3. For each instance of L2, we apply the protocol of cost O(k′d−ε) to decide it.

If one of the instances is YES, then we accept, otherwise we reject.

The correctness of the protocol is straightforward from the equivalence of

kernelization and d-composition. If the initial instances was a YES instance of

OR(L1), then one of the composed instances will be a YES instance. On the other

hand, if one of the composed instances is a YES instance, then at least one of the

instances of L2 and hence, of L1 had to be a YES instance, and the initial instance

was a YES instance of OR(L1).

Also, since the number of composed instances is bounded by s, the cost of

the protocol is bounded by O(s. k′d−ε). We also know that k′ ≤ t(s)1/dkc for some

constant c, hence we get

O(s. k′d−ε) = O(s. (t(s)1/dkc)d−ε)

= O(s. t(s)1−ε/d. kc(d−ε))

= O(s. t(s)1−ε/d. sc(d−ε)) (as k ≤ s)

= O(s1+cd−cε. t(s)1−ε/d)

= O(t(s)) (for t(s) sufficiently large, such that t(s)ε/d = Ω(s1+cd−cε))

= O(t(s) log t(s))

35


So, OR(L1) has a oracle communication protocol of cost O(t(s) log t(s)), and

so L1 is in coNP/poly by Complementary Witness Lemma. But, we also know

that L1 is NP-hard, hence we get NP ⊆ coNP/poly, which concludes the proof of

lemma.

5.3 An example of weak composition: RAMSEY

Ramsey theory asks questions like “how many elements of some structure must

there be to guarantee that a particular property will hold?”. One such question

is finding dimensional of a grid you need so that the game of tic-tac-toe never

results in s draw, irrespective of the strategies of the players. In graph theory, the

question is about the number of vertices necessary to force certain size of clique

or independent set.

Ramsey’s theorem says that any graph on very large number of vertices will

have either a clique, or independent set of certain size. For defining out problem,

we frame our question on similar notion and ask whether a given graph has

clique or independent set of certain size. We call this problem RAMSEY and

define it formally as following-

RAMSEY(k)

Instance: A graph G = (V,E), an integer k

Parameter: k

Problem: Does G have a clique or independent set of size k?

It is easy to see that the problem is NP-complete.

Lemma 4. RAMSEY is hard for NP.

Proof. We prove this by giving a reduction from clique. Given an instance of G, k

of CLIQUE, we construct an instance G′, k′ of RAMSEY in polynomial time such

that G, k is a YES instance if and only if G′, k′ is a YES instance.

We first define the operation join of two graphs formally as below, which adds

all possible edges between two graphs.

Definition 16. A graph H obtained by the join of two graphs G = (V,E) andG′ = (V ′, E ′) is a graph on vertex set V ∪ V ′ and edge set E ∪ E ′ ∪ Eacross, whereEacross = (u, v) : u ∈ V, v ∈ V ′.

36


Let G has n vertices. To obtain G′, we take a clique C of size n+1 and perform

a join with G. We set k′ = k + n + 1. If (G, k) is a YES instance, then G′ has a

clique on k+ n+ 1 vertices, consisting of n+ 1 vertices of C and k vertices of the

clique in G. Conversely, let G′, k′ be a YES instance of Ramsey. Clearly, G′ can

not have an independent set on more than n vertices. It is easy to see since the

independent set can not have vertices from C as well as G. Moreover, it can have

at most one vertex from C, hence the size of independent set is bounded by n.

So, to be a YES instance of RAMSEY, G′ must have a clique on n+ k + 1 vertices,

out of which, only n + 1 can come from C, so G has a cliques on k vertices and

hence (G, k) is a YES instance of CLIQUE.

Looking at the fixed parameterized setting, we know that both CLIQUE and

INDEPENDENT SET are [W ]-hard, and hence are not expected to be FPT. But if

we combine the two problems as RAMSEY, it is trivially FPT by any upper bound

on Ramsey Number Rk,k. Let Rk,k ≤ f(k). If the given graph has more than

f(k) vertices, then we say YES, otherwise the number of vertices is bounded by

f(k), and we can do a brute force to check whether the graph has a clique or an

independent set of size k.

A valid question to ask is whether RAMSEY has a polynomial kernel. We will

take RAMSEY as an example and see that it can admit kernels of size k2−ε. In

the next chapter, we will extend this result and show that it does not have a

polynomial size kernel. The proof here is due to Kratsch [14].

For doing the Weak composition, we will define a problem similar to RAMSEY.

In this, we give a guarantee that the graph has a clique and independent set of

size k − 1 and ask for whether it has a clique or independent set on k vertices.

Formally, we define the problem as following.

REFINEMENT RAMSEY(k)

Instance: A graph G = (V,E), an integer k, such that G has both a clique

and independent set of size k − 1.

Parameter: k

Problem: Does G have a clique or independent set of size k?

