why some problems are incompressible

On the Infeasibility of

Obtaining Polynomial Kernels

PROBLEM KERNELS

whatwhyhow

when

no polynomial kernels

Examples

WorkarouNds

satisfiability disjoint factors

many kernels

vertex-disjoint cycles

PROBLEM KERNELS

whatwhyhow

when

no polynomial kernels

Examples

WorkarouNds

satisfiability disjoint factors

many kernels

vertex-disjoint cycles

ere are people at a professor’s party.

e professor would like to locate a group of six people who are popular.

i.e, everyone is a friend of at least one of the six.

Being a busy man, he asks two of his students to find such a group.

e first grimaces and starts making a list of(256

)possibilities.

e second knows that no one in the party has more than three friends.

Academicians tend to be lonely.

ere are people at a professor’s party.

e professor would like to locate a group of six people who are popular.

i.e, everyone is a friend of at least one of the six.

Being a busy man, he asks two of his students to find such a group.

e first grimaces and starts making a list of(256

)possibilities.

e second knows that no one in the party has more than three friends.

Academicians tend to be lonely.

ere is a graph on n vertices.

e professor would like to locate a group of six people who are popular.

i.e, everyone is a friend of at least one of the six.

Being a busy man, he asks two of his students to find such a group.

e first grimaces and starts making a list of(256

)possibilities.

e second knows that no one in the party has more than three friends.

Academicians tend to be lonely.

ere is a graph on n vertices.

Is there a dominating set of size at most k?

i.e, everyone is a friend of at least one of the six.

Being a busy man, he asks two of his students to find such a group.

e first grimaces and starts making a list of(256

)possibilities.

e second knows that no one in the party has more than three friends.

Academicians tend to be lonely.

ere is a graph on n vertices.

Is there a dominating set of size at most k?

(Every vertex is a neighbor of at least one of the k vertices.)

Being a busy man, he asks two of his students to find such a group.

e first grimaces and starts making a list of(256

)possibilities.

e second knows that no one in the party has more than three friends.

Academicians tend to be lonely.

ere is a graph on n vertices.

Is there a dominating set of size at most k?

(Every vertex is a neighbor of at least one of the k vertices.)

e problem is NP–complete,

e first grimaces and starts making a list of(256

)possibilities.

e second knows that no one in the party has more than three friends.

Academicians tend to be lonely.

ere is a graph on n vertices.

Is there a dominating set of size at most k?

(Every vertex is a neighbor of at least one of the k vertices.)

e problem is NP–complete,

but is trivial on “large” graphs of bounded degree,

e second knows that no one in the party has more than three friends.

Academicians tend to be lonely.

ere is a graph on n vertices.

Is there a dominating set of size at most k?

(Every vertex is a neighbor of at least one of the k vertices.)

e problem is NP–complete,

but is trivial on “large” graphs of bounded degree,

as you can say NO whenever n > kb.

Pre-processing is a humble strategy for coping with hard problems,almost universally employed.

Mike Fellows

We would like to talk about the “preprocessing complexity” of a problem.

To what extent may we simplify/compress a problem before we begin tosolve it?

Note that any compression routine has to run efficiently to look attractive.

is makes it unreasonable for any NP–complete problem to admitcompression algorithms.

However, (some) NP–complete problems can be compressed.

We extend our notion (and notation) to understand them.

Notation

We denote a parameterized problem as a pair (Q, κ) consisting of a clas-sical problem Q ⊆ {0, 1}∗ and a parameterization κ : {0, 1}∗ → N.

.

.Size of the Kernel

.

.Kernel

Data reduction

involves pruning down

a large input

into an equivalent,

significantly smaller object,

quickly.

.

.Size of the Kernel

.

.Kernel

Data reduction

is a function f : {0, 1}∗ × N → {0, 1}∗ × N, such that

a large input

into an equivalent,

significantly smaller object,

quickly.

.

.Size of the Kernel

.

.Kernel

Data reduction

is a function f : {0, 1}∗ × N → {0, 1}∗ × N, such that

for all (x, k), |x| = n

into an equivalent,

significantly smaller object,

quickly.

.

.Size of the Kernel

.

.Kernel

Data reduction

is a function f : {0, 1}∗ × N → {0, 1}∗ × N, such that

for all (x, k), |x| = n

f(x), k ′ ∈ L iff (x, k) ∈ L,

significantly smaller object,

quickly.

.

.Size of the Kernel

.

.Kernel

Data reduction

is a function f : {0, 1}∗ × N → {0, 1}∗ × N, such that

for all (x, k), |x| = n

f(x), k ′ ∈ L iff (x, k) ∈ L,

|f(x)| = g(k),

quickly.

.

.Size of the Kernel

.

.Kernel

Data reduction

is a function f : {0, 1}∗ × N → {0, 1}∗ × N, such that

for all (x, k), |x| = n

f(x), k ′ ∈ L iff (x, k) ∈ L,

|f(x)| = g(k),

and f is polytime computable.

