optimum location of point to minimize total distance using ternary algorithm code example
Example: how to find closest distance for given points
# A divide and conquer program in Python3
# to find the smallest distance from a
# given set of points.
import math
import copy
# A class to represent a Point in 2D plane
class Point():
def __init__(self, x, y):
self.x = x
self.y = y
# A utility function to find the
# distance between two points
def dist(p1, p2):
return math.sqrt((p1.x - p2.x) *
(p1.x - p2.x) +
(p1.y - p2.y) *
(p1.y - p2.y))
# A Brute Force method to return the
# smallest distance between two points
# in P[] of size n
def bruteForce(P, n):
min_val = float('inf')
for i in range(n):
for j in range(i + 1, n):
if dist(P[i], P[j]) < min_val:
min_val = dist(P[i], P[j])
return min_val
# A utility function to find the
# distance beween the closest points of
# strip of given size. All points in
# strip[] are sorted accordint to
# y coordinate. They all have an upper
# bound on minimum distance as d.
# Note that this method seems to be
# a O(n^2) method, but it's a O(n)
# method as the inner loop runs at most 6 times
def stripClosest(strip, size, d):
# Initialize the minimum distance as d
min_val = d
# Pick all points one by one and
# try the next points till the difference
# between y coordinates is smaller than d.
# This is a proven fact that this loop
# runs at most 6 times
for i in range(size):
j = i + 1
while j < size and (strip[j].y -
strip[i].y) < min_val:
min_val = dist(strip[i], strip[j])
j += 1
return min_val
# A recursive function to find the
# smallest distance. The array P contains
# all points sorted according to x coordinate
def closestUtil(P, Q, n):
# If there are 2 or 3 points,
# then use brute force
if n <= 3:
return bruteForce(P, n)
# Find the middle point
mid = n // 2
midPoint = P[mid]
# Consider the vertical line passing
# through the middle point calculate
# the smallest distance dl on left
# of middle point and dr on right side
dl = closestUtil(P[:mid], Q, mid)
dr = closestUtil(P[mid:], Q, n - mid)
# Find the smaller of two distances
d = min(dl, dr)
# Build an array strip[] that contains
# points close (closer than d)
# to the line passing through the middle point
strip = []
for i in range(n):
if abs(Q[i].x - midPoint.x) < d:
strip.append(Q[i])
# Find the closest points in strip.
# Return the minimum of d and closest
# distance is strip[]
return min(d, stripClosest(strip, len(strip), d))
# The main function that finds
# the smallest distance.
# This method mainly uses closestUtil()
def closest(P, n):
P.sort(key = lambda point: point.x)
Q = copy.deepcopy(P)
Q.sort(key = lambda point: point.y)
# Use recursive function closestUtil()
# to find the smallest distance
return closestUtil(P, Q, n)
# Driver code
P = [Point(2, 3), Point(12, 30),
Point(40, 50), Point(5, 1),
Point(12, 10), Point(3, 4)]
n = len(P)
print("The smallest distance is",
closest(P, n))
# This code is contributed
# by Prateek Gupta (@prateekgupta10)