As shown in the results of Bodlaender et al. [2],these problems are very help-

ful in arriving at composition algorithms. It is easy to see that REFINEMENT RAMSEY

37


is no harder to kernelize than RAMSEY, hence, for getting a kernelization lower

bound for RAMSEY, it is sufficient to show it for REFINEMENT RAMSEY. But before

giving a composition for REFINEMENT RAMSEY, we must show that it’s classical

version is NP-complete. For it, we give a reduction from RAMSEY.

Lemma 5. REFINEMENT RAMSEY is hard for NP.

Proof. Given an instance (G, k) of RAMSEY, we construct an instance (G′, k′) of

REFINEMENT RAMSEY such that (G, k) is a YES instance if and only if (G′, k′) is a

YES instance. For that, we construct G′ as first taking a disjoint union of G and a

clique C on k − 1 vertices and then joining this structure (both C and G with an

independent set I of size k. Then we set k′ = k + 1.

We first show that G′ has both a clique and independent set of size k: the

clique C combined with any one vertex of I is a clique of size k, and I itself is an

independent set of size k.

If (G, k) is a YES instance of RAMSEY, then

• Case 1: G has a clique of size k. In this case, we get a clique of size k + 1

in G′ by combining the clique in G with any one vertex in I.

• Case 2: G has an independent set of size k. In this case, we combine it

with one of the vertices in C to get an independent set in G′ of size k + 1.

Conversely, if (G′, k′) is a YES instance of REFINEMENT RAMSEY, then

• Case 1: G′ has a clique of size k + 1. We observe that the clique can not

come from I ∪ C alone, since the largest clique in I ∪ C is of size k. So, it

has at least one vertex from G, which forces at most one vertex from I and

no vertices from C, since there are no edges between G and C. Hence, all

but one vertex of the k+ 1 sized clique have to come from G and G has a k

sized clique.

• Case 2: G′ has an independent set of size k + 1. We again observe that

the independent set can not come from I ∪ C alone, since the largest in-

dependent set in I ∪ C is of size k. So, it has at least one vertex from G,

which forces at most one vertex from C and no vertices from I, since all

the vertices of G are adjacent to all the vertices of I. Hence, all but one

38


vertex of the k+ 1 sized independent set have to come from G and G has a

k sized independent set.

Observation 3. In the above proof, we saw that an instance (G, k) of RAMSEY canbe transformed to an instance (G′, k′) of REFINEMENT RAMSEY in polynomial timewhere k′ = k+ 1. Also, both the problems are NP-complete. Hence, by the definitionof Polynomial Parametric Transformation from Chapter 2, we have RAMSEY(k) ≤pptREFINEMENT RAMSEY(k).

Weak composition for REFINEMENT RAMSEY

Now, we give a weak d-composition from REFINEMENT RAMSEY(k) to RAMSEY(k).

For that, we take t instances (G1, k), . . . (Gt, k) of REFINEMENT RAMSEY(k), where

t is a perfect square. If not, we repeat some instances to do that, which increases

the number of instances by at most a constant factor.

Now, we divide the t instances arbitrarily into√t groups of

√t size each. Now,

we perform join (recall Definition 16 from Chapter 2 of every two instances,

which are in the same group to get graphs Y1, . . . , Y√t. We call disjoint union

of these√t graphs G′. We output (G′, k′) as an instance of RAMSEY where k′ =

(k − 1)√t+ 1.

Proposition 1. (G′, k′) is a YES instance of RAMSEY if and only if (G, k) was a yesinstance of REFINEMENT RAMSEY for some i ∈ [t].

Proof. (⇐) Let (Gi, k) be a YES instance of REFINEMENT RAMSEY.

• Case 1: Gi has a clique of size k. In this case, we get a clique of size

k′ = (k − 1)√t + 1 in G′ by combining this k-sized clique with all the

cliques of different graphs in Yj in which Gi belongs. This is possible

since all the graphs have a clique of size k − 1 (they are instances of

REFINEMENT RAMSEY) and all possible edges between any two graphs are

present.

• Case 2: Gi has an independent set of size k. In this case, we take√t − 1

independent sets of size k − 1 from all Yj ’s except the one in which Gi

belongs. Then we combine it with the independent set of size k in Gi to

39


get an independent set of size k′ = (k − 1)√t + 1 in G′. This is because all

the independent sets come from different Yj ’s and there is no edge between

them.

(⇒) if (G′, k′) is a YES instance of RAMSEY, then

• Case 1: G′ has a clique of size (k − 1)√t+ 1. This clique has to come from

one Yi since there is no edge between vertices of two different Yi.’s. Since

there are√t graphs in Yi, there exists one graph (Gj) which contributes

k vertices to the clique. Since the adjacencies inside any of the original

instances is not changed, (Gj, k) is a YES instance of REFINEMENT RAMSEY.