.

.Size of the Kernel

.

.Kernel

A kernelization procedure

is a function f : {0, 1}∗ × N → {0, 1}∗ × N, such that

for all (x, k), |x| = n

f(x), k ′ ∈ L iff (x, k) ∈ L,

|f(x)| = g(k),

and f is polytime computable.

.

.Size of the Kernel

.

.Kernel

A kernelization procedure

is a function f : {0, 1}∗ × N → {0, 1}∗ × N, such that

for all (x, k), |x| = n

(f(x), k ′) ∈ L iff (x, k) ∈ L,

|f(x)| = g(k),

and f is polytime computable.

.

.Size of the Kernel

.

.Size of the Kernel

A kernelization procedure

is a function f : {0, 1}∗ × N → {0, 1}∗ × N, such that

for all (x, k), |x| = n

f(x), k ′ ∈ L iff (x, k) ∈ L,

|f(x)| = g(k),

and f is polytime computable.

eorem

Having a kernelization procedure implies, and is implied by,parameterized tractability.

Definition

A parameterized problem L is fixed-parameter tractable if there exists analgorithm that decides in f (k) · nO(1) time whether (x, k) ∈ L, wheren := |x|, k := κ(x), and f is a computable function that does not dependon n.

eorem

Having a kernelization procedure implies, and is implied by,parameterized tractability.

Definition

A parameterized problem L is fixed-parameter tractable if there exists analgorithm that decides in f (k) · nO(1) time whether (x, k) ∈ L, wheren := |x|, k := κ(x), and f is a computable function that does not dependon n.

eorem

A problem admits a kernel if, and only if,it is fixed–parameter tractable.

Definition

A parameterized problem L is fixed-parameter tractable if there exists analgorithm that decides in f (k) · nO(1) time whether (x, k) ∈ L, wheren := |x|, k := κ(x), and f is a computable function that does not dependon n.

eorem

A problem admits a kernel if, and only if,it is fixed–parameter tractable.

Given a kernel, a FPT algorithm is immediate (even brute–force on thekernel will lead to such an algorithm).

eorem

A problem admits a kernel if, and only if,it is fixed–parameter tractable.

On the other hand, a FPT runtime of f(k)·nc gives us a f(k)–sized kernel.

We run the algorithm for nc+1 steps and either have a trivial kernel if thealgorithm stops, else:

nc+1 < f(k) · nc

eorem

A problem admits a kernel if, and only if,it is fixed–parameter tractable.

On the other hand, a FPT runtime of f(k)·nc gives us a f(k)–sized kernel.

We run the algorithm for nc+1 steps and either have a trivial kernel if thealgorithm stops, else:

n < f(k)

“Efficient” Kernelization

What is a reasonable notion of efficiency for kernelization?

e smaller, the better.

In particular,

Polynomial–sized kernels <arebetter than Exponential–sized Kernels

“Efficient” Kernelization

What is a reasonable notion of efficiency for kernelization?e smaller, the better.

In particular,

Polynomial–sized kernels <arebetter than Exponential–sized Kernels

“Efficient” Kernelization

What is a reasonable notion of efficiency for kernelization?e smaller, the better.

In particular,

Polynomial–sized kernels <arebetter than Exponential–sized Kernels

e problem of finding Dominating Set of size k on graphs where thedegree is bounded by b, parameterized by k, has a linear kernel. is is anexample of a polynomial–sized kernel.

PROBLEM KERNELS

whatwhyhow

when

no polynomial kernels

Examples

WorkarouNds

satisfiability disjoint factors

many kernels

vertex-disjoint cycles

A composition algorithm A for a problem is designed to act as a fastBoolean OR of multiple problem-instances.

It receives as input a sequence of instances.x = (x1, . . . , xt) with xi ∈ {0, 1}∗ for i ∈ [t], such that

(k1 = k2 = · · · = kt = k)

It produces as output a yes-instance with a small parameter if and only ifat least one of the instances in the sequences is also a yes-instance.

κ(A(x)) = kO(1)

Running time polynomial in Σi∈[t]|xi|

A composition algorithm A for a problem is designed to act as a fastBoolean OR of multiple problem-instances.

It receives as input a sequence of instances.x = (x1, . . . , xt) with xi ∈ {0, 1}∗ for i ∈ [t], such that

k1 = k2 = · · · = kt = k

It produces as output a yes-instance with a small parameter if and only ifat least one of the instances in the sequences is also a yes-instance.

κ(A(x)) = kO(1)

Running time polynomial in Σi∈[t]|xi|

A composition algorithm A for a problem is designed to act as a fastBoolean OR of multiple problem-instances.

