Joint work with Hans Bodlaender Bart Jansen Independent Set Kernelization for a Refined Parameter: Upper and Lower bounds Joint work with Hans Bodlaender Algorithm seminar, Bergen January 7th, 2011

Independent Set Kernelization for a Refined Parameter: Upper and Lower bounds Introduction Independent Set Parameters Kernelization Upper bounds Small kernel for parameter P3 cover Reduction rules Analysis Ideas for for parameter Feedback Vertex Set Lower bounds Effect of introducing vertex weights Conclusion

Our target problem independent set

Independent Set Input: Graph G, integer q Question: Is there a set S of ≥ q vertices which are pairwise non-adjacent? NP-complete, even on planar graphs max degree 3 Not approximable We show how to attack the problem if some measure of “graph complexity” is low Data reduction

Solutions to vertex deletion problems as complexity measures Parameters

Vertex Deletion Problems Vertex Cover Input: Graph G, integer q Question: Is there a set S of ≤ q vertices such that G-S is edgeless? Equivalent question: Is there an Independent Set of size ≥ n – q? Vertex Cover Edgeless Graphs

Vertex Deletion Problems P3 Cover Input: Graph G, integer q Question: Is there a set S of ≤ q vertices such that G-S is a collection of paths on at most 2 vertices? P3 cover Paths ≤2 nodes Vertex Cover Edgeless Graphs

Vertex Deletion Problems Feedback Vertex Set Input: Graph G, integer q Question: Is there a set S of ≤ q vertices such that G-S is a forest? (Acyclic) Feedback vtx Set Forests P3 cover Paths ≤2 nodes Vertex Cover Edgeless Graphs

Graph Complexity Measures We can use the minimum sizes of these vertex deletion sets as measures of the complexity of a graph Every edgeless graph is a collection of paths on ≤ 2 nodes Every collection of paths on ≤ 2 nodes is a forest Difference between the parameters can be unbounded Feedback vtx Set Forests P3 cover Paths ≤2 nodes Vertex Cover Edgeless Graphs ≤ ≤

Collections of paths on ≤ 2 vertices Graph families All graphs Forests Collections of paths on ≤ 2 vertices Edgeless graphs

Attacking hard problems with small parameters kernelization

Graph problems with structural parameters Consider a computational decision problem on graphs Input: encoding x of a question about graph G, integer k. Question: does graph G have a (…)? Parameter:k Parameter value k expresses some measure of the complexity of the graph size of a minimum Vertex Cover, P3 Cover, Feedback Vertex Set, etc.

Kernelization for graph problems A kernelization algorithm takes (x, k) as input and computes an instance (x’, k’) of same problem in polynomial time, such that Answer to x is YES  answer to x’ is YES k’ ≤ k |x’| ≤ f(k) for some function f The function f is the size of the kernel We want f to be a (small) polynomial Kernelization reduces the size of the graph to something which depends only on the complexity measure of the input, not on the size of the input Afterwards solve the smaller instances by some other method

Perspective for this talk We want to solve the Independent Set problem We use the solution values of the vertex deletion problems as complexity measures (parameters) of the input instances Previous state of the art: “Does graph G with vertex cover of size k have an independent set of size q?” can be transformed in polynomial time into: “Does graph G’ with vertex cover of size k’ have an independent set of size q’ ?” where |G’| ≤ 2 k, and k’ ≤ k. Complexity-theoretic evidence that the factor 2 is optimal

Our results: upper bounds “Does graph G with feedback vertex set of size k have an independent set of size q?” can be transformed in polynomial time into: “Does graph G’ with feedback vertex set of size k’ have an independent set of size q’ ?” where |G’| ≤ O(k3), and k’ ≤ k. Our new bound uses more units of a smaller measure |G’| ≤ O(|MinFVS|3)  |G’| ≤ 2 |MinVC| Compare: “1000 ants weigh less than 3 horses” Refined parameter For simplicity we present the following result: Transformation such that |G’| ≤ O(|MinP3Cover(G)|3). The Independent Set problem parameterized by the size of a feedback vertex set admits a cubic-vertex kernel

Cubic-vertex kernel for parameter p3cover

Independent Set with P3-cover Input: Graph G, modulator X such that G – X is a collection of paths on at most 2 vertices, integer q. Question: Does G have an Independent Set of size q? Parameter: k := |X|. X G - X

Canonical solution structure G - X The maximum independent set (MIS) of G – X contains 1 vertex from each path in G – X We call this a canonical solution for graph G It uses no vertices of X Poly-time computable Vertices from X are only useful if they allow for a larger IS than the canonical solution X

