LOCALIZACIÓN DE ESTRUCTURAS EN GRAFOSelena/web/PresentacionesBaeza/PDFsPresentacione… · RED...

RED ESPAÑOLA LOCALIZACIÓN 2007 LOCALIZACIÓN DE ESTRUCTURAS EN GRAFOS - p. 1

LOCALIZACIÓN DE ESTRUCTURAS ENGRAFOS

J. PuertoUniversidad de Sevilla.

[email protected]

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Part 0: Introduction

Part II: The model

Part III: Obtaining the

CENTER-MEDIAN Pareto set

Part IV: CENTER-MEDIAN

Pareto with length constraint

Part V: RANGE

(MAXIMUM-MINIMUM) PATH

PROBLEM


???

Gracias, Gracias, Gracias, Gracias, Gracias,...

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Part II: The model





Part V: RANGE


PROBLEM


???


Elena por tus esfuerzos

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Part II: The model





Part V: RANGE


PROBLEM


???


Elena por tus esfuerzos

Alfredo por arriesgarte

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● In this talk:

● Scope

● The papers

● Model of Computation

● Time analysis

● The problem: Naïve analysis

● Literature

● Resume of results

● Results in this talk:


● Complexities

● Complexities (Cont.)

Part II: The model





Part V: RANGE


PROBLEM



● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


In this talk:

The goals:■ Framework space: Tree network■ Locating extensive facilities (paths, trees), (Boffey & Mesa, 1996)■ Objective functions: Center, Median, Hurwicz, Range■ Pareto set of outcomes and some related optimization proble ms

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


In this talk:


A path P is assigned a pair, (f1(P ), f2(P )), which are the f1 and f2

functions of distances from vertices. The total cost function,f(f1(P ), f2(P )) is assumed to be monotone in its two arguments.

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


In this talk:


A path P is assigned a pair, (f1(P ), f2(P )), which are the f1 and f2

functions of distances from vertices. The total cost function,f(f1(P ), f2(P )) is assumed to be monotone in its two arguments.

The goal is then to minimize f over PL

PL is the set of all the paths with a bounded length L.

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


Scope

■ Good news: Highly efficient algorithms.

■ Bad news (??): The scope is limited to

some classes of problems.

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


Scope

■ Good news: Highly efficient algorithms.

■ Bad news (??): The scope is limited to

some classes of problems.

Minimax problems with an additional constraint

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


The papers

■ J. Puerto, A. Tamir and D. Perez-Brito. The centdian subtree ontree networks, Discrete Applied Mathematics 118, (2002),263-278.

■ J. Puerto, A. Tamir. Locating tree-shaped facilities using theordered median objective. Mathematical Programming , vol.102: 313-338 (2004).

■ J. Puerto, A. Tamir, J.A. Mesa, A.M. Rodríguez-Chía.Conditional location of path and tree shaped facilites on tr ees.J. of Algorithms , vol. 38: 31-44 (2005).

■ J. Puerto, A. Tamir, A.M. Rodríguez-Chia, D. Perez-Brito. T heBi-Criteria Doubly Weighted Center-Median Path Problem on ATree, Networks , 47(4), 237-247 (2006).

■ J. Puerto, F. Ricca, A. Scozzari. Path location problems und erequity measures. Second revision in Discrete AppliedMathematics .

■ J. Puerto, F. Ricca, A. Scozzari. The continuous-and-discr etePath location problems on trees. Discrete Applied MathematicsTo appears .

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


Model of Computation

Our goal is to find efficient algorithms for some problems.What does “efficient algorithm” mean?

In our framework we looks for algorithms that run as fast as possible.

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● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM





The computation model must be computer independent so that timeis referred to elemental operations.

■ Access to memory is an operation (reading, writing)■ Real arithmetic operations (sum, substraction, multiplication,

division) and comparisons are elemental operations.■ Some elemental function evaluations (trigonometric, n-th radical ...)

can be considered elemental operations.

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM





The computation model must be computer independent so that timeis referred to elemental operations.

■ Access to memory is an operation (reading, writing)■ Real arithmetic operations (sum, substraction, multiplication,

division) and comparisons are elemental operations.■ Some elemental function evaluations (trigonometric, n-th radical ...)

can be considered elemental operations.

We are only interested in:■ Worst case behavior (there are other ways to analyze problem

complexity: average, best case ...)■ Asymptotic analysis when the data size goes to infinity: O(|size|)

notation.