It receives as input a sequence of instances.x = (x1, . . . , xt) with xi ∈ {0, 1}∗ for i ∈ [t], such that

k1 = k2 = · · · = kt = k

It produces as output a yes-instance with a small parameter if and only ifat least one of the instances in the sequences is also a yes-instance.

k ′ = kO(1)

Running time polynomial in Σi∈[t]|xi|

A composition algorithm A for a problem is designed to act as a fastBoolean OR of multiple problem-instances.

It receives as input a sequence of instances.x = (x1, . . . , xt) with xi ∈ {0, 1}∗ for i ∈ [t], such that

k1 = k2 = · · · = kt = k

It produces as output a yes-instance with a small parameter if and only ifat least one of the instances in the sequences is also a yes-instance.

k ′ = kO(1)

Running time polynomial in Σi∈[t]|xi|

e Recipe for Hardness

Composition Algorithm + Polynomial Kernel

⇓Distillation Algorithm

⇓PH = Σ

p3

eorem. Let (P, k) be a compositional parameterized problem such thatP is NP-complete. If P has a polynomial kernel, then P also has a distil-lation algorithm.

e Recipe for Hardness

Composition Algorithm + Polynomial Kernel

⇓Distillation Algorithm

⇓PH = Σ

p3

eorem. Let (P, k) be a compositional parameterized problem such thatP is NP-complete. If P has a polynomial kernel, then P also has a distil-lation algorithm.

Transformations

Let (P, κ) and (Q, γ) be parameterized problems.

We say that there is a polynomial parameter transformation from P to Q ifthere exists a polynomial time computable function f : {0, 1}∗ −→ {0, 1}∗,and a polynomial p : N → N, such that, if f(x) = y, we have:

x ∈ P if and only if y ∈ Q,

and

γ(y) 6 p(κ(x))

Transformations

Let (P, κ) and (Q, γ) be parameterized problems.

We say that there is a polynomial parameter transformation from P to Q ifthere exists a polynomial time computable function f : {0, 1}∗ −→ {0, 1}∗,and a polynomial p : N → N, such that, if f(x) = y, we have:

x ∈ P if and only if y ∈ Q,

and

γ(y) 6 p(κ(x))

eorem: Suppose P is NP-complete, and Q ∈ NP. If f is a polynomialtime and parameter transformation from P to Q and Q has a polynomialkernel, then P has a polynomial kernel.

..

.f(x) = y

Instance of Q

.K(y)

.z ∈ P

.PPT Reduction

.NP–completeness

Reduction

.Kernelization

.x

Instance of P

eorem: Suppose P is NP-complete, and Q ∈ NP. If f is a polynomialtime and parameter transformation from P to Q and Q has a polynomialkernel, then P has a polynomial kernel.

..

.f(x) = y

Instance of Q

.K(y)

.z ∈ P

.PPT Reduction

.NP–completeness

Reduction

.Kernelization

.x

Instance of P

PROBLEM KERNELS

whatwhyhow

when

no polynomial kernels

Examples

WorkarouNds

satisfiability disjoint factors

many kernels

vertex-disjoint cycles

Recall

A composition for a parameterized language (Q, κ) is required to“merge” instances

x1, x2, . . . , xt,

into a single instance x in polynomial time, such that κ(x) is polynomialin k := κ(xi) for any i.

..e output of the algorithm belongs to(Q,κ(x))

if, and only ifthere exists at least one i ∈ [t] for which xi ∈ (Q,κ).

..A Composition Tree

.ρ(a, b)

.a

.ρ(x1, x2)

.x1 .x2

.ρ(x3, x4)

.x3 .x4

.

.b

.ρ(xt−3, xt−2)

.xt−3 .xt−2

.ρ(xt−1, xt)

.xt−1 .xt

General Framework For Composition

..A Composition Tree

.ρ(a, b)

.a

.ρ(x1, x2)

.x1 .x2

.ρ(x3, x4)

.x3 .x4

.. . .

.b

.ρ(xt−3, xt−2)

.xt−3 .xt−2

.ρ(xt−1, xt)

.xt−1 .xt

General Framework For Composition

..A Composition Tree

.ρ(a, b)

.a

.ρ(x1, x2)

.x1 .x2

.ρ(x3, x4)

.x3 .x4

.…

.b

.ρ(xt−3, xt−2)

.xt−3 .xt−2

.ρ(xt−1, xt)

.xt−1 .xt

General Framework For Composition

..

Most composition algorithms can be statedin terms of a single operation, ρ,

that describes the dynamic programmingover this complete binary tree on t leaves.

Weighted Satisfiability

Given a CNF formula φ,Is there a satisfying assignment of weight at most k?

Parameter: b+k.When the length of the longest clause is bounded by b,

there is an easy branching algorithm with runtime O(bk · p(n)).

Weighted Satisfiability

Given a CNF formula φ,Is there a satisfying assignment of weight at most k?

Parameter: b+k.When the length of the longest clause is bounded by b,

there is an easy branching algorithm with runtime O(bk · p(n)).

