# Introduction to Kernel Lower Bounds Daniel Lokshtanov.

## Presentation on theme: "Introduction to Kernel Lower Bounds Daniel Lokshtanov."— Presentation transcript:

Introduction to Kernel Lower Bounds Daniel Lokshtanov

What? Kernelization is a mathematical framework to analyze the quality of polynomial time pre- processing Until recently: Many upper bounds known. No ”non-trivial” lower bounds. This talk: Survey of recent lower bounds.

Part I Introduction to Kernelization

Parameterization Hard to analyze pre-processing for NP-hard problems within classical complexity. Reason: poly-time compression = poly-time solution. We consider parameterized problems. Each instance I comes with a parameter k ≤ |I| that is supposed to reflect how hard the instance is. Small k = easier instance.

Parameterization: Example Point Line Cover IN: n points in the plane, integer k. PARAMETER: k QUESTION: Can the points be covered by k straight lines? Notice – easier to solve when k is small.

Kernelization A f(k)-kernel for a problem P is an algorithm that: Takes as input an instance (I,k) Runs in time poly(|I|) Outputs an equivalent instance (I’,k’) with – |I’| ≤ f(k) – k’ ≤ f(k)

Point Line Cover IN: n points in the plane, integer k. PARAMETER: k QUESTION: Can the points be covered by k straight lines?

Point Line Cover TASK: Shoot the little devils, with only 3 shots. If some line covers 4 devils, must use it. Otherwise need 4 shots.

k 2 - kernel for Point Line Cover R1: If some line covers more than k points  delete all points on the line and decrease k by 1. R2: If no line covers at least n/k points, answer ”NO” If neither R1 nor R2 can be applied  n ≤ k 2

Edge Clique Cover IN: Graph G, integer k. PARAMETER: k QUESTION: Can the edges of G be covered by k cliques?

4 k - Kernel for Edge Clique Cover R1: If u and v are adjacent and have same neighbours, delete v. R2: If R1 can’t be applied and n > 2 k, answer NO. If R1, R2 can’t be applied, then n < 2 k and m < 4 k.

Recap A k 2 kernel for Point Line Cover  polynomial kernel A 4 k kernel for Edge Clique Cover  exponential kernel Which all parameterized problems have f(k)-kernels for some function f? Which parameterized problems have poly(k)- kernels?

Which problems have f(k) - kernels? Theorem[Folklore]: A decidable parameterized problem P has an f(k)-kernel for some f  P is fixed parameter tractable (FPT), i.e. solvable in time g(k)n O(1) for some g.

Kernelization Complexity Q1: Does P have an f(k) kernel? P is FPT  YES P is W-hard  NO, unless FPT=W[1] Q2: Does P have a poly(k) kernel. poly(k) kernel  YES How to say NO?

Part II Framework for ruling out polynomial kernels

Longest Path IN: Graph G, integer k PARAMETER: k QUESTION: Does G have a path of length k? Known: 2 k n c time algorithm [Williams 09] Does Longest Path have a polynomial kernel?

Poly kernel for Longest Path? Suppose Longest Path has a k c kernel. Set t = kc + 1 and consider t instances with the same parameter k: (G 1,k), (G 2,k)... (G t,k) The instance (G 1 U G 2... U G t, k) is a yes instance iff some (G i, k) is. Kernelize this instance – the kernel has k c < t bits. Less than one bit per original instance, was at least one of the instances ”solved”?

Poly Kernel for Longest Path? G 1,k G 2,k G t,k... Disjoint union G’,k’... G,k Polynomial kernel

OR-Distillation Algorithms Detour back to classical problems. An OR-distillation algorithm for a problem L Takes as input instances I 1... I t. Runs in polynomial time Outputs an instance O of L’ such that – |O| ≤ max poly(|I i |) – O is ”yes”  some I i is “yes”.

OR-Distillation Algorithms Intuition: A distillation algorithm looks at several problem instances and pics the one ”most likely” to be a yes instance. Should not exist for NP-hard problems. Theorem [FS08]: Unless coNP ⊆ NP/poly, no NP- hard problem has an OR-distillation algorithm.

