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First solution in 3rd party category for Similar Triangles by amvasiliev80
import numpy as np
from typing import List, Tuple
Coords = List[Tuple[int, int]]
def similar_triangles(coords_1: Coords, coords_2: Coords) -> bool:
f_norm = lambda d1, d2: np.linalg.norm(np.array(d1) - np.array(d2))
tr1 = sorted(
(
f_norm(coords_1[1], coords_1[0]) ,
f_norm(coords_1[2], coords_1[0]) ,
f_norm(coords_1[2], coords_1[1]) ,
))
tr2 = sorted(
(
f_norm(coords_2[1], coords_2[0]) ,
f_norm(coords_2[2], coords_2[0]) ,
f_norm(coords_2[2], coords_2[1])
))
return len(set(np.array(tr1) / np.array(tr2)))==1
if __name__ == '__main__':
print("Example:")
print(similar_triangles([(0, 0), (1, 2), (2, 0)], [(3, 0), (4, 2), (5, 0)]))
# These "asserts" are used for self-checking and not for an auto-testing
assert similar_triangles([(0, 0), (1, 2), (2, 0)], [(3, 0), (4, 2), (5, 0)]) is True, 'basic'
assert similar_triangles([(0, 0), (1, 2), (2, 0)], [(3, 0), (4, 3), (5, 0)]) is False, 'different #1'
assert similar_triangles([(0, 0), (1, 2), (2, 0)], [(2, 0), (4, 4), (6, 0)]) is True, 'scaling'
assert similar_triangles([(0, 0), (0, 3), (2, 0)], [(3, 0), (5, 3), (5, 0)]) is True, 'reflection'
assert similar_triangles([(1, 0), (1, 2), (2, 0)], [(3, 0), (5, 4), (5, 0)]) is True, 'scaling and reflection'
assert similar_triangles([(1, 0), (1, 3), (2, 0)], [(3, 0), (5, 5), (5, 0)]) is False, 'different #2'
print("Coding complete? Click 'Check' to earn cool rewards!")
Oct. 31, 2021
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