Weighted Satisfiability

.

.|bp(n)|||d|)

Given a CNF formula φ,Is there a satisfying assignment of weight at most k?

Parameter: b+k.When the length of the longest clause is bounded by b,

there is an easy branching algorithm with runtime O(bk · p(n)).

Input: α1, α2, . . . , αt.

κ(αi) = b + k.n := maxi∈[n] ni

If t > bk, then solve every αi individually.

Total time = t · bk · p(n)

If not, t < bk – this gives us a bound on the number of instances.

Input: α1, α2, . . . , αt.κ(αi) = b + k.

n := maxi∈[n] ni

If t > bk, then solve every αi individually.

Total time = t · bk · p(n)

If not, t < bk – this gives us a bound on the number of instances.

Input: α1, α2, . . . , αt.κ(αi) = b + k.n := maxi∈[n] ni

If t > bk, then solve every αi individually.

Total time = t · bk · p(n)

If not, t < bk – this gives us a bound on the number of instances.

Input: α1, α2, . . . , αt.κ(αi) = b + k.n := maxi∈[n] ni

If t > bk, then solve every αi individually.

Total time = t · bk · p(n)

If not, t < bk – this gives us a bound on the number of instances.

Input: α1, α2, . . . , αt.κ(αi) = b + k.n := maxi∈[n] ni

If t > bk, then solve every αi individually.

Total time = t · bk · p(n) Total time = t · bk · p(n)

If not, t < bk – this gives us a bound on the number of instances.

Input: α1, α2, . . . , αt.κ(αi) = b + k.n := maxi∈[n] ni

If t > bk, then solve every αi individually.

Total time = t · bk · p(n) Total time < t · t · p(n)

If not, t < bk – this gives us a bound on the number of instances.

Input: α1, α2, . . . , αt.κ(αi) = b + k.n := maxi∈[n] ni

If t > bk, then solve every αi individually.

Total time = t · bk · p(n) Total time < t · t · p(n)

If not, t < bk – this gives us a bound on the number of instances.

..Level 1

.(β1 ∨ b0) ∧ . . . ∧ (βn ∨ b0) .(β ′1 ∨ b̄0) ∧ . . . ∧ (β ′

m ∨ b̄0)

is is the scene at the leaves, where the βjs are the clauses in αi forsome i and the β ′

js are clauses of αi+1.

..Level j

.(β1 ∨ b) ∧ . . . ∧ (βn ∨ b) .(β ′1 ∨ b̄) ∧ . . . ∧ (β ′

m ∨ b̄)

is is the scene at the leaves, where the βjs are the clauses in αi forsome i and the β ′

js are clauses of αi+1.

..(β1 ∨ b) ∧ (β2 ∨ b) ∧ . . . ∧ (βn ∨ b) ∧

(β ′1 ∨ b̄) ∧ (β ′

2 ∨ b̄) ∧ . . . ∧ (β ′m ∨ b̄)

.(β1 ∨ b) ∧ . . . ∧ (βn ∨ b) .(β ′1 ∨ b̄) ∧ . . . ∧ (β ′

m ∨ b̄)

Take the conjunction of the formulas stored at the child nodes.Take the conjunction of the formulas stored at the child nodes.

..(β1 ∨ b ∨ bj) ∧ (β2 ∨ b ∨ bj) ∧ . . . ∧ (βn ∨ b ∨ bj)∧

(β ′1 ∨ b̄ ∨ bj) ∧ (β ′

2 ∨ b̄ ∨ bj) ∧ . . . ∧ (β ′m ∨ b̄ ∨ bj)

.(β1 ∨ b) ∧ . . . ∧ (βn ∨ b) .(β ′1 ∨ b̄) ∧ . . . ∧ (β ′

m ∨ b̄)

where the βjs are the clauses in αi for some i and the.if the parent is a “left child”.

..(β1 ∨ b ∨ b̄j) ∧ (β2 ∨ b ∨ b̄j) ∧ . . . ∧ (βn ∨ b ∨ b̄j)∧

(β ′1 ∨ b̄ ∨ b̄j) ∧ (β ′

2 ∨ b̄ ∨ b̄j) ∧ . . . ∧ (β ′m ∨ b̄ ∨ b̄j)

.(β1 ∨ b) ∧ . . . ∧ (βn ∨ b) .(β ′1 ∨ b̄) ∧ . . . ∧ (β ′

m ∨ b̄)

where the βjs are the clauses in αi for some i and the.if the parent is a “right child”.

adding a suffix to “control the weight”

α

∧

(c̄0 ∨ b̄0) ∧ (c0 ∨ b0)∧

(c̄1 ∨ b̄1) ∧ (c1 ∨ b1)∧

. . .

(c̄i ∨ b̄i) ∧ (ci ∨ bi)∧

. . .

