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Lower Bounds for Polynomial Kernelization

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Lower Bounds for Polynomial Kernelization Lower bounds for polynomial kernelization Micha l Pilipczuk Institute of Informatics, University of Warsaw, Poland Parameterized Complexity Summer School Vienna, September 2nd, 2017 Micha l Pilipczuk Kernelization lower bounds 1/33 k P-time instance of L instance of L size 6 f (k) Kernelization | recap Micha l Pilipczuk Kernelization lower bounds 2/33 k P-time instance of L instance of L size 6 f (k) Kernelization | recap instance of L Micha l Pilipczuk Kernelization lower bounds 2/33 k P-time instance of L instance of L size 6 f (k) Kernelization | recap k instance of L Micha l Pilipczuk Kernelization lower bounds 2/33 k instance of L instance of L size 6 f (k) Kernelization | recap k P-time instance of L Micha l Pilipczuk Kernelization lower bounds 2/33 k instance of L Kernelization | recap k P-time instance of L instance of L size 6 f (k) Micha l Pilipczuk Kernelization lower bounds 2/33 Parameterized problems , Sets of pairs (x; k), where x 2 Σ? and k is a nonnegative integer Unparameterized variant: k is appended to x in unary. Kernelization algorithm takes on input an instance (x; k), and outputs an instance (x 0; k0) such that 0 0 0 0 (x; k) 2 L , (x ; k ) 2 L and jx j + k 6 f (k) for some computable function f . Background in complexity theory Unparameterized problems , Languages over Σ, for a finite alphabet Σ , Subsets of Σ? Micha l Pilipczuk Kernelization lower bounds 3/33 Unparameterized variant: k is appended to x in unary. Kernelization algorithm takes on input an instance (x; k), and outputs an instance (x 0; k0) such that 0 0 0 0 (x; k) 2 L , (x ; k ) 2 L and jx j + k 6 f (k) for some computable function f . Background in complexity theory Unparameterized problems , Languages over Σ, for a finite alphabet Σ , Subsets of Σ? Parameterized problems , Sets of pairs (x; k), where x 2 Σ? and k is a nonnegative integer Micha l Pilipczuk Kernelization lower bounds 3/33 Kernelization algorithm takes on input an instance (x; k), and outputs an instance (x 0; k0) such that 0 0 0 0 (x; k) 2 L , (x ; k ) 2 L and jx j + k 6 f (k) for some computable function f . Background in complexity theory Unparameterized problems , Languages over Σ, for a finite alphabet Σ , Subsets of Σ? Parameterized problems , Sets of pairs (x; k), where x 2 Σ? and k is a nonnegative integer Unparameterized variant: k is appended to x in unary. Micha l Pilipczuk Kernelization lower bounds 3/33 Background in complexity theory Unparameterized problems , Languages over Σ, for a finite alphabet Σ , Subsets of Σ? Parameterized problems , Sets of pairs (x; k), where x 2 Σ? and k is a nonnegative integer Unparameterized variant: k is appended to x in unary. Kernelization algorithm takes on input an instance (x; k), and outputs an instance (x 0; k0) such that 0 0 0 0 (x; k) 2 L , (x ; k ) 2 L and jx j + k 6 f (k) for some computable function f . Micha l Pilipczuk Kernelization lower bounds 3/33 Let (x; k) be the input instance. If jxj 6 f (k), then we already have a kernel. Otherwise f (k) · jxjc = O(jxjc+1). Any FPT problem admits a kernelization algorithm: Question of existence of any kernel is equivalent to being FPT. We are interested in polynomial kernels, where f is a polynomial. Before 2008, no tool to classify FPT problems wrt. whether they have polykernels or not. Kernelization and FPT If a decidable problem has a kernelization algorithm, then it is FPT. Micha l Pilipczuk Kernelization lower bounds 4/33 Let (x; k) be the input instance. If jxj 6 f (k), then we already have a kernel. Otherwise f (k) · jxjc = O(jxjc+1). Question of existence of any kernel is equivalent to being FPT. We are interested in polynomial kernels, where f is a polynomial. Before 2008, no tool to classify FPT problems wrt. whether they have polykernels or not. Kernelization and FPT If a decidable problem has a kernelization algorithm, then it is FPT. Any FPT problem admits a kernelization algorithm: Micha l Pilipczuk Kernelization lower bounds 4/33 If jxj 6 f (k), then we already have a kernel. Otherwise f (k) · jxjc = O(jxjc+1). Question of existence of any kernel is equivalent to being FPT. We are interested in polynomial kernels, where f is a polynomial. Before 2008, no tool to classify FPT problems wrt. whether they have polykernels or not. Kernelization and FPT If a decidable problem has a kernelization algorithm, then it is FPT. Any FPT problem admits a kernelization algorithm: Let (x; k) be the input