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Presentation Transcript

Outline

- Distance from triviality
- Previous work/Introduction
- Concrete example: Clique problem
- Ideas/techniques for other problems
- Conclusion

Outline

- Parameterizing above or below guaranteed values
- Introduction
- Parameterizing above or below guaranteed values
- Tight lower and upper bounds
- Hard above or below guarantee problems
- Parameterizing sufficiently above or below guaranteed values
- open problems

Previous work

Motivation: Many NP-hard graph problems (such as Vertex Cover) become easy (linear-time solvable) on trees.

Treewidth measures the distance from the triviality “tree”, hence using it as a parameter (if small) solves many hard graph problems efficiently.

- Hence these problems are FPT with respect to this parameter.

Distance from triviality

Idea: Consider tractable special cases of generally hard problems and introduce parameters that measure the distance from these special cases.

Parameterizing away from triviality: The parameter is the number of “steps” away from an instance that is polynomially solvable.

Vertex Cover

Note: A tree with k edges added has treewidth bounded by k+1.

- Vertex Cover is trivially solvable on trees.
- Let parameter k be the number of edges that need to be deleted to get a tree.
- Since tree has treewidth bounded by k+1, use tree decomposition approach to solve problem.
- Vertex Cover is FPT when parameterized by k.

Clique

Input: A graph G = (V,E) and a nonnegative integer s.

Question: Does G contain a clique, i.e., a complete subgraph, of size s?

- Trivial case: Cluster graphs which are a disjoint union of cliques.
- Solve Clique on graphs which are cluster graphs with k edges added.
- Finding the added k edges is FPT with respect to k.

Clique

How to solve Clique on “almost cluster graph” G after identifying the k edges added and the cluster graph G’?

- If largest clique in G not in G’ then each of its vertices gained in degree by at least 1.
- Hence can be formed by subset of up to 2k vertices (endpoints) of added edges.
- Solve Clique on subgraph of G induced by those vertices.

Power Dominating Set

Task: put PMUs (monitoring devices) on vertices to observe all edges and vertices.

- 1. A PMU in a vertex v observes v and all incident edges and neighbors of v.
- 2. Any vertex that is incident to an observed edge is observed.
- 3. Any edge joining two observed vertices is observed.
- 4. If a vertex is incident to a total of i > 1 edges and if i−1 of these edges are observed, then all iedges are observed.

Power Dominating Set

Input: A graph G = (V,E) and a nonnegative integer k.

Question: Does G have a power dominating set of size at most k, that is, a subset M ⊆ V of vertices such that by placing a PMU in every v ∈ M, all vertices in V are observed?

- Trivial case: Trees. Work bottom-up by putting PMUs at vertices that have at least 2 unobserved children.

Power Dominating Set

Simple algorithm in case of one edge added:

- Graph G has one cycle and a collection of trees Titouching the cycle at their roots.
- Use above algorithm on each tree Ti and when at root ri either:
- Root needs to be in M so we can remove it.
- Root not in M but observed. At most one child unobserved so remove Ti except for root and unobserved child.
- Root still needs to be observed.Exactly one child unobserved so remove Ti except for root and unobserved child.

Power Dominating Set

After these reductions, either

- One edge is observed then break cycle.
- Add vertex to M, try all possibilities, choose optimal.

In case of adding k edges:

- Use the above simple algorithm and in same manner, deal with vertices with degree > 2.
- Power Dominating Set is FPT when parameterized by k.

Tree-like Weighted Set Cover

Input:

- Given a base set S = {s1, s2, . . . , sn} and a tree-like collection C of subsets of S, C = {c1, c2, . . . , cm}.
- We say C is a tree-like subset collection of S if we can organize the subsets in C in an unrooted tree T such that every subset one-to-one corresponds to a node of T and, for each element sj ∈ S, 1 ≤ j ≤ n, all nodes in T corresponding to the subsets containing sj induce a subtree of T .
- Each element of S can be in at most d subsets for a fixed d ≥ 1.
- Each subset in C has a positive real weight w(ci) > 0 for 1 ≤ i ≤ m.
- The weight of a subset collection is the sum of the weights of all subsets in it.

Tree-like Weighted Set Cover

Task: Find C’ ⊆ C with minimum weight which covers all elements in S.

Trivial case: The number of leaves of the subset tree functions as the distance parameter from the path-like case.

- Idea: Divide the tree-like instance into several “independent” path-like instances and solve these instances separately by using thealgorithm that solves this problem when the underlying tree is a path.

Longest Common Subsequence

Input: Given a set of k strings X1,X2, . . . , Xkover an alphabet Σ and a positive integer m.