(c̄l−1 ∨ b̄l−1) ∧ (cl−1 ∨ bl−1)

Claim

e composed instance α has a satisfying assignment of weight 2k

⇐⇒at least one of the input instances admit a satisfying assignment of

weight k.

Proof of Correctness

... . . ∧ (α ∨ (b0 ∨ b1 ∨ b̄2)) ∧ . . .

.. . . α ∨ b0 ∨ b1∨b̄2. . .

.. . . α ∨ b0∨ b1. . .

.. . . α ∨ b0 . . .

Disjoint Factors

Let Lk be the alphabet consisting of the letters {1, 2, . . . , k}.

A factor of a word w1 · · ·wr ∈ L∗k is a substring wi · · ·wj ∈ L∗

k, with1 6 i < j 6 r, which starts and ends with the same letter.

Disjoint factors do not overlap in the word.Are there k disjoint factors?

Does the word have all the k factors, mutually disjoint?

Disjoint FactorsLet Lk be the alphabet consisting of the letters {1, 2, . . . , k}.

A factor of a word w1 · · ·wr ∈ L∗k is a substring wi · · ·wj ∈ L∗

k, with1 6 i < j 6 r, which starts and ends with the same letter.

123235443513

Disjoint factors do not overlap in the word.Are there k disjoint factors?

Does the word have all the k factors, mutually disjoint?

Disjoint FactorsLet Lk be the alphabet consisting of the letters {1, 2, . . . , k}.

A factor of a word w1 · · ·wr ∈ L∗k is a substring wi · · ·wj ∈ L∗

k, with1 6 i < j 6 r, which starts and ends with the same letter.

123235443513

Disjoint factors do not overlap in the word.Are there k disjoint factors?

Does the word have all the k factors, mutually disjoint?

Disjoint FactorsLet Lk be the alphabet consisting of the letters {1, 2, . . . , k}.

A factor of a word w1 · · ·wr ∈ L∗k is a substring wi · · ·wj ∈ L∗

k, with1 6 i < j 6 r, which starts and ends with the same letter.

123235443513

Disjoint factors do not overlap in the word.Are there k disjoint factors?

Does the word have all the k factors, mutually disjoint?

Disjoint FactorsLet Lk be the alphabet consisting of the letters {1, 2, . . . , k}.

A factor of a word w1 · · ·wr ∈ L∗k is a substring wi · · ·wj ∈ L∗

k, with1 6 i < j 6 r, which starts and ends with the same letter.

123235443513

Disjoint factors do not overlap in the word.Are there k disjoint factors?

Does the word have all the k factors, mutually disjoint?

Disjoint FactorsLet Lk be the alphabet consisting of the letters {1, 2, . . . , k}.

A factor of a word w1 · · ·wr ∈ L∗k is a substring wi · · ·wj ∈ L∗

k, with1 6 i < j 6 r, which starts and ends with the same letter.

123235443513

Disjoint factors do not overlap in the word.Are there k disjoint factors?

Does the word have all the k factors, mutually disjoint?

Disjoint FactorsLet Lk be the alphabet consisting of the letters {1, 2, . . . , k}.

A factor of a word w1 · · ·wr ∈ L∗k is a substring wi · · ·wj ∈ L∗

k, with1 6 i < j 6 r, which starts and ends with the same letter.

123235443513

Disjoint factors do not overlap in the word.Are there k disjoint factors?

Does the word have all the k factors, mutually disjoint?

Disjoint FactorsLet Lk be the alphabet consisting of the letters {1, 2, . . . , k}.

A factor of a word w1 · · ·wr ∈ L∗k is a substring wi · · ·wj ∈ L∗

k, with1 6 i < j 6 r, which starts and ends with the same letter.

123235443513

Disjoint factors do not overlap in the word.Are there k disjoint factors?

Does the word have all the k factors, mutually disjoint?

Disjoint FactorsLet Lk be the alphabet consisting of the letters {1, 2, . . . , k}.

A factor of a word w1 · · ·wr ∈ L∗k is a substring wi · · ·wj ∈ L∗

k, with1 6 i < j 6 r, which starts and ends with the same letter.

123235443513

Disjoint factors do not overlap in the word.Are there k disjoint factors?

Does the word have all the k factors, mutually disjoint?

Disjoint FactorsLet Lk be the alphabet consisting of the letters {1, 2, . . . , k}.

A factor of a word w1 · · ·wr ∈ L∗k is a substring wi · · ·wj ∈ L∗

k, with1 6 i < j 6 r, which starts and ends with the same letter.

123235443513

Disjoint factors do not overlap in the word.Does the word have all the k factors, mutually disjoint?Does the word have all the k factors, mutually disjoint?

Disjoint FactorsLet Lk be the alphabet consisting of the letters {1, 2, . . . , k}.

A factor of a word w1 · · ·wr ∈ L∗k is a substring wi · · ·wj ∈ L∗

k, with1 6 i < j 6 r, which starts and ends with the same letter.

