Kernel Lower Bounds

Lower Bounds on Kernelization
Venkatesh Raman
Institiue of Mathematical Sciences, Chennai

March 6, 2014

Venkatesh Raman


Some known kernelization results

Linear: MaxSat – 2k clauses, k variables

Venkatesh Raman



Quadratic: k-Vertex Cover – 2k vertices but O(k 2 )
edges

Venkatesh Raman



edges
Cubic: k-Dominating Set in graphs without C4 – O(k 3 )
vertices

Venkatesh Raman



edges
vertices
Exponential: k-Path – 2O(k)

Venkatesh Raman



edges
vertices
No Kernel: k-Dominating Set is W-hard. So is not
expected to have kernels of any size.

Venkatesh Raman



edges
vertices
No Kernel: k-Dominating Set is W-hard. So is not
expected to have kernels of any size.
In this lecture, we will see some techniques to rule out
polynomial kernels.

Venkatesh Raman


OR of a language

Definition
Let L ⊆ {0, 1}∗ be a language. Then define
Or(L) = {(x1 , . . . , xp ) | ∃i such that xi ∈ L}
Definition
Let t : N → N {0} be a function. Then define
Ort (L) = {(x1 , . . . , xt(|x1 |) ) | ∀j |xj | = |x1 |, and ∃i such that xi ∈ L}

Venkatesh Raman


Distillation

Let L, L ⊆ {0, 1}∗ be a pair of languages and let t : N → N {0}
be a function. We say that L has t-bounded distillation algorithm
if there exists
a polynomial time computable function f : {0, 1}∗ → {0, 1}∗
such that
f ((x1 , . . . , xt(|x1 |) )) ∈ L if and only if
(x1 , . . . , xt(|x1 |) ) ∈ Ort (L), and
|f ((x1 , . . . , xt(|x1 |) )| ≤ O(t(|x1 |) log t(|x1 |)).

Venkatesh Raman


Fortnow-Santhanam

Theorem (FS 09)
Suppose for a pair of languages L, L ⊆ {0, 1}∗ , there exists a
polynomially bounded function t : N → N {0} such that L has a
t-bounded distillation algorithm. Then L ∈ NP/poly. In particular,
if L is NP-hard, then coNP ⊆ NP/poly.

Venkatesh Raman


Outline of proof of Fortnow Santhanam theorem

NP-complete problem L with A, a t-bounded distillation
algorithm.

Venkatesh Raman



algorithm.
Use A to design NDTM that, with a “polynomial advice”, can
decide L in P-time.

Venkatesh Raman



algorithm.
Use A to design NDTM that, with a “polynomial advice”, can
decide L in P-time.
L ∈ NP/poly ⇒ coNP ⊆ NP/poly and we get the theorem!

Venkatesh Raman


Filling in the details
For the proof, we deﬁne the notions needed and the requirements.
Let |xi | = n ∀i ∈ [t(n)].

Venkatesh Raman


Let |xi | = n ∀i ∈ [t(n)].
Let α(n) = O(t(n) log(t(n))).

Venkatesh Raman


Let |xi | = n ∀i ∈ [t(n)].
Ln = {x ∈ L : |x| ≤ n}.

Venkatesh Raman


Let |xi | = n ∀i ∈ [t(n)].
Ln = {x ∈ L : |x| ≤ n}.
given any (x1 , x2 , · · · , xt(n) ) ∈ Or(L) (ie, xi ∈ Ln ∀i ∈ [t(n)])
/
A maps it to y ∈ L ≤α(n)

Venkatesh Raman


Let |xi | = n ∀i ∈ [t(n)].
Ln = {x ∈ L : |x| ≤ n}.
/
we want to obtain a Sn ⊆ L α(n) with |Sn | polynomially
bounded in n such that

Venkatesh Raman


Let |xi | = n ∀i ∈ [t(n)].
Ln = {x ∈ L : |x| ≤ n}.
/
If x ∈ Ln - ∃ strings x1 , · · · , xt(n) ∈ Σ n with xi = x for some i
such that A(x1 , · · · , xt(n) ) ∈ Sn

Venkatesh Raman


Let |xi | = n ∀i ∈ [t(n)].
Ln = {x ∈ L : |x| ≤ n}.
/
If x ∈ Ln - ∃ strings x1 , · · · , xt(n) ∈ Σ n with xi = x for some i
such that A(x1 , · · · , xt(n) ) ∈ Sn
If x ∈ Ln - ∀ strings x1 , · · · , xt(n) ∈ Σ n with xi = x for some i,
/
A(x1 , · · · , xt(n) ) ∈ Sn
/

Venkatesh Raman


How will the nondeterministic algorithm work?

Having Sn as advice gives the desired NDTM which when given x
such that |x| = n, checks whether x ∈ L in the following way.
Guesses t(n) strings, x1 , · · · , xt(n) ∈ Σ n

Venkatesh Raman



Checks whether one of them is x

Venkatesh Raman



Checks whether one of them is x
Computes A(x1 , · · · , xt(n) ) and accepts iﬀ output is in Sn .

