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First solution in Clear category for Break Rings by SergeyKhentov
def break_rings(connections):
if len(connections) == 0:
return 0
r1, r2 = connections[0]
return min(
break_rings([c for c in connections if r1 not in c]) + 1,
break_rings([c for c in connections if r2 not in c]) + 1)
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"
Feb. 27, 2015
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