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Brute force with itertools.combinations() solution in Clear category for Break Rings by M157q
import itertools
def break_rings(rings):
all_rings = set.union(*rings)
for num_of_broken_rings in range(1, len(all_rings)+1):
for rings_to_break in itertools.combinations(all_rings, num_of_broken_rings):
# if left rings are all seperate, problem solved.
if all(len(pair - set(rings_to_break)) <= 1 for pair in rings):
return num_of_broken_rings
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. 3, 2015