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Bart Jansen Polynomial Kernels for Hard Problems on Disk GraphsPowerPoint Presentation

Bart Jansen Polynomial Kernels for Hard Problems on Disk Graphs

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### Bart JansenPolynomial Kernels for Hard Problems on Disk Graphs

Accepted for presentation at SWAT 2010

Overview

- Introduction
- Kernelization
- Graph classes

- Kernels
- Triangle Packing, Kt-matching, H-matching
- Red/Blue Dominating Set
- Connected Vertex Cover

- Conclusion

Kernelization for graph problems

- Consider a computational decision problem on graphs
- Input: encoding x of a question about graph G and integer k.
- Question: does graph G have a (…)?
- Parameter:k

- Parameter expresses some property of the question (size of what we are looking for, treewidth of graph, etc.)
- A kernelization algorithm takes (x, k) as input and computes instance (x’, k’) of same problem in polynomial time, such that
- Answer to x is YES answer to x’ is YES
- k’ ≤ g(k) for some function g
- |x’| ≤ f(k) for some function f

- The function f is the size of the kernel
- We want f to be a (small) polynomial

Recent kernelization results

Bad news

Good news

If we require G to be planar, lots of problems have linear or quadratic kernels

Even if we relax planarity to bounded genus, H-minor-free, …

- Many parameterized problems are W[1]-hard and have no kernels
- Several easier parameterized problems only have kernels where f is exponential

Expanding the range of good news

- The frameworks giving general good news about small kernels only apply under restrictions that make the graph G sparse: |E| ≤ c |V|
- Dense graphs without special structure make the problem hard, implying non-existence of kernels
- We consider graphs that exhibit structure, but are not sparse: (unit)disk graphs
- Yields good news:
- Red-Blue Dominating Set, H-Matching, Connected Vertex Cover
- Do not have polynomial kernels in general graphs
- Have polynomial kernels in (unit)disk graphs

- And the problems are still hard on disk graphs

- Red-Blue Dominating Set, H-Matching, Connected Vertex Cover

Graph classes

Linear edge count

Meta-theorems

Quadratic edge count

planar

unit-disk

bounded-genus

bounded-genus

H-minor-free

disk

Our kernels

Ki,j-subgraph-free

Subquadratic edge count

Kernels for Dominating Set

general

Disk graphs

- Consider a set S of closed disks in the plane
- The intersection graph of S:
- has a vertex v for every disk D(v),
- has an edge between u and v iff. the disks D(v) and D(u) intersect.
- (touching disks also intersect)

Properties of disk graphs

- If all disks have the same radius, their intersection graph is a unit disk graph
- All planar graphs are disk graphs (varying radii)
- Any clique is a (unit)disk graph
- Compare with K5 which is not planar
- So there are disk graphs with

- Class of (unit)disk graphs
- Closed under vertex deletion
- Not closed under edge deletion
- Not closed under edge contraction

Triangle packing and h-matching

Triangle Packing

- Input: Graph G, integer k
- Question: Are there k vertex-disjoint triangles in G?
- Parameter: k
- NP-complete, even on planar graphs
- In FPT on general graphs with a O(k2)-vertex kernel

Triangle Packing

- Input: Graph G, integer k
- Question: Are there k vertex-disjoint triangles in G?
- Parameter: k
- Single reduction rule
- Try all O(n3) sets of size 3, and test if they form a triangle
- Mark vertices that occur in a triangle
- Delete all vertices that were not marked

Kernelization algorithm

- Greedily build a maximal triangle packing
- Suppose the greedy packing contains k* copies

Neighborhood Clique Lemma

- Let v be a vertex in a unit-disk graph G. Then there is a clique of size ⌈deg(v) / 6⌉ among the neighbors of G.
- G[N(v)] has a clique of size ⌈deg(v) / 6⌉

