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## Ice Cream And Wedge Graph

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**Ice Cream And Wedge Graph**Eyal Ackerman TsachikGelander Rom Pinchasi**Table of content**• Introduction • Wedge-graph • Overview • Main theorem • Ice-cream lemma • Intuition for the lemma • Explanation of the main theorem • Proofs • main theorem • Ice-cream lemma**Wedge-graph**• Given 3 -directional antenna a c b**The wedge-graph we got:**a c b**Overview**• [1] I. Caragiannis, C. Kaklamanis, E. Kranakis, D. Krizanc and A. Wiese, Communication in wireless networks with directional antennas, Proc. 20th Symp. on Parallelism in Algorithms and Architectures, 344{351, 2008. • formulated a different model of directed communication graph • [2] P. Carmi, M.J. Katz, Z. Lotker, A. Rosen, Connectivity guarantees for wireless networks with directional antennas, Computational Geometry: Theory and Applications, to appear. • [3] M. Damian and R.Y. Flatland, Spanning properties of graphs induced by directional antennas, Electronic Proc. 20th Fall Workshop on Computational Geometry, Stony Brook University, Stony Brook, NY, 2010. • formulated a different model of directed communication graph • [4] S. Dobrev, E. Kranakis, D. Krizanc, J. Opatrny, O. Ponce, and L. Stacho, Strong connectivity in sensor networks with given number of directional antennae of bounded angle, Proc. 4th Int. Conf. on Combinatorial Optimization and Applications, 72{86, 2010. • formulated a different model of directed communication graph**Overview**• [1] I. Caragiannis, C. Kaklamanis, E. Kranakis, D. Krizanc and A. Wiese, Communication in wireless networks with directional antennas, Proc. 20th Symp. on Parallelism in Algorithms and Architectures, 344{351, 2008. • [2] P. Carmi, M.J. Katz, Z. Lotker, A. Rosen, Connectivity guarantees for wireless networks with directional antennas, Computational Geometry: Theory and Applications, to appear. • found the minimum so that no metter what finit set P of location, it is always posible to position them so the wedge graph is connected • [3] M. Damian and R.Y. Flatland, Spanning properties of graphs induced by directional antennas, Electronic Proc. 20th Fall Workshop on Computational Geometry, Stony Brook University, Stony Brook, NY, 2010. • [4] S. Dobrev, E. Kranakis, D. Krizanc, J. Opatrny, O. Ponce, and L. Stacho, Strong connectivity in sensor networks with given number of directional antennae of bounded angle, Proc. 4th Int. Conf. on Combinatorial Optimization and Applications, 72{86, 2010.**Ice-cream lemma**• S is a compact convex set in the plane. • . • There exist a point O in the plane and 2 rays, q and r. • q and r apex is O • q and r are touching S at X and Y exclusively • r and q satisfy • the angle bounded by r and q is**Intuition for the lemma**Let S be a Compact convex set and fix S**Intuition for the lemma**There exist a point O O S**Intuition for the lemma**And 2 rays q and r q O S r**Intuition for the lemma**Rays are touching S in X and Y and creating an angle of X q O S r Y**Intuition for the lemma**O was selected to satisfy X q O S r Y**Intuition for the lemma**The ice-cream lemma X q O S r Y**Main Theorem**• P is a set of n points in the plain (general position). • CH(p) is the convex hull of p. • h is the number of vertices in CH(p). • It takes O(n log h) - time to find n wedges (apexes are in p) of angle • The wedge-graph is connected. • The wedge-graph has a path of length 2 and each of the other vertices in the graph is connected by an edge to one of the three vertices of the path.**Explanation of the main theorem**• The angle is best possible as written by Carmi et al.**A specific case for the example, an equilateral triangle.**We’ll try to position the –directional antennas with a m m c m b**And antenna b**a c b**Antenna c will never be connected in the wedge-graph of a,b**and c a c b**The wedge-graph we got in any setting of antenna c is:**a c b**proof of the main theorem**• Let p be:**proof of the main theorem**• CH(p) is: This can be found in o(nlogh)**proof of the main theorem**• <X,Y> Is a “good pair” if exists O- point, and rays q,r X q O r Y**proof of the main theorem**• Given CH(P) and 2 vretices X,Y, it takes O(1)-time to check whether X and Y are a “good pair”. • for any pair of points, there are only 2 possible location for O, and you only need test whether their neighbors are in • The ice cream lemma guarantees that CH(P) has a good pair**proof of the main theorem**• How to find a good pair: • For any edge (x,x’) of CH(P) we can find in O(logn) time (binary search) the Y point that satisfies this: X X’ q O r Y**proof of the main theorem**• How to find a good pair: • For any edge (x,x’) and point Y in CH(P) we can test in constant time if (X,Y) or (X’,Y) are “good pair”: O X X’ q r Y**proof of the main theorem**• How to find a good pair: • We got that a good pair can be found in O(nlogh) time O X X’ q r Y**proof of the main theorem**• is a line creating angle of with q and r • There is a point on • A and B are the points of the intersections of ,q and r • and the triangle ABO is covering p • Z Can be found in O(logh) O X Y A B Z**proof of the main theorem**• X’ is a point on such that is equilateral and Y’ is a point on such that is equilateral. • there are 2 general cases for this: O O case2 case1 X Y X Y A B A B Z Y’ X’ Z X’ Y’**proof of the main theorem**• Case 1: • Z is in and in O X Y A B Z Y’ X’**proof of the main theorem**• Case 1: • . O X Y A B Z Y’ X’**proof of the main theorem**• Case 1: • is a wedge of containing X and Y • Note that the wedge graph of X,Y and Z is connected O Y X A B Z Y’ X’**proof of the main theorem**• Case 2: • The general case where Z is notin • We’ll find Z’ to be the point on r, and is equilateral O Z’ X Y A B Z X’ Y’ q r**proof of the main theorem**• Case 2: • is • Note that the wedge graph of X,Y and Z is connected O Z’ X Y A B Z X’ Y’ q r**proof of the main theorem**• Finally we can see that the wedge graph contains 2-path on X,Y,Z. • , and covere , which means that any other point in p is connected to X,Y or Z in the wedge-graph O O Y X Z’ X Y A B A B Z X’ Y’ Z Y’ X’ q r**proof of the ice-cream lemma**• Consider the point O such that the 2 tangents of S through O create and angle of • And the area of is maximum. • From compactness -> such point exists X q O S r Y**proof of the ice-cream lemma**• We’ll show that X q O S r Y**proof of the ice-cream lemma**• X’ can be equal to X and Y’ can be equal to Y. • and X X’ q O S r Y’ Y**proof of the ice-cream lemma**• Claim: • and • This means that : • X X’ q O S r Y’ Y**proof of the ice-cream lemma**• Proof of the claim: • Assuming to the contrary that • Let small positive number X X’ q O S r Y’ Y**proof of the ice-cream lemma**• Proof of the claim: • obtained from XO’ and is from Y’O’ • Let small positive angle of rotation X’ X q O’ S O r Y’ Y**proof of the ice-cream lemma**• Proof of the claim: • The area of is greater than the area of , • The difference is . • This contradict the choice of the point O. X’ X q O’ S O r Y’ Y