• Case 2: G′ has an independent set of size (k − 1)√t + 1. In this case, by

pigeonhole principle, at least k vertices come from some Yi. Now, inside Yi,

no two vertices of different Gj ’s can be part of the independent set since the

join operation adds all edges between two vertices belonging to different

graphs. Hence, all these k vertices come from a graph Gj and (Gj, k) is a

YES instance of REFINEMENT RAMSEY.

It is easy to see that the composition runs in time polynomial in the input size.

Also, we see that k′ = t1/2kO(1). Hence, we have given a weak d-composition from

REFINEMENT RAMSEY(k) to RAMSEY(k) for d = 2, hence by Lemma 3, we get the

following theorem-

Theorem 12. RAMSEY(k) has no kernel of size O(k2−ε) for all ε > 0 unless NP ⊆coNP/poly.

5.4 Summary

We first saw the notion of weak d-composition defined by Hermelin and Wu [12]

and saw how that can be used to give polynomial lower bounds on kernels. To

illustrate that, we gave an example of RAMSEY and saw that it does not admit

kernels of size O(k2−ε). In that, we used the promise problem reframsey cru-

cially. The notion of weak composition is further utilized to give polynomial

40


lower bounds on kernels. Some examples are d-SET PACKING, d-SET COVER,

d-EXACT SET COVER, HITTING SET WITH d-BOUNDED OCCURRENCES and some

packing problems like Kd-MATCHING. For details the reader may refer to [12]

and [6].

41

6Co-nondeterminism in Composition

In the last two chapters we saw how the results given by Dell and Melkebeek

[7] and Hermelin and Wu [12] help us ruling out kernels of certain size. In this

chapter, we will look at how nondeterminism can be exploited to arrive at lower

bounds for kernelization. In particular, we will rule out any polynomial kernel

for RAMSEY(k), which we proved that it does not have a kernel of size O(k2−ε) in

the previous chapter.

6.1 Introduction

One aspect of the result Dell and Melkebeek [7] came up with, was the extension

of the result of Fortnow and Santhanam [10] by allowing co-nondeterminism. It

means that PH collapses even when OR of a NP-complete problem has a small

cost protocol, even when the first player is allowed to be conondeterministic.

(see Complementary Witness Lemma 1 in Chapter 4). This gives more power to

the composition algorithm, which was not made use of in any composition until

Kratsch [14] used it to give lower bounds for RAMSEY, which asks whether the

given graph has clique or independent set of certain size. It exploits the freedom

of having co-nondeterminism and a property of Ramsey numbers to rule out

polynomial kernels for the problem.

42

Chapter 6. Co-nondeterminism in Composition

6.2 coNP-Composition

Analogous to definition of Bodlaender et al. [2], we first define co-NP composi-

tion as following.

Definition 17. [14] Let Q ⊆ Σ∗ × N. A co-nondeterministic polynomial-timealgorithm C is a coNP-composition forQ if there is a polynomial p such that on inputof t instances (x1, k), . . . , (xt, k) ∈ Σ∗ × N, the algorithm C takes time polynomialin∑t

i=1 |xi| and outputs on each computation path an instance (y, k) ∈ Σ∗×N withk ≤ to(1)p(k) and such that the following hold:

• If at least one instance (xi, k) is a yes-instance of of Q then all computationpaths lead to the output of a yes-instance (y, k) of Q.

• Otherwise, if all instances (xi, k) are no-instances of Q, then at least onecomputation path leads to the output of a no-instance of Q.

Now we prove the following lemma which establishes the connection be-

tween coNP-composition and kernelization lower bounds. For it, we recall that

the unparameterized version Q of a parameterized problem Q is defined as

Q := x#1k|(x, k) ∈ Q.

Lemma 6. [14] Let Q ⊆ Σ∗ × N be a parameterized problem such that Q is NP-hard. IfQ has a coNP-composition then it does not admit a polynomial kernelizationunless NP ⊆ coNP/poly and the polynomial hierarchy collapses to its third level.

Proof. We assume that Q has a kernel of size O(kc) for some constant c. By

Observation 2 of Chapter 5 Q has a communication protocol of cost O(kc).

We have to show that OR(Q) (see Definition 12 in Chapter 3 for definition of


t(s), where the OR(Q) has t(s) instances and the largest instance size is s, and


Now, given an instance (x, . . . , xt(s)) of OR(Q), we first convert it to t instances

(x1, k1), . . . , (xt(s), kt(s)) of Q. Now, we do the following sequence of operations-

1. We divide the instances of into groups, such that problem instances in each

group has the same parameter value. Let Y1, . . . , Yr be the groups. Clearly,

r ≤ maxi ki ≤ s.

43


2. For each of the subsequences, we apply the co-NP-composition. This will

result in at most s instances of Q, with parameter of each bounded by

k ≤ to(1)p(k) which is at most k ≤ to(1)p(s), since k ≤ s.

3. For each of the instances, output by the coNP-composition, we apply the

protocol of cost O(k′c) to decide it. If one of the instances is YES, then we

accept, otherwise we reject.

