P-Center & The Power of Graphs A part of the facility location problem set
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Transcript of P-Center & The Power of Graphs A part of the facility location problem set
P-Center & The Power of Graphs
A part of the facility location problem set
By Kiril Yershov and Alla Segal
For Geometric Optimizations courseFall 2010
The P-Center problemGiven n cities and the distances
between all pairs of cities, we aim to choose p cities as centers so that the furthest distance of a city from a center is minimized.
A problem from the facility location problems set.
What is facility locating?
‘GOOD’
SERVICE
Client needs this kind of service.
Facility provides some kind of
service.
Location problems in most general form can be stated as follows: a set of clients originates demands for some kind
of goods or services. the demands of the customers must be supplied
by one or more facilities. the decision process must establish where to
locate the facilities. issues like cost reduction, demand capture,
equitable service supply, fast response time etc drive the selection of facility placement.
The basic elements of location models:
a universe, U, from which a set C of client input positions is selected,
a distance metric, d: U x U → R+, defined over the universe R+,
an integer, p ≥ 1, denoting the number of facilities to be located, and
an optimization function g that takes as input C a set of client positions and p facility positions and returns a function of their distances as measured by the metric d.
Problem statement
• Select a set F of p facility positions in universe U that minimizes g(F,C).
Models of the universe of clients and facilities
• Continuous space
Universe is defined as a region, such that clients and facilities may be placed anywhere within the continuum, and the number of possible locations is uncountably infinite.
• Discrete space
Universe is defined by a discrete set of predefined positions.• Network space
Universe is defined by an undirected weighted graph. Client positions are given by the vertices. Facilities may be located anywhere on the graph.
Models of the universe of clients and facilities
Where should the radio tower be located?
• Tower may be positioned anywhere – (continuous)
• Tower may be positioned in some available slots – (discrete)
• Tower may be located on the roadside – (network)
Distance metric
Distance metric between two elements of the universe further differentiates between specific problems• Euclidean, Manhattan, etc.• Network distance
P-Centers:Optimization function is maximum
The objective function is the most significant characterization of a facility location problem.• p-center problem
Given a universe U, a set of points C, a metric d, and a positive integer p, a p-center of C is a set of p points F of the universe U that minimizes
maxjεC {miniεF dij}.
p-centerP=2
p-centerP=2
Applications of p-center (1/2)
Find appropriate location of branch offices.◦ It could be cost-effective if the branch offices place
as close as possible to any offices.
3-center
Euclidean p-center
ReminderNP – “nondeterministic polynomial time”NP-Problem is a problem that its possible
solution can be verified in polynomial time.
NP-Hard Problem is at least as hard as the hardest problems in NP.
NP-Complete problem is both NP and NP-Hard.
Summary of known complexities
Time complexities of algorithmic solutions to the Euclidean p-center problem
• Drezner 1984, Hoffmann 2005 (arbitrary p and d = 1)• Megiddo 1983 (p=1, d=2), Linear programming.• Agarwal et al 1993, Chazelle et al. 1995 (p = 1 and d fixed)• Chan 1999 (p=2, d=2)• Agarwal and Sharir 1998 (p=2, d arbitrary)• Agarwal and Procopuic 1998 (p fixed, d arbitrary)
P-Center is NP-HardKariv and Hakimi proved in ‘79 that
p-center problem is NP-Hard.
We will show a stronger claim: P-center best possible approximation is of factor 2.
ApproximationPTAS: a C-approximation, where C=1+e
for any e>0We saw some problems with PTAS.
Does every problem has a PTAS?
Best Possible ApproximationA polynomial c-approximation algorithm
is called “Best Possible” if there is no algorithm with a better approximation value.
Formally: c-approximation is best possible if the problem of approximation within c-e is NP-Complete for any e>0, but that limit can be reached in polynomial time.
Best Possible ApproximationSuch best possible results were found
only for bottleneck problems.
Bottleneck ProblemsThe solution has some value that needs
to be minimized or maximized.The value of the solution is the weight of
one of the edges of the graph.We can add one smallest edge at a time
and examine the resulting graph for the required condition
P-Center: Factor 2 is best possibleLEMMA: The problem of approximating
P-Center (with triangle inequality) within a factor of (2-e) is NP-Complete for any e>0.
Proof: By reduction from the dominating set problem.
Factor 2 is best possible - ProofDominating set problem:
◦To determine whether in a given graph there is a set of vertices DS of size ≤ P that “dominates” all vertices, i.e. such that all vertices are in DS or adjacent to DS.
◦ It is well known and proven that Dominating Set is NP-Hard.
