![]() ![]() ![]() If we find the closest pair across the subarrays, the pair of points should exist within distance d from the line dividing the array into subarrays. First, we will make a subset of the input, which consists of points within d distance from the midpoint in terms of X coordinate d is the shortest distance between the pair within subarrays. The last step, looking at the pairs across subarrays, needs some tricks to keep the cost at a linear level (e.g. The for loop is actually the ideal place to get the value looked up. def find_closest_brute_force(array): result = result = array result = array result = np.sqrt((array-array)**2 +(array-array)**2) for i in range(len(array)-1): for j in range(i+1, len(array)): distance = np.sqrt((array-array)**2 +(array-array)**2) if distance array: return np.array(, array]) else: return array elif length > 2: array_l = array array_r = array array_l_sorted = merge_sort(array_l, coordinate) array_r_sorted = merge_sort(array_r, coordinate) l_length = len(array_l) r_length = len(array_r) l = 0 r = 0 sorted_list = for i in range(length): if r = r_length: sorted_list.append(array_l_sorted) l += 1 elif l = l_length: sorted_list.append(array_r_sorted) r += 1 elif array_l_sorted > array_r_sorted: sorted_list.append(array_r_sorted) r += 1 elif array_l_sorted dict_r: dict_both = dict_r else: dict_both = dict_l Y_list = for i in range(length): if X-dict_both dict_y: dict_final = dict_y else: dict_final = dict_both else: for i in range(len(Y)-7): dict_y = find_closest_brute_force(Y) if dict_both > dict_y: dict_final = dict_y else: dict_final = dict_both return dict_final It will certainly be faster if you vectorize the distance calculations: def closestnode (node, nodes): nodes np.asarray (nodes) dist2 np.sum ( (nodes - node)2, axis1) return np. Passing xyarr i to the twopointdistance () function is inefficient you should lookup the value from the array once and save it in a local variable, instead of looking up the same value each time through the loop. With two loops, the code for this algorithm can be written as follows. For n number of points, we would need to measure n(n-1)/2 distances and the cost is square to n, or Θ(n²). The brute-force way is, like one that counts inversions in an array, to calculate the distances of every pair of points in the universe. Brute-Force Method - Finding the Closest Pair The split-conquer method works well in this challenge, in a similar way with the algorithm to count inversions. ![]() Now you can use function composition to combine them to d(f(x)) to find distance of each point. Now your curve is a function (x) (x, y), so that f(x) (x, x2). The cost increase is more than linear to the number of points, and we try to write an algorithm which contains the cost growth as low as possible. Distance from point (1,0) is the function: d((x, y)) (x 1)2 + (y 0)2 It's a function (x, y) (d). It is no different for computers when they determine the closest pairs of points on a plane, the more points the dataset has, the longer it takes for algorithms to find the couple with the least distance. If you are in the middle of nowhere, the infinite number of stars in the dark night sky would make it impossible to determine the pair. Looking up at the sky at Times Square or downtown Tokyo, we would find the closest pair of stars quite easily, because you can see only a few stars there. ![]()
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