CSCI 4260MATH 4150

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CSCI 4260-MATH 4150

• Lecture 20
• Approximation Algorithms for NP-Complete Problems

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Why approximation?

• Many optimization problems on graphs are
  NP-Complete.
• This means that, in general, we can not solve
  these problems in polynomial time.
• Approximation algorithms provide one way of to
  deal with such problems without sacrificing too
  much from optimality

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Definition

• Algorithm A is said to be a c-approximation
  algorithm for a minimization problem
• A runs in polynomial time
• And
• For all possible inputs ??

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Problem 1 Vertex Cover

• Input G (V,E)
• Find a minimum cardinality S ?V such that every
  edge in E is adjacent to a vertex in S.
• Vertex Cover is NP-Complete

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2-approximation algorithm for vertex cover

• S ? empty set
• Repeat
• Pick an edge (u,v)
• Add u and v to S
• Remove every edge adjacent to either u or v from
  G
• Until no more edges remain

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Why is this a 2-approximation algorithm?

• Clearly it runs in polynomial time
• Note that we output a maximal matching
• We know that the size of any matching M is less
  than or equal to the size of any vertex cover.
• Let OPT be the cardinality of the optimal
  vertex-cover for our input.
• What is the number of vertices in S?

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Is this analysis tight?

• That is, is there an instance where our algorithm
  outputs as many as 2xOPT vertices
• K(n,n)

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Problem 2 Dominating Set

• Input G (V,E)
• Find a minimum cardinality S ?V such that every
  vertex in V is adjacent to a vertex in S.
• Dominating Set is NP-Complete

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Recall the Set-cover problem

• Given a set X a,b,c,d,e,f,g
• And a collection of subsets of X
• S S1,S2,S3,S4, S5, S6
• S1 a, b, c
• S2 c, d, e
• S3 c, e, f, g
• S4 a, f, g
• S5 f
• S6 g
• Cover a collection of subsets from S whose union
  gives X
• Example S1, S2, S5, S6 is a cover of S
• Minimum Set Cover Problem Find a cover of
  minimum cardinality (with minimum number of
  subsets)

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From Dominating Set to Set Cover

• We can reduce the dominating set problem to a set
  cover problem as follows
• Base set X ? Vertex Set V
• Subsets ? For each vertex v, add v ?? N(v) as
  a subset in the set system S
• Every set cover of (X,R) is a dominating set of G
  and vice versa

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Greedy Set Cover Algorithm

• Repeat
• Pick a subset that covers most number of
  uncovered elements and add it to the solution.
• Until all elements of X are covered

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Analysis

• Let SOL be our solution and OPT be the optimal
  solution
• The number of subsets in SOL is equal to the
  number of iterations

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Analysis

• At every state we pick the element that covers
  most uncovered elements.
• The optimal algorithm covers all elements using
  OPT elements
• Therefore at every stage there must be a subset
  that covers at least
• (total number of uncovered elements)/OPT
• Uncovered elements

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Analysis (cont)

• Let r(i) denote the number of uncovered elements
  after ith iteration

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Analysis

• The last inequality comes from (1-x) ? e-x

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Analysis

• Therefore, no more uncovered elements remain
  after OPT ln(n) iterations.
• Hence, greedy gives a ln(n) approximation for the
  set-cover problem

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Is this analysis tight?

• Is there an instance where greedy is log(n) times
  worse than OPT?

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Problem 3 Metric k-center

• Input G(V,E) complete undirected graph with
  edge costs satisfying the triangle inequality
• For any set S ? V and a vertex v, define
  connect(v,S) cost of the cheapest edge from v to
  a vertex in S
• Output S ? V with S k that minimizes
• max_v connect(v,S)

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A 2-approximation algorithm

• Suppose that we knew the cost achieved by the
  optimal solution.
• This is no big deal because we know that it will
  be equal to the weight of some edge and we can
  try all possibilities.
• Sort the edges in nondecreasing coste1, e2, ,
  em
• Let Gi be the unweighted graph obtained by
  deleting all edges that are costlier than ei from
  G

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Crucial observation

• If OPT cost(ei) for some i
• Then
• Gi must have a dominating set of size k
• Of course, we dont know how to compute
  dominating sets optimally
• But we can compute a lower bound on OPT and
  utilize this lower bound to test Gi

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Lemma

• Given a graph H, let H2 denote a graph
  containing an edge (u,v) whenever H has a path of
  length 2 between u and v
• Given a graph H, let I be an independent set in
  H2. Then
• I ? dom(H) size of the minimum dominating set
  in H

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Proof

• Let D be the dominating set in H
• H contains D stars
• Each of these stars will be a clique in H2
• Any independent set can pick at most one vertex
  from each clique

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K-center algorithm

• Construct G12, G22, , Gm2
• Compute a maximal independent set Mi for each
  Gi2
• Find the smallest index i such thatMi ? k
• Call this index j
• Return Mj

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Analysis

• cost(ej) ? OPT
• ej index chosen by the algorithm
• OPT Cost incurred by the optimum k-center
  solution
• Proof
• For all i lt j, we have Mi gt k
• But then dom(Gi) ? Mi gt k
• So OPT is more than cost(ei)

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2 approximation proof

• First observe that every maximal independent set
  is also a dominating set
• Suppose not. Then there must be a non-dominated
  vertex. This vertex can be added to the
  independent set contradicting maximality
• So, our independent set in Gi2 is also a
  dominating set in Gi2
• But not necessarily in Gi. However,

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But what is the cost?

• Suppose u is a vertex in Mj and v is adjacent to
  u
• distance(u,v) lt 2cost(ej)
• by triangle inequality

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What have we achieved

• OPT achieves cost(ej) using k centers
• We achieve 2cost(ej) using k centers

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Is the analysis tight?

• red edges have cost 1, black edges cost 2
• Optimal solution for k 1 is (cost) 1