1. Data reduction lower bounds: Problems without polynomial kernels Hans L. Bodlaender Joint work with Downey, Fellows, Hermelin, Thomasse, Yeo

2. No Polynomial Kernels2 This talk Kernelisation Distillation conjecture and problems without polynomial kernels –Example: Long Path –Composability Proofs using transformations –Example: Disjoint cycles Conclusions If you want: sketch of Fortnow-Santhanam theorem

3. No Polynomial Kernels3 Why kernelisation? Approach for solving NP-hard problem –Data reduction: transform input to equivalent, smaller input –Solve smaller input (e.g., ILP, branch and reduce, …) –Transform solution back to solution of original problem

4. No Polynomial Kernels4 Parameterised problem Input: pair (X,k) Parameter: k Question: Q(X,k)? Fixed parameter theory (Downey and Fellows) helps to analyze –NP-complete for fixed k –O(n f(k) ) –O(f(k)* n c

5. No Polynomial Kernels5 Kernelisation A kernelisation algorithm –Receives an input to a parameterized problem Q –Transforms (X,k) to (X’,k’) with k  g(k) such that Q(X,k)  Q(X’,k’) (Equivalent) –In polynomial time –|X’|  f(k) (Size bounded by function of parameter)

6. No Polynomial Kernels6 Small kernels Given a parameterized problem Q, we can ask: –Does it have a kernel? –If so, what is the best size of the kernel that we can obtain Polynomial kernel: kernelisation algorithm that yields reduced instances of size bounded by polynomial in k

7. No Polynomial Kernels7 Kernel iff FPT A problem belongs to FPT, iff it can be solved in O(n c f(k)) time Proposition: a problem belongs to FPT, if and only if it has a kernelisation algorithm Consequence: hardness for W[1] or larger classes shows that a problem has no kernel, unless the Exponential Time Hypothesis does not hold

8. No Polynomial Kernels8 Kernel sizes O(1): problem belongs to P –NP-hardness gives negative evidence Polynomial in k: can be shown with polynomial kernelisation algorithm –This talk: negative evidence Any function of k: can be shown with FPT- algorithm –W[1]-hardness gives negative evidence

9. No Polynomial Kernels9 Examples Dominating set : no kernel (unless …) Dominating set on planar graphs : O(n) (Alber et al, 2002) Feedback vertex set : O(n 2 ) (Thomasse, 2008) Feedback vertex set on planar graphs : O(n) (Penninkx, B. 2008) Disjoint cycles : exponential kernel, no polynomial kernel (unless …) Disjoint cycles on planar graphs : O(n) (Penninkx, B, 2008) Long path : exponential kernel, no polynomial kernel (unless …) Long path on planar graphs : exponential kernel, no polynomial kernel (unless …)

10. No Polynomial Kernels10 This talk Arguments why a problem does not have a kernel of polynomial size Techniques that are used: –Composability (and-composable, or- composable) –Distillation conjectures –Result by Fortnow and Santhanam –Polynomial time and parameter transformations

11. No Polynomial Kernels11 Example problems Long path Instance: Undirected graph G, integer k Parameter k Question: Does G have a simple path of length at least k? Disjoint cycles Instance: Undirected graph G, integer k Parameter k Question: Does G have at least k vertex disjoint cycles? Both are FPT, but no kernels of polynomial size are known Both are FPT, but no kernels of polynomial size are known

12. No Polynomial Kernels12 Example theorem The Long Path problem has no polynomial kernel, unless NP  coNP/poly. Intuition: –Consider a graph with many connected components. It has a path of length k, if and only if at least one of its components has a path of length k. –If we have >> p(k) connected components, it seems unlikely that we can kernelise to a kernel of size p(k)…

13. No Polynomial Kernels13 Or-distillation conjecture There is no algorithm, that given instances X(1), …, X(r) of an NP-complete problem, finds an instance X’ of the problem, such that –X’ has a solution, if and only if there is an i with X(i) has a solution –The time of the algorithm is polynomial in the sum of the sizes of the X(j)’s –The size of X’ is bounded by a polynomial in the maximum size of the X(j)’s. Note: if it holds for one NP-complete problem, then for all. We use Satisfiability.

