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First solution in Clear category for Break Rings by francis.schmidt23
# migrated from python 2.7
import itertools
from functools import reduce
def break_rings(rings_union):
return min([len(broken_rings) for broken_rings in possible_broken_rings(sorted(reduce(set.union, rings_union))) if rings_are_free(broken_rings,rings_union)])
def rings_are_free(broken_rings,rings_union):
for ring_union in rings_union:
ring_union = list(ring_union)
if (not ring_union[0] in broken_rings) and (not ring_union[1] in broken_rings):
return False
return True
def possible_broken_rings(rings):
broken_rings = []
for i in range(len(rings)+1):
broken_rings = broken_rings + list(itertools.combinations(rings, i))
return 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"
July 10, 2015