Question: Is there a string X ∈ Σ∗of length at least m that is a subsequence of Xifor i= 1, . . . , k?

Simple case: Let n be the maximum length of input string and siabe the number of occurrences of a in Xi. Parameterization with k and s.

WLOG, let all strings have length n and s=1.

Longest Common Subsequence

- Construct a directed graph G with n×kvertices, vi,jrepresents the letter at position j of string Xi.
- Add directed edge between vi,j and vi+1,liff the letters in position j of Xi and in position l of Xi+1 are the same.
- Alongest common subsequence one-to-one corresponds to a maximum set of directed paths in G which do not cross each other.

Longest Common Subsequence

- Create “path-compatibility” graph with a vertex Pa for each directed path where a is the position in X1where path starts. Add directed edge (Pa,Pb) if Pa and Pb don’t cross and a < b.
- Use DFS to find longest path hence longest common subsequence.
- General case uses the same idea and is FPT in s and k.

Conclusion

A natural way of parameterizing problems was proposed: the parameter measures some distance from triviality. The approach consists of two fundamental steps. Assume that we study a hard problem X:

- Determine efficiently solvable special cases of X: the triviality.
- Identify useful distance measures from the triviality: the (structure) parameter.

Motivation

- Some problems have some nontrivial lower bound for the optimum value which is exploited to give a trivial fixed-parameter algorithm.
- But the parameter value can be considerably large compared to the input size so the algorithm isn’t necessarily practical.
- Different parameterization: difference between the optimum value and the guaranteed lower (upper) bound.

Introduction

- SNP class: Contains a limited subset of NP based on its logical characterization in terms of graph-theoreticalproperties.
- MAX SNP: The class of problems having constant-factor approximation algorithms, but no approximation schemes unless P=NP.
- All optimization problems in MAX SNP have a non-trivial lower bound for the optimum value.

Parameterized above or below guaranteed values

Let Q1 and Q2 be two optimization (maximization or minimization) problems. We say that Q1L-reduces to Q2 if there exist polynomial-time computable functions f , g, and constants α,β >0 such that for each instance I1 of Q1:

1. f (I1) = I2 is an instance of Q2, such that opt(I2) ≤α · opt(I1).

2. Given any solution y2 of I2, g maps (I2, y2) to a solution y1 of I1 such that |V(I1, y1) − opt(I1)| ≤β · |V(I2, y2) − opt(I2)| (V is a polynomial computable function)

Parameterized above or below guaranteed values

Need to show ∃ broad class of NPO problems with non-trivial lower or upper bound on the optimum solution size.

- MAX 3-SAT ϵ MAX SNP and let Q ϵ MAX SNP, so use L-reduction from Q to MAX 3-SAT.
- For an instance I of Q , f(x) is an instance of MAX 3-SAT such that opt(f(x)) ≤ α · opt(x).
- If f(x) is a formula with m clauses, then ⌈≤ opt(f(x)) and therefore opt(x) is bounded below by . This proves that each instance x of Q has a lower bound (which depends on problem to which it is reduced to).

Parameterized above or below guaranteed values

Need to show that the above or below guarantee question with respect to these bounds is FPT.

Above-guarantee parameterized version L of MAX 3-SAT:

- L = { (f,k): f is a MAX 3-SAT instance and ∃ an assignment satisfying at least k+clauses of the formula f }.
- This problem is FPT with respect to parameter k.

Parameterized above or below guaranteed values

Theorem 1: Let Q be a maximization problem in MAX SNP and (f, g,α,β) an L-reduction from Q to MAX 3-SAT. For an instance x of Q , let lowxrepresent the MAX 3-SAT-lowerbound of x. Then the following problem is in FPT:

LQ= { (x,k): x is an instance of Q and opt(x) ≥ lowx+k }

Parameterized above or below guaranteed values

- There exists an FPT algorithm A for the above-guarantee parameterized version of MAX 3-SAT.
- Consider an instance (x,k) of LQ. Then f(x) is an instance of MAX 3-SAT. Let f(x) have m clauses. Then the guaranteed lower bound for the instance x of Q , lowx= , and opt(f(x)) ≤α · opt(x).
- Apply algorithm A on input (f(x),kα):

Parameterized above or below guaranteed values

- If A outputs yes, then opt(f(x)) ≥+kα, implying opt(x) ≥+k = lowx+k. Thus (x,k) ∈ LQ.
- If A outputs no, then ⌈≤ opt(f(x)) < ⌈+ kα . Apply algorithm A on the inputs (f(x),1), (f(x),2), . . . , (f(x),kα), one by one, to obtain opt(f(x)). Let c’= opt(f(x)).
- Use function g of the L-reduction to obtain a solution to x with cost c. So we have |c −opt(x)| ≤β · |c’−opt(f(x))| (by reduction)
- So c = opt(x). Compare c with lowx+kto check whether (x,k) ∈ LQ.
- This is an FPT algorithm for LQ.

