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Lower Bounds on Kernelization
Venkatesh Raman
Institiue of Mathematical Sciences, Chennai
March 6, 2014
Venkatesh Raman Lower Bounds on Kernelization
Some known kernelization results
Linear: MaxSat – 2k clauses, k variables
Quadratic: k-Vertex Cover – 2k vertices but O(k2)edges
Cubic: k-Dominating Set in graphs without C4 – O(k3)vertices
Exponential: k-Path – 2O(k)
No Kernel: k-Dominating Set is W-hard. So is notexpected to have kernels of any size.
In this lecture, we will see some techniques to rule outpolynomial kernels.
Venkatesh Raman Lower Bounds on Kernelization
Some known kernelization results
Linear: MaxSat – 2k clauses, k variables
Quadratic: k-Vertex Cover – 2k vertices but O(k2)edges
Cubic: k-Dominating Set in graphs without C4 – O(k3)vertices
Exponential: k-Path – 2O(k)
No Kernel: k-Dominating Set is W-hard. So is notexpected to have kernels of any size.
In this lecture, we will see some techniques to rule outpolynomial kernels.
Venkatesh Raman Lower Bounds on Kernelization
Some known kernelization results
Linear: MaxSat – 2k clauses, k variables
Quadratic: k-Vertex Cover – 2k vertices but O(k2)edges
Cubic: k-Dominating Set in graphs without C4 – O(k3)vertices
Exponential: k-Path – 2O(k)
No Kernel: k-Dominating Set is W-hard. So is notexpected to have kernels of any size.
In this lecture, we will see some techniques to rule outpolynomial kernels.
Venkatesh Raman Lower Bounds on Kernelization
Some known kernelization results
Linear: MaxSat – 2k clauses, k variables
Quadratic: k-Vertex Cover – 2k vertices but O(k2)edges
Cubic: k-Dominating Set in graphs without C4 – O(k3)vertices
Exponential: k-Path – 2O(k)
No Kernel: k-Dominating Set is W-hard. So is notexpected to have kernels of any size.
In this lecture, we will see some techniques to rule outpolynomial kernels.
Venkatesh Raman Lower Bounds on Kernelization
Some known kernelization results
Linear: MaxSat – 2k clauses, k variables
Quadratic: k-Vertex Cover – 2k vertices but O(k2)edges
Cubic: k-Dominating Set in graphs without C4 – O(k3)vertices
Exponential: k-Path – 2O(k)
No Kernel: k-Dominating Set is W-hard. So is notexpected to have kernels of any size.
In this lecture, we will see some techniques to rule outpolynomial kernels.
Venkatesh Raman Lower Bounds on Kernelization
Some known kernelization results
Linear: MaxSat – 2k clauses, k variables
Quadratic: k-Vertex Cover – 2k vertices but O(k2)edges
Cubic: k-Dominating Set in graphs without C4 – O(k3)vertices
Exponential: k-Path – 2O(k)
No Kernel: k-Dominating Set is W-hard. So is notexpected to have kernels of any size.
In this lecture, we will see some techniques to rule outpolynomial kernels.
Venkatesh Raman Lower Bounds on Kernelization
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 Lower Bounds on Kernelization
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 algorithmif 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 Lower Bounds on Kernelization
Fortnow-Santhanam
Theorem (FS 09)
Suppose for a pair of languages L, L′ ⊆ {0, 1}∗, there exists apolynomially bounded function t : N→ N \ {0} such that L has at-bounded distillation algorithm. Then L ∈ NP/poly. In particular,if L is NP-hard, then coNP ⊆ NP/poly.
Venkatesh Raman Lower Bounds on Kernelization
Outline of proof of Fortnow Santhanam theorem
NP-complete problem L with A, a t-bounded distillationalgorithm.
Use A to design NDTM that, with a “polynomial advice”, candecide L in P-time.
L ∈ NP/poly ⇒ coNP ⊆ NP/poly and we get the theorem!