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


Time analysis

Assume that one elemental operation takes 10−6 seconds.

100 1000 10000 100000

O(n) 10−4s. 10−3s. 10−2 s. 10−1 s.

O(n log n) 2×10−4s. 3×10−3s. 4×10−2 s. 0.5 s.

O(n2) 10−2 s. 1 s. 1 m. 40 s. 2.77 h.

O(n3) 1 s. ≈ 17 m. 11.5 d. 31,7 y.

O(n4) 1m. 40 s. 11.5 d. 31.7 y. 3 × 106 y.

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


Time analysis

Our goal is to prove that “ALL” the

nondominated points in the planar set

{(f1(P ), f2(P )) : P ∈ PL} are at most O(n).

Moreover, we compute them all in O(n log n).

Then, apply the results to several importantproblems that are solved in linear time. (After thepreprocessing)

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


The problem: Naïve analysis

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● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


The problem: Naïve analysis

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Quadratic number of paths

Evaluating f1 and f2 in our case is linear per path O(n3)

Pairwise comparison of values O(n4)

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


Literature

Subquadratic complexity for weighted median andweighted center path location models (Alstrup et al. 1997;

Becker et al. 2001, 2002, 2003; Minieka and Patel 1983; Morga n and

Slater 1980; Peng and Lo 1996; Peng et al. 1993; Shioura and

Shigeno 1997; Shioura and Uno 2002; Tamir et al. 2001; Wang

2000, 2002.)

Pareto set for UNWEIGHTED center-median in O(n log n)

(Averbakh and Berman 1999).

Weighted median or center subject to length constraints inO(n log2 n) (Becker et al. 2002, Becker et al. 2004).

Path variance without length constraint O(n2 log2 n)

(Caceres et al. 2004)

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


Literature

■ Other very interesting problems: Subtree location .

■ One open problem: Path location on trees with the ordered

median function

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


Resume of results

Unweighted modelsPROBLEM AUTHORS COMPLEXITY

Pareto (MAX, SUM) Aberbakh & Berman (1999) O(n log n)

min SUM

MAX ≤ αAberbakh & Berman (1999) O(n)

min MAX

SUM ≤ αAberbakh & Berman (1999) O(n)

Weighted modelsmin SUM

MAX ≤ α

L(P ) ≤ l

Becker et al. (2002-4) O(n log2 n)

min MAX

SUM ≤ α

L(P ) ≤ l

Becker et al. (2002-4) O(n log2 n)

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


Results in this talk:

Double weighted modelsPROBLEM COMPLEXITY

Pareto (MAX, SUM) O(n log n)

min SUM

MAX ≤ αO(n log n)

min MAX

SUM ≤ αO(n log n)

min SUM

MAX ≤ α

L(P ) ≤ l

O(n log n)

min MAX

SUM ≤ α

L(P ) ≤ l

O(n log n)

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


Results in this talk:

Range-type Problems Hurwicz-type ProblemsP1 min R(P ) O(n) P4 min H(P ) O(n)

P2min MAX(P )

s.t. MIN(P ) ≥ γO(n) P5

min MAX(P )

s.t. MIN(P ) ≤ γO(n)

P3max MIN(P )

s.t. MAX(P ) ≤ γO(n) P6

min MIN(P )

s.t. MAX(P ) ≤ γO(n)

R(P) = MAX(P) − MIN(P) (1)

H(P) = αMAX(P) + (1− α)MIN(P). (2)

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


Complexities

P1-P6 are NP-hard on general networks.

Problems P2 and P5 contain as a special case the problem offinding a minimax path (Hakimi et al. 1993).Problem P4 contains as a special case the minimax path whenα = 1.

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


Complexities

P1-P6 are NP-hard on general networks.

Problems P2 and P5 contain as a special case the problem offinding a minimax path (Hakimi et al. 1993).Problem P4 contains as a special case the minimax path whenα = 1.

Problem P1. Given an arbitrary graph G = (V,E) and a nonnegative number R0, decide if there exists a path P such thatR(P ) ≤ R0. We show that the Hamiltonian Path problem can bereduced to this problem. Let |V | = n, and suppose that a lengthequal to one is assigned to each edge e ∈ E. For each vi ∈ V ,i = 1, ..., n, consider two additional vertices vi1 and vi2, andconstruct a new graph G′ = (V ′, E ′) such thatV ′ = V ∪

⋃n

i=1{vi1, vi2} and E ′ = E ∪

⋃n

i=1{(vi, vi1), (vi, vi2)}.