Also, since the number of composed instances is bounded by s, the cost of

the protocol is bounded by O(s. k′c). Let p(k) = O(kd). We also know that

k′ ≤ to(1)p(k), hence we get

O(s. k′c) = O(s. (to(1)kd)c)

= O(s. t(s)o(1). kcd)

= O(. t(s)o(1). scd) (as k ≤ s)

= O(t(s)o(1). scd+1)

= O(t(s)) (putting t(s) = (scd+2))

= O(t(s) log t(s))

Now, to show the correctness, we have to show that if one of the xi’s was a

YES instance at the beginning, then the communication protocol will say YES on

all paths. From the definition of co-NP composition, it follows that if one of the

inputs to it was a YES instance, it will lead to a YES instance on all paths, and

hence, at least one of the composed instances will be YES instance on all paths,

and the protocol will give the right answer. On the other hand, if all the xi’s were

NO instances, then we have to show that there exists one path which leads to

NO answer. Again, from the definition of coNP-composition, it is clear that on

at least one path, all of the composed instances will be NO instances, and hence

the protocol will say NO.

So, OR(Q) has a oracle communication protocol of cost O(t(s) log t(s)) where

the first player is allowed to be co-nondeterministic, and so is in coNP/poly by

Complementary Witness Lemma. But, we also know that Q is NP-hard, hence we

get NP ⊆ coNP/poly, which concludes the proof of lemma.

44


6.3 Application of coNP-composition: RAMSEY

In the previous chapter, we saw that RAMSEY(k) does not admit kernels of size

O(k2−ε) for all ε > 0 unless NP ⊆ coNP/poly. But it does not rule out all polyno-

mial kernels. For that, we will give a coNP-composition for REFINEMENT RAMSEY(k)

and then use the PPT reduction to establish that RAMSEY(k) does admit polyno-

mial kernel.

For the composition algorithm, we will need the following result about the

gaps between Ramsey Numbers.

Lemma 7. For every integer t > 3 there exists an integer l ∈ 1, . . . , d8 log(t)esuch that R(l + 1) > R(l) + t.

Proof. Let us assume that the statement of the lemma is not true. Then, we have

R(d8 log(t)e + 1) ≤ R(d8 log(t)e) + t. If we keep expanding the RHS terms, then

we get R(d8 log(t)e+1) ≤ td8 log(t)e+R(1). We use Erdos’ classical upper bound

on Ramsey Numbers, which says R(n) ≥ 2(n−1)/2 for all n ∈ N. Putting n =

d8 log(t)e+ 1, we get R(d8 log(t)e+ 1) ≥ 2(d8 log(t)e+1)/2 ≥ 24 log(t) ≥ t4. Combining

the two inequalities, we get, t4 ≤ td8 log(t)e+R(1), which is false for t > 3, since

we know that R(1) = 1. This concludes the proof of the lemma.

Now we are ready to give the coNP-composition algorithm for REFINEMENT RAMSEY.

6.4 coNP-composition for REFINEMENT RAMSEY

For the coNP-composition, given t instances of REFINEMENT RAMSEY, namely

(x1, k), . . . , (xt, k), we want to generate another instance (G′, k) of REFINEMENT RAMSEY

co-nondeterministically, such that with k ≤ to(1)p(k) and the following hold.

• If at least one instance (xi, k) is a yes-instance of of REFINEMENT RAMSEY

then all computation paths lead to the output of a yes-instance (y, k) of

REFINEMENT RAMSEY.

• Otherwise, if all instances (xi, k) are no-instances of RAMSEY, then at least

one computation path leads to the output of a no-instance of RAMSEY.

For that, we first define an embedding construction and explain how it will

be used in the composition.

45


6.4.1 Embedding Construction

Given t instances (G1, k), . . . , (Gt, k) of REFINEMENT RAMSEY(k) and a host graph

H with t′ vertices, such that t′ ≥ t, we construct G′ = Embed(H, k;G1, ..., Gt) as

following.

1. We first take a clique C of size k − 2 and an independent set I of size k − 1

and perform a join of them to form a graph D.

2. Now, for each Gi, we perform the join of Gi and D, and call it F .

3. We perform the join of Gi and D for each i, and call the graph obtained F ′.

Let G′i be disjoint union of F and F ′.

4. We associate each G′i with at least one vertex of H arbitrarily. Let the vertex

be called vG′i. No two G′i’s get mapped to the same vertex of H.

5. We replace each vertex vGiof H with corresponding G′i. In the new graph

G′, (u, v) is an edge, where u ∈ Gi, v ∈ Gj, if and only if vGi, vGj

) was an

edge in H. Adjacencies of vertices inside G′i are preserved.

We observe that we may obtain different embeddings by associating different

vertices of H to a graph Gi, but we will see that this will have no effect on the

result.