123235443513

Disjoint factors do not overlap in the word.Does the word have all the k factors, mutually disjoint?

Parameter: k

A k! · O(n) algorithm is immediate.

A 2k · p(n) algorithm can be obtained by Dynamic Programming.

Let t be the number of instances input to the composition algorithm.Again, the non-trivial case is when t < 2k.

Let w1, w2, . . . , wt be words over L∗k.

A k! · O(n) algorithm is immediate.A 2k · p(n) algorithm can be obtained by Dynamic Programming.

Let t be the number of instances input to the composition algorithm.Again, the non-trivial case is when t < 2k.

Let w1, w2, . . . , wt be words over L∗k.

A k! · O(n) algorithm is immediate.A 2k · p(n) algorithm can be obtained by Dynamic Programming.

Let t be the number of instances input to the composition algorithm.Again, the non-trivial case is when t < 2k.

Let w1, w2, . . . , wt be words over L∗k.

A k! · O(n) algorithm is immediate.A 2k · p(n) algorithm can be obtained by Dynamic Programming.

Let t be the number of instances input to the composition algorithm.Again, the non-trivial case is when t < 2k.

Let w1, w2, . . . , wt be words over L∗k.

..Leaf Nodes

.b0w1b0 .b0w2b0.. . . .b0wt−1b0 .b0wtb0

Level j.

..bjbj−1ubj−1vbj−1bj

.bj−1ubj−1 .bj−1vbj−1

Claim

e composed word has all the 2k disjoint factors

⇐⇒at least one of the input instances has all the k disjoint factors.

Proof of Correctness

..b2b1b0pb0qb0b1b0rb0sb0b1b2b1b0wb0xb0b1b0yb0zb0b1b2

.b2b1b0pb0qb0b1b0rb0sb0b1b2

.b1b0pb0qb0b1

.b0pb0 .b0qb0

.b1b0rb0sb0b1

.b0rb0 .b0sb0

.b2b1b0wb0xb0b1b0yb0zb0b1b2

.b1b0wb0xb0b1

.b0wb0 .b0xb0

.b1b0pyb0zb0b1

.b0yb0 .b0zb0

Input words: p, q, r, s, w, x, y, z.

Proof of Correctness

..b2b1b0pb0qb0b1b0rb0sb0b1b2b1b0wb0xb0b1b0yb0zb0b1b2

.b2b1b0pb0qb0b1b0rb0sb0b1b2

.b1b0pb0qb0b1

.b0pb0 .b0qb0

.b1b0rb0sb0b1

.b0rb0 .b0sb0

.b2b1b0wb0xb0b1b0yb0zb0b1b2

.b1b0wb0xb0b1

.b0wb0 .b0xb0

.b1b0pyb0zb0b1

.b0yb0 .b0zb0

Input words: p, q, r, s, w, x, y, z.

Proof of Correctness

..b2b1b0pb0qb0b1b0rb0sb0b1b2b1b0wb0xb0b1b0yb0zb0b1b2

.b2b1b0pb0qb0b1b0rb0sb0b1b2

.b1b0pb0qb0b1

.b0pb0 .b0qb0

.b1b0rb0sb0b1

.b0rb0 .b0sb0

.b2b1b0wb0xb0b1b0yb0zb0b1b2

.b1b0wb0xb0b1

.b0wb0 .b0xb0

.b1b0pyb0zb0b1

.b0yb0 .b0zb0

Input words: p, q, r, s, w, x, y, z.

Proof of Correctness

..b2b1b0pb0qb0b1b0rb0sb0b1b2b1b0wb0xb0b1b0yb0zb0b1b2

.b2b1b0pb0qb0b1b0rb0sb0b1b2

.b1b0pb0qb0b1

.b0pb0 .b0qb0

.b1b0rb0sb0b1

.b0rb0 .b0sb0

.b2b1b0wb0xb0b1b0yb0zb0b1b2

.b1b0wb0xb0b1

.b0wb0 .b0xb0

.b1b0pyb0zb0b1

.b0yb0 .b0zb0

Input words: p, q, r, s, w, x, y, z.

Vertex-Disjoint CyclesInput: G = (V, E)

Question: Are there k vertex–disjoint cycles?Parameter: k.

Related problems:FVS, has a O(k2) kernel.Edge–Disjoint Cycles, has a O(k2 log2

k) kernel.

In contrast, Disjoint Factors transforms into Vertex–Disjoint Cycles inpolynomial time.

Vertex-Disjoint CyclesInput: G = (V, E)

Question: Are there k vertex–disjoint cycles?

Parameter: k.

Related problems:FVS, has a O(k2) kernel.Edge–Disjoint Cycles, has a O(k2 log2

k) kernel.

In contrast, Disjoint Factors transforms into Vertex–Disjoint Cycles inpolynomial time.