Assume that the edges added to G have length equal to 1/2. SetR0 = 0. It is easy to see that problem P1 has a solution in G′ ifand only if G has a Hamiltonian Path.

● ???


● In this talk:

● Scope

● The papers


● Time analysis


● Literature




● Complexities


Part II: The model





Part V: RANGE


PROBLEM


Complexities (Cont.)

Problem P3: Given two non negative numbers µ0 andM , decide if there exists a path P such that µ(P ) ≥ µ0

and E(P ) ≤ M . We refer to the same reduction asabove. We assign length equal to one to the originaledges e ∈ E, and length equal to M to the new edges.It is easy to see that, by setting µ0 = 2 and M ≥ 2,problem P3 has a solution in G′ if and only if G has aHamiltonian Path.

Finally, Problem P6 also reduces to Hamiltonian Path.

● ???


Part II: The model

● The model: Elements

● Notation

● Representation

● Example

● Our approach

● The set of admissible

MAX(P) values

● The set A∗∗

● Generating A∗∗





Part V: RANGE


PROBLEM


Part II: The model

● ???


Part II: The model


● Notation

● Representation

● Example

● Our approach


MAX(P) values

● The set A∗∗






Part V: RANGE


PROBLEM


The model: Elements

T = (V, E) is an undirected tree network with node set V = {v1, ..., vn}and edge set E = {e2, ..., en}.

Each node vi ∈ V is associated with a pair of nonnegative weights,(ui, wi). ui and wi are the center-weight and median-weight.

For each pair of nodes vi and vj let Pij be the path connecting the pair.We are interested in the following three objective functions:1. Weighted path center problem

MAX(Pij) := maxvk∈V

ukd(vk, Pij).

2. Weighted path median problem

SUM(Pij) :=n∑

k=1

wkd(vk, Pij).

3. The unweigthed minimum

µ(P ) = minvk∈V \Pij

d(u, Pij).

● ???


Part II: The model


● Notation

● Representation

● Example

● Our approach


MAX(P) values

● The set A∗∗






Part V: RANGE


PROBLEM


Notation

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vs

vj

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vq

v1

T s T t

vp

P [vp, vq]: Path Center

S(vj) : Children of vj

N(vj) : Neighbors of vj

Fj : Forest induced by vj

Figure 1: Notation.

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Part II: The model


● Notation

● Representation

● Example

● Our approach


MAX(P) values

● The set A∗∗






Part V: RANGE


PROBLEM


Representation

To simplify the notation we introduce three functions which map eachpath in P∗ to a point in the plane:

ϕ : P∗ −→ R2

Pij −→(MAX(Pij), SUM(Pij)

).

Our goal in this paper is to construct PARETO(P∗), defined as thesubset of all the Pareto points of the planar setϕ(P∗) = {(MAX(Pij), SUM(Pij)) : Pij ∈ P∗}.

We consider a superset Wi, i = 1, 2 that includes the set of thePareto-Optimal paths in the outcome space (MAX(·), µ(·)) along withsome extra paths, and we generate the following representation setsφ(Wi), i = 1, 2:

φ(Wi) = {(MAX(P ), µ(P )) ⊂ R2|P ∈ Wi} i = 1, 2. (3)

We note that in general a Pareto planar point can correspond toseveral paths in P∗.

For each Pareto point we will find only one path in P∗ which is mappedinto this point. We call such a path a representative path.

● ???


Part II: The model


● Notation

● Representation

● Example

● Our approach


MAX(P) values

● The set A∗∗






Part V: RANGE


PROBLEM


Example

Notice that a Pareto-Optimal path - both for �1 and �2 - does notnecessarily connect two leaves of the tree. An example for �1 isshown in Figure 1.

1

100 100 100 100

100

1 1

1 1

a b

Figure 2: The path P = P (a, b) has MAX(P ) = 101 and µ(P ) = 100.Any path connecting two leaves is dominated by P w.r.t. �1.

● ???


Part II: The model


● Notation

● Representation

● Example

● Our approach


MAX(P) values

● The set A∗∗






Part V: RANGE


PROBLEM


Our approach

1. In all the cases we compute a super set W :

Pareto(P) ⊂ W ⊂ ϕ(P), and |W | = O(n).