Venkatesh Raman Lower Bounds on Kernelization
Outline of proof of Fortnow Santhanam theorem
NP-complete problem L with A, a t-bounded distillationalgorithm.
Use A to design NDTM that, with a “polynomial advice”, candecide L in P-time.
L ∈ NP/poly ⇒ coNP ⊆ NP/poly and we get the theorem!
Venkatesh Raman Lower Bounds on Kernelization
Outline of proof of Fortnow Santhanam theorem
NP-complete problem L with A, a t-bounded distillationalgorithm.
Use A to design NDTM that, with a “polynomial advice”, candecide L in P-time.
L ∈ NP/poly ⇒ coNP ⊆ NP/poly and we get the theorem!
Venkatesh Raman Lower Bounds on Kernelization
Filling in the details
For the proof, we define the notions needed and the requirements.
Let |xi | = n ∀i ∈ [t(n)].
Let α(n) = O(t(n) log(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)
we want to obtain a Sn ⊆ L′α(n) with |Sn| polynomiallybounded in n such that
If x ∈ Ln - ∃ strings x1, · · · , xt(n) ∈ Σn with xi = x for some isuch that A(x1, · · · , xt(n)) ∈ SnIf x /∈ Ln - ∀ strings x1, · · · , xt(n) ∈ Σn with xi = x for some i ,A(x1, · · · , xt(n)) /∈ Sn
Venkatesh Raman Lower Bounds on Kernelization
Filling in the details
For the proof, we define the notions needed and the requirements.
Let |xi | = n ∀i ∈ [t(n)].
Let α(n) = O(t(n) log(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)
we want to obtain a Sn ⊆ L′α(n) with |Sn| polynomiallybounded in n such that
If x ∈ Ln - ∃ strings x1, · · · , xt(n) ∈ Σn with xi = x for some isuch that A(x1, · · · , xt(n)) ∈ SnIf x /∈ Ln - ∀ strings x1, · · · , xt(n) ∈ Σn with xi = x for some i ,A(x1, · · · , xt(n)) /∈ Sn
Venkatesh Raman Lower Bounds on Kernelization
Filling in the details
For the proof, we define the notions needed and the requirements.
Let |xi | = n ∀i ∈ [t(n)].
Let α(n) = O(t(n) log(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)
we want to obtain a Sn ⊆ L′α(n) with |Sn| polynomiallybounded in n such that
If x ∈ Ln - ∃ strings x1, · · · , xt(n) ∈ Σn with xi = x for some isuch that A(x1, · · · , xt(n)) ∈ SnIf x /∈ Ln - ∀ strings x1, · · · , xt(n) ∈ Σn with xi = x for some i ,A(x1, · · · , xt(n)) /∈ Sn
Venkatesh Raman Lower Bounds on Kernelization
Filling in the details
For the proof, we define the notions needed and the requirements.
Let |xi | = n ∀i ∈ [t(n)].
Let α(n) = O(t(n) log(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)
we want to obtain a Sn ⊆ L′α(n) with |Sn| polynomiallybounded in n such that
If x ∈ Ln - ∃ strings x1, · · · , xt(n) ∈ Σn with xi = x for some isuch that A(x1, · · · , xt(n)) ∈ SnIf x /∈ Ln - ∀ strings x1, · · · , xt(n) ∈ Σn with xi = x for some i ,A(x1, · · · , xt(n)) /∈ Sn
Venkatesh Raman Lower Bounds on Kernelization
Filling in the details
For the proof, we define the notions needed and the requirements.
Let |xi | = n ∀i ∈ [t(n)].
Let α(n) = O(t(n) log(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)
we want to obtain a Sn ⊆ L′α(n) with |Sn| polynomiallybounded in n such that
If x ∈ Ln - ∃ strings x1, · · · , xt(n) ∈ Σn with xi = x for some isuch that A(x1, · · · , xt(n)) ∈ SnIf x /∈ Ln - ∀ strings x1, · · · , xt(n) ∈ Σn with xi = x for some i ,A(x1, · · · , xt(n)) /∈ Sn
Venkatesh Raman Lower Bounds on Kernelization
Filling in the details
For the proof, we define the notions needed and the requirements.