Factor 2 is best possible - ProofReduction:
◦Given a Dominating set problem G=(V,E), we construct a complete graph with the following weights: w(e) = 1 if e is from E w(e) = 2 if e is not from E
◦ In this complete graph the existence of p-center is equivalent to the existence of a dominating set of size ≤ p in G.
INTERMISSION
Metric p-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 cheapest edge from v to a
vertex in S
Output: S V with |S| = p that minimizes max_v{connect(v,S)}
u v
x w
),(),(),(,,
wvcvucwucVwvu
Triangle InequalityCost Function Satisfies
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A 2-approximation algorithmThe value of the solution is the weight of one of
the edges of the graph.
Sort the edges in nondecreasing coste1, e2, …, em
Approach
Let Gi be the subgraph obtained by deleting all edges that are costlier than ei from G
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Dominating set
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Input
Equivalence between…p-center
◦ Any weight of edge between a vertex and some chosen vertex is not more than OPT.
Dominating set of Gi
◦ Any vertex is connected to some chosen vertex by a edge with weight less or equal than OPT.
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cost (ei*) = OPT.
Crucial observationIf OPT = cost(ei) for some iThen
◦Gi must have a dominating set of size pOf course, we don’t know how to
compute dominating sets optimallyBut we can compute a lower bound on
OPT and utilize this lower bound to test Gi
Power of GraphsA major technique for finding best possible
approximations for bottleneck problems.Given a graph G=(V,E)the square graph G2=(V,E2)E2 is the set of all edges between vertices
that have a path of length 1 or 2 between them.
Weight is set accordingly.Gt is similar with edges between vertices
that have a path of length up to t.
LemmaGiven a graph H, let H^2 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 H^2. Then
|I| dom(H) = size of the minimum dominating set in H
Square of a graph
H H2
H’ H’2
H’’ H’’2
ProofLet D be the dominating set in HH contains |D| starsEach of these stars will
be a clique in H^2Any independent set can pick at most one
vertex from each clique
H D star clique
p-center algorithmConstruct G1^2, G2^2, …, Gm^2Compute a maximal independent set Mi
for each Gi^2Find the smallest index i such that
|Mi| pCall this index jReturn Mj
Algorithm
Input: graph G, pos. int. p.
1. Construct G12, …,
Gm2.
2. Find a maximal independent set Mi
for each Gi2.
3. Find the smallest index j s.t. |Mi| ≦ p.
4. Return Mj.G
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AlgorithmInput: graph G, pos.
int. p.1. Construct G1
2, …, Gm
2.2. Find a maximal
independent set Mi
for each Gi2.
3. Find the smallest index j s.t. |Mj| ≦ p.
4. Return Mj. G
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p=2
Analysiscost(ej) OPT
◦ej : index chosen by the algorithm◦OPT: Cost incurred by the optimum p-center
solutionProof:
◦For all i < j, we have |Mi| > p◦But then dom(Gi) |Mi| > p◦So OPT is more than cost(ei)
2 approximation proofFirst 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 Gi^2 is also a dominating set in Gi^2
But not necessarily in Gi. However,
2 approximation proofDominating set = set of centers of stars in Gi
2.
Gi2 Gi
Some edgesdoes notappear in Gi.
The edgeexists in Gi
2.
In Gi, these twovertices are connected with a path length=2.
Gi2
But what is the cost?Suppose u is a vertex in Mj and v is
adjacent to udistance(u,v) <= 2cost(ej) <= 2*OPT
◦by triangle inequality
Conclusion: Cost of our solution is at most twice the cost of optimal solution.
Weighted p-center
p =2
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W = 30.
p-center Weighted p-center
Weightson vertices
Weighted p-centerWeighted p-center is a generalization of
p-center.
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Weighted p-center
Not exactly the same p-center
◦# vertices in an independent set in H2 # vertices ≦in a minimal dominating set of H
Weighted p-center◦Sum of weights of vertices in an independent set in
H2 Sum of weights of vertices in a minimal ≦dominating set of H
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HH2
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Adjacent vertices with min. weights(u): vertex adjacent to u in H with the
minimum weight.◦u can be s(u).◦Not in H2.
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c d
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s(a)=a,
s(b)=a,
s(c)=c,
s(d)=c.
H
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c d
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H2
Now the approach is similarCan be proved that weight(I) <= weight(D).
Weight(I) = {s(u)|u in I}, I is for H^2 Weight(D) = sum of weights of nodes in D. D is for H
1. Construct G12, …, Gm
2. 2. Compute a maximal independent set, Mi , in each
graph Gi2.
3. Si={si(u): u belongs to Mi}4. Find the minimum index j s.t. w(Si) <= W.5. Return Sj.GIVES 3 APPROXIMATION
Questions?