2. Remove dominated outcomes constructing the rectilinear lowerenvelope (Kapoor 2000).

● ???


Part II: The model


● Notation

● Representation

● Example

● Our approach


MAX(P) values

● The set A∗∗






Part V: RANGE


PROBLEM


Our approach




HOW TO COMPUTE W?

● ???


Part II: The model


● Notation

● Representation

● Example

● Our approach


MAX(P) values

● The set A∗∗






Part V: RANGE


PROBLEM


Our approach




HOW TO COMPUTE W?

Every path is identified by the value of the node that attains themaximum distance “critical node”

● ???


Part II: The model


● Notation

● Representation

● Example

● Our approach


MAX(P) values

● The set A∗∗






Part V: RANGE


PROBLEM


The set of admissible MAX(P ) values

Define Ps, Pt and P byPs = {Pij ∈ P∗ such thatPij ⊆ T s}

Pt = {Pij ∈ P∗ such thatPij ⊆ T t}

P = {Pij ∈ P∗ such thatvi ∈ T s, vj ∈ T t and vi, vj 6= v1}.

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● ???


Part II: The model


● Notation

● Representation

● Example

● Our approach


MAX(P) values

● The set A∗∗






Part V: RANGE


PROBLEM






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CASE 1

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● ???


Part II: The model


● Notation

● Representation

● Example

● Our approach


MAX(P) values

● The set A∗∗






Part V: RANGE


PROBLEM






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CASE 2

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● ???


Part II: The model


● Notation

● Representation

● Example

● Our approach


MAX(P) values

● The set A∗∗






Part V: RANGE


PROBLEM






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CASE 2

● ???


Part II: The model


● Notation

● Representation

● Example

● Our approach


MAX(P) values

● The set A∗∗






Part V: RANGE


PROBLEM


The set A∗∗

The set of admissible values can be further refined since the uniquecritical nodes for paths in P̄ are those in P [vp, vq].Thus, if for vk 6= v1 set

γk = maxvb∈Va(k)

ubd(vb, vk),

The only admissible values for MAX(Pij) if Pij ∈ P̄ are:

Ast = {γk : vk ∈ P [vp, vq]} ∪ {ǫs, ǫt}.

● ???


Part II: The model


● Notation

● Representation

● Example

● Our approach


MAX(P) values

● The set A∗∗






Part V: RANGE


PROBLEM


The set A∗∗


γk = maxvb∈Va(k)

ubd(vb, vk),



CONSEQUENCES1) we do not have to consider critical nodes of paths Pij ∈ P outsideP [vp, vq].2) we only need to compute maximum weighted distances to nodes inP [vp, vq] from nodes in their descendant subtrees.

● ???


Part II: The model


● Notation

● Representation

● Example

● Our approach


MAX(P) values

● The set A∗∗






Part V: RANGE


PROBLEM


The set A∗∗


γk = maxvb∈Va(k)

ubd(vb, vk),



CONSEQUENCES1) we do not have to consider critical nodes of paths Pij ∈ P outsideP [vp, vq].2) we only need to compute maximum weighted distances to nodes inP [vp, vq] from nodes in their descendant subtrees.

A∗∗ = {γk : vk ∈ P[vp,vq]} ∪ {ǫs, ǫt} ∪ {δk : vk ∈ V}.

● ???


Part II: The model


● Notation

● Representation

● Example

● Our approach


MAX(P) values

● The set A∗∗






Part V: RANGE


PROBLEM


Generating A∗∗

1. δk is the value of the weighted (point) 1-center function f(x),evaluated at vk.Using the centroid decomposition we can compute the weighted1-center objective value at all nodes of a tree network in O(n log n)time (see Tamir (2004)).

2. Computation of γk : Each node vk ∈ P [vp, v1] is a point, say v′k, onthe real line, where v′p = 0, and v′k = d(vp, vk).For each vb ∈ T s, let vi(b) be the closest node on P [vp, v1] to vb, andlet gb(x) be defined by gb(x) = 0 if x < v′

i(b), and

gb(x) = ub[d(vb, vi(b)) + (x − v′i(b))] when x ≥ v′i(b).

ThenG(x) = max

vb∈T sgb(x).