Let |xi | = n ∀i ∈ [t(n)].
Let α(n) = O(t(n) log(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)
we want to obtain a Sn ⊆ L′α(n) with |Sn| polynomiallybounded in n such that
If x ∈ Ln - ∃ strings x1, · · · , xt(n) ∈ Σn with xi = x for some isuch that A(x1, · · · , xt(n)) ∈ Sn
If x /∈ Ln - ∀ strings x1, · · · , xt(n) ∈ Σn with xi = x for some i ,A(x1, · · · , xt(n)) /∈ Sn
Venkatesh Raman Lower Bounds on Kernelization
Filling in the details
For the proof, we define the notions needed and the requirements.
Let |xi | = n ∀i ∈ [t(n)].
Let α(n) = O(t(n) log(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)
we want to obtain a Sn ⊆ L′α(n) with |Sn| polynomiallybounded in n such that
If x ∈ Ln - ∃ strings x1, · · · , xt(n) ∈ Σn with xi = x for some isuch that A(x1, · · · , xt(n)) ∈ SnIf x /∈ Ln - ∀ strings x1, · · · , xt(n) ∈ Σn with xi = x for some i ,A(x1, · · · , xt(n)) /∈ Sn
Venkatesh Raman Lower Bounds on Kernelization
How will the nondeterministic algorithm work?
Having Sn as advice gives the desired NDTM which when given xsuch that |x | = n, checks whether x ∈ L in the following way.
Guesses t(n) strings, x1, · · · , xt(n) ∈ Σn
Checks whether one of them is x
Computes A(x1, · · · , xt(n)) and accepts iff output is in Sn.
Venkatesh Raman Lower Bounds on Kernelization
How will the nondeterministic algorithm work?
Having Sn as advice gives the desired NDTM which when given xsuch that |x | = n, checks whether x ∈ L in the following way.
Guesses t(n) strings, x1, · · · , xt(n) ∈ Σn
Checks whether one of them is x
Computes A(x1, · · · , xt(n)) and accepts iff output is in Sn.
Venkatesh Raman Lower Bounds on Kernelization
How will the nondeterministic algorithm work?
Having Sn as advice gives the desired NDTM which when given xsuch that |x | = n, checks whether x ∈ L in the following way.
Guesses t(n) strings, x1, · · · , xt(n) ∈ Σn
Checks whether one of them is x
Computes A(x1, · · · , xt(n)) and accepts iff output is in Sn.
Venkatesh Raman Lower Bounds on Kernelization
How to get Sn
A : (Ln)t → L′≤α(n)
y ∈ L′≤α(n) covers a string x ∈ Ln — ∃x1, · · · , xt ∈ Σn withxi = x for some i and A(x1, · · · , xt(n)) = y
We construct Sn by iteratively picking the string in L′≤α(n)which covers the most number of instances in Ln till there areno strings left to cover.
Let us consider one step of the process. Let F be the set ofuncovered instances in Ln at the start of step.
By PHP there exists a string y ∈ L′≤α(n) such that A maps atleast
|F |t(n)
|L′≤α(n)|
tuples in F t(n) to y .
Venkatesh Raman Lower Bounds on Kernelization
How to get Sn
A : (Ln)t → L′≤α(n)
y ∈ L′≤α(n) covers a string x ∈ Ln — ∃x1, · · · , xt ∈ Σn withxi = x for some i and A(x1, · · · , xt(n)) = y
We construct Sn by iteratively picking the string in L′≤α(n)which covers the most number of instances in Ln till there areno strings left to cover.
Let us consider one step of the process. Let F be the set ofuncovered instances in Ln at the start of step.
By PHP there exists a string y ∈ L′≤α(n) such that A maps atleast
|F |t(n)
|L′≤α(n)|
tuples in F t(n) to y .