Venkatesh Raman


How to get Sn
A : (Ln )t → L ≤α(n)

Venkatesh Raman


How to get Sn
A : (Ln )t → L ≤α(n)
y ∈ L ≤α(n) covers a string x ∈ Ln — ∃x1 , · · · , xt ∈ Σ n with
xi = x for some i and A(x1 , · · · , xt(n) ) = y

Venkatesh Raman


How to get Sn
A : (Ln )t → L ≤α(n)
We construct Sn by iteratively picking the string in L ≤α(n)
which covers the most number of instances in Ln till there are
no strings left to cover.

Venkatesh Raman


How to get Sn
A : (Ln )t → L ≤α(n)
Let us consider one step of the process. Let F be the set of
uncovered instances in Ln at the start of step.

Venkatesh Raman


How to get Sn
A : (Ln )t → L ≤α(n)
Let us consider one step of the process. Let F be the set of
uncovered instances in Ln at the start of step.
By PHP there exists a string y ∈ L ≤α(n) such that A maps at
least
|F |t(n)
|L ≤α(n) |
tuples in F t(n) to y .
Venkatesh Raman


How to get Sn (Cont.)
At least

|F |t(n)
|L ≤α(n) |

1/t(n)

=

|F |
|L

≤α(n) |

1/t(n)

strings in F are

covered by y in each step.

Venkatesh Raman


At least

|F |t(n)
|L ≤α(n) |

1/t(n)

=

|F |
|L

≤α(n) |

1/t(n)

strings in F are

We can restate the above statement, saying that at least ϕ(s)
fraction of the remaining set is covered in each iteration,
where
1
1
= (α(n)+1)/t(n)
ϕ(n) =
1/t(n)
2
|L ≤α(n) |

Venkatesh Raman


At least

|F |t(n)
|L ≤α(n) |

1/t(n)

=

|F |
|L

≤α(n) |

1/t(n)

strings in F are

where
1
1
= (α(n)+1)/t(n)
ϕ(n) =
1/t(n)
2
|L ≤α(n) |
There were 2n strings to cover at the starting. So, the number
of strings left to cover after p steps is at most
(1 − ϕ(n))p 2n ≤

2n
e ϕ(n)·p

which is less than one for p = O(n/ϕ(n)).

Venkatesh Raman


At least

|F |t(n)
|L ≤α(n) |

1/t(n)

=

|F |
|L

≤α(n) |

1/t(n)

strings in F are

where
1
1
= (α(n)+1)/t(n)
ϕ(n) =
1/t(n)
2
|L ≤α(n) |
There were 2n strings to cover at the starting. So, the number
of strings left to cover after p steps is at most
(1 − ϕ(n))p 2n ≤

2n
e ϕ(n)·p

which is less than one for p = O(n/ϕ(n)).
So, the process ends after O(n/ϕ(n)) ≤ n · 2(α(n)+1)/t(n) steps,
which is polynomial in n since α(n) = O(t(n) log(t(n))).
Venkatesh Raman


Take away
A few comments about the theorem

coNP ⊆ NP/poly implies PH = Σ3 .
p
The theorem gives us the collapse even if the distillation
algorithm is allowed to be in co-nondeterministic.
Main message is, that if you have t(n) instances of size n, you
can not get an instance equivalent to the Or of them in
polynomial time of size O(t(n) log t(n))

Venkatesh Raman


How to use the theorem to prove kernel lower bounds
We know that NP-complete problems can not have a
distillation algorithm unless coNP ⊆ NP/poly.

Venkatesh Raman


We want to deﬁne some analogue of distillation to produce an
instance (x, k) of a parameterized problem L , starting from
many instances of an NP-complete language L.

Venkatesh Raman


We call such an algorithm a composition algorithm. We will
deﬁne it formally in the next slide.

Venkatesh Raman


The goal is that composition of an NP-complete language L
into L , combined with a kernel of certain size for L , gives us
distillation L.

Venkatesh Raman


The goal is that composition of an NP-complete language L
into L , combined with a kernel of certain size for L , gives us
distillation L.
So, if we can show that a composition algorithm exists from L
to L with desired properties, then L can not have a kernel of
certain size.

Venkatesh Raman


Weak d-Composition
˜
(Weak d-composition). Let L ⊆ Σ ∗ be a set and let
∗ × N be a parameterized problem. We say that L weak
Q⊆Σ
d-composes into Q if there is an algorithm C which, given t strings
x1 , x2 , . . . , xt , takes time polynomial in t |xi | and outputs an
i=1
instance (y , k) ∈ Σ∗ × N such that the following hold:
k ≤ t 1/d (maxt |xi |)O(1)
i=1
The output is a YES instance of Q if and only if at least one
˜
instance xi is a YES-instance of of L.
Theorem
˜
˜
Let L ⊆ Σ ∗ be a set which is NP-hard. If L weak d-composes into
the parameterized problem Q, then Q has no kernel of size
O(k d− ) for all > 0 unless NP ⊆ coNP/poly.

