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Second solution in Clear category for Similar Triangles by book1978
from typing import List, Tuple
Coords = List[Tuple[int, int]]
def similar_triangles(coords_1: Coords, coords_2: Coords) -> bool:
dist = lambda a, b: ((b[0] - a[0]) ** 2 + (b[1] - a[1]) ** 2) ** 0.5
t1 = sorted(dist(coords_1[i], coords_1[i - 1]) for i in range(3))
t2 = sorted(dist(coords_2[i], coords_2[i - 1]) for i in range(3))
return len({round(x / y, 6) for x, y in zip(t1, t2)}) == 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!")
May 31, 2023
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