Conflicts induced by a vertex in X G - X Consider vertex v in X Compute a maximum independent set in G-X which avoids neighbors of v Compare to the canonical solution (MIS in G-X) Call the difference cf(v) the number of conflicts induced by v Intuitively: the price we pay in G-X for using vertex v in an independent set We can only improve on the canonical solution if the number of vertices we gain in X, is more than the number we lose in G-X X

Reduction rule 1 Deleting single vertices in X G - X If cf(v) ≥ |X| then delete v There is always an optimal IS without v Consider an IS using v Might use |X| within X Solution inside G-X at least |X| worse than canonical Compare to: Don’t use anything in X Use optimum in G – X (Canonical solution) X

Conflicts induced by pairs of vertices in X G - X Consider non-adjacent vertices {u,v} in X Compute a maximum independent set in G-X which avoids neighbors of {u,v} Compare to canonical solution Call the difference cf({u,v}) the number of conflicts induced by{u,v} Intuitively: the price we pay in G-X for using vertices {u,v} in an independent set X

Reduction rule 2 Adding edges in X G - X If cf({u,v})≥|X| then add edge {u,v} There is always an optimal IS that avoids one of {u,v} Consider an IS using {u,v} Compared to the canonical solution it uses at least |X| less in G-X So the canonical solution is at least as large Does not use any vertices from X X

Reduction rule 3 Deleting P1 components from G-X If there is an isolated vertex v in G – X which does not have any neighbors in X, then delete v and decrease q by 1 We can always use v in an independent set “Does G have an independent set of size q?” now reduces to “Does G – v have an independent set of size q-1?” X

Reduction rule 4 Deleting P2 components from G-X If there is a P2 in G-X on vertices {x,y} such that no single vertex in X sees {x,y}, no pair of non-adjacent vertices in X together sees {x,y} then delete {x,y} and decrease q by 1 We can always use one of {x,y} in an independent set No independent set in X contains neighbors of x and y simultaneously “Does G have an independent set of size q?” now reduces to “Does G - {x,y} have an independent set of size q-1?” X Observe: P2’s in G – X that survive this rule have restricted structure!

Analysis To prove: after exhausting these reduction rules we have |V| ≤ |X| + 2|X|2 + 2|X|3. Count how many paths we have in G – X. Type 1: All P1 and the P2 whose vertices have a common neighbor Type 2: The P2 whose vertices have no common neighbor Claim: # Type 1 ≤ |X|2. Claim: # Type 2 ≤ |X|3. G - X X

Type 1 ≤ |X|2 Type 1: All P1 and the P2 whose vertices have a common neighbor A P1 path of Type 1 must be adjacent to a vertex in X (Rule 3) A P2 path of Type 1 must be adjacent to a vertex in X (definition) Claim: no vertex in X is adjacent to more than |X| paths of Type 1 If v in X is adjacent to |X| paths of Type 1, then cf(v) ≥ |X| If we use v in IS, then we cannot use any vertices on adjacent Type 1 paths IS size in G-X decreases by at least |X| if we use v But by Rule 1 there are no vertices in X with cf(v) ≥ |X| So we can charge all Type 1 paths to a (common) neighbor in X We charge less than |X| to each vertex in X Total charge = number of Type 1 paths ≤ |X|2 X G - X

Type 2 ≤ |X|3 Type 2: The P2 whose vertices have no common neighbor Claim: for Type 2 paths on {x,y} there are non-adjacent u,v in X such that u sees x, and v sees y From definition of Type 2: no vertex in X sees both If there is no such pair u,v then the path is deleted by Rule 4 Claim: if u,v in X are non-adjacent, then there are less than |X| P2’s in G – X such that u sees the left endpoint and v sees the right endpoint If there are at least |X| such P2’s then cf({u,v}) ≥ |X| and we would add the edge {u,v} by Rule 2 So we can charge each P2 of Type 2 to some non-adjacent pair in X which sees the endpoints of this P2 We charge less than |X| to each pair Total charge = number of Type 2 paths ≤ |X|2 • |X| = |X|3 X G - X X G - X

Summing it up To prove: after exhausting these reduction rules we have |V| ≤ O(|X|3). We proved bounds on the number of paths: # Type 1 ≤ |X|2. # Type 2 ≤ |X|3. Number of vertices on a path is at most 2 Besides vertices on paths, graph contains only X. |V|=|X| + |V(Type 1)| + |V(Type 2)| ≤ |X|+2(|X|2+|X|3). Reduction rules can be applied in polynomial time What is left of X forms a P3 Cover for the resulting graph Complexity of final instance is not greater than of input instance

Cubic-vertex kernel for parameter FEEDBACK VERTEX SET A sketch of the general result Cubic-vertex kernel for parameter FEEDBACK VERTEX SET

