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Bart Jansen

Independent Set Kernelization for a Refined

Parameter: Upper and Lower bounds

ALGORITMe Staff Colloquium, UtrechtSeptember 10th, 2010

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 SETOur target problem

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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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PARAMETERSSolutions to vertex deletion problems as complexity measures

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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 families

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KERNELIZATIONAttacking hard problems with small parameters

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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| 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

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CUBIC-VERTEX KERNEL FOR PARAMETER P3COVER

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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

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

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.

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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

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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

Type 2 ≤ |X|3

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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 Vertices on # Type 1 paths ≤ 2|X|2. Vertices on # Type 2 paths ≤ 2|X|3.

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

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CUBIC-VERTEX KERNEL FOR PARAMETER FEEDBACK VERTEX SET

A sketch of the general result

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Independent Set with Feedback Vertex Set

Input: Graph G, modulator X such that G – X is a forest, integer k.

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

No problem: we can use a constant-factor approximation algorithm to find some deletion set

Still allows us to bound the size of reduced instances in the minimum deletion set Size increases by a constant-factor

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NO POLYNOMIAL KERNEL FOR PARAMETER P3COVER

The weighted variant of the problem

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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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CONCLUSION AND DISCUSSION

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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

Cluster Deletion Distance

Feedback Vertex Set

Bipartite Deletion Distance

Outerplanar Deletion Distance

Treewidth

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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?