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First solution in Clear category for Break Rings by yarmak_vladislav
# migrated from python 2.7
from functools import reduce
def break_rings(links):
links, rings = set(frozenset(l) for l in links), reduce(set.union, links)
broken = 0
while True:
least_linked = min(rings, key=lambda r: sum(1 for l in links if r in l))
neighbours = set(r for l in links for r in l if least_linked in l)
broken += len(neighbours) - 1
links -= set(l for l in links if (l & neighbours))
if not links:
return broken
rings = reduce(frozenset.union, links)
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"
Aug. 12, 2015
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