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Functional style solution solution in Clear category for Similar Triangles by DenisKnyazev
from itertools import combinations
from math import dist
from typing import List, Tuple
Coords = List[Tuple[int, int]]
def similar_triangles(coords_1: Coords, coords_2: Coords) -> bool:
sides_1 = sorted(dist(x[0], x[1]) for x in combinations(coords_1, 2))
sides_2 = sorted(dist(x[0], x[1]) for x in combinations(coords_2, 2))
scales = [x[0] / x[1] for x in zip(sides_1, sides_2)]
return len(scales) == scales.count(scales[0])
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 30, 2021
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