Venkatesh Raman Lower Bounds on Kernelization
How to get Sn
A : (Ln)t → L′≤α(n)
y ∈ L′≤α(n) covers a string x ∈ Ln — ∃x1, · · · , xt ∈ Σn withxi = x for some i and A(x1, · · · , xt(n)) = y
We construct Sn by iteratively picking the string in L′≤α(n)which covers the most number of instances in Ln till there areno strings left to cover.
Let us consider one step of the process. Let F be the set ofuncovered instances in Ln at the start of step.
By PHP there exists a string y ∈ L′≤α(n) such that A maps atleast
|F |t(n)
|L′≤α(n)|
tuples in F t(n) to y .
Venkatesh Raman Lower Bounds on Kernelization
How to get Sn
A : (Ln)t → L′≤α(n)
y ∈ L′≤α(n) covers a string x ∈ Ln — ∃x1, · · · , xt ∈ Σn withxi = x for some i and A(x1, · · · , xt(n)) = y
We construct Sn by iteratively picking the string in L′≤α(n)which covers the most number of instances in Ln till there areno strings left to cover.
Let us consider one step of the process. Let F be the set ofuncovered instances in Ln at the start of step.
By PHP there exists a string y ∈ L′≤α(n) such that A maps atleast
|F |t(n)
|L′≤α(n)|
tuples in F t(n) to y .
Venkatesh Raman Lower Bounds on Kernelization
How to get Sn
A : (Ln)t → L′≤α(n)
y ∈ L′≤α(n) covers a string x ∈ Ln — ∃x1, · · · , xt ∈ Σn withxi = x for some i and A(x1, · · · , xt(n)) = y
We construct Sn by iteratively picking the string in L′≤α(n)which covers the most number of instances in Ln till there areno strings left to cover.
Let us consider one step of the process. Let F be the set ofuncovered instances in Ln at the start of step.
By PHP there exists a string y ∈ L′≤α(n) such that A maps atleast
|F |t(n)
|L′≤α(n)|
tuples in F t(n) to y .
Venkatesh Raman Lower Bounds on Kernelization
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.
We can restate the above statement, saying that at least ϕ(s)fraction of the remaining set is covered in each iteration,where
ϕ(n) =1
|L′≤α(n)|1/t(n)=
1
2(α(n)+1)/t(n)
There were 2n strings to cover at the starting. So, the numberof strings left to cover after p steps is at most
(1− ϕ(n))p2n ≤ 2n
eϕ(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 Lower Bounds on Kernelization
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.
We can restate the above statement, saying that at least ϕ(s)fraction of the remaining set is covered in each iteration,where
ϕ(n) =1
|L′≤α(n)|1/t(n)=
1
2(α(n)+1)/t(n)
There were 2n strings to cover at the starting. So, the numberof strings left to cover after p steps is at most
(1− ϕ(n))p2n ≤ 2n
eϕ(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 Lower Bounds on Kernelization
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.
We can restate the above statement, saying that at least ϕ(s)fraction of the remaining set is covered in each iteration,where
ϕ(n) =1
|L′≤α(n)|1/t(n)=
1
2(α(n)+1)/t(n)
There were 2n strings to cover at the starting. So, the numberof strings left to cover after p steps is at most
(1− ϕ(n))p2n ≤ 2n
eϕ(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 Lower Bounds on Kernelization
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.
We can restate the above statement, saying that at least ϕ(s)fraction of the remaining set is covered in each iteration,where
ϕ(n) =1
|L′≤α(n)|1/t(n)=
1
2(α(n)+1)/t(n)
There were 2n strings to cover at the starting. So, the numberof strings left to cover after p steps is at most
(1− ϕ(n))p2n ≤ 2n
eϕ(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 Lower Bounds on Kernelization
Take awayA few comments about the theorem
coNP ⊆ NP/poly implies PH = Σ3p.
The theorem gives us the collapse even if the distillationalgorithm is allowed to be in co-nondeterministic.