Lemma 8. Let H be a host graph on t′ vertices and (G1, k), ..., (Gt, k) be legalinputs for REFINEMENT RAMSEY(k) such that t ≤ t′. Suppose every vertex of H iscontained in a clique of size l or an independent set of size l but H neither containsa clique nor an independent set of size l + 1, then Embed(H, k;G1, . . . , Gt) has aclique and an independent set of size l· (2k − 2). Furthermore, it contains a cliqueor an independent set of size l· (2k − 2) + 1 if and only if (Gi, k) is a YES instancefor some i ∈ 1, . . . , t.

Proof. The graph D has both a clique and independent set of size k − 1. So,

each of F and F ′ have independent sets of size k − 2, combining which, we get

an independent set of size 2k − 2 in G′i . Also, since (Gi, k) is an instance of

REFINEMENT RAMSEY. there is a clique of size k − 1 in Gi, which along with the

clique C of size k − 2 and one vertex of the independent set I, gives a clique of

46


size 2k − 2 in G′i. Also, if (Gi, k) is a YES instance of REFINEMENT RAMSEY, then

Gi contains either an independent set or clique of size k. If Gi contained a clique

of size k, combining same way as we did earlier, we see that G′i has both clique

and independent set of size 2k − 1. Similar argument goes through if Gi has an

independent set of size k. For the converse, let G′i has a clique of size 2k−1, then

these vertices have to come from either F or F ′, out of which D can contribute

only k − 1 vertices. So, Gi or Gi has a clique of size k and hence (Gi, k) is a YES

instance of REFINEMENT RAMSEY. Similar argument goes through if G′i has an

independent set of size 2k− 1 So, we have shown that G′i has a both a clique and

independent set of size 2k− 2, and has a clique or independent set of size 2k− 1

if and only if (Gi, k) was a YES instance of REFINEMENT RAMSEY.

Now, to show that G′ has a clique and independent set of size l· (2k − 2), if

H has any clique of size l, then we combine the cliques of G′i’s mapped to these

vertices to get a clique of size l· (2k− 2), otherwise we combine the independent

sets. Now, if one of the (Gi, k) was a YES instance, G′i has 2k − 1 vertices sized

clique and independent set. Again, if the vertex vG′i was part of a clique, we

combine the cliques of all other G′i’s, which are mapped to the vertices of clique

in H to get a clique of size l· (2k− 2) + 1, otherwise we combine the independent

sets.

Conversely, if G′ has a clique or independent set of size l· (2k− 2) + 1, then at

least one of theG′i’s contributes 2k−1 vertices to it (since there is no clique of size

bigger than l in H), which in turn makes one of the instances (Gi, k) a YES in-

stance or REFINEMENT RAMSEY. Similarly we argue when G′ has an independent

set of size l· (2k−2)+1. Hence, we have shown thatG′ = Embed(H, k;G1, . . . , Gt)

has both cliques and independent sets of size l· (2k − 2), and also that it has a

clique or independent set of size l· (2k − 2) + 1 if and only if (Gi, k) was a YES

instance of REFINEMENT RAMSEY for some i ∈ i, . . . , t. This concludes proof of

the lemma.

We see that if we can find a suitable graph H (every vertex is part of a clique

or independent set of size l) on more than t vertices, then we can embed t in-

stances of REFINEMENT RAMSEY into it, such that the obtained graph with pa-

rameter l· (2k − 2) + 1 is YES instance of REFINEMENT RAMSEY is and only if one

of the original instances was a YES instance of RAMSEY. So, we have almost got

47


our compostition algorithm, modulo the restriction on parameter. Clearly, if we

can obtain the graph H such that every vertex is part of a clique or independent

set of size l, where l ∈ o(t), then we will have a composition algorithm, which

will rule out polynomial kernels for REFINEMENT RAMSEY.

6.4.2 Construction of Host Graph

We use lemma 7 to obtain a host graph co-nondeterministically. For that, we give

perform following steps.

1. Guess integers T ∈ 1, . . . , (d8 log te+ 1)· t and l ∈ 1, . . . , d8 log(t)e.

2. Guess a graph H ′ on T vertices. Initiate a subset of vertices of H ′ as empty

set.

3. Guess a subset of vertices (not already guessed) of size l from H ′. If the

graph induced on these vertices is not an independent set or a clique, say

YES and abort, otherwise add these vertices to A and delete from H ′.

4. If |A| ≥ t, stop and return H = H ′[A], otherwise go to step 3.

Clearly, if the algorithm outputs a graph H, then every vertex is part of a

clique or independent set of size l ∈ 1, . . . , d8 log(t)e, or l ∈ o(t).

Lemma 9. The algorithm outputs a graph H on at least one computational path,such that every vertex is part of a clique or independent set of size l, but the graphH does not contain a clique or independent set of size l + 1, l ∈ 1, . . . , d8 log(t)e.

