upper tree but the claim is that their absence cannot increase the complexity

Upper tree but the claim is that their absence cannot

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upper tree, but the claim is that their absence cannot increase the complexity of this operation, only decrease it. However, a careful proof needs to take this into account.) Since the zone of each line in the arrangement has complexity O ( n ), all n zones have total complexity O ( n 2 ). Thus, the total time spent in updating the UHT and LHT trees is O ( n 2 ). Lecture 28: Ham-Sandwich Cuts Ham Sandwich Cuts of Linearly Separated Point Sets: In this short lecture, we consider an appli- cation of duality and arrangements, namely computing a Ham-Sandwich cut of two linearly separable point sets. We are given n red points A , and m blue points B , and we want to compute a single line that simultaneously bisects both sets. If the cardinality of either set is odd, then the line passes through one of the points of the set (see Fig. 147(a)). It is a well-known theorem from mathematics Lecture Notes 155 CMSC 754
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that such a simultaneous bisector exists for any pair of sets (even for shapes, where bisection is in terms of area). (a) (b) A B Fig. 147: Ham sandwich cuts (a) general and (b) linearly-separable. This problem can be solved in O ( n 2 ) time through the use of duality and line arrangements, but we will consider a restricted version that can be solved much faster. In particular, let us assume that the two sets can be separated by a line (see Fig. 147(b)). We may assume that the points have been translated and rotated so the separating line is the y -axis. Thus all the red points (set A ) have positive x -coordinates, and all the blue points (set B ) have negative x -coordinates. As long as we are simplifying things, let’s make one last simplification, that both sets have an odd number of points. This is not difficult to get around, but makes the pictures a little easier to understand. Ham-Sandwich Cuts in the Dual: Consider one of the sets, say A . Observe that for each slope there exists one way to bisect the points. In particular, if we start a line with this slope at positive infinity, so that all the points lie beneath it, and drop in downwards, eventually we will arrive at a unique placement where there are exactly ( n - 1) / 2 points above the line, one point lying on the line, and ( n - 1) / 2 points below the line (assuming no two points share this slope). This line is called the median line for this slope. What is the dual of this median line? Suppose that we dualize the points using the standard dual transformation, where a point p = ( p a , p b ) is mapped to the line p * : y = p a x - p b . We obtain n lines in the plane. By starting a line with a given slope above the points and translating it downwards, in the dual plane we moving a point from -∞ upwards in a vertical line. Each time the line passes a point in the primal plane, the vertically moving point crosses a line in the dual plane. When the translating line hits the median point (see Fig. 148(a)), in the dual plane the moving point will hit a dual line such that there are exactly ( n - 1) / 2 dual lines above this point and ( n - 1) / 2 dual lines below this point (see Fig. 148(b)). We define a point to be at level k , L k
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