Main message is, that if you have t(n) instances of size n, youcan not get an instance equivalent to the Or of them inpolynomial time of size O(t(n) log t(n))
Venkatesh Raman Lower Bounds on Kernelization
How to use the theorem to prove kernel lower bounds
We know that NP-complete problems can not have adistillation algorithm unless coNP ⊆ NP/poly.
We want to define some analogue of distillation to produce aninstance (x , k) of a parameterized problem L′, starting frommany instances of an NP-complete language L.
We call such an algorithm a composition algorithm. We willdefine it formally in the next slide.
The goal is that composition of an NP-complete language Linto L′, combined with a kernel of certain size for L′, gives usdistillation L.
So, if we can show that a composition algorithm exists from Lto L′ with desired properties, then L′ can not have a kernel ofcertain size.
Venkatesh Raman Lower Bounds on Kernelization
How to use the theorem to prove kernel lower bounds
We know that NP-complete problems can not have adistillation algorithm unless coNP ⊆ NP/poly.
We want to define some analogue of distillation to produce aninstance (x , k) of a parameterized problem L′, starting frommany instances of an NP-complete language L.
We call such an algorithm a composition algorithm. We willdefine it formally in the next slide.
The goal is that composition of an NP-complete language Linto L′, combined with a kernel of certain size for L′, gives usdistillation L.
So, if we can show that a composition algorithm exists from Lto L′ with desired properties, then L′ can not have a kernel ofcertain size.
Venkatesh Raman Lower Bounds on Kernelization
How to use the theorem to prove kernel lower bounds
We know that NP-complete problems can not have adistillation algorithm unless coNP ⊆ NP/poly.
We want to define some analogue of distillation to produce aninstance (x , k) of a parameterized problem L′, starting frommany instances of an NP-complete language L.
We call such an algorithm a composition algorithm. We willdefine it formally in the next slide.
The goal is that composition of an NP-complete language Linto L′, combined with a kernel of certain size for L′, gives usdistillation L.
So, if we can show that a composition algorithm exists from Lto L′ with desired properties, then L′ can not have a kernel ofcertain size.
Venkatesh Raman Lower Bounds on Kernelization
How to use the theorem to prove kernel lower bounds
We know that NP-complete problems can not have adistillation algorithm unless coNP ⊆ NP/poly.
We want to define some analogue of distillation to produce aninstance (x , k) of a parameterized problem L′, starting frommany instances of an NP-complete language L.
We call such an algorithm a composition algorithm. We willdefine it formally in the next slide.
The goal is that composition of an NP-complete language Linto L′, combined with a kernel of certain size for L′, gives usdistillation L.
So, if we can show that a composition algorithm exists from Lto L′ with desired properties, then L′ can not have a kernel ofcertain size.
Venkatesh Raman Lower Bounds on Kernelization
How to use the theorem to prove kernel lower bounds
We know that NP-complete problems can not have adistillation algorithm unless coNP ⊆ NP/poly.
We want to define some analogue of distillation to produce aninstance (x , k) of a parameterized problem L′, starting frommany instances of an NP-complete language L.
We call such an algorithm a composition algorithm. We willdefine it formally in the next slide.
The goal is that composition of an NP-complete language Linto L′, combined with a kernel of certain size for L′, gives usdistillation L.
So, if we can show that a composition algorithm exists from Lto L′ with desired properties, then L′ can not have a kernel ofcertain size.
Venkatesh Raman Lower Bounds on Kernelization
Weak d-Composition
(Weak d-composition). Let L̃ ⊆ Σ∗ be a set and letQ ⊆ Σ∗ × N be a parameterized problem. We say that L weakd-composes into Q if there is an algorithm C which, given t stringsx1, x2, . . . , xt , takes time polynomial in
∑ti=1 |xi | and outputs an
instance (y , k) ∈ Σ∗ × N such that the following hold:
k ≤ t1/d(maxti=1 |xi |)O(1)
The output is a YES instance of Q if and only if at least oneinstance xi is a YES-instance of of L̃.
Theorem
Let L̃ ⊆ Σ∗ be a set which is NP-hard. If L̃ weak d-composes intothe parameterized problem Q, then Q has no kernel of sizeO(kd−ε) for all ε > 0 unless NP ⊆ coNP/poly.