14. No Polynomial Kernels14 Fortnow-Santhanam result If the Or-distillation conjecture does not hold, then NP  coNP/poly Some problems (e.g., Treewidth ) need And- Distillation conjecture. No such result is known for and-distillation

15. No Polynomial Kernels15 Proof for Long Path Suppose we have a kernelisation algorithm for Long Path. Note that Long Path is NP-complete. We build an or-distillation algorithm for Satisfiability, as follows: 1.Take formulas F(1), …, F(r) (inputs of SAT) 2.Use polynomial time transformation (implied by NP- completeness), and transform each to an equivalent instance of Long Path : (G(1),k(1)), …, (G(r), k(r))

16. No Polynomial Kernels16 Proof continued 3.Group these in sets with the same parameter: (G(1,1),1), (G(1,2),1), … (G(1,r 1 ),1), (G(2,1),2), (G(2,2),1), … (G(2,r 2 ),2), … (G(s,1),s), (G(s,2),s), … (G(s,r s ),s). 4.For each group, build one instance of Long Path by taking the disjoint union of the graphs (G(1,1)  G(1,2)  …  G(1,r 1 ), 1), (G(2,1)  G(2,2)  …  G(2,r 2 ), 2), …, (G(s,1)  G(s,2)  …  G(s,r s ), s).

17. No Polynomial Kernels17 Proof further continued 5.Apply the kernelization algorithm to each group: (G’(1),k 1 ), …, (G’(s),k s ) 6.Use polynomial time transformation (implied by NP-completeness), and transform each to an equivalent instance of Satisfiability : F’(1),.., F’(s) 7.“Or” the formulas: F=F’(1) or F’(2) or … or F’(s)

18. No Polynomial Kernels18 Correctness Satisfiability : easy check Size: –Say maximum size of formula F(i) is n. –Maximum parameter s is bounded by polynomial in n –Each of the F’(i)’s has size bounded by polynomial in value at most s QED

19. No Polynomial Kernels19 Technique works for many problems If problem has or-compositionality: –If we have several instances x(1), …, x(r) with the same parameter, we can build in polynomial time one instance that holds, if and only if at least one x(i) holds Several graph problems are or-compositional, often just using disjoint union of connected components (does G contain a certain substructure?) Some graph problems need and-compositionality: –If we have several instances x(1), …, x(r) with the same parameter, we can build in polynomial time one instance that holds, if and only if all x(i) hold –Example: Treewidth

20. No Polynomial Kernels20 Theorem If problem is or-compositional and NP-complete, it has no polynomial kernel, unless NP  coNP/poly. If problem is and-compositional and NP-complete, it has no polynomial kernel, unless and-distillation conjecture does not hold. Extending this with transformations…

21. No Polynomial Kernels21 Transformations Theorem: Disjoint Cycles has no polynomial kernel, unless NP  coNP/poly. Note: unexpected: related problems ( Feedback Vertex Set, Edge Disjoint Cycles ) have polynomial kernels Transformation via Disjoint Factors –Disjoint Factors –Input: string s in {1,…,k}* –Parameter: k –Question: find disjoint substrings in s, starting and ending with 1, with 2, …, with k, each of length at least 2 41412323141343433

22. No Polynomial Kernels22 Or-compositionality of Disjoint factors Suppose we have inputs s(1), …,s(r)  {1, …,k}* If r > 2 k, we can solve with dynamic programming each input in O(2 k |s(i)|) time which is polynomial Suppose r  2 k. Adding at most k extra letters, we build one input as follows, e.g., for k=3: 6 5 4 s(1) 4 s(2) 4 5 4 s(3) 4 s(4) 4 5 6 5 4 s(5) 4 s(6) 4 5 4 s(7) 4 s(8) 4 5 6 (2k) recursive string with first half (2k) recursive string with second half (2k) … if and only if …

