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Simple Permutations solution in Clear category for Break Rings by zookatron
import itertools
def break_rings(links):
links = list(map(list, links))
num_rings = max(itertools.chain(*links))
for num_breaks in range(num_rings):
for breaks in itertools.combinations(range(1, num_rings+1), num_breaks):
if not [link for link in links if link[0] not in breaks and link[1] not in breaks]:
return num_breaks
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"
July 19, 2015
