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On the Problem of Computing Zookeeper Routes. Hakan Jonsson & Sofia Sundberg. 2004.07.01 ISSN 1402-1528 / ISRN LTU-FR--04/10--SE / NR 2004:10. What’s Zookeeper’s Problem. Introduced by Chin & Ntafos 1.A simple polygon(zoo) with a disjoint set of k convex polygons(cage)
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Hakan Jonsson & Sofia Sundberg 2004.07.01 ISSN 1402-1528 / ISRN LTU-FR--04/10--SE / NR 2004:10
What’s Zookeeper’s Problem • Introduced by Chin & Ntafos • 1.A simple polygon(zoo) with a disjoint set of k convex polygons(cage) • 2.Every cage shares one edge with the zoo • 3.Find shostest route in the interior of the zoo without cross any cages.
FIXED • The route is forced to pass through a start point-s and s on the boundary of the zoo. • If zookeeper’s is non-fixed,it’s NP-hard
Def • Z:the zoo,a simple polygon and remain the edge of cages • K:the number of cages • N: the size of zoo • P:all the simple polygon • Zopt: the shortest zookeeper route • Zapp: the approximation zookeeper route
Def2 Zopt: the shortest zookeeper route Zc: the common part of Zopt and Zapp Zo,Za: the unique part of Zopt and Zapp
The Algorithms • Chin & Ntafos:O(n^2) exact solution • Jonsson:O(n) approximate solution • P contain a set C of k edges denoted C1,C2,….Ck,and start point s on the boundary but not in any cage
Exact Algorithms • They use the Reflection Principle,from a mirror to dash-b to find the shortest path (a to b) We unfolding it into an hourglass,then after adjustment ,can get the Zopt, it cost O(n)
Approximate solution-1 • Jonsson use a simpler approach,during a clockwise traversal of the boundary of the zoo,we gives each cage a unique first and last vertex • supporting chain is shortest path connect the first and last vertices of two consecutive cages.
Approximate solution-2 • For each cage Ci has one supporting chain Si ,if two supporting intersect we give a signpost for the cage • The touch point of a cage Ci is the point on the boundary of the cage that lies closest to the sign post of the cage.
Properties of zookeeper • If we chose different vertex of the cage, we will get the different length with other route. • Obstacle • Changing the Zoo
Obstacle Obervation 1:If /(Za/Zc)/ is a constant, When /Zc/ increases,then (/Za/+ /Zc/)/ (/Zo/+ /Zc/) decreases. Geometric terms:to achieve a worst case for (/Za/+ /Zc/)/ (/Zo/+ /Zc/) , it should probably be a minimum of common parts between the routes and the zoo.
Changing the Zoo • Obervation 2:Touch point ti of Zapp on a given cage Ci is unaffected by changes in cages other then Ci-1,Ci,Ci+1 • Obervation 3: Zopt remains that same as all tangents li are not changed
Observed worst case-2 • DEF:A isoceles zoo is a zoo, with starting point and two cages, as an isosceles triangle with height h and top a • Lemma 1:In a isoceles zoo the Zapp is • Dapp(a,h) = {2hsina 0<a< /2 • 2h /2<a<
Observed worst case-3 Lemma 2:In an isosceles zoo the length of Zapp is Dapp(a,h) =[2hsin(a/2)][(1+cos(a/2))] Lemma 3: In an isosceles zoo the quotient q(a,h)=Zapp/Zopt=Dapp/Dopt is maximized to for a = 2/3 and any h.