50 lines
1.5 KiB
Python
50 lines
1.5 KiB
Python
from typing import Type
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class Point2D:
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def __init__(self, x: int, y: int):
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self.x = x
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self.y = y
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def copy(self):
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return Point2D(self.x, self.y)
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def coordinates(self):
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return (self.x, self.y)
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def __repr__(self):
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return f"Point2D(x: {self.x}, y: {self.y})"
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def is_in_triangle(self, xy0: Type[Point2D], xy1: Type[Point2D], xy2: Type[Point2D]):
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"""Returns True is the point is in a triangle defined by 3 others points.
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From: https://stackoverflow.com/questions/2049582/how-to-determine-if-a-point-is-in-a-2d-triangle#:~:text=A%20simple%20way%20is%20to,point%20is%20inside%20the%20triangle.
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Args:
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xy0 (Type[Point2D]): Point of the triangle.
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xy1 (Type[Point2D]): Point of the triangle.
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xy2 (Type[Point2D]): Point of the triangle.
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Returns:
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bool: False if the point is not inside the triangle.
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"""
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dx = self.x - xy0.x
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dy = self.y - xy0.y
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dx2 = xy2.x - xy0.x
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dy2 = xy2.y - xy0.y
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dx1 = xy1.x - xy0.x
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dy1 = xy1.y - xy0.y
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s_p = (dy2 * dx) - (dx2 * dy)
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t_p = (dx1 * dy) - (dy1 * dx)
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d = (dx1 * dy2) - (dy1 * dx2)
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if d > 0:
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return (s_p >= 0) and (t_p >= 0) and (s_p + t_p) <= d
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else:
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return (s_p <= 0) and (t_p <= 0) and (s_p + t_p) >= d
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def distance(self, point: Type[Point2D]):
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return sqrt((point.x - self.x) ** 2 + (point.y - self.y) ** 2)
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