G(x) piecewise linear and can be generated in O(n log n) time(Hershberger (1989)). Moreover, for each vk ∈ P [vp, v1], vk 6= v1,γk = G(v′k), and γs

1 = G(v′1).

● ???


Part II: The model



● Computing the set W

● The set Pareto(P∗ )

● Solving some related

problems

● Case 1: Computing W1● Case 1: (Continuation)

● Case 2: Computing W2● Case 2 (Continuation)

● Case 2 (Continuation)



Part V: RANGE


PROBLEM


Part III: Obtaining the CENTER-MEDIANPareto set

● ???


Part II: The model






problems






Part V: RANGE


PROBLEM


Computing the set W

For each value a ∈ A∗∗, we will find a path of T which minimizes theweighted sum function among the subset of paths whose weightedcenter function is equal to a.

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● ???


Part II: The model






problems






Part V: RANGE


PROBLEM


Computing the set W


Case 1. The value of a is in the set {δk : vk ∈ V }, and denote by W 1

the corresponding elements of W . From the above discussion thecorresponding paths are contained either in Ps or Pt. In this case, acritical node of a path is the closest node, vk, of this path to theweighted 1-center point, c̄u = v1.A Pareto path must be such that it extends from vk in the directions oftwo distinct descendants of vk with maximum decrease in the sumobjective. Complexity O(n)

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● ???


Part II: The model






problems






Part V: RANGE


PROBLEM


Computing the set W


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CASE 1

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● ???


Part II: The model






problems






Part V: RANGE


PROBLEM


Computing the set W


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Case 2. In this case we consider points in W corresponding to thepaths in P. Specifically, from the previous section, we focus oninstances where the value of a is either in {ǫs, ǫt} or in{γk : vk ∈ P [vp, vq]}. Complexity O(n)

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● ???


Part II: The model






problems






Part V: RANGE


PROBLEM


Computing the set W


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vs vt

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CASE 2

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Part II: The model






problems






Part V: RANGE


PROBLEM


Computing the set W


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■ In case 1 , a critical node of a path is the closest node, vk, of thispath to the weighted 1-center point, c̄u = v1.A Pareto path must be such that it extends from vk in the directionsof two distinct descendants of vk with maximum decrease in the sumobjective. Complexity O(n)

■ In case 2 we consider points in W corresponding to the paths in P.Specifically, from the previous section, we focus on instances wherethe value of a is either in {ǫs, ǫt} or in {γk : vk ∈ P [vp, vq]}.Complexity O(n)

● ???


Part II: The model






problems






Part V: RANGE


PROBLEM


The set Pareto( P∗)

To summarize, the total effort needed to compute the supersetW = W 1 ∪ W 2 is O(n log n).

With the above scheme we can also record for each point in W arepresentative path corresponding to the MAX and SUM values ofthat point.

Given the planar set W , we can then identify the Pareto set itself intime O(|W | log |W |), (Kapoor 2000).Therefore, the overall complexity of identifying PARETO(P∗), thecomplete set of Pareto optimal points, and the respectiverepresentative Pareto paths of the doubly weighted center-medianpath problem on a tree graph is O(n log n).

● ???


Part II: The model






problems






Part V: RANGE


PROBLEM


Solving some related problems

■ Minimizing the α-cent-dian for any α ∈ [0, 1].■ Minimizing the weighted sum of distances subject to some maximum

distance constraint.■ Minimizing the weighted maximum of distances subject to some sum

of the distances constraint.

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Part II: The model






problems






Part V: RANGE


PROBLEM


Solving some related problems

■ Minimizing the α-cent-dian for any α ∈ [0, 1].■ Minimizing the weighted sum of distances subject to some maximum

distance constraint.■ Minimizing the weighted maximum of distances subject to some sum

of the distances constraint.

Scan in linear time the setW.

Total complexity O(n)

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Part II: The model






problems






Part V: RANGE


PROBLEM


Case 1: Computing W1

Let vi, vj be a pair of distinct nodes such that (vi, vj) ∈ E:

SUM0(vi) =∑

vb∈Vi

wbd(vb, vi), (4)

SUM1(vi, vj) =∑

vb∈T (vi,vj)

wbd(vb, vi), (5)

SUM2(vi, vj) = minvb∈T (vi,vj)

∑

vd∈T (vi,vj)

wdd(vd, P [vi, vb]), (6)

SUM3(vi, vj) = SUM1(vi, vj) − SUM2(vi, vj). (7)

Note that for any node vi the weighted sum of the distances from itsdescendants satisfies:

SUM0(vi) =∑

vj∈S(vi)

SUM1(vi, vj).

