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Duality. line. point. point. line. primal plane. dual plane. Collinear Points. collinear on line. Point. dual lines. concurrent at. iff. lies above. iff. lies above. Dual of a Line Segment?. Point. Relation to a Parabola. Dual of is the tangent

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## Duality

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**Duality**line point point line primal plane dual plane**Collinear Points**collinear on line Point dual lines concurrent at iff lies above iff lies above**Dual of a Line Segment?**Point**Relation to a Parabola**Dual of is the tangent line of the parabola at . Point Proof Let . Slope of the tangent: . But dual is line , which has the same slope. Since , is lies on the dual line.**Relation to a Parabola (2)**Point The dual line and it passes through .**More on Duality**Construct thedual line of q without measuring distances: Point • Through draw two tangent • lines to the parabola. • Let and be the points • of tangency, respectively. 3) is the line through and . The two tangent lines are and . and intersect at passes through and .**Back to the Discrepancy Problem**Point # points below = #lines strictly above Efficient algorithm exists!**Arrangement of Lines**L: a set of n lines. Point A(L): planar subdivision induced by L. with unbounded edges and faces no two lines are parallel. no three lines are concurrent. Simple arrangement if edge face vertex**Reduction to Line Arrangment**Point Problem on points problem on an arrangement of dual lines. dual plane primal plane Structure of a line arrangement is more apparent than that of a point set.**Combinatorial Complexity**#vertices + #edges + #faces Point Theorem #vertices #edges #faces Equality holds if and only if A(L) is simple.**Proof of Complexity**We first show that #vertices, #edges, #faces are maximal when A(L) is simple and not otherwise. Point 1) Let be parallel to one or more lines. small enough rotation to result new vertex in one change in A(L) Complexity increases in this case. (Hence such configuration cannot be maximal.)**Proof (Cont’d)**2) Supppose passes through a vertex. Point translate slightly 2 new vertices 3 new edges 1 new face Since complexity increases, such configuration cannot be maximal either. The arrangement with maximal complexity must be simple.**Exact Size of a Simple Alignment**# vertices Point # edges # faces Any pair of lines intersect . #edges on one line = 1 + #intersections on the line**Number of Faces**Point 1. Add a vertex at infinity. 2. Connect the two endpoints of every edge to . No edge crossing will be generated. A planar graph, by Euler’s theorem,**Storage of Line Arrangement**Point Doubly-connected edge list. Add a bounding box to contain all vertices in interior.**Construction of DCEL for A(L)**Plane sweep? Point pairwise intersection Not optimal!**Incremental Algorithm**Preprocessing Point Compute the bounding box B(L). - intersections. - leftmost, rightmost, top, bottom intersections. Add lines one by one. Update the DCEL after each addition.**Updating the Subdivision**subdivision due to Point Case 1enter a face f through edge e. • walk along boundary of f using the • the Next pointer. • find exit edge e’. • use its Twin() pointer to enter face g.**Updating the Subdivision**Case 2leave a face (g) through a vertex (u). Point • walk around u to find the next face • (h) intersected by Alternatively use Next() and Twin() pointers,**First Edge of Intersection**How to find the first edge (leftmost edge) intersected by ? Point It must be an edge on the bounding box B(L). Just test all edges on B(L). In case is vertical, locate the bottom intersection to start off traversal. has edges on B(L) since each edge intersects it twice. First intersection edge can be found in O(i) time.

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