Tight lower and upper bounds

- Tight lower bound:Let Q = {F, S, V, opt} be an NP-optimization problem and let f :N→N. We say that f is a tight lower bound for Q if the following conditions hold:
- 1. f(|I|) ≤opt(I) for all I ∈ F.
- 2. There exists an infinite family of instances F’ ⊆F such that opt(I) = f(|I|) for all I ∈F’.

Tight lower and upper bounds

Example:

Max Planar Independent Set

Instance: A planar graph G with n vertices and m edges.

Question: Find a maximum independent set of G.

Bounds: tlb= ; tub =n.

A planar graph is 4-colorable, and in any valid 4-coloring of the graph, the vertices that get the same color form an independent set. By the pigeonhole principle, there exists an independent set of size at least . A disjoint set of K4’s can be use to construct arbitrary sized instances with independence number exactly .

Tight lower and upper bounds

Min Dominating Set-B

Instance: A graph G with n vertices such that the degree of each vertex is bounded by B.

Question: Find a minimum dominating set of G.

Tight lower and upper bounds

Min Dominating Set-B

Instance: A graph G with n vertices such that the degree of each vertex is bounded by B.

Question: Find a minimum dominating set of G.

Bounds: tlb=; tub =n.

Observe that any vertex can dominate at most B + 1 vertices (including itself) and thus any dominating set has size at least . A set of r disjoint copies of KB+1 has a minimum dominating set of size exactly r =. The upper bound of n is met by the class of empty graphs.

Hard above or below guarantee

Bounded Degree Min Spanning Tree

Input: A connected graph G = (V , E) with edge costs w : E →Z+and nonnegative integers c and k.

Parameter: The integer k.

Question: Does there exist a spanning subgraphT of total edge-weight at most k such that each vertex in T has degree at most c?

This problem is FPT with respect to parameter k.

Hard above or below guarantee

- The total weight of any spanning tree is at least n −1.
- The above guarantee version (Does G have a spanning tree with total weight at most n − 1 + k and vertex degrees bounded by c?) is not fixed-parameter tractable unless P = NP as the case c = 2 and k = 0 reduces to solving the Hamiltonian Path problem.

Parameterizing sufficiently above or below guaranteed values

Given an NP-maximization problem Q with a tight lower and a tight upper bound, what is the parameterized complexity of the following questions?

Qa,max(ϵ) = { (I,k): max(I) ≥tlb(I)+ϵ · |I| +k },

Qb,max(ϵ) = { (I,k): max(I) ≥tub(I)−ϵ · |I| −k },

|I| denotes the input size, ϵis some fixed positive rational, k is the parameter and a and b denote, respectively, the above and below-guarantee version of the problem.

Parameterizing sufficiently above or below guaranteed values

Overview of the proof for maximization problems:

Assume that for some ϵin a specified range, Qa,max(ϵ) is indeed in FPT.

Consider an instance (I, s) of the underlying decision version of Q .

Here is a P-time procedure for deciding it:

- If s ≤tlb, then the answer is trivially yes.

Parameterizing sufficiently above or below guaranteed values

- If s lies between tlb and tlb + ϵ· |I|, then “add” a gadget of suitable size corresponding to the tub, to obtain an equivalent instance (I’, s’).
- This increases the input size, but since we are adding a gadget whose optimum value matches the upper bound, the increase in the optimum value of I is more than proportional, so that now s’exceeds tlb+ϵ· |I’| and we handle this next case.

Parameterizing sufficiently above or below guaranteed values

- If s already exceeds tlb + ϵ· |I|, then “add” a gadget of suitable size corresponding to the tlb, to obtain an equivalent instance (I’, s’).
- This increases the input size faster than it boosts the optimum value of I , so that now s’exceeds tlb+ϵ· |I’| by only a constant, say c1.
- Use the hypothesized FPT algorithm for Qa,max(ϵ) with input (I’, c1) to correctly decide the original question.

Open Problems

- Is there a characterization for the class of problems for which the above or below-guarantee question with respect to a tight lower or upper bound is in FPT (or W[1]-hard)?
- Planar Independent Set: Given an n-vertex planar graph and an integer parameter k, does G have an independent set of size at least +k?

Open Problems

- Perfect Vertex Deletion: Given a graph G on n vertices and m edges and an integer k, does there exist a vertex-induced subgraph on n−k vertices that is perfect? A similar question can be framed for the edge version.

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