Venkatesh Raman Lower Bounds on Kernelization
Proof of the theorem
Theorem
Let L̃ ⊆ Σ∗ be a set which is NP-hard. If L̃ weak d-composes intothe parameterized problem Q, then Q has no kernel of sizeO(kd−ε) for all ε > 0 unless NP ⊆ coNP/poly.
Proof. Let xi = n ∀i ∈ [t(n)] for the input of composition. Afterapplying the kernelization on the composed instance, the size ofthe instance we get is
O(t(n)1/dnc)d−ε) = O(t(n)1−(ε/d)nc(d−ε))
= O(t(s)) (for t(s) sufficiently large)
= O(t(s) log t(s))
Venkatesh Raman Lower Bounds on Kernelization
Some comments about composition
In composition, we asked for the parameter k to be at mostt1/d(n)O(1). That ruled out kernels of size kd−ε.
What if we can output an instance with k = to(1)(n)O(1)?Then we can rule out kernels of kd−ε for ALL d!
We call such an algorithm just “composition”.
Since theorem of Fortnow-Santhanam allowsco-nondeterminism, so that allows using coNP compositionsfor proving lower bounds.
Sometimes getting composition from arbitrary instances of alanguage can be difficult.
Some structure on the input instances helps to get acomposition (next slide).
Venkatesh Raman Lower Bounds on Kernelization
Some comments about composition
In composition, we asked for the parameter k to be at mostt1/d(n)O(1). That ruled out kernels of size kd−ε.
What if we can output an instance with k = to(1)(n)O(1)?Then we can rule out kernels of kd−ε for ALL d!
We call such an algorithm just “composition”.
Since theorem of Fortnow-Santhanam allowsco-nondeterminism, so that allows using coNP compositionsfor proving lower bounds.
Sometimes getting composition from arbitrary instances of alanguage can be difficult.
Some structure on the input instances helps to get acomposition (next slide).
Venkatesh Raman Lower Bounds on Kernelization
Some comments about composition
In composition, we asked for the parameter k to be at mostt1/d(n)O(1). That ruled out kernels of size kd−ε.
What if we can output an instance with k = to(1)(n)O(1)?Then we can rule out kernels of kd−ε for ALL d!
We call such an algorithm just “composition”.
Since theorem of Fortnow-Santhanam allowsco-nondeterminism, so that allows using coNP compositionsfor proving lower bounds.
Sometimes getting composition from arbitrary instances of alanguage can be difficult.
Some structure on the input instances helps to get acomposition (next slide).
Venkatesh Raman Lower Bounds on Kernelization
Some comments about composition
In composition, we asked for the parameter k to be at mostt1/d(n)O(1). That ruled out kernels of size kd−ε.
What if we can output an instance with k = to(1)(n)O(1)?Then we can rule out kernels of kd−ε for ALL d!
We call such an algorithm just “composition”.
Since theorem of Fortnow-Santhanam allowsco-nondeterminism, so that allows using coNP compositionsfor proving lower bounds.
Sometimes getting composition from arbitrary instances of alanguage can be difficult.
Some structure on the input instances helps to get acomposition (next slide).
Venkatesh Raman Lower Bounds on Kernelization
Some comments about composition
In composition, we asked for the parameter k to be at mostt1/d(n)O(1). That ruled out kernels of size kd−ε.
What if we can output an instance with k = to(1)(n)O(1)?Then we can rule out kernels of kd−ε for ALL d!
We call such an algorithm just “composition”.
Since theorem of Fortnow-Santhanam allowsco-nondeterminism, so that allows using coNP compositionsfor proving lower bounds.
Sometimes getting composition from arbitrary instances of alanguage can be difficult.
Some structure on the input instances helps to get acomposition (next slide).
Venkatesh Raman Lower Bounds on Kernelization
Some comments about composition
In composition, we asked for the parameter k to be at mostt1/d(n)O(1). That ruled out kernels of size kd−ε.