instance. Micha l Pilipczuk Kernelization lower bounds 4/33 Otherwise f (k) · jxjc = O(jxjc+1). Question of existence of any kernel is equivalent to being FPT. We are interested in polynomial kernels, where f is a polynomial. Before 2008, no tool to classify FPT problems wrt. whether they have polykernels or not. Kernelization and FPT If a decidable problem has a kernelization algorithm, then it is FPT. Any FPT problem admits a kernelization algorithm: Let (x; k) be the input instance. If jxj 6 f (k), then we already have a kernel. Micha l Pilipczuk Kernelization lower bounds 4/33 Question of existence of any kernel is equivalent to being FPT. We are interested in polynomial kernels, where f is a polynomial. Before 2008, no tool to classify FPT problems wrt. whether they have polykernels or not. Kernelization and FPT If a decidable problem has a kernelization algorithm, then it is FPT. Any FPT problem admits a kernelization algorithm: Let (x; k) be the input instance. If jxj 6 f (k), then we already have a kernel. Otherwise f (k) · jxjc = O(jxjc+1). Micha l Pilipczuk Kernelization lower bounds 4/33 We are interested in polynomial kernels, where f is a polynomial. Before 2008, no tool to classify FPT problems wrt. whether they have polykernels or not. Kernelization and FPT If a decidable problem has a kernelization algorithm, then it is FPT. Any FPT problem admits a kernelization algorithm: Let (x; k) be the input instance. If jxj 6 f (k), then we already have a kernel. Otherwise f (k) · jxjc = O(jxjc+1). Question of existence of any kernel is equivalent to being FPT. Micha l Pilipczuk Kernelization lower bounds 4/33 Before 2008, no tool to classify FPT problems wrt. whether they have polykernels or not. Kernelization and FPT If a decidable problem has a kernelization algorithm, then it is FPT. Any FPT problem admits a kernelization algorithm: Let (x; k) be the input instance. If jxj 6 f (k), then we already have a kernel. Otherwise f (k) · jxjc = O(jxjc+1). Question of existence of any kernel is equivalent to being FPT. We are interested in polynomial kernels, where f is a polynomial. Micha l Pilipczuk Kernelization lower bounds 4/33 Kernelization and FPT If a decidable problem has a kernelization algorithm, then it is FPT. Any FPT problem admits a kernelization algorithm: Let (x; k) be the input instance. If jxj 6 f (k), then we already have a kernel. Otherwise f (k) · jxjc = O(jxjc+1). Question of existence of any kernel is equivalent to being FPT. We are interested in polynomial kernels, where f is a polynomial. Before 2008, no tool to classify FPT problems wrt. whether they have polykernels or not. Micha l Pilipczuk Kernelization lower bounds 4/33 Suppose for a moment that k-Path admits a kernelization algorithm that, say, produces kernels with at most k3 vertices. 7 Take t = k instances (G1; k); (G2; k);:::; (Gt ; k). Let H be a disjoint union of G1; G2;:::; Gt . Then the answer to (H; k) is YES if and only if the answer to any (Gi ; k) is YES. Apply kernelization to (H; k) obtaining an instance with k3 vertices, encodable in k6 bits. Intuition The final number of bits is much less than the number input instances. Most of the instances have to be discarded completely. Motivating intuition Consider the k-Path problem: verify whether the input graph contains a simple path on k vertices. Micha l Pilipczuk Kernelization lower bounds 5/33 7 Take t = k instances (G1; k); (G2; k);:::; (Gt ; k). Let H be a disjoint union of G1; G2;:::; Gt . Then the answer to (H; k) is YES if and only if the answer to any (Gi ; k) is YES. Apply kernelization to (H; k) obtaining an instance with k3 vertices, encodable in k6 bits. Intuition The final number of bits is much less than the number input instances. Most of the instances have to be discarded completely. Motivating intuition Consider the k-Path problem: verify whether the input graph contains a simple path on k vertices. Suppose for a moment that k-Path admits a kernelization algorithm that, say, produces kernels with at most k3 vertices. Micha l Pilipczuk Kernelization lower bounds 5/33 Let H be a disjoint union of G1; G2;:::; Gt . Then the answer to (H; k) is YES if and only if the answer to any (Gi ; k) is YES. Apply kernelization to (H; k) obtaining an instance with k3 vertices, encodable in k6 bits. Intuition The final number of bits is much less than the number input instances. Most of the instances have to be discarded completely. Motivating intuition Consider the k-Path problem: verify whether the input graph contains a simple path on k vertices.
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