Proof. To prove that, let us assume l to be the smallest integer for which R(l +

1) > R(l)+t. We know that from Lemma 7 that l ≤ d8 log(t)e. Again, since l is the

smallest such integer, for all the other integers l′ < l, R(l′+ 1) ≤ R(l′) + t. Hence

we get R(l) ≤ t· (l − 1) + R(1) ≤ d8 log(t)e· t. So, guessing a graph H ′ on T =

d8 log(t)e· t+ t vertices guarantees that for some choice of l ∈ 1, . . . , d8 log(t)e,T = R(l) + t < R(l + 1). So, there exists a graph on T vertices which does not

contain a clique or independent set of size l+1, which will be guessed on at least

one computation path.

After guessing this graph H ′ on T vertices, we know that it must contain a

clique or independent set of size L since T ≥ R(l) + t. We guess the clique or

48


independent set and put it in A. If |A| < t, then |H ′| > R(l) and hence there is

one clique or independent set of size l in H ′ which we can guess. We continue

this process till A ≥ t. Notice that H ′[A] will not have a clique or independent

set of size l + 1, as it is a subgraph of H ′.

6.4.3 Composition

The coNP-composition of REFINEMENT RAMSEY(k), given input (G1, k), . . . , (Gt, k),

consists of following steps-

• Construct a host graph H.

• Let G′ = Embed(H, k;G1, . . . , Gt).

• Output (G′, l· (2k − 2) + 1), where l is the integer guessed in Step 2 of

construction of the graph H.

It is easy to see that the above steps can be performed in polynomial time,

and we also observe that k′ = l· (2k − 2) + 1) which is to(1)poly(k).

To prove the correctness, assume that one of the instances (Gi, k) was a YES

instance. We have to prove that on all the computational paths, we return YES. If

the construction of the host graph fails, then we return YES in step 3. Otherwise,

each vertex of H is part of a clique or independent set of size l. So, by Lemma 8,

we get that (G′, l· (2k − 2) + 1) is a YES instance.

On the other hand, if all the (Gi, k) are NO instances, then by Lemma 9, on at

least one computational path, the algorithm outputs a graph H, which does not

contain a clique or independent set of size l + 1. Then, by Lemma 8, it follows

that (G′, l· (2k − 2) + 1) is a NO instance.

Since REFINEMENT RAMSEY is NP-hard and REFINEMENT RAMSEY(k) has a coNP-

composition, hence by Lemma 6, we get the following theorem-

Theorem 13. REFINEMENT RAMSEY(k) does not admit a polynomial kernelizationunless NP ⊆ coNP/poly and the polynomial hierarchy collapses to its third level.

But we know from Observation 3 of Chapter 3 that RAMSEY(k) ≤ppt REFINEMENT RAMSEY(k).

Also, we know that RAMSEY and REFINEMENT RAMSEY are NP-cmplete. Hence

from Theorem 2, we get the following corollary.

49


Corollary 4. RAMSEY(k) does not admit a polynomial kernelization unless NP ⊆coNP/poly and the polynomial hierarchy collapses to its third level.

6.5 Summary

In this chapter we saw how the co-nondeterminism allowed in the communi-

cation protocol can be used to prove lower bounds on kernels. We first for-

mally saw the notion of co-NP composition defined by Kratsch [14]. We had

another look at the RAMSEY problem and used a property of gaps between Ram-

sey numbers along with the co-nondeterminism to rule out polynomial kernels.

Co-nondeterminism in compositions can further utilized to rule out polynomial

kernels for problems where we are looking for a induced subgraph satisfying

certain hereditary properties. For details, the reader may refer to [15].

50

Part III

Summing Up

51

7Unified View

7.1 Introduction

In the previous chapters, we have seen how the theory of kernelization lower

bounds have evolved. Cross composition gives an easy way to use the compo-

sition framework given by Bodlaender et al. [2] by allowing non parameterized

problems as input and also by allowing output instance of a different language.

Weak composition generalizes it by allowing some depemdendence on the num-

ber of input instances to rule out polynomial kernels of certain size, and coNP-

composition makes use of co-nondeterminism to rule out polynomial kernels.

They all look somewhat different, but since all of them are based on similar re-

sults of Fornow and Santhanam [10] and Dell and Melkebeek [7], it is a natural

question to ask whether they all are special cases of a generalized version of

composition. In this chapter, we will try to build one such framework, which

takes into consideration all the essential aspects of the techniques presented in

previous chapters.

7.2 The Idea

On building towards a generalized composition framework, we make the follow-

ing observations:

1. The existing types are Cross Composition (subsumes the classical com-

position model by Bodlaender et al. [2]), Weak Composition and coNP-

52

Chapter 7. Unified View

composition.

2. Cross composition is the only framework which allows the inputs to be non-

parameterized problems by the notion of polynomial equivalence relation.

All other compositions, where the parameters of input instances are asked

to be the same, can be easily incorporated in this method by making the

polynomial equivalence relation based on the parameter.

