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Bart Jansen
Independent Set Kernelization for a Refined
Parameter: Upper and Lower bounds
TACO Day, UtrechtJanuary 12th, 2011
Joint work with Hans Bodlaender
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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
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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
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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?
Vertex Cover
Edgeless Graphs
Equivalent question:Is there an Independent
Set of size ≥ n – q?
Equivalent question:Is there an Independent
Set of size ≥ n – q?
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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?
Vertex Cover
Edgeless Graphs
P3 cover
Paths ≤2 nodes
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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)
Vertex Cover
Edgeless Graphs
P3 cover
Paths ≤2 nodes
Feedback vtx Set
Forests
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Graph Complexity Measures
Vertex Cover
Edgeless Graphs
P3 cover
Paths ≤2 nodes
Feedback vtx Set
Forests
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
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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.
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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
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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
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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| 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
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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|.
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Canonical solution structure
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
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Conflicts induced by a vertex in 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
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Reduction rule 1Deleting single vertices in 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)
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Conflicts induced by pairs of vertices in 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
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Reduction rule 2Adding edges in 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
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Reduction rule 3Deleting 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?”
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Reduction rule 4Deleting P2 components from G-X
If there is a P2 in G-X on vertices {x,y} such that both
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?”
Observe:P2’s in G – X that survive this rule
have restricted structure!
Observe:P2’s in G – X that survive this rule
have restricted structure!
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Analysis
After exhausting the reduction rules: each single vertex induces at most |X|
conflicts each non-adjacent pair induces at most |X|
conflicts Total number of conflicts at most |X|2 + |X|3
Not hard to show that each path in G – X contributes to the number of induced conflicts
# vertices per path is ≤ 2 # vertices in G – X is ≤ 2(|X|2 + |X|3)
|V| ≤ |X| + 2(|X|2 + |X|3) = O(|X|3)
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Summing it up
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
Independent Set parameterized by the size of a P3 Cover admits a kernel with O(k3) vertices
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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
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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
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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)
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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|.
Weight 12Weight 12
Weight 30Weight 30
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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!
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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!
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Summary of kernelization results
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
Independent SetWeighted
Independent Set
Parameter Vertex Cover
2k * 2k *
Parameter P3 Cover O(k3) No poly(k)
Parameter Feedback Vertex Set
O(k3) No poly(k)
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Kernelizability of (Unweighted) Independent Set
Vertex Cover
Edgeless Graphs
P3 cover
Paths ≤2 nodes
Feedback vtx Set
Forests
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Kernelizability of (Unweighted) Independent Set
Vertex Cover
P3 Cover
Clique Deletion Distance
Feedback Vertex Set
Bipartite Deletion Distance
Outerplanar Deletion Distance
Treewidth
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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 X is given)
BUT: There is no polynomial kernel for this parameterized problem (unless …) Proof using cross-composition
[With Hans Bodlaender and Stefan Kratsch]
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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 …)
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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?