What if we can output an instance with k = to(1)(n)O(1)?Then we can rule out kernels of kd−ε for ALL d!
We call such an algorithm just “composition”.
Since theorem of Fortnow-Santhanam allowsco-nondeterminism, so that allows using coNP compositionsfor proving lower bounds.
Sometimes getting composition from arbitrary instances of alanguage can be difficult.
Some structure on the input instances helps to get acomposition (next slide).
Venkatesh Raman Lower Bounds on Kernelization
Polynomial Equivalence Relation
(Polynomial Equivalence Relation). An equivalence relation Ron Σ∗ is called a polynomial equivalence relation if the followingtwo conditions hold:
1 There is an algorithm that given two strings x , y ∈ Σ∗ decideswhether x and y belong to the same equivalence class in(|x |+ |y |)O(1) time.
2 For any finite set S ⊆ Σ∗ the equivalence relation R partitionsthe elements of S into at most (maxx∈S |x |)O(1) classes.
Venkatesh Raman Lower Bounds on Kernelization
What to do with Polynomial Equivalence Relation
The equivalence relation can partition the input on the basisof different parameters. These equivalence classes can be usedto give the input to the composition a nice structure.
The helpful choices are often partitions which have the samenumber of vertices, or the asked solution size etc.
Then all we need to do, is to come up with a compositionalgorithm 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 Micha lPilipczuk.
Venkatesh Raman Lower Bounds on Kernelization
Proof
OR-SAT
OR-SAT
L̃
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
L
NP
-hrd
1
NP
-hrd
1
NP
-hrd
1
NP
-hrd
2
NP
-hrd
2
NP
-hrd
2NP
-hrdk′
NP
-hrd
k′
NP
-hrd
k′
L
cmp
poly(k)
cmp
poly(k)
cmp
poly(k)
L
kern
kern
kern
OR-L
OR-L
Michał Pilipczuk No-poly-kernels tutorial 11/31
ProofOR-SAT
OR-SAT
L̃
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
L
NP
-hrd
1
NP
-hrd
1
NP
-hrd
1
NP
-hrd
2
NP
-hrd
2
NP
-hrd
2NP
-hrdk′
NP
-hrd
k′
NP
-hrd
k′
L
cmp
poly(k)
cmp
poly(k)
cmp
poly(k)
L
kern
kern
kern
OR-L
OR-L
Michał Pilipczuk No-poly-kernels tutorial 11/31
ProofOR-SAT
OR-SAT
L̃
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
L
NP
-hrd
1
NP
-hrd
1
NP
-hrd
1
NP
-hrd
2
NP
-hrd
2
NP
-hrd
2NP
-hrdk′
NP
-hrd
k′
NP
-hrd
k′
L
cmp
poly(k)
cmp
poly(k)
cmp
poly(k)
L
kern
kern
kern
OR-L
OR-L
Michał Pilipczuk No-poly-kernels tutorial 11/31
ProofOR-SAT
OR-SAT
L̃
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
L
NP
-hrd
1
NP
-hrd
1
NP
-hrd
1
NP
-hrd
2
NP
-hrd
2
NP
-hrd
2NP
-hrdk′
NP
-hrd
k′
NP
-hrd
k′
L
cmp
poly(k)
cmp
poly(k)
cmp
poly(k)
L
kern
kern
kern
OR-L
OR-L
Michał Pilipczuk No-poly-kernels tutorial 11/31
ProofOR-SAT
OR-SAT
L̃
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
L
NP
-hrd
1
NP
-hrd
1
NP
-hrd
1
NP
-hrd
2
NP
-hrd
2
NP
-hrd
2NP
-hrdk′
NP
-hrd
k′
NP
-hrd
k′
L
cmp
poly(k)
cmp
poly(k)
cmp
poly(k)
L
kern
kern
kern
OR-L
OR-L
Michał Pilipczuk No-poly-kernels tutorial 11/31
ProofOR-SAT
OR-SAT
L̃
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
L
NP
-hrd
1
NP
-hrd
1
NP
-hrd
1
NP
-hrd
2
NP
-hrd
2
NP
-hrd
2NP
-hrdk′
NP
-hrd
k′
NP
-hrd
k′
L
cmp
poly(k)