OR-Composition algorithms: Intuition OR-Composition = ”formalization of disjoint union” OR-Composition + Kernel = OR-Distillation

OR-Composition Algorithms Back to parameterized problems. An OR-composition algorithm for a problem P Takes as input instances I 1... I t with parameter k Runs in polynomial time Outputs an instance (O,k’) of P such that – k’ ≤ poly(k) – (O,k’) is ”yes”  some (I i,k) is “yes”.

OR-Composition for Longest Path G 1,k G 2,k G t,k... Disjoint union... G,k

Ruling Out Polynomial Kernels Theorem [BDFH08]: If a parameterization P of an NP-hard * problem L has a composition algorithm, then P has no polynomial kernel unless coNP ⊆ NP/poly. Corollary [BDFH08]: Longest Path has no polynomial kernel unless coNP ⊆ NP/poly. * Originally proved only for NP-complete. New statement/proof by Holger Dell

Proof of [BDFH08]-Theorem Given OR-Composition + Kernel for P we give an OR-distillation for L into OR(L). By [FS08] this implies that coNP ⊆ NP/poly.

I1I1 I2I2 I3I3 ItIt t instances of size n... I 1,1 I 2,1 I 3,2 I t,n... Parameterization Group by parameter OR-Composition O 1,k 1 O 2,k 2 O n,k n... n instances instead of t. k i ≤ poly(n)

O 1,k 1 O 3,k 2 O n,k n... n instances instead of t. k i ≤ poly(n) Kernelization O’ 1,k’ 1 O’ 2,k’ 2 O’ n,k’ n... n instances of size poly(n) each. Forget parameter O’ 1 O’ 2 O’ n... n instances of size poly(n) each. This is one instance to OR(L) of size poly(n)

Recap II NP-hard + OR-composition = no poly kernel. Longest Path has no polynomial kernel Longest Cycle has no polynomial kernel...

AND-Distillations / Compositions We can define AND-Distillation / Composition similarly to the OR case AND-Composition + Kernel = AND-Distillation Conjecture [BDFH08]: No NP-hard problem has an AND-Distillation.

AND-Compositions Some interesting problems have AND- compositions; – treewidth – pathwidth –...width – vertex ranking Under ”AND-Distillation Conjecture” they have no polynomial kernel.

Open Problem Relate the ”AND-Distillation” conjecture to a reasonable assumption in classical / parameterized complexity

Part III Kernel lower bounds for more problems

Next Polynomial Parameter Transformations: Reductions to show kernel lower bouds ”Non-trivial” OR-Composition algorithms

k-k-Paths IN: Graph G, integer k PARAMETER: k QUESTION: Does G contain k vertex-disjoint k- paths? Disjoint union doesnt work as OR-composition. Other way to show no poly kernel?

Polynomial Parameter Transformations A Polynomial Parameter Transformation (PPT) from A to B is an algorithm that: Takes as input an instance (I,k) of A Runs in polynomial time Outputs an instance (O,k’) of B such that – k’ ≤ poly(k) – (O,k’) is ”yes” for B  (I,k) is “yes” for A.

Reduction between problems Theorem [BTY09]: If there is a PPT from A to B, and a P-time reduction from B to A* then: B has a poly(k) kernel  A has a poly(k) kernel *If B is NP and A is NP-hard, a trivial p-time reduction exists.

Proof of Theorem [BTY09]: I,k I’,k’ O’,k’ O*,k* PPT Kernel P-time reduction A A B B

Back to k-k-Paths Theorem [L09]: to k-k-Paths have no polynomial kernel unless coNP ⊆ NP/poly G,k k-Path G,k k-k-Paths k-1 paths of length k NP-completeness gives reduction back.

Non-trivial Compositions? Next, excluding polynomial kernels for: – Bounded Universe Set Cover – Connected Vertex Cover (2-approximable!) – Steiner Tree

Bounded Universe Set Cover IN: Set family F={S 1...S m } over a universe U of size k, integer t PARAMETER: k QUESTION: Is there a subfamily F’ ⊆ F of size ≤ t such that F’ covers U? Theorem [DLS09]: Bounded Universe Set Cover has no poly(k) kernel unless coNP ⊆ NP/poly.

Steiner Tree IN: Graph G=(V,E), subset S ⊆ V of size k, integer t PARAMETER: t QUESTION: Is there a subtree T on ≤ t vertices of G, containing S?