The values of SUMt(vi, vj), t = 1, 2, 3, for all (vi, vj) can be computedin O(n) time using a straightforward modification of the algorithm inMorgan and Slater (1980).

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Part II: The model






problems






Part V: RANGE


PROBLEM


Case 1: (Continuation)

To find P we identify vi, vj ∈ S(vk), vi 6= vj , (i.e., two children of vk)such that SUM3(vk, vi) ≥ SUM3(vk, vj) ≥ SUM3(vk, vb) for anyvb ∈ S(vk) \ {vi, vj}. Then,

SUM(P ) =∑

vb∈N(vk)

SUM1(vk, vb) − SUM3(vk, vi) − SUM3(vk, vj).

For each δk the effort to compute SUM(P ) above is O(|N(vk)|).

Given the set {δk : vk ∈ V }, the additional total time spent to generateall the points in W 1 corresponding to Case 1 is O(n), since∑

vk∈V |N(vk)| = 2n − 2.

● ???


Part II: The model






problems






Part V: RANGE


PROBLEM



For each vk ∈ P [vp, v1], let

Psk = {P [vi, v1] : vi ∈ T s, P [vi, v1] ∩ P [vp, v1] = P [vk, v1]},

Ss(k) = minP [vi,v1]∈Ps

k

∑

vr∈T s

wrd(vr, P [vi, v1]).

Similar notation for vk ∈ P [v1, vq].

● ???


Part II: The model






problems






Part V: RANGE


PROBLEM




Psk = {P [vi, v1] : vi ∈ T s, P [vi, v1] ∩ P [vp, v1] = P [vk, v1]},

Ss(k) = minP [vi,v1]∈Ps

k

∑

vr∈T s

wrd(vr, P [vi, v1]).

Similar notation for vk ∈ P [v1, vq].

bs(k) = Ss(k) + minvb∈P [v1,vq],γb≤γk

St(b).

For vk ∈ P [v1, vq] set,

bt(k) = St(k) + minvb∈P [vp,v1],γb≤γk

Ss(b).

Define

W 2 = {(γk, bs(k)) : vk ∈ P [vp, v1]} ∪ {(γk, bt(k)) : vk ∈ P [v1, vq]}.

● ???


Part II: The model






problems






Part V: RANGE


PROBLEM


Case 2 (Continuation)


fsk = min

vb∈P [v1,vq],γb≤γk

St(b),

and for each vk ∈ P [v1, vq], let

f tk = min

vb∈P [vp,v1],γb≤γk

St(b).

● ???


Part II: The model






problems






Part V: RANGE


PROBLEM




fsk = min

vb∈P [v1,vq],γb≤γk

St(b),

and for each vk ∈ P [v1, vq], let

f tk = min

vb∈P [vp,v1],γb≤γk

St(b).

The sequence {γk} is decreasing when we move vk along P [vp, v1]from v1 to vp, and when we move vk along P [v1, vq] from v1 to vq.Therefore, the sets {fs

k : vk ∈ P [vp, v1]} and {f tk : vk ∈ P [v1, vq]} can

be computed in O(n) time when the terms {Ss(k)}, {St(k)}, areavailable.

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Part II: The model






problems






Part V: RANGE


PROBLEM



How to compute Ss(k)?Define SUM4(vk) for any vk ∈ P [vp, v1], as the sum of the weighteddistances to P [vk, v1] from all the nodes vr such thatd(vr, P [vp, v1]) = d(vr, P [vk, v1]),

SUM4(vk) =∑

vr∈T s\Va(k)

wrd(vr, P [vk, v1]).

● ???


Part II: The model






problems






Part V: RANGE


PROBLEM




SUM4(vk) =∑

vr∈T s\Va(k)


SUM4(v1) = 0,

SUM4(vk) = SUM4(vk−1) + SUM0(vk) − (SUM0(vk+1) − Wk+1d(vk+1,vk)), k

where Wk =∑

v∈Vkwv. ({Wk}

mk=1 and the values SUM4(vk) for all vk,

k = 1, . . . , m − 1, can clearly be computed in O(n).)