Vertex-Disjoint CyclesInput: G = (V, E)

Question: Are there k vertex–disjoint cycles?Parameter: k.

Related problems:FVS, has a O(k2) kernel.Edge–Disjoint Cycles, has a O(k2 log2

k) kernel.

In contrast, Disjoint Factors transforms into Vertex–Disjoint Cycles inpolynomial time.

Vertex-Disjoint CyclesInput: G = (V, E)

Question: Are there k vertex–disjoint cycles?Parameter: k.

Related problems:FVS, has a O(k2) kernel.Edge–Disjoint Cycles, has a O(k2 log2

k) kernel.

In contrast, Disjoint Factors transforms into Vertex–Disjoint Cycles inpolynomial time.

Vertex-Disjoint CyclesInput: G = (V, E)

Question: Are there k vertex–disjoint cycles?Parameter: k.

Related problems:FVS, has a O(k2) kernel.Edge–Disjoint Cycles, has a O(k2 log2

k) kernel.

In contrast, Disjoint Factors transforms into Vertex–Disjoint Cycles inpolynomial time.

.

.w = 1123343422

.v1 .v2 .v3 .v4 .v5 .v6 .v7 .v8 .v9 .v10

.1 .1 .2 .3 .3 .4 .3 .4 .2 .2

.1 .2 .3 .4

Disjoint Factors 4ppt Disjoint Cycles

.

.w = 1123343422

.v1 .v2 .v3 .v4 .v5 .v6 .v7 .v8 .v9 .v10

.1 .1 .2 .3 .3 .4 .3 .4 .2 .2

.1 .2 .3 .4

Claimw has all k disjoint factors ⇐⇒ Gw has k vertex–disjoint cycles.

PROBLEM KERNELS

whatwhyhow

when

no polynomial kernels

Examples

WorkarouNds

satisfiability disjoint factors

many kernels

vertex-disjoint cycles

No polynomial kernel?

Look for the best exponential or subexponential kernels...

... or build many polynomial kernels.

No polynomial kernel?

Look for the best exponential or subexponential kernels...

... or build many polynomial kernels.

No polynomial kernel?

Look for the best exponential or subexponential kernels...

... or build many polynomial kernels.

Many polynomial kernels give us better algorithms:(k2

k

)2klogk

versus

p(n)ck

p(n) ·(

ck

k

)

We say that a subdigraph T of a digraph D is an out–tree if T is anoriented tree with only one vertex r of in-degree zero (called the root).

e D M L O-B problem is to find anout-branching in a given digraph with the maximum number of leaves.

Parameter: k

We say that a subdigraph T of a digraph D is an out-branching if T is anoriented spanning tree with only one vertex r of in-degree zero.

e D M L O-B problem is to find anout-branching in a given digraph with the maximum number of leaves.

Parameter: k

We say that a subdigraph T of a digraph D is an out-branching if T is anoriented spanning tree with only one vertex r of in-degree zero.

e D M L O-B problem is to find anout-branching in a given digraph with the maximum number of leaves.

Parameter: k

We say that a subdigraph T of a digraph D is an out-branching if T is anoriented spanning tree with only one vertex r of in-degree zero.

e D k–L O-B problem is to find anout-branching in a given digraph with at least k leaves.

Parameter: k

We say that a subdigraph T of a digraph D is an out-branching if T is anoriented spanning tree with only one vertex r of in-degree zero.

e D k–L O-B problem is to find anout-branching in a given digraph with at least k leaves.

Parameter: k

R k–L O–B admits a kernel of size O(k3).

k–L O–B does not admit a polynomial kernel (proofdeferred).

However, it clearly admits n polynomial kernels (try all possible choicesfor root, and apply the kernel for R k–L O-B.

R k–L O–B admits a kernel of size O(k3).

k–L O–B does not admit a polynomial kernel (proofdeferred).

However, it clearly admits n polynomial kernels (try all possible choicesfor root, and apply the kernel for R k–L O-B.

R k–L O–B admits a kernel of size O(k3).

k–L O–B does not admit a polynomial kernel (proofdeferred).

However, it clearly admits n polynomial kernels (try all possible choicesfor root, and apply the kernel for R k–L O–B.

R k–L O–T admits a kernel of size O(k3).

k–L O–B does not admit a polynomial kernel (proofdeferred).

However, it clearly admits n polynomial kernels (try all possible choicesfor root, and apply the kernel for R k–L O-B.

R k–L O–T admits a kernel of size O(k3).

k–L O–T does not admit a polynomial kernel (composition viadisjoint union).

However, it clearly admits n polynomial kernels (try all possible choicesfor root, and apply the kernel for R k–L O-B.

R k–L O–T admits a kernel of size O(k3).

k–L O–T does not admit a polynomial kernel (composition viadisjoint union).

However, it clearly admits n polynomial kernels (try all possible choicesfor root, and apply the kernel for R k–L O–T.

