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Second solution in Clear category for Break Rings by mortonfox
from itertools import combinations
def break_rings(rings):
vertices = set().union(*rings)
for nbreak in range(1, len(vertices)):
for break_set in combinations(vertices, nbreak):
if all(any(r in break_set for r in link) for link in rings):
return nbreak
if __name__ == '__main__':
# These "asserts" using only for self-checking and not necessary for auto-testing
assert break_rings(({1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}, {4, 6})) == 3, "example"
assert break_rings(({1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4})) == 3, "All to all"
assert break_rings(({5, 6}, {4, 5}, {3, 4}, {3, 2}, {2, 1}, {1, 6})) == 3, "Chain"
assert break_rings(({8, 9}, {1, 9}, {1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}, {6, 7}, {8, 7})) == 5, "Long chain"
assert break_rings([[1, 2], [2, 3], [3, 4], [4, 5], [5, 2], [1, 6], [6, 7], [7, 8], [8, 9], [9, 6], [1, 10], [10, 11], [11, 12], [12, 13], [13, 10], [1, 14], [14, 15], [15, 16], [16, 17], [17, 14]]) == 8, 'A1'
May 26, 2017