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Part II: The model






problems






Part V: RANGE


PROBLEM




SUM4(vk) =∑

vr∈T s\Va(k)


SUM4(v1) = 0,

SUM4(vk) = SUM4(vk−1) + SUM0(vk) − (SUM0(vk+1) − Wk+1d(vk+1,vk)), k

where Wk =∑

v∈Vkwv. ({Wk}

mk=1 and the values SUM4(vk) for all vk,

k = 1, . . . , m − 1, can clearly be computed in O(n).)Thus,

Ss(k) = SUM0(vk) − max

vr∈S(vk)\P[vp,vk]SUM3(vk,vr) + SUM4(vp(k)),

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Part II: The model





● The approach

Part V: RANGE


PROBLEM


Part IV: Weighted CENTER-MEDIAN Paretoset with length constraint

● ???


Part II: The model





● The approach

Part V: RANGE


PROBLEM


The approach

The approach is similar although computing theelements in the analysis with similar complexitiesto those in the case without length constraint ismore involved.

● ???


Part II: The model





Part V: RANGE


PROBLEM

● Formulas

● Cont.

● Formulas 2

● Case 2

● Scalar problems

● Hurwicz type problem

● Concluding Remark


Part V: RANGE (MAXIMUM-MINIMUM) PATHPROBLEM

● ???


Part II: The model





Part V: RANGE


PROBLEM

● Formulas

● Cont.

● Formulas 2

● Case 2

● Scalar problems




Formulas

CASE 1

We are given a critical node v and want to identify the extension of thepath in Tv with the largest minimum distance.

Let us define:β(v) = max

Pv∈P(Tv)µTv

(Pv) of a path Pv ∈ Tv that has v as an end

vertex.

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Part II: The model





Part V: RANGE


PROBLEM

● Formulas

● Cont.

● Formulas 2

● Case 2

● Scalar problems




Formulas

CASE 1



Pv∈P(Tv)µTv


vertex.

Let d1(v) = minw∈S(v)

ℓ(v, w), with w1 ∈ arg min{ℓ(v, w)|w ∈ S(v)}.

Moreover, let d2(v) = minw∈S(v)|w 6=w1

ℓ(v, w), with

w2 ∈ arg min{ℓ(v, w)|w 6= w1}, and d3(v) = minw∈S(v)|w 6∈{w1,w2}

ℓ(v, w).

We set d1(v) = +∞ when v is a leaf of Tc.

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Part II: The model





Part V: RANGE


PROBLEM

● Formulas

● Cont.

● Formulas 2

● Case 2

● Scalar problems




Formulas

CASE 1



Pv∈P(Tv)µTv


vertex.

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3

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7β(v)

=7

v 1

CA

SE1

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Part II: The model





Part V: RANGE


PROBLEM

● Formulas

● Cont.

● Formulas 2

● Case 2

● Scalar problems




Formulas

CASE 1



Pv∈P(Tv)µTv


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β(v) = 7

v1

CASE 1

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● ???


Part II: The model





Part V: RANGE


PROBLEM

● Formulas

● Cont.

● Formulas 2

● Case 2

● Scalar problems




Formulas

CASE 1



Pv∈P(Tv)µTv


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β(v) = 7

β(u) = 6

63

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v1

CASE 1

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Part II: The model





Part V: RANGE


PROBLEM

● Formulas

● Cont.

● Formulas 2

● Case 2

● Scalar problems




Formulas

CASE 1



Pv∈P(Tv)µTv


vertex.

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β(v) =

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:

d1(v) if flag(v) = 0 and |S(w1)| = 0

max{d1(v), β(w1)} if |S(v)| = 1

d2(v) if |S(v)| > 1 and flag(w1) = 0

max{d1(v), min{d2(v), β(w1)}} if |S(v)| > 1 and flag(w1) = 1,

(8)

while we set β(v) = +∞ when v is a leaf.

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● ???


Part II: The model





Part V: RANGE


PROBLEM

● Formulas

● Cont.

● Formulas 2

● Case 2

● Scalar problems




Formulas

CASE 1



Pv∈P(Tv)µTv


vertex.

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vP [w1, w2]

w2w1

3 5 7

v1

CASE 1

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Part II: The model





Part V: RANGE


PROBLEM

● Formulas

● Cont.

● Formulas 2

● Case 2

● Scalar problems




Cont.