3. Weak composition is the only framework which allows us to rule out kernels

of polynomial size by allowing dependence on t. Hence, the generalized

framework must have this feature.

4. coNP compositions are different only in the sense that they allow non-

determinism, and it is not hard to incorporate this feature in any of the

previous frameworks, as they are also based on Dell and Melkebeek [7]

result, which allows non-determinism.

So, essentially, we already have the raw material for the generalized compo-

sition in the form of the previous definitions. We just need to take out the most

generalized feature of each one of them.

7.3 Definition

We first recall the definition of polynomial equivalence relation from Chapter 3.

Definition 18. (Polynomial Equivalence Relation). An equivalence relation R

on Σ∗ is called a polynomial equivalence relation if the following two conditionshold:

1. There is an algorithm that given two strings x, y ∈ Σ∗ decides whether x andy belong to the same equivalence class in (|x|+ |y|)O(1) time.

2. For any finite set S ⊆ Σ∗ the equivalence relation R partitions the elementsof S into at most (maxx∈S |x|)O(1) classes.

We define the notion of generalized composition. We call it G-composition.

53


Definition 19. (Gen d-composition). Let L ⊆ Σ∗ be a set and let Q ⊆ Σ∗×N bea parameterized problem. We say that L (the parameterized language, correspond-ing to L) Gen d-composes into Q if there is a polynomial equivalence relation R

and an algorithm C (possibly nondeterministic) which, given t strings x1, x2, . . . , xtbelonging to the same equivalence class of R, takes time polynomial in

∑ti=1 |xi|

and outputs on each computation path an instance (y, k) ∈ Σ∗ × N such that thefollowing hold:

• k ≤ t1/d(maxti=1 |xi|+ log t)O(1)

• If at least one instance xi is a YES-instance of of L then all computation pathslead to the output of a YES-instance (y, k) of Q.

• Otherwise, if all instances xi are NO-instances of L, then at least one compu-tation path leads to the output of a NO-instance of Q.

Now we prove the following theorem.

Theorem 14. Let L ⊆ Σ∗ be a set which is NP-hard under Karp reductions. If LGen d-composes into the parameterized problem Q, then Q has no kernel of sizeO(kd−ε) for all ε > 0 unless NP ⊆ coNP/poly.

Proof. We assume that Q has a kernel of size O(kd−ε) for some ε > 0. By Obser-

vation 2 of Chapter 5 Q has a communication protocol of cost O(kd−ε).

We have to show that OR(L) (see Definition 12 in Chapter 3 for definition of


t(s), where the OR(L) has t(s) instances and the largest instance size is s, and


Define s := maxt1=1(s)|xi|. If, t(s) > (|Σ|+ 1)s, then there must be duplicate

inputs, since the number of distinct inputs of length s′ ≤ s is |Σ|s′. We discard the

duplicate instances from the input and make sure that t(s) ≤ (|Σ|+ 1)s, which

gives us log(t(s)) ∈ O(s).

As the definition of cross-composition guarantees a polynomial equivalence

relation R, we apply it on the t(s) instances of L, namely x1, . . . , xt(s), by com-

paring them pairwise and partitioning them into r groups. Let the groups be

Y1, . . . , Yr such that the instances inside any Yi are equivalent under R. Note that

54


r is polynomial in s and this partitioning takes time polynomial in the input size

by the definition of polynomial equivalence relation.

We apply the Gen d-composition algorithm C on each of the Yi’s. It is possible

to do so since all the instances in Yi are equivalent under R. Let the output of

the cross-composition algorithm be (zi, ki) for Yi, where (zi, ki) is an instance

of Q. By the definition of cross-composition, for all i ∈ [r], ki is bounded by

t(s)1/d(maxt(s)i=1 |xi| + log t(s))O(1). But we know that log t(s) ∈ O(s), and also for

all i ∈ [t(s)], |xi| is bounded s. Hence, ki is bounded by t(s)1/dsO(1) for all i ∈ [r].

For each of the r instances of Q, we apply the protocol of cost O(kd−εi ) to

decide it. If one of the instances is YES, then we accept, otherwise we reject.

Since r is bounded by a polynomial in s (say O(sa)), the cost of the protocol is

bounded by O(sa. (maxri=1 ki)d−ε). We also know that ki is bounded by t(s)1/d· sc

for some constant c for all i ∈ [r], hence we get

O(sa. (r

maxi=1

ki)d−ε) = O(sa. (t(s)1/d· sc)d−ε)

= O(sa. t(s)1−ε/d. sc(d−ε))

= O(sa+cd−cε. t(s)1−ε/d)

= O(t(s)) (for t(s) sufficiently large, such that t(s)ε/d = Ω(sa+cd−cε))

= O(t(s) log t(s))

To prove the correctness, we see that non-determinism is involved only in

the Gen d-composition step. We have to show that if one of the xi’s was a YES

instance at the beginning, then the communication protocol will say YES on all

paths. We look at the group in which this particular YES instance belongs. From

the definition of Gen d-composition, it follows that if one of the inputs to it was a

YES instance, it will lead to a YES instance on all paths, and hence, the instance

corresponding to Yi will be YES instance on all paths (kernelization preserves

equivalence), and the protocol will give the right answer.