..(D,k)

.(D1, v1, k)

.(D ′1, v1, k)

.(W1, b1)

.(W1, bmax)

.(D2, v2, k)

.(D ′2, v2, k)

.(W2, b2)

.(W2, bmax)

.(D3, v3, k)

.(D ′3, v3, k)

.(W3, b3)

.(W3, bmax)

.· · ·

.· · ·

.· · ·

.(Dn, vn, k)

.(D ′n, vn, k)

.(Wn, bn)

.(Wn, bmax)

.(D ′, bmax + 1)

.(D ′′, k ′′)

.(D∗, k∗)

.(D ′, bmax + 1)

.Decompose k–Leaf Out–Tree into n

instances of Rooted k–Leaf Out–Tree

. Apply Kernelization(Known for Rooted k–Leaf Out-Tree)

.Reduce to instances of k–Leaf Out–Branching on Nice Willow Graphs(Using NP–completeness)

. WLOG(After Simple Modifications)

. Composition(produces an instance of k–Leaf Out Branching)

. Kernelization(Given By Assumption)

.NP–completeness reduction from k–Leaf Out Branchingto k–Leaf Out Tree)

Figure: e No Polynomial Kernel Argument for k-L O-B.

..(D,k)

.(D1, v1, k)

.(D ′1, v1, k)

.(W1, bmax)

.(W1, bmax)

.(D2, v2, k)

.(D ′2, v2, k)

.(W2, bmax)

.(W2, bmax)

.(D3, v3, k)

.(D ′3, v3, k)

.(W3, bmax)

.(W3, bmax)

.· · ·

.· · ·

.· · ·

.(Dn, vn, k)

.(D ′n, vn, k)

.(Wn, bmax)

.(Wn, bmax)

.(D ′, bmax + 1)

.(D ′′, k ′′)

.(D∗, k∗)

.(D ′, bmax + 1)

.Decompose k–Leaf Out–Tree into n

instances of Rooted k–Leaf Out–Tree

. Apply Kernelization(Known for Rooted k–Leaf Out-Tree)

.Reduce to instances of k–Leaf Out–Branching on Nice Willow Graphs(Using NP–completeness)

. WLOG(After Simple Modifications)

. Composition(produces an instance of k–Leaf Out Branching)

. Kernelization(Given By Assumption)

.NP–completeness reduction (from k–Leaf Out Branchingto k–Leaf Out Tree)

Figure: e No Polynomial Kernel Argument for k-L O-B.

.

.(D ′, bmax + 1)

.(D ′′, k ′′)

.(D∗, k∗)

.(D ′, bmax + 1)

. Composition(produces an instance of k–Leaf Out Branching)

. Kernelization(Given By Assumption)

.NP–completeness reduction from k–Leaf Out Branchingto k–Leaf Out Tree)

concluding remarks

A polynomial pseudo–kernelization K pro-duces kernels whose size can be bounded by:

h(k) · n1−ε

where k is the parameter of the problem,and h is polynomial.

concluding remarks

Analogous to the composition framework,there are algorithms (called Linear OR)whose existence helps us rule out polynomialpseudo–kernels.

Most compositional algorithms can be ex-tended to fit the definition of Linear OR.

concluding remarks

Analogous to the composition framework,there are algorithms (called Linear OR)whose existence helps us rule out polynomialpseudo–kernels.

Most compositional algorithms can be ex-tended to fit the definition of Linear OR.

concluding remarks

A strong kernel is one where the parameterof the kernelized instance is at most the pa-rameter of the original instance.

Mechanisms for ruling out strong kernels aresimpler and rely on self–reductions and thehypothesis that P 6= NP.

Example: k–Path is not likely to admit astrong kernel.

concluding remarks

A strong kernel is one where the parameterof the kernelized instance is at most the pa-rameter of the original instance.

Mechanisms for ruling out strong kernels aresimpler and rely on self–reductions and thehypothesis that P 6= NP.

Example: k–Path is not likely to admit astrong kernel.

concluding remarks

A strong kernel is one where the parameterof the kernelized instance is at most the pa-rameter of the original instance.

Mechanisms for ruling out strong kernels aresimpler and rely on self–reductions and thehypothesis that P 6= NP.

Example: k–Path is not likely to admit astrong kernel.

Open Problems

ere are no known ways of inferring that a problem is unlikely to have akernel of size kc for some specific c.

Open Problems

Can lower bound theorems be proved under some “well-believed”conjectures of parameterized complexity - for instance - FPT 6= XP, or,

FPT 6= W[t] for some t ∈ N+?

Open Problems

A lower bound framework for ruling out p(k) · f(l)–sized kernels forproblems with two parameters (k, l) would be useful.

Open Problems

“Many polynomial kernels” have been found only for the directedoutbranching problem. It would be interesting to apply the technique toother problems that are not expected to have polynomial–sized kernels.

FIN