Partition Into Triangles on Bounded Degree Graphs

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Partition Into Triangleson Bounded Degree Graphs

Johan M. M. van RooijMarcel E. van Kooten Niekerk

Hans L. Bodlaender

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Problem Statement andOverview of Results

Partition Into Triangles Input: A graph G=(V,E). Question: Can V be partitioned into 3-element sets

S1, S2, ..., Sn/3 such that for each Si

the graph G[Si] is a triangle? We consider this problem

on bounded degree graphs. Maximum degree three. Linear time solvable. Maximum degree four. Equivalence with Exact SAT. Hard under ETH. O(1.02220n) algorithm.

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Some Simple Observations to Start with:Vertices of Degree One or Two

Degree zero and one: Cannot be in any triangle. No-instance.

Degree two: Unique triangle. Reduction rule: remove this triangle from the instance. Resulting instance Yes-instance ⇔ original instance Yes-instance.

Simple reduction rules that allow us to assume minimum degree three.

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On Graphs of Maximum Degree Three

We can assume maximum and minimum degree three. A vertex v can have four possible local neighbourhoods

(different induced subgraphs G[N[v]]).

Linear time algorithm! Reduce if vertices of degree ≤2, otherwise do as shown here!

no triangles Removing any triangle leads to vertices of degree ≤1

unique triangle

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On Graphs of Maximum Degree Four: Overview

If any vertex has degree at most three. Reduction rules.

11 possible local neighbourhoods for a vertex of degree four. Direct reduction rules for 8 of them. One can only form connected components in which every vertex

has the same local neighbourhood - additional reduction rule. Two remaining: they form ‘clouds’ and ‘fans’.

Equivalent to Exact 3-Satisfiability: A cloud is a variable. A fan is a clause.

Corollaries: NP-Complete. No subexponential-time algorithms under ETH. Very fast exponential time algorithms.

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Reduction Rules for Vertices ofDegree at Most Three

Degree at most two: No-instance or unique triangle. Degree three: four possible local neighbourhoods.

no triangles Any triangle leads to degree

≤1 vertices

unique triangle

Different in max degree four

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Reduction Rule fora Vertex of Degree Three

One local neighbourhood remaining. Taking any of the two triangles cannot give degree ≤1 vertices. Hence, top and bottom right vertex of degree four! Additional edge incident to bottom left vertex irrelevant.

Some case analysis required to ensure that no new triangles are created, i.e., these are the only two possible partitionings.

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On Graphs of Maximum Degree Four:An Overview






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Eleven Possible Local Neighbourhoods

Eleven possible local neighbourhoods for a vertex of degree four.

contain edges that are not in a triangle: remove and reduce (degree ≤ three rules)

similar edge! Any triangle leads to degree ≤1 vertices

next slide!

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Two Specific Local Neighbourhoods

Consider the red edge. Any triangle containing this edge leads to

vertices of degree ≤ 1. Remove and use the reduction rule for the

resulting degree ≤ 3 vertex.

Consider the red edge. It can be in one or two triangles: with the

yellow vertex or (possibly) the green vertex. If in a triangle with the yellow vertex: other two

vertices need a common neighbour. All four symmetric edges: endpoints or

opposite vertices need common neighbour. By maximum degree four: there must be an

edge that cannot be in a triangle. Remove and use the reduction rule.

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also contain edges that cannot be in a solution

next slide!


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Only Three Possible Local Neighbourhoods Remaining

Consider the local neighbourhood below. What local neighbourhoods can the blue

vertices have? Consider the top right (green) vertex. It can only have the same local neighbourhood

as it neighbouring yellow vertex. By induction: all vertices in the connected

component must have this local neighbourhood! Result, only components like this one exist:

can be partitioned into triangles ⇔

number of vertices is a multiple of three

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also contain edges that cannot be in a solution

done!


these are the only two local neighbourhoods that

we cannot reduce

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What Structures Can We Build Using These Two Local Neighbourhoods?