- Proof.
- Consider centers of v and its neighbors in a disk realization
- Divide the plane into 6 equal sectors around v
- Some sector contains ⌈deg(v) / 6⌉ sectors (Pigeonhole Principle)

v

Neighbors in each sector form a clique

- Assume every disk has radius ½
- If v has a neighbor x then distance |xv| ≤ 1

y

v

x

v

Neighbors in each sector form a clique

- Assume every disk has radius ½
- If v has a neighbor x then distance |xv| ≤ 1
- Consider two neighbors x,y in the same sector
- By adjacency to v: |xv| ≤ 1, |yv| ≤ 1
- Sector definition: angle xvy ≤ 60o
- By law of Cosines: |xy| ≤ 1
- So x,y adjacent
- Neighbors within sector form a clique

y

x

v

Analysis of kernel size

- If there is a maximal triangle packing with k* copies in G, then |V| is O(k*)
- Proof.
- We divide V in two subsets:
- set S with vertices that are used in a selected copy
- set W with the remainder

- Since all triangles are vertex-disjoint, there are exactly 3k* vertices in S (every triangle uses 3 vertices)
- We bound the size of W
- Every vertex in W must be adjacent to vertex in S
- Every vertex in S has at most 12 neighbors in W

- So |W| ≤ 12 |S| ≤ 12(3 k*) ∈ O(k*)

- We divide V in two subsets:

Extension to Kt-matching

- We get a kernel with O(k) vertices for Triangle Packing in unit-disk graphs
- Current best kernel for general graphs has O(k2) vertices

- Generalizes to Kt-matching for every fixed t
- Pack vertex-disjoint complete subgraphs of size t
- Important properties still hold:
- Every vertex that is not selected in a maximal packing must be adjacent to a selected vertex
- Every selected vertex has O(t) neighbors in W

Extension to H-Matching

- H-matching problem
- Pack vertex-disjoint copies of a fixed connected graph H
- Kernel with O(k|H|-1) vertices by H. Moser [SOFSEM ‘09]
- No kernel polynomial in |H| + k

- H-matching on unit-disk graphs
- H can be arbitrary
- Graph G in which we find the copies is a unit-disk graph

- Our result
- O(k)-vertex kernel for every fixed graph H
- Constant is exponential in the diameter of H

- Properties of maximal H-matching in reduced graph
- Every unused vertex has distance ≤ diameter(H) to a used vertex
- Every vertex has O(|H|) unused neighbors

Red/blue dominating set

Red/Blue Dominating Set

- Input: Graph G with red vertices R, blue vertices B, integer k
- Question: Is there a set of ≤ k red vertices that dominate all blue vertices?
- Parameter: min(|R|,|B|)

Background

- min(|R|,|B|) as parameter since parameter k is W[1] hard, even on unit-disk graphs
- In FPT on general graphs, no polynomial kernel
- Usually assume G is bipartite with R and B as color classes
- We do not assume this here; bipartite disk graphs are planar

- Our results:
- O(min(|R|,|B|))-vertex kernel on planar graphs
- O(min(|R|,|B|)2)-vertex kernel on unit-disk graphs
- O(min(|R|,|B|)4)-vertex kernel on disk graphs

Reduction Rules

- Red vertices r1, r2 such that N(r1) ∩ B ⊆ N(r2) ∩ B
- Delete r1

- Blue vertices b1, b2 such that N(b1) ∩ R ⊆ N(b2) ∩ R
- Delete b2

Balance

- After exhaustive application of reduction rules, the color classes must be balanced
- Number of vertices in the classes must be polynomially related

- Easy for planar graphs: |R| ≤ 5|B| (and vice versa)
- Contribution:
- |R| ∈ O(|B|2) (and vice versa) for unit-disk graphs
- |R| ∈ O(|B|4) (and vice versa) for disk graphs

- These structural results immediately yield kernels

Balance in colored unit-disk graphs

- Usual model: two vertices adjacent iff their disks intersect
- Double the radius of disks
- Now: two vertices adjacent iff the disk of one contains the center of the other, and vice versa

- We prove: if no two red vertices see the same blue vertices, then |R| ∈ O(|B|2).

