Approximation Algorithms
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Motivation and Definitions
TSP
Vertex Cover
Scheduling
Motivation
• Consider an NP-hard optimization problem like TSP
– Input: n cities, pairwise distances d(i,j)
– Task: find a tour of minimum length
• Given this is NP-hard, we are unlikely to find an optimal
solution in polynomial time.
• What are our options?
– Give up
– Try to build the fastest possible algorithm that returns an optimal
solution
– Find a polynomial-time algorithm that returns good solutions that
are “approximately” optimal.
• This is the option we focus on today.
Definitions
• For any input instance I,
– let OPT(I) denote both the optimal solution and the value of the
optimal solution
– let A(I) denote both an approximate solution and the value of the
approximate solution generated by running polynomial-time
algorithm A.
• So what do we mean by a good solution that is
“approximately” optimal (for a minimization problem)?
• Additive approximation
– For any input I, A(I) ≤ OPT(I) + c for some constant c.
• Multiplicative approximation
– For any input I, A(I) ≤ c OPT(I) for some constant c.
Are both approximations possible?
• Consider the TSP problem
– Suppose c = 3.
– Additive: if OPT(I) = 10, then APPROX(I) ≤ 13. If
OPT(I2) = 100, then APPROX(I) ≤ 103.
– Multiplicative: if OPT(I) = 10, then APPROX(I) ≤ 30.
If OPT(I2) = 100, then APPROX(I) ≤ 300.
– Argue that at least one of these approximations is
impossible unless we can solve TSP optimally in
polynomial time.
– Hint: think about scaling the integers in your input.
– Is the other approximation possible for TSP?
Definitions II
• Because we typically cannot achieve additive
approximation due to scaling, we try to get multiplicative
approximation
• For a minimization problem, an algorithm is a capproximation algorithm if for all inputs I,
– A(I) ≤ c OPT(I)
• For a maximization problem, an algorithm is a capproximation algorithm if for all inputs I,
– A(I) ≥ 1/c OPT(I)
• Approximation goal: for an NP-hard optimization problem,
find a polynomial time algorithm that is a c-approximation
algorithm for the smallest c possible.
Example Problems
• TSP
– general TSP
– metric TSP
• Vertex cover
• Scheduling
TSP
• We showed earlier there can be no additive
approximation algorithm unless P=NP
• Show now that for any constant c, there can be no
c-approximation algorithm for TSP unless P=NP
• Hint: show that if there is a c-approximation
algorithm for some constant c, then Hamiltonian
circuit can be solved in polynomial time
Metric TSP
• In the metric TSP, the city distances must satisfy a
triangle inequality.
– For any 3 cities i, j, k, it must be the case that d(i,j) ≤
d(i,k) + d(k,j)
• Observe how our previous argument violates this
triangle inequality
• Show how to come up with a c-approximation
algorithm for TSP based on a minimum spanning
tree
• What value of c can you come up with?
Cristofides Improvement
• Start with MST T as before
• Identify nodes with odd degree
• Find a minimum weight matching M on
these nodes
• Now compute an Euler tour of the graph of
T union M (with shortcuts to prevent
visiting an edge twice)
• This solution is guaranteed to have length at
most 3/2.
Vertex Cover
• Input: Graph G = (V,E)
• Task: Find C subset of V of minimum size such
that for each edge (u,v) in E, either u is in C or v is
in C
• This can be as bad as Θ(lg n)
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Hint: make a bipartite graph
Make x nodes in one set all have the same max degree
May y nodes in other set have varying degrees
Greedy picks all the y nodes (with bad tie-breaking),
optimal picks all the x nodes
Vertex Cover
• Better solution
– Find a maximal matching M in G
• Matching: set of edges in G that do not share any
common vertices
• Maximal matching: No edges can be added to the
matching to produce a larger matching
– Return as C all the nodes in edges in M
• What approximation ratio does this
guarantee and why?
Scheduling
• Input:
– Set of m identical machines
– Set of n jobs with length pi
• Task: Assign the n jobs to the m machines with the goal of
minimizing the maximum total length of jobs assigned to
any one machine
• Greedy strategy
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Take the jobs one by one in any order
Assign the current job to the currently least loaded machine
What approximation bound can be derived?
What if we sort the jobs first? Should we do longest or shortest
first?
Scheduling
• Greedy strategy
– Take the jobs one by one in any order
– Assign the current job to the currently least
loaded machine
• What approximation bound can be derived?
– Lower bounds on OPT
• What bounds can we derive on the best schedule?
• How can we relate this algorithm’s schedule to
OPT’s?