I need help to solve the question 2.2 in the following exercise , I add my solution for 2.1 it might help . This problem is at least not very easy to solve exactly. In the followingwe therefore propose a rough but rather quick and simple approximationalgorithm: Instead of the Euclidean distance, take the Manhattan distanceand minimizenΣwi · (|x − xi| + |y − Yi|).i=12.1. Explain how the point that minimizes the weigthed sum of Manhattandistances can be found in polynonial time. That is: Sketch an algorithm,argue why it is correct, and explain your time bound. Try to keep the timebound as low as possible.2.2. Show that this algorithm has an approximation ratio √√2. More specifi-cally: The point found by the algorithm has a weighted sum of Euclidean(!)distances that is at most √2 times the optimal sum.Advice: In 2.1, split the problem in two “independent” one-dimensionalproblems, that is, work separately in x- and y-direction. To get a correctidea where to find the optimal coordinate x (and y), it might be good tostudy examples with very small n first. In 2.2, first write down in generalterms what the approximation ratio of the proposed algorithm is, then dothe necessary geometry. Finally check whether you have really achieved thegoal of proving the claimed approxiamtion ratio. (It is easy to forget thisgoal and stop half-way.) 2.1 The task requires finding a point (x,y) that minimizes that weighted sum ofManhattan distances from a set of points with given Cartesian coordinates (xi, yi) andpositive weight wi.To solve this in polynomial time, we can take advantage of the linearity of theManhattan distance in each coordinate direction. This means we can find the optimalcoordinates independently.Firstly, we separate the problem into one-dimensional problems, one for the x-coordinates and one for the y-coordinates.For x-coordinates we sort the points by their x-coordinate, and we calculate a weightedmedian that minimize the sum of weighted absolute differences Σï=₁ w₁|x − x¡]. Thiscan be done by accumulating the weights from one end and finding the point where thesum of the weights before it is as close as possible to sum of weights after it, we repeatthe same for the y-coordinates.The algorithm works correctly because the weighted median minimizes the sum ofweighted absolute distances in one dimension. This is de to the fact that any weightedside that it would decrease on the lighter side.The sorting step can be done in O(n log n) time.Finding the weighted median can be done in O(n) time after sorting, hence the totaltime complexity is O(n log n) due to the sorting step.

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In this problem, we are dealing with a set of points (xi,yi) for i=1,2,....,n in a 2D plane, each…

2.2. Show that this algorithm has an approximation ratio √2. More specifi- cally: The point found by the algorithm has a weighted sum of Euclidean(!) distances that is at most √2 times the optimal sum.