Venkatesh Raman


Proof of the theorem

Theorem
˜
˜
Let L ⊆ Σ ∗ be a set which is NP-hard. If L weak d-composes into
the parameterized problem Q, then Q has no kernel of size
O(k d− ) for all > 0 unless NP ⊆ coNP/poly.
Proof. Let xi = n ∀i ∈ [t(n)] for the input of composition. After
applying the kernelization on the composed instance, the size of
the instance we get is
O(t(n)1/d nc )d− ) = O(t(n)1−(
= O(t(s))

/d) c(d− )

n

)

(for t(s) suﬃciently large)

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

Venkatesh Raman


Some comments about composition
In composition, we asked for the parameter k to be at most
t 1/d (n)O(1) . That ruled out kernels of size k d− .

Venkatesh Raman


What if we can output an instance with k = t o(1) (n)O(1) ?
Then we can rule out kernels of k d− for ALL d!

Venkatesh Raman


We call such an algorithm just “composition”.

Venkatesh Raman


Since theorem of Fortnow-Santhanam allows
co-nondeterminism, so that allows using coNP compositions
for proving lower bounds.

Venkatesh Raman


Sometimes getting composition from arbitrary instances of a
language can be diﬃcult.

Venkatesh Raman


Sometimes getting composition from arbitrary instances of a
language can be diﬃcult.
Some structure on the input instances helps to get a
composition (next slide).

Venkatesh Raman


Polynomial Equivalence Relation

(Polynomial Equivalence Relation). An equivalence relation R
on Σ ∗ is called a polynomial equivalence relation if the following
two conditions hold:
1

2

There is an algorithm that given two strings x, y ∈ Σ ∗ decides
whether x and y belong to the same equivalence class in
(|x| + |y |)O(1) time.
For any ﬁnite set S ⊆ Σ ∗ the equivalence relation R partitions
the elements of S into at most (maxx∈S |x|)O(1) classes.

Venkatesh Raman


What to do with Polynomial Equivalence Relation

The equivalence relation can partition the input on the basis
of diﬀerent parameters. These equivalence classes can be used
to give the input to the composition a nice structure.
The helpful choices are often partitions which have the same
number of vertices, or the asked solution size etc.
Then all we need to do, is to come up with a composition
algorithm for instances belonging to same equivalence class.
Since there are only polynomial number of equivalence classes,
in the end we can just output an instance of Or(L )
Next slide is a nice illustration of this method by Michal
Pilipczuk.

Venkatesh Raman


Proof

Michał Pilipczuk

No-poly-kernels tutorial

11/31

OR-SAT

OR-SAT

Proof

Michał Pilipczuk


11/31

OR-SAT
NP-hrd

NP-hrd

NP-hrd

NP-hrd

NP-hrd

11/31
Michał Pilipczuk

NP-hrd

NP-hrd

NP-hrd

OR-SAT

NP-hrd

˜
L

Proof

NP-hrd

NP-hrd

NP-hrd

NP-hrd

NP-hrd

NP-hrd

NP-hrd

1

2

2

2

k

k

k

OR-SAT

NP-hrd

1

11/31
Michał Pilipczuk

OR-SAT

NP-hrd

1
L

Proof

NP-hrd

NP-hrd

NP-hrd

NP-hrd

NP-hrd

NP-hrd

NP-hrd

L

NP-hrd

1

1

1

2

2

2

k

k

OR-SAT

NP-hrd

OR-SAT

Proof

k

cmp

cmp

cmp

poly(k)

poly(k)

L

poly(k)

Michał Pilipczuk


11/31

NP-hrd

NP-hrd

NP-hrd

NP-hrd

NP-hrd

1

2

2

2

k

k

k

OR-SAT

NP-hrd

1
L

NP-hrd

kern

kern

kern

11/31
Michał Pilipczuk

poly(k)
poly(k)
poly(k)

cmp

cmp

cmp

L

NP-hrd

1

OR-SAT

NP-hrd

L

Proof

NP-hrd

NP-hrd

NP-hrd

NP-hrd

1

2

2

2

k

k

k

OR-SAT

NP-hrd

1
L

NP-hrd

kern

kern

kern

11/31
Michał Pilipczuk

poly(k)
poly(k)
poly(k)

cmp

cmp

L

NP-hrd

cmp

L

NP-hrd

1

OR-SAT

NP-hrd

OR-L

Proof

Take away

We use compositions to rule out polynomial kernels.

Venkatesh Raman


Take away

A composition from NP-hard problem L to parameterized
problem L gives kernelization hardness for for L .

Venkatesh Raman


Take away

k = t o(1) nc ⇒ No polynomial kernel.

Venkatesh Raman


Take away

k = t 1/d nc ⇒ No kernel of size k d− .

Venkatesh Raman


Take away

We can make use of equivalence classes to give structure to
input of the composition.

Venkatesh Raman


Take away

We can make use of equivalence classes to give structure to
input of the composition.
Examples on the board!

Venkatesh Raman


Thank You!

Venkatesh Raman


Kernel Lower Bounds

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