cmp
poly(k)
cmp
poly(k)
L
kern
kern
kern
OR-L
OR-L
Michał Pilipczuk No-poly-kernels tutorial 11/31
ProofOR-SAT
OR-SAT
L̃
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
L
NP
-hrd
1
NP
-hrd
1
NP
-hrd
1
NP
-hrd
2
NP
-hrd
2
NP
-hrd
2NP
-hrdk′
NP
-hrd
k′
NP
-hrd
k′
L
cmp
poly(k)
cmp
poly(k)
cmp
poly(k)
L
kern
kern
kern
OR-L
OR-L
Michał Pilipczuk No-poly-kernels tutorial 11/31
ProofOR-SAT
OR-SAT
L̃
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
NP
-hrd
L
NP
-hrd
1
NP
-hrd
1
NP
-hrd
1
NP
-hrd
2
NP
-hrd
2
NP
-hrd
2NP
-hrdk′
NP
-hrd
k′
NP
-hrd
k′
L
cmp
poly(k)
cmp
poly(k)
cmp
poly(k)
L
kern
kern
kern
OR-L
OR-L
Michał Pilipczuk No-poly-kernels tutorial 11/31
Take away
We use compositions to rule out polynomial kernels.
A composition from NP-hard problem L to parameterizedproblem L′ gives kernelization hardness for for L′.
k = to(1)nc ⇒ No polynomial kernel.
k = t1/dnc ⇒ No kernel of size kd−ε.
We can make use of equivalence classes to give structure toinput of the composition.
Examples on the board!
Venkatesh Raman Lower Bounds on Kernelization
Take away
We use compositions to rule out polynomial kernels.
A composition from NP-hard problem L to parameterizedproblem L′ gives kernelization hardness for for L′.
k = to(1)nc ⇒ No polynomial kernel.
k = t1/dnc ⇒ No kernel of size kd−ε.
We can make use of equivalence classes to give structure toinput of the composition.
Examples on the board!
Venkatesh Raman Lower Bounds on Kernelization
Take away
We use compositions to rule out polynomial kernels.
A composition from NP-hard problem L to parameterizedproblem L′ gives kernelization hardness for for L′.
k = to(1)nc ⇒ No polynomial kernel.
k = t1/dnc ⇒ No kernel of size kd−ε.
We can make use of equivalence classes to give structure toinput of the composition.
Examples on the board!
Venkatesh Raman Lower Bounds on Kernelization
Take away
We use compositions to rule out polynomial kernels.
A composition from NP-hard problem L to parameterizedproblem L′ gives kernelization hardness for for L′.
k = to(1)nc ⇒ No polynomial kernel.
k = t1/dnc ⇒ No kernel of size kd−ε.
We can make use of equivalence classes to give structure toinput of the composition.
Examples on the board!
Venkatesh Raman Lower Bounds on Kernelization
Take away
We use compositions to rule out polynomial kernels.
A composition from NP-hard problem L to parameterizedproblem L′ gives kernelization hardness for for L′.
k = to(1)nc ⇒ No polynomial kernel.
k = t1/dnc ⇒ No kernel of size kd−ε.
We can make use of equivalence classes to give structure toinput of the composition.
Examples on the board!
Venkatesh Raman Lower Bounds on Kernelization
Take away
We use compositions to rule out polynomial kernels.
A composition from NP-hard problem L to parameterizedproblem L′ gives kernelization hardness for for L′.
k = to(1)nc ⇒ No polynomial kernel.
k = t1/dnc ⇒ No kernel of size kd−ε.
We can make use of equivalence classes to give structure toinput of the composition.
Examples on the board!
Venkatesh Raman Lower Bounds on Kernelization
Thank You!
Venkatesh Raman Lower Bounds on Kernelization