∙

∙

∙

∙

radius 1

radius ½

Proof

- We prove: if no two red vertices see the same blue vertices, then |R| ∈ O(|B|2)
- Look at arrangement of the plane induced by blue circles
- Each region contains at most one red center
- Complexity of the arrangement is O(|B|2)

∙

∙

∙

∙

∙

∙

∙

∙

∙

∙

∙

∙

Balance in colored disk graphs

- Reconsider usual model: vertices adjacent iff disks intersect
- We prove: if no red disk sees a subset of the blue vertices seen by another red disk, then |R| ∈ O(|B|4)

[A,B]

[A,B,C]

[B,A]

[B,A,C]

A

B

C

Balance in colored disk graphs

- A face in the arrangement of bisector curves determines a unique order of encountering blue disks
- The blue neighbors of a red disk are a prefix of the string determined by the face containing its center
- So any face contains at most one red disk

[A,B,C]

[B,A,C]

A

B

[B,C,A]

[A,C,B]

C

[C,A,B]

[C,B,A]

Balance in colored disk graphs

- Given n curves for which each pair intersects O(1) times, the complexity of the arrangement is O(n2)
- We have O(|B|2) curves, hence complexity is O(|B|4)
- Total number of red disks is O(|B|4)

[A,B,C]

[B,A,C]

[B,C,A]

[A,C,B]

[C,A,B]

[C,B,A]

Summary of kernels for Red/Blue Dominating Set

- By applying the reduction rules we find in polynomial time an equivalent instance such that no red vertex sees a subset of what another red vertex sees
- Same for the blue vertices

- Structural theorems show that in such colored graphs the sizes of the color classes are polynomially related
- So size of the largest class is polynomial in the size of smallest class
- Hence |V| = |R| + |B| ≤ min(|R|+|B|) + max(|R|,|B|) is O(min(|R|+|B|)c)

Connected vertex cover

Connected Vertex Cover

- Input: Graph G, integer k
- Question: Is there a vertex cover of ≤ k vertices that induces a connected subgraph?
- Parameter: k
- FPT on general graphs, no polynomial kernel
- Trivial linear-vertex kernel on unit-disk graphs
- Any vertex cover for a unit-disk graph must have size ≥ n/12 (Erik-Jan’s thesis)

Annotated Connected Vertex Cover

- Input: Graph G, set of marked vertices S, integer k
- Question: Is there a vertex cover of ≤ k vertices that induces a connected subgraph, and which contains all marked vertices?
- Parameter: k
- Unmarked vertex v is dead if all its neighbors are marked, if not then v is live
- Reduction rules
- Unmarked vertex v with degree > k: mark v
- Distinct dead vertices u,v such that N(u) ⊆ N(v): delete u

Analysis

- Call an edge covered if it’s incident on a marked vertices
- Otherwise an edge is uncovered
- > k2 uncovered edges: output NO
- > k marked vertices: output NO
- In remaining cases ≤ k2 uncovered edges
- ≤ 2k2 live vertices since each live vertex is incident on an uncovered edge
- ≤ k marked vertices
- Remains to bound the dead vertices

- # Dead vertices can be bounded in # marked vertices by the balance argument, gives #dead is O(k4)
- More intricate argument gives O(k2) bound
- Annotation can be undone

Conclusion and discussion

- Several parameterized problems without polynomial kernels on general graphs, do allow polynomial kernels on dense (unit)disk graphs
- Colored Ki,j-subgraph-free graphs also have the “polynomial balance property”
- Polynomial kernels for Red/Blue Dom. Set and Connected V.C.

- Open problems
- Poly kernel for H-matching in disk graphs?
- Poly kernel for unit-disk Edge Clique Cover?
- Poly kernel for unit-disk Partition (Vertex Set) Into Cliques?
- Improve the quartic bound for balance in disk graphs
- Find other problems where colored graph balance implies poly kernels

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