Note we have defined py as the first in the pair, so set will be sorted by y coordinates.ģ. Then we inserted the first point in the pnts array to the set box. First ,we have sorted the array of points on x coordinates.Ģ.
#Matlab line style sweep code
The red region in the image is the region containing points which are to be evaluated with the current point.The points left to this region are removed from the set.įollowing is the C++ code for above algorithm:įor(typeof(box.begin()) it=box.lower_bound(make_pair(pnts.py-best, pnts.px-best)) it!=box.end() & pnts.py+best>=it->py it++)īest = min(best, sqrt(pow(pnts.py - it->py, 2.0)+pow(pnts.px - it->px, 2.0))) ġ.
#Matlab line style sweep update
Initialize h as the distance between first two points Update the value of h if present distance between points is less than h Okay, that's a lot of theory, let's see this algo running One thing to note is that at any instance, the number of points which are active events is O(1)(there can be atmost 5 points around a point which are active excluding the point itself). Īfter this processing, we'll add the Nth point to the set. All points in the set with x coordinates less than $$x_N-h$$ are to be deleted.
So, all such points whose x coordinate lie in and y coordinates lie in are what we are concerned with and these form the active events of the set. Now, we know we can only go till h distance from $$x_N$$ to find such point, and in the y direction we can go in h distance upwards and h distance downwards. For Nth point, we want to find points whose distance from Nth point is less than or equal to h. Now, suppose we have processed the points from 1 to N-1, and let h be the shortest distance we have got so far. So, first we sort the points in x direction as we want our line to move towards right. Here, we'll discuss it using line sweep technique.įor this problem, we can consider the points in the array as our events.Īnd in a set, we store the already visited points sorted by y coordinate. This problem can be solved by comparing all pairs of points, but then its complexity is $$O(N^2)$$ Problem: Find the closest pair of points in the given array of $$N$$ distinct points Generally, we can use set in C++ but sometimes we require some extra information to be stored, so we go for balanced binary tree. One other thing to note is that the efficiency of this technique depends on the data structures we use. At any instance, the data structure stores only the active events. Other than events, we maintain a data structure which stores the events generally sorted by y coordinates (the criteria for ordering of data structure may vary sometimes) which is helpful in the processing when we encounter some event. The events are based on the problem we are considering, we'll see them in the algorithms discussed below. We sweep the line based on some events, in order to discretize the sweep. That's why, the algorithms based on this concept are sometimes also called plane sweep algorithms.
In this article, we'll learn some algorithms based on the simple tools of computational geometry.Ī sweep line is an imaginary vertical line which is swept across the plane rightwards.
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