Steiner Tree Theorem [DLS09]: Steiner Tree has no poly(k) kernel unless coNP ⊆ NP/poly. Proof: PPT from Bounded Universe Set Cover Universe  Terminals Sets  Non-Terminals

Connected Vertex Cover IN: Graph G=(V,E) integer k. PARAMETER: k QUESTION: Is there a set S of at most k vertices such that G[S] is connected and every edge if G has at least one endpoint in S.

Connected Vertex Cover Theorem [DLS09]: Connected Vertex Cover has no poly(k) kernel unless coNP ⊆ NP/poly. Proof: PPT from Steiner Tree Terminals Non-Terminals

Bounded Universe Set Cover Theorem [DLS09]: Bounded Universe Set Cover has no poly(k) kernel unless coNP ⊆ NP/poly. Proof plan: – Composition for ”Colored Bounded Universe Set Cover” – PPT from Colored Bounded Universe Set Cover to Bounded Universe Set Cover.

Colored Bounded Universe Set Cover IN: t set families F 1 ={A 1...A a }, F 2 ={B 1...B b }, F t ={X 1...X c } over a universe U of size k, integer t PARAMETER: k QUESTION: Is there a family F’ = {A i,B j,... X l } of size t containing one set of each color, such that F’ covers U?

Composition, recap An OR-composition algorithm for a problem P Takes as input instances I 1... I t with parameter k Runs in polynomial time Outputs an instance (O,k’) of P such that – k’ ≤ poly(k) – (O,k’) is ”yes”  some (I i,k) is “yes”.

Composition for CBUSC Task: Given t instances of CBUSC all of size ≤ n and parameter k, output in polynomial time one ”equivalent” CBUSC instance. Theorem [FKW04]: CBUSC instances with |U|=k can be solved in time O(2 k |F|). Trick: If t ≥ 2 k then t2 k |F| is polynomial, so wlog t < 2 k.

Composition for CBUSC Plan: Glue the instances together on the universe. BABCCA Universe Sets BABCCA Universe Sets B A BC A Universe Sets C B A BC A C

Composition for CBUSC GOOD: If one input is YES  YES BAD: Can have NO + NO  YES Need to make sure: A solution picks sets from the same instance.

ID’s and boxes ID’s: Every instance gets a unique identification number from 0 to 2 k -1, written in binary (k bits!) Identification Check: Will check that for every pair of colors, the two solution verties of these colors come from the same instance = have the same ID.

Boxes and ID’s A box is a gadget containing k elements. RED-BLUE box BLUE-RED box 101100 101000 The red-blue and blue-red boxes together make sure that the blue and red solution vertices come from the same instance

Composition for CBUSC Modified plan: Glue the instances together on the universe. Add two boxes for every pair of colors. Universe size increases to O(k 3 ), still poly(k). Theorem [DLS09]: Colored Bounded Universe Set Cover has no poly(k) kernel unless coNP ⊆ NP/poly.

No kernel for Bounded Universe Set Cover Theorem [DLS09]: Bounded Universe Set Cover has no poly(k) kernel unless coNP ⊆ NP/poly. PPT from CBUSC to BUSC BABCCA Universe Sets Universe Sets More Universe

Epilogue Compositions and Polynomial Parameter Transformations are tools to show kernel lower bounds. Longest Path and Connected Vertex Cover are FPT but have no polynomial kernel unless coNP ⊆ NP/poly.

List of FPT problems with no poly(k) kernels unless coNP ⊆ NP/poly. [HN06+FS08] k-Variable CNF-SAT [BDFH08] Longest Path, Longest Cycle [BTY09] Vertex Disjoint Paths, Cycles [DLS09] Bounded Universe Hitting Set, Bounded Universe Set Cover, Connected Vertex Cover, Steiner Tree, Capacitated Vertex Cover [KW09] Windmill-free Edge-Deletion [KW09’] Cases of MinOnesSat [JLS??] Dogson Score

List of FPT problems with no poly(k) kernels unless AND-Distillation fails. [BDFH08] Treewidth, Pathwidth, Cutwidth, your-favourite width, and all sorts of stuff parameterized by them. [Z09] Vertex Ranking