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upper tree, but the claim is that their absence cannot increase the complexity of this operation, onlydecrease it. However, a careful proof needs to take this into account.) Since the zone of each line inthe arrangement has complexityO(n), allnzones have total complexityO(n2). Thus, the total timespent in updating the UHT and LHT trees isO(n2).Lecture 28: Ham-Sandwich CutsHam 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 separablepoint sets.We are givennred pointsA, andmblue pointsB, and we want to compute a singleline that simultaneously bisects both sets. If the cardinality of either set is odd, then the line passesthrough one of the points of the set (see Fig. 147(a)). It is a well-known theorem from mathematicsLecture Notes155CMSC 754
that such a simultaneous bisector exists for any pair of sets (even for shapes, where bisection is interms of area).(a)(b)∈A∈BFig. 147: Ham sandwich cuts (a) general and (b) linearly-separable.This problem can be solved inO(n2) time through the use of duality and line arrangements, but wewill consider a restricted version that can be solved much faster.In particular, let us assume thatthe two sets can be separated by a line (see Fig. 147(b)). We may assume that the points have beentranslated and rotated so the separating line is they-axis. Thus all the red points (setA) have positivex-coordinates, and all the blue points (setB) have negativex-coordinates. As long as we are simplifyingthings, let’s make one last simplification, that both sets have an odd number of points. This is notdifficult to get around, but makes the pictures a little easier to understand.Ham-Sandwich Cuts in the Dual:Consider one of the sets, sayA. Observe that for each slope thereexists 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 uniqueplacement 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 themedianlinefor this slope.What is the dual of this median line?Suppose that we dualize the points using the standard dualtransformation, where a pointp= (pa, pb) is mapped to the linep*:y=pax-pb. We obtainnlines inthe plane. By starting a line with a given slope above the points and translating it downwards, in thedual plane we moving a point from-∞upwards in a vertical line. Each time the line passes a pointin the primal plane, the vertically moving point crosses a line in the dual plane. When the translatingline hits the median point (see Fig. 148(a)), in the dual plane the moving point will hit a dual line suchthat 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 atlevelk,Lk