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First solution in Clear category for Break Rings by v0id
from itertools import combinations
def break_rings(rings):
all_rings = set.union(*rings)
for n in range(1, len(all_rings)):
# it is NP-hard minimum vertex cover problem
# can only be solved with brute force search
for broken_rings in combinations(all_rings, n):
remaining_rings = [pair.difference(broken_rings) for pair in rings]
if all(len(rings) < 2 for rings in remaining_rings):
return n
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"
Nov. 14, 2015