Yellow vertex and green vertex have the same local neighbourhood: they occur as pairs. We call these pairs ‘fans’. Example:

The second structure can be made into chains and loops. We call these ‘clouds’. Example:

(blue vertices arefan vertices)

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Equivalence to Exact 3-Satisfiabilitya Fan is a Clause!

Exact 3-Satisfiability Input: A set of variables X and a set of clauses C

each of size at most three. Question: Does there exists a truth assignment of the

variables in X such that each clause contains exactly one literal that is set to true?

A fan: pick exactly one of the tree triangles!

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Equivalence to Exact 3-Satisfiabilitya Cloud is a Variable!

One vertex cloud: variable with one positive and one negative occurrence.

Larger clouds: points adjacent to fans form the literals.

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Partition Into Triangles Interpreted as an Exact 3-Satisfiability Formula!

Consider the following graph: Cloud vertices are yellow. Fan vertices are blue.

We name the clouds w, x, y, z. Each fan is a clause!

Formula:(w, z, ¬x) ⋀ (x, z, ¬y) ⋀ (y, z, ¬w)

Satisfying assignment: True: w, x, y. False: z.

w yz

z

x

z

(w, z, ¬x)

(x, z, ¬y)

(y, z, ¬w)

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NP-Completeness

Property of Exact 3-SAT instances obtained in this way: For any variable: #positive literals = #negative literals (mod 3). Any such variable can be represented by a cloud. Every clause is represented by 2 vertices (fan). Any cloud that represents a variable that occurs x times is

represented by 2x-3 vertices.

The problem is NP-complete. Exact 3-SAT is NP-complete. Given an instance of Exact 3-SAT, we copy each clause 3 times. #positive literals = #negative literals (mod 3), for any variable. Hence, we can construct an equivalent Partition Into Triangles

instance on graphs of maximum degree four. New instance has size linear in the number of clauses.

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No Subexponential-Time Algorithm Exist Under the Exponential Time Hypothesis

Exponential Time Hypothesis (Impagliazzo et al. 2001): There is no algorithm for 3-SAT that runs in O(2εn) for all ε>0: no

subexponential-time algorithm. Assuming ETH, also no algorithm that runs in O(2εm) for all ε>0

by the Sparsificiation Lemma.

Assuming ETH, there is no subexponential-time algorithm for Partition Into Triangles on graphs of maximum degree four. From a given 3-SAT formula to Exact 3-SAT:

Then, same linear transformation as for NP-completeness. A, for all ε>0, O(2εm) time algorithm for Partition Into Triangles on

graphs of maximum degree four implies a, for all ε>0, O(2εn) time algorithms for 3-SAT!

SAT(x,y,z) = XSAT(x,a,b) ⋀ XSAT(y,b,c) ⋀ XSAT(a,c,d) ⋀ XSAT(¬z,b,e)

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Fast Exponential-Time Algorithms

First attempt: use the current fastest algorithms for Exact Satisfiability and Exact 3-Satisfiability. Exact 3-SAT: O(1.0984n) due to Wahlström. Exact SAT: O(1.1749n) due to Byskov, Madsen, and Skjernaa.

This algorithm removes variables with only one positive and one negative occurrence by reduction rules: They are not counted in the time bound.

If many such variables use Exact SAT algorithm, otherwise use Exact 3-SAT algorithm. Balancing gives an O(1.02445n) time algorithm.

Second attempt: algorithm for Exact 3-SAT measured by number of vertices used to create it. Extensive case analysis gives an O(1.02220n) time algorithm.

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Conclusion

Partition Into Triangles on graphs of maximum degree 3: is linear time solvable.

Partition Into Triangles on graphs of maximum degree 4: is closely related to Exact 3-Satisfiability is NP-complete. admits no subexponential-time algorithm under the ETH. is solvable in O(1.02220n) time.

Questions?

Open problem: Can you find a problem that admits no

subexponential-time algorithm and that is solvable faster than in O(1.02220n) time?

No constructions such as Independent Set in graphs where 99% of the vertices have maximum degree two allowed!(These instances can be reduced directly)

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Partition Into Triangles on Bounded Degree Graphs