Property 2 Let P f,gv be a path in Tv passing through v, with |S(v)| ≥ 3,

and connecting the descendants of two sons f and g of v. Then, wehave- if f = w1 and g 6= w2 (or viceversa), then any best path Pv starting at

v and connecting v with the descendants of w1 hasµTv

(Pv) ≥ µTv(P w1,g

v );- if f = w2 and g 6= w1 (or viceversa), then any best path Pv starting at

v and connecting v with the descendants of w2 hasµTv

(Pv) ≥ µTv(P w2,g

v );

- if f, g 6= w1, w2, then µTv({v}) ≥ µTv

(P f,gv ).

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Part II: The model





Part V: RANGE


PROBLEM

● Formulas

● Cont.

● Formulas 2

● Case 2

● Scalar problems




Cont.

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vP [w1, w2]

w2w1

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v1

CASE 1

● ???


Part II: The model





Part V: RANGE


PROBLEM

● Formulas

● Cont.

● Formulas 2

● Case 2

● Scalar problems




Cont.

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vP [w1, w2]

w2w1

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CASE 1

After Property 2, we need to evaluate only the paths connecting v andthe descendants of w1 and w2, while all the other paths passingthrough v, can be ignored.

● ???


Part II: The model





Part V: RANGE


PROBLEM

● Formulas

● Cont.

● Formulas 2

● Case 2

● Scalar problems




Formulas 2

Then, the maximum of the minimum distances MP(Tv) of a (best) pathPv ∈ P(Tv) (passing through or starting at v) with respect to the wholetree Tc can be computed as follows

MP(Tv) =

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:

min{ℓ(v, p(v)), β(v)} if |S(v)| ≤ 1

min{ℓ(v, p(v)),max{β(v),min{β(w1), β(w2)}}} if |S(v)| = 2

min{ℓ(v, p(v)),max{β(v),min{β(w1), β(w2), d3(v)}}} if |S(v)| ≥ 3

(9)

where we set ℓ(v, p(v)) = +∞ if v is the root of T .

● ???


Part II: The model





Part V: RANGE


PROBLEM

● Formulas

● Cont.

● Formulas 2

● Case 2

● Scalar problems




Case 2

For a path P (u1, u2) we define the function mil(P (u1, u2)), i.e., theminimum incident length, as the minimum length of an edge notbelonging to P (u1, u2), but incident to P (u1, u2) in one of the vertices t

with t 6= u1 and t 6= u2:mil(P (u1, u2)) = min

(s, t)|t ∈ P (u1, u2)\{u1, u2}

s 6∈ P (u1, u2)

ℓ(s, t). (10)

We denote by M̂P(p1p2) the maximum of the minimum distances from abest path Pp1p2 of type 2, to all the other vertices of the tree. On thebasis of formulas we have:

M̂P(p1p2) = min{β̂(p1), β̂(p2), mil(P (p1, p2))}. (11)

● ???


Part II: The model





Part V: RANGE


PROBLEM

● Formulas

● Cont.

● Formulas 2

● Case 2

● Scalar problems




Scalar problems

Once the set φ(W1) is available, taking into account Proposition 1 andProposition 4, problems P1-P3 can be solved in O(n) time as follows:

Problem P1 :Among all the pairs (MAX(P ), µ(P )) ∈ φ(W1), find the minimum ofR(P ) = MAX(P ) − µ(P ).

Problem P2 :For a given γ ≤ max{ℓ(e)|e ∈ E}, find the minimum of MAX(P )among all the pairs (MAX(P ), µ(P )) ∈ φ(W1) such that µ(P ) ≥ γ.

Problem P3 :For a given γ, find the maximum of µ(P ) among all the pairs(MAX(P ), µ(P )) ∈ φ(W1) such that E(P ) ≤ γ.

In addition, we notice that the set φ(π1) can be extracted from φ(W1) intime O(n log n) by finding the rectilinear lower envelop of the set φ(W1)with the algorithm provided by Kapoor (2000).

● ???


Part II: The model





Part V: RANGE


PROBLEM

● Formulas

● Cont.

● Formulas 2

● Case 2

● Scalar problems




Hurwicz type problem

We follow a similar analysis butwith a different partial ordering

(MAX (↓), MIN (↓)).

Similar complexities

● ???


Part II: The model





Part V: RANGE


PROBLEM

● Formulas

● Cont.

● Formulas 2

● Case 2

● Scalar problems




Concluding Remark

Efficient algorithms for minimaxproblem with an additional

constraint.

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