updated closest pair of points (n*(logn)^2) to (n*logn)

Author: Dharni0607

divide_and_conquer/closest_pair_of_points.py

@@ -1,25 +1,24 @@
 """
 The algorithm finds distance btw closest pair of points in the given n points.
 Approach used -> Divide and conquer 
-The points are sorted based on Xco-ords 
+The points are sorted based on Xco-ords and 
+then based on Yco-ords separetely.
 & by applying divide and conquer approach, 
 minimum distance is obtained recursively.
 
 >> closest points lie on different sides of partition
 This case handled by forming a strip of points 
 whose Xco-ords distance is less than closest_pair_dis
-from mid-point's Xco-ords.
+from mid-point's Xco-ords. points sorted based on Yco-ords 
+are used in this step to reduce sorting time.
 Closest pair distance is found in the strip of points. (closest_in_strip)
 
 min(closest_pair_dis, closest_in_strip) would be the final answer.
  
-Time complexity: O(n * (logn)^2)
+Time complexity: O(n * logn)
 """
 
 
-import math 
-
-
 def euclidean_distance_sqr(point1, point2):
     return pow(point1[0] - point2[0], 2) + pow(point1[1] - point2[1], 2)
 
@@ -66,7 +65,7 @@ def dis_between_closest_in_strip(points, points_counts, min_dis = float("inf")):
     return min_dis
 
 
-def closest_pair_of_points_sqr(points, points_counts):
+def closest_pair_of_points_sqr(points_sorted_on_X, points_sorted_on_Y, points_counts):
     """ divide and conquer approach
 
     Parameters : 
@@ -79,35 +78,41 @@ def closest_pair_of_points_sqr(points, points_counts):
 
     # base case
     if points_counts <= 3:
-        return dis_between_closest_pair(points, points_counts)
+        return dis_between_closest_pair(points_sorted_on_X, points_counts)
 
     # recursion
     mid = points_counts//2
-    closest_in_left = closest_pair_of_points(points[:mid], mid)
-    closest_in_right = closest_pair_of_points(points[mid:], points_counts - mid)
+    closest_in_left = closest_pair_of_points_sqr(points_sorted_on_X, 
+                                                 points_sorted_on_Y[:mid], 
+                                                 mid)
+    closest_in_right = closest_pair_of_points_sqr(points_sorted_on_Y, 
+                                                  points_sorted_on_Y[mid:], 
+                                                  points_counts - mid)
     closest_pair_dis = min(closest_in_left, closest_in_right)
 
     """ cross_strip contains the points, whose Xcoords are at a 
     distance(< closest_pair_dis) from mid's Xcoord
     """
 
     cross_strip = []
-    for point in points:
-        if abs(point[0] - points[mid][0]) < closest_pair_dis:
+    for point in points_sorted_on_X:
+        if abs(point[0] - points_sorted_on_X[mid][0]) < closest_pair_dis:
             cross_strip.append(point)
 
-    cross_strip = column_based_sort(cross_strip, 1)
     closest_in_strip = dis_between_closest_in_strip(cross_strip, 
                      len(cross_strip), closest_pair_dis)
     return min(closest_pair_dis, closest_in_strip)
 
 
 def closest_pair_of_points(points, points_counts):
-    return math.sqrt(closest_pair_of_points_sqr(points, points_counts))
+    points_sorted_on_X = column_based_sort(points, column = 0)
+    points_sorted_on_Y = column_based_sort(points, column = 1)
+    return (closest_pair_of_points_sqr(points_sorted_on_X, 
+                                       points_sorted_on_Y, 
+                                       points_counts)) ** 0.5
 
 
-points = [(2, 3), (12, 30), (40, 50), (5, 1), (12, 10), (0, 2), (5, 6), (1, 2)]
-points = column_based_sort(points)
+points = [(2, 3), (12, 30), (40, 50), (5, 1), (12, 10), (3, 4)] 
 print("Distance:", closest_pair_of_points(points, len(points)))