Independent Set with Feedback Vertex Set Input: Graph G, modulator X such that G – X is a forest, integer q. Question: Does G have an Independent Set of size q? Parameter: k := |X|. Solve in 2|X|(|V| + |E|) time Try all subsets S of X Skip if S is not independent Otherwise compute MIS in G-X which avoids neighbors of S Solve MIS in G – X – N(S) This is a forest! Return maximum value of |S| + MIS G - X X

Outline We can still compute a canonical solution (MIS of G – X) in polynomial time since G – X is a forest As before, number of conflicts induced by vertex v in X, or a non-adjacent pair {u,v} in X, is the decrease in the size of the solution within G – X, when using those vertices Rule 1: Delete v in G – X with cf(v) ≥ |X| Rule 2: Add edge between non-adjacent u,v in X if cf({u,v}) ≥ |X| Rule 3: Delete a tree T in G – X if there are no non-adjacent vertices {u,v} in X which induce a conflict on T Decrease q by MIS(T) Not obvious that checking for pairs is enough Rule 4, 5: Simplify structure of trees in G – X Analysis: charge vertices in a tree to neighbors in X total charge cannot be too big without triggering reduction rules 20 pages of proof for the analysis

The modulator X in the input We have assumed that we get the modulator X (the deletion set) as part of the input Might not be the case in practice Kernelization claims do not rely on X being a minimum set; the size of the reduced instance is bounded in |X| So we compute a 2-approximation X, use it instead |G’| is bounded in O(|X|3) |X| is bounded by 2 |MinFVS(G)| Hence |G’| is bounded by O(|MinFVS(G)|3)

no polynomial kernel for parameter p3cover The weighted variant of the problem no polynomial kernel for parameter p3cover

Weighted Independent Set with P3-cover Input: Vertex-weighted graph G, modulator X such that G – X is a collection of paths on at most 2 vertices, integer q. Question: Does G have an Independent Set of total weight at least q? Parameter: k := |X|. G - X X Weight 12 Weight 30

Contrasting result Weighted Independent Set with P3-cover does not admit a polynomial kernel (assuming a widely-believed conjecture from complexity theory) Proof uses a variation of many-one reductions Intuition: There is no answer-preserving polynomial-time procedure that reduces an instance of Weighted Independent Set to some instance whose size is bounded by the size of a P3 cover Independent Set parameterized by P3 cover is the first example where the use of vertex weights does not affect fixed-parameter tractability, but does affect kernelizability Compare: for Independent Set with parameter Vertex Cover both the weighted and unweighted problem admit small kernels!

Why vertex weights make the problem harder to kernelize Main idea: Build a graph G which contains adjacent pairs of vertices inside the modulator X If you select exactly one from each pair, then the rest of the independent set behaves in some nice way But any maximum cardinality independent set would not use any vertices from X at all Give the vertices in these pairs high weight! G - X X

conclusion and discussion

Summary of kernelization results Independent Set Weighted Independent Set Parameter Vertex Cover 2k * Parameter P3 Cover O(k3) No poly(k) Parameter Feedback Vertex Set Table shows number of vertices in reduced graphs * marks existing results Our results can be combined with existing kernelization Ensures reduces instances using new technique are not bigger than using old technique

Kernelizability of (Unweighted) Independent Set Feedback vtx Set Forests P3 cover Paths ≤2 nodes Vertex Cover Edgeless Graphs ≤ ≤

Kernelizability of (Unweighted) Independent Set Vertex Cover Increasing size P3 Cover Cluster Deletion Distance Feedback Vertex Set ? ? Bipartite Deletion Distance Outerplanar Deletion Distance Treewidth

Kernel lower bounds for Unweighted Independent Set with structural parameters Consider some graph class F such that F is hereditary (closed under vertex deletion) F contains all complete graphs Maximum Independent Set can be solved in polynomial time for graphs in F The independent set problem parameterized by the minimum number of vertices which have to be deleted to obtain a graph in class F, is in FPT (assuming the deletion set S is given) BUT: There is no polynomial kernel for this parameterized problem (unless …) Proof using cross-composition

Implications The Maximum Independent Set problem parameterized by the number k of vertices which have to be deleted to obtain a Perfect graph, Chordal graph, Interval graph, Cograph, Etc …, is in the class FPT but does not admit a polynomial kernel (unless …)

Conclusion We have studied Independent Set parameterized by different measures of graph complexity Size of a Vertex Cover, P3 Cover, Feedback Vertex Set Usage of vertex weights affects kernelizability Hierarchy of parameters (complexity measures) which we can explore Open problems Deletion distance to bipartite/outerplanar graphs Improve the degree of the polynomial: cubic to quadratic?