On the other hand, if all the xi’s were no instance, then the Gen d-composition

algorithm on each Yi will output a NO instance on at least one computation path.

After kernelizing the instances, all the instances will remain to be NO instances,

and the protocol will reject.

So, OR(L) has a oracle communication protocol of cost O(t(s) log t(s)) where

55


the first player is allowed to be co-nondeterministic, and so is in coNP/poly by

Complementary Witness Lemma. But, we also know that L is NP-hard, hence we

get NP ⊆ coNP/poly, which concludes the proof of theorem.

Also, we see from Theorem 14 that if the parameter of the instance output

by the Gen d-composition algorithm is bounded by to(1)(maxti=1 |xi| + log t)O(1),

instead of t1/d(maxti=1 |xi| + log t)O(1), then it satisfies the definition of Gen d-

composition for any d, since to(1) ∈ O(t1/d) for all d ∈ N \ 0. We call this special

case (where the output parameter to(1)(maxti=1 |xi| + log t)O(1)) Gen Composition

because of it’s independence on d. Hence we get the following corollary.

Corollary 5. Let L ⊆ Σ∗ be a set which is NP-hard under Karp reductions. If LGen Composes into the parameterized problem Q, then Q has no polynomial kernelunless NP ⊆ coNP/poly.

Proof. As described earlier, the definition of Gen Composition satisfies the defi-

nition of Gen d-composition for every d ∈ N \ 0. Hence, by Theorem 14, Q can

not have kernel of size kd−ε for any ε for any d unless NP ⊆ coNP/poly. Hence, Q

does not admit polynomial kernel unless NP ⊆ coNP/poly.

7.4 Relation to Earlier Techniques

In this section, we show how all the earlier techniques can be written as special

case of Gen d-composition or Gen Composition.

7.4.1 Cross Composition

We take the Gen composition algorithm, and restrict the algorithm to be deter-

ministic only. Also, we remove t1/d dependence on k, and allow it to be only

k ≤ (maxti=1 |xi|+ log t)O(1). This way, the restricted Gen Composition is same as

Cross Composition as described in Chapter 3, and rules out polynomial kernels

of any size.

7.4.2 Weak Composition

To show weak d-composition as special case of Gen d-composition, we first con-

vert the parameterized instances into classical instances. Now, we set the poly-

56


nomial equivalence relation R to be such that all the instances with same param-

eter value in the parameterized instances (with same number of 1′s at the end

followed by #) are in the same equivalence class.

We restrict the algorithm to be deterministic and remove the log t dependence

from k. Also, we replace maxti=1 |xi| by the parameter value of the instances in

the equivalence class (say k). So, we allow the bounds on the new parameter, k′

to be k′ ≤ t1/dkO(1). In this restricted setting, the Gen d-composition is the same

as weak d-composition and rules out polynomial kernels of certain size.

7.4.3 coNP-composition

To show coNP-composition as special case of Gen Composition, we first convert

the parameterized instances into classical instances. Now, we set the polynomial

equivalence relation R to be such that all the instances with same parameter

value in the parameterized instances (with same number of 1′s at the end fol-

lowed by #) are in the same equivalence class.

Now, instead of allowing k′ ≤ to(1)(maxti=1 |xi|+log t)O(1), we replace maxti=1 |xi|by k, as in the case of weak composition, and remove the term having log t. So,

we allow the bounds on the parameter of output instance to be k′ ≤ to(1)kO(1). In

this restricted setting, Gen Composition is same as coNP-composition and rules

out polynomial kernels.

57

8Conclusion

In this thesis, we had a look at all the recent techniques for arriving at lower

bounds on kernels. We presented all the frameworks defined for arriving at both

super polynomial and polynomial lower bounds on kernels. We described the

techniques of cross-composition, weak d-composition and coNP-composition. We

looked at a few examples of these techniques, which included CLIQUE, VERTEX

COVER, RAMSEY and CLIQUE parameterized by the size of vertex cover.

We note that cross-composition relies on the results on results of Fortnow

and Santhanam, the later techniques use the slightly more general result of Dell

and Melkebeek. Observing that all these techniques rely on similar results, we

give a unified framework, which is an easy consequence of combining all the

techniques.

The techniques have been widely used in recent times to arrive at kernel

lower bounds. We conclude the thesis by mentioning that we have not yet ex-

ploited the result by Dell and Melkebeek in all possible ways, since all the frame-

works use only one way communication protocol. It will be interesting to see if

more rounds can be used to arrive at lower bounds on kernelization.

58

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