MSUStudentWork/ComputationalGeometry/6Ex/PythonVersion/main.py

175 lines
4.1 KiB
Python

# Python3 program to find the minimum enclosing
# circle for N integer points in a 2-D plane
from math import sqrt
from random import randint, shuffle
# Defining infinity
INF = 1e18
MAX_POINT_COORD = 100
# Structure to represent a 2D point
class Point:
def __init__(self, X=0, Y=0) -> None:
self.X = X
self.Y = Y
# Structure to represent a 2D circle
class Circle:
def __init__(self, c=Point(), r=0) -> None:
self.C = c
self.R = r
def print(self) -> None:
print(f"(x - {self.C.X})^2 + (y - {self.C.Y})^2 = {self.R}^2")
# Function to return the euclidean distance
# between two points
def dist(a, b):
return sqrt(pow(a.X - b.X, 2)
+ pow(a.Y - b.Y, 2))
# Function to check whether a point lies inside
# or on the boundaries of the circle
def is_inside(c, p):
return dist(c.C, p) <= c.R
# The following two functions are used
# To find the equation of the circle when
# three points are given.
# Helper method to get a circle defined by 3 points
def get_circle_center(bx, by,
cx, cy):
B = bx * bx + by * by
C = cx * cx + cy * cy
D = bx * cy - by * cx
return Point((cy * B - by * C) / (2 * D),
(bx * C - cx * B) / (2 * D))
# Function to return the smallest circle
# that intersects 2 points
def circle_from1(A, B):
# Set the center to be the midpoint of A and B
C = Point((A.X + B.X) / 2.0, (A.Y + B.Y) / 2.0)
# Set the radius to be half the distance AB
return Circle(C, dist(A, B) / 2.0)
# Function to return a unique circle that
# intersects three points
def circle_from2(A, B, C):
I = get_circle_center(B.X - A.X, B.Y - A.Y,
C.X - A.X, C.Y - A.Y)
I.X += A.X
I.Y += A.Y
return Circle(I, dist(I, A))
# Function to check whether a circle
# encloses the given points
def is_valid_circle(c, P):
# Iterating through all the points
# to check whether the points
# lie inside the circle or not
for p in P:
if (not is_inside(c, p)):
return False
return True
# Function to return the minimum enclosing
# circle for N <= 3
def min_circle_trivial(P):
assert (len(P) <= 3)
if not P:
return Circle()
elif (len(P) == 1):
return Circle(P[0], 0)
elif (len(P) == 2):
return circle_from1(P[0], P[1])
# To check if MEC can be determined
# by 2 points only
for i in range(3):
for j in range(i + 1, 3):
c = circle_from1(P[i], P[j])
if (is_valid_circle(c, P)):
return c
return circle_from2(P[0], P[1], P[2])
# Returns the MEC using Welzl's algorithm
# Takes a set of input points P and a set R
# points on the circle boundary.
# n represents the number of points in P
# that are not yet processed.
def welzl_helper(P, R, n):
# Base case when all points processed or |R| = 3
if (n == 0 or len(R) == 3):
return min_circle_trivial(R)
# Pick a random point randomly
idx = randint(0, n - 1)
p = P[idx]
# Put the picked point at the end of P
# since it's more efficient than
# deleting from the middle of the vector
P[idx], P[n - 1] = P[n - 1], P[idx]
# Get the MEC circle d from the
# set of points P - :p
d = welzl_helper(P, R.copy(), n - 1)
# If d contains p, return d
if (is_inside(d, p)):
return d
# Otherwise, must be on the boundary of the MEC
R.append(p)
# Return the MEC for P - :p and R U :p
return welzl_helper(P, R.copy(), n - 1)
def welzl(P):
P_copy = P.copy()
shuffle(P_copy)
return welzl_helper(P_copy, [], len(P_copy))
def generate(num: int) -> list[Point]:
data: list[Point] = list()
for i in range(num):
data.append(Point(randint(-MAX_POINT_COORD, MAX_POINT_COORD), randint(-MAX_POINT_COORD, MAX_POINT_COORD)))
return data
def printPoints(data: list[Point]) -> None:
for i in range(len(data)):
print(f"{chr(i + 65)} = ({data[i].X}, {data[i].Y})")
# Driver code
if __name__ == '__main__':
num = int(input("Enter number of points: "))
data = generate(num)
printPoints(data)
mec = welzl(data)
mec.print()