23. No Polynomial Kernels23 Disjoint cycles theorem proved Disjoint factors is NP-complete (proof omitted), and or-compositional, so has no polynomial kernel unless NP  coNP/poly. Suppose we have a polynomial kernelisation algorithm for Disjoint Cycles. We build one for Disjoint Factors : –Take string s in {1, …, k}* –Build graph: take path a vertex for each letter in s –Take k new vertices v i, one for each element in {1,…,k} –Make v i adjacent to each letter i

24. No Polynomial Kernels24 1234 414223131 Graph has k disjoint cycles: each uses one green vertex: string has k disjoint factors

25. No Polynomial Kernels25 Proof with reductions Polynomial time and parameter reduction: Reduction from a parameterized problem to another parameterized problem: –Uses time, polynomial in input size + k –Maps to equivalent instances –New parameter is bounded by polynomial in old parameter Similar to classic notion from NP-completeness theory, but now also parameter bound

26. No Polynomial Kernels26 Use of reductions Suppose that we have parameterized problems P and Q. Let P’ and Q’ be the corresponding non- parameterized problems, assuming that parameters are given in unary. If Q’ is NP-complete, P’ is in NP, and we have a polynomial time and parameter reduction from P to Q. Then, if Q has a polynomial kernel, then P has a polynomial kernel.

27. No Polynomial Kernels27 Examples Or-compositional: Long Path, Long Cycle, Cycle of Length exactly k, k-Clique minor, several problems parameterized by treewidth of graph And-compositional: Treewidth, Pathwidth, Cutwidth, Branchwidth, Transformations: Disjoint Cycles, Disjoint Paths (Linkage), Hamiltonian Circuit parameterized by Treewidth …???

28. No Polynomial Kernels28 Conclusions Data reduction and kernelisation: interesting research topic Compositionality is easy to use, and gives easy arguments that problems are not likely to have polynomial kernel Kernelisation gives interesting insights in problems, as data reduction is used in practical settings Transformations extend applicability of results Many angles for interesting further research

29. No Polynomial Kernels29 Fortnow-Santhanam Theorem. If the Or-distillation conjecture does not hold, then NP  coNP/poly

30. No Polynomial Kernels30 Fortnow Santhanam proof Suppose L  {0,1}* is NP-complete problem with Or-distillation algorithm A Let L C be the complement of L Let L C (n) all strings in L C of length at most n A maps a sequence x(1), …, x(t)  L C (n) to a string in L C (n c ) with c not depending on t

31. No Polynomial Kernels31 Claim (part of proof) Claim. If n, t large enough, there is a set S(n)  L C (n) such that –|S(n)| is polynomially bounded in n –If x  L C (n), then there exists strings x(1),.., x(t), each of length at most n with x(i)=x for some i, such that A maps x(1),.., x(t) to an element in S(n) –If x  L C (n), then for all strings x(1),.., x(t), each of length at most n with x(i)=x for some i, A maps x(1),.., x(t) to an element not in S(n)

32. No Polynomial Kernels32 Proof of claim (sketch) If x  L C (n) and there exists strings x(1),.., x(t), each of length at most n with x(i)=x for some i, such that A maps x(1),.., x(t) to an element y in S(n), we say that y covers x. We need polynomial size set that cover all strings in L C (n). Counting argument: there is a string that covers a constant fraction of the not yet covered strings, if we take t = O(n c ). Repeating this gives the desired set

33. No Polynomial Kernels33 Using the claim and ending the proof coNP/poly algorithm to check if given string belongs to L C (n) Given a string x of length n: –Guess t strings of length at most n. –If neither is x, reject. –Otherwise, apply A, say we get y –If y  S(n), then accept, otherwise reject QED