Python/divide_and_conquer/closest_pair_of_points.py

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"""
The algorithm finds distance between closest pair of points
in the given n points.
Approach used -> Divide and conquer
The points are sorted based on Xco-ords and
then based on Yco-ords separately.
And by applying divide and conquer approach,
minimum distance is obtained recursively.
>> Closest points can 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. 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 * log n)
"""
def euclidean_distance_sqr(point1, point2):
return (point1[0] - point2[0]) ** 2 + (point1[1] - point2[1]) ** 2
def column_based_sort(array, column = 0):
return sorted(array, key = lambda x: x[column])
def dis_between_closest_pair(points, points_counts, min_dis = float("inf")):
""" brute force approach to find distance between closest pair points
Parameters :
points, points_count, min_dis (list(tuple(int, int)), int, int)
Returns :
min_dis (float): distance between closest pair of points
"""
for i in range(points_counts - 1):
for j in range(i+1, points_counts):
current_dis = euclidean_distance_sqr(points[i], points[j])
if current_dis < min_dis:
min_dis = current_dis
return min_dis
def dis_between_closest_in_strip(points, points_counts, min_dis = float("inf")):
""" closest pair of points in strip
Parameters :
points, points_count, min_dis (list(tuple(int, int)), int, int)
Returns :
min_dis (float): distance btw closest pair of points in the strip (< min_dis)
"""
for i in range(min(6, points_counts - 1), points_counts):
for j in range(max(0, i-6), i):
current_dis = euclidean_distance_sqr(points[i], points[j])
if current_dis < min_dis:
min_dis = current_dis
return min_dis
def closest_pair_of_points_sqr(points_sorted_on_x, points_sorted_on_y, points_counts):
""" divide and conquer approach
Parameters :
points, points_count (list(tuple(int, int)), int)
Returns :
(float): distance btw closest pair of points
"""
# base case
if points_counts <= 3:
return dis_between_closest_pair(points_sorted_on_x, points_counts)
# recursion
mid = points_counts//2
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_sorted_on_x:
if abs(point[0] - points_sorted_on_x[mid][0]) < closest_pair_dis:
cross_strip.append(point)
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):
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
if __name__ == "__main__":
points = [(2, 3), (12, 30), (40, 50), (5, 1), (12, 10), (3, 4)]
print("Distance:", closest_pair_of_points(points, len(points)))