minimum diameter spanning trees and related problems
TRANSCRIPT
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Minimum Diameter Spanning Trees and Related Problems
Jan-Ming Ho, D. T. Lee, Chia-Hsiang Chang, and C. K. Wang
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MDST Problem
• Given a graph G=(V, E) and a cost function W(e)Z+ for all eE, find a spanning tree T for G such that
is minimized.
pe
TpeW )(max
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Goal of the paper
• In this paper we shall study the problem of finding a minimum diameter spanning tree of a special graph, called Euclidean graph, induced by a set of n points in the Euclidean space, referred to as a GMDST problem.
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Main Results
• Lemma 2: There exists a GMDST of a set S of n points which is either monopolar (n≧ 3) or diapolar (n≧ 4).
• Theorem 1: Given a set S of n points, the GMDST for S can be found in θ(n3) time and O(n) space.
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monopolardipolar
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Definitions
• A spanning tree of an n-points set S is said to be monopolar if there exists a point called the monopole such that all the remaining points are connected to it.
• And it is said to be dipolar if there exists two points called the dipole such that all the remaining points are directly connected to one of the two points in the dipole.
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Definition 1
• An edge (Ai-1, Ai) is a center edge of a path P = (A0, A1 ,…, Ak) of points if max{distp (A
0, Ai-1) , distp (Ai, Ak) } is minimized.
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Lemma 1
• Let (Ai-1, Ai) be a center edge of a path P = (A0, A1 ,…, Ak) of points. Then
(1) distp (A0, Ai-1) ≦ distp (Ai-1, Ak) and
(2) distp (Ai, Ak) ≦ distp (A0, Ai)
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Lemma 2
• There exists a GMDST of a set S of n points which is either monopolar (n≧ 3) or diapolar(n≧ 4).
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Case1: If there exists a diameter of T, which has only two edges.
distT’(P, Q)
= |P, A1 |+ |A1, Q|
≦ distT(P, A1) + distT(A1, Q)
≦ | A0, A1 |+ |A1, A2 |
= DT
T
P
Q
A1
A0
A2
T’
P
Q
A1
A0
A2
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Case2: If every diameter of T contains more than two edges.
By deleting the center edge (Ai-1, Ai) from T, we obtain two subtree Ti-1, Ti .
(1) P, Q belong to different subtrees
(2) P, Q belong to the same subtrees
distT’’(P, Q)
= |P, Ai |+ |Ai, Q|
≦ distT(P, Ai) + distT(Ai, Q)
≦ distT(Ai, Ak) + distT(Ai, Ak)
≦ | A0, Ai |+ |Ai, Ak |
= DT
T P
Q
A2
A0
A3
A1
T’’ P
Q
A2
A0
A3
A1
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Algorithm find_min_of_DPST
• O(nlogn)
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Algorithm find_min_of_MPST
• For each piG, points other than pi are sorted acc
ording to their distances with respect to pi.O(n2logn)
• For each pi, pjG, find the best dipolar diameter
with dipole pi and pj. O(n3)
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pi
pj
pi
pj
pi
pj
pipj
pipj
pi
pj
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Theorem 1
• Given a set S of n points, the GMDST for S can be found in O (n3) time and O(n) space.
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Conclusion
• The result actually is applicable to any complete graph whose edge weight satisfies the triangle inequality.
• Whether or not the O (n3) time bound for finding a minimum diameter spanning tree can be improved is of great interest.
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On the minimum diameter spanning tree problem
Refael Hassin, Arie Tamir
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Introduction
• In this note we consider the general case where the edge lengths do not necessarily satisfy the triangle inequality.
• We then observe that the MDST problem is identical to the well studied absolute 1-center problem. As such, one can apply existing algorithms and solve the MDST problem on a general graph in O(mn+n2logn)
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Definition
Given graph G(V, E)• A(G): the continuum set of point on the edges of
G.
• dG(x, y): the length of a shortest path in A(G) connecting x and y.
• For each x in A(G) let T(x) denote a shortest path tree connecting x to all the nodes in V.
• For each x in A(G) define F(x) = ),(max vxdGVv
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Absolute 1-center problem(A1CP)
• The absolute 1-center problem(A1CP) on G is to minimize the function F(x). A point x * in A(G) is an absolute 1-center of G if the function F attains its minimum at x * .
• There are several efficient algorithms to locate an absolute 1-center. The most efficient known algorithm can be implemented in O(mn+n2logn).
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Equivalence of A1CP and MDSTTheorem 1: Let x * be an absolute 1-center of G and let
T(x * ) be a shortest path tree connecting x * to all nodes in V. Then T (x * ) is a minimum diameter spanning tree of G.
Proof: Let T be an arbitrary spanning tree of G. Let y*(T) be the absolute 1-center